What the figures prove — page 3

• in dimension 1 the sum is 1 − (−1)^d ×8
• the alternating sum of the 5-cell is 0 ×5
• cyclic, 6 corners: cells = edges − corners ×4
• cyclic, 6 corners: faces = 2 × (edges − corners) ×4
• cyclic, 6 corners: the alternating sum is 0 ×4
• 5-cell: cells = edges − corners ×3
• 5-cell: faces = 2 × (edges − corners) ×3
• 5-cell: the alternating sum is 0 ×3
• V − E + F is 0 for this solid ×3
• with exact readings the 40 by 36 grid counts the targets ×3
• the whole boundary of the 3-cube has alternating sum 1 − (−1)^3 ×2
• ∫ χ(h ≥ s) ds is the pieces' lifetimes less the holes'
• a bipyramid lies on the line F = 2V − 4
• a convex solid has exactly two critical points in every direction
• a deeper overlap tolerates more smoothing
• a honeycomb on the torus has V − E + F = 0
• a prism lies on the line V = 2F − 4
• a pyramid has as many corners as faces
• a spanning tree has one fewer edge than the graph has vertices
• a wide gap separates the targets' features from the mistakes'
• about one sensor's width of smoothing is needed to swallow the mistakes
• and 720° is 360° times V − E + F
• and its twelve edges
• and nothing he forbids is reached
• and six sides to every face
• below the squared field's floor the region is the whole torus, with χ = 0
• between 20 and 400 surfaces
• between one and three holes are punched
• corner cut off: V − E + F = 2
• counting and discarding the edge pieces extrapolate to the same constant
• cube: each face has three sides and each corner three edges
• cube: Euler
• cutting a three-edged corner adds two corners and one face
• dodecahedron: each face has three sides and each corner three edges
• dodecahedron: Euler
• each ring has χ = 0
• each shape follows its own exact curve
• each unfolded face closes back on the corner
• every count Steinitz allows is reached by a pyramid and the two moves
• every detection region is a disc, χ = 1
• every detection region is a ring, χ = 0
• every edge lies between exactly two faces
• every edge of the cube is shared by two faces
• every edge of the dodecahedron is shared by two faces
• every edge of the icosahedron is shared by two faces
• every edge of the octahedron is shared by two faces
• every edge of the surface is shared by exactly two faces
• every edge of the tetrahedron is shared by two faces
• every game stays within the bounds
• every partial boundary before the last facet sums to 1
• every random game of Brussels sprouts on n crosses lasts exactly 5n − 2 moves
• four levels
• gluing on a tetrahedron adds one corner and two faces
• icosahedron: each face has three sides and each corner three edges
• icosahedron: Euler
• lives remaining are 3n minus the number of moves
• no leftover edge closes a loop among the faces
• no mistake lives longer than twice the largest change smoothing left in the readings
• octahedron: each face has three sides and each corner three edges
• octahedron: Euler
• on a surface with no edges, maxima minus saddles plus minima is zero
• once divided by the formula's prefactor, every scale gives the same curve
• pentagonal bipyramid: each face has three sides and each corner three edges
• pentagonal bipyramid: Euler
• pentagonal prism: each face has three sides and each corner three edges
• pentagonal prism: Euler
• pentagonal pyramid: each face has three sides and each corner three edges
• pentagonal pyramid: Euler
• pieces minus holes matches the exact Euler characteristic
• pieces minus holes, counted separately, follows the exact density
• rounding the lifetimes counts the targets
• sixteen times the sensors, about sixteen times the error
• smoothed, with each lifetime rounded, it counts them again
• tetrahedron glued on: V − E + F = 2
• tetrahedron: each face has three sides and each corner three edges
• tetrahedron: Euler
• tetrahedron: V − E + F = 2
• the 5-sided bipyramid has V − E + F = 2
• the 5-sided prism has V − E + F = 2
• the 5-sided pyramid has V − E + F = 2
• the alternating sum is not 2, on a solid with flat faces and straight edges
• the alternating sum of the tesseract is 0
• the average Euler characteristic follows the formula
• the blind rules tilt the balance of signs and the exact rule keeps it level
• the constant is below the percolation prediction by more than two standard errors
• the counted pieces minus holes agree with the exact density
• the counts obey Euler's formula
• the crossing curves steepen as the square grows
• the cube's faces fall short of hexagons by exactly 12
• the defects on the cube come to 720°
• the defects on the dodecahedron come to 720°
• the defects on the icosahedron come to 720°
• the defects on the octahedron come to 720°
• the defects on the tetrahedron come to 720°
• the dodecahedron's faces fall short of hexagons by exactly 12
• the error follows p/2 times the sensors that see nothing
• the faces at a corner of the cube do not close up flat
• the faces at a corner of the dodecahedron do not close up flat
• the faces at a corner of the icosahedron do not close up flat
• the faces at a corner of the octahedron do not close up flat
• the faces at a corner of the tetrahedron do not close up flat
• the faces short of six sides fall short by 12 in all
• the flattened cube keeps its eight corners
• the flattened cube still gives 2
• the game ends with 4n regions, one free arm in each
• the hexagonal prism's faces fall short of hexagons by exactly 12
• the index matches the table
• the indices add up to the Euler characteristic counted directly
• the large pieces follow a power law near percolation's exponent
• the leftover edges are what the tree did not use
• the leftover edges join every face to every other
• the leftover edges number one fewer than there are faces
• the length runs from 2n to 3n − 1
• the level is between −3 and 3
• the median gives way first; the rounded lifetimes hold to three in ten
• the mistaken readings alone throw the integer count far off
• the mode is one the family draws
• the option is one this view reads
• the rings overlap in two separate pieces
• the rounded count holds until the overlaps' lifetimes fall below a half
• the search agrees with the pattern
• the solid is one of the five regular ones
• the solid is one the family builds
• the squared field breaks the symmetry the Gaussian one keeps
• the sum never changes
• the sum over the level sets equals the sum of the regions' own χ
• the sweep ends at the torus's Euler characteristic
• the targets' regions are discs or rings
• the tesseract has 32 edges
• the tesseract's counts are the 16-cell's, reversed
• the tetrahedron's faces fall short of hexagons by exactly 12
• the tree is grown from one of the graph's own vertices
• the tree reaches every vertex
• the triangular prism's faces fall short of hexagons by exactly 12
• the truncated icosahedron's faces fall short of hexagons by exactly 12
• the truncated octahedron's faces fall short of hexagons by exactly 12
• the truncated tetrahedron's faces fall short of hexagons by exactly 12
• the two trees together use every edge exactly once
• the unrounded integral only falls as the smoothing widens
• the window closes as the overlap gets shallow
• three edges at every corner
• three faces meet at every corner of the cube
• three faces meet at every corner of the dodecahedron
• three faces meet at every corner of the hexagonal prism
• three faces meet at every corner of the tetrahedron
• three faces meet at every corner of the triangular prism
• three faces meet at every corner of the truncated icosahedron
• three faces meet at every corner of the truncated octahedron
• three faces meet at every corner of the truncated tetrahedron
• up to one reading in ten, both repairs count exactly
• V − E + F = 2 for the cube
• V − E + F = 2 for the dodecahedron
• V − E + F = 2 for the hexagonal prism
• V − E + F = 2 for the tetrahedron
• V − E + F = 2 for the triangular prism
• V − E + F = 2 for the truncated icosahedron
• V − E + F = 2 for the truncated octahedron
• V − E + F = 2 for the truncated tetrahedron
• V − E + F is -2 for this solid
• V − E + F is 2 for the cube
• V − E + F is 2 for the dodecahedron
• V − E + F is 2 for the icosahedron
• V − E + F is 2 for the octahedron
• V − E + F is 2 for the tetrahedron
• which is V − E + F = 2
• with every region a disc, the integral counts the targets
• with every region a ring, the integral sees nothing
• with exact readings the integral counts the targets
• χ above u is maxima minus saddles plus minima above u
• χ of a disc is 1
• χ of a disc with two holes is -1
• χ of a ring is 0
• χ of two discs is 2
• χ(A ∪ B) = χ(A) + χ(B) − χ(A ∩ B)

• the search agrees with Stahl's formula at n = 5, m = 2 ×19
• globe 11 has an odd number of agreeing pairs ×10
• χ(K(4, 1)) is n − 2k + 2 ×8
• removing any vertex of SG(5, 2) lets 2 colours suffice ×6
• the stable graph SG(5, 2) needs as many colours as K(5, 2) ×6
• the shortest odd cycle in K(5, 2) has length 5 ×4
• at 6 × 6 the boundary condition holds ×3
• at 6 × 6 there is a complementary edge ×3
• every necklace can be shared by 3 thieves with 4 cuts ×3
• some necklace needs all 4 cuts ×3
• 3 colours do not suffice
• 3 to 30 region centres
• a degree from 1 to 5
• a nearly fixed point exists
• a quarter-turn reverses the imbalance
• a rotation's iterates stay at their starting distance
• a sector of 120° has diameter √3
• a seed from 1 to 999
• a triangle without opposite labels does not wind
• a two-by-two bimatrix
• all 1,024 antipodal labellings of the 2 × 2 grid were tried
• an even grid of at most 40 cells a side
• and at those mixtures the two pure replies pay exactly the same
• and has no direction to choose there
• and it breaks the boundary condition
• and it is inside the disc
• and it lies inside the disc
• and it sits exactly on the rim
• and its equator winds an odd number of times
• and n at k colours
• and second
• and sits on it
• and so does every bisection
• and some necklace needs all of them
• and the crossing found is a genuine zero
• and the rule with the hole has none, at any of the sampled points
• and the second
• and the sweep finds points that barely move, near it
• at each pair both readings agree to nine decimal places
• at least ((r − 1)!)ᵈ = 4 splits work, as Sierksma conjectured
• at n = 8 the largest family with no orthogonal pair has eight members, and eight pieces suffice
• at the left end the map moves the point right or not at all
• at the right end it moves it left or not at all
• between 0 and 4 turns
• between 4 and 16 globes
• between 8 and 120 arrows are drawn per panel
• bisection's cost does not depend on the turns
• boards of 2 to 14 fit the picture
• both choosers have a mixture of the other's that leaves them indifferent
• both find the fixed point
• cutting costs a few readings per factor of ten in accuracy
• dipole has the index it claims
• disjoint pairs share no colour
• each quadrant holds a quarter
• each region reaches arccos(1/3) from its centre
• each set is swept at between 200 and 20,000 points
• every agreeing point comes with its opposite point
• every arrow is tangent to the sphere
• every corridor ends at the fixed point's square or one touching it
• every intersecting family of pairs sits in a star or a triangle
• every necklace can be shared fairly with no more cuts than colours
• every one of the random sets has at least four working splits
• every one of them has a complementary edge
• every open half-circle holds at least two of the points
• every orbit runs into the fixed point
• every outer vertex has an inner partner
• every pair gets its m colours
• every pair is covered at least m times
• every point of every random set lies on an odd number of halving lines, so there are at least n/2
• every rectangle kept winds once round
• every sampled antipodal labelling has a complementary edge
• exactly one colour wins
• exactly one cut falls inside the blue block
• exactly one cut falls inside the green block
• exactly one cut falls inside the orange block
• exactly one cut falls inside the purple block
• few enough arrangements to check every one
• for a map that expands, the fixed point can lie on the side the cut discards
• for a rotation, which neither shrinks nor stretches, the cuts still keep the fixed point
• for λ near one the bisector passes almost through the fixed point
• four equal quarters
• half a turn along the equator reverses the difference exactly
• in both of its readings
• iteration needs far more readings near λ = 1
• no boundary edge is itself complementary in the labelling drawn
• no cut through the centroid keeps more than five-ninths of the area
• no face region of the tetrahedron holds a point together with its opposite point
• no point of a ring, turned stays where it is, and none comes close
• no point of the whole plane, shifted stays where it is, and none comes close
• no six points in general position split into three groups whose hulls meet
• no three points on a line
• odd regular polygons need three pieces and even ones two
• on the board of four exactly the short diagonal wins
• one of the two has convex values and the other does not
• one to four colours a vertex
• one to three colours a vertex
• opposite boundary vertices carry opposite labels
• outside the circle the twisted and gentle maps agree
• random fields without the boundary condition often escape
• saddle has the index it claims
• seven points always have a split into three groups whose hulls meet
• shrinking hard is far cheaper than bisection, and shrinking barely is dearer
• so both values are within one step's change of zero
• so every direction is marked by some colour
• so every point of the open disc moves
• so it has opposite signs at the two ends of a half turn
• so the sweep crosses zero at least twice
• so the winding must be carried by triangles holding opposite labels
• some direction and its opposite are marked by the same colour
• some edge joins a label to its negative
• some first move wins, as strategy stealing requires
• some half-circle holds fewer than two points
• some line halves both halves
• some necklace of these beads needs as many cuts as there are colours
• some region holds a point together with its opposite point
• some turn of the frame cuts four equal quarters
• source has the index it claims
• Stahl's formula is Lovász's at one colour
• ten random points have at least (3!)² = 36 splits into four
• the 64 points split into as many pieces as the count forces
• the angle between the two lines passes through a right angle at least twice in half a turn
• the annulus is turned by between 5 and 175 degrees
• the boundary winds an odd number of times round the diamond
• the colouring is proper
• the control's fixed point is fixed, first coordinate
• the convex-valued rule has a point inside its own value
• the corridor ends in the square that holds the fixed point
• the corridor on maps twisted in proportion to n grows like n squared
• the count first beats the dimension at n = 44, in 946 dimensions
• the count is a Lucas number minus two
• the count rises by 2 into even m and by n − 2 into odd m
• the cuts keep the fixed point
• the cutting method gets cheaper, not dearer, as λ approaches one
• the difference reverses sign between opposite points
• the equator's difference winds an odd number of times
• the extra readings grow in proportion to the number of turns
• the fairest division of this necklace uses one cut per colour
• the farthest complementary edge from a zero gets closer as the grid is refined
• the field has a zero in the square
• the field is one the figure knows
• the field vanishes at the pole
• the fixed point is unique
• the fixed point is within (1 + λ)/2 of the distance to c from the bisector
• the gentle corridor and both bisections grow like n
• the grouped necklace needs one cut for every colour
• the halving map does have a fixed point
• the imbalance changes sign somewhere in half a turn
• the iterations needed are log(distance/ε) / log(1/λ)
• the last rectangle is the fixed point's square
• the line halves the first shape
• the map is one the figure knows
• the map keeps the interval inside itself
• the map sends the disc into itself
• the middle two points trade places once for each halving line
• the opposite pairs where both readings agree are odd in number
• the option is one this view reads
• the pair found really does take the same value
• the periodic points are all found
• the point found really is fixed
• the ring is turned by between 5 and 175 degrees
• the row chooser's best reply is one of the two rows
• the search finds ten points with the most halving lines possible, thirteen
• the search never exceeds the published maximum and never falls below n/2
• the shape is one the family knows
• the signed count of fixed points is one minus the degree
• the signed count stays at minus one
• the smallest-element colouring is proper
• the solved point is fixed, first coordinate
• the solved point is fixed, second
• the sphere is shown in three or four pieces
• the step and the gap are the same rule away from the jump
• the sweep takes between 36 and 2000 samples
• the sweep takes between four hundred and twenty thousand samples
• the tetrahedral pieces have diameter 2√((3 + √3)/6)
• the twisted map's corridor is several times longer
• the two cells of the walk stay adjacent
• the two colours win equally often, by reflecting the board
• the two sets found are disjoint — a monochromatic edge
• the two shapes cut have between 3 and 12 sides each
• the view is one the family draws
• the walk ends at the corner of the winning colour
• the walk reaches a corner of the board
• thief A gets half the blue beads
• thief A gets half the green beads
• thief A gets half the orange beads
• this labelling has no complementary edge
• three closed sets covering the sphere hold a pair of opposite points in one of them
• three colours do
• three segments through one point give six points that do split
• turning the line through half a turn reverses the imbalance
• twelve points in convex position have exactly six halving lines
• two colours do not suffice
• two colours never suffice
• two correspondences the family knows are compared
• two cuts give thief A exactly half of each colour
• two sets covering the circle always hold a pair of opposite points in one of them
• two to four colours with an even number of beads each
• values in opposite cones make an angle of at least a right angle
• which the open disc does not contain
• without the boundary condition some labellings have no complementary edge
• z to the d has d − 1 roots of unity and nought and infinity as fixed points

