Where the order and the sum disagree

The simplest game above both found something the subject had no obligation to provide. The twenty-two values born by day two are ordered by a relation with a fourth answer in it — two values can be neither above nor below one another — and an order like that has no business supplying least upper bounds. It supplies them anyway. Every pair of day-two values has a join and a meet inside day two, all 253 of them, and the essay closed by naming the question that leaves standing:

what the joins and meets do to sums, where the lattice and the group meet.

They do nothing at all. That is the finding, and it is sharper than a failure usually is.

The identity, and why it is the one to ask #

Day two carries two structures and they were built by different arguments.

The order comes from comparison: G ≥ H when Left wins G − H moving second, which is a search rather than a look. The addition comes from the disjunctive sum, and it makes the values into a group — every value has a negative, and the negative is the position with the colours swapped.

A structure with both is not automatically a structure where the two agree. The agreement has a name and a shortest statement, and the statement is an identity:

$$(a \vee b) + (a \wedge b) = a + b.$$

In words: the larger-of-the-two and the smaller-of-the-two, added, come to the same as the two themselves. Nothing is created by taking the join and nothing is lost by taking the meet; between them they account for exactly the pair they were taken of. An ordered group in which that holds for every pair is a lattice-ordered group, and the integers are one, and the rationals are one, and so is every ordered vector space anybody has met.

The identity is worth asking rather than some other because it is the weakest thing that could reasonably be called agreement. It does not ask the join to be computable, or the order to be total, or the addition to be anything but a group law. It asks only that the two operations, applied to a pair, come back to where they started.

The answer, in one line #

Two hundred and fifty-three pairs of day-two values. The identity holds on 201 of them and fails on 52.

Two hundred and one is the number of pairs in which one value is above the other. Fifty-two is the number in which neither is.

So the identity holds exactly where the two values can be compared, and fails exactly where they cannot. There is no pair that is comparable and breaks it, and no pair that is incomparable and satisfies it anyway. Both of those are asserted in the code rather than reported: a run that found either would stop the build, because the sentence above would then be false and the figure would be repeating it.

The comparable half is forced, and saying so is the difference between a finding and an arithmetic identity dressed up.

If a ≥ b, then a is an upper bound for the pair and nothing above both is below it, so the join is a; and b is a lower bound and nothing below both is above it, so the meet is b. The identity then reads a + b = a + b. It cannot fail, and it does not fail, and the two hundred and one is a control rather than a result — a count that says the machinery computing joins and meets is computing the right thing.

The measurement is the other fifty-two.

What a failure looks like #

Take zero and star. Neither is above the other: star is confused with zero, which is what it means to be a first-player win. Inside day two the simplest value above both is a half, and the simplest below both is minus a half.

Their sum is nought. Their join plus their meet is also nought — and yet the identity fails on this pair, because 0 + ∗ is not nought. It is star. The discrepancy is exactly ∗.

That is as small as a discrepancy gets and it is not nothing. Star is the value that decides a game rather than scoring one: a position worth star is a first-player win, and a position worth nought is a second-player win, and swapping them swaps the winner. So the identity is not off by a rounding error on this pair. It is off by the whole answer.

At the other end the errors are large. ∗ against {1 | −1} has join 1 | 0 and meet 0 | −1, whose sum is nought, against a pair sum of {1∗ | −1∗} — a fight over two points that the join and the meet have quietly settled at zero.

No two of the fifty-two differ by the same amount. Fifty-two failures and fifty-two distinct discrepancies, which is worth a sentence because the obvious guess about a failure this systematic is that it has one cause with one size.

Thirteen, thirteen and twenty-six #

The failures split by direction, and the split is exact:

• on thirteen pairs the join plus the meet is strictly above the pair’s own sum;
• on thirteen it is strictly below;
• on twenty-six it is confused with it — neither above nor below.

