42

Post your shortest code, by character count, to check if a player has won, and if so, which.

Assume you have an integer array in a variable b (board), which holds the Tic Tac Toe board, and the moves of the players where:

  • 0 = nothing set
  • 1 = player 1 (X)
  • 2 = player 2 (O)

So, given the array b = [ 1, 2, 1, 0, 1, 2, 1, 0, 2 ] would represent the board

X|O|X
-+-+-
 |X|O
-+-+-
X| |O

For that situation, your code should output 1 to indicate player 1 has won. If no-one has won you can output 0 or false.

My own (Ruby) solution will be up soon.

Edit: Sorry, forgot to mark it as community wiki. You can assume the input is well formed and does not have to be error checked.

Update: Please post your solution in the form of a function. Most people have done this already, but some haven't, which isn't entirely fair. The board is supplied to your function as the parameter. The result should be returned by the function. The function can have a name of your choosing.

4

35 に答える 35

37

Crazy Python solution - 79 characters

max([b[x] for x in range(9) for y in range(x) for z in range(y)
    if x+y+z==12 and b[x]==b[y]==b[z]] + [0])

However, this assumes a different order for the board positions in b:

 5 | 0 | 7
---+---+---
 6 | 4 | 2
---+---+---
 1 | 8 | 3

That is, b[5] represents the top-left corner, and so on.

To minimize the above:

r=range
max([b[x]for x in r(9)for y in r(x)for z in r(y)if x+y+z==12and b[x]==b[y]==b[z]]+[0])

93 characters and a newline.

Update: Down to 79 characters and a newline using the bitwise AND trick:

r=range
max([b[x]&b[y]&b[z]for x in r(9)for y in r(x)for z in r(y)if x+y+z==12])
于 2010-02-11T16:44:54.877 に答える
22

C, 77 (83) characters

This is a variant of dmckee's solution, except that each pair of digits in the Compact Coding is now the base-9 digits of the ASCII characters.

The 77-char version, does not work on MSVC:

// "J)9\t8\r=,\0" == 82,45,63,10,62,14,67,48,00 in base 9.
char*k="J)9 8\r=,",c;f(int*b){return(c=*k++)?b[c/9]&b[c%9]&b[*k--%9]|f(b):0;}

This 83-char version, should work on every C compiler:

f(int*b){char*k="J)9    8\r=,",s=0,c;while(c=*k++)s|=b[c%9]&b[c/9]&b[*k%9];return s;}

(Note that the spaces between the 9 and 8 should be a tab. StackOverflow converts all tabs into spaces.)

Test case:

#include <stdio.h>  
void check(int* b) {
    int h0 = b[0]&b[1]&b[2];
    int h1 = b[3]&b[4]&b[5];
    int h2 = b[6]&b[7]&b[8];
    int h3 = b[0]&b[3]&b[6];
    int h4 = b[1]&b[4]&b[7];
    int h5 = b[2]&b[5]&b[8];
    int h6 = b[0]&b[4]&b[8];
    int h7 = b[2]&b[4]&b[6];
    int res = h0|h1|h2|h3|h4|h5|h6|h7;
    int value = f(b);
    if (value != res)
        printf("Assuming f({%d,%d,%d, %d,%d,%d, %d,%d,%d}) == %d; got %d instead.\n", 
            b[0],b[1],b[2], b[3],b[4],b[5], b[6],b[7],b[8], res, value);
}
#define MAKEFOR(i) for(b[(i)]=0;b[(i)]<=2;++b[(i)])

int main() {
    int b[9];

    MAKEFOR(0)
    MAKEFOR(1)
    MAKEFOR(2)
    MAKEFOR(3)
    MAKEFOR(4)
    MAKEFOR(5)
    MAKEFOR(6)
    MAKEFOR(7)
    MAKEFOR(8)
        check(b);

    return 0;
}
于 2010-02-11T19:29:04.420 に答える
12

Python 80 (69) char

Not the shortest Python solution, but I like how it introduces "DICE" into a game of tic-tac-toe:

