13 into 12
pigeonhole 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
Langford pairings of order 3, 4 and 7
Which orders the parity allows, and what the search finds
How many Langford pairings and Skolem sequences there are
All twenty-six Langford pairings of order 7
How much searching a Langford count takes
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 pairing of order 3 puts k numbers between the two ks ×3
- a grid of 3 to 9 boxes a side ×1
- a Langford pairing exists exactly when the parity allows ×1
- and a Skolem sequence exactly when its parity allows ×1
- each case really does overflow its boxes ×1
- each is a Langford pairing ×1
- each number alone keeps q‖qα‖ above a quarter ×1
- each order multiplies the search by between five and ten ×1
- each puts the two ks k places apart ×1
- every best simultaneous approximation brings both numbers within about 1/√q ×1
- every item is in a box ×1
- few enough runners to read ×1
- more items than boxes forces a box with two ×1
- six Skolem sequences of order 4 ×1
- the fullest box holds ⌈items/holes⌉ ×1
- the Langford counts are 1, 1, 26, 150, 17,792 and 108,144 ×1
- the least gap is exactly the conjectured value ×1
- the option is one this view reads ×1
- the product has fallen below a hundredth ×1
- the search runs to one, five or twenty million ×1
- the view is one the family draws ×1
- their difference brings both multiples within 1/n of whole numbers ×1
- there are at least as many things as boxes ×1
- there is at least one box ×1
- twenty-six pairings of order 7, up to reversal ×1
- two of the points share a box ×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.
A sum that forbids half the pairings
Put two 1s, two 2s, …, two ns in a row so that between the two ks there are exactly k other numbers. For n = 3 there is one way, 312132; for n = 4 one way; for n = 5 and 6 none at all, and the reason is a single sum — adding up the positions of every entry in two ways forces n(3n − 1)/2 to be even. A search confirms the count forbids nothing that exists and finds 26 arrangements at n = 7 and 108,144 at n = 12, but the sum is all anyone knows about why.
DiscreteAlone on the track
Runners set off together at different constant speeds. Is each of them, at some moment, far from all the others? A pigeonhole on time gives half the conjectured distance in two lines, and sixty years of work have not closed the other half. The speeds 1, 2, …, k are exactly at the limit, and so, surprisingly, are 1, 3, 4, 7 and 1, 4, 5, 6, 7, 11, 13.
NumberHow close a fraction can get
Drop eight points into seven boxes and two of them share. That one line, applied to the multiples of an irrational number, proves that every irrational has infinitely many astonishingly good rational approximations — and no construction is needed anywhere.
DiscreteMore things than boxes
If there are more objects than containers, some container holds two. That is the entire principle, it is impossible to disagree with, and it settles questions that look nothing like it.
DiscreteOne bottleneck and nothing else
A set of jobs can be filled by distinct people unless some group of jobs has too few candidates between them — and that single obstruction is the only one there is, which is what makes the theorem worth having.
DiscreteSix people at a party
Among any six people, three are mutual acquaintances or three are mutual strangers. Five is not enough, and the arrangement that saves five is a pentagon. Beyond that the numbers become unknowable.
DiscreteThree in a row on the number line
Colour the numbers one to eight in two colours and it can be arranged that no three equally spaced numbers agree. Add the ninth and it cannot. The structure being forced is arithmetic rather than graphical, and the proof is a different proof.
ProbabilityTwenty-three people
A room needs 253 people before someone probably shares a birthday with you. It needs 23 before two of them probably share one with each other. The gap between those numbers is the whole problem.
DiscreteTwo numbers approximated at once
Thirty-seven points in thirty-six boxes force a multiple of √2 and a multiple of √3 to land near whole numbers together, and the same count in more boxes makes q times the two distances less than one infinitely often. Littlewood guessed that the product can be made as small as anyone likes. Each number alone has a floor; the product finds a new record eight times below twenty million, and nobody can prove it will keep going.