Sums and comparison

A position is the sum of its parts. Four separate Hackenbush sprigs. A move is a move in one of them, so the position is their disjunctive sum, and its value is the sum of their values. Which part to play in is the entire decision, and the values are what makes it decidable.

The sum is the object

Real positions come apart into independent regions, and a move happens in exactly one of them. That operation — the disjunctive sum — is what the whole theory is built to survive, and it is the reason values exist at all.

Comparing two positions is playing their difference. To decide whether one position is worth at least another, subtract and see who wins moving second. It is the only definition of comparison the subject has, and it produces a partial order — some pairs come out confused, which no comparison of numbers ever does.

Comparing positions

One position is worth at least another when the second player wins their difference. That is the only definition there is, it is a computation rather than a judgement, and it produces an order in which some pairs are simply not comparable.

Knowing who wins is not enough. Three pairs of positions, every one of which is in outcome class N on its own. Their sums are not all the same, and not all in the same outcome class — so the outcome of a sum cannot be worked out from the outcomes of its parts, and that is why the theory needs values.

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.

Which part to move in

The value of a sum is the sum of the values. The move in a sum is not the move in any part, and there is no rule that reads it off the values — in the smallest interesting example, the only winning move is in the component worth nothing.

How many ups

When every component of a position is smaller than every positive number, no number can decide it. What decides it is a count of ups — and comparison can pin that count down exactly, except when a star is present, when it cannot.

Every position has an exact opposite. A position beside its negative, which is the same game with the players exchanged, and the sum of the two. The sum is worth zero every time — a second-player win — because the second player can answer each move with its mirror image. It is the fact that makes values a group, and it is what lets one position be subtracted from another.

Turn the board through a right angle

A two-by-four Domineering board is worth something no number can express, and Right is ahead on it. Turn a second board through a right angle, put the two side by side, and the total is exactly zero. Every position has an exact opposite, and that single fact is what makes subtraction — and therefore comparison — possible at all.

Three ways to add the same games

A move in exactly one component is a choice, not a law. Move in every component at once and the game is different; move in any set of them and it is different again. The same two positions, added three ways, give three different answers — and only one of the three has values that add.

Comparing two positions means playing a third

There is no way to look at two games and see which is better. The question "is G at least H?" is answered by building G − H and asking who wins it — so the most basic operation in the theory is a decision problem, and every canonical form is built out of them.

Three partizan positions against every nimber, and not one match. Sprague and Grundy give every impartial position a single number that is complete: two positions with the same value are interchangeable everywhere. The three positions here are partizan — the two players have different moves — and each is compared against every nimber up to eight. Nothing is equal to anything. The magenta cells are worse than inequality: a position confused with a nimber is not above it or below it either, so no ordering could rescue the substitution.

Where the impartial theory stops

Sprague–Grundy gives every impartial position one number, and the number is complete. The moment the two players have different moves no number works at all — not a harder one to compute, none — and three positions here are compared against every nimber to show it.

A boundary drawn, and a boundary there. One Domineering board split two ways. Above, a line imagined down the middle: the two halves are evaluated separately and their sum is not the value of the board, because every horizontal domino that would have crossed the line has been thrown away. Below, the same column blocked out: the halves are then genuinely independent and the sum is exact. Every value is computed from its own board.

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.

Confused is not the same as unknown

Two positions can be neither greater, nor smaller, nor equal. That is a fourth relation with its own symbol, it is a fact about the pair rather than a limit of the method, and it is what makes a game worth playing — a position is a first-player win exactly when it is confused with zero.

The picture is the numeral. Blue-red Hackenbush strings and their values. Left may cut a blue edge, Right a red one, and everything above the cut falls. The value of each string is a number, and reading the string from the ground upward gives the binary expansion of exactly that number.

The other sum, the one that nests

A move in one part wipes the other out entirely. That is the ordinal sum, it is what a Hackenbush stalk actually is — 1 : (−1) is a half, and 1 : (−1) : 1 is three quarters — and it is not an operation on values at all: three positions all worth zero give three different answers under it.

The context that tells them apart. Two positions put into the same company, one context at a time. Each column is a game X; each cell is the outcome class of that side added to X. Equality means every column agrees, for every X there is — so a single disagreeing column is a disproof, and agreement across a bounded list of contexts is evidence rather than proof. The proof is that the difference is zero.

Equal in every company

Two games are equal when no third game can tell them apart — a quantifier over every position there is, discharged by one finite test. A search over 184 contexts separates all 5,790 unequal pairs it is handed and still calls two different games the same, which is exactly why G − H = 0 is a theorem and an exhaustive search is not.

