Counterexample — where it appears
Named by 160 essays across 9 fields — each of them below, with the objects they name alongside it.
Outcomes do not add
Knowing who wins each part of a position tells almost nothing about who wins the whole. Counted over every sum of two values born by day two, six of the outcome table's ten entries are settled and four are not — and every settled one is settled by the order rather than by anything about outcomes. Two first-player wins reach all four classes between them.
The theorem that names a winner and no move
Strategy stealing proves that the first player wins Hex and wins Chomp, on every board, in about four lines. It exhibits no move, contains nothing a move could be extracted from, and is not going to. The moves have to come from somewhere else, and where they come from runs out almost immediately.
Independence is a claim
Splitting a position into parts and adding the values is the whole method of this subject, and the splitting step is a claim about the position rather than a fact about the drawing. Where it is false the two answers differ — and the failures that matter are the ones that keep the same winner and change the value, because nothing reports those.
The period is small and the proof does not say so
Every subtraction game repeats eventually — that is a theorem, and its proof gives a bound of sixteen thousand for a three-move set. Over 112 sets the longest period measured is twenty-two. The proof and the fact are four orders of magnitude apart, and the rule of thumb that closes the gap is broken by one set in the sweep.
Maundy Cake
Cutcake with one word added: a cut must divide the piece into equal parts. The values are still whole numbers, and the rule this site has been repeating about them is false — over all 1,296 cakes to 36×36 the largest-odd-divisor account has 946 counterexamples. What survives is a count of prime factors, and it says who wins without saying by how much.
Three players and no answer
Every theorem here is about two players, and the reason is not convenience. With two players the game is zero-sum, so 'play well' needs no further explanation. Add a third and the winner of a Nim position becomes a fact about the convention: two reasonable ones disagree on 56 of the 71 positions swept. The one question no convention touches — can a player force a win against the other two together — is answered 'nobody' in 65 of the 71.
A rule with no promise at all
Playing in a hottest component comes with a bound: never more than the largest single temperature below the mean of the board. Over 220 sums the bound holds 220 times — and so does the bound for a rule with nothing behind it, which scores exactly what perfect play scores on 205 sums against the hottest rule's 196. The control that shows the bound is doing work is the rule that plays the coldest component, which breaks it 74 times and loses up to eleven points.
How long it lasts
Move in every component at once and the game ends the moment any one of them does. Grundy values say nothing about that game; what decides it is the remoteness, a second number computed from the same tree that measures how long a component can be made to last. Over 2,268 positions the rule is right every time, and the two numbers determine each other in neither direction.
How cold a sum of hot games can be
The temperature of a sum is at most the largest temperature in it, and the bound leaves the whole interval below it open. Over 1,035 pairs the sums do not use that interval: 864 sit exactly at the maximum, 160 are frozen outright, and eleven land anywhere in between — every one of them with a component whose wall bends.
How much a list of options can lose
Deleting a dominated option is the reduction with no surprises, and how many options it takes is decided by the shape of the order rather than by the values in it: the survivors are the maximal elements, and the count is the length of the list less the number of them. The essay separating the two reductions closed by predicting that the longest chain would give the number. It is a lower bound, exact on 3,859 of the 7,315 four-option lists and wrong on the rest.
The strip where every number is a whole one
Delete the hop from Toads and Frogs and the halves, quarters and ups vanish completely: over 9,801 strips, every value that is a number is an integer, without a single exception. The guess that the hopless game therefore has a formula reading the gaps is half right and exactly wrong — 1,460 strips of eight squares are switches, and three strips with the same counts of toads, frogs and empty squares are worth 1, {2 | 1} and 2.
A pool built to punish greed
The rung below found a rule with no theorem behind it beating the rule with one, and predicted that a pool of deliberate traps would reverse the result. It does not. The traps miss, because the rule called greedy scores a move by the stop it leaves and a stop already contains the follow-up — and the rule the traps do catch, losing sixteen points where the guaranteed rule loses five, had to be written to make the point.
What a wider pool rescues
The misère outcome table has sixteen cells, and over a pool of ten positions fifteen of them hold fewer than four outcomes — which looks like structure and might be a shortage of positions. Thirteen values further on there is nothing left: every pair of outcome classes takes every outcome, so the near-misses were the pool, and the prediction the rung below made was right.
What a component has to carry
Three impartial games — Fibonacci Nim, Top Entails and Nim with a held pass — break the sum, and they break it for the same reason: a component cannot say what its own legal moves are. Measured with one instrument — one number per part, exclusive-ored — the failure rate runs from a quarter to nearly half, against a control where the same recipe is a theorem and is never wrong.
When the regions add
The rung below described the NoGo boards whose regions add as the ones with symmetric walls, and said the description was a guess made from six examples. It is wrong: fourteen symmetric boards do not add and sixteen that add are not symmetric. What replaces it is a criterion about liberties — sound on all 117 boards, provable in a line, and complete on only nine of the twenty-four.
The number nobody needs
The compound theory carries a third quantity — the suspense number — computed by the remoteness recursion with both preferences reversed, for the compound that stops as soon as any component stops. It governs that compound correctly. So does remoteness, so does the plain Grundy value, and the shortening does not change the winner on any of 1,176 positions.
Where the order and the sum disagree
Day two is a lattice, and day two is a group, and it is not a lattice-ordered group. The one identity that would join the two structures — the join plus the meet equals the pair — holds on exactly the 201 pairs where it cannot fail and on none of the other 52, and the errors split thirteen high, thirteen low and twenty-six confused.
A fight with no midpoint
The mean of {a | b} is the midpoint and the temperature is half the gap — on the twenty-one values of day three where a and b are numbers. One hundred and forty-six others have the same shape and not the hypothesis, and the repair that suggests itself, reading the two stops instead of the two options, holds on about three quarters of them and no more.
The answer that starts another fight
A local move is answered while the ambient temperature stays below the follow-up's — and that rule, which this site has carried since the anchor opened, is exact only when the answer ends the fight. When the answer starts another one the crossover is exactly half the follow-up's temperature, on every position tested, and a third level of fight does not halve it again.
Which shapes are worth fighting over
Forty-four of the 104 Domineering regions of at most six squares are worth numbers and the rest are not, and the rung below said no visible property of a shape predicts which. Half of that is wrong: a region only one orientation fits in is a whole number, on all eleven of them, for a reason a reader can supply in a sentence. The other half stands, and thirty-three shapes are what makes it stand.
The reading that survives too much
Counting the empty squares in front of each coin gets a Push position right half the time, and the rung below said the failures were exactly the positions with two coins of opposite colour side by side. Sixty-six of the 1,072 failures have no such pair, the smallest is five squares long, and the condition that does decide it is not about the board at all — it is about every position the board can reach.
