Twenty-six other values

The mirror construction takes a set $S$ of values and forms $\{S \mid -S\}$, which is its own negative whatever $S$ is. What identifies two subsets reduced the 1,793 subsets of day two to 96 antichains, and the case that was supposed to be hard proved the rule that reads 66 of them: the value is $\ast k$ for the least $k$ such that no element of the antichain is at least $\ast k$.

That page closed on what its own argument makes askable:

the argument above works because $\ast k$ is its own negative and the sum’s outcome is decidable from one comparison, and any self-negative value has the first property. Whether some other self-negative value plays the role $\ast k$ plays here, for an antichain the mex is silent on, is the question the argument makes askable.

Twenty-six of them do, and they do not replace the nimbers — they continue them.

The family is not a choice #

The rule is silent on thirty antichains, and it is silent for a definite reason: each of them holds an element at least as large as every small nimber, so no least $k$ exists.

That is worth reading as a positive statement rather than a failure. An antichain that dominates every nimber is one containing a value large enough to beat $\ast$, $\ast2$, $\ast3$ and the rest — which is not an obscure condition; a single element worth 1 does it, since $1 \ge \ast k$ for every $k$. So the rule’s silence is not about pathological antichains. It is about the ones containing anything large, which is most of them once the antichain has three members.

What are those thirty worth? Every one of them is a self-negative value — which had to be true, since $\{S \mid -S\}$ always is — and every one is among the thirty self-negative values born by day three. That second fact did not have to be true and it is what makes the question answerable: the family a wider mex should run over is not something to be designed, it is the family the answers already live in.

So the candidates are fixed before any rule is written. Thirty self-negative values, of which $0$, $\ast$, $\ast2$ and $\ast3$ are four.

Two of those thirty are worth naming for the shape they have. $\{1 \mid -1\}$ and $\{2 \mid -2\}$ are switches, and a switch about a symmetric pair of numbers is its own negative for the obvious reason — turn the board round and it is the same switch. So the family is not exotic: it runs from nought and the nimbers through the plain symmetric switches and out into forms with several options a side, and its thirty members are simply everything born by day three that survives being negated.

The same mex, over a longer list #

A mex needs an ordered family and a rule: take the first member of the family the antichain does not dominate. The nimbers are one such family and give 66 of 66 on the antichains they reach.

Run the identical rule over the thirty self-negative values, in a suitable order, and it is right on all ninety-six — including on all thirty the nimbers are silent about.

And the order it needs begins $0$, $\ast$, $\ast2$, $\ast3$. So the nimber rule is not a different rule that happens to work on a subset; it is this rule truncated after its fourth member, and the thirty antichains it cannot reach are exactly the ones whose answer lies further down the list.

That is the strongest form the answer to the rung’s question could take. It asked whether some other self-negative value plays $\ast k$’s role. Twenty-six of them do, in the same rule, in a list that starts with the nimbers.

It is worth checking the shape of the result against the alternative that would have been less interesting. The rule could have been exact by being vacuous — a family long enough that some member is always the first undominated one, giving an answer on every antichain whether or not it is the right answer. The middle column guards against that: the birthday order also fires on all ninety-six and is right on seventy-four, so firing is easy and being right is not. Exactness at ninety-six of ninety-six is doing work.

What the order costs #

The qualification in that paragraph is in a suitable order, and it is where the result stops being a formula.

For the mex to give the right answer on an antichain, every family member placed before that antichain’s value must be one the antichain dominates. Collect that requirement over all ninety-six antichains and it is 2,050 constraints of the form this value comes before that one. They are consistent — the constraint graph has no cycle, which is why an order exists at all — and they force 342 of the 435 pairs, leaving 93 free.

So a suitable order exists, is far from unique, and is obtained here by sorting the constraints rather than by looking at the values. The obvious order does not work: sorting the same thirty values by birthday, which is exactly what indexes the nimbers, gives 74 of 96 and only 9 of the 30.

What is established, then, is that the rule’s shape is right — a mex over an ordered family of self-negative values accounts for every antichain — and that its index is not derived from anything. The nimbers come with an index; the other twenty-six do not, and nothing here supplies one.

The 93 free pairs are worth taking seriously rather than treating as slack. They are pairs of self-negative values that no antichain in the pool ever forces into an order, which means the pool cannot see a difference between them — and a larger pool, from day three into day four, would either force them or leave them free for a reason. Either outcome would be informative and neither is available here.

And the 2,050 constraints are not evenly distributed. Every antichain contributes one constraint for every family member it fails to dominate, so an antichain that dominates almost nothing constrains almost everything, and the singleton antichains — {0}, {∗}, and the rest — are doing most of the ordering work. That is the same concentration the zero fibre has, where forty-eight of the ninety-six antichains are worth nought, and it is the same cause: an antichain of small elements dominates little.

Why the nimbers have an index and the others do not #

That asymmetry is worth a paragraph, because it is the whole difference between the rung below’s result and this one.

$\ast k$ is indexed by $k$, and $k$ is a fact about the value: $\ast k$ is the Grundy value of a heap of $k$, born on day $k$, and its position in the list is its own birthday. So the nimber mex is a rule stated entirely in terms of the values it produces — the answer can be read without a lookup table.

That is what makes the rung below’s result a theorem rather than a table. Its argument runs: the antichain fails to dominate $\ast k$ and dominates every $\ast j$ below it, so Left moving first cannot reach anything at least $\ast k$ and can reach something at least $\ast j$ for each smaller $j$ — which is exactly the outcome condition for the sum to be $\ast k$. Every step of that mentions $k$ as a number, and every step of it would need rewriting for a family whose members have no number attached.

