The ledger

What the figures prove — page 3

Page 3 of 6, continuing through the fields in the same order. Every claim here was checked as the drawing was made rather than typed beside it.

Topology

9 families

`euler-solid`

85 kinds of claim · 29 placements

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
the whole boundary of the 3-cube has alternating sum 1 − (−1)^3 ×2
a honeycomb on the torus has V − E + F = 0
a spanning tree has one fewer edge than the graph has vertices
and 720° is 360° times V − E + F
and its twelve edges
and six sides to every face
between one and three holes are punched
each unfolded face closes back on the corner
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 partial boundary before the last facet sums to 1
no leftover edge closes a loop among the faces
the alternating sum is not 2, on a solid with flat faces and straight edges
the alternating sum of the tesseract is 0
the counts obey Euler's formula
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 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 hexagonal prism's faces fall short of hexagons by exactly 12
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 mode is one the family draws
the solid is one of the five regular ones
the solid is one the family builds
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
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
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

`fixed-point`

123 kinds of claim · 55 placements

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 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 second
and sits on it
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 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 4 and 16 globes
between 8 and 120 arrows are drawn per panel
both choosers have a mixture of the other's that leaves them indifferent
dipole has the index it claims
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 necklace can be shared fairly with no more cuts than colours
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 sampled antipodal labelling has a complementary edge
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
half a turn along the equator reverses the difference exactly
in both of its readings
no boundary edge is itself complementary in the labelling drawn
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
one of the two has convex values and the other does not
opposite boundary vertices carry opposite labels
random fields without the boundary condition often escape
saddle has the index it claims
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 half-circle holds fewer than two points
some necklace of these beads needs as many cuts as there are colours
some region holds a point together with its opposite point
source has the index it claims
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 difference reverses sign between opposite points
the equator's difference winds an odd number of times
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 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 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 opposite pairs where both readings agree are odd in number
the pair found really does take the same value
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 smallest-element colouring is proper
the solved point is fixed, first coordinate
the solved point is fixed, second
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 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
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 colours do
turning the line through half a turn reverses the imbalance
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 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

`gluing`

97 kinds of claim · 48 placements

enumeration and recurrence agree at n 1, genus 0 ×15
a sphere with 3 cross-caps has characteristic -1 ×3
the enumeration agrees with Harer–Zagier at genus 0 ×3
a surface with 2 handles has characteristic -2 ×2
the classification is drawn up to 3 of them in this view ×2
a drawn random gluing has 6 to 60 edges
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 torus has characteristic 0
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 no edge to itself
and no face to itself
and no vertex is its own opposite, so the map moves every point
and so is the sidedness
and ten faces
and the quotient's is a projective plane's
and thirty edges
and twenty faces
at least one move applies to the word given
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
each drawn pairing is glued head to tail
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
every pairing is drawn for polygons of 4 to 8 edges
every sampled vertex count has the parity the genus formula allows
every vertex has its opposite among the vertices
few enough vertices to colour each one differently
fifteen edges
genus 0 is the Catalan number
no letter appears more than twice
no move raises the measure
no two of the surfaces share a characteristic
one vertex, for even n, is 1/(n + 1)
one-vertex gluings are 1/(n + 1) of all
the antipodal map carries every edge to an edge
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 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 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 polygon has an even number of edges
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 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 table runs to between 3 and 10 pairs
the two numbers name the surface without anything else being consulted
the two surfaces with the same characteristic are told apart by orientability
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

`knot`

88 kinds of claim · 44 placements

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
a colouring is drawn on a diagram that has crossings
a colouring uses three, five or seven colours
a crossing joins two different Seifert circles
a diagram with k crossings is cut into k arcs
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
and Gauss's integral, which sees no disc at all, gives it too
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 three and five deformation sizes, none of them large
between two and four links the family knows
each crossing has exactly one of its two strands going underneath
every row of the matrix vanishes at t = 1, as a presentation of a knot must
few enough circles to try every ordering
following the smoothing from a segment comes back to that segment
knots sharing a determinant are told apart by the polynomial
no single arc passes over every crossing
one boundary circle, so the genus is (1 − χ)/2
the (2, 4) torus link: the diagram's count and Gauss's integral agree for components 1 and 2
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: |Δ(−1)| is the determinant
the cinquefoil: on an alternating diagram the two bounds agree
the constant colourings are always allowed, so there are at least p
the discs and bands form one connected surface
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: |Δ(−1)| is the determinant
the figure-eight knot: on an alternating diagram the two bounds agree
the first component of this link is a flat circle, so its disc is the obvious one
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 knot is one the family draws, with crossings
the knot is one the figure knows
the knots in the table have different determinants
the link is one the family knows
the linking number is the same at every deformation
the minor is not zero
the polynomial at t = −1 is the determinant found from the colourings
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 seven-crossing torus knot has the crossings it claims
the seven-crossing torus knot: |Δ(−1)| is the determinant
the seven-crossing torus knot: on an alternating diagram the two bounds agree
the strand is broken where it goes underneath, not somewhere else
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 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 three arcs
the trefoil takes ± its determinant at θ = π
the trefoil takes the value 1 at θ = 0
the trefoil: |Δ(−1)| is the determinant
the trefoil: on an alternating diagram the two bounds agree
the two computations agree at this deformation
the two loops stay clear of each other at every amplitude
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
twice the over-strand equals the two under-strands added, at every crossing
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 or three links whose first component is a flat circle
while the number of crossings is not