Where the nimbers run out
A single End-Nim heap is a Nim heap and every palindromic row is worth a nimber, so the impartial theory looks as though it might get a long way into a partizan game. It gets one row in thirteen. Five nimbers occur in five and a half thousand rows, the palindromes account for two fifths of them, and the rows worth something else run to 2,693 distinct values.
Looking for the symmetry
Answering every move with its mirror image wins Cram on a board with both sides even, which is the argument everybody meets. Asked of every connected shape of at most eight squares instead of of thirteen rectangles, it wins twelve — and accounts for a sixth of the second-player wins there are, because 852 of the 1,042 shapes have no symmetry to answer with in the first place.
The rule the symbols follow
Two genus symbols make a third by three lines and no lookup table: the base exclusive-ors, the sum is fickle only when every component is, and the symbol follows. Checked on 252 pairs across nine games it is right on 238 — and the fourteen failures are exactly the fourteen pairs with a wild heap in them, which is the boundary the genus is defined up to arriving as a measurement.
Fifty-two errors and seven sizes
Day two is a lattice and a group and not a lattice-ordered group, and the fifty-two incomparable pairs it fails on leave fifty-two different error terms. Measured rather than listed, the fifty-two collapse: seven pairs of stops, three means, three temperatures, and a rule that predicts the temperature from the pair on forty-four of them.
The cheapest way to show a value
Eleven thousand positions from fifteen rulesets reach 1,193 values, and for each of them there is a smallest board that shows it. Set against the birthday the two measures agree hardly at all — until the numbers are taken out, at which point they agree rather well, and the whole apparent independence turns out to be a fact about integers.
Which option the reduction keeps
Domination deletes an option when another is at least as good, so what survives is the top of an order. On a board that order is made of moves, and two descriptions of the surviving move suggest themselves. Over 1,586 Domineering option lists one of them is right 47% of the time and the other 90%, and the one that wins is not the one a player would guess.
Counting the moves each side has
How many dominoes could each player still place? Subtract, and there is a whole number computable from the drawing with no game theory in it. Over 1,042 regions it is the value on 141 of the 315 worth numbers, lands between the stops on 619 of the other 727, and its failures are two different kinds — one of which was inevitable and one of which is a fact about the game.
The rows that are their own mirror
Four hundred and ten End-Nim rows are worth nimbers and 168 of them are palindromes, so a condition covering the other 242 was outstanding. It is that the row is equal to its own negative — and on this game that condition is not merely sufficient but exact, which is more than the group law promises and is a fact about End-Nim rather than about games.
How thick a wall has to be
A single stone between two empty stretches of a NoGo board couples them, and the obvious repair is a thicker wall. Over 590 walled strips a thicker wall does help — and splitting the same 590 by the colour of the stones shows that thickness was never the variable. A wall of four one colour couples the sides exactly as one stone does.
The family the Fibonacci numbers belong to
Fibonacci Nim lets a player take at most twice what the last one took, and the heaps the opener loses are the Fibonacci numbers. Two is an arbitrary number. At one the losing heaps are the powers of two, at three and four and five they are four more sequences, each with a linear recurrence whose lag is twice one less than the factor — until the factor is six, where the pattern stops.
The symmetry one move away
A pairing argument proves the second player wins and names no move to do it with. Asked of every shape of up to eight squares it settles twelve boards. Asked one move later — can the first player reach a position a half-turn pairs? — it settles 288, and which boards those are is decided by parity before anything about their outline is looked at.
What the colon respects
The ordinal sum reads the form of its base rather than its value, which is why the colon principle is stated for positions and not for values. Built over 9,604 forms it turns out to read the value on 636 of the 640 values that have more than one form, and the four it can tell apart are zero, one, minus one and star — the values born by day one, and no others.
A schedule instead of a number
Playing in the hottest component loses at most the largest temperature on the board — the classical guarantee, stated against one number. Sorting the temperatures and reading the guarantee one step further down gives a promise 47 per cent smaller that is never breached over 1,734 sums. Two steps down it fails 120 times, so the schedule has exactly one step of slack in it.
What the halving is a function of
A move is answered while the board is cooler than the follow-up's temperature — or half of it, depending which of two classes the position is in, and the classes were stated in terms of forms. They are a feature of one wall: whether the follow-up's right wall rises straight up before it leans. Fifty-five positions, no exception, and the crossover becomes something a reader can see.
When to leave the environment
A Go player's question is not which fight to take but when to stop taking the small stuff. Put two fights beside a stack of coupons and the orthodox answer — leave when the coupon falls to the hottest temperature on the board — is exact on sixty of eighty-one pairs. All twenty-one departures have a fight with a follow-up in them, and every pair of plain switches leaves on time.
How wrong a nearly-independent split is
Treating a connected board as a sum of two halves is a claim, and the rung below counted how often it fails. This one prices it: over every vertical cut of every small Domineering rectangle the error is a game rather than a number, it is never in Right's favour, and it is bounded below by twice the height of the cut — a bound the height alone does not supply.
What the numerals knew
Every factor in the Fibonacci Nim family gives a numeral system, and the rung below predicted its separation condition would be the lag of the recurrence the losing heaps satisfy. It is not. The gap is the factor — one at c = 1, Zeckendorf's two at c = 2, and c at every factor to eight — while the lag goes 1, 2, 4, 6, 8, 11, 14, 17 and leaves its own pattern at six. The numerals then solve every one of 194,480 states, cap and all.
A code that climbs by three
Five hexadecimal codes were known to repeat with a constant added, and every one of the five constants was a power of two — either a fact about exclusive-or or a coincidence over five cases. Sweeping all 255 two-digit codes settles it: seventy-one climb, seventy of them by 1, 2, 4 or 16, and one by three. The exception is ·3f, whose values are 3⌊n/6⌋ + (n mod 3) on every heap to twelve hundred.
The rule a smaller move breaks
Moore's Nim lets a player take from at most k heaps, and its winning condition is the binary columns summed modulo k + 1. Cap the amount as well and the obvious repair — reduce each heap modulo the cap plus one, then read the columns — is exact at every cap when k is one and wrong at every cap when k is two or three. The reason is stronger than a broken rule: at k ≥ 2 the residues do not determine the outcome at all, so nothing of that shape can work.
The moves a player can be talked out of
The difference of the two players' largest domino packings is the value of a Domineering region on 141 of the 315 worth numbers. The count is optimistic for its owner and pessimistic for the other, and one number cannot be both — so it becomes an interval, from what a player can be reduced to against what the opponent can achieve. The interval contains the value on 209, is a single point on 505 of the 1,042 regions, and never exceeds two moves wide.
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.
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.