Thirteen and thirteen is not a coincidence and the reason is the day’s own symmetry. Negation reverses the order, so it carries joins to meets and meets to joins; and it is an automorphism of the addition. Apply it to a pair on which the join-and-meet sum comes out high and the image is a pair on which it comes out low. The two counts are therefore forced to be equal, and the twenty-six in the middle are the pairs that negation sends to themselves or to each other within the confused class.

The twenty-six are the more interesting half. A discrepancy that is confused with zero is one where the join-and-meet reading and the honest sum are not merely different numbers but incomparable positions — so there is no sense in which the lattice reading is an over-estimate or an under-estimate. It is a different answer.

What the identity would have bought #

It is worth being concrete about what is lost, because “day two is not a lattice-ordered group” is a sentence with no consequences until the consequences are named.

A distributive law over sums. In an ordered group with the identity, (a ∨ b) + c = (a + c) ∨ (b + c) — the join commutes with adding a fixed component. That much survives here, and survives trivially: adding c to everything is an order isomorphism of the values onto their translates, so it carries the join of a pair to the join of the translated pair. It is the other direction, the one that mixes two joins, that the identity is the hinge of.

A decomposition into a positive and a negative part. In a lattice-ordered group every element splits as a = a⁺ − a⁻ with a⁺ = a ∨ 0 and a⁻ = (−a) ∨ 0, and the two parts are disjoint. That construction is the reason such groups are tractable, and the identity is what makes the two parts add back up. Here they do not: ∗ ∨ 0 is a half and ∗ ∨ 0 again — star is its own negative — so the recipe returns 1/2 − 1/2 = 0 for a value that is not nought.

An absolute value. |a| = a ∨ (−a) is a well-defined thing to write and it is not a well-behaved thing to have. For star it is a half, which is a number, which is exactly what star is not.

So the failure is not a technicality about one identity. It removes the whole apparatus that makes ordered groups usable, and it removes it on the pairs that carry the subject’s characteristic phenomenon.

Where it comes from #

The identity fails precisely on the confused pairs, so the question of why is the question of what confusion does to a sum.

A confused pair is one where the difference a − b is a first-player win. Its join is a value above both and its meet a value below both, and neither of those has to be near either. 0 ∨ ∗ = 1/2 is a half a unit above both of two values that are within an infinitesimal of each other, because inside day two there is nothing closer above both. The join has been forced to overshoot, and the meet has been forced to undershoot by the same amount, and the two errors cancel in the sum — which is why the discrepancy comes out at ∗ rather than at a half.

That cancellation is the reason the failures are as small as they are. It is also the reason they exist: the two errors cancel exactly, and what is left over is the part of the pair that no pair of numbers could have carried.

It also explains why the effect is not going to go away in a larger universe. The join is a fact about the day it is taken in, and a later day supplies more candidates, so both the join and the meet move inward. What does not move is the pair. The identity asks the two moved values to add back to the unmoved pair, and there is no reason the inward movements should cancel to zero rather than to something — and on day two they cancel to fifty-two different somethings.

The count that is not a proportion #

Fifty-two of 253 is a fifth, which sounds like a minority result. It is not, and the reason is that the fraction is about day two rather than about values.

How rare it is to be bigger counted comparability one day further out and found it collapsing: on day two four pairs in five can be compared, and the share falls sharply on day three. The identity holds exactly on the comparable pairs, so the identity’s success rate falls with it. On a day where almost no pair is comparable, almost no pair satisfies it.

So the honest statement is not that the identity mostly holds. It is that the identity holds on the pairs that behave like numbers and fails on the pairs that make the subject worth having, and the second class is the one that grows.

What a lattice-ordered group would have been worth #

The identity that fails is worth pricing, because the reason it is the one to ask is that a great deal follows from it and none of it is available here.

In a lattice-ordered group the join distributes over addition, and that single fact carries a small library with it. Every element decomposes into a positive and a negative part with $G = G^+ - G^-$ and $G^+ \\wedge G^- = 0$. Absolute values exist and satisfy a triangle inequality. The order determines the group up to the obvious things, so an order-preserving map is automatically a homomorphism. None of those are exotic: they are the reason lattice-ordered groups are a subject and why anybody would want the game values to form one.

