The collection

Every essay — page 19

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

Particular games

Hackenbush, Nim, Domineering, Toads and Frogs — the specific games the general theory was built to explain.

Two conditions, one of which survives. Two candidate conditions on a pair of subtraction lists, scored over all 961 pairs drawn from one to five. Translation holds on 83 pairs and every one of them repeats; all-odd holds on 49 and four of them do not.

The condition that survived the wider sweep

Which pairs of subtraction lists have a value sequence that repeats? Over the 49 pairs drawn from one, two and three, two conditions answer it identically — a translation and all-odd — and both are exactly right. Over the 961 pairs drawn from one to five, all 83 translations still repeat with no exception and four all-odd pairs do not, at heap ninety with a period as long as forty-two. Neither condition is necessary: 104 pairs repeat that satisfy neither.

6 figures · Partizan subtraction
Three strips a criterion cannot tell apart. Three Push strips identical in length, reading, coin counts and run structure, whose readings are wrong by a quarter, a half and a quarter more than one. The order of the colours inside the run is the only thing separating them.

The criterion that cannot exist

The rung below asked for a quantitative version of its condition — turn 'the reading survives mixing three quarters of the time' into a statement about the strip. Three strips of four squares settle it. `.LLR`, `.LRL` and `.RLL` have the same length, the same reading, the same coins and the same single run, and their readings are wrong by 1¼, ¼ and ½. The error is a fact about the order of the colours, and 207 of 805 statistical classes carry more than one of them.

6 figures · Push
What survives being added. The packing count and the packing interval on boards of one to four regions. The count is exact on 45 per cent of single regions and 11 per cent of four-region boards; the interval contains the value 67 per cent and 74 per cent of the time.

Two errors that cancel

Replacing the packing count with an interval left a doubt that the pessimistic half would add across a board. It adds, for a one-line reason. What is worth measuring is what the reading is then worth: over boards of one to four regions the count decays from exact on 45 per cent to exact on 11, and the interval's containment does not decay at all — it rises from 67 per cent to 74, because the interval's width adds and its error does not.

6 figures · Domineering
Seven, ten, thirteen and sixteen. The four values the rung below named, each read off the prime factorisation: one plus the largest prime plus the product of the two largest.

The size of a cake

Ω gives the sign of a Maundy Cake and says nothing about the size, and the rung below left four values — 7, 10, 13 and 16 — unaccounted for. For a one-row cake they are a formula: write the prime factors largest first and add up their running products. The rule behind it is greedy — cut by the largest prime — and it is exact on every one-row cake to two hundred and wrong on a fifth of the two-sided ones.

6 figures · Cutcake
A numeral in the empty squares. Runs of coins with one to five empty squares in front, and the value of each. Every row is a binary expansion converging on a fraction the colours determine.

A numeral in the empty squares

The rung below ruled out a quantitative criterion for Push and asked for a numeral over the coins combined with a count over the gaps. The two ingredients are the right way round: the colours pick a fraction — −1, −1/3, −1/7, −1/15 — and the empty squares give the binary precision, so a run of k coins before one of the other colour with g gaps is worth exactly (1 − 2^(−kg)) ÷ (2^k − 1). And it does not compose: a strip of two runs is not the sum of them, on any pair tried.

6 figures · Push
Still a count of squares. One-sided Amazons regions on two-dimensional boards, swept exhaustively. Every one is worth exactly the number of free squares in it.

A region one player owns

On a strip, a region containing only one player's amazons is worth exactly its free-square count, on all 45,057 positions of the rung below's sweep — and it predicted the exactness would fail in two dimensions, where an amazon can be short of room in one direction and not another. It does not fail. Over 2,412 two-dimensional regions there is no exception, and the reason is one clause: an amazon may shoot back at the square it has just left.

6 figures · Amazons
The count is a count of odd runs. The largest packing of dominoes a player can hope for in a region, written as a formula in the region's own lines. Every run of odd length wastes one cell, so the packing is half of what is left, and the count is half the difference between the two directions' odd runs.

Half the difference in odd runs

The rung below asked what the regions the packing reading fails on have in common, and whether it is something a player could see. It is: the reading itself. The count has a closed form — half the difference between the region's odd horizontal runs and its odd vertical runs — and it is exact seven times in ten when it claims one move of advantage, on none of the largest regions where it claims two, and it exaggerates four times in five when it is wrong at all.

