Series

Universes — the series

8 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. How much company equality needs. Each row restricts the quantifier in the definition of equality to the games named, and counts how many of the 22 values born by day two survive as distinct. The bar is the same number drawn; the jump from nine numbers to four games is the whole argument.

    Equal in this company

    Equality quantifies over every game there is, and the quantifier can be made smaller. Restricted to a company of nine numbers, the twenty-two values born by day two collapse to seventeen; restricted to four games — nought, one, minus one and star — they stay twenty-two, and no three of the four will do. The company that decides equality is tiny, and it has to contain a star.

    part 1 · limits
  2. What a finite closed company is made of. The finite closed companies found by the search, counted by the properties they share. Every one of them consists of games equal to their own negatives and has a size that is a power of two, and not all of them are made of nimbers.

    The company that is closed

    Restricted equality licenses substitution only inside a company closed under addition, and none of the five companies this site computes in is closed — day two keeps a quarter of its own sums. Searching for companies that are closed finds seven, at one, two, four and eight members, and every member of every one of them is its own negative.

    part 2 · limits
  3. The second closure picks out the nimbers. The seven finite companies closed under addition, tested for closure under forming options. The four that are groups of nimbers keep every option; the three containing plus-or-minus one lose theirs.

    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 this site computes in 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.

    part 3 · limits
  4. A product against a sum. The mean cost ratio on two components and on three. The saving from substituting grows with the board rather than staying a fixed factor.

    A product against a sum

    A company closed under both addition and options licenses a solver to rewrite any subposition, and the rung below found that exactly the nimber groups have both closures. Priced on Cram boards, that licence is the difference between walking a product of position sets and walking their sum — 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 because their class representative is a heap rather than a form.

    part 4 · limits
  5. Four solvers on one sum. The states each solver has to distinguish on a three by four board plus a three by five, with one more substitution allowed at each step. A million and a half becomes fourteen.

    Half a licence is nearly all of it

    The rung below priced the substitution licence a restricted universe gives a solver and asked what half of one is worth — the licence to rewrite components but not subpositions. It is worth nearly the whole saving. Rewriting components collapses a million and a half states to three thousand six hundred; rewriting subpositions collapses those to eight hundred and eighty-four, and splitting the pieces takes it to fourteen.

    part 5 · limits
  6. A gap that widens without bound. Both savings as the number of components grows, enumerated where possible and given by the closed forms beyond.

    One half multiplies, the other adds

    The rung below priced the two halves of a substitution licence on sums of two Cram boards and predicted that the first half's saving would grow with the number of components while the second's would not. It is right, and both halves have closed forms: the component licence saves s^(k−1)/k and the subposition licence k·s over a shape count that never moves.

    part 6 · limits
  7. Two readings of one sequence. The three licences with their savings and their tables, which order them oppositely.

    The licence that weighs nothing

    The third substitution licence is constant in the number of components, exactly as predicted, and it saves under two times where the first saves seventy-six million. Priced by its table instead of by its saving it is the only one of the three whose cost does not run away — which reverses the order three rungs of this anchor have put them in.

    part 7 · limits
  8. One licence, five prices. The third licence measured by saving, by table size, by expansions avoided, and by work under two implementations.

    The price of asking what the parts are

    The third licence lets a solver look up a region rather than a position, and the rung below priced it by the entries it stores. Priced by the work it costs, it saves between a third and two thirds of the expansions and pays for them with a flood fill at every node — six times the total. A square would have to be ten times cheaper than a table probe before it broke even.

    part 8 · limits

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