The failure removes all of it at once, and it removes it structurally rather than approximately. There is no nearly here: an identity either holds or it has counterexamples, and the counterexamples come with a companion result — a lattice-ordered group has no elements confused with nought, since $G \\vee 0$ and $G \\wedge 0$ would have to reconstruct $G$. Games confused with nought are the entire subject.

So the two facts on this page are one fact. The order is a lattice because every pair has a simplest game above it, which is a statement about how much structure the construction supplies; and the lattice does not respect the group because the group contains star, which is a statement about what the games are. A theory with a fourth relation cannot have a distributive join, and the fifty-two incomparable pairs are where the two halves collide.

Which reframes what the failure is evidence about. It is not that the game values are a slightly defective lattice-ordered group. It is that the confusion relation, which is the subject’s whole reason for existing, is incompatible with the structure a reader coming from ordered algebra would expect — and the incompatibility was decided the moment the fourth relation was admitted.

What the census does not say #

Three limits, and the first is the one a reader is most likely to overrun.

The joins and meets are taken inside day two. That is a choice and it is the choice the rung below made, for the reason it gave: a least upper bound is least among the candidates available, and the candidates are what the day supplies. Taking them inside day three gives different joins on some pairs, and whether the identity does better there is a separate computation this page has not made. What can be said is that it cannot do better on the pair 0, ∗, since a join and a meet that add to the pair’s own sum would have to add to star, and the join is above both while the meet is below both.

Nothing here says the order is bad. It is a lattice, which is a great deal more than a partial order has to be, and the structure it does carry is what domination, the gift horse principle and the simplicity rule are all arguments about. The finding is that the lattice and the group are two structures on one set rather than one structure, and a reader who has been treating them as one has been reasoning about the integers.

And the identity is one identity. A lattice-ordered group satisfies a family of laws and this is the shortest of them; failing it rules the family out, and does not say which of the others fail worst or whether some weaker agreement survives. A version stated with an error term — how far from equal, as a function of how confused the pair is — is a real question and is not one this page asks.

The convention, named #

Normal play throughout. The values are the canonical forms of the twenty-two values born by day two, computed by the recursion rather than listed; the order is the comparison relation, computed by playing out the difference; and the join of a pair is the unique value above both that is below every value above both, found by scanning the day rather than by a formula. Where no such unique value existed the census would refuse to run, which is the check that the day is a lattice at all.

The addition is the disjunctive sum. The negation used in the symmetry argument is the one that swaps the two players, and it is an anti-automorphism of the order and an automorphism of the group, which is exactly the combination that forces the thirteen-and-thirteen.

Where the ladder goes next #

lattice has two rungs to here: that the order on day two is a lattice, and now that the lattice and the group do not respect each other.

The rung above takes the fifty-two incomparable pairs this page ends on and refuses to list them. Fifty-two errors and seven sizes measures the error terms instead, and the fifty-two collapse into something much smaller: seven distinct pairs of stops, three means and three temperatures between them, and a rule that predicts the temperature of the error from the pair on forty-four of the fifty-two.

That is the difference between a counterexample set and a phenomenon. Fifty-two pairs where a law fails is a list; fifty-two failures taking seven shapes is a structure, and it says the failure of $(G \vee H) + K$ to be $(G+K) \vee (H+K)$ is not fifty-two accidents but a small number of configurations recurring.

It also puts a bound on how much a repair could ever buy. If the error takes three temperatures and a rule predicts which on forty-four of the fifty-two, then a corrected law would have three cases and eight exceptions — which is a worse object than the two clean theorems it would be replacing, and knowing that is worth more than trying.

The neighbour worth the trip is the simplest game above both, where the join is established and where the same fifty-two incomparable pairs appear as the reason the order needed a join at all. Reading the two together shows one set of pairs doing two jobs: creating the lattice structure, and then breaking its compatibility with the group.