A map of the interval must fix a point

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

The whole library · What the figures prove