Perform one value-iteration update for a Markov decision process. For every state, evaluate each action using its immediate reward and expected discounted next-state value, then keep the best action value.
Vnew(s)=amax[R(s,a)+γs′∑T(s,a,s′)V(s′)]Here:
Return the updated value of every state as a list of floats.
Input: values = [0, 0], transitions = [[[0.8, 0.2], [0.3, 0.7]], [[0.5, 0.5], [0.1, 0.9]]], rewards = [[1, 2], [-1, 0]], gamma = 0.9
Output: [2.0, 0.0]
Explanation: With zero current values, only immediate rewards contribute, so the best rewards are 2 and 0.
Input: values = [0, 0, 0], transitions = [[[0, 1, 0], [0, 0, 1]], [[1, 0, 0], [0, 0, 1]], [[0, 1, 0], [1, 0, 0]]], rewards = [[1, 2], [3, 0], [-1, 5]], gamma = 0.9
Output: [2.0, 3.0, 5.0]
sum(probability * value for probability, value in zip(action_transitions, values)) computes one expected next-state value.
Use max() over the action values computed for each state.
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Perform one value-iteration update for a Markov decision process. For every state, evaluate each action using its immediate reward and expected discounted next-state value, then keep the best action value.
Vnew(s)=amax[R(s,a)+γs′∑T(s,a,s′)V(s′)]Here:
Return the updated value of every state as a list of floats.
Input: values = [0, 0], transitions = [[[0.8, 0.2], [0.3, 0.7]], [[0.5, 0.5], [0.1, 0.9]]], rewards = [[1, 2], [-1, 0]], gamma = 0.9
Output: [2.0, 0.0]
Explanation: With zero current values, only immediate rewards contribute, so the best rewards are 2 and 0.
Input: values = [0, 0, 0], transitions = [[[0, 1, 0], [0, 0, 1]], [[1, 0, 0], [0, 0, 1]], [[0, 1, 0], [1, 0, 0]]], rewards = [[1, 2], [3, 0], [-1, 5]], gamma = 0.9
Output: [2.0, 3.0, 5.0]
sum(probability * value for probability, value in zip(action_transitions, values)) computes one expected next-state value.
Use max() over the action values computed for each state.
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