Bipartite graph
Named by 9 essays across 3 fields — each of them below, with the objects they name alongside it.
The walk through the middle levels
On seven places, the words with three ones and the words with four number thirty-five each. Is there a closed walk through all seventy, changing one place at a time and never leaving those two levels? On five places the answer is 24 walks, on seven and nine a search finds one in moments — and whether one exists for every odd length was open for thirty years, until Torsten Mütze proved in 2016 that it always does.
Enough partners in every finite group
Hall's theorem says a matching exists unless some group has too few partners between them. On an infinite graph that is false: one vertex that can be paired with anybody, and every other with exactly one, passes the test on every finite group and has no matching at all. The failure needs a vertex with infinitely many choices. Forbid that, and the theorem comes back, by an argument about trees rather than about matchings.
Which graphs let the tokens go anywhere
A sliding puzzle is a graph with a token on every vertex but one. Richard Wilson found in 1974 what every such puzzle can reach, and the answer has a surprise in it — the half the tray is stuck with is not a fact about permutations at all, but about the board being two-coloured — and one exception, a graph of seven vertices that reaches exactly 120 of 720.
Every pair side by side, once
Seat an odd number of guests at round tables for as many nights as it takes, the same table sizes every night, so that every two guests sit side by side on exactly one night. For a single table a zigzag turned a notch each night does it for any number of guests. For other table plans the answer is almost always yes — and for six guests at two tables of three, nine at tables of four and five, and eleven at three, three and five, an exhaustive search proves it is no.
Matching when the arrivals are shuffled
An adversary who chooses the order in which applicants arrive can hold any fixed matching rule to half the best. Take the order away and let it be random, and the plainest rule of all — give each arrival its first free place in a fixed list — rises to 1 − 1/e, because shuffling the arrivals turns out to be exactly the trick RANKING plays with the places, seen from the other side. Shuffle both and the guarantee rises again, to somewhere between 0.696 and 0.727, and where in that interval it lies is not known.
Half the neighbours forces a tour
Deciding whether a graph has a closed tour through every point is one of the hardest problems computers are asked to solve. Gabriel Dirac found a condition that settles it in one glance: if every point is joined to at least half the others, a tour exists. The proof is a counting argument on a single path, it runs as an algorithm, and two small graphs show that the half cannot be lowered — while random graphs show how much stronger than necessary it is.
The boards a knight can tour
A knight can visit every square of a chessboard once and land a move from where it began. On a 5 × 5 board it cannot, on a 4 × 100 board it cannot, and on a 3 × 8 board it cannot, though on 3 × 10 it can in sixteen ways. Allen Schwenk found in 1991 the complete list of rectangles without a closed tour, and the reasons on it are of two kinds — a colouring that counts squares, and a second colouring that a tour would have to make agree with the first.
Signs that make a determinant count
Hall's theorem says whether a board can be covered by dominoes. How many ways it can be covered is a different question, and for a general graph a hopeless one — the count is a permanent, and nobody expects an efficient way to compute permanents. On a flat board it is a determinant. Put a minus sign on the vertical pairs in every second column and the thirty-six coverings of a four-by-four board stop cancelling and add up; the chessboard has 12,988,816. A hole in the board breaks the rule in a way a cut repairs, and on K₃,₃, which cannot be drawn flat, no choice among its 512 signings works at all.
The diamond that freezes its corners
A square board's domino tilings grow like 1.3385 choices per square. Arrange the same squares in a staircase-edged diamond and the count is exactly 2 to the n(n+1)/2 — only 1.1892 choices per square, at every size. The boundary has reached into the middle. Pick one of the diamond's tilings at random and the reason is visible: outside the circle inscribed in the diamond the dominoes are frozen into solid brickwork, and only inside it is the tiling free.
Named alongside it
The objects these essays reach for when they reach for this one.
Exhaustive searchMatchingParityCounting argumentHamiltonian cycleDeterminantGreedy algorithmPigeonhole principleAxiom of choiceBinomial coefficientCombinationsCompactness