Every essay — page 12
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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