The birthday is a floor
The rung below measured a correlation of 0.73 between a value's birthday and the size of its cheapest exhibit, and asked which values are dearer than the birthday suggests. The relation is not a trend. Over all 728 non-number values the exhibit is never smaller than the birthday and is exactly the birthday on 476 of them, and the excess on the other 252 belongs to the game rather than to the value — the ruleset accounts for 40 per cent of its variance.
The bend is the condition
The rung below offered a description of the class its stop reading is exact on — neither wall bends below the meeting point — and a route to proving it: that a bend happens precisely when some option is neither a number nor an infinitesimal. The first is exact on all 138 values, both directions, no exception. The second is half right: every bent value has such an option and 49 unbent ones do too. And the eight apparent exceptions to the first turn out to be a bookkeeping convention.
How many moves are worth making
A value answers who wins and by how much, and the anchor below names the quantities it discards. This is the first of them counted. Over 1,034 Domineering regions and 125 values, 63 values have two regions disagreeing about how many placements are worth making and 52 disagree over whether there is any choice at all — and the count of good moves stays near one and a half however large the region gets.
A self-negative value costs a day
The rung below placed the thirty values equal to their own negatives on the temperature scale and asked whether being self-negative forces anything about when a value can be born. It does, exactly: the earliest self-negative value of temperature t is born the day after t itself, which accounts for the four temperatures that carry one and the four that carry none. The guess it offered — that the first value of each temperature is a self-negative one — holds at four temperatures out of five and is not the shape of the answer.
The closure that picks the nimbers
Closure under addition lets a sum be rewritten and turned out to admit companies that are not nimbers at all. Closure under forming options lets a subposition be rewritten, and it pulls the other way: every company in use for these computations has it and none has the first, and among the seven finite addition-closed companies, keeping every option is exactly being a group of nimbers — four of seven, both directions, no exception. Demand both at once and nineteen of twenty-two day-two values generate nothing finite.
No fifth value
The colon reads a form rather than a value, and the rung below found the forms of a value disagreeing at exactly four of them — the values born by day one. It could only check forms whose options came from day two. Built one day deeper, by adding day-three gift horses to day-three values, eighteen thousand forms give no disagreement at all, while the same treatment still splits nought four ways. The class is about the width of the base's form and not the depth of its options.
The margin a count needs
Leaving the opponent fewest replies names only surviving options nine times in ten, which leaves the question of what a bound stated in that count would have to be weakened to. It is a margin. Over 57,879 pairs of Domineering options, the one leaving the opponent fewer replies is the worse of the two 1,052 times at a margin of one and 72 times at a margin of two — and at a margin of three, never.
A subtraction, not a factor
The crossover factor was a half on fights whose answer starts another fight, measured on a pool with two three-deep positions in it. A pool built to be deep gives twenty, and the factor does not survive them: the crossover is the follow-up's temperature less a half on eighteen of the twenty, and a factor of a half agrees with that only where the temperature is one — which nearly every position in the earlier pool had.
What the class does not buy
Dead-ending is the hypothesis several modern misère results are stated under, and the rung below sorted this site's games into it without running the comparison those results are about. Running it: a quarter to a half of ordered pairs compare inside a ruleset's own universe, which is a great deal — and the ruleset that is not dead-ending sits in the middle of that range. Ten comparisons are lost when a universe is enlarged, and every one is lost to a dead-ending company.
The weight that blunts the count
The rung below found that counting the opponent's replies gets the direction of a comparison right once the gap reaches three, and proposed a repair: weigh each reply by whether it leaves the opponent anything. Weighing it makes the count worse. The threshold goes from three to four, the failures from 1,124 to 1,320, and all seventy-two of the pairs the repair was written for come through it unchanged.
Wider costs less
The rung below found the cheapest exhibit of a value never smaller than its birthday, exactly equal on two thirds, and the ruleset explaining 40 per cent of the rest. The variable it proposed for the remainder was the width of the form. Width and excess correlate at −0.39: the wider the value, the closer to its birthday it is exhibited, and inside a ruleset the relation cannot even agree on a sign.
The wider move is the easier game
An earlier essay ruled out every rule that reduces the heaps and reads the residues, and asked for a two-part statistic: the residues plus one more count. Four second parts are tested here and none of them decides. What turns up instead contradicts the premise the request was made under — a move that may reach three heaps is more predictable than one that may reach two, on every cap, every candidate rule, and after the change in the base rate is taken out.
Half of the smaller temperature
The correction to the sente crossover has been priced twice — first as a factor of a half, then as a subtraction of a half — each time on a pool whose answers were all about the same size. Over 128 fights with answers from a number up to a temperature of six, the correction is half the answer's temperature, saturating at half the fight's own. Both earlier readings are regions of that one law.
A bound with one number too many
The rung below bounded how far a hot addend can drag a stop — twice the smaller of the two temperatures — over sums whose addends were all plain switches, and conjectured that an addend with a follow-up would need twice the smaller of three numbers. Over 1,440 sums with bent addends the two-number bound holds everywhere and is attained 358 times, and the three-number version fails on 66.
The third digit
The rung below found 71 of the 255 two-digit hexadecimal codes repeating with a constant added rather than exactly, and asked whether the same share holds one digit wider. It rises. Of the 4,095 three-digit codes, 1,433 climb and 617 repeat exactly — seven in ten of the settled ones — so a saltus is the ordinary way a hexadecimal game settles and the exact repetition the octal survey was built to find is the special case.
At least five hundred and seventy-one
The rung below dated the self-negative values — the earliest of temperature t is born the day after t — and left the count to a day-four census nobody can run. The construction settles it instead: a value is its own negative exactly when its form is a mirror, so the subgroup can be built from subsets of the day below rather than sifted out of the day above. Day four supplies at least 571 against day three's 26, and the share of a day that is self-negative keeps falling.
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.
The quantity that does not order a board
The rung below found the players leaving an environment at the larger of a position's two temperatures, and proposed that a board should therefore be played in the order of that quantity. Over 220 boards of three components it plays exactly on 124 against playing-in-the-hottest's 196, loses 85 of the 97 disagreements, breaks Hotstrat's guarantee on six boards, and costs nine points on its worst one.
A function with no formula
The rule that composes genus symbols is exact on tame pairs and wrong on all fourteen wild ones, which looked like an exact boundary. Two heaps further it is wrong on 34 of 35 and right on one — Kayles' five and nine — so the boundary was a boundary of the pool. What survives is stronger and stranger: the pair of symbols still determines the sum on the wild side, and no rule of that shape describes it.
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.
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.
The threshold was a fact about the census
Two rungs failed to account for the seventy-two pairs where a mobility count gets the direction of a comparison wrong, and the third looks at them one at a time. They are not a class of shapes. All seventy-two are on the largest board in the census, at two depths, and sixteen positions up to symmetry — and one board larger the count fails at a margin of three, which the ladder has been quoting as the point at which it never does.
