最终版的Dice of Doom
这个游戏终于结束了，基本上这本书的主要内容也算看完了。实际上，我根本就写不出这些东西，源码也读的云里雾里。但这是一本很有趣的书，我想看完htdp之后用htdp推荐风格重新处理下这本书的代码，顺便也复习一下。
让Dice of Doom 更加的有趣
我们上回完成的版本虽然很棒，但总缺乏些什么。因为之前程序的设计，我们可以很轻易的扩展它，让它变得更有趣
先加载我们之前的游戏^1
1 | (load "dice-v3"); |
增加玩家数
因为我们之前的架构，增加玩家数很简单。为了增加玩家数后计算机能反应敏捷，让它变“傻”一点。
1 | (defparameter *num-players* 4) (defparameter *die-colors* '((255 63 63) (63 63 255) (63 255 63) (255 63 255))) (defparameter *max-dice* 5) (defparameter *ai-level* 2) |
建立概率节点
我们本质上是向游戏树中加入几率节点，先升级attacking-moves函数，增加个board-attack-fail分支。
1 | (defun attacking-moves (board cur-player spare-dice) (labels ((player (pos) (car (aref board pos))) (dice (pos) (cadr (aref board pos)))) (lazy-mapcan (lambda (src) (if (eq (player src) cur-player) (lazy-mapcan (lambda (dst) (if (and (not (eq (player dst) cur-player)) (> (dice src) 1)) (make-lazy (list (list (list src dst) (game-tree (board-attack board cur-player src dst (dice src)) cur-player (+ spare-dice (dice dst)) nil) (game-tree (board-attack-fail board cur-player src dst (dice src)) cur-player (+ spare-dice (dice dst)) nil)))) (lazy-nil))) (make-lazy (neighbors src))) (lazy-nil))) (make-lazy (loop for n below *board-hexnum* collect n))))) |
board-attack-fail函数遍历所有位置。如果这个位置是攻击方就只能留下一个骰子
1 | (defun board-attack-fail (board player src dst dice) (board-array (loop for pos from 0 for hex across board collect (if (eq pos src) (list player 1) hex)))) |
让骰子滚动起来
一下程序让一堆骰子滚动起来，并判断谁赢了
1 | ;rolling the dice (defun roll-dice (dice-num) (let ((total (loop repeat dice-num sum (1+ (random 6))))) (fresh-line) (format t "On ~a dice rolled ~a. " dice-num total) total)) (defun roll-against (src-dice dst-dice) (> (roll-dice src-dice) (roll-dice dst-dice))) |
我们还需要在游戏引擎中调用让骰子滚动的代码,选出正确的子树
1 | (defun pick-chance-branch (board move) (labels ((dice (pos) (cadr (aref board pos)))) (let ((path (car move))) (if (or (null path) (roll-against (dice (car path)) (dice (cadr path)))) (cadr move) (caddr move))))) |
更新handle-computer函数让它调用选择几率分支的函数
1 | (defun handle-human (tree) (fresh-line) (princ "choose your move:") (let ((moves (caddr tree))) (labels ((print-moves (moves n) (unless (lazy-null moves) (let* ((move (lazy-car moves)) (action (car move))) (fresh-line) (format t "~a. " n) (if action (format t "~a -> ~a" (car action) (cadr action)) (princ "end turn"))) (print-moves (lazy-cdr moves) (1+ n))))) (print-moves moves 1)) (fresh-line) (pick-chance-branch (cadr tree) (lazy-nth (1- (read)) moves)))) |
同样我们可以处理handle-computer函数
1 | (defun handle-computer (tree) (let ((ratings (get-ratings (limit-tree-depth tree *ai-level*) (car tree)))) (pick-chance-branch (cadr tree) (lazy-nth (position (apply #'max ratings) ratings) (caddr tree))))) |
升级AI
但现在还有个问题，AI还没有考虑骰子的滚动，依然认为骰子多的一定获胜。因此需要更新AI。
根据概率论原理，我们可以得到这样一个矩阵来表征攻击成功的概率
| 防守方\攻击方 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|
| 1 | 0.84 | 0.97 | 1.0 | 1.0 |
| 2 | 0.44 | 0.78 | 0.94 | 0.99 |
| 3 | 0.15 | 0.45 | 0.74 | 0.91 |
| 4 | 0.04 | 0.19 | 0.46 | 0.72 |
| 5 | 0.01 | 0.06 | 0.22 | 0.46 |
现在把这个概率交给计算机，更新get-ratings函数,给评分加权
1 | (defun get-ratings (tree player) (let ((board (cadr tree))) (labels ((dice (pos) (cadr (aref board pos)))) (take-all (lazy-mapcar (lambda (move) (let ((path (car move))) (if path (let* ((src (car path)) (dst (cadr path)) (odds (aref (aref *dice-odds* (1- (dice dst))) (- (dice src) 2)))) (+ (* odds (rate-position (cadr move) player)) (* (- 1 odds) (rate-position (caddr move) player)))) (rate-position (cadr move) player)))) (caddr tree)))))) |
为了让修剪游戏树的函数能工作，还需要做些修改让它能处理新的游戏树。同时为了简化不再使用alpha-beta算法。
1 | (defun limit-tree-depth (tree depth) (list (car tree) (cadr tree) (if (zerop depth) (lazy-nil) (lazy-mapcar (lambda (move) (cons (car move) (mapcar (lambda (x) (limit-tree-depth x (1- depth))) (cdr move)))) (caddr tree))))) ;We mapcar across the tail of each move, so trimming is performed on ;both branches of any chance nodes. |
改进增援方案
之前是依据消灭敌人多少来决定增援骰子数。现在我们换一种方法：依据占据的最大领地来决定增援多少。这使游戏的地缘性更加复杂。
回想我们之前在wumpus游戏里写过的相似函数
1 | (defun get-connected (board player pos) (labels ((check-pos (pos visited) (if (and (eq (car (aref board pos)) player) (not (member pos visited))) (check-neighbors (neighbors pos) (cons pos visited)) visited)) (check-neighbors (lst visited) (if lst (check-neighbors (cdr lst) (check-pos (car lst) visited)) visited))) (check-pos pos '()))) |
获得最大的集群,最后添加援军，注意，spare-dice仍然作为参数传递但已经不再被add-new-dice使用了。
1 | (defun largest-cluster-size (board player) (labels ((f (pos visited best) (if (< pos *board-hexnum*) (if (and (eq (car (aref board pos)) player) (not (member pos visited))) (let* ((cluster (get-connected board player pos)) (size (length cluster))) (if (> size best) (f (1+ pos) (append cluster visited) size) (f (1+ pos) (append cluster visited) best))) (f (1+ pos) visited best)) best))) (f 0 '() 0))) (defun add-new-dice (board player spare-dice) (labels ((f (lst n) (cond ((zerop n) lst) ((null lst) nil) (t (let ((cur-player (caar lst)) (cur-dice (cadar lst))) (if (and (eq cur-player player) (< cur-dice *max-dice*)) (cons (list cur-player (1+ cur-dice)) (f (cdr lst) (1- n))) (cons (car lst) (f (cdr lst) n)))))))) (board-array (f (coerce board 'list) (largest-cluster-size board player))))) |
享受你的劳动成果吧
1 | (serve #'dod-request-handleer) |
疑问
我发现我可以选中任何自己的领地，只要上面的骰子数大于2。我可以攻击任何强大的敌人，而且一定能成功。不知道哪里的更改或错误！！