W=lambda b:max([b[c/5-9]&b[c/5+c%5-9]&b[c/5-c%5-9]for c in map(ord,"DICE>3BQ")])

69 chars for the simpler expression:

max([b[c/5-9]&b[c/5+c%5-9]&b[c/5-c%5-9]for c in map(ord,"DICE>3BQ")])
于 2010-02-12T16:29:37.457 に答える
10

Perl、8785文字

もちろん、正規表現を使用して0、1、または2を返す関数（スクロールバーを回避するために改行が存在するだけです）。

sub V{$"='';$x='(1|2)';"@_"=~
/^(...)*$x\2\2|^..$x.\3.\3|$x..\4..\4|$x...\5...\5/?$^N:0}

V(@b)たとえば、と呼ぶことができます。

于 2010-02-12T00:45:55.653 に答える
10

J, 50 chars

w=:3 : '{.>:I.+./"1*./"1]1 2=/y{~2 4 6,0 4 8,i,|:i=.i.3 3'
于 2010-02-15T10:08:26.853 に答える
9

I'm not happy with repeating myself (horizontal/vertical, and the diagonals), but I think it's a fair start.

C# w/LINQ:

public static int GetVictor(int[] b)
{
    var r = Enumerable.Range(0, 3);
    return r.Select(i => r.Aggregate(3, (s, j) => s & b[i * 3 + j])).Concat(
        r.Select(i => r.Aggregate(3, (s, j) => s & b[j * 3 + i]))).Aggregate(
        r.Aggregate(3, (s, i) => s & b[i * 3 + i]) | r.Aggregate(3, (s, i) => s & b[i * 3 + (2 - i)]),
        (s, i) => s | i);
}

Strategy: Bitwise AND each element of a row/column/diagonal with the other elements (with 3 as a seed) to obtain a victor for that subset, and OR them all together at the end.

于 2010-02-11T17:07:45.343 に答える
8

Ruby, 115 chars

Oops: Somehow I miscounted by a lot. This is actually 115 characters, not 79.

def t(b)[1,2].find{|p|[448,56,7,292,146,73,273,84].any?{|k|(k^b.inject(0){|m,i|m*2+((i==p)?1:0)})&k==0}}||false end

# Usage:
b = [ 1, 2, 1,
      0, 1, 2,
      1, 0, 2 ]
t(b) # => 1

b = [ 1, 1, 0,
      2, 2, 2,
      0, 2, 1 ]
t(b) # => 2

b = [ 0, 0, 1,
      2, 2, 0,
      0, 1, 1 ]
t(b) # => false

And the expanded code, for educational purposes:

def tic(board)
  # all the winning board positions for a player as bitmasks
  wins = [ 0b111_000_000,  # 448
           0b000_111_000,  #  56
           0b000_000_111,  #   7
           0b100_100_100,  # 292
           0b010_010_010,  # 146
           0b001_001_001,  #  73
           0b100_010_001,  # 273
           0b001_010_100 ] #  84

  [1, 2].find do |player| # find the player who's won
    # for the winning player, one of the win positions will be true for :
    wins.any? do |win|
      # make a bitmask from the current player's moves
      moves = board.inject(0) { |acc, square|
        # shift it to the left and add one if this square matches the player number
        (acc * 2) + ((square == player) ? 1 : 0)
      }
      # some logic evaluates to 0 if the moves match the win mask
      (win ^ moves) & win == 0
    end
  end || false # return false if the find returns nil (no winner)
end

I'm sure this could be shortened, especially the big array and possibly the code for getting a bitmask of the players's moves--that ternary bugs me--but I think this is pretty good for now.