• every cuff of every pair of pants is used once, genus 0 with 3 boundary ×21
• enumeration and recurrence agree at n 1, genus 0 ×15
• country 0 borders six different countries ×7
• a genus-2 decomposition cuts along 3g − 3 circles ×3
• a sphere with 3 cross-caps has characteristic -1 ×3
• at most 2g − 3 circles separate, genus 2 ×3
• the enumeration agrees with Harer–Zagier at genus 0 ×3
• a surface with 2 handles has characteristic -2 ×2
• a two-sheeted cover of a sphere with 3 cross-caps has twice its characteristic ×2
• all 12 are one triangulation relabelled ×2
• the classification is drawn up to 3 of them in this view ×2
• the decompositions of genus 2 come in the counted number of types ×2
• a drawn random gluing has 6 to 60 edges
• a move never leaves the list of types
• a non-orientable surface and an orientable one share the characteristic and differ
• a pairing makes a sphere exactly when no two chords cross
• a sphere has characteristic 2
• a sphere with 1 cross-cap has characteristic 1
• a sphere with 2 cross-caps has characteristic 0
• a sphere with k cross-caps is covered by a surface with k − 1 handles
• a torus has characteristic 0
• a two-sheeted cover of a Klein bottle has twice its characteristic
• a two-sheeted cover of a projective plane has twice its characteristic
• a two-sheeted cover of a surface with 2 handles has twice its characteristic
• a two-sheeted cover of a torus has twice its characteristic
• aba⁻¹b⁻¹ gives the characteristic of a torus
• aba⁻¹b⁻¹ has the Euler characteristic a torus has
• aba⁻¹b⁻¹ is orientable
• abab gives the characteristic of a projective plane
• abab has the Euler characteristic a projective plane has
• abab is not orientable
• abab⁻¹ gives the characteristic of a Klein bottle
• abab⁻¹ has the Euler characteristic a Klein bottle has
• abab⁻¹ is not orientable
• abb⁻¹a⁻¹ has the Euler characteristic a sphere has
• abb⁻¹a⁻¹ is orientable
• an orientable pairing has a whole, non-negative genus
• and every face to a face
• and it is the one that crosses on exactly the letters used twice the same way
• and no edge to itself
• and no face to itself
• and no vertex is its own opposite, so the map moves every point
• and not one Klein bottle
• and one move on the dumbbell's middle circle gives the theta back
• and one side: the projective plane
• and so is the sidedness
• and ten faces
• and the graph has as many independent cycles as the surface has handles
• and the quotient's is a projective plane's
• and thirty edges
• and twenty faces
• and two sides: a torus
• at least one move applies to the word given
• at most four letters, so at most fifteen covers are listed
• between one and five cross-caps are drawn
• between one and five gluings are drawn side by side
• between three and twelve steps are drawn
• between two and four words, each of at least three edges
• capping the boundaries gives a closed surface with characteristic at most two
• decompositions are enumerated for genus 2 to 4
• each drawn pairing is glued head to tail
• each lattice triangle is one of the fourteen
• each pairing carries one side onto its partner's geodesic
• each sliver is one of the ten triangles
• each type is met as often as its symmetries predict
• edge a is glued to exactly one other edge
• edge b is glued to exactly one other edge
• edge c is glued to exactly one other edge
• edge d is glued to exactly one other edge
• edge e is glued to exactly one other edge
• edge f is glued to exactly one other edge
• eight corners do make a Klein bottle
• every pairing is a hyperbolic translation, its trace above two
• every pairing is drawn for polygons of 4 to 8 edges
• every sampled vertex count has the parity the genus formula allows
• every type is reachable from every other by moves
• every vertex has its opposite among the vertices
• every vertex of the surface has exactly two vertices above it, so nothing is pinched
• exactly one connected double cover is two-sided
• few enough vertices to colour each one differently
• fifteen edges
• genus 0 is the Catalan number
• genus four has seventeen
• genus three has five
• genus two has two types
• Heawood's bound allows this many vertices
• no letter appears more than twice
• no move raises the measure
• no two of the surfaces share a characteristic
• none of them two-sided
• one move on the theta gives the dumbbell
• one vertex, for even n, is 1/(n + 1)
• one-vertex gluings are 1/(n + 1) of all
• only letters of the word can be glued across
• opposite points of the rim carry the same label
• seven labelled vertices carry a hundred and twenty tori
• six labelled vertices carry twelve projective planes
• the angle measured from the drawn arcs is the one the formula gives
• the antipodal map carries every edge to an edge
• the bound is the least n with n² − 7n + 6χ ≥ 0
• the camera angles are numbers
• the census runs to between 6 and 24 edges
• the characteristic is unchanged by the cancel move
• the characteristic is unchanged by the collapse move
• the characteristic is unchanged by the crosscap move
• the characteristic is unchanged by the pair move
• the cover's characteristic is twice the surface's, because every cell is doubled
• the cycle count agrees with the pooled corners
• the drawing scale is sane
• the drawn range reaches at least one characteristic that both sides hold
• the drawn vertex counts hold nearly all of the distribution
• the first decomposition has no separating circle
• the fourteen triangles close up into a surface
• the genera share out all (2n − 1)!! pairings
• the icosahedron has twelve vertices
• the icosahedron's characteristic is a sphere's
• the mean vertex count sits just above the harmonic number H(2n), by less than 1/n
• the measured area is 4π(g − 1), minus 2π times the Euler characteristic
• the name follows from the number
• the name follows from the two numbers
• the orientable row is on or off
• the pairings number (2n − 1)!!
• the pants' characteristics, −1 each, add up to the surface's
• the polygon has an even number of edges
• the polygon's corners all become one vertex, so every choice of letters to cross gives a cover
• the quotient has six vertices
• the recurrence divides exactly
• the reduction reaches a standard form
• the sample mean sits within four standard errors of the exact mean
• the seven-vertex torus has 42
• the side pairings satisfy the surface's relation [a, b][c, d] = 1
• the six-vertex plane has 60 symmetries
• the sphere's share is Catalan over (2n − 1)!!
• the spheres are the Catalan number
• the standard form names the surface the original word named
• the surface is one-sided, so it has an orientation cover to single out
• the table runs to between 3 and 10 pairs
• the ten triangles close up into a surface
• the tiles within distance R number about the disc's area over 4π
• the two copies join up exactly when some letter is used twice the same way round
• the two numbers name the surface without anything else being consulted
• the two surfaces with the same characteristic are told apart by orientability
• the type census is drawn for eight corners
• the types are drawn side by side for genus 2 and 3
• the view is one the family draws
• the word aab leaves at least one edge unpaired
• the word abab⁻¹c leaves at least one edge unpaired
• the word abc leaves at least one edge unpaired
• the word abcb⁻¹ leaves at least one edge unpaired
• the word glues every edge to exactly one other
• the word has at least four edges
• the words drawn give different triples, so the figure is separating them
• three fresh letters are available for the collapse
• using all fifteen pairs of six vertices as edges
• using all twenty-one pairs of seven vertices as edges
• with characteristic one
• with characteristic zero
• with the copies given opposite senses of turning, every pair glued in the cover meets with opposite arrows

