Trees, and what each is worth
A row of blue-red Hackenbush trees with the value the recursion returns under each. Every one is a number, and none of them is the binary reading of anything a reader can see in the picture.
2 essays call
bush-tree. The drawing above is what it returns with no arguments at all; every
call below passes it something, because a placement that passes nothing draws whichever
member of the family the generator happens to default to rather than the one its essay argues
about.
The positions it draws
11 distinct positions, harvested by drawing the figure again at the options each essay asked for.
| Position | Worth | Outcome | Drawn in |
|---|---|---|---|
above the trunk |
0 |
P | A tree is still a number |
cherry |
1 |
L | A tree is still a number · Where the numeral stops |
fan |
1/8 |
L | A tree is still a number · Where the numeral stops |
ladder |
3/2 |
L | A tree is still a number · Where the numeral stops |
LR |
1/2 |
L | A tree is still a number |
LRL |
3/4 |
L | A tree is still a number · Where the numeral stops |
red-cherry |
−1 |
R | A tree is still a number · Where the numeral stops |
RL |
−1/2 |
R | A tree is still a number |
tree 1 |
−1/4 |
R | A tree is still a number |
two-trunks |
0 |
P | A tree is still a number · Where the numeral stops |
whole tree |
1 |
L | A tree is still a number |
Where it is called
Every one of these figures is a drawing of the same construction, so a correction to it corrects them all.
Particular games
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.
Particular games
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 whole library · The position index · The figures that play back