于 2010-02-11T22:25:11.537 に答える
4

Perl, 76 char

sub W{$n=$u=0;map{$n++;$u|=$_[$_-$n]&$_[$_]&$_[$_+$n]for/./g}147,4,345,4;$u}

There are three ways to win horizontally:

0,1,2   ==>   1-1, 1, 1+1
3,4,5   ==>   4-1, 4, 4+1
6,7,8   ==>   7-1, 7, 7+1

One way to win diagonally from lower left to upper right:

2,4,6   ==>   4-2, 4, 4+2

Three ways to win vertically:

0,3,6   ==>   3-3, 3, 3+3
1,4,7   ==>   4-3, 4, 4+3
2,5,8   ==>   5-3, 5, 5+3

One way to win diagonally from upper left to lower right:

0,4,8   ==>   4-4, 4, 4+4

Read the middle columns to get the magic numbers.

于 2010-02-12T16:36:29.267 に答える
4

Octave/Matlab, 97 characters, including spaces and newlines. Outputs 0 if no winner, 1 if player 1 won, 2 if player 2 won, and 2.0801 if both players "won":

function r=d(b)
a=reshape(b,3,3)
s=prod([diag(a) diag(fliplr(a)) a a'])
r=sum(s(s==1|s==8))^(1/3)

If we change the specification and pass in b as a 3x3 matrix from the start, we can remove the reshape line, getting it down to 80 characters.

于 2010-02-12T23:32:08.930 に答える
3

because nobody wins at tictactoe when properly played i think this is the shortest code

echo 0;

7 chars

Update: A better entry for bash would be this:

86 characters or 81 excluding function definition(win()).

win()for q in 1 28 55 3 12 21 4 20;{ [[ 3*w -eq B[f=q/8]+B[g=q%8]+B[g+g-f] ]]&&break;}

But, This is code from by tic-tac-toe program in bash so it does not quite meet specification.

# player is passed in caller's w variable. I use O=0 and X=2 and empty=8 or 9
# if a winner is found, last result is true (and loop halts) else false
# since biggest test position is 7 I'll use base 8. could use 9 as well but 10 adds 2 characters to code length
# test cases are integers made from first 2 positions of each row
# eg. first row (0 1 2) is 0*8+1 = 1
# eg. diagonal (2 4 6) is 2*8+4 = 20
# to convert test cases to board positions use X/8, X%8, and X%8+(X%8-X/8)
# for each test case, test that sum of each tuplet is 3*player value
于 2010-02-11T21:14:09.063 に答える
2

Haskell, Assuming the magic squares above. 77 Characters

77 excludes imports and defining b.

import Data.Bits
import Data.Array

b = listArray (0,8) [2,1,0,1,1,1,2,2,0]
w b = maximum[b!x.&.b!y.&.b!z|x<-[0..8],y<-[x+1..8],z<-[12-x-y],z<8,z>=0,z/=y]

Or 82 assuming the normal ordering:

{-# LANGUAGE NoMonomorphismRestriction #-}
import Data.Bits
import Data.Array

b = listArray (0,8) [1,2,1,0,1,2,1,0,2]
w b = maximum[b!x.&.b!y.&.b!z|x<-[0..8],d<-[1..4],y<-[x+d],z<-[y+d],d/=2||x==2,z<9]
于 2010-02-11T22:01:47.653 に答える
2

(Iron)python, 75 characters

75 characters for a full function

T=lambda a:max(a[b/6]&a[b/6+b%6]&a[b/6+b%6*2]for b in[1,3,4,9,14,15,19,37])

66 characters if you leave out the function definition like some others have done

r=max(a[b/6]&a[b/6+b%6]&a[b/6+b%6*2]for b in[1,3,4,9,14,15,19,37])

The 8 different directions are represented by starting value + incrementor, compressed into a single number that can be extracted using division and modula. For example 2,5,8 = 2*6 + 3 = 15.

Checking that a row contains three equal values is done using the & operator. (which results in zero if they aren't equal). max is used to find the possible winner.

于 2010-02-12T13:19:42.987 に答える
2

Ruby, 85 char

def X(b)
u=0
[2,6,7,8,9,13,21,-9].each do|c|u|=b[n=c/5+3]&b[n+c%5]&b[n-c%5]end
u
end

If the input has both players winning, e.g.