A second level of stops
The rung below found the stop reading's error to be exactly half the follow-up's temperature and asked whether the correction survives a wider pool, survives two bends, and can be stated without a thermograph. It survives eleven times the pool, missing two values in 1,459. It needs no thermograph — the follow-up's temperature is half its own stop gap. And it does not survive two bends, because day three contains no value with two of them.
The entry fee was the cap
Two rungs measured how much bigger a position has to be than the value it exhibits, and attributed what was left to the ruleset — Toads and Frogs paying 2.25 squares on everything, green Hackenbush paying nothing. Neither number is a property of the rules. Inside every ruleset the excess falls as the birthday rises, because the sweep's size cap censors exactly the values that would pay most — and three squares past the cap, Toads and Frogs exhibits values born later than the strip is long.
Three rules and a tie-break
An exhaustive table of what a Domineering strategy has to remember is 3,308 lines. Three rules applied in order answer 94.5 per cent of it — leave the opponent fewest replies, then keep the region whole, then take whichever placement comes first — and the fourth and fifth rules answer not one more. The residue is 181 decisions in which every rule scores the candidates the same and one of them is worse.
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.
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.
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.
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.
The option nothing names
The rung below found the arithmetic on reduced forms to be add and reduce again, needing the second pass on 431 of its sums, and asked whether the option that pass deletes can be named from the parts. Eight rules were scored and the best reaches four in five — and on a pool closed under negation it falls to under half, which says the near-miss is a property of the population. What the second pass does have is a shape and a cheap test that rules it out.
What identifies two subsets
Every subset of a day gives a self-negative value by mirroring it, and 1,793 subsets of day two give thirty values. The collapse happens in two stages with different characters: domination takes the 1,793 to 96 antichains and is a theorem, and the rest of the reduction takes 96 to 30 and is concentrated almost entirely on two values — nought, which has an exact description, and star, which has none.
The count of odd heaps
The rung below refused a family of two-part rules for bounded Moore's Nim and asked what the 364 losing positions have in common as a set. They have an invariant, and it is a statistic of the whole position rather than of a heap: how many heaps hold an odd number. Every all-even position is lost, at every width of move, by a restoring strategy — and the count settles every position at one heap a move and at four, and a little over half at two.
The only way to split into three
Nineteen three-digit hexadecimal codes climb by three, and the rung below asked whether they share a form and what digits they have in common. The digits are exact: on eighteen of them the only way to split a heap into three is by taking exactly three counters, and taking three counters can do nothing else. The form is not shared — the eighteen carry four distinct sequences, and exactly one of the four counts in base three.
The check that was not a check
The rung below asked for a depth-conditioned solver and named the board to measure it on. Building it found two things first. The pairing check is unsound at interior positions — on a four by five Cram board it fires on 8,613 positions and 1,026 of them are losses — and the board it named has twenty-five squares, so the check can never fire there at all. Repaired, the check is right everywhere, and the policy that pays is the root alone.
The wild side does not close
The rung below asked for the wild composition table and for two things about it: whether the wild genus symbols form a small closed set, and whether that set is a misère quotient in disguise. Building the table needed a wider sweep — nine counters a heap gives a diagonal rather than a table — and both answers are no. Not one of the twelve entries is a symbol any wild heap carries, and two wild heaps added together are tame two thirds of the time.
A threshold is a detection limit
The rung below had two points — a margin of three at fifteen squares, four at eighteen — and asked whether the mobility rule's threshold grows with the board. Eleven more sweeps say no property of a board orders the thresholds, that the same board at two depths gives two of them, and that a tenth of the sweep which produced the four reports three instead. What does move, on every board measured twice, is the depth.
Wrong in one direction only
The rung below asked for the simplified comparison test the dead-ending hypothesis is supposed to license, and predicted it would agree with the quantifier on the dead-ending rulesets and not on Toads and Frogs. Written three ways and scored on 492 pairs, it agrees best on the ruleset that is not dead-ending — and never once refuses a comparison that holds, which makes it a sound filter and not a test.
Two strips that end the same way
If a Push strip's sensitivity is governed by its last run, then two strips agreeing at the far end should behave the same however different their fronts. On 78 of 80 tails they do, exactly. On two of them a single empty square in the prefix reaches across a gap that grows without bound and halves the rate — and the run reading turns out to be sound in one direction only.
A bend that never reaches the surface
How many levels of the recursion a thermograph needs before its stops suffice is a number attached to a position, and the rung below conjectured it was the depth of the deepest bend in the tree. It is not: on 124 values a bend one level down costs nothing at all. What the number counts is the longest unbroken chain of bends running down from the top, exact on 2,400 of 2,403.
The mirror was the floor
Toppling Dominoes' share of genuinely new values had fallen from one to a half over eight sizes, and the rung below could not tell a floor from a slow fall. Four more sizes settle it: the distance above a half halves every two sizes. And the half is not a shortage of values but a symmetry — a row played from the other end is the same game, and the ruleset is as injective as that allows.
A heuristic that becomes a theorem
The mobility rule's failure rate had been measured at two depths on each of five boards and found to fall. Swept at every depth it does not merely fall — it accelerates, and it reaches exactly nought before the endgame. From four to eight empty squares onwards the rule has no exceptions at all, which turns a rule of thumb into a guarantee for the last few moves.
The price of taking the maximum
The seventy-two decisions where a Domineering strategy's rules name the wrong placement are never wrong by more than one reply, and a third of them are the second rule's fault rather than the mobility count's. The repair that follows — keep every placement within one reply of the best — retains a best placement every time and costs thirty decisions for every one it saves.
Where the runs meet
A Domineering region's value interval is a function of its run lengths and its value is not — twenty-one groups of shapes share a multiset and disagree. The crossing count separates none of them, and neither do ten other local statistics, fifteen sets of which agree on everything and differ in value. What separates nineteen of the twenty-one is where along its runs each crossing sits.
An effect that changes sign
Which squares two Amazons share turns out to matter about as much as how many — three shared squares in a line run at 0.63 where three scattered run at 2.51. But the effect of clumping is hotter at one distance and colder at the next, so the arrangement predicts well and describes nothing, which is not what the four rungs below it produced.
A pool built to have an answer
The coefficient in the rule score a component by t − λa scored identically for every λ in the unit interval, because the rule reads an ordering and that pool's orderings changed at three places. A pool designed to have twelve crossings turns the interval into thirteen different rules, and all three board sizes agree on one cell: between a quarter and a third.
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 clause that turns the class off
Three rungs failed to find the dead-ending class doing measurable work, and each time the population was blamed. Toads and Frogs with and without the jump is the matched pair the anchor wanted — the same board with the class switched on and off — and on it the test the class licenses gains less from the class than a control that has never heard of it.