The bracket of a sum, against the sum of the brackets. Two all-small positions, the interval of multiples of ↑ each lies between, those two intervals added coordinatewise, and the interval the sum actually lies between. The added one always contains the computed one — greater-than survives addition — so the bracket never widens under a sum. Where it narrows, the parts were each too vague to pin down and the sum is not.

When the ups add

Atomic weight brackets do not add over a sum — they bound it. Over all 120 pairs from a fifteen-game family the sum's bracket came out exactly the sum of the parts' brackets 56 times, strictly narrower 64 times, and wider never; and the rule separating the two is one line long, because every one of the 54 pairs with a pinned part is exact and only 2 of the other 66 are.

The values born by day three that are their own negatives. Every game satisfies G + (−G) = 0, so a game equal to its own negative satisfies G + G = 0 — it has order two. The nimbers do, and they are not the only ones: a switch symmetric about zero is unchanged by negation, and so is anything whose Left options are the negatives of its Right options. Each row carries the value, whether it is a nimber, and its outcome.

The values that are their own negatives

Every game satisfies G + (−G) = 0, so a game equal to its own negative satisfies G + G = 0 — it has order two in a group whose elements otherwise have infinite order. The nimbers do. So does ±1, on sight. Over the 1,474 values born by day three there are 30 of them and only four are nimbers, every one of the 900 sums of two is another, and the equality test and a symmetry of the written form agree 1,474 times out of 1,474.

Swapping a branch for another of the same value. The ordinal sum of a base with a branch, and the same sum with the branch replaced by a heap of a different game carrying the same Grundy value. The two are compared by playing their difference, not by inspection — and they agree every time, which is what the colon principle claims and what the partizan case denies.

When the nested sum only sees the value

The ordinal sum reads the form and not the value: three positions all worth zero, placed under a star, give three different answers. On impartial games it reads the value after all — 72 substitutions of an equal-valued heap from a different game, and every ordinal sum comes back unchanged. That difference is the whole reason a green Hackenbush tree can be collapsed one branch at a time.

Every day-three value, moved by every quarter. Four claims counted over 3,000 translations — 120 values, each moved by every quarter from −3 to 3: that adding a number leaves the temperature alone, that it shifts both stops by exactly itself, that the interval between the negated stops is where the two players are confused, and that the only failures of the last are on its endpoints.

What a number does to a fight

Adding a number to a position moves everything and changes nothing: over three thousand translations the temperature never once shifted and both stops moved by exactly the number added, every time. What the number decides is whether the fight is worth having — and the interval where the two players are confused is exactly the open interval between the negated stops, right in all 2,890 cases away from its endpoints and wrong in 110 that are all on them.

How often one value is above another. The partial order counted on two successive days. The proportion of pairs that can be compared at all falls sharply, and so does the proportion of values that can be compared with zero — which is the proportion of positions whose winner does not depend on who moves.

How rare it is to be bigger

Values are partially ordered, and 'partially' does most of the work. On day two, 179 of 231 pairs can be compared and 13 of the 22 values can be compared with zero. One day later the shares are 60% and 29%, and the largest set of mutually incomparable values found rises from four to at least twenty-three. Comparison is the exception; confusion is what values normally do to one another.

What the reduction collapses. Each reduced form with the values that reduce to it. The largest class is the one that reduces to zero and it holds every infinitesimal on the list, which is exactly what the reduction is for — against a hot background, none of them is distinguishable from nothing.

What is left when the small change is thrown away

Canonical form answers a demanding question: which positions are interchangeable inside every sum whatever. A player with a hot board does not have every sum — an infinitesimal difference cannot decide anything against a genuine fight — so there is a coarser question with an exact answer. The reduced canonical form takes the 1,474 values born by day three to 61, with 292 of them collapsing to zero, and it is a homomorphism on all 8,100 pairs tested only when a second pass is made.

Cancellation, by exhaustion. The law checked on every triple of values born by day two, and then put to work: two Domineering regions compared directly and compared again inside a larger board. The comparison never changes, which is the licence every decomposition on this site is drawn under.

What can be struck out

From G + X = H + X it follows that G = H, in one line, by adding −X to both sides. It is the shortest theorem here and the most used: it is what makes comparing two boards region by region legitimate. Over 10,648 triples the hypothesis fires 484 times and the conclusion holds 484 times — and the licence expires in three separate directions, each of which loses the same axiom in a different way.

Two numbers from the same tree. Four subtraction games, each with its Grundy sequence and its remoteness sequence. The Grundy value decides a disjunctive sum and the remoteness decides a conjunctive one; the only thing they always agree about is which heaps are losses for the player to move.

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.