     X | O | X
    ---+---+---
     X | O | O
    ---+---+---
     X | O | X

then the output is 3.

于 2010-02-12T16:25:42.037 に答える
2

C, 99 chars

Not a winner, but maybe there's room for improvement. Never did this before. Original concept, first draft.

#define l w|=*b&b[s]&b[2*s];b+=3/s;s
f(int*b){int s=4,w=0;l=3;l;l;l=2;--b;l=1;b-=3;l;l;return l;}

Thanks to KennyTM for a few ideas and the test harness.

The "development version":

#define l w|=*b&b[s]&b[2*s];b+=3/s;s // check one possible win
f( int *b ) {
        int s=4,w=0; // s = stride, w = winner
        l=3;     // check stride 4 and set to 3
        l;l;l=2; // check stride 3, set to 2
        --b;l=1; // check stride 2, set to 1
        b-=3;l;l; return l; // check stride 1
}
于 2010-02-13T02:51:29.557 に答える
1

Python-75文字（64）

それぞれ64文字の2つの式を考え出しました。

max(a[c/8]&a[c/8+c%8]&a[c/8-c%8]for c in map(ord,'\t\33$#"!+9'))

と

max(a[c/5]&a[c/5+c%5]&a[c/5+c%5*2]for c in[1,3,4,8,12,13,16,31])

「W=lambdab：」を追加して関数にすると、75文字になります。これまでで最短のPython？

于 2010-05-18T08:54:35.533 に答える
1

ルア、130文字

130文字は関数サイズのみです。一致するものが見つからない場合、関数は何も返しません。これは、Luaではfalseを返すのと似ています。

function f(t)z={7,1,4,1,1,3,2,3,3}for b=1,#z-1 do
i=z[b]x=t[i]n=z[b+1]if 0<x and x==t[i+n]and x==t[i+n+n]then
return x end end end

assert(f{1,2,1,0,1,2,1,0,2}==1)
assert(f{1,2,1,0,0,2,1,0,2}==nil)
assert(f{1,1,2,0,1,2,1,0,2}==2)
assert(f{2,1,2,1,2,1,2,1,2}==2)
assert(f{2,1,2,1,0,2,2,2,1}==nil)
assert(f{1,2,0,1,2,0,1,2,0}~=nil)
assert(f{0,2,0,0,2,0,0,2,0}==2)
assert(f{0,2,2,0,0,0,0,2,0}==nil)

assert(f{0,0,0,0,0,0,0,0,0}==nil)
assert(f{1,1,1,0,0,0,0,0,0}==1)
assert(f{0,0,0,1,1,1,0,0,0}==1)
assert(f{0,0,0,0,0,0,1,1,1}==1)
assert(f{1,0,0,1,0,0,1,0,0}==1)
assert(f{0,1,0,0,1,0,0,1,0}==1)
assert(f{0,0,1,0,0,1,0,0,1}==1)
assert(f{1,0,0,0,1,0,0,0,1}==1)
assert(f{0,0,1,0,1,0,1,0,0}==1)
于 2010-02-12T00:44:09.403 に答える
1

A solution in C (162 Characters):

This makes use of the fact that player one value (1) and player two value (2) have independent bits set. Therefore, you can bitwise AND the values of the three test boxes together-- if the value is nonzero, then all three values must be identical. In addition, the resulting value == the player that won.

Not the shortest solution so far, but the best I could do:

void fn(){
    int L[]={1,0,1,3,1,6,3,0,3,1,3,2,4,0,2,2,0};
    int s,t,p,j,i=0;
    while (s=L[i++]){
        p=L[i++],t=3;
        for(j=0;j<3;p+=s,j++)t&=b[p];
        if(t)putc(t+'0',stdout);}
}