• linking of fibres 0 and 1 ×21
• 3_1 is not the unknot ×14
• 3_1's braid gives the determinant in the tables ×14
• both checkerboard colourings give 3_1 changed's signature ×14
• both checkerboard colourings give 7_4's signature ×14
• the changes found for 3_1 are the tables' unknotting number ×14
• both checkerboard colourings give T(2,7)'s signature ×4
• in every placement of G7 the linking numbers over disjoint cycle pairs add to an odd number ×3
• in G7 every two disjoint edges are separated by an even number of cycle pairs, so a crossing change keeps the parity ×3
• the Borromean rings: the diagram's count and Gauss's integral agree for components 1 and 2 ×3
• the figure-eight knot: 3 colours are available exactly when 3 divides its determinant ×3
• the figure-eight knot: the count with 3 colours ×3
• the trefoil: 3 colours are available exactly when 3 divides its determinant ×3
• the trefoil: the count with 3 colours ×3
• the unknot: 3 colours are available exactly when 3 divides its determinant ×3
• the unknot: the count with 3 colours ×3
• three rings in a chain: the diagram's count and Gauss's integral agree for components 1 and 2 ×3
• 2π times the average number of maxima is the total curvature
• a colouring is drawn on a diagram that has crossings
• a colouring uses three, five or seven colours
• a crossing change moves the signature by at most two
• a crossing joins two different Seifert circles
• a diagram with k crossings is cut into k arcs
• a knot and its mirror image have the same Alexander polynomial
• a knot's Goeritz form has no zero eigenvalue
• a knot's signature is even
• a loop that dips through and back: punctures and crossings agree
• a loop that dips through and back: the diagram's count and Gauss's integral agree for components 1 and 2
• a pentagon with five crossings in its picture is found
• a polygon of each kind is found among the first four thousand
• a regular fibre links the exceptional circles q and p times, and they link once
• a seven-stick polygon is an unknot, a trefoil or a figure-eight
• a switched crossing is one the diagram has, named once
• a torus knot squeezed onto its core turns 2π per circuit of the axis
• almost every position along the path is in general position
• an unknot with one twist has the crossings it claims
• an unknot with one twist: an alternating diagram with a crossing that untwists has a bracket narrower than 4c
• an unknot with one twist: an alternating diagram's extreme states meet themselves exactly at its untwistable crossings
• an unknot with one twist: an alternating diagram's two extreme states have c + 2 circles between them
• an unknot with one twist: the bracket gives the Jones polynomial the handedness requires
• and a Lissajous curve with each of their polynomials is in the census
• and Gauss's integral, which sees no disc at all, gives it too
• and it is unknotted
• and its determinant
• and its determinant is 1 or 7 modulo 8
• and one whose linking number is bigger than one
• and the one found uses all three
• and their signs add to an even number
• and they reach zero in two different ways — one with no punctures, one with two that cancel
• at every crossing the three arcs are all alike or all different
• at least two of the links have linking number zero
• between −3 and 4 extra full turns
• between three and five deformation sizes, none of them large
• between two and four links the family knows
• Călugăreanu: the linking number is twist plus writhe
• changing a negative crossing never raises it
• changing a positive crossing never lowers the signature
• Conway and Gordon: the a₂ of the 360 cycles add to an odd number
• Conway and Gordon: the ten linking numbers add to an odd number
• count is read only by the stickears, stickcensus, randdets, k7many, k6knots views
• crossings grow like n for walks and n² in a box; the mean square linking number like n² in a box and far more slowly than the crossings for walks
• each crossing has exactly one of its two strands going underneath
• each drawn polygon's full polynomial agrees with its identification
• every curve stays above 4π
• every disjoint pair in the Petersen graph is two pentagons
• every edge survives trimming at both ends
• every embedding has an odd sum of linking numbers
• every five-pointed star a pentagon makes has twists three one way and two the other
• every knot's determinant is odd
• every knotted curve exceeds 4π
• every Lissajous knot's polynomial is a square modulo 2
• every one has fifteen edges
• every orbit's genus is a whole number
• every pair of orbits up to length 8 links at least once
• every pair of trefoil fibres links six times
• every placement has a trefoil among its paths
• every placement has an odd number of knotted paths
• every polygon of seven or fewer sticks has determinant 1, 3 or 5
• every power of A in the normalised bracket is a multiple of four, as a knot's must be
• every row of the matrix vanishes at t = 1, as a presentation of a knot must
• every state leaves at least one circle
• few enough circles to try every ordering
• following the smoothing from a segment comes back to that segment
• from 100 to 300 sticks the double trefoil comes within a fifth of the rule
• Gauss's integral lands on a whole number
• in every direction the knotted curve has at least two maxima
• in every placement of K₃,₃,₁ the linking numbers over disjoint cycle pairs add to an odd number
• in every placement of K₄,₄ minus an edge the linking numbers over disjoint cycle pairs add to an odd number
• in every placement of K₆ the linking numbers over disjoint cycle pairs add to an odd number
• in every placement of Petersen graph the linking numbers over disjoint cycle pairs add to an odd number
• in K₃,₃,₁ every two disjoint edges are separated by an even number of cycle pairs, so a crossing change keeps the parity
• in K₄,₄ minus an edge every two disjoint edges are separated by an even number of cycle pairs, so a crossing change keeps the parity
• in K₆ every two disjoint edges are separated by an even number of cycle pairs, so a crossing change keeps the parity
• in Petersen graph every two disjoint edges are separated by an even number of cycle pairs, so a crossing change keeps the parity
• its determinant is 3, so it is knotted
• its determinant is 5
• its polynomial is the figure-eight's, −t + 3 − t⁻¹, from four directions
• its polynomial is the trefoil's, t² − t + 1, from four directions
• knots sharing a determinant are told apart by the polynomial
• Levine: four divides the signature exactly when the determinant is 1 mod 4
• longer polygons are knotted far more often
• loops far apart are rarely linked
• Milnor's identity holds for every curve
• most placements of six points have no knotted path
• no hexagon is a figure-eight
• no knot is undone with fewer changes than its signature allows
• no orbit of period four or less is knotted
• no pentagon had two corners' triangles pierced
• no pentagon is knotted
• no single arc passes over every crossing
• one boundary circle, so the genus is (1 − χ)/2
• one crossing left is the unknot; three is not
• reflection negates the signature
• seen from above it has 2ab − a − b crossings
• signed crossings between two trefoil fibres
• six points have 60 closed paths through all of them
• six sticks make only unknots and trefoils
• so the number of knotted paths is odd too
• so the number of linked pairs is odd too, since every linking number here is 0 or ±1
• some number of extra turns keeps the linking number
• squeezed onto the core, the curvature nears 2π per turn about the axis
• sticks is read only by the randone, randclose views
• the (2, 4) torus link: the diagram's count and Gauss's integral agree for components 1 and 2
• the (2, c) torus knot has signature 1 − c
• the a₂ sum is odd
• the a₂ sum is odd for every placement
• the a₂ sum stays odd as the point moves
• the average of the directional counts is the writhe
• the bracket lies inside the window its two extreme states allow
• the bracket's span is at most four times the crossings
• the cinquefoil admits a colouring using more than one of the 5 colours
• the cinquefoil has the crossings it claims
• the cinquefoil takes ± its determinant at θ = π
• the cinquefoil takes the value 1 at θ = 0
• the cinquefoil, one crossing switched: a diagram that does not alternate has a bracket narrower than 4c
• the cinquefoil: |Δ(−1)| is the determinant
• the cinquefoil: a reduced alternating diagram's bracket spans exactly 4c
• the cinquefoil: an alternating diagram's extreme states meet themselves exactly at its untwistable crossings
• the cinquefoil: an alternating diagram's two extreme states have c + 2 circles between them
• the cinquefoil: on an alternating diagram the two bounds agree
• the cinquefoil: the bracket gives the Jones polynomial the handedness requires
• the classification is checked on at least a hundred and fifty small diagrams
• the composite's slope is about twice the trefoil's
• the composites' share rises with length
• the constant colourings are always allowed, so there are at least p
• the core curve is one the family knows
• the corrected steps return to the start
• the crossing count and Gauss's integral agree
• the curve is the knot 5₂ (polynomial 2t − 3 + 2t⁻¹)
• the curve stays clear of the projection point
• the decay length is a few hundred sticks
• the discs and bands form one connected surface
• the drawn view is the same knot
• the family contains the Petersen graph, K₃,₃,₁ and K₄,₄ minus an edge
• the figure-eight knot admits a colouring using more than one of the 5 colours
• the figure-eight knot has the crossings it claims
• the figure-eight knot takes ± its determinant at θ = π
• the figure-eight knot takes the value 1 at θ = 0
• the figure-eight knot, one crossing switched: a diagram that does not alternate has a bracket narrower than 4c
• the figure-eight knot: |Δ(−1)| is the determinant
• the figure-eight knot: a reduced alternating diagram's bracket spans exactly 4c
• the figure-eight knot: an alternating diagram's extreme states meet themselves exactly at its untwistable crossings
• the figure-eight knot: an alternating diagram's two extreme states have c + 2 circles between them
• the figure-eight knot: on an alternating diagram the two bounds agree
• the figure-eight knot: the bracket gives the Jones polynomial the handedness requires
• the first component of this link is a flat circle, so its disc is the obvious one
• the five twists are three one way and two the other
• the four ends at a crossing alternate over, under, over, under
• the Gauss integral of the computed orbits agrees with the template
• the heptagon: the polynomial is the same from every direction
• the hexagon: the polynomial is the same from every direction
• the Hopf link: punctures and crossings agree
• the Hopf link: the diagram's count and Gauss's integral agree for components 1 and 2
• the Jones polynomial of a knot takes the value 1 at t = 1
• the knot is one of 3_1, 4_1, 5_1, 5_2, 6_1, 6_2, 6_3, 7_1, 7_2, 7_3, 7_4, 7_5, 7_6, 7_7
• the knot is one the family draws, with crossings
• the knot is one the figure knows
• the knot's Jones polynomial is one the family knows
• the knots in the table have different determinants
• the knotted curve turns through more than two full turns
• the link is one the family knows
• the linking number has heavier tails than a bell
• the linking number is the same at every deformation
• the minor is not zero
• the mirror image of the trefoil has the crossings it claims
• the mirror image of the trefoil: a reduced alternating diagram's bracket spans exactly 4c
• the mirror image of the trefoil: an alternating diagram's extreme states meet themselves exactly at its untwistable crossings
• the mirror image of the trefoil: an alternating diagram's two extreme states have c + 2 circles between them
• the mirror image of the trefoil: the bracket gives the Jones polynomial the handedness requires
• the mirror's Jones polynomial is the original's with t replaced by 1/t
• the modular computation agrees with the symbolic one
• the modular determinant is the exact one
• the number of embeddings is a whole number between 200 and 5000
• the number of linked pairs changes by an even amount
• the orbit closes after one period
• the orbit winds round the lobes in the order its name says
• the orbits are among LR, LLR, LRR, LLLR, LLRR, LRRR, LLLRR, LLRLR, LLRRR, LRLRR
• the pentagon: the polynomial is the same from every direction
• the picture from above has three crossings
• the polygon formula gives the integer once the polygons are fine enough to stay disjoint
• the polynomial at t = −1 is the determinant found from the colourings
• the polynomial is symmetric
• the polynomial is symmetric — the same read backwards
• the polynomial takes the value ±1 at t = 1
• the polynomial's bound never exceeds the surface's genus
• the punctures added with signs give the linking number
• the ribbon closes up
• the seed is a whole number between 1 and 100000
• the seven-crossing torus knot has the crossings it claims
• the seven-crossing torus knot, one crossing switched: a diagram that does not alternate has a bracket narrower than 4c
• the seven-crossing torus knot: |Δ(−1)| is the determinant
• the seven-crossing torus knot: a reduced alternating diagram's bracket spans exactly 4c
• the seven-crossing torus knot: an alternating diagram's extreme states meet themselves exactly at its untwistable crossings
• the seven-crossing torus knot: an alternating diagram's two extreme states have c + 2 circles between them
• the seven-crossing torus knot: on an alternating diagram the two bounds agree
• the seven-crossing torus knot: the bracket gives the Jones polynomial the handedness requires
• the signed crossings add to twice the linking number
• the signed crossings of two closed curves add to an even number
• the strand is broken where it goes underneath, not somewhere else
• the sum stays odd at every position
• the surface's Euler characteristic gives a whole genus
• the table contains a link whose components cross and whose linking numbers are all zero
• the table is ordered by crossing number
• the table's columns are small primes
• the table's diagrams alternate until a crossing is switched
• the total is odd in this placement too
• the trefoil admits a colouring in more than one colour
• the trefoil admits a colouring using more than one of the 3 colours
• the trefoil has the crossings it claims
• the trefoil is the commonest knot at every length from 100
• the trefoil is three arcs
• the trefoil takes ± its determinant at θ = π
• the trefoil takes the value 1 at θ = 0
• the trefoil, one crossing switched: a diagram that does not alternate has a bracket narrower than 4c
• the trefoil: |Δ(−1)| is the determinant
• the trefoil: a reduced alternating diagram's bracket spans exactly 4c
• the trefoil: an alternating diagram's extreme states meet themselves exactly at its untwistable crossings
• the trefoil: an alternating diagram's two extreme states have c + 2 circles between them
• the trefoil: on an alternating diagram the two bounds agree
• the trefoil: the bracket gives the Jones polynomial the handedness requires
• the trefoil's share of knotted loops falls with length
• the trefoil's share rises with length to 400 sticks
• the tube radius is between 0.05 and 0.9
• the two computations agree at this deformation
• the two loops stay clear of each other at every amplitude
• the two polynomial values identify the knot the full polynomial names
• the two strands at a crossing are not parallel
• the unknot is a single arc
• the view is one the family draws
• there are nine three-colourings of the trefoil in all
• there are three moves, and Reidemeister proved there are no others
• triangle–star exchanges from K₆ reach exactly seven graphs
• turning the third phase alternates between 5₂ and the unknot
• twice the over-strand equals the two under-strands added, at every crossing
• twist and writhe add to the held integer at every shape
• two circles side by side: punctures and crossings agree
• two circles side by side: the diagram's count and Gauss's integral agree for components 1 and 2
• two closed curves cross an even number of times
• two edges never meet in space
• two independent runs agree on the mean square linking number
• two or three links whose first component is a flat circle
• two projections agree after the shift
• two projections give the same linking number
• two projections give the same polynomial
• two projections give the same whole number
• two regular fibres link pq times
• two triangles' signed crossings add to an even number
• two words in L and R of length 2 to 8, each using both letters
• up to seven crossings exactly 5₂, 6₁, 7₄ and 7₅ pass both tests
• while the number of crossings is not
• with one edge deleted, some random placement links nothing
• x is unchanged
• y changes sign
• z changes sign