Seven infinitesimals, added to a whole day. Each row adds one infinitesimal to every value born by day three. The stops never move — that is what being smaller than every number means — the temperature moves a handful of times, and the outcome class moves in a quarter of the additions.

What an infinitesimal does to a fight

Adding a number moves both stops by exactly itself. Adding something smaller than every number moves neither — across 10,318 additions to a whole day of values, not once — and the outcome class changes anyway, 2,622 times. It changes at exactly one kind of position: the ones with a stop sitting on zero, which is where the numbers have run out of things to say.

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.

How hot a background has to be

The reduced canonical form throws away infinitesimals, and the rung below asked for a bound: how hot must the rest of the board be for the discarded part not to matter? There is no such bound. Safety is not monotone in the background's temperature — an eighth is safe, a quarter is not, two is safe again — and the quantity that does decide it is not a temperature but a stop.

Four rules, asked of compounds made of two different games. Compounds whose two components come from different subtraction games, solved in full and compared with what each rule predicts. The three rules the compound theory supplies are exact on every position; the shortcut a reader carries instead is not.

A compound of two different games

Every rule the compound theory has survives mixing exactly — the minimum-remoteness rule is right on all 5,184 mixed pairs and all 7,560 triples — and the reason is not that the rules are strong. It is that each of them reads one number per component, and a number does not remember which ruleset produced it. The thing mixing damages is the shortcut a reader carries instead.

Where the thirty sit on the scale. How many of the values equal to their own negatives carry each temperature. Fifteen sit at nought, fourteen are hot, and one is a number — so the subgroup runs the whole length of the scale rather than living at the cold end of it.

The thirty that cancel themselves

Thirty values born by day three are equal to their own negatives, and every one of them has a mean of exactly nought and two stops that are exact opposites. Neither property comes close to picking them out — 496 values of the day have a mean of nought — and half of the thirty are hot, one of them the hottest value the day produces.

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.

Four candidate bounds, and the one that holds. Each candidate bound tested against every failing cut. One domino and the height of the cut both fail on six; twice the height holds on all twenty-two; the whole board's temperature fails on eighteen.

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.

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 stops stop working. Adding a switch to each of 120 day-three values, and asking whether the two stops of the sum are the sums of the two stops. They are on 330 of 720.

What a fight does to a fight

A number added to a position shifts both its stops by itself; an infinitesimal moves neither. A hot game does neither: over 720 sums the two stops add on 330 and are wrong on the rest. What survives is the mean, which adds on every one of the 720 — and the failure has a bound, since no stop is ever out by more than twice the smaller of the two temperatures, a bound 222 of the sums attain exactly.

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 bound that needed no third number. The rung below's bound and the conjectured replacement, scored over 1,440 sums whose addend has a follow-up. The two-number bound holds everywhere and is attained; the three-number one fails.

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.

Self-negative values, day by day. How many values each day of the construction supplies and how many of them are their own negatives. Day four cannot be counted; a corner of it supplies 571.

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.

Add, then reduce again

The homomorphism promises that a sum's reduced form can be computed from its parts', and says nothing about what the operation is. It is addition followed by a second reduction — needed on 431 of 3,600 pairs of day-three values, and on not one of the 1,751 pairs with a cold part. What the second pass removes is an option that only becomes dominated once the two fights are side by side.

Which end, in four lines. The complete rule for where a sum's error lands, exact on every pair in the census. Three of its four cases are decided by the value being translated alone.

Which end a sum lands at

The rung below found the errors in a translated stop clustered at the two ends of the range its bound allows — 660 at nought and 358 exactly on the bound — and asked for a rule saying which end a given pair lands at. There is one, in four lines, exact on all 1,440 sums. Three of the four cases are decided by the value being translated alone, and the property that decides them is the bend the switches ladder found for a different question entirely.

Eight ways to name it, and none of them works. Candidate rules for which option the second reduction deletes, scored on every pair where it deletes exactly one. The best reaches four in five and none is exact.

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.

One quantity, two currencies. Each bent value with how far its temperature falls short of its stop reading and how far its stop falls short of the translation bound. The second is exactly twice the first.

The same number in two currencies

The rung below found bent-walled values falling strictly inside the translation bound and asked how far. The shortfall is the value's own hottest follow-up's temperature — exactly, on 400 of 408 pairs, and twice it on the other eight — which makes the whole error one expression. And it is the switches ladder's constant: a half there and a whole here, because a temperature is half a stop gap.

A mex with no impartial game in it

The rung below described the zero fibre of the mirror map and left star's fourteen undescribed. Star's fibre is 'some element is at least nought, and none is at least star' — and the two rules are one rule: the mirror value is the least nimber no element of the set reaches. That is a mex, in a construction built entirely from partizan values.

