井字游戏 - Minimax
Tic Tac Toe - Minimax
我正在尝试使用带有 Rust 的极小极大算法构建一个井字游戏。我卡住了。我试图根据维基百科页面上的伪代码编写一个 Rust 代码。 https://en.wikipedia.org/wiki/Minimax。但是,它没有用。 Ai 总是先下手为强。如果你能帮助我,我会很高兴。
在main.rs
fn main() {
let mut g = Game::new();
while g.game_state() == Game_State::Continuous {
g.print();
println!("{}", minimax(&g));
if g.turn == Player::Cross {
g.take_input();
}
else {
g = best_move(&g);
}
}
g.print();
if let Game_State::Win(Player::None) = g.game_state() {
println!("Draw");
}
else {
g.print_winner();
}
}
在ai.rs
pub fn child_nodes(game: &Game) -> Vec<Game> {
let mut children: Vec<Game> = Vec::new();
for r in 0..3 {
for c in 0..3 {
if game.grid[r][c] == Player::None {
let mut child = game.clone();
child.grid[r][c] = game.turn;
child.turn = reverse_player(child.turn);
children.push(child);
}
}
}
return children;
}
pub fn minimax(game: &Game) -> isize {
match game.game_state() {
Game_State::Win(winner) => to_scor(winner),
Game_State::Continuous => {
use std::cmp::{min, max};
let children_vec = child_nodes(&game);
let mut score: isize;
if game.turn == Player::Cross {
score = -2;
for i in &children_vec {
score = max(score, minimax(i));
}
}
else {
score = 2;
for i in &children_vec {
score = min(score, minimax(i));
}
}
return score;
}
}
}
pub fn best_move(game: &Game) -> Game {
let children = child_nodes(game);
let mut values: Vec<isize> = Vec::new();
for i in 0..children.len() {
values.push(minimax(&children[i]));
}
let mut index: usize = 0;
let iter = values.iter().enumerate();
if game.turn == Player::Cross {
if let Option::Some(t) = iter.max() {
index = t.0;
}
}
else if game.turn == Player::Circle {
if let Option::Some(t) = iter.min() {
index = t.0;
}
}
let best_pos = children[index];
best_pos
}
pub fn to_scor(x: Player) -> isize {
match x {
Player::Cross => 1,
Player::Circle => -1,
Player::None => 0
}
}
.enumerate()
returns 元组迭代器,.max()
和 .min()
元组迭代器将比较元组 - 即 (1, x)
对于 x
和 y
的任何值,始终被认为小于 (2, y)
。这可以用这个片段来证明:
fn main() {
let v = vec![3, 1, 2, 5, 3, 6, 7, 2];
println!("{:?}", v.iter().enumerate().min());
println!("{:?}", v.iter().enumerate().max());
}
打印:
Some((0, 3))
Some((7, 2))
它们只是列表的第一个和最后一个元素(而不是最小或最大元素)。
但是,如图,您可以max_by
使用您自己的函数来比较元组。
我正在尝试使用带有 Rust 的极小极大算法构建一个井字游戏。我卡住了。我试图根据维基百科页面上的伪代码编写一个 Rust 代码。 https://en.wikipedia.org/wiki/Minimax。但是,它没有用。 Ai 总是先下手为强。如果你能帮助我,我会很高兴。
在main.rs
fn main() {
let mut g = Game::new();
while g.game_state() == Game_State::Continuous {
g.print();
println!("{}", minimax(&g));
if g.turn == Player::Cross {
g.take_input();
}
else {
g = best_move(&g);
}
}
g.print();
if let Game_State::Win(Player::None) = g.game_state() {
println!("Draw");
}
else {
g.print_winner();
}
}
在ai.rs
pub fn child_nodes(game: &Game) -> Vec<Game> {
let mut children: Vec<Game> = Vec::new();
for r in 0..3 {
for c in 0..3 {
if game.grid[r][c] == Player::None {
let mut child = game.clone();
child.grid[r][c] = game.turn;
child.turn = reverse_player(child.turn);
children.push(child);
}
}
}
return children;
}
pub fn minimax(game: &Game) -> isize {
match game.game_state() {
Game_State::Win(winner) => to_scor(winner),
Game_State::Continuous => {
use std::cmp::{min, max};
let children_vec = child_nodes(&game);
let mut score: isize;
if game.turn == Player::Cross {
score = -2;
for i in &children_vec {
score = max(score, minimax(i));
}
}
else {
score = 2;
for i in &children_vec {
score = min(score, minimax(i));
}
}
return score;
}
}
}
pub fn best_move(game: &Game) -> Game {
let children = child_nodes(game);
let mut values: Vec<isize> = Vec::new();
for i in 0..children.len() {
values.push(minimax(&children[i]));
}
let mut index: usize = 0;
let iter = values.iter().enumerate();
if game.turn == Player::Cross {
if let Option::Some(t) = iter.max() {
index = t.0;
}
}
else if game.turn == Player::Circle {
if let Option::Some(t) = iter.min() {
index = t.0;
}
}
let best_pos = children[index];
best_pos
}
pub fn to_scor(x: Player) -> isize {
match x {
Player::Cross => 1,
Player::Circle => -1,
Player::None => 0
}
}
.enumerate()
returns 元组迭代器,.max()
和 .min()
元组迭代器将比较元组 - 即 (1, x)
对于 x
和 y
的任何值,始终被认为小于 (2, y)
。这可以用这个片段来证明:
fn main() {
let v = vec![3, 1, 2, 5, 3, 6, 7, 2];
println!("{:?}", v.iter().enumerate().min());
println!("{:?}", v.iter().enumerate().max());
}
打印:
Some((0, 3))
Some((7, 2))
它们只是列表的第一个和最后一个元素(而不是最小或最大元素)。
但是,如图max_by
使用您自己的函数来比较元组。