• rows 0 and 1 of the character table are orthogonal ×42
• on the 1-sheeted cover, the lift of a loop winding 0 times closes exactly when it should ×24
• generator 1 acts as a permutation of the 3 sheets ×14
• point 1's entry is a permutation of the 4 sheets ×13
• edge 1's entry is a permutation of the 3 sheets ×10
• the rank comes out at 1 + 3(2 − 1), which is the index times one less than the rank below ×9
• classes within 1 letters once the letters commute ×8
• elements within 1 letters in the free group ×8
• a loop winding 0 times lifts to a closed path exactly when 0 is a multiple of 3 ×7
• rows 0 and 0 of the character table are normalised ×7
• the class sizes of S2 add to its order ×4
• vertex 1 has at most one a leaving it ×4
• vertex 1 has at most one b leaving it ×4
• face 1 closes up on its own sheet ×3
• the loop drawn for class 0 is a whole number ×3
• 4 sheets, two 4-cycles and a double swap: the character sum and the enumeration agree ×2
• the loop drawn for class -2 is a whole number ×2
• the symmetry returns to the identity first at the 4th power ×2
• 2 sheets, four swaps: the character sum and the enumeration agree
• 3 sheets, three 3-cycles: the character sum and the enumeration agree
• 4 sheets, three double swaps: the character sum and the enumeration agree
• 4 sheets, two double swaps and a 3-cycle: the character sum and the enumeration agree
• a class on the torus is a pair of whole numbers, not both zero
• a closed lift on the grid encloses a whole number of squares, counted with sign
• a free group of rank two has four words of length one
• a list containing a single d-cycle — a polynomial — is always realised
• a meander crossing the line 6 to 14 times
• a meander is found
• a spanning tree of a connected graph has one edge fewer than it has vertices
• a straight loop of a coprime class never crosses itself
• a three-crossing curve whose one non-trivial symmetry moves no region gives five plane curves, one for each region
• a twist is a homeomorphism, so a loop that did not cross itself still does not
• and are not under the other one
• and at most one arriving
• and at the same height
• and by the end the loop has been dragged through the hole
• and consecutive ones differ by exactly one turn
• and every list that is not realised passes the count and the parity anyway
• and exactly one arriving
• and in the second, so it closes up on the torus
• and inside the outer boundary
• and it goes round each hole a net zero times
• and its degree is even
• and its degree is odd
• and its exponent sums vanish
• and on exactly two, once in each direction
• and the answer is a whole number, as a count of lists must be
• and the losses add to an even number, as the sign of a product of permutations forces
• and the same word cancels to nothing on the wedge, where it stays four letters long
• and the top and bottom
• and the top and bottom as many times as its second
• and their number divides the number of sheets
• and they are those multiples in order
• and twelve of length two
• and up
• between one and five words to be refused
• between one and four generator words in a, b and their inverses
• between three and six of the shape lists the family knows are tabulated
• between two and eight branch points
• between two and five sheets
• between two and six sheets
• beyond 3 the small windings are commoner than the big ones
• both loops start at the same point of the ring
• Brownian windings have the Cauchy tail at 3
• by thirty sheets nearly every covering is connected and nearly the whole count comes from the two one-dimensional characters
• carrying the loop along either path leaves its class alone
• commuting pairs of permutations of n things number n! times the partitions of n
• crossings plus turning number is always odd
• curve is read only by the twist views
• curves is read only by the slopes and simple views
• each curve is counted once for every way of starting and orienting it, divided by its symmetries
• each loop in the orbit is in its class, across
• each loop on the torus is a pair of whole numbers, not both zero
• each pair: one to four pairs of whole numbers, not both zero
• each shape names a conjugacy class of the group
• each sheet has exactly one edge of each label leaving it
• eight failures at ten sheets contain no shape with all cycles equal
• every branch point's cycles fit side by side inside its column
• every class drawn here is coprime
• every class on the way is coprime
• every coprime class walks down to (1, 0)
• every cover's counted rank matches the formula
• every curve along the way turns twice
• every letter is used as often forwards as backwards
• every lifted edge appears on some face
• every marked point actually branches
• every pair is counted once, and exactly Cₙ pairs make n loops — the pairs of identical systems
• every point of the fibre projects onto the marked point of the circle
• every sheet is reached by some word, so the cover is connected
• every turning number is a whole number
• every two-letter loop nought in homology is the trivial loop
• every word a curve realises satisfies Gauss's even-interlacing condition
• exceptions by degree 3 to 13
• four crossings give nineteen curves on the sphere
• Frobenius's formula counts the solutions of [a, b][c, d] = 1 in the symmetric group exactly
• from is read only by the twistpath view
• Gauss's condition first admits a word no curve realises at five crossings
• going round all the branch points returns every sheet, so this is a covering of the sphere
• going round the face returns every sheet to itself, so the permutations describe a covering
• Hall's recursion agrees with counting every pair
• in the disc every stage of the shrink stays inside
• lattice windings have a tail at 3 far thinner than the Cauchy law's
• longer loops are less often unwound
• longer loops hide more and have longer words
• no sample of the loop lands on the point it is wound about
• no step of the walk turns more than a quarter of a turn
• no two cycles drawn in one lane overlap
• no two deck transformations agree on the base vertex
• of the 36 four-letter closed lifts, 28 are trivial words
• one permutation for each of the 2g edges
• one to four pairs of loops
• order 6 to 9
• over the line, the only loop that lifts to a loop is the one that goes nowhere
• pairs is read only by the meet view
• running one loop then the other adds the counts
• so the two maps are not the same map
• so the whole fibre sits over one point
• so there are at most as many symmetries as sheets
• span is read only by the primitive view
• symmetry is read only by the twistorder view
• the a's in the word add up to the winding number about that hole
• the b's in the word add up to the winding number about that hole
• the base circle itself does not lift to a loop
• the bouquet below has two or three circles
• the bouquet has between one and four circles
• the bouquet has two to four circles
• the cells count what the Riemann–Hurwitz formula says
• the chance a random pair is not transitive is about 1/n
• the character sum and the enumeration agree
• the character table of this symmetric group is one the family carries
• the class is coprime with both counts positive
• the class walked down: one to four pairs of whole numbers, not both zero
• the classes drawn are whole numbers no bigger than four
• the classes that lift to loops are the multiples of the sheet count
• the classes weighed by one over their symmetries add to the labelled codes over 8n
• the commutator does not cancel down any further
• the commutator has exponent sum zero in the first generator
• the composition of two deck transformations is one
• the conjugacy classes 1, 3, 7, 26, 97
• the cover has between two and five sheets
• the cover has between two and four sheets
• the cover has genus d(g − 1) + 1
• the cover is connected, so its deck group acts on one object
• the cover is connected, so the index argument applies
• the cover is connected, so the sheets are one orbit
• the cover's Euler characteristic is even, as a closed orientable surface's must be
• the covering is one of torus, genus2, cuberoot, power, simple
• the curve about the left hole is a whole number
• the curve about the right hole is a whole number
• the curve round one handle reads a reduced commutator
• the curve: one to four pairs of whole numbers, not both zero
• the curves: one to four pairs of whole numbers, not both zero
• the degree of the map that is not odd is a whole number
• the degree of the odd map is a whole number
• the disc lifts to one face on every sheet
• the drawing shows between two and five turns
• the edges outside the tree are exactly the free generators the graph carries
• the eight four-letter loops that are nought in homology and not trivial are the commutators
• the Euler characteristic is that of a closed orientable surface
• the Euler characteristic multiplies by the number of sheets
• the example winds zero times round each puncture and is still a word of length four or more
• the extrapolated growth rate lies between 11 and 13.5
• the first loop is a whole number
• the first map really does send opposite points to opposite points
• the folded graph has at least one independent loop
• the free group on two letters has 1, 3, 13, 71, 461 subgroups of index 1 to 5
• the free group on two letters has thirteen subgroups of index three
• the free group outgrows the abelian group of the same rank
• the gallery's turning numbers are 1, 0, 2, 3, 4 and −2
• the generator aa is accepted by the folded graph
• the generator ab is accepted by the folded graph
• the generator abab is accepted by the folded graph
• the generator abaB is accepted by the folded graph
• the generator ba is accepted by the folded graph
• the generator bb is accepted by the folded graph
• the hole is between a fifth and three fifths of the ring
• the identity is a deck transformation
• the index of the subgroup is the number of sheets
• the integers have one subgroup of each index
• the lattice runs from 4 to 12 each way
• the lattice walks are already much narrower than the Cauchy law, whose quartiles are 2 apart
• the lift ends a whole number of turns above where it started
• the lift ends at the letters' exponent sums
• the limaçon's tangent turns through exactly two full turns
• the loop at p is a whole number
• the loop being lifted is a whole number
• the loop crosses itself one time fewer than the common factor
• the loop crosses the side edges as many times as its first count
• the loop drawn at p has the class claimed
• the loop drawn for a class really has that class
• the loop is in the class it claims, across
• the loop lifted goes round between one and four times
• the loop lifted is the one asked for
• the loop lifted winds between zero and six times
• the loop stays clear of the hole
• the loops composed have counts no bigger than three
• the loops tabulated wind between zero and eight times
• the marked point is on the base circle
• the meanders of each order match the published counts
• the monodromy table is drawn over two generators
• the number of sheets is what one lap advances by
• the number of subgroups of index 30 is within one per cent of 2n (n!)^(2g−2)
• the pairs sorted by the orbit of the first sheet account for all of them
• the path crosses the side edges as many times as its first count says
• the permutations connect every sheet to every other
• the permutations reach every sheet from every other, so the covering is connected
• the plane counts are 1, 2, 5, 20, 82, 435, 2645
• the printed terms add to the printed answer
• the probe's two images are exactly opposite under the odd map
• the second loop is a whole number
• the second map does not, which is what makes it a control
• the seven-sheet list (4,3), (3,3,1), (2,2,2,1) is realised
• the shape list is one the family knows
• the share falls with every two letters
• the share of coprime classes is near 6/π²
• the shrink is legal to begin with and stops being legal
• the slide is drawn in three to five stages
• the speed stays well away from nought along the whole deformation
• the speed vanishes exactly at s = ½ and nowhere else nearby
• the sphere counts are 1, 1, 2, 6, 19, 76, 376
• the subgroup counts 1, 3, 13, 71, 461, 3447
• the surface has genus 1, 2 or 3
• the symmetries reach every sheet exactly when there are as many of them as sheets
• the symmetry is ab or aba
• the table holds a list that is realised and a list that is not
• the table runs to between three and six sheets
• the table runs to between two and six sheets
• the three permutations multiply to the identity
• the torus has σ(n) subgroups of index n
• the turning number is 1 before the cusp and 0 after it
• the twist is done one to three times
• the twisted loop crosses the side edges as the matrix predicts
• the two run one after the other is a whole number
• the two squares never overlap during the slide
• the view is one the family draws
• the walk ends
• the winding number is a whole number
• the word a is refused
• the word aa is refused
• the word ab is refused
• the word b is refused
• the word is one this family realises
• the word is reduced and not empty, so the loop is not trivial in the group
• the word read off the curve is the word the curve was built from
• the words that come back are closed under composition
• there are exactly eight meanders of order three
• three crossings give six curves on the sphere and twenty in the plane
• three letters give 7 and 97 subgroups of index two and three
• times is read only by the twist view
• two alternating squares multiply to four swaps
• two fixed-point-free involutions realise only the even split
• two matchings forming one alternating cycle multiply to two equal cycles
• two non-trivial loops lift to closed paths
• two or three circles below
• two straight loops meet as often as the determinant of their classes says
• two words land on the same sheet exactly when one undoes the other into the subgroup
• up to eight sheets nineteen data fail, none of them in prime degree
• while a word that really is trivial reduces away, so the test can fail
• Whitney's formula holds on every random curve
• with the last point's permutation removed, going round no longer returns every sheet
• words of length up to two to four are listed

• the ledger's bound for dimension 1 is one less than the next power of two ×32
• at dimension 2 the lower bound meets Whitney's ×5
• after half a turn the first arrow points the opposite way
• and between 24 and 140 the long way
• and so does the second, which is exactly what the double flip at the join demands
• and the band's most economical section is zero exactly once
• between four and nine stations
• between none and sixteen cross-sections are outlined
• between one and four sections are drawn
• between one and three half-twists
• between three and nine frames are drawn along the band
• both rows the figure is about are drawn
• carrying the surface round once brings it back on the other face
• each folded band has length k tan(π/2k), above π/2
• each section meets itself at the join
• each section meets itself at the join with its sign changed
• every section of the cylinder crosses the centre line an even number of times
• every section of the Möbius band crosses the centre line an odd number of times
• going once round the long way returns to the point with the cross-section reversed
• going once round the tube returns to the same point
• on its centre circle the curvature is −1/4
• the band is drawn at a whole number of half-twists
• the cylinder has a section that is never zero
• the edge runs once round the triangle
• the folded strip closes with a flip
• the folded strip lies in an equilateral triangle of height 1
• the frame comes back the same size — only its sense can change
• the frame returns with a reversed sign exactly when the band is one-sided
• the gluing arrow matches the twist the figure is about
• the inverse ledger is the row of Pascal's triangle the next power of two predicts
• the lengths fall as the folds increase
• the mesh is between 16 and 96 round the tube
• the section's sign at the join is fixed by the parity of the half-twists
• the self-intersection circle is on or off
• the single edge has length twice the strip's
• the tangent and normal ledgers multiply to the room's, which is 1
• the textbook band is curved like a saddle at every point sampled
• the tube's distance from the axis leaves the figure-eight room to close
• the turning frame never degenerates and never changes its sense
• the view is one the family draws
• three and nine folds join with a flip, six folds join straight
• three folds at sixty degrees close a strip of length √3 with a flip
• u = 0 and u = π are two different places on the surface and one place in space
• which is not the same point unless it is one the reversal fixes

• and one lap leaves one loop while two laps leave two
• the cut is inside the band
• the scissors take one lap down the middle and two off-centre

• none of the 594 pairs of non-neighbouring edges meet ×5
• a pair of horns at stage 1 is clasped, of linking number ±1 ×4
• stage 3 has 2^3 − 1 clasped pairs ×4
• ear clipping: a polygon of 20 corners is cut into 18 triangles ×2
• the image triangulation: a polygon of 20 corners is cut into 18 triangles ×2
• the source triangulation: a polygon of 20 corners is cut into 18 triangles ×2
• a fill spreading through corners escapes through a diagonal step of the outline
• a fill spreading through edges stays inside and covers about the enclosed area
• a line with a point removed falls into two pieces
• a plane with a line removed falls into two pieces
• a plane with a point removed stays in one piece
• a triangulation of a polygon has one diagonal fewer than it has triangles
• all 16 rays agree on the parity
• an acute triangle carries three squares, a right-angled two, an obtuse one
• and a ray from the centre crosses it an even number of times
• and both insides keep a substantial part of the frame
• and is not linked with that horn's partner
• and the clasp shows as at least two crossings of the projection
• and the steepest chord anywhere grows geometrically as the scale shrinks
• and they disagree on the count, which is the point
• at every angle tried the chords steepen by more than fifteenfold over the scales measured
• at every proportion sampled the octagon carries a rectangle
• away from squares the count of rectangles is even at all but isolated proportions
• away from the instants of contact the count changes by two
• between 2 and 36 rays are drawn
• consecutive squares in the visiting order are neighbours
• each found rectangle has the requested proportions
• each square opens into two rectangles as the proportion leaves one
• each square's side is at most 2⁻ⁿ, so the diameters go to zero
• ear clipping: no two triangles overlap
• ear clipping: the triangles account for exactly the polygon's own area
• every cell of the grid is visited exactly once
• every direction gives the same parity
• every direction of ray gives the same verdict at this point
• every polygon in the sample carries an odd number of squares
• every simple closed curve of eight-neighbours leaves exactly two pieces of four-neighbours
• every simple closed curve of four-neighbours, larger than a 2 × 2 block, leaves exactly two pieces of eight-neighbours
• every simple polygon of four or more corners has an ear to clip
• halving the pitch more than halves the largest disc that fits
• horns from different clasps are not linked with each other
• no simple closed curve of eight-neighbours separates anything when the rest also uses eight
• no two of the kept squares overlap
• on the honeycomb every simple closed curve leaves exactly two pieces
• read with eight neighbours the same pixels are connected but not a simple curve
• some sampled point is inside the curve
• some simple closed curve of four-neighbours leaves three pieces of four-neighbours
• space with a line removed stays in one piece
• the arms of the spiral are between 22 and 70 units apart
• the clipping terminates
• the count is odd along the path except at isolated instants
• the counts themselves differ, which is why the parity is the claim
• the curve goes twice round its own centre
• the disc drawn touches no edge of the curve
• the drawn outline is one piece when corners count
• the drawn ray crosses the surface more than once
• the five-pointed loop crosses itself five times
• the horns shrink geometrically, so their tips converge
• the image triangulation: no two triangles overlap
• the image triangulation: the triangles account for exactly the polygon's own area
• the infinite product is bounded away from zero
• the kept area is the product of the stages' own factors
• the loop drawn goes once round one horn
• the magnification of the second panel is between 3 and 27
• the marked point is inside the frame, given as fractions of it
• the number of stages drawn is between 1 and 4
• the number of stages of horns is between 1 and 4
• the number of terms summed is between 2 and 6
• the octagon carries rectangles of each of these proportions
• the outside point is clear of the tube
• the plane comes apart into exactly two pieces, no more and no fewer
• the point is put inside or outside
• the radius stays positive, so no two angles share a point
• the resolution of the flood fill is between 24 and 90
• the same curve leaves three pieces of four-neighbours and two of eight
• the sample grid is between 8 and 40 columns
• the sampled inside agrees with the polygon's own area to within six per cent
• the second corridor is the narrower one
• the source triangulation: no two triangles overlap
• the source triangulation: the triangles account for exactly the polygon's own area
• the spiral turns between 1 and 6 times
• the terms roughen faster than they shrink
• the tube is closed up out of two triangles per patch
• the two triangles either side of a diagonal send a point on it to the same place
• the view is one the family draws
• this octagon carries exactly five inscribed squares
• two pitches, both inside the range the family draws
• under the mixed rule, parity along each row marks exactly the piece the curve cuts off
• with eight neighbours for both, the inside leaks out through the corners
• with eight neighbours for the curve and four for the rest, the diamond is one piece and leaves two
• with four neighbours for both, the diamond falls into twelve single pixels and still leaves two pieces

