Generator

A map of the interval must fix a point

A generator in the topology library, called 83 times across 18 essays. Below: what it draws with nothing chosen and at each mode an essay asks for, what it checks while drawing, and everywhere it is used.

fixed-point is one function. Everything below came out of it during this build, at parameters taken from the essays rather than invented for this page — so a figure here is the same figure a reader meets in an essay, and if the generator changes, this page changes with it.

With nothing chosen

A map of the interval must fix a point. a continuous map of the interval, drawn with the diagonal. Every continuous map of the interval into itself meets the diagonal somewhere; this one does so at x = 0.6944.

Cutting the square down to a fixed point

Iterating against cutting as the map stops shrinking

Readings per digit of accuracy, iterating and cutting

Every cut through the centre keeps at most five-ninths

The cut is safe while the map does not stretch

What it checks while it draws

Collected by running the family and recording what it asserted, not written here. The count is how many separate times the claim was put to the test while these drawings were made.

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 quarter-turn reverses the imbalance ×1
a rotation's iterates stay at their starting distance ×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 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
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 rectangle kept winds once round ×1
every sampled antipodal labelling has a complementary edge ×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
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 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 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 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 opposite pairs where both readings agree are odd in number ×1
the pair found really does take the same value ×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 shape is one the family knows ×1
the smallest-element colouring is proper ×1
the solved point is fixed, first coordinate ×1
the solved point is fixed, second ×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 twisted map's corridor is several times longer ×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
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 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
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 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

Where it is called

Every figure on this list is drawn by the same rule, so a change to the rule changes all of them at once. That is why the list is published.

Topology

A cut that gets sharper as the map gets lazier

Iterating a map that shrinks distances by 0.9999 takes 141,254 steps to reach its fixed point to a millionth. Cutting takes six: read the map at one point, and the fixed point must lie on the far side of the line halfway between the point and its image. The cut gets better exactly where iteration gets worse, because a map that barely shrinks puts its fixed point almost on every such line.

Algebra

A loop that cannot miss the middle

Feed a circle into a polynomial and a closed loop comes out. A small circle gives a loop that does not enclose the origin; a large one gives a loop that goes round it as many times as the degree. Something has to happen in between, and that something is a root.

Topology

A map that offers a choice

Brouwer's theorem needs a function, and the object it was most wanted for is not one — a best reply is a whole set whenever a chooser is indifferent. Allow a point to be sent to a set and the fixed point survives, provided the sets are convex, and the convexity is the entire hypothesis.

Analysis

A map that shrinks everything

One extra hypothesis — that every distance is shortened by at least a fixed factor — turns the existence of a fixed point into its uniqueness, an algorithm for finding it, and a bound on the error after any number of steps.

Dynamics

A twist that cannot avoid two points

Turn the two edges of a ring in opposite directions without changing any area, and something in between must stay exactly where it is — not one point, but at least two, and the reason is that two loops enclosing the same area have to cross.

Topology

As many cuts as colours

Two thieves steal a necklace and want half of every colour of bead each. However the beads are strung, they never need more cuts than there are colours — three cuts for three colours, four for four — and sometimes they need every one. The guarantee is the Borsuk–Ulam theorem again, with a point on a sphere read as a way of cutting the necklace, and every necklace of several small kinds has been checked against it.

Topology

Four equal quarters with two lines

Any flat shape, however lopsided, can be cut into four pieces of equal area by two perpendicular straight lines. The proof turns the pair of lines like the hands of a clock: the area in one quadrant, minus a quarter, reverses its sign every quarter-turn, so somewhere it is zero. The same kind of argument cuts a solid into eight equal pieces with three planes. It stops working in five dimensions, where some masses cannot be cut into thirty-two equal pieces by five hyperplanes — and in four, nobody knows.

Topology

Nothing on a sphere can be combed flat

Point an arrow along the surface at every place on a sphere, continuously, and somewhere an arrow has to vanish. On a doughnut it can be done. The difference between the two is a number that was already known from counting corners.

Topology

One line that halves them both

Two shapes lying anywhere on a page, of any sizes and any shapes at all. There is always a single straight line that cuts both of them into two equal halves at once — and finding it needs no cleverness, only the observation that a quantity which reverses sign has to pass through zero.

Topology

Opposite labels that have to meet

Cut a square into triangles, label every corner +1, −1, +2 or −2, and insist only that opposite points of the edge get opposite labels. Somewhere inside, an edge must join a label to its negative. The proof counts quarter-turns round a diamond — an odd number on the boundary, zero in any triangle that avoids opposites — and making the triangles smaller turns the count back into the theorem about opposite points on the Earth.

Topology

Seven points split three ways

Any seven points in the plane can be divided into three groups whose convex hulls share a point, and six points in general position never can. The theorem is Helge Tverberg's; its topological version, in which straight lines may bend, is true when the number of groups is a power of a prime and false otherwise — and the proof that decides which is the same antipodal argument that halves a sandwich.

Topology

Several colours on every vertex

Give every pair from six points three colours, so that pairs with nothing in common share no colour. Counting says nine colours might do; ten are needed. Stahl conjectured in 1976 exactly how many colours every such problem needs — a formula that meets Lovász's topological answer at one colour a vertex and the obvious answer at k — and a search over stars and triangles confirms it in every case small enough to run.

Topology

Something always stays put

Stir a cup of coffee however violently and let it settle. Some molecule is exactly where it started. Crumple a map and drop it on the region it depicts, and one point lies over the place it names.

Topology

The colours a circle forces

Take every pair from five things and join two pairs when they share nothing. Three colours are enough to colour the result so joined pairs differ, and two are not — but no triangle, no dense cluster and no counting argument explains why. The reason is five points on a circle and a direction that cannot be told apart from its opposite, and the same reason, one sphere at a time, settles Kneser's question for every size.

Discrete

Three colours force a triangle

Cut a triangle into small ones and colour the corners under one restriction. However the cutting and the colouring are done, some small triangle ends up with all three colours — and the number of them is always odd.

Topology

Two opposite points that agree twice

At any moment there are two points on opposite sides of the Earth with the same temperature and the same pressure. On a seeded globe they sit at 11.9°N 44.6°E and 11.9°S 135.4°W. The reason is the circle argument that halved two shapes, run one dimension up: the differences between opposite readings, walked round the equator, wind round zero an odd number of times — and an odd number cannot be zero.

Topology

What it costs to find the point that stays put

Brouwer's theorem promises a point a map leaves where it is, and Sperner's corridor walks to it. On a gentle map the walk is short; twist the map about its fixed point and the corridor follows every turn, so its cost grows like the square of the grid. Halving the square by winding number finds the same point in a number of readings proportional to the grid's side whatever the map does — and no method that only reads the map can do fundamentally better.

Topology

Where the fixed point escapes

The theorem asks for a set that is closed, bounded and free of holes. Drop any one of the three and a map appears that moves every single point — and in each case the point that should have stayed still can be seen leaving.

The whole library · What the figures prove