Which top is the top
The crossover law's proof rests on the walls above the crossover being governed by the top two options, and the check was never run. Run on 23,586 heights it holds exactly — but only when the options are ranked by mean value. Ranked by the temperatures the law is stated in, it fails on a fifth of them.
The ceiling was a plateau
Three halves of a move looked like a ceiling on a Domineering region's temperature: it held at eight squares, at nine and at ten, and the rise that had been a quarter every two sizes stopped. At eleven squares four regions reach seven quarters — and they contain the hottest eight-square shapes and are hotter than them, so the extra material is not cold.
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 wall that bends
On a NoGo strip, two empty stretches add when no group breathes into both — a wall of two stones of different colours does it, and the criterion explains nine of ninety-three boards and all nine that it covers. On a three-row board it explains none of 227, and not because it is less accurate. A wall across a board has to bend, a stone at the bend sees empty squares on both sides by itself, and every one of the 227 has a group breathing into both regions. The condition is unsatisfiable.
One king, and two files to be in
The whole apparatus needs the files to be independent, and a king is what makes them not. With the kings unable to move the sum of the parts is exact on every configuration; give each king a single waiting move and the sum names the wrong winner on one configuration in six, and on a hundred and twenty-six of two hundred and forty-three with three files.
Two ways to count a finished board
Territory scoring and area scoring are both in daily use and they are not variants of one rule. Over seventy-nine small endgames they agree exactly on the thirty-nine with an even number of neutral points and on none of the forty with an odd number — sixteen name a different winner, and twelve are played differently, which is not something a convention is supposed to do.
The second dimension is not the deep end
The rung below says a row of eight reaches every corner of the vocabulary and goes far into none of them, and that the narrowness is a fact about the board. So the obvious next move is a rectangle — and nine squares in a square hold twenty-five values where nine squares in a line hold fifty-eight. The geometry says why before any stone is placed.
A hypothesis has to hold all the way down
Milnor's bound is proved by induction over the play, so the condition it needs has to hold at every position the play can reach. Checked on the row instead, ninety-two pairs pass the test and twenty-four of them break the bound. Checked at every subposition, twenty-eight pairs pass and none breaks it.
The sentence that solved the other convention
Bouton's paper solves misère Nim too, in one line, and it is the only misère result in the subject that fits on one. Transplanted the way the normal criterion is, it fails differently — the normal one calls losses wins and never the reverse, and this one errs in both directions on every game tried, because the clause it adds is about counters rather than about moves.
A point with three neighbours
The switching game on links is settled by counting — enough links, arranged as two trees. Played on points instead, it is the game Hex belongs to, and the count is gone. The link game turns out to be the point game in which every contested point has exactly two neighbours; give one a third, and two graphs with the same points, the same links and the same number of separate routes can have opposite winners.
Where the needle has a sentence
Strategy stealing proves the first player wins every Chomp bar and names no square to take. On two families the square can be said in a sentence — a square bar and a bar two rows deep — and in both the sentence is a pairing that names every later move too. Three rows deep the needle wanders, and the observation that every bar has exactly one needle survives ninety-four rectangles and fails on the ninety-fifth.
A board one column wider
Strategy stealing proves the first player wins Hex, and it needs three things: no draws, an extra stone never hurting, and rules that treat the two players alike. Add one column to the board and the third goes. The player whose edges are now nearer together wins whoever moves first — and does it with a table of pairs that names every reply, checked against every line to a board of twenty cells.
Every move closes the largest gap
A census of Sylver Coinage by genus finds a parity that nearly decides the game and asks whether any known property of a numerical semigroup predicts the outcome. One does, completely: a semigroup whose gaps pair off around the largest one is never lost for the player to move — none of 583 up to genus sixteen. The reason is strategy stealing, and it is the same reason the top-right square decides Chomp: every move from such a position closes the largest gap.
One bit of memory
Judge an infinite play by whether one position keeps recurring and every winner can play from a table of one move per position, with nothing remembered. Ask for two positions to keep recurring and that stops being true. At a hub with two spokes a player has to alternate, and a table cannot alternate: over every three-node game, 49,487 position-and-mover pairs are won with one bit of memory and lost without it.
The capture that has to be made
Dawson's chess is quoted as the octal game ·137, and the step from a pawn diagram to a row of counters has been taken on trust. Searched as a chess position, the diagram agrees with ·137 on every board from one file to twelve, under both endings, and every exchange it can start is an odd number of moves that lands on one of ·137's options. The whole reduction rests on one rule of the diagram that the octal code never mentions: a capture, when one is available, must be made. Make it optional and the winner changes on two, three, six and seven files.
A potential that names every move
Strategy stealing names no move, and the pairings that do name moves need a board with the right symmetry. The Erdős–Selfridge potential needs neither: Down, moving second in Hex, takes the empty cell through which Across's unfinished chains weigh most. Its guarantee reaches only boards two rows deep. It wins far past the guarantee — on every board of three rows that Down can win — and then, on a four-by-five board that a table of pairs wins for Down with certainty, it answers Across's first stone in a different cell and loses along the bottom edge.
The rule decides who has to remember
Whether a winner needs memory is a property of the winning condition and not of the board, and the property everybody reaches for is the wrong one. Of the 128 conditions on which of three positions a play keeps returning to, 32 demand memory and 19 of those are closed under union. What separates them is measured two independent ways and the two agree on all 128: a condition needs no memory exactly when it can be rewritten as a number on each position.
Two graphs a rule cannot tell apart
The repair proposed for the matching criterion was to read the cycle as well. Over every connected graph with exactly one cycle up to six vertices — 21 graphs, 114 starts — eleven such rules reach at most 91, and the winner is not a function of the matching, the cycle's length, the start's distance from it, its degree and the edge count together: six cells of that table hold both verdicts, the smallest a pair of five-edge graphs.
The winning reply is the fourth choice
The repair proposed for the potential was to weigh an edge chain more heavily. Fifty-five weightings later, none holds the four-by-five board, and an edge bonus costs Down four boards it was already holding. The reason is not the numbers: over 393,660 turns of the pairing that does hold that board, the potential would take the same cell 26.1% of the time, and the winning cell is its 3.7th choice on average and as low as its seventeenth.
The bound names the hottest part and the cost does not
Moving in the hottest component costs at most the largest temperature on the board, and that bound is attained: 100 lines of 4,240 pay exactly it. It is still the wrong quantity. Across four pools and boards of two, three and four parts the cost is nothing on 90.8% of lines and otherwise takes one of two values — half a point or one — on boards whose largest temperature runs to three, and it exceeds the coolest component on 13 lines and twice it on none.
Two things to hold at once, or three
Whether a condition makes a winner remember has been settled over every condition on three positions; how much it makes them remember has not. A tree built out of the condition alone, with no board in it anywhere, prices all 128: sixty-one cost nothing, fifty-eight cost two states and nine cost three. It also names the property that was nearly right — closure under union of the sets a condition rejects decides it exactly, where being writable as numbers is sufficient and reaches twenty-six.