• the image of 0 lies on the circle ×14
• the ray to 0 passes through the pole ×14
• the image of -2 lies on the circle ×13
• the ray to -2 passes through the pole ×13
• a circle missing the pole projects to a circle, to within a fitted residual
• a circle through the pole projects to a straight line
• and no two of them meet
• and the south chart does have one there
• at least two finite points are carried across the projection
• at the equator the horizon passes through the south pole and projects to a straight line
• between 2 and 24 rays are drawn — one ray is not a correspondence
• between a hundred and four thousand points checked
• between three and seven fibres, each at a latitude strictly inside the sphere
• dim and rays are read by the ray views alone
• each altitude circle has centre cos φ/(sin φ + sin a) and radius cos a/(sin φ + sin a)
• each fibre lies on the unit three-sphere
• each transformation is normalised to determinant one
• every engraved circle of the plate — horizon, altitude, azimuth, ecliptic — is a circle to within rounding
• every patch has the same area on the sphere
• every two fibres are linked exactly once, counted from the drawn curves
• for every star, inside the horizon on the plate is the same as above the horizon by spherical trigonometry
• nearly every sample lay in the overlap
• none of the circles in this list passes through the pole
• only the parabolic map has a repeated fixed point
• the crossing angle is the same on the sphere and in the plane
• the fitted transformation is invertible
• the horizon's circle shrinks as the latitude rises
• the hour angle of sunrise read off the plate equals arccos(−tan φ tan δ)
• the image is on the ray from the pole through the point
• the image lands on the plane
• the north chart has no value at its own pole
• the projected patches are wildly unequal even though the originals are equal
• the projection is drawn from a circle or from a sphere
• the rotation, read in the plane, is exactly the fitted Möbius transformation
• the sixth hour of the night is the lower meridian, a straight line
• the three chosen points determine the transformation
• the trace classifies the transformation as the label claims
• the two charts differ by the reciprocal at every point of the overlap
• the two charts miss different points
• the unequal-hour lines are not circles, and the three-point arc misses each by under two per cent of the plate
• the view is one the family draws
• there are pairs to link

• 0.3 and 0.7 are equally far from a fair coin
• a coin with bias 0.6 reaches twenty to one almost always
• a halfer told it is the first day moves above one half
• a p-value of 0.05 is worth at most about 2.46 to one
• a positive result is evidence for and a negative against
• a prior with thinner tails has a finite expected amount
• a thirder told it is the first day comes back to one half
• adding decibels is multiplying odds
• and at seven days it is 13/27
• and at three doors it is two thirds
• and it is above a half for every pair
• and the boy–boy block holds 2d − 1 of them
• at a hundred million tosses even odds need about four standard errors
• at a trillion tosses the Bayes boundary sits more than two standard errors above the iterated-logarithm envelope
• at every middle amount the other envelope is worth 5/4 as much on average
• at one awakening every account agrees on one half
• at p = 0.05 and a million tosses the fair coin is favoured by more than a hundred to one
• at p = 0.05 even the best single bias makes the result only about 6.8 times likelier than a fair coin
• at p = 0.05 the uniform alternative and the fair coin are at even odds somewhere between 50 and 100 tosses
• at the same z-score the best bias beats a fair coin by the same factor, about 6.8, at every sample size
• at η near (1 − ρ)/(1 + ρ) the coverage is back to 95%
• averaged over what might be seen, switching gains exactly nothing
• Broome's partial sums grow without bound
• combining p-values while choosing whether to continue inflates the error well past five per cent
• each test alone keeps its false-positive rate
• every order of the evidence ends at the same odds
• every run ends almost certain of the bias 0.6, which is false
• everybody has it, so everybody positive has it
• fewer than one fair coin in twenty ever shows twenty to one against fairness
• however the spread is chosen, p = 0.05 favours the biased coin by at most about two to one
• Jahn's 104,490,000 trials: p below 0.0005 and the fair coin favoured about twelve to one
• no fixed bet grows faster than the true likelihood ratio
• nobody has it, so nobody who tests positive has it
• one of the two strategies wins, and only one
• one ticket a run is fair at one half
• one ticket an awakening is fair at 1/(wakes + 1)
• one to five positive likelihood ratios
• prior is a probability, between 0 and 1
• seeing 1, the other is surely 2
• seeing any larger amount, the other is double with probability 2/5
• so its expected value is 1.1 times the amount seen
• some fair coin leaves the 5% band at least four times
• specificity is a probability, between 0 and 1
• staying wins as often as the first pick was right
• switching gets better with more doors
• switching wins two thirds against a host who knows, and half against one who does not
• tails brings between 1 and 8 awakenings
• the answer falls as the detail becomes commoner
• the answer rises with the base rate
• the area under the likelihood ratio is the Bayes factor for the biased coin
• the asked families settle near 1/3
• the average over a uniform bias is many times smaller at the larger sample
• the awakenings settle near one share in wakes + 1
• the bound is the best calibrator
• the calibration −1/(e p ln p) puts p = 0.05 at about 2.5 to one
• the chance of the larger is a half plus half the chance the threshold falls between the two
• the closest hypothesis in Kullback–Leibler divergence is 0.6
• the count is the formula (2 − p)/(4 − p)
• the coverage is what the inflated variance predicts
• the curve agrees with the grid count
• the error rate is near Wald's bound 1/(1 + 10^(threshold/10))
• the formula is the kept area
• the four cells fill the square
• the game needs at least three doors
• the halfer's square gives heads one half
• the ignorant host's story throws two of the six worlds away
• the ignorant-host variant is only drawn at three doors, where switching has one meaning
• the iterated-logarithm envelope has passed 1.96
• the kept area gives the protocol's answer
• the largest pair is 2^N and 2^(N+1), N between 2 and 10
• the learned product passes 20, and a fresh prior in every study falls behind it
• the marked row and column share one cell
• the met-a-child families settle near 1/2
• the mixture's rate rises towards the best rate
• the more the errors are shared, the less the second test adds
• the only news drawn is that it is the first day
• the paths differ on the way
• the product of e-values passes 20 at most one time in twenty, whatever the continuation rule
• the product of learned e-values is the Bayes factor of all the tosses pooled
• the question is any, older or met
• the runs settle near one half
• the same coin's Bayes factor never reaches twenty to one
• the shaded fraction is Bayes' theorem
• the share of fair coins whose Bayes factor ever reaches k stays below 1/k
• the single bet beats the uniform prior near its own guess and is blind to a coin that favours tails
• the spread that suits the biased coin best shrinks with the sample
• the thirder's square gives heads one share in wakes + 1
• the trait takes between 1 and 12 values
• the two curves are mirror images about one half
• the two strategies exhaust the possibilities
• the typical Bayes factor falls steadily as tosses accumulate
• the typical stopping point is of the same order as the fixed sample a 5% test with 80% power needs
• the view is one of sweep, monty, monty-scale, children, tuesday, trait, mention, children-sim, beauty, beauty-sim, beauty-bets, beauty-many, oddsladder, oddsorders, oddsdependent, oddssequential, oddsworth, envbounded, envbroome, envexpect, envsim, envthreshold, misclosest, miskl, misdrift, mistie, miscover, mistemper, lindhero, lindlike, lindbound, lindthresh, lindprior, oshero, ospath, osville, osmean, osboundary, ospower, evhero, evcont, evgrow, evcost, evconvert, evprior
• the walk's slope is the difference of the two divergences
• the worlds fill the square
• the z-score for even odds is close to √ln(2n/π)
• under a fair coin the Bayes factor for a coin of unknown bias averages exactly one at every number of tosses
• watched for 200 tosses, a fair coin passes 20 at most one time in twenty
• watching after every toss, more than half of fair coins reach p < 0.05 within 100,000 tosses
• watching for twenty to one needs more tosses than a single test at five per cent
• with no shared cause the product is exact
• with the number of studies fixed in advance, Fisher's method holds its level

• the near-birthday chance at k = 0 is a distinct-birthday chance on a shorter year ×7
• 4 lists find solutions at 12 bits in most runs ×4
• at the predicted sample size the measured typical error is near a tenth for 1000 ×3
• 22 is still under a half
• 23 is over it
• a block is between 2 and 365 days
• a few thousand draws pin ten thousand kinds to within several per cent
• a group smaller than the calendar
• a larger first sample lets the smoothed sum reach further
• a match within a day needs 14
• a run with a solution is found among the first fifty
• a seasonal swing is between nothing and the whole mean
• a spike holds more than its fair share and less than everything
• a tail and a cycle short enough to draw
• a weekend day is thinned, not removed or boosted
• an exact match needs 23 people
• an uneven calendar has at most as many effective days as it has days
• and its mean length is within ten per cent of √(πp/2)
• and no calendar needs a bigger room than the even year
• and the four cancel
• and the threshold only falls
• and where it misses, the calendar needs one MORE person than its effective days predict, never one fewer
• at 512 strings a list, four lists nearly always succeed and two nearly never do
• at t = 1 the plain estimate is within ten per cent
• at t = 1.5 the spread is more than ten billion times the answer
• at t = 2 the smoothed mean is within ten per cent
• at t = 6 the smoothed sum falls short of the truth
• at the last step the two agree modulo the factor found
• averaged over two hundred samples the sum is within two per cent of the truth for t ≤ 1
• before the last step they agree modulo neither prime
• between 4 and 24 slots, every one possible
• c = 0 and c = −2 run far longer
• c = 1 and c = 3 average √(πp/2)
• Chapman's form of the estimate is nearly unbiased
• doubling the sample shrinks the typical error
• each merge keeps the list size near 2^(n/4)
• early repeats come at the rate a random function gives
• evening the two days raises the chance that nobody matches
• every product was split
• every solution XORs to zero
• for c = 0 every tail is a vanishing share of √p
• n is a product of two primes
• n is a whole number below a trillion
• nobody is alone at 3,064
• nobody is alone in a room of 3,064, more likely than not
• one person shares with nobody
• somebody shares at 23
• T(T − 1)/2 is right on average
• the chance of a match only rises as the calendar grows more uneven
• the curve bends downward
• the curve is symmetric about the even split
• the cycle length is the order of 2 modulo the odd part of the start's order, every time
• the day-probabilities add to one
• the effective number is far below the true count
• the effective number of days places the threshold within one person
• the even year's threshold is 23
• the expected number of cancelling quadruples
• the expected number of loners peaks when the room is as large as the year
• the first pair agrees on the low bits
• the half-way point is 23 people
• the highest point is the even split
• the largest share at theta one is near the Golomb–Dickman constant
• the last new value leads back into the cycle
• the mean length is √(πn/2) divided by the spread of the number of preimages
• the modulus is a prime of at most 5000
• the option is one this view reads
• the pair count is k choose 2
• the pair-count estimate settles on the effective number, not the count
• the plain estimate is off by more than the whole answer soon after t = 1
• the sample holds the primes asked for
• the search ends with a proper factor within forty steps
• the second differences are constant, so the curve is a parabola
• the second pair agrees on the low bits
• the simulation agrees with the exact walk
• the simulation agrees with the formula
• the steps grow as roughly the square root of the smaller prime
• the sweep is seasonal or weekly
• the sweep starts at the even year
• the symmetric-polynomial count agrees with the product on an even year
• the three thresholds come in this order
• the threshold follows the square-root estimate to within a couple of people
• the two pointers meet within tail plus cycle steps
• the uneven calendar is never less likely to produce a match
• the view is one of pairs, uneven, merge, sweep, effective, rho, rholength, pollard, cost, near, circle, spacing, nearscale, strong, loners, thresholds, smallyear, rhoc, tailcycle, squaring, rtoone, early, wagnertree, wagnersuccess, wagnerwork, wagnercount, estfirst, estpairs, estcapture, estuneven, wagnersolution, wagnerlevels, estcost
• the walk's survival stays within ten points of the birthday curve everywhere
• they meet at a multiple of the cycle length, once both are on the cycle
• two different slots
• two lists grow like a half power and four like a third
• what it returns divides n
• with only rare kinds the plain sum is within eight per cent at t = 3

• at least one needle is dropped
• every needle claimed was drawn
• so a needle crosses with probability 2/π
• the area under half a sine wave is 1
• the box has area π/2
• the estimate is in a plausible range for pi
• the printed estimate is 2Ln / dc