Not a domination, in the order the rung below meant. The second pass's deletions scored as dominations in two orders: the partial order on games, and the order on stops.

Not a domination, in that order

The rung below asked which pair the second reduction acts on, taking for granted that the operation is a domination. It is not: on none of the 525 deletions is a surviving option greater than or equal to the deleted one. In the order the reduced form actually works in — both stops at least as good — every deletion with a survivor is a domination, the dominator is unique on all but twelve, and it always comes from the other part.

The census, closed. The three classes of the translation census with the expression each obeys, scored over all 1,440 pairs.

Where the value stops mattering

Fourteen straight-walled pairs missed the bound on a translated stop and had only a threshold to explain them. Their stops move by exactly the addend's temperature — a formula with the value nowhere in it — which turns the threshold into the boundary between two lines and closes a census of 1,440 pairs that has been open for four rungs.

Four moves, three arguments. Every first move in the sum, with what answers it and how many cases of each the census holds.

The case that was supposed to be hard

The mex rule for the mirror construction was to be proved by induction, and the step flagged as needing care was the one where an option is incomparable with the nimber. There is no induction: the argument is four lines, and incomparability is what makes two thirds of the cases go through — because a fuzzy sum is a first-player win and the first player is the opponent.

The test, scored. The recognition test run on every deletion the second reduction makes, against what actually happens.

A side about to lose its move

A fifth of the second reduction's work removes the last option a player had on a side, and no rule on the ladder had looked at one — because a deletion with no survivor has no pair in it. The recognisable object is not which option goes but whether the side is one an option can go from, and two comparisons on the parts decide it on all 525.

The bound survives and the rule does not. The census against the widened sweep, with the bound and the two-line rule scored separately on each.

One expression proved, and one withdrawn

The census closed with three expressions exact on 1,440 pairs, and the rung above asked for derivations. The cold one has a four-line proof. The straight one has a threshold the census cannot determine — any constant between 4/3 and 3/2 fits it — and eight more addends of the same family break it on 38 pairs while leaving the bound above it untouched.

A cross in the table

Which pairs need a second reduction had never been asked. Sorted by the two temperatures the answer is a cross — the whole gap-of-a-quarter column and the whole colder-is-three-quarters row — and it is exactly necessary on all 431 with no exception, and wrong 477 times in the other direction.

The proposed case, scored. The gift-horse theorem and the domination argument proposed for it, each scored over every gift horse added.

The proof needs both reductions

The gift-horse theorem was to be proved by showing the added option dominated. It is, on 97.8 per cent — and the other 232 are reversible instead, with nothing left over. The case the proposal missed is almost entirely one follower: none under a positive number, 190 under a negative one.

The fall stops. Comparability on days two, three and four, the last built and corrected.

A floor, and not a decline

Comparability fell eighteen points from day two to day three and the next day cannot be enumerated. It can be built — and the construction's bias measured one day lower, where the truth is known. Corrected, day four comes to 60.6 per cent against day three's 59.7: the fall was a one-day event.

One number, two statements. The smallest true bound on the cost of splitting a board, in both of the currencies it can be stated in.

One number, stated two ways

Twice the height of the cut held and was loose; the height alone failed. The smallest true constant is three halves — exact and attained as a bound on how far the value can fall, and an infimum attained nowhere as a bound on the value. The gap between the two is one move.

Twenty-six other values

The mex rule accounts for sixty-six of the ninety-six antichains and is silent on the other thirty. Every one of those thirty is worth a self-negative value born by day three — and the same mex, run over that family instead of over the nimbers, is exact on all ninety-six. The nimber rule is this one cut short after its fourth member.

The reversing move is the follower's. For each follower under which some gift horse escapes domination, the number of escapes, how many are reversed by Right's move inside the follower, and how many by a Right move in the base part. The follower's move reverses every one.

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.

What survived, and what did not. The gift-horse theorem, the two-case proof and the one-line description of the reversal case, each scored on the day-three sweep and its mirror and on the day-four sweep and its mirror.

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.

The day-four figure, twenty draws at a time. Corrected day-four comparability from twenty seeds of each of two constructions, as dots on a percentage axis, with the range of day three's four slices shaded and the single figure the earlier essay reported marked. The seeds spread over about twelve points and the two constructions agree.

Twenty draws and a second recipe

Day four's comparability was reported as 60.6 per cent against day three's 59.7, from one built sample and one calibration. Built twenty times with each of two recipes whose biases differ by six points, and calibrated against all 1,474 day-three values rather than a quarter of them, the corrected figure spreads over twelve points from seed to seed and the two recipes agree within one standard error. The floor survives; the decimal was one draw.