The cheap fights make the rule cheaper
A conjecture stands that playing the hottest part costs at most the coolest temperature times the number of parts sharing it — proposed on a range where that number never exceeds two. Swept to five-part boards over 10,410 lines it is false, and false the other way round: every line costing more than the coolest part has one or two parts at that temperature, and over the 3,230 lines with three or more, not one does.
The auction never gets to the money
The critical fraction is computed and never played. Played out with a countable pool of chips — twelve positions, four pool sizes, every split of the chips, every bid answered — the verdict does not move with the money on a single one of the forty-eight sweeps, and the rule for equal bids settles all forty-eight. The reason is one line long: declining every auction wins, and bidding nothing declines.
Left always wins, and loses more often than not
Alternating play answers with one of four classes and the coin answers with a chance, and the two do not have to agree. Over the twenty-two values born by day two they never disagree and the margin is exactly nothing — the lowest chance on a position Left wins whoever moves is a half. Over the 1,474 born by day three, seven of them sit at seven sixteenths, and seven mirror them on the other side.
The best chance is the wrong move
Maximising a probability and denying an opponent a reply are different objectives, and on 189 of the 904 day-three positions where Left has a choice and a winning move, the option the coin prefers is one that loses the alternating game outright. The smallest case is two options and one line of arithmetic: five eighths beats a half, and a half is the move that wins.
A thousand positions and no exception
The parity law was fitted to constructed bags of chains and loops inside a string budget. A board's positions are a different population — the sizes are what the geometry allows, the components come correlated, and a six-box board holds exactly one position that is a loop of six. Tested on all 1,032 of them and all 160 of the four-box board's, the law is right every time, against a verdict computed from the strings by a walk that has never heard of a component.
A coin with three strings is worth something
Every chain and every loop is worth nought on its own, whatever its size, and that is exactly what makes their nim-sum useless. A coin with three strings on it is worth nought, one, two or three depending on its arms — 31 of the 35 measured are not nought, and the four that are are the ones whose arms are all long. A coin with four strings is back to nought every time.
The picture Bouton's proof leaves behind
His argument is two closure properties of one set, and the Sprague–Grundy theorem is the same two sentences with nought replaced by a variable — checked here on five games and every value in range, with no move staying inside a class and no class failing to be reachable from above. What the argument also leaves behind is a picture in which the values descend, and that is false: 99 of 444 moves here raise a value, and none of them is in Nim.
The names are not built out of the old ones
The guess was that a three-position region's missing names would be sums of two loopy ones — on plus over, and that family. Built and tried, every pair of the six stoppers covers none of the 4,931 regions that need one, and so does every two-position stopper there is, all seventy-nine of them with small games added. Thirteen names have to be invented, and forty-eight cover the whole census against ten at two positions.
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.
Three heaps and a pass
Nim with a single pass that may not end the game is easy on one heap and on two: a heap swaps each odd size with the even one above it, and two heaps lose exactly at (2k − 1, 2k). On three heaps the losses are known only as a list. Fix the smallest heap and each slice of the list settles into a pattern after an irregular start — period 4, 8, 10, then 160 at a smallest heap of ten, and nothing visible from eleven.
What a component would have to carry
For a held pass to be decided by a summary of each component, the summary must separate every pair of components some company tells apart. The Grundy value does not — Nim 1 and Kayles 8 are equal games that a held pass separates beside a single Nim heap of two. Nor does the Grundy value with the component's own held-pass value: Kayles 3 and Kayles 6 agree on both and are split by a company of two Nim heaps. Over twenty-four components, fifteen classes against fourteen pairs, and the gap widens as the pool grows.
The follower does the reversing
The gift-horse theorem for the ordinal sum needs two cases, and the second — the added option is reversible — was counted and not described. Recorded move by move, the reversing answer is always Right's move inside the follower: on all 410 escapes under five followers, and on every one of the 2,628 gift horses under every follower that gives Right a move at all. The case split is by follower, not by horse.
The split slips one day deeper
The reversal case of the gift-horse theorem was described in one line — the follower's own move reverses every gift horse, whenever the follower has one — and tested only where it was found. In the mirror it holds exactly, with 1 and −1 trading places. One day deeper it fails: under ↑ and ½, three gift horses on built day-four bases are not reversed by the follower's move. All three are dominated, so the theorem stands; the clean split by follower does not.
Cancelling is not pairing
The coin rows that cancel against their own negatives looked like the rows whose coins pair off as nested equal pairs, and on rows of four they are exactly those. From six coins the description fails in both directions — twenty rows pair off perfectly and do not cancel, and one coin set has a hundred and thirty-six that cancel with no pairing at all — and a row of five coins cancels, though an odd row can never pair off. What does hold, on every row swept, is that the first player in a row plus its negative never finishes behind.
A cancelling pair is a zero
Two cancelling coin rows side by side cancel against their two negatives — all 190 pairs from rows of two, four and five coins, the odd row that cancels without pairing off included. And a cancelling row beside its negative is invisible next to any other row: in 741 tests against every row of one to three coins, neither score of the context moves. A non-cancelling pair moves a score in 452 of 780. Scoring games have no inverses in general; this class has them, and they behave as inverses must.
Even rows always reward the move
Milnor's mean-value theory needs an incentive to move — the player to move must do at least as well as if the opponent moved first. On a coin row with an even number of coins that is not a hypothesis but a theorem: the first player can collect one whole parity class of coins, and one of the two classes holds at least half the total. So the condition excludes no even row whatever the coins, the class the earlier table called 'incentive at the top' was every row of four, and the hereditary condition is a condition on odd intervals alone.
Two heaps of testing are enough
A misère quotient is computed by testing positions against positions, and the universe used to find twelve classes of Dawson's chess was every position of up to four heaps tested against every other — 511,225 outcomes. Varied one size at a time, the count stops growing at tests of two heaps and positions of three: 12,100 outcomes find the same twelve classes. The narrower universe the earlier essay drew did not merge anything; it held fewer positions. And the corner that is enough moves: for Kayles at heap twelve, two-heap tests miss a class.
The cheapest fight is not the yardstick
Hottest-first play was found to cost at most twice the coolest temperature on the board, on pools whose coolest parts sat at a quarter or a half. Put a fight worth an eighth on the board and the rule costs seven times it. On one board the cost falls as the cool part warms, which is the opposite of a bound in the coolest. The number that holds on all 14,946 lines of seven pools is the second-largest temperature.