• the wait for a new one at 0 seen is 6/6 ×21
• after 5 shuffles the distance is Bayer and Diaconis's 0.924 ×6
• the 6 waits add to 6 times the harmonic sum ×3
• a deck of 4 to 520 cards
• a pattern losing both two-way races need not come last three-way
• a run of the game reproduces the expected length
• a three-toss pattern that beats both others two-way also wins the three-way race
• a triple with no circle and no pattern beating both others holds a tie
• a typical run sits just above the even split, the learner's worst case
• a walk of 40000 steps spends about the solved share of its time at A
• a walk of 40000 steps spends about the solved share of its time at B
• a walk of 40000 steps spends about the solved share of its time at C
• after five thousand tosses Kelly is ahead of every rival on most runs
• and agree better than the two smallest do
• and at most twice the conductance
• and comes back to A about as often as the solve says
• and comes back to B about as often as the solve says
• and comes back to C about as often as the solve says
• and equal chances are the best case — any unevenness makes the wait longer
• and it is still climbing at the same rate
• and it really is stationary
• and its narrowest door is half the ring, conductance 1/m
• and no better than the true minimum
• and on the expected length of the race
• and so does a four-toss one
• and so does the riffle shuffle's
• and that sum is bounded, so the shares can be normalised
• and the chance of ending at the top satisfies B = R + Q B
• and the chance of ever coming back is less than one
• and the chance of winning is the share of the stake held
• and the share of wins reproduces the computed chance
• and the traffic round the cycle is not equal in the two directions
• and they are the shares the rule leaves alone
• and they do not all take the same time, although every one of them has the same chance in any given window
• as much traffic goes from A to B as comes back
• as much traffic goes from A to C as comes back
• as much traffic goes from A to D as comes back
• as much traffic goes from A to E as comes back
• as much traffic goes from B to C as comes back
• as much traffic goes from B to D as comes back
• as much traffic goes from B to E as comes back
• as much traffic goes from C to D as comes back
• as much traffic goes from C to E as comes back
• as much traffic goes from D to E as comes back
• at an even split the learner holds about √(π/2n)
• at even odds the expected length is k times N minus k
• at least two chains, so the comparison says something
• away from even odds the chance of winning is the ratio the odds give
• betting Laplace's estimate toss by toss gives 2ⁿ H! T!/(n + 1)!
• between six and sixty steps are drawn
• between two and four step-up chances are drawn
• between two and six patterns of two to five coin tosses
• cube dimensions between 4 and 2048
• each larger cube falls over a smaller share of its own mixing time
• each law's curve ends at (1 + CV²)/2
• every chain named is one the family carries
• every distribution's tail follows 1/(n + 1) to within four standard deviations
• every measured constant lies on (1 + CV²)/2
• every point is inside the drawn axes
• every run's average is larger at a million waits than at a hundred
• every simulated average is within 0.03 of the exact one
• every start can go either way
• every state has somewhere to go
• every state holds a share
• every vertex of the graph has a neighbour
• for a die the waits that end average 137/50
• half Kelly falls to half on about an eighth of the runs
• HH: the chain's answer and the overlap sum agree
• HH's wait is between the window size and twice it less two
• HHH against HHT is settled by the toss after the first HH
• HHH: the chain's answer and the overlap sum agree
• HHH's three-way chance is its chance against TTH alone
• HHH's wait is between the window size and twice it less two
• HHHH: the chain's answer and the overlap sum agree
• HHHH's wait is between the window size and twice it less two
• HHT and THH wait alike
• HHT beats HTH with chance 1/(1 + q)
• HHT: the chain's answer and the overlap sum agree
• HHT's wait is between the window size and twice it less two
• hindsight beats knowing p, and knowing p beats learning it
• HT: the chain's answer and the overlap sum agree
• HT's wait is between the window size and twice it less two
• HTH: the chain's answer and the overlap sum agree
• HTH's wait is between the window size and twice it less two
• HTHT: the chain's answer and the overlap sum agree
• HTHT's wait is between the window size and twice it less two
• HTT and TTH wait alike
• HTT: the chain's answer and the overlap sum agree
• HTT's wait is between the window size and twice it less two
• HTTH: the chain's answer and the overlap sum agree
• HTTH's wait is between the window size and twice it less two
• HTTT: the chain's answer and the overlap sum agree
• HTTT's wait is between the window size and twice it less two
• iterating from A reaches the same share for A
• iterating from A reaches the same share for B
• iterating from A reaches the same share for C
• Kelly falls to half its starting wealth on about half the runs
• Kelly's growth equals ln 2 less the entropy at every bias
• laziness leaves no eigenvalue below zero
• lumping by weight loses nothing
• no three three-toss patterns beat one another in a circle
• no two edge weights are drawn on top of one another
• no two of the three are tied
• on a fair coin the best reply to HHH is THH
• on a fair coin the best reply to HHT is THH
• on a fair coin the best reply to HTH is HHT
• on a fair coin the best reply to HTT is HHT
• on a fair coin the best reply to THH is TTH
• on a fair coin the best reply to THT is TTH
• on a fair coin the best reply to TTH is HTT
• on a fair coin the best reply to TTT is HTT
• on a fair coin the first player can guarantee a third and no more
• on dumbbells the gap falls as the conductance itself
• on rings the fall takes the same share of the clock at every size
• on rings the gap falls as the square of the conductance
• on this sequence Kelly ends ahead of half and twice Kelly, and staking everything is ruined
• one of the three always appears first
• one to four matchups
• p³ passes a half at the cube root of a half's reciprocal
• rings run with the lower line and dumbbells with the upper
• row A holds probabilities
• row B holds probabilities
• row C holds probabilities
• row D holds probabilities
• row E holds probabilities
• row F holds probabilities
• row G holds probabilities
• row H holds probabilities
• shuffling again never makes a deck less random
• some four-toss triples do
• TH: the chain's answer and the overlap sum agree
• TH's wait is between the window size and twice it less two
• the alternating sum over subsets and the integral of one minus the product agree
• the average regret is ½ ln n plus a constant
• the best growth rate is ln 2 less the entropy of the coin
• the best multiple is close to mε²/(1 + mε²)
• the best sweep cut is within Cheeger's guarantee √(2·gap)
• the break-even sample is of the order of one over the edge squared
• the chain and the gamblers agree on HHH
• the chain and the gamblers agree on HHHT
• the chain and the gamblers agree on HTT
• the chain and the gamblers agree on HTTH
• the chain and the gamblers agree on TTH
• the chain and the gamblers agree on TTHH
• the chain is one of mixing, cycle, reducible, lazy
• the chain is one the family carries
• the chain with the smaller gap really does take longer
• the chance of a win is strictly between nothing and everything
• the chance of the most likely value keeps swinging
• the chances of leaving A add to one
• the chances of leaving B add to one
• the chances of leaving C add to one
• the chances of leaving D add to one
• the chances of leaving E add to one
• the chances of leaving F add to one
• the chances of leaving G add to one
• the chances of leaving H add to one
• the coin favours heads
• the coin favours the bettor
• the collection is one the family knows
• the corroborating run is between 500 and 40,000 games
• the crossing step averages 2/3 and the overshoot 1/3
• the cube's window shrinks against its mixing time
• the cycle is the discretised extreme-value law
• the cycle's forward chance is between a half and one
• the cycle's shares are stationary even so
• the cycle's stationary shares are equal
• the distance never exceeds the chance a coordinate is unpicked
• the distance to stationarity never rises
• the drawing runs out to between 10 and 60 states
• the error changes sign about every 0.42
• the Eulerian numbers count every ordering once
• the expected number of steps satisfies t = 1 + Q t
• the expected return to A is one over its share
• the expected return to B is one over its share
• the expected return to C is one over its share
• the family is rings or dumbbells
• the first player is favoured only when the coin is very biased
• the game is played over between four and nine totals
• the gap is at least half the square of the conductance
• the graph is connected, so one is a simple eigenvalue
• the graph is one of cycle, dumbbell, grid, cube, complete, path, twogrids, lollipop
• the graph is small enough to search every vertex set
• the grid solution of the renewal equation agrees with the exact uniform formula
• the learner's wealth is the average over q of the wealth of staking 2q − 1
• the line is at a whole number from 1 to 6
• the log of the 30th record's position averages about 29
• the long-run shares are drawn only for a chain that has them
• the loss from estimating is about 1/(2m) a toss
• the mean count of 1s follows (n/2)(1 − (1 − 1/n)^t)
• the mean number of records is the harmonic number
• the mean ranks the fractions one way and the median the other
• the mean settles at log₂ n + γ/ln 2 − 3/2
• the measured decay rate is the second eigenvalue
• the measured mixing time follows the eigenvalue's prediction
• the number of kinds to collect is between 2 and 16
• the odds formula for HHH against HHT
• the odds formula for HHH against HTH
• the odds formula for HHH against HTT
• the odds formula for HHH against THH
• the odds formula for HHH against THT
• the odds formula for HHH against TTH
• the odds formula for HHH against TTT
• the odds formula for HHT against HHH
• the odds formula for HHT against HTH
• the odds formula for HHT against HTT
• the odds formula for HHT against THH
• the odds formula for HHT against THT
• the odds formula for HHT against TTH
• the odds formula for HHT against TTT
• the odds formula for HTH against HHH
• the odds formula for HTH against HHT
• the odds formula for HTH against HTT
• the odds formula for HTH against THH
• the odds formula for HTH against THT
• the odds formula for HTH against TTH
• the odds formula for HTH against TTT
• the odds formula for HTT against HHH
• the odds formula for HTT against HHT
• the odds formula for HTT against HTH
• the odds formula for HTT against THH
• the odds formula for HTT against THT
• the odds formula for HTT against TTH
• the odds formula for HTT against TTT
• the odds formula for THH against HHH
• the odds formula for THH against HHT
• the odds formula for THH against HTH
• the odds formula for THH against HTT
• the odds formula for THH against THT
• the odds formula for THH against TTH
• the odds formula for THH against TTT
• the odds formula for THT against HHH
• the odds formula for THT against HHT
• the odds formula for THT against HTH
• the odds formula for THT against HTT
• the odds formula for THT against THH
• the odds formula for THT against TTH
• the odds formula for THT against TTT
• the odds formula for TTH against HHH
• the odds formula for TTH against HHT
• the odds formula for TTH against HTH
• the odds formula for TTH against HTT
• the odds formula for TTH against THH
• the odds formula for TTH against THT
• the odds formula for TTH against TTT
• the odds formula for TTT against HHH
• the odds formula for TTT against HHT
• the odds formula for TTT against HTH
• the odds formula for TTT against HTT
• the odds formula for TTT against THH
• the odds formula for TTT against THT
• the odds formula for TTT against TTH
• the overlap sum is the expected wait for HHH
• the overlap sum is the expected wait for HHT
• the overlap sum is the expected wait for HTH
• the overlap sum is the expected wait for HTT
• the overlap sum is the expected wait for THH
• the overlap sum is the expected wait for THT
• the overlap sum is the expected wait for TTH
• the overlap sum is the expected wait for TTT
• the panels drawn are genuinely different cases
• the patterns compared are all the same length
• the points to fit a slope through are not all at one place
• the recurrence agrees with counting every sequence
• the return chances satisfy the rule that defines them
• the ring's does not
• the ring's gap is (1 − cos 2π/m)/2
• the rows never move further apart as the power rises
• the run is between 2,000 and 400,000 steps
• the run spends its time in A in the computed share
• the run spends its time in B in the computed share
• the run spends its time in C in the computed share
• the same current flows across every edge of the cycle
• the seed is a whole number from 1 to 9999
• the sequence is between 50 and 5000 draws
• the series is added over between 60 and 4,000 terms
• the share read off the weights is the share the equations give
• the shares add to one
• the shortfall reaches one standard deviation at ½ n ln n
• the simulated rate is close to the formula
• the solved share for A survives a step
• the solved share for B survives a step
• the solved share for C survives a step
• the stationary distribution adds to one
• the system and the chain agree
• the system is not singular
• the top eigenvalue of a walk is one
• the traffic between neighbours balances
• the two largest cubes' curves agree on this clock
• the variance closes in on π²/(6 ln² 2) + 1/12 from below
• the view is one the family draws
• the wait for k heads in a row averages 2^(k+1) − 2
• the walk is between a thousand and four hundred thousand steps
• the weighted Conway odds give the chain's answer
• the weights add to the geometric sum they are
• the weights are a square table of between three and six states
• the weights are all one, so the sum grows without bound
• the weights are symmetric and never negative
• the weights grow, so there is nothing to normalise
• the worst sequence is all one face, and costs ln(n + 1)
• THH beats HHH unless the first three tosses are heads
• THH beats HHT unless the first two tosses are heads
• three patterns of one length
• TT: the chain's answer and the overlap sum agree
• TT's wait is between the window size and twice it less two
• TTH wins the spoiler race only by opening with TT
• while the wait for the rarest kind alone is already a lower bound

