Adversarial Monte Carlo Tree Search Asymmetry

Question

Monte Carlo Tree Search with UCT is praised for it's asymmetric tree growth, growing promising subtrees more than non-promising ones.

But in a 2-player adversarial game, when a win at one node is a loss at the node below it, wouldn't the tree growth be extremely favourable for the current player at each node, and result in a not so asymmetric tree growth?

score 1 · Answer 1 · edited Jan 12 '17 at 09:38

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The most explored path will be the Nash Equilibrium, because at each step we choose the best move for each player.
The asymmetry happens at each level towards the best move for the current player, thus overall the tree will not grow asymmetrically.

edited Jan 12 '17 at 09:38

Eli Stevens

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answered Mar 21 '16 at 13:22

NightRa

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score 0 · Answer 2 · answered Dec 22 '17 at 16:51

With each new rollout, considering games with many traps like chess, the random playouts converge super slowly if at all. Usually based on very high rollouts and on nodes with greater exploration.. Good game play for me happens around +100000 rollouts but dependant on the game. For one rollout you are correct, but with each rollout things get incrementally better. Backpropagation over many rollouts solve this in the UCB.

Adversarial Monte Carlo Tree Search Asymmetry

2 Answers2