Solution
A Garden of Hackenbush Stalks
Show the problem again
In Blue-Red Hackenbush each stalk is a vertical chain of edges rooted in the ground, each edge coloured blue or red. Blue (Left) chops a blue edge on their turn; Red (Right) chops a red edge. Chopping an edge removes it and everything above it. A player with no legal move loses. Three stalks stand in a row; reading each from the ground upward, they are: blue-red-blue; red-blue-red-red; blue-blue-red. The value of a position is the number v such that, after adding a stalk worth -v to it, the second player wins however play begins; a lone blue edge has value 1, a lone red edge has value -1, and positive values favour Blue. Give the value of the whole position as a reduced fraction.
Worked solution
A stalk's value is computed from the ground: a run of n same-coloured edges is worth ±n, and every later edge adds ±1/2, ±1/4, ±1/8, ... in turn (blue positive, red negative). Blue-red-blue = 1 - 1/2 + 1/4 = 3/4. Red-blue-red-red = -1 + 1/2 - 1/4 - 1/8 = -7/8. Blue-blue-red = 2 - 1/2 = 3/2. Values of independent components add, so the position is worth 3/4 - 7/8 + 3/2 = 6/8 - 7/8 + 12/8 = 11/8. A direct game-tree search of the sum plus a stalk worth -11/8 confirms it is a second-player win for either starter.
Source: Hackenbush values are from Berlekamp, Conway and Guy, Winning Ways. Statement written for AxiomIQ.