• after 6 rolls the sum stopped after 1 terms is above the truth ×105
• after 6 rolls the sum stopped after 2 terms is below the truth ×105
• for 1 people the chance lies between the two stopped sums ×60
• after 6 rolls the full sum agrees with the chain on faces seen ×35
• on the staircase of size 2, 0 rooks fit in S(2, 2) ways ×27
• the band of width 1 at 2: rook numbers and search agree ×17
• starting from event 1, the formula and every pattern of successes agree ×14
• with chances 1/k, starting at 1 is the classical rule passing over 0 ×14
• at 5 couples the chance is e^−2·(1 − 1/n) to within 1/n² ×12
• the ménage count at 3 by formula and by search ×10
• the 1-th factorial moment is 1 ×9
• the arrangements of 8 with exactly 0 in place, listed and counted ×9
• the count for threshold 0 agrees with the formula ×8
• the cycles of length 1, over all arrangements, number 40320/1 ×8
• r0 is Kaplansky's 2n/(2n − k)·C(2n − k, k) ×6
• Riordan's formula counts the 3 × 3 Latin rectangles with a fixed first row ×5
• every one of the 5040 orders was tried ×4
• the counted best for 5 matches the formula ×4
• the share with 0 matches is within sampling noise of the Poisson ×4
• the alternating sum lands on the 265 the search found ×3
• width 1 approaches e^−1 ×3
• with 2 choices among 8, every order counted agrees with the plan ×3
• between 3 and 9 objects, few enough to list ×2
• between three and twelve sizes, each between 2 and 20000 ×2
• 200 to 5000 trials
• 9 of them leave nothing where it started
• a 4 × 4 square has rook numbers 1, 16, 72, 96, 24
• a die of 3 to 12 faces
• a group of up to 20 to 120 people
• a longer horizon never lowers the guaranteed value
• a placement leaving a gap is drawn
• a run of rolls starting at the number of faces
• a single threshold keeps at least 1 − 1/e
• a worse first value makes the rule stricter at the second
• a year of 50 to 1000 days
• after any first night the pairs left contain no table of three
• all 24 arrangements are drawn
• among all deterministic strategies for four prisoners the best wins 10 of 24 shuffles
• an exhaustive point is at most 8 objects
• an odd number of guests from 5 to 13
• and by a larger factor each time
• and comes within a few per cent of it
• and it does better than taking one at random
• and its height there is the same number
• and sends no two objects to the same place
• and settles above the 0.5802 the limit is, rather than falling to 1/e
• and so forbid the same number of permutations
• and so is the chance of success
• and so is the variance
• and stays below the family's limit of about 2.332
• and stays below the limit it is climbing towards
• and the best single threshold secures at least half of it
• and the half is never breached
• and with three, 0.6842
• another choice always helps
• arrangements of 20 to 5000 objects
• at least as many rolls as faces
• between 3 and 12 objects
• between 3 and 14 events, so every pattern can be enumerated
• between one and three rooks
• between three and eight couples
• between three and ten couples
• between three and ten sizes
• each correction crosses the answer rather than approaching it from one side
• each distribution is one the view knows
• each extra person brings the distribution closer to the Poisson
• each proportion is within one over the next factorial of 1/e
• each stop is on the side the rule says
• each stop lands on its own side of the truth, or on it
• equal rook numbers exactly when the multisets hᵢ − i agree
• every arrangement is counted once
• every arrangement is visited once
• every guest dines every night
• every pair is side by side on exactly one night
• from five couples on the chance rises towards its limit
• inclusion–exclusion over rook numbers equals the count by search
• more choices in hand never mean a later start
• N is from 50 to 100000
• no man in the drawing sits beside his partner
• no online rule beats the oracle
• one probability per event, each strictly between nought and one
• one to five increasing numbers of choices, up to 6
• one to twenty thousand placements a point
• playing the rule 200,000 times pays its computed value
• simulation agrees with the exact sum
• six guests in three couples leave twelve allowed pairs
• six to thirty arcs
• sizes up to at most sixteen
• sizes up to at most twenty
• sizes up to between 6 and 16
• skipping the certain one is worth exactly one, whatever the setting
• so does the threshold set at the median of the maximum
• some permutation avoids the board
• stop after one head, two heads in two, three in three, or three heads in four
• stopping at the first lead pays exactly π/4
• tables of three or more, eleven guests at most
• the alternating sum ends at the chance of winning
• the alternating sum reaches the same number
• the arrangement is a permutation of its own places
• the arrangement names every object exactly once
• the best achievable rank rises with the size of the field
• the best rule does at least as well as one threshold
• the best rule for three values has expected rank about 1.3915
• the best rule keeps at least 0.745 of the prophet's expected maximum
• the best rule looks at some of them and not all of them
• the best rule with one bar per position has expected rank about 1.4009
• the best threshold is heading for one over e
• the best threshold rises with the field
• the board is one the family draws
• the case is secretary, constant or custom
• the case is uniform or tight
• the chance falls as the field grows
• the chance of no match is 0.016232727…
• the deals sorted by number of matches account for every deal
• the distribution is one the view knows
• the exhaustive check is over 4 to 8 candidates
• the exhaustive count is drawn for between 3 and 8 objects
• the expected number of arcs approaches Janson's approximation
• the expected number of matches is exactly four
• the expected rank rises with the number of values
• the field has between 4 and 400 candidates
• the first value is kept only below a third, so a later value is kept only when it leads
• the full alternating sum counts the numbers up to 1000 free of 2, 3, 5, 7, 11, 13
• the full sum is the chance, found without it
• the grid of every arrangement is drawn for at most 5 objects
• the index rule attains the exact optimum
• the largest field drawn is already close to that limit
• the largest gap's excess over ln n averages Euler's constant
• the limit curve peaks at one over e
• the lower bounds stay below the value Häggström and Wästlund bracketed
• the mean number of matches is 1
• the number of objects is in the range this view can enumerate
• the option is one this view reads
• the oracle is never beaten
• the partial sums lie alternately above and below the probability
• the placements drawn are all of them
• the plans with no seating are exactly 3+3, 4+5 and 3+3+5
• the probability is strictly between nought and one
• the product of the failures times the sum of the odds is the same chance
• the rank the rule will accept gets stricter as more remain
• the ratio is drawn up to between 2 and 10 objects
• the recursion agrees with the formula above half
• the rolls that show every face are the onto maps, counted by the same sum
• the rook polynomial of disjoint blocks is the product of theirs
• the rule is exact with no switching cost and loses with one
• the running total ends at the count by search
• the schedule seats every allowed pair side by side exactly once
• the settings are between nought and one
• the share stays above 0.745 however heavy the tail
• the shortfall approaches the half as the setting shrinks
• the sizes drawn are between 3 and 400
• the square-root rule beats the first lead and is beaten by the best
• the standard demanded rises with the number still to come
• the standard for acceptance falls as the end approaches
• the start the odds rule picks is the best start
• the threshold divided by √n settles near Shepp's constant
• the two boards drawn have the same rook numbers
• the value grid has between 200 and 2000 steps
• the value kept is the first below the line
• the view is one the family draws
• this plan has a seating
• Touchard's formula equals the count by search
• Touchard's sum equals the count by search
• two choices approach e⁻¹ + e^(−3/2)
• two forbidden cells attack exactly when they are neighbours on the cycle
• two hundred thousand seeded deals agree with the exact chance
• two to eight primes
• using the first value at the second step does better than a fixed bar
• whenever the odds reach one, the rule wins at least one time in e
• which is far above what the relative-ranking rule achieves
• with 1 choice among 8, every order counted agrees with the plan
• with many ranks the chance tends to e^(−d)
• with one choice the plan is the classical rule
• with one object there is nothing to find, and the search finds nothing
• with three it is 3/2
• with two candidates the best expected rank is 5/4
• with two it wins three times in four
• with two values the best threshold rule has expected rank 5/4

• a path bounces once per row
• and are centred under the funnel
• and ends in a bin the board has
• the exact bin shares add to one
• the path ends over the bin its right-turns name
• the row count is a whole number between 1 and 24

• the connected graphs on 4 points, counted twice by different means ×4
• above the threshold the measured share matches the equation's root
• and below it the largest piece is a vanishing share
• and below the threshold the equation has only the zero root
• and it is nonetheless not connected
• and it stays put to within half
• and the chance of at least one never exceeds the expected count, as Markov requires
• and the triangle's stays wider than it
• and well above it a triangle is nearly certain
• at ratio 3 every formula is satisfiable and at 6 none is
• being connected is at most as likely as having no isolated point
• between four and forty samples at each size
• between one and four sizes, each at most six points
• between three and five sizes, each between fifty and twelve hundred
• between twenty and two hundred samples per point
• between two and four edge chances, each strictly between zero and one
• between two and four sizes, each between twenty and a hundred and forty
• both rules fail by 3.5, far below the threshold, and the shortest-clause rule mostly succeeds at 2
• connectivity's transition narrows relative to its threshold as the graph grows
• dividing by the two-thirds power is the division that stays put
• each sweep runs from mostly-absent to mostly-present
• every edge present is connected with probability one
• every point is in exactly one piece
• for not-all-equal 3-SAT the pairs at half agreement dominate at low ratios
• for plain 3-SAT the second moment is dominated by pairs agreeing on more than half the variables
• most formulas at this ratio have no solution
• no edges is connected with probability zero
• the chance of being connected rises with p
• the comparison runs on between three and six points
• the component grows with the size, as it must
• the count of isolated points follows n·e⁻ᶜ
• the drawn graphs have between six and twenty points
• the enumeration runs on between two and six points
• the first-moment bound lies above every threshold
• the half-way crossing at eighty variables sits near 4.27
• the hardest formulas sit near the threshold
• the measured crossings sit near the predicted thresholds
• the measured mean count is near the exact expectation
• the measured triangle count matches the expectation it is supposed to
• the number of points is between three and twenty
• the peak stands well above both ends
• the predicted share solves its own equation
• the range drawn reaches the connectivity threshold
• the sampled graphs have between sixty and three thousand points
• the sampled graphs have between two hundred and three thousand points
• the seed is a whole number the graphs can be drawn from
• the symmetric second moment holds exactly up to three halves, where its curvature at one half vanishes
• the view is one the family draws
• well below the threshold a triangle is rare
• well past average degree one, most of the graph is in one piece

• row 0 of the transition matrix adds to one ×20
• every box of width 1/2^0 and height 1/2^6 holds exactly one point ×16
• the grid's error in 4 dimensions falls like N^(−2/4) ×5
• in 2 dimensions the share crossed falls like N to the power −1/2 ×4
• every one of the 16 boxes of side 1/4 holds exactly one point ×3
• in 2 dimensions the grid is the faster ×2
• in 8 dimensions the grid is the slower ×2
• a Fibonacci lattice between 13 and 377 points
• a power of two between 16 and 256
• a tilted line and a circle both near −3/4
• and a run of the chain reproduces it to within a twentieth
• and at four dimensions the two rates meet
• and holds no negative probability
• and is between 5 and 400 sharp
• and is positive everywhere on the interval, so no weight is unbounded
• and is repeated between 40 and 800 times
• and is run for between 2,000 and 200,000 steps
• and so does every row
• and so is the weighted one — neither is biased
• and the Halton points' falls faster than that
• and the random one falls like N^(−1/2), whatever the dimension
• and the weighted one is appreciably quieter
• as much probability flows from i to j as back again
• at sixteen thousand points the smoothed estimate is over a hundred times better
• away from the axes the advantage over random points grows only like N to the quarter
• between 32 and 2,048 points are drawn
• both intervals are consistent with the true value
• each dimension is a whole number between 1 and 12
• each estimate uses between 50 and 5,000 draws
• every column of width 1/N holds exactly one lattice point
• in 1 dimension the grid is the faster
• lines from the radius procedure look the same inside any smaller circle
• most single estimates fall below the true size, and the median is below it
• on an integrand that is not periodic the lattice loses most of its advantage
• Owen scrambling beats a random digital shift and approaches N to the minus three halves
• plain Monte Carlo falls like N to the minus one half
• plain sampling near −1/2
• random points' error grows like the radius for a small circle
• random sampling's error falls like one over the square root of N
• smooth near −3/2, a jump along an axis near −1
• so a step leaves the target distribution where it was
• the axes are far cheaper than every angle between them
• the chain has between 5 and 24 states
• the discarded start is shorter than the run
• the discrepancy is probed on a grid of between 40 and 300 boxes a side
• the grid's error in 1 dimension falls like N^(−2/1)
• the Halton points are more evenly spread than the random ones on this measure
• the integrand is periodic, plain, or periodised
• the jump costs half the exponent, and integrating it out restores it
• the lattice rule's error on a smooth periodic integrand falls about like 1/N²
• the line crosses between √N and 2√N of the N boxes
• the mean of the estimates matches the exact size at every board
• the option is one this view reads
• the peak sits inside the interval
• the plain average is centred on the answer
• the points to fit a slope through are not all at one place
• the proposal is a probability density
• the random points' discrepancy falls like one over the square root
• the rule sees a Fourier mode exactly when its frequency is on the dual lattice
• the sampled mean matches the exact mean
• the scrambled error grows like the square root of the circle's length
• the seed is a whole number the drawing can be made from
• the shifted lattice's interval is far narrower
• the spread is measured over between 4 and 200 runs
• the sweep runs over between 3 and 8 sizes
• the sweep runs to between 2^8 and 2^18 points
• the typical estimate misses the deep subtree by a factor of more than a thousand, and only a handful of paths reach it
• the view is one the family draws
• twenty thousand probes come within five per cent of the true size

