The company that is closed

Two games are equal when they play the same in every company. That quantifier is what makes the theory work and what makes it expensive, so equal in this company replaced every with these and made the relation computable — a company is a finite set of games, two positions are indistinguishable in it when no member separates them, and the whole thing is a search rather than a theorem.

It closed by naming the property the restriction actually needs:

Closure. Which of the small companies is closed under addition, and what the smallest closed company containing a star is. That is the version of the question that licenses substitution, and it is a search rather than an argument.

The search has two answers. The first is that the companies in use are not closed. The second is that the ones that are closed are a much stranger family than the question suggests.

Why closure is the property that matters #

Restricted equality says: nothing in this company can tell A from B. What a reader wants from that is a licence — the right to replace A by B inside a larger position and know that nothing has changed.

The licence needs one more step and the step is closure. Suppose A and B are indistinguishable in company U, and a position A + X is to be rewritten as B + X with X in U. To know the two behave alike in a further company member Y, the test needed is A + X + Y against B + X + Y — and that is a test of A against B in the company member X + Y. If X + Y is not in U, the indistinguishability was never checked against it, and the substitution is unlicensed.

So a company that is not closed under addition licenses one substitution and not two. That is the whole reason the rung above asked the question, and it is a real limitation rather than a technicality: a game of any length is a sum of many components, and a licence that survives one is worth little.

Not one of them is closed #

The site computes in five companies, and they are the ones every separation figure on the site uses.

The company of nothing but zero is closed, trivially and uselessly: 0 + 0 = 0, and a company with one member separates nothing.

Every other one leaks. The numbers from −2 to 2 in halves keep 61 of their 81 sums; the sums that leave are the ones past the ends, since 2 + 2 is 4. The all-small values of day two keep 31 of 49. Day one keeps 10 of 16 — 1 + 1 = 2, and 2 is not in day one. And day two, the largest and the one most sweeps use, keeps 118 of 484 and leaves through 162 distinct values.

That last figure is the one to carry. A quarter of the sums stay inside; three quarters do not; and the escapes are not a handful of edge cases but 162 different games.

The census asserts it: a company other than the trivial one turning out to be closed would stop the build, because the essay would then be about a different set of companies.

What a closed company looks like #

The other half of the question — what the smallest closed company containing a star is — is answered by building companies rather than testing them. Start from a handful of games, add every pair, add every pair of the answers, and keep going.

The smallest closed company containing a star has two members: zero and star. ∗ + ∗ = 0, so nothing else is reachable, and the answer is as small as it could be.

Adding ∗2 gives four: {0, ∗, ∗2, ∗3}. Adding ∗3 and ∗4 gives eight. The sizes are 1, 2, 4, 8, and the pattern is not a coincidence about nimbers.

Zero and ±1 is closed at two. ±1 — the switch {1 | −1} — is not a nimber, is not impartial, and is a perfectly ordinary hot position; and ±1 + ±1 = 0, so {0, ±1} is a closed company with a fight in it.

Star and ±1 together generate four, and the fourth member is {1∗ | −1∗}. Star, ∗2 and ±1 generate eight.

The condition is not being a nimber #

The characterisation the sizes point at is a single property and it is not the one the nimbers suggest.

Every member of every finite closed company found here is equal to its own negative.

That follows from closure directly, once stated. A closed company is finite, so adding a fixed member G to itself repeatedly must eventually repeat — and since the values form a group, repetition means nG = mG for some n > m, so (n − m)G = 0 and G has finite order. A group in which every element has finite order and which is generated by finitely many such elements is finite; the ones here all have order two, so 2G = 0, which is exactly G = −G.

And that is why the sizes are powers of two: a group in which every element satisfies 2G = 0 is a vector space over the field of two elements, and a finite one has size 2ᵏ.

So the condition is being one’s own negative, which is the two-torsion of the group of short games. The nimbers are inside it and are not the whole of it — day three alone holds thirty self-negative values of which four are nimbers — so there are closed companies with switches in them, and companies of nimbers are a special case rather than the definition.

Both claims are asserted. A finite company with a member that is not its own negative, or one whose size is not a power of two, stops the build.

Why a number ruins it #

The four companies that do not close all start from something of infinite order.

Zero and one generates every integer. 1 + 1 = 2, 2 + 1 = 3, and there is no stopping point; each sum is a new value and the company is ℤ.