A more readable version:

void fn2(void)
{
    // Lines[] defines the 8 lines that must be tested
    //  The first value is the "Skip Count" for forming the line
    //  The second value is the starting position for the line
    int Lines[] = { 1,0, 1,3, 1,6, 3,0, 3,1, 3,2, 4,0, 2,2, 0 };

    int Skip, Test, Pos, j, i = 0;
    while (Skip = Lines[i++])
    {
        Pos = Lines[i++];   // get starting position
        Test = 3;           // pre-set to 0x03 (player 1 & 2 values bitwise OR'd together)

        // search each of the three boxes in this line
        for (j = 0; j < 3; Pos+= Skip, j++)
        {
            // Bitwise AND the square with the previous value
            //  We make use of the fact that player 1 is 0x01 and 2 is 0x02
            //  Therefore, if any bits are set in the result, it must be all 1's or all 2's
            Test &= b[Pos];
        }

        // All three squares same (and non-zero)?
        if (Test)
            putc(Test+'0',stdout);
    }
}
于 2010-02-11T17:34:34.637 に答える
1

Python, 102 characters

Since you didn't really specify how to get input and output, this is the "raw" version that would perhaps have to be wrapped into a function. b is the input list; r is the output (0, 1 or 2).

r=0
for a,c in zip("03601202","11133342"):s=set(b[int(a):9:int(c)][:3]);q=s.pop();r=r if s or r else q
于 2010-02-11T21:15:23.780 に答える
1

Visual Basic 275 254 (with loose typing) characters

 Function W(ByVal b())

    Dim r

    For p = 1 To 2

            If b(0) = b(1) = b(2) = p Then r = p
            If b(3) = b(4) = b(5) = p Then r = p
            If b(6) = b(7) = b(8) = p Then r = p
            If b(0) = b(3) = b(6) = p Then r = p
            If b(1) = b(4) = b(7) = p Then r = p
            If b(2) = b(5) = b(8) = p Then r = p
            If b(0) = b(4) = b(8) = p Then r = p
            If b(6) = b(4) = b(2) = p Then r = p

    Next

    Return r

End Function
于 2010-02-12T15:49:35.587 に答える
1

JavaScript - function "w" below is 114 characters

<html>   
<body>
<script type="text/javascript">

var t = [0,0,2,0,2,0,2,0,0];

function w(b){
    i = '012345678036147258048642';
    for (l=0;l<=21;l+=3){
        v = b[i[l]];
        if (v == b[i[l+1]]) if (v == b[i[l+2]]) return v;   
    }
}

alert(w(t));

</script>
</body>
</html>
于 2010-02-12T16:29:38.923 に答える
1

J, 97 characters.

1+1 i.~,+./"2>>(0 4 8,2 4 6,(],|:)3 3$i.9)&(e.~)&.>&.>(]<@:#"1~[:#:[:i.2^#)&.>(I.@(1&=);I.@(2&=))

I was planning to post an explanation of how this works, but that was yesterday and now I can't read this code.

The idea is we create a list of all possible winning triples (048,246,012,345,678,036,147,258), then make the powerset of the squares each player has and then intersect the two lists. If there's a match, that's the winner.

于 2010-04-15T18:57:23.007 に答える
1

Python, 285 bytes

b,p,q,r=["."]*9,"1","2",range
while"."in b:
 w=[b[i*3:i*3+3]for i in r(3)]+[b[i::3]for i in r(3)]+[b[::4],b[2:8:2]]
 for i in w[:3]:print i
 if["o"]*3 in w or["x"]*3 in w:exit(q)
 while 1:
  m=map(lambda x:x%3-x+x%3+7,r(9)).index(input())
  if"."==b[m]:b[m]=".xo"[int(p)];p,q=q,p;break

...Oh, this wasn't what you meant when you said "Code Golf: Tic Tac Toe"? ;) (enter numpad numbers to place x's or o's, i.e. 7 is north-west)

