Combinatorial Game Theory
who wins, and by how much
Series

Ordinal sum — the series

7 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · sums
  2. 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.

    part 2 · sums
  3. How much of the value the colon respects. Every form whose option lists are antichains of day-two values, grouped by the value it reduces to, and each group asked whether all its forms give the same ordinal sum with star. On 636 of the 640 groups they do.

    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.

    part 3 · sums
  4. No fifth value. Forms of day-three values built by adding day-three gift horses, and the ordinal sums they give. Over eighteen thousand forms and four followers, no value's forms disagree.

    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.

    part 4 · sums
  5. 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.

    part 5 · sums
  6. 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.

    part 6 · sums
  7. 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.

    part 7 · sums

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