The same parts, different wholes
Equality is worth having because equal parts make equal wholes: a game may be replaced by an equal one inside any sum and any larger position. The relation that 184 contexts define — no context tells the two apart — is exact on day three and merges a third of a sample of day four. Put 567 of its merged pairs through negation, addition and forming options, and it survives two of the three. Adding the same game to both separates them 1,527 times.
Three pairings and no order
The potential lost four-by-five Hex for Down, and fifty-five re-weightings lost it too, which left the question whether anything that orders the cells can hold the board. The plainest ordering — one fixed list, whatever the position — holds none of the boards the pairing is known for, and on four by five the best it can do is the left column. At the other end, every table of replies that wins a small board is a pairing, and four by five has exactly three pairings, each using all twenty cells.
A rare class exclusive-or respects
The rare values of ·007 were measured by rank — everything outside the twenty-two commonest — and found getting commoner. A sparse-space argument needs a different split: a rare class closed under nothing and a common class closed under exclusive-or, which is to say a parity of bits. Over every parity of every bit, ·007 and ·6 have no class that thins. Across the 79 growing octal codes of up to three digits, one does: ·354, whose odd values fall from 288 in its first two thousand heaps to seven in its last ten thousand.
What survives a coalition
Three-player Nim has no winner without a convention and no sum theory at all. Fix one player against the other two and both come back: every position has a winner, the winner is read off the number of heaps of one modulo three and the number of larger heaps, and forty-five positions fall into six interchangeable classes — fewer than Nim's sixteen. Fix a scoring rule instead and the winner comes back, identical to one of the conventions, and nothing else does: no two positions are interchangeable.
One value more than Nim
Top Entails broke the nim-sum: nine of thirty-six two-heap positions came out wrong when a move could compel the reply. The repair is one extra value. Read each heap off the heap itself, and every heap of Top Entails and of a second entailing game is either a nimber or loony — a win for the mover in any company. One rule then decides every sum of those games and Nim together: 4,016 positions, and it is never wrong. A loony heap is a pass, and the compulsion is what buys it.
What a part would have to report
Outcome classes leave four entries of the addition table open. Give each part its two stops as well and four inequalities settle 176 of the 369 open day-two sums, never wrongly — and every sum they leave has a bound sitting exactly on nought. Used as a lookup table instead, even the whole thermograph with the outcome beside it leaves 96 sums undecided. Four outcomes do what no thermograph can: who wins a day-two position alone, beside −1, beside 1 and beside a star names it, and no three questions do.
Composition at every size
Twenty-seven of the fifty new four-position sides were written by no sum of two earlier games. Sums of three write sixteen of them — seven plain finite games born after day two, nine built on the names three positions forced — and leave eleven, met seventeen times in nine thousand regions. Run the same test one size down and the story the counts told falls apart: eight of the thirteen names three positions were thought to invent are sums of up to three older parts, two of them just two finite values. Every size composes. What each invents is a residue — five, then eleven.
The seven were the order
A Sylver Coinage proof told its depth beat the search that finishes on seven positions of 1,766, and on all seven the winning move came fifth when moves were tried smallest first. Run both searches under five orders and the seven vanish under every other one; try the largest move first and there are fifty-five new ones instead. Give both searches the winning move first, free, and the proof is cheaper on none. The exceptions were never positions where proving is easy. They were positions where one order made the finishing search stumble — and what is left when the order is perfect is the price of asking a proof's question.
The plays do not say who chose them
Every way a game of edge geography can go is a maximal trail, and the parity of each trail says who moved last. So the list of their lengths ought to decide the game, and over every connected graph to six vertices it nearly does — 780 of 810 starts, 108 of 114 on one cycle, where no rule on the matching and the cycle reached 91. It still fails, and the smallest failure is two trees: the same two plays from each start, and a different player choosing between them. Keep that choice one move at a time and the ambiguity falls from eighteen profiles to seven, one and none. The list costs sixteen times the search.
The price of a tie
The second-largest temperature bounded hottest-first play on 14,946 lines, and the pool that should have broken it puts a threat on top. It does not break: over 2,254 more lines the cost never passes the second temperature and meets it on 97. What the pool finds instead is where the rule goes wrong — one move into the game about as often as at its start — and why. Almost every line that meets the bound is a tie: two parts equally hot and the rule taking the first on the board. Break ties well and the lines meeting it fall from 636 to 51 on the earlier pools, and to none on any board with the threat on top.
Two rows still reward the move
A single coin row with an even number of coins always rewards the player to move, because the first player can take a whole parity class. Two such rows side by side break that argument, and the conclusion survives anyway: on every one of 16,380 sums of two rows the first player is never behind, and there is a proof — open the row worth more, then answer in whichever row the opponent moves in — that guarantees the difference of the two rows' margins, which is what two switches would give. The sum meets that floor on three pairs in four and beats it on the rest, once by both margins together. At three rows the switch arithmetic stops being a floor. The incentive does not.
The odd values keep coming
·354 is the one growing octal code of up to three digits whose values have a class that thins: 288 odd values in the first two thousand heaps, seven between ten and twenty thousand, the last of those at 18,908. A sparse-space argument needs that class to stop. Taken to a hundred thousand heaps it does not: 48 more odd values arrive after twenty thousand, every block of ten thousand holds some, and the last in range is at 99,600. They keep no schedule a remainder can see, their gaps are spread like arrivals, and most of them are one number, 37 — each one a heap where every split that could have reached 37 misses it.
The pairs stand still and the order spreads
The share of pairs of game values the order can compare falls from 77.5 per cent on day two to 60 on day three and then stops: built twenty times under two recipes and calibrated, day four sits within two standard errors of day three. That looked like a floor under the order. Measured beyond pairs it is not. Chains of three keep falling, triples with no comparable pair keep rising, and the largest set of mutually confused values in a sample of 120 grows from about 20 to about 30 — every one of forty seeds wider than day three's mean. The floor belongs to one statistic, not to the order it summarises.
A current through the board
In the early 1950s Shannon and Moore built a machine that played Hex by wiring the board as a network of resistors and reading its move off the electrical potential. There is no theorem behind that rule, and on every Hex board small enough to solve it can be scored against perfect play. On boards to twelve cells it takes a winning cell on more than 98 per cent of the positions where one exists, far above chance. And it loses: every board of three rows or more it has to defend, and four by four moving first, because an opponent needs only one of the two per cent and can steer the game to it.
A circuit asked once
The Hex machine's circuit rule loses every board past two rows that it has to defend, and the strategy that holds those boards is a pairing — a partner for every cell, fixed before the game begins. So make the circuit commit: consult it once per cell on an empty board and never again. Its table answers twelve openings with four cells, and loses every board the running circuit lost. Forced to choose a whole pairing, it picks a losing one under seven of eight ways of reading its scores; the eighth finds the winning pairing of three by four by a margin of rescaling and gives up two by four for it.