`loops`

161 kinds of claim · 81 placements

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
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
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 free group of rank two has four words of length one
a spanning tree of a connected graph has one edge fewer than it has vertices
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 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 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
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
both loops start at the same point of the ring
carrying the loop along either path leaves its class alone
each loop on the torus is a pair 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
every branch point's cycles fit side by side inside its column
every cover's counted rank matches the formula
every lifted edge appears on some face
every marked point actually branches
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
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
in the disc every stage of the shrink stays inside
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
one permutation for each of the 2g edges
over the line, the only loop that lifts to a loop is the one that goes nowhere
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
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 character sum and the enumeration agree
the character table of this symmetric group is one the family carries
the classes drawn are whole numbers no bigger than four
the classes that lift to loops are the multiples of the sheet count
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 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 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 Euler characteristic is that of a closed orientable surface
the Euler characteristic multiplies by the number of sheets
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 outgrows the abelian group of the same rank
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 lift ends a whole number of turns above where it started
the loop at p is a whole number
the loop being lifted is a whole number
the loop drawn at p has the class claimed
the loop drawn for a class really has that class
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 monodromy table is drawn over two generators
the number of sheets is what one lap advances by
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 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 shape list is one the family knows
the shrink is legal to begin with and stops being legal
the slide is drawn in three to five stages
the surface has genus 1, 2 or 3
the symmetries reach every sheet exactly when there are as many of them as sheets
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 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 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 read off the curve is the word the curve was built from
the words that come back are closed under composition
two or three circles below
two words land on the same sheet exactly when one undoes the other into the subgroup
while a word that really is trivial reduces away, so the test can fail
with the last point's permutation removed, going round no longer returns every sheet
words of length up to two to four are listed

`mobius`

16 kinds of claim · 16 placements

and between 24 and 140 the long way
between none and sixteen cross-sections are outlined
between three and nine frames are drawn along the band
carrying the surface round once brings it back on the other face
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
the band is drawn at a whole number of half-twists
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 mesh is between 16 and 96 round the tube
the self-intersection circle is on or off
the tube's distance from the axis leaves the figure-eight room to close
the view is one the family draws
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

`mobius-cut`

3 kinds of claim · 4 placements

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

`parity`

66 kinds of claim · 28 placements

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 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
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
between 2 and 36 rays are drawn
consecutive squares in the visiting order are neighbours
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 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 two of the kept squares overlap
some sampled point is inside the curve
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 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 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 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 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
two pitches, both inside the range the family draws

`stereographic`

34 kinds of claim · 36 placements

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
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 fibre lies on the unit three-sphere
each transformation is normalised to determinant one
every patch has the same area on the sphere
every two fibres are linked exactly once, counted from the drawn curves
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 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 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 view is one the family draws
there are pairs to link

Probability

10 families

`bayes`

42 kinds of claim · 25 placements

a halfer told it is the first day moves above one half
a thirder told it is the first day comes back to one half
and at seven days it is 13/27
and at three doors it is two thirds
and the boy–boy block holds 2d − 1 of them
at one awakening every account agrees on one half
everybody has it, so everybody positive has it
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)
prior is a probability, between 0 and 1
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 asked families settle near 1/3
the awakenings settle near one share in wakes + 1
the count is the formula (2 − p)/(4 − p)
the curve agrees with the grid count
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 kept area gives the protocol's answer
the marked row and column share one cell
the met-a-child families settle near 1/2
the only news drawn is that it is the first day
the question is any, older or met
the runs settle near one half
the shaded fraction is Bayes' theorem
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 view is one of sweep, monty, monty-scale, children, tuesday, trait, mention, children-sim, beauty, beauty-sim, beauty-bets, beauty-many
the worlds fill the square

`birthday`

57 kinds of claim · 29 placements

the near-birthday chance at k = 0 is a distinct-birthday chance on a shorter year ×7
22 is still under a half
23 is over it
a block is between 2 and 365 days
a group smaller than the calendar
a match within a day needs 14
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 threshold only falls
and where it misses, the calendar needs one MORE person than its effective days predict, never one fewer
at the last step the two agree modulo the factor found
before the last step they agree modulo neither prime
between 4 and 24 slots, every one possible
evening the two days raises the chance that nobody matches
every product was split
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
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 day-probabilities add to one
the effective number of days places the threshold within one person
the even year's threshold is 23
the expected number of loners peaks when the room is as large as the year
the half-way point is 23 people
the highest point is the even split
the last new value leads back into the cycle
the modulus is a prime of at most 5000
the pair count is k choose 2
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 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
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
what it returns divides n

