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我在學習案例中遇到問題。 我對gridworld模型的強化學習感興趣。 模型是運動的7x7領域的迷宮。 考慮一個領域的迷宮。有四個方向：上，下，左和右（或N，E，S，W）。所以最多的政策是。在碰撞牆上使用直接懲罰時，許多人可以被排除在外。 另外採用抑制回報原則，通常更少的行爲是可以接受的。許多政策僅在目標之後的部分或者具有同等效力。如何在R程序中獲得SARSA代碼爲gridworld模型？
▼國家：障礙物 ▼獎勵：如果r = 1，如果S = G，否則r = 0的任何容許的舉動和else R = -100 ▼初始化：Q0（A，S）〜N（0 ，0.01）
要解決這個模型，我做了一個R代碼，但它不能正常工作。
模式：7x7的，S：啓動狀態，G：終端狀態，O：存取狀態，X：牆壁
[O,O,G,X,O,O,S]
[O,X,O,X,O,X,X]
[O,X,O,X,O,O,O]
[O,X,O,X,O,X,O]
[O,X,O,O,O,X,O]
[O,X,O,X,O,X,O]
[O,O,O,X,O,O,O]
所以我想知道如何能爲這個gridworld模型正確的代碼（不uppon代碼）並且想知道如何通過SARSA模型來解決這個模型。
actions <- c("N", "S", "E", "W")
x <- 1:7
y <- 1:7
rewards <- matrix(rep(0, 49), nrow=7)
rewards[1, 1] <- 0
rewards[1, 2] <- 0
rewards[1, 3] <- 1
rewards[1, 4] <- -100
rewards[1, 5] <- 0
rewards[1, 6] <- 0
rewards[1, 7] <- 0
rewards[2, 1] <- 0
rewards[2, 2] <- -100
rewards[2, 3] <- 0
rewards[2, 4] <- -100
rewards[2, 5] <- 0
rewards[2, 6] <- -100
rewards[2, 7] <- -100
rewards[3, 1] <- 0
rewards[3, 2] <- -100
rewards[3, 3] <- 0
rewards[3, 4] <- -100
rewards[3, 5] <- 0
rewards[3, 6] <- 0
rewards[3, 7] <- 0
rewards[4, 1] <- 0
rewards[4, 2] <- -100
rewards[4, 3] <- 0
rewards[4, 4] <- -100
rewards[4, 5] <- 0
rewards[4, 6] <- -100
rewards[4, 7] <- 0
rewards[5, 1] <- 0
rewards[5, 2] <- -100
rewards[5, 3] <- 0
rewards[5, 4] <- 0
rewards[5, 5] <- 0
rewards[5, 6] <- -100
rewards[5, 7] <- 0
rewards[6, 1] <- 0
rewards[6, 2] <- -100
rewards[6, 3] <- 0
rewards[6, 4] <- -100
rewards[6, 5] <- 0
rewards[6, 6] <- -100
rewards[6, 7] <- 0
rewards[7, 1] <- 0
rewards[7, 2] <- 0
rewards[7, 3] <- 0
rewards[7, 4] <- -100
rewards[7, 5] <- 0
rewards[7, 6] <- 0
rewards[7, 7] <- 0
values <- rewards # initial values
states <- expand.grid(x=x, y=y)
# Transition probability
transition <- list("N" = c("N" = 0.8, "S" = 0, "E" = 0.1, "W" = 0.1),
"S"= c("S" = 0.8, "N" = 0, "E" = 0.1, "W" = 0.1),
"E"= c("E" = 0.8, "W" = 0, "S" = 0.1, "N" = 0.1),
"W"= c("W" = 0.8, "E" = 0, "S" = 0.1, "N" = 0.1))
# The value of an action (e.g. move north means y + 1)
action.values <- list("N" = c("x" = 0, "y" = 1),
"S" = c("x" = 0, "y" = -1),
"E" = c("x" = 1, "y" = 0),
"W" = c("x" = -1, "y" = 0))
# act() function serves to move the robot through states based on an action
act <- function(action, state) {
action.value <- action.values[[action]]
new.state <- state
if(state["x"] == 1 && state["y"] == 7 || (state["x"] == 1 && state["y"] == 3))
return(state)
#
new.x = state["x"] + action.value["x"]
new.y = state["y"] + action.value["y"]
# Constrained by edge of grid
new.state["x"] <- min(x[length(x)], max(x[1], new.x))
new.state["y"] <- min(y[length(y)], max(y[1], new.y))
#
if(is.na(rewards[new.state["y"], new.state["x"]]))
new.state <- state
#
return(new.state)
}
rewards
bellman.update <- function(action, state, values, gamma=1) {
state.transition.prob <- transition[[action]]
q <- rep(0, length(state.transition.prob))
for(i in 1:length(state.transition.prob)) {
new.state <- act(names(state.transition.prob)[i], state)
q[i] <- (state.transition.prob[i] * (rewards[state["y"], state["x"]] + (gamma * values[new.state["y"], new.state["x"]])))
}
sum(q)
}
value.iteration <- function(states, actions, rewards, values, gamma, niter, n) {
for (j in 1:niter) {
for (i in 1:nrow(states)) {
state <- unlist(states[i,])
if(i %in% c(7, 15)) next # terminal states
q.values <- as.numeric(lapply(actions, bellman.update, state=state, values=values, gamma=gamma))
values[state["y"], state["x"]] <- max(q.values)
}
}
return(values)
}
final.values <- value.iteration(states=states, actions=actions, rewards=rewards, values=values, gamma=0.99, niter=100, n=10)
final.values