Zero and a half is the same story one denominator down. Zero and up generates ⇑, 3·↑, and every multiple, none of which is any of the others — ups do not repeat, which is the whole reason atomic weight is a measure. Zero and up-star goes the same way.

That is not four separate observations. It is one: a value of infinite order generates an infinite company, and the values of infinite order are everything except the two-torsion. So the finite closed companies and the two-torsion are the same subject, and the question which small companies are closed has the answer the ones made entirely of games that cancel themselves.

What this does to the restricted equality programme #

The consequences for the machinery the rung below built are worth stating plainly, because they are mixed.

The bad news is that the useful companies are unusable as licences. Day two separates a great deal precisely because it contains numbers, switches and infinitesimals; those are what give it discriminating power, and every one of them is what stops it closing.

The good news is that the useful companies were never being used as licences. Equal in this company uses them to measure — how much of the real equality a bounded test recovers — and a measurement needs no substitution theorem. So nothing on this site is broken by the finding; what is established is that the obvious next step, promoting a measurement to a licence, is not available with these companies.

And where the licence is genuinely wanted, the closed companies are the wrong shape. A misère quotient is a licence-bearing structure and it is built by taking a game’s own positions, closing under addition, and quotienting — which produces an infinite set with a finite quotient rather than a finite closed set. Misère quotients is the worked version of that and it is a different construction from anything on this page.

Set the two properties beside each other on the five working companies and the natural conclusion draws itself: the company that keeps every one of its sums sees four things, and the company that sees all twenty-two keeps a quarter of them.

The reading that suggests itself is that the two pull against each other, and that a company chosen for one cannot have much of the other. It is worth writing down because it is wrong, and the companies that refute it are the ones this page has just built.

Closure is free, and blindness to numbers is not #

Every closed company found above keeps every sum by construction, so each of them licenses a substitution. Asking the rung below’s question of them — how many of the twenty-two does this company separate? — is a computation of one line, and it has not been run.

A closed company of four separates the whole day. So closure costs nothing here at all: the best any company of any kind manages on this pool is twenty-two, the four games born by day one reach it and are not closed, and {0, ∗, ±1, {1∗ | −1∗}} reaches it and is. A licence to substitute is available at full discriminating power.

What the table separates instead is nimbers from everything else. {0, ∗} sees nine of the twenty-two; adding ∗2 and ∗3 takes it to twelve; adding four more nimbers takes it to twelve again, because a company of nimbers can only ever report whether a position’s infinitesimal part cancels against one, and every number in the pool looks alike to all of them. {0, ±1} — two members, one of them a switch — also sees nine, and adding a star to that is what reaches twenty-two.

So the ingredient is the same one the rung below identified, arriving from the other side. Equal in this company found that a company of numbers, however large, cannot separate an infinitesimal from zero, and that the company has to contain a star. Here the mirror holds: a company of nimbers, however large, cannot separate one number from another, and a company that separates everything has to contain something that is not a nimber. ±1 is that something, it is its own negative, and being its own negative is what let it into a closed company in the first place.

That is the exchange rate the rung below asked for, and the rate is zero. What a substitution licence costs is not discriminating power; it is the freedom to pick the company from the values a reader finds natural, since every member has to cancel itself.

There is a real exchange here and it is with the other closure rather than with this one. The rung above finds that among these same seven, keeping every option is exactly being a group of nimbers — and the groups of nimbers are exactly the ones that never pass twelve. So no company on this page is both fully discriminating and licensed to rewrite a subposition, and the two things a solver wants are held apart by the same property that lets a switch into a closed company at all.

What the closed companies are, as objects #

It is worth naming what the seven companies found here are, because the name is standard and it says immediately what else is true of them.

A finite group in which every element satisfies 2G = 0 is a vector space over the two-element field, and its size is 2ᵏ where k is the number of basis elements. {0, ∗} is one-dimensional with basis {∗}; {0, ∗, ∗2, ∗3} is two-dimensional with basis {∗, ∗2}; {0, ∗, ±1, {1∗ | −1∗}} is two-dimensional with basis {∗, ±1}.

So a closed company is decided entirely by its basis, and choosing one is choosing which independent games are in it. That gives the search a shape it did not obviously have: rather than testing sets for closure, pick any set of self-negative games and the company they generate is closed, finite, and of size 2 to the power of however many of them are independent.