`buffon`

7 kinds of claim · 9 placements

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

`chain`

131 kinds of claim · 52 placements

the wait for a new one at 0 seen is 6/6 ×21
the 6 waits add to 6 times the harmonic sum ×3
a run of the game reproduces the expected length
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
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 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 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
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
every chain named is one the family carries
every start can go either way
every state has somewhere to go
every state holds a share
HH: the chain's answer and the overlap sum agree
HH's wait is between the window size and twice it less two
HHH: the chain's answer and the overlap sum agree
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: the chain's answer and the overlap sum agree
HHT's wait is between the window size and twice it less two
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: 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
no two edge weights are drawn on top of one another
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
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 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 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 collection is one the family knows
the corroborating run is between 500 and 40,000 games
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 to stationarity never rises
the drawing runs out to between 10 and 60 states
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 game is played over between four and nine totals
the long-run shares are drawn only for a chain that has them
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 panels drawn are genuinely different cases
the patterns compared are all the same length
the return chances satisfy the rule that defines them
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 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 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 is not singular
the traffic between neighbours balances
the view is one the family draws
the walk is between a thousand and four hundred thousand steps
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
TT: the chain's answer and the overlap sum agree
TT's wait is between the window size and twice it less two
while the wait for the rarest kind alone is already a lower bound

`derange`

127 kinds of claim · 65 placements

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 die of 3 to 12 faces
a group of up to 20 to 120 people
a run of rolls starting at the number of faces
a year of 50 to 1000 days
all 24 arrangements are drawn
an exhaustive point is at most 8 objects
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 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 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
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
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
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 looks at some of them and not all of them
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 exhaustive check is over 4 to 8 candidates
the exhaustive count is drawn for between 3 and 8 objects
the field has between 4 and 400 candidates
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 largest field drawn is already close to that limit
the limit curve peaks at one over e
the mean number of matches is 1
the number of objects is in the range this view can enumerate
the oracle is never beaten
the placements drawn are all of them
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 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 running total ends at the count by search
the settings are between nought and one
the shortfall approaches the half as the setting shrinks
the sizes drawn are between 3 and 400
the standard demanded rises with the number still to come
the start the odds rule picks is the best start
the two boards drawn have the same rook numbers
the value grid has between 200 and 2000 steps
the view is one the family draws
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 to eight primes
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 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

`galton`

6 kinds of claim · 12 placements

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

`rgraph`

38 kinds of claim · 33 placements

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
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
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
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 measured triangle count matches the expectation it is supposed to
the number of points is between three and twenty
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 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

`sample`

36 kinds of claim · 21 placements

row 0 of the transition matrix adds to one ×20
the grid's error in 4 dimensions falls like N^(−2/4) ×5
in 2 dimensions the grid is the faster ×2
in 8 dimensions the grid is the slower ×2
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 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
between 32 and 2,048 points are drawn
each dimension is a whole number between 1 and 12
each estimate uses between 50 and 5,000 draws
in 1 dimension the grid is the faster
so a step leaves the target distribution where it was
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 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 seed is a whole number the drawing can be made from
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 view is one the family draws

`spread`

138 kinds of claim · 51 placements

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
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
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
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 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
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 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 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
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 2,000 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
dividing by n shrinks the spread like 1/√n
each distribution is one the family knows
each n is between 2 and 200
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 distribution swept is one the family knows
every level is between the mean and the largest value
every run finishes inside four standard errors of the mean
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
it is centred at zero
no histogram is more likely than exp(−nD)
one input can move the answer at all
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
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 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 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 probability stays under the exponential the rate predicts
the exhaustive sweep is feasible
the extremal distribution has probabilities adding to one
the extremal distribution has some spread to speak of
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 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 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 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 one-sided answer is Cantelli's
the one-sided bound is strictly better than the two-sided one from the same two numbers
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 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 two-sided answer is Chebyshev's bound
the view is one the family draws
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
three face values
three positive chances
three positive chances adding to one
while a light-tailed average narrows by the square root of the count

`walk`

54 kinds of claim · 35 placements

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
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
an endpoint has the parity of the step count
an even number of steps between 8 and 200
and impossible with everything
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
between 4 and 16 steps are drawn, since every path of them is enumerated
each exact distribution is of between 2 and 400 steps
each step moves one place along one axis
every game drawn ends at a barrier
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 walk is counted once
folding leaves the endpoint alone
folding matches the touching paths with the paths from the mirrored start one for one
ruin is certain with nothing left
some walk goes outside the envelope, because √n is typical and not maximal
the distribution sums to one
the drawn path really does touch the level
the drawn walk's mean square displacement per step is near one in every window
the durations satisfy their own recurrence
the enumerated check runs at an even length between 4 and 16
the extremes are more likely than the middle, which is the whole surprise
the least likely outcome is an even split
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 quoted constant is Pólya's
the ruin probabilities satisfy their own recurrence
the spread of the endpoints is of the order of the square root of the steps
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 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 win probability is a probability

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