AI运动100%喜欢C64上的MAFIA游戏

更新一位回答我的用户链接到C#解决方案项目：https://github.com/TedDawkin/C64_AI_Movement

更新

我发现了——如果有人感兴趣，只需使用C64论坛中的线程链接：http://www.lemon64.com/forum/viewtopic.php?p=712902#712902在那里我（和自己）讨论了这个话题。有趣的是，它最后是多么简单。一点也不好笑，我花了两个多星期的时间。

稍后我将发布BASIC到C#代码，并"回答"我自己的问题。

如果有人关心并且不想去论坛链接，下面是它在C64-basic中的工作原理。

30110 IF KS(S)=0 THEN GOSUB 30400 : GOTO30105
30400 POKE 211,20: POKE214,18: SYS CS: PRINT"SPIELER"F
30405 Y=INT(KP(S,F)/40)
  SYS CR, KP(S,F)-40*Y, Y
  X=PEEK(UA)-1
  Y=PEEK(UA+1)-40
30410 IF PEEK(UA+2)=1 OR PEEK (UA+3)=1 GOTO 30420
30415 IFINT(RND(1)*2)=0ORGW(KS(S),F)<4GOTO30450
30420 IFX=0THENX=Y:GOTO30215
30421 IFY<>0GOTO30450
30425 GOTO30215
30450 IFX<>0ANDRI(F)<>(1+2*(X=1))THENP=X:GOSUB30490:IFP=0THENRETURN
30451 IFY<>0ANDRI(F)<>(40+80*(Y=1))THENP=Y:GOSUB30490:IFP=0THENRETURN
30455 IFX<>0ANDRI(F)<>(1+2*(X=1))GOTO30460
30456 IFRI(F)<>-1THENP=1:GOSUB30490:IFP=0THENRETURN
30457 IFRI(F)<>1THENP=-1:GOSUB30490:IFP=0THENRETURN
30458 GOTO30465
30460 IFY<>0ANDRI(F)<>(40+80*(Y=1))GOTO30465
30461 IFRI(F)<>-40THENP=40:GOSUB30490:IFP=0THENRETURN
30462 IFRI(F)<>40THENP=-40:GOSUB30490:IFP=0THENRETURN
30465 RETURN
30490 Q=KP(S,F)+P:IFQ<0ORQ>520OR(PEEK(BR+Q)<>32ANDPEEK(BR+Q
<>96)THENRETURN
30491 POKEBR+KP(S,F),32:KP(S,F)=Q:POKEBR+KP(S,F),193:POKEFR+KP(S,F),6
30492 RI(F)=P:P=0:RETURN

一些提示：

X = moved right (1), left (-1), up (-40) or down(40)
P = Position. There is no Y because the next/prev line is
40 characters away. (C64 Screen =  40 Columns and 25 Rows)
S = switches between 0 and 2 to determine if its human or ai turn
KP(S,F) = Offset-Position in Video-Memory-Adress
BR = Start-Adress of Video-Memory
32 = 0x20 = Space to clear old position
193 = Ascii-Character used as pawn for player and ai
6 = Mark field as AI (Human position is marked with 0x0b)
F = Playernumber
RI(F) = dont know yet

原始问题：

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正在进行重制版并试图理解C64代码，我正在与人工智能运动作斗争。花了一个多星期的时间，我无法再现100%相同的行为。