The other twenty-six self-negative values have birthdays too, and their birthdays are the wrong index: the birthday order gives 74 of 96. Whatever decides where a self-negative value belongs in this list, it is not how early the value was born, and the constraint graph does not say what it is — it says only that some order works and pins down four fifths of it.

There is one place to look and this page has not looked. The nimbers are totally ordered by an index while being mutually incomparable as games — $\ast \parallel \ast2$ — so the index cannot come from the game order. Where it does come from is Sprague–Grundy: $k$ is a mex, computed from the heap’s options. The other self-negative values are built as $\{S \mid -S\}$ and have an $S$ of their own, so an index computed from $S$ the way $k$ is computed from a heap’s options is the obvious candidate, and testing it is a definite piece of work.

The circularity in that has to be faced squarely. The index would be computed from the value’s own option set, and the rule computes the value from an antichain’s option set, so a rule of that shape reads a mex to produce a value whose position is itself a mex. That is not vicious — Sprague–Grundy is exactly such a recursion and terminates because options are simpler than positions — but it means the index cannot be looked up before the family is built, and the family is what the rule is stated over. Whether the recursion bottoms out is the first thing a candidate index would have to establish.

There is also a cheaper thing to try first, and it is worth naming because it is one line. The constraints force 342 pairs; any candidate index can be scored against those 342 immediately, without building anything, and a candidate that fails one of them is dead. That turns a search over indices into a filter, and the filter is available now.

What this does to the counting #

At least five hundred and seventy-one turned the mirror construction into a counting tool: the number of self-negative values born by a day is at least the number of distinct values the construction produces, and the construction’s output was only partly described.

The rule here does not immediately improve that floor, and it is worth saying why not. A floor comes from counting distinct outputs, and this rule is a way of computing an output from an antichain rather than a way of counting how many outputs there are. What it would take to turn it into a count is the index — knowing which position in the family an antichain’s value occupies, without computing the value, would turn a count of antichains into a count of values.

So the two things this page leaves open are the same thing. The index is what a formula needs and what a count needs, and it is the one part of the nimber rule that did not generalise.

That is a tidier position than the anchor has been in for several rungs. The construction produces self-negative values; the map from antichains to values is now completely described by a rule; and exactly one ingredient of the rule is a lookup rather than a computation. Before this page the description stopped at sixty-six of ninety-six with no account of the rest at all.

What the rule looks like, written out #

For a reader who wants the rule rather than the account of it, it is three lines.

Take a subset $S$ of the day-two values. Reduce it to its antichain $M$ of maximal elements — everything else is dominated and $\{S \mid -S\}$ does not see it. Then walk the family of thirty self-negative values in its order, and the value of $\{S \mid -S\}$ is the first member $V$ such that no element of $M$ is at least $V$.

The nimber version is the same three lines with the family cut to $0, \ast, \ast2, \ast3$, and it works precisely when the answer is one of those four. Nothing else changes — not the reduction, not the walk, not the test.

What a reader cannot do from that description is compute the family’s order, and so cannot use the rule without the list. That is the whole gap between this and the rung below, and it is why the page above says shape rather than formula.

What is measured #

The exactness is on the pool the rule was fitted to, and that is a real limitation. The order was constructed from the constraints the ninety-six antichains impose, so being right on those ninety-six is guaranteed by construction rather than tested. What is not guaranteed and is tested is that the constraints are consistent at all — a family for which they were not would admit no order, and the answer to the rung’s question would be no.

Ninety-six antichains, thirty family members, one construction. All of it is day two into day three: the antichains are subsets of the 22 day-two values and the family is the self-negative values born by day three. Whether the pattern continues — day three into day four, with a larger family — is not asked here, and the rung below’s reason applies: day four is out of reach.

The order is fitted and the page says so throughout. A rule with a fitted parameter is a weaker object than a rule without one, and the honest description of what is established is: the mex over an ordered family of self-negative values is the right shape, plus some suitable order exists, plus four fifths of it is forced. It is not: here is the rule.

The rule is checked on antichains, not on subsets. The 1,793 subsets collapse to 96 antichains because domination removes the rest, and that collapse is the rung two below’s result rather than this page’s. Everything here is a statement about the 96, and it inherits the 1,793 through that reduction.

And the constraint graph being acyclic is the load-bearing fact. If it had a cycle, no order would work and the answer to the rung’s question would be no. That it does not is checked rather than assumed, and the check would fail loudly — the topological sort would place fewer than thirty members and the sweep refuses to draw.

What it means for a mex to be silent #

A rule that accounts for two thirds of a population and says nothing about the rest is in a particular and recognisable position, and it is worth naming.

It has not failed. Nothing in the thirty contradicts the mex; the mex simply does not reach them, because the family it is stated over does not contain the values they need. A rule that was wrong on thirty would be evidence that the mex is the wrong instrument. A rule that is silent on thirty is evidence that it was pointed at the wrong family, which is a repair rather than a refutation — and the repair is to run the same mex over the family the silent thirty actually live in.

Where the ladder goes next #

The negation anchor has ten rungs, and the tenth says the ninth’s rule was a fragment.

The rung above is the index. Everything here points at one question: is there a quantity computable from a self-negative value that gives its position in the family, agreeing with the birthday on the nimbers and disagreeing with it elsewhere? The constraint graph already forces 342 of the 435 pairs, so a candidate index has 342 tests it must pass before anybody has to think about the other 93 — which is an unusually well-specified search, and the pool for it is thirty values.

Two neighbours are worth the trip. The values that are their own negatives is where this family was first counted, and it is the page whose thirty values turn out to be the alphabet the rule is written in. And Sprague–Grundy is where a mex gets an index by being computed rather than looked up, which is precisely what this rule is missing.