The collection

Every essay — page 12

One idea per essay, ordered so that the earlier ones set up the later ones — but nothing here depends on being read in sequence.

Impartial games

Both players have the same moves. Every such position is a Nim heap, and the theorem that says so is the field's first.

Thirty-two words, four of them lost. Every position of five heaps grouped by the parities of its heaps in decreasing order of size. Each word is uniform, and four of the thirty-two are losing.

The parities, in size order

The rung below settled four of six parity classes in bounded Moore's Nim and asked whether the sizes pick out the losing positions in the two it could not. They do — but only through the order they put the parities in. Sort the heaps largest first, read off their parities, and that five-bit word settles the whole game at every width of move, with the losing words forming a subspace.

6 figures · Moores-nim
Two symmetries, the same two clauses. The half-turn pairing and the reflection pairing written side by side, with the fixed squares and self-paired dominoes each has to exclude.

A pairing, and the pairing

The rung below repaired the half-turn check and asked whether a reflection would fire where it does not. It does — forty positions of 58,830 on the largest board — and it is sound, and it is worth one node in a thousand to a solver. It can never fire on an empty rectangle at all, which is why the ladder's whole subject is the half turn.

6 figures · Pairing
Two counters, not one. The four periods of the odd-saltus class against the two base-three counters, with which each follows.

Two counters, and one displaced term

The rung below found four Grundy sequences in the odd-saltus class and asked which term each displaces and whether the digits predict it. They do — but there are two base-three counters and not one, chosen by whether a heap of one can be taken away. And there are three sequences rather than four: the fourth is the third with three isolated values, and was counted separately because its period had not settled.

6 figures · Hexadecimal
Four conditions. The four linear conditions whose kernels are the losing sets, with the cases each covers.

The parameter was the difference

The losing words of bounded Moore's Nim form a linear subspace and no map was known whose kernel they are. The equations exist, four conditions cover all thirteen cases at three to six heaps, and they are indexed not by the heap count but by the heaps less the width of a move — which turns the failure at six heaps into a prediction about seven.

6 figures · Moores-nim
The clause that was free. Five requirements on a pairing strategy, with which of them each map meets.

A symmetry that is not a pairing

The quarter turn was the last symmetry a Cram pairing argument had not tried, and the one a square board seemed to offer. It fires on the empty four by four and it settles nothing the half turn misses — and the reason is a clause four rungs of this anchor never had to write down, because every map tried so far was its own inverse.

6 figures · Pairing
Not rare at all. How many hexadecimal codes whose sequence settles have a stretch of heaps before the pattern begins.

A pattern that has not started yet

A pre-period was supposed to be rarer in this family than a defect. Two hexadecimal codes in five have one, 321 have a pre-period longer than their own period, and the code the rung below found slow takes fifty-four heaps to settle rather than two blocks — which is also the account of three defects the rung below recorded and could not explain.

6 figures · Hexadecimal
Five of six. The six predictions made for seven heaps by the difference reading, each scored against the sweep that was declined at the time.

The family with two witnesses

Six predictions about seven heaps were written down and deliberately not run. Five of them held. The one that broke is the condition that had been checked against two cases when it was proposed — the fewest of the four — and at seven heaps it does not merely give the wrong answer, it asks a question the parity word has stopped being able to answer.

7 figures · Moores-nim
A pairing no motion of the square gives. The smallest Cram shape carrying a pairing that is not a rigid motion, with its three pairs drawn as lines between the squares they join.

A pairing that is not a symmetry

Every pairing strategy this ladder has found is a rigid motion of the square, and the requirement mentions no geometry at all. Searching all 8.8 million fixed-point-free involutions instead of the eight maps more than doubles what a pairing explains — and the smallest new one turns out to be a reflection with its two fixed squares swapped.

6 figures · Pairing
One of the two quantities is inert. The candidate predictors of a pre-period's length, each scored by correlation against the measured length.

The quantity that carried nothing

The rung below proposed predicting a pre-period's length from the saltus and the period. The saltus correlates with it at −0.03, which is nothing; the period correlates at 0.77 with a coefficient of one, so a pre-period is about one period long. The digit that predicts whether there is one predicts nothing at all about how long.

6 figures · Hexadecimal
The dual is the value table's span. The dual code against the span of the bit-planes of the one-coin Grundy values, on every game measured.

The dual was the value table

A coin-turning game's losing rows form a linear code, and a code has a dual that nothing in the game appeared to read. It reads it constantly: the dual is spanned by the bit-planes of the Grundy values — the parity checks are the value table stood on end — and on Mock Turtles over eight coins the losing rows are exactly the span of the table that decides them.

6 figures · Codes
Two statements, two routes. The two measured identities the rung below left unproved, with the argument each was expected to need.

One proof, and one wrong lemma

Two measured identities were left for a proof: the move rule by induction, the gap condition from the reply bound. The induction is exact on 31,731 heaps at eight factors. The reply bound holds at c = 2 and on one index pair in twenty-seven at c = 3 — and the inequality that does the work is a third one nobody proposed.

6 figures · Fibonacci nim
Four readings, one game. The closed form, the strict mating, the cancelling matching and the parity term, with how much of the game each accounts for.

The pairing the formula hides

Welter's closed form sums a function over every pair of coins and needs an extra term when the count is odd, which the rung below called a surprise. Read as a matching it is not: an odd number of coins cannot be paired, the left-over coin contributes its own square, and some matching gives the value on every position measured.

6 figures · Welter
What the theorem replaces. Every grid a brute-force solve can reach, valued both ways, with the grid the theorem is normally drawn at underneath. Twelve squares is four thousand arrangements against twelve products; sixty-four squares is eighteen quintillion against sixty-four.

Four hundred and seventy steps

The tartan theorem replaces a search with a multiplication. Measured on every grid a brute-force solve can reach, the two agree on all of them — and the ratio doubles with every square added. On the 8 × 8 grid the theorem is normally drawn at, the search would have to value eighteen quintillion arrangements; the theorem needs twenty-six different nimber products, and computing all of them by the rule that defines them looks at four hundred and seventy pairs.

6 figures · Nim
Lasker's Nim in sixteen cells. A four-by-four table. Each row and column is a residue mod 4 of one part of a split heap, with the residue of that part's Grundy value beside it; each cell is the residue mod 4 of the split's value, the nim-sum of the two parts. Every split of every heap to four hundred lands in the cell its residues name.

The proof is sixteen cells

Lasker's Nim has a four-clause formula that was checked on two thousand heaps and never proved. The proof fits in a four-by-four table: the last two bits of a split's value are fixed by the last two bits of its parts, so no split can land in its own heap's class — except at 3 mod 4, where it lands exactly on the one value the takes leave missing and pushes the answer up by one.

7 figures · Lasker
One split is enough, and some are not. Lasker's Nim beside five versions of it that allow only some splits, over the first twenty-four heaps, with every cell that leaves the formula outlined. Allowing only the split that takes one counter off reproduces the whole sequence; allowing only equal halves turns it back into Nim.

One split is enough

A heap of n in Lasker's Nim offers ⌊n/2⌋ ways to split, and the values use at most one of them. Allow only the split that takes a single counter off and every heap to six hundred keeps its value; of all sixty-three sets of split sizes up to six, a set keeps the formula exactly when it contains 1 or 2. Equal halves alone give back plain Nim, because a split into equal parts is a move to nought.

7 figures · Lasker

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