Hackenbush — where it appears
Named by 34 essays across 5 fields — each of them below, with the objects they name alongside it.
Hackenbush is a numeral
Draw a stalk of coloured edges. Read it as a string, blue for one and red for zero, and the string is the binary expansion of what the position is worth. Not approximately — exactly, and the site computes it both ways and refuses to build if they disagree.
Nim, and the nim-sum
Three heaps of counters, take as many as you like from one of them, and the player who takes the last counter wins. The winning condition is not a search, not a table, and not a heuristic — it is the bitwise exclusive-or of the heap sizes, and it was found in 1901.
The sum is the object
Real positions come apart into independent regions, and a move happens in exactly one of them. That operation — the disjunctive sum — is what the whole theory is built to survive, and it is the reason values exist at all.
Who moves last
The player who cannot move loses. That single convention generates the whole theory — and it produces four outcomes rather than three, because a position can be confused with zero rather than greater, smaller or equal to it.
Domineering
One player places dominoes vertically, the other horizontally, on a shared grid. The rules take one line, the values are a mess, and that mess is the point — this is what the theory looks like applied to a game nobody designed for it.
The simplicity rule
When both players' options are numbers, the position is worth the simplest number strictly between them. Not the midpoint, not the average, and the difference between "simplest" and "middle" is the entire content of the rule.
Toads and Frogs
Toads shuffle right, frogs shuffle left, and either may jump over one of the other. A strip six cells long is worth exactly up. Another six-cell strip is worth exactly down. Nobody has a formula for which.
Infinitesimals
Some positions are positive — Left wins them whoever moves first — and smaller than every positive number, including a millionth and a millionth of that. They are the values that decide close games, and the smallest of them is a single move's worth of nothing.
The day a number is born
Start with a position in which neither player can move, apply one rule, and the numbers appear — but only the fractions with a power of two underneath, and only in a particular order. That order is what "simplest" means.
The numbers came out of the game
The construction is always taught numbers first and games second, and the discovery ran the other way. Conway arrived at the number system from positions, which is why the definition quantifies over sets of previously built objects rather than over cuts — and why it produces a genuinely different collection at every finite stage.
Squash every loop to a point
Colour every Hackenbush edge green and the game becomes impartial, so the whole picture is worth a single Nim heap. Two principles find which one without playing anything — fuse the cycles, then run one pass up the tree — and a nine-vertex lattice that costs 1,283 positions to solve costs twelve steps to read.
A green edge on a blue one
Blue over green and green over blue are the same two edges in the other order. One is worth 1∗ and the other ↑∗ — a number with a star on it against something smaller than every positive number — so a stalk with all three colours in it stops being a numeral and starts being a position whose value depends on what is underneath.
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.
When the ups add
Atomic weight brackets do not add over a sum — they bound it. Over all 120 pairs from a fifteen-game family the sum's bracket came out exactly the sum of the parts' brackets 56 times, strictly narrower 64 times, and wider never; and the rule separating the two is one line long, because every one of the 54 pairs with a pinned part is exact and only 2 of the other 66 are.
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.
Nothing worth fighting over
Shove is a strip of coins beside a cliff, and both players have completely different moves. Every one of its 728 positions is worth a number, so nobody ever wants to move; the winner is the owner of the coin furthest from the cliff, in all 728; and the number the board is worth is not the sum of its coins — that reading is exact on 126 strips and wrong on 588 of the other 602.
A tree is still a number
A Hackenbush string spells its own value in binary. Put a fork in it and the numeral has nothing to read — there is no leftmost anything. The value is still a number, in all 10,066 forests up to six edges; it is still computable, by the ordinal sum, in all 3,238 single-trunk trees; and the reading is right on 762 of them, of which 126 are the strings it was written for.
Topple it from either end
A row of blue and red is the picture this site opens with, and under Hackenbush's rules it is always a number. Knock the pieces over instead of cutting them — everything on the chosen side falls — and 480 of the 510 rows up to eight pieces stop being numbers. The two games agree on sixteen rows, every one of a single colour, and the temperature of the hottest row climbs by exactly a half for each domino added.
The values nobody's game produces
The construction hands down 1,474 values by day three. Seventeen rulesets on this site, swept to eleven thousand positions, produce 1,193 — and only 116 of those are on the construction's list. Two of the twenty-two values born by day two are produced by no position of any game here, and 1,077 of the values that are produced are born later than day three. A value's birthday and a value's reachability have almost nothing to do with each other.