The nimbers are the standard example and their arithmetic is exactly this structure — the nim-sum is addition of binary digits without carrying, which is what addition in a vector space over two elements looks like when the basis is ∗1, ∗2, ∗4, ∗8. Every fact a reader knows about the nim-sum is a fact about closed companies, and the surprise is only that the same structure is available with ±1 in it.

One of the seven repays a second look, because it carries an operation this page has not asked for. {0, ∗, ∗2, ∗3} is closed under addition by the argument above, and it is closed under the nimber multiplication as well — which makes it a field rather than merely a group, and the smallest interesting one there is.

And the family has one member for every finite set of independent self-negative games, which on day three alone is a set of thirty to choose from. Nobody has enumerated the resulting companies here and it would be a finite computation.

It would also be a computation that leaves the day almost at once, and that is worth knowing before anybody starts it. The thirty are closed under addition as a subgroup of all games and are not closed inside day three: of the 465 unordered pairs among them, self-pairs included, only 68 have a sum still born by day three and 397 do not. $\ast$ is born by day two and $\{2 \mid -2\}$ by day three, and their sum $\{2\ast \mid -2\ast\}$ is not born until day four.

So a company generated by a handful of day-three values is finite and small — $2^k$ for however many of them are independent — and its members are mostly not day-three values. That is the right way round for this page’s purpose, since a company is a set of games to add rather than a slice of the construction, but it means the search this section proposes cannot be run inside an enumerated day. It has to be run in the group, one generated subspace at a time.

What the census does not say #

Four limits.

Eleven starting sets is not a survey. The search builds companies from eleven seeds chosen to make the two claims visible. It does not enumerate all closed companies — there are infinitely many, one per finite subgroup of the two-torsion — and the characterisation above is an argument checked on seven cases rather than a theorem proved here.

Sixty-four is a search bound, not a proof of infinity. A company reported as infinite is one whose closure exceeded 64 members before terminating. Each of the four is infinite for a reason given in words above, and the census records the bound rather than the argument.

Closure under addition is not the whole licence. Substitution inside a game also needs the company to survive forming options, which is a stronger requirement and one this page has not tested. A company closed under addition and not under options licenses rewriting a sum and not rewriting a subposition.

And the negatives are not asked for. A company closed under addition need not contain the negatives of its members, and the two-torsion companies contain them automatically because every member is its own. A company of positive values closed under addition would be a different object and there are none here.

The convention, named #

Normal play, and every game is a canonical form. A company is a finite set of games; two games are indistinguishable in it when no member separates their outcomes; a company is closed when the sum of any two of its members, reduced to canonical form, is a member.

Closure is computed on identity rather than on the written name. Games are interned here, so two forms are the same object exactly when they are the same value — and the first version of this computation keyed on the printed name, which elides deep forms and reported that the company generated by up has ten members. It has infinitely many.

The two-torsion is the set of games satisfying G = −G, tested by computing the negative and comparing canonical forms rather than by any structural shortcut.

Where the ladder goes next #

The universes anchor has two rungs to here: restricted equality made computable, and now which restrictions license a substitution.

The rung above takes the second closure this page names and finds it pulling the opposite way. The closure that picks the nimbers asks which companies are closed under forming options — which is what lets a recursive computation rewrite a subposition rather than a whole component — and the answer is unexpectedly clean: among the seven finite addition-closed companies, keeping every option is exactly being a group of nimbers, four of the seven, both directions and no exception. Demand both closures at once and nineteen of the twenty-two day-two values generate nothing finite at all.

The three rungs after that price the licence rather than characterising it. A product against a sum measures what a solver gains on Cram boards: four to twenty-five times on two components, twenty-four to a hundred and sixty-one on three — and it is available to impartial games precisely because their class representative is a heap rather than a form. Half a licence is nearly all of it then splits the two halves and finds the component half doing nearly everything: rewriting components takes a million and a half states to three thousand six hundred, and rewriting subpositions takes those to eight hundred and eighty-four.

One half multiplies the other adds closes it with the two closed forms that name the difference: the component licence saves $s^{k-1}/k$ and the subposition licence saves $k \cdot s$, over a shape count that never moves. So the first half’s benefit grows with the number of components and the second’s does not, which is why a board that falls into many pieces is where a restricted universe pays.

Two neighbours are worth the trip. Equal in every company is the unrestricted theorem this page’s companies are approximations to, and it is where the substitution argument this page needs closure for is stated properly. And the values that are their own negatives is the subgroup every finite closed company lives in, counted on day three, where thirty values qualify and four of them are nimbers.