A repetition is a guess
Read an octal game's values one heap at a time and report the first pattern that fits, and the report is a guess that changes. Over the 291 short octal codes whose period can be proved, the guess is wrong 4,684 times before it settles; on ·45 it changes 181 times, is right from heap 538 and cannot be proved until heap 1,038. No false guess ever reaches the window that would certify it — the furthest gets 74 per cent of the way — and no fixed number of repetitions is a safe substitute, because the window asks for as many as the period's late start requires: two on some codes, twenty-seven on ·45.
A universe grows by days, not by sums
A relation built from a sample of contexts can be repaired two ways: allow sums of more values, or allow values born a day later. On a sample of day four the first stalls — 147 pairs of different games stay merged from five summands to seven, through 25,696 contexts — while the 985 values of day three, a seventh as many, separate every pair. The reason is four lines long: the context that tells two games apart is always minus one of their options.
A pairing is checked, not measured
The Hex machine's circuit chose pairings near the winning ones and never at them, which left a finite question: does any score of a pair of cells, computed from the board alone, have a winning pairing as its best? The obvious scores count the chains a pair shares. All four hold two by three; on three by four the plain count cannot choose between a winning pairing and a losing one, and the corrected counts choose the loser; on four by five every one loses. The reason is arithmetic — a sum will give up one chain to cover two others twice. Told which chains it missed, the count finds the diagonal pairing at one rate in three, and by then it is checking, not measuring.
Two neighbours keep a pairing, three lose it
In the switching game on points, Short wins moving second by a pairing when every smallest set of points that cuts A from B holds both points of some pair. On every graph of seven points — 971,104 of them — every Short win has one, at every degree. In random link games, where each contested point has two neighbours, every win has one too, although the winning strategy follows trees that change as the game goes. The first wins with no pairing appear at nine points with four neighbours and at ten points with three: a cubic graph Short wins whoever moves, which no fixed set of answers explains.
A foot is a menu, not a number
When a sum's fight ends level on nought, the winner is decided by what is left at the foot of each part's fight, and the obvious number to report from there is the atomic weight. Over 32,428 sums of a day-two and a day-three value it leaves 14,924 undecided, and on day two it does worse than a single outcome letter. What sits at a foot is a choice, made by one player, with someone to move when it is made. Report all three and every sum of two cold parts is decided; 3,164 sums are still not, and each holds a hot part whose own fight is not the one played in company.
The weights were not the argument
A chain count corrected by its own misses held the four-by-five Hex board with the diagonal pairing, the one pairing there with a sentence behind it, and the hope was that its weights were the sentence in numbers. They are not. One rate in twelve holds the board and one start in twenty; the weights it ends on survive five per cent of noise and not twenty; and weighting exactly the chains the argument relies on finds no winning pairing at all. Told which pairing to prefer, a measurement reaches any of the three in a dozen corrections. What chain weights can do is rule a pairing out, and on three by four they rule out one that wins.
A pairing that changes once
The cubic ten-point graph is a win for Short that no fixed pairing explains, and the full strategy that does explain it is a tree of 68 decisions. One change is enough: after Cut's first deletion Short claims a stated point, and from then on a fixed pairing of what is left holds every cutting set. Four such plans, eight pairs between them, answer every opening. Of the 23 pairless wins in the samples, 21 need one change and two need a second, and in both the second is a forced claim. Unless the hard problems collapse, no fixed number of changes can be enough for every graph.
The teacher's game
A machine that weighs Across's chains could be taught every winning Hex pairing on four by five, and on three by four it could never be taught the one pairing another winner covers chain by chain. Whether anything else could hide from it — a winner covered not by one rival but by a mixture of several — is a question about a game between a teacher choosing weights and a rival choosing pairings, and Jean Ville's 1938 theorem of the alternative, written for von Neumann's games, answers it in general. On five by six and six by seven every winner is taught in sixteen corrections or fewer, and nothing hides at all.
The rows that cancel
Every one of Dawson's diagrams that failed to be the sum of its rows had an even number of rows, and twenty-one with an odd number all added. One file past the edge of that sweep, four files beside three beside one fails: three rows, worth ∗ against ∗3 by its rows. What the parity was tracking is two deletions a sum allows — a row worth nought, and two identical rows — which the compulsory board turns out to allow as well. Each of 44 diagrams is worth exactly what is left after them, and in the fifty-diagram sweep that remainder happened to have as many rows, odd or even, as the diagram it came from.
Two losers that average to a winner
On Hex boards to 6 × 7, a machine that weighs the opponent's winning sets could be taught every winning pairing except one, which another single winner covered. Ville's alternative allowed a second way to be out of reach — covered only by a mixture — and the switching game on points, where winners are many, is where it turns up. None in all graphs of seven points or in 4,000 of eight; 35 covers at nine and ten points. Every one is the even average of two losing pairings, and on six graphs Short wins and no weighting can teach the machine any winning pairing at all.
A compulsion that points elsewhere
Top Entails and entailing Nim are decided by one value more than Nim because the compelled reply stays in the heap that compelled it. Point it at a different heap instead and the two games part company. Entailing Nim does not notice — its heaps are worth exactly what Nim's are, because Nim's winning reply was never in the heap just moved. Top Entails loses its loony heaps and then every value of the Nim kind: no assignment of nimbers explains its sums, because a heap of three beside two heaps of one is a win and beside two heaps of two is a loss.
The list that cannot run out
Give each cell of an n × n square its own list of n colours, and ask for a colouring from the lists with no repeat in any row or column. Dinitz asked in 1979 whether that is always possible, and reading the cells in order fails about four times in ten. Galvin's proof of 1994 needs one page. Use an ordinary Latin square to turn each row and column into a preference, so that every cell has exactly n − 1 moves, and peel off a stable set for one colour at a time. A list of n cannot run out against n − 1 moves.
Blocking is not enough
Everybody knows noughts and crosses is a draw, and the usual explanation is that the second player can always block. That explanation is false. Against an opponent who can only block, the first player wins, from the centre or any corner. The game is a draw only because the second player's blocks build threats of their own. On fifteen boards solved outright, it is the only one where those two things come apart: everywhere else a draw is a draw because blocking is enough.
A win priced in stones
Change the rate of play and a yes-or-no question becomes a number. Against a Maker placing one stone a turn, Breaker needs two stones to stop three in a row on every board up to 4 × 4 and six by three, and three on 5 × 4. Against a Breaker placing one, a Maker needs two stones on most boards drawn at one, and three to make four in a row on 4 × 3. The extra stone is enough to erase noughts and crosses altogether, and the biased sum is exact there and high everywhere else.
Named alongside it
The objects these essays reach for when they reach for this one.
EnumerationExhaustive searchInvariantHeuristicDisjunctive sumValueStrategyImpartialTemperatureCanonical formApproximationDecomposition