6 figures · Domineering
Moving them apart does not make them independent. The value of a two-run Push strip as the gap between the runs widens. Each row converges, and none of them converges to the sum of its two runs.

The cliff a cut invents

The rung below asked for a correction term in the gap between two Push runs. There is none, because the gap's contribution vanishes: widen it and the strip's value converges geometrically, at a rate set by the back run's length alone, to a limit that is not the sum. And Shove — whose reading is exact everywhere — fails at the same cut, which says the broken thing is the cut and not the game.

6 figures · Push
Cut small unless you are behind. The complete rule for the best cut in a Maundy Cake, in three cases decided by the two sides' counts of prime factors. It is exact on every cake in a sixty by sixty grid.

Cut small unless you are behind

The rung below found the greedy rule — cut at the largest prime — wrong on 104 of 552 Maundy Cakes and asked for a description of them. On all 104 the best cut is at the smallest prime, the exact opposite. A middle divisor is never needed on any cake in a sixty by sixty grid, and which of the two extremes wins is decided by Ω alone: cut small when Ω(m) + 1 ≥ Ω(n), large otherwise, and that is exact on all 3,540.

6 figures · Cutcake
Almost none of them is a number. Every three by three Amazons region holding one amazon of each colour, sorted by what kind of value it carries. Fifty-six of the two thousand are fractions and the great majority are hot positions.

The fractions that were not there

The rung below counted 1,452 fractions among the shared Amazons regions and asked which fractions they are. Fifty-six of them are fractions. The other 1,396 are hot positions with a fraction somewhere inside their options, counted by a regular expression looking for a slash — and the quantity the separation of the two amazons actually sets is not a denominator but a temperature.

6 figures · Amazons
Running products, and where to stop. Six Maundy Cakes with the prime factors of the longer side, the running products those primes make, and the value the sum of them gives.

The short side only says how many

The rung below settled which cut to make in a Maundy Cake and left the value open. With the cut settled the recursion is a walk, the walk unrolls, and what it unrolls into is the running products of the long side's prime factors, largest first. The short side never enters the products at all — it decides how many of them there are and nothing else, so sixty-two different short sides give one value.

6 figures · Cutcake
Every gap dies at the last run's rate. Three-run Push strips with each gap widened in turn, and which of four candidate rates the convergence matches. The rearmost run's rate wins every family and the compound rate wins none.

Read from the back forwards

The rung below found a two-run Push strip converging at a rate set by the back run and asked what a third run does — whether the rate is still the rearmost run's, or whether the rates compound. It is the rearmost run's, and for every gap: widen the front gap of a three-run strip, two whole runs away, and the value still dies at the last run's rate. Shove, the game one clause away, compounds.

6 figures · Push
Four counts and an interval. One Domineering region with its run lengths in each direction and both packing counts written as sums over the runs.

One domino every three cells

The rung below gave the optimistic packing count as a formula in odd runs and asked for the other end of the interval, expecting a formula in the even ones. Parity is the wrong arithmetic: the smallest maximal packing is a sum of ⌈(len−1)/3⌉ over the runs, exact on all 1,042 shapes. That makes the whole interval readable off a drawing — and shows it can never reach the value, because regions with the same runs have different values.

6 figures · Domineering
The same distance, different room. Two shared Amazons regions with the amazons the same distance apart, differing in how many squares both of them can still reach. The one sharing more is the colder.

Room pulls two ways

The rung below found the distance between two amazons setting a shared region's temperature and asked for something finer — the squares each can reach, or the squares both can. Neither beats the distance on its own. Together they beat it by half as much again, and the reason is that they pull opposite ways: further apart is hotter, and sharing more reachable squares is colder.

6 figures · Amazons
Term by term. Every cut of one long side, with its value written as running products beside the terms of the largest-prime cut.

The short side is not in the lemma

The closed form for a two-sided Maundy Cake rested on one unproved statement: that no divisor beats the largest prime. Written out, that statement never mentions the short side — it is an inequality between a multiset of primes and a term count — and once it is stated that way it has a two-line proof, term by term. The ladder ends in a theorem rather than a grid.

6 figures · Cutcake

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