How long a row a value needs
Add a third colour that either player may topple and a row of seven dominoes reaches 1,047 distinct values where two colours reach 149. That makes the length of the shortest row worth a value into a measure of the value's complexity — one a reader can hold in their hand — and it is not the birthday: 1↑ is born on day three and needs seven dominoes.
The reading that survives too much
Counting the empty squares in front of each coin gets a Push position right half the time, and the rung below said the failures were exactly the positions with two coins of opposite colour side by side. Sixty-six of the 1,072 failures have no such pair, the smallest is five squares long, and the condition that does decide it is not about the board at all — it is about every position the board can reach.
The cheapest way to show a value
Eleven thousand positions from fifteen rulesets reach 1,193 values, and for each of them there is a smallest board that shows it. Set against the birthday the two measures agree hardly at all — until the numbers are taken out, at which point they agree rather well, and the whole apparent independence turns out to be a fact about integers.
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.
The criterion that cannot exist
The rung below asked for a quantitative version of its condition — turn 'the reading survives mixing three quarters of the time' into a statement about the strip. Three strips of four squares settle it. `.LLR`, `.LRL` and `.RLL` have the same length, the same reading, the same coins and the same single run, and their readings are wrong by 1¼, ¼ and ½. The error is a fact about the order of the colours, and 207 of 805 statistical classes carry more than one of them.
The birthday is a floor
The rung below measured a correlation of 0.73 between a value's birthday and the size of its cheapest exhibit, and asked which values are dearer than the birthday suggests. The relation is not a trend. Over all 728 non-number values the exhibit is never smaller than the birthday and is exactly the birthday on 476 of them, and the excess on the other 252 belongs to the game rather than to the value — the ruleset accounts for 40 per cent of its variance.
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.
Wider costs less
The rung below found the cheapest exhibit of a value never smaller than its birthday, exactly equal on two thirds, and the ruleset explaining 40 per cent of the rest. The variable it proposed for the remainder was the width of the form. Width and excess correlate at −0.39: the wider the value, the closer to its birthday it is exhibited, and inside a ruleset the relation cannot even agree on a sign.
A numeral in the empty squares
The rung below ruled out a quantitative criterion for Push and asked for a numeral over the coins combined with a count over the gaps. The two ingredients are the right way round: the colours pick a fraction — −1, −1/3, −1/7, −1/15 — and the empty squares give the binary precision, so a run of k coins before one of the other colour with g gaps is worth exactly (1 − 2^(−kg)) ÷ (2^k − 1). And it does not compose: a strip of two runs is not the sum of them, on any pair tried.
The entry fee was the cap
Two rungs measured how much bigger a position has to be than the value it exhibits, and attributed what was left to the ruleset — Toads and Frogs paying 2.25 squares on everything, green Hackenbush paying nothing. Neither number is a property of the rules. Inside every ruleset the excess falls as the birthday rises, because the sweep's size cap censors exactly the values that would pay most — and three squares past the cap, Toads and Frogs exhibits values born later than the strip is long.
The rate was the alphabet
The rung below asked for a quantity a size cap cannot censor and proposed the rate: how many new values a ruleset produces per extra square. The rate is honest and it measures the notation — every ruleset grows at close to the number of symbols its positions are written in, and the seven span less than a factor of two. What separates them is the yield, which spans a hundred and nineteen.
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.
Three groups, and three yields
The conjecture was that each ruleset's yield tends to the reciprocal of its symmetry group's order. Three rulesets have a trivial group and predicted yields of one; they measure 1.000, 0.531 and 0.204. And every colliding value in Push and Shove — all 175 of them — has two rows no symmetry relates.
Where the numeral stops
A Hackenbush string is a numeral and a tree is a trunk with a forest on it, so the obvious next question is a graph with a cycle in it. Green Hackenbush answers that by fusing the cycle to a point. In blue and red the fusion is right on every three-edge cycle, on fewer than half of the six-edge ones, and the smallest thing it gets wrong has four edges.
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.
Named alongside it
The objects these essays reach for when they reach for this one.
EnumerationCanonical formExhaustive searchBirthdayValueCounterexampleInfinitesimalPartizanBinaryNormal playOrdinal sumNumbers