• the ratio at n = 1 ×100
• coin summed 2 times has probabilities adding to one ×7
• coin summed 2 times has some spread to speak of ×7
• skew summed 2 times has probabilities adding to one ×7
• skew summed 2 times has some spread to speak of ×7
• die summed 2 times has probabilities adding to one ×6
• die summed 2 times has some spread to speak of ×6
• the sum of 4 divided by 4 has probabilities adding to one ×6
• the sum of 4 divided by 4 has some spread to speak of ×6
• the sum of 4 divided by the root of 4 has probabilities adding to one ×6
• the sum of 4 divided by the root of 4 has some spread to speak of ×6
• every one of the 6561 inputs at n = 8 is counted ×5
• the average of 1 draws still has the distribution of one draw ×5
• pairs of edges positively correlated in a uniform random connected on 3 vertices ×4
• pairs of edges positively correlated in a uniform random forest on 3 vertices ×4
• pairs of edges positively correlated in a uniform random tree on 3 vertices ×4
• the margin for a town of 1250 ×4
• the sum of 4 has probabilities adding to one ×4
• the sum of 4 has some spread to speak of ×4
• and doubling the draws moves it toward the computed one at 0.3 ×3
• at 1.5 standard deviations the mass outside is within the bound ×3
• the bound holds at every deviation the distribution reaches, n = 8 ×3
• the measured rate at 0.3 is never below the computed one ×3
• there are exactly 91 possible histograms of 12 draws over three faces ×3
• four products of between 1 and 60 factors ×2
• the largest sum is a power of two between 16 and 128 ×2
• the support is an odd number of points between 21 and 121 ×2
• ±½, even odds has mean 0
• ±½, even odds has variance 1/4
• −1, 0 or 1 has mean 0
• −1, 0 or 1 has variance 1/4
• 0 nine times in ten, 10 otherwise has probabilities adding to one
• 0 nine times in ten, 10 otherwise has some spread to speak of
• 0 nine times in ten, 10 otherwise stays under the bound at every width
• a bound that says nothing belongs to a function one input nearly controls
• a die has probabilities adding to one
• a die has some spread to speak of
• a distribution the figure knows
• a fair coin, ±1 has probabilities adding to one
• a fair coin, ±1 has some spread to speak of
• a fair coin, ±1 stays under the bound at every width
• a fair die has probabilities adding to one
• a fair die has some spread to speak of
• a fair die stays under the bound at every width
• a sample that divides into k equal blocks
• a sample with exactly one rare jump exists among the seeds tried
• a uniform factor's coefficient is 1/√(1 + t²)
• and a distribution that is actually bell-shaped is far inside both of them
• and a useful bound to one where it does not
• and dividing by the root of n leaves it exactly where it started
• and increasing away from the mean
• and it comes within half the bound somewhere
• and it reaches it, so the bound is attained rather than merely true
• and its average against any such distribution is exactly the bound
• and its mean is exactly the level asked about
• and none is smaller than that divided by a polynomial in n
• and scaled so that one standard deviation is one unit
• and the binomial one stays under Hoeffding's lemma
• and the median of block means ignores it where the mean does not
• and the more lopsided summand is further from the bell at every n
• and the rate at the mean itself is zero
• and the rate is the information distance from the original to the tilted law
• and the runs are further apart early than late
• and the second panel's settle on the height of the bell curve
• and the typical value settles at √(2/π)
• and the variance adds too
• asking about one tail rather than two is worth something
• at least one of the useful rows is not a sum, which is the point
• Benford's value lies inside each digit's swing
• Bennett's closed form
• Bennett's exponent is never below Bernstein's
• between 100 and 3,000 powers
• between 2 and 12 runs are drawn
• between 2 and 3 summands are compared
• between 2 and 4 averages, each of at most 64 draws
• between 2 and 40 dice
• between 2,000 and 60,000 averages are formed
• between 3 and 40 draws
• between 3 and 6 sums, each of at most 200 copies
• between 6 and 14 inputs
• between two and four input counts, each from 6 to 16
• between two and four levels are marked
• both have the same mean
• dividing by n shrinks the spread like 1/√n
• each averaging narrows the swing and none removes it
• each distribution is one the family knows
• each n is between 2 and 200
• each quantile's spread matches √(p(1 − p))/(f √n)
• each run is between 50 and 5000 draws
• each sum is closer to the bell curve than the last
• each sum is of between 1 and 400 copies
• each window is between half a standard deviation and six
• every bound's exponent is at most the true one
• every distribution swept is one the family knows
• every extra factor brings the digits closer to the law
• every level is between the mean and the largest value
• every run finishes inside four standard errors of the mean
• every shape stays above Bickel's bound
• every tightest forest ratio is below one
• exact < Bennett ≤ Bernstein < Hoeffding
• exact ≤ Bennett ≤ Bernstein at every threshold
• five values, unevenly weighted has probabilities adding to one
• five values, unevenly weighted has some spread to speak of
• five values, unevenly weighted stays under the bound at every width
• Hoeffding's bound holds both ways
• in 100 dimensions the coordinate's density is within 3% of the normal curve with standard deviation 1/√n
• it is centred at zero
• it never falls inside the gap
• length of the drawn common subsequence
• matchings of the path with four edges
• more dimensions, more pairs within ten degrees of a right angle
• more letters, shorter common subsequences
• no histogram is more likely than exp(−nD)
• on a 2-connected graph no pair is positively correlated
• on normal data the plain mean is the more accurate at every confidence
• on the ordinary sphere a band's share is its width: Archimedes' hat-box theorem
• one input can move the answer at all
• Serfling's sharper bound holds without replacement
• sixteen dice has probabilities adding to one
• sixteen dice has some spread to speak of
• so one scaling is collapsing
• some distribution meets the constraints
• some pair of edges is positively correlated in a uniform random matching
• t with 3 degrees, Cauchy or Laplace
• the average of two Cauchys has the density of one Cauchy, by the convolution integral
• the bell curve convolved with itself is the bell curve again, widened by √2
• the bell curve's exponent differs from the rate by enough to show over the range drawn
• the best distribution does not beat the bound
• the block holding the jump has the most extreme block mean
• the bound holds for n times the first flip
• the bound holds for the longest non-decreasing run of choices
• the bound holds for the longest run of heads in n flips
• the bound holds for the total of n coin flips
• the bound is met at between 1.2 and 5 standard deviations
• the certificate lies above the indicator at every support point
• the constrained region is not empty
• the constrained set has a probability strictly between nothing and everything
• the constraint asks for a mean above the true one and below the largest face
• the digit probabilities of a product of uniforms add to one
• the Dirichlet share tends to log₁₀ 2
• the distance never exceeds the Fourier bound
• the distribution has probabilities adding to one
• the distribution has some spread to speak of
• the distribution is one the family knows
• the event has positive probability at every n drawn
• the exact chance that half the blocks fail is under e^(−k/8)
• the exact law of the median is within two hundredths of the normal approximation
• the exact mean lengths for strings of one and two digits are 1/2 and 18/16
• the exact probability stays under the exponential the rate predicts
• the exhaustive sweep is feasible
• the expected length is superadditive
• the extremal distribution has probabilities adding to one
• the extremal distribution has some spread to speak of
• the far cap is never larger than e^(−nε²/2)
• the first and third edges of the path are positively correlated in a random matching
• the first panel's curves grow taller as they narrow
• the function is one of sum, longest, distinct, rising, dictator
• the functions are among sum, longest, distinct, rising, dictator
• the gap for 0 nine times in ten, 10 otherwise falls like one over the square root of n
• the gap for a fair coin, ±1 falls like one over the square root of n
• the gap for a fair die falls like one over the square root of n
• the histograms' chances add to one
• the hypergeometric probabilities add to one
• the level asked about is above the mean and below the largest value the summand can take
• the level is between the mean and the largest value the summand takes
• the logarithmic share closes in on Benford's value
• the mass at exactly k standard deviations is the whole of the bound
• the mean grows with the number of inputs, so the panels differ
• the mean of a sum is the sum of the means
• the mean share rises towards its limit
• the mean's error sits under Chebyshev's guarantee at every confidence
• the measured decay rate agrees with the rate function computed from the summand alone
• the measured rate moves towards the smallest relative entropy in the region
• the measured rates come from between 16 and 96 draws
• the measured ratio of variances matches 4 f(m)² σ²
• the median falls on either side of the gap equally often
• the median spreads like n^(−1/4) when the density vanishes linearly at the middle, n^(−1/2) otherwise
• the median's spread matches 1/(2f(m)√n)
• the natural share of primes beginning with 1 keeps swinging
• the one-sided answer is Cantelli's
• the one-sided bound is strictly better than the two-sided one from the same two numbers
• the option is one this view reads
• the powers of two follow the law closely
• the product of the dice is closer to the law than their sum
• the rate at a level away from the mean is positive
• the rate function is convex all the way along
• the scaled values are increasing
• the share swings between 1/9 and 5/9
• the spread shrinks as the dimension grows
• the standardised lengths are close to symmetric
• the support runs to between 3 and 12 deviations
• the sweep includes a lopsided distribution
• the sweep is feasible
• the sweep runs to between 24 and 200 draws
• the table separates the functions the bound says something about from the ones it does not
• the tilted distribution has probabilities adding to one
• the tilted distribution has some spread to speak of
• the tilted weights are a distribution
• the tree's edges in the left half vary less than independent edges with the same chances would
• the two edges are negatively correlated
• the two-point variable has the largest generating function
• the two-sided answer is Chebyshev's bound
• the uniform products approach the law by a factor |E Xⁱᵗ| per factor
• the variance grows faster than a square root of n and no faster than n
• the variance respects the Efron–Stein bound n/2
• the variance shrinks by exactly (N − n)/(N − 1)
• the view is one the family draws
• the weighted share keeps rising
• the window is between 1.2 and 4 deviations wide
• the window is between 1.5 and 4 deviations
• the window is two-sided or one-sided
• the wrapped density flattens as factors are added
• three face values
• three positive chances
• three positive chances adding to one
• two opposite edges of K4 are independent in a uniform spanning tree
• two-point: 1 or −¼ has mean 0
• two-point: 1 or −¼ has variance 1/4
• while a light-tailed average narrows by the square root of the count
• with rare jumps the median of means is far more accurate at 1-in-1000
• without replacement the generating function is never larger (Hoeffding 1963)
• γₖ√k is approaching 2 at large alphabets

• the enumeration finds the known number of 0-step walks ×20
• the measured exponent is the larger of ½ and ρ (a = 0) ×9
• the formula agrees with the enumeration at 0 steps above ×7
• the sum of 4 steps has probabilities adding to one ×4
• the sum of 4 steps has some spread to speak of ×4
• the share of red after 1000 draws follows the beta law with parameters 1 and 1 ×3
• between 2 and 5 steps ×2
• in 2 dimensions the measured range is within 25% of the classical law at 10,000 steps ×2
• 10 to 200 matches a box
• A and B each lose over 300 rounds and the random mixture wins
• A B B is the best of the sixteen
• a bell below the threshold, two humps at ρ = 3/4, uniform for Pólya
• a walk in space escapes, and the simulation sees it
• a walk on a line comes back, and the horizon barely hides it
• a walk on a plane comes back too, but more slowly
• a walk takes a whole positive number of steps
• above it like n to the power ρ (Pólya, ρ = 1)
• above it like n to the power ρ (ρ = 3/4)
• above one half it stays well away from nought
• an endpoint has the parity of the step count
• an even number of steps between 8 and 200
• and impossible with everything
• and is then exactly fair
• and moves the start to its mirror image
• and so are the reflected ones
• and the distribution is symmetric
• and the gap to the bell curve closes at every step
• and the variance of its position is exactly the number of steps
• and their spread is the square root of the number of steps
• at the threshold slightly faster than √n, by a logarithm
• at ε = 0 game B spends exactly 5/13 of its time on the bad coin
• at ρ = 1/3 every start heads for the bell
• at ρ = 3/4 the shape is set by the start and barely moves with n
• B's capital avoids remainder 1, and mixing widens the distribution
• below one half the excess kurtosis dies away
• below the threshold the spread grows like √n (red adds a blue)
• below the threshold the spread grows like √n (ρ = 1/3)
• between 4 and 16 steps are drawn, since every path of them is enumerated
• dim is read only by the range view
• each exact distribution is of between 2 and 400 steps
• each game alone loses
• each step moves one place along one axis
• every game drawn ends at a barrier
• every order of the same draws has the same chance
• every path from below the level has to cross it
• every path is counted once
• every step is drawn
• every step is one place, either way
• every step is to a neighbour
• every walk is counted once
• folding leaves the endpoint alone
• folding matches the touching paths with the paths from the mirrored start one for one
• from one of each, every count of reds has chance exactly 1/(n + 1)
• in 1 dimension the measured range is within 6% of the classical law at 10,000 steps
• in one dimension the range is the width of the explored interval
• lattice is read only by the sawcount view
• mixing the two losing games wins
• most urn walks make their last return to the start within the first hundred steps
• playing only B or only A loses and some mixture wins
• ruin is certain with nothing left
• simulated urns agree with the exact law
• some coin settings give the paradox
• some walk goes outside the envelope, because √n is typical and not maximal
• steps is not read by the views that fix their own lengths
• ten draws decide most of the spread above one half and little of it below
• the chances add to one
• the distribution sums to one
• the drawn path really does touch the level
• the drawn walk is in one or two dimensions
• the drawn walk's mean square displacement per step is near one in every window
• the drawn walks are exactly the known number
• the durations satisfy their own recurrence
• the enumerated check runs at an even length between 4 and 16
• the enumeration finds the known counts
• the extremes are more likely than the middle, which is the whole surprise
• the full distributions agree with the residue chain's expected capital
• the lattice is the square grid or the honeycomb
• the least likely outcome is an even split
• the mixture wins at ε = 0 and stops winning at some handicap below 0.02
• the mixture wins over a wide range of γ
• the one-dimensional survival matches C(2m, m)/4^m
• the option is one this view reads
• the paths ending where they end are counted by a binomial coefficient
• the player starts with something and not everything
• the position has the parity of the step count
• the probabilities of every remainder add to one
• the quoted constant is Pólya's
• the random mixture wins for every ε below about 0.0131
• the range is the number of distinct sites on the path
• the ruin probabilities satisfy their own recurrence
• the sampled walk never revisits a site
• the share decided grows with m and stays between nought and one
• the spread of the endpoints is of the order of the square root of the steps
• the squared distance grows faster than the simple walk's n, with exponent near 3/2
• the table holds between 4 and 40 units
• the three windows are whole numbers of steps
• the three windows are whole numbers of steps inside the walk
• the time spent above is always even
• the urn is balanced: every draw adds the same number of balls
• the urn's expected returns are half the logarithm of n plus a constant
• the view is one the family draws
• the walk can reach its endpoint in the steps allowed
• the walk comes back to where it started at least once
• the walk has no drift
• the walk is between 400 and 60,000 steps long
• the walk starts above the level it may touch
• the walks are centred on where they started
• the walks leave at very different speeds
• the win probability is a probability
• three-dimensional walks that have not returned by step 2,000 are about 1 − 0.3405 of them
• with no handicap game B is exactly fair

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