Long Version

board = ["."]*9   # the board
currentname = "1" # the current player
othername = "2"   # the other player

numpad_dict = {7:0, 8:1, 9:2, # the lambda function really does this!
               4:3, 5:4, 6:5,
               1:6, 2:7, 3:8}

while "." in board:
    # Create an array of possible wins: horizontal, vertical, diagonal
    wins = [board[i*3:i*3+3] for i in range(3)] + \ # horizontal
           [board[i::3]      for i in range(3)] + \ # vertical
           [board[::4], board[2:8:2]]               # diagonal

    for i in wins[:3]: # wins contains the horizontals first,
        print i        # so we use it to print the current board

    if ["o"]*3 in wins or ["x"]*3 in wins: # somebody won!
        exit(othername)                    # print the name of the winner
                                           # (we changed player), and exit
    while True: # wait for the player to make a valid move
        position = numpad_dict[input()] 
        if board[position] == ".": # still empty -> change board
            if currentname == "1":
                board[position] = "x"
            else:
                board[position] = "o"
            currentname, othername = othername, currentname # swap values
于 2010-06-20T00:26:05.240 に答える
0

Common Lisp、171文字

ゴルフモード：

(defun x(b)(find-if-not 'null(mapcar(lambda(r)(let((v(mapcar(lambda(c)(elt b c))r)))(if(apply '= v)(car v))))'((0 1 2)(3 4 5)(6 7 8)(0 3 6)(1 4 7)(2 5 8)(0 4 8)(2 4 6)))))

読み取り可能モード：

(defun ttt-winner (board)
  (find-if-not 'null
          (mapcar (lambda (row)
                    (let ((vals (mapcar (lambda (cell) (elt board cell)) row)))
                      (if (apply '= vals) (car vals))))
              '((0 1 2) (3 4 5) (6 7 8) (0 3 6) (1 4 7) (2 5 8) (0 4 8) (2 4 6)))))
于 2010-04-16T07:45:56.347 に答える
0

LinQ 236

関数宣言なしでC＃ではおそらく少なくなる可能性があります;）

Function P(ByVal b())
    Dim s() = "012.048.036.147.258.345.678".Split(".")
    If (From x In s Where b(Val(x(0))) & b(Val(x(1))) & b(Val(x(2))) = "111").Any Then Return 1
    If (From x In s Where b(Val(x(0))) & b(Val(x(1))) & b(Val(x(2))) = "222").Any Then Return 2
    Return 0
End Function
于 2010-03-31T19:00:48.570 に答える
0

C＃、148だと思います。

 int[] m={0,1,3,1,6,1,0,3,1,3,2,3,0,4,2,2};int i,s,w,r=0,o;for(i=0;i<16;i+=2){s=m[i];w=m[i+1];o=v[s];if((o==v[w+s])&&(o==v[s+(w*2)])){r=o;}}return r;
于 2010-02-13T00:28:31.043 に答える
0

I'm sure there's a shorter way to do this but... Perl, 141 characters (134 inside the function)

sub t{$r=0;@b=@_;@w=map{[split//]}split/,/,"012,345,678,036,147,258,048,246";for(@w){@z=map{$b[$_]}@$_;$r=$z[0]if!grep{!$_||$_!=$z[0]}@z;}$r;}
于 2010-02-11T16:47:51.317 に答える
0

Ruby, 149 characters

def s(b)(0..8).to_a+[0,3,6,1,4,7,2,5,8,0,4,8,2,4,6].each_slice(3){|m|if b.values_at(*m).uniq.length<2&&b[m[0]]!=0;return b[m[0]];end}return false;end

It's a reasonably straightforward solution, I'm sure I'll be able to reduce it some more. Here is a readable version:

def someone_won(b)
    helper = (0..8).to_a + [ 0, 3, 6, 1, 4, 7, 2, 5, 8, 0, 4, 8, 2, 4, 6]
    helper.each_slice(3) { |m|
        if b.values_at(*m).uniq.length < 2 && b[m[0]] != 0
            return b[m[0]]
        end
    }

    return false
end
于 2010-02-11T17:21:45.403 に答える
0

C#, 154 163 170 177 characters

Borrowing a couple of techniques from other submissions. (didn't know C# let you init arrays like that)

static int V(int[] b)
{
   int[] a={0,1,3,1,6,1,0,3,1,3,2,3,0,4,2,2};
   int r=0,i=-2;
   while((i+=2)<16&&(r|=b[a[i]]&b[a[i]+a[i+1]]&b[a[i]+a[i+1]*2])==0){}
   return r;
}
于 2010-02-11T18:37:34.783 に答える
0

c -- 144 characters

Minified:

#define A(x) a[b[x%16]]
int c,b[]={4,8,0,1,2,4,6,0,3,4,5,2,8,6,7,2};int
T(int*a){for(c=0;c<16;c+=2)if(A(c)&A(c+1)&A(c+2))return A(c);return 0;}

Both returns count (one necessary and the other would need replacing with a space).

The array codes for the eight ways to win in triplets starting from even positions and taken mod 16.

Bitwise and trick stolen from Eric Pi.

More readable form:

#define A(x) a[b[x%16]]

// Compact coding of the ways to win.
//
// Each possible was starts a position N*2 and runs through N*2+2 all
// taken mod 16
int c,b[]={4,8,0,1,2,4,6,0,3,4,5,2,8,6,7,2};

int T(int*a){
  // Loop over the ways to win
  for(c=0;c<16;c+=2)
    // Test for a win
    if(A(c)&A(c+1)&A(c+2))return A(c);
  return 0;
}

Testing scaffold:

#include <stdlib.h>
#include <stdio.h>

int T(int*);

int main(int argc, char**argv){
  int input[9]={0};
  int i, j;
  for (i=1; i<argc; ++i){
    input[i-1] = atoi(argv[i]);
  };
  for (i=0;i<3;++i){
    printf("%1i  %1i  %1i\n",input[3*i+0],input[3*i+1],input[3*i+2]);
  };
  if (i = T(input)){
    printf("%c wins!\n",(i==1)?'X':'O');
  } else {
    printf("No winner.\n");
  }
  return 0;
}
于 2010-02-11T19:22:30.893 に答える
0

Probably could be made better, but I'm not feeling particularly clever right now. This is just to make sure Haskell gets represented...

Assuming that b already exists, this will put the result in w.

import List
a l=2*minimum l-maximum l
z=take 3$unfoldr(Just .splitAt 3)b
w=maximum$0:map a(z++transpose z++[map(b!!)[0,4,8],map(b!!)[2,4,6]])

Assuming input from stdin and output to stdout,

import List
a l=2*minimum l-maximum l
w b=maximum$0:map a(z++transpose z++[map(b!!)[0,4,8],map(b!!)[2,4,6]])where
 z=take 3$unfoldr(Just .splitAt 3)b
main=interact$show.w.read
于 2010-02-11T20:12:58.267 に答える
0

C#, 180 characters :

var s=new[]{0,0,0,1,2,2,3,6};
var t=new[]{1,3,4,3,2,3,1,1};
return(s.Select((p,i)=>new[]{g[p],g[p+t[i]],g[p+2*t[i]]}).FirstOrDefault(l=>l.Distinct().Count()==1)??new[]{0}).First();

(g being the grid)

Could probably be improved... I'm still working on it ;)

于 2010-02-12T00:06:40.927 に答える
0

Python, 140 chars

My first code golf, weighing in at a hefty 140 chars (import statement, I deny you!):

import operator as o

def c(t):return({1:1,8:2}.get(reduce(o.mul,t[:3]),0))
def g(t):return max([c(t[x::y]) for x,y in zip((0,0,0,1,2,2,3,6),(1,3,4,3,3,2,1,1))])

Slightly less obscure g:

def g(t):return max([c(t[x::y]) for x,y in [[0,1],[0,3],[0,4],[1,3],[2,3],[2,2],[3,1],[6,1]]])
于 2010-02-12T15:45:59.873 に答える
0

C# Solution.

Multiply the values in each row, col & diagonal. If result == 1, X wins. If result == 8, O wins. 

int v(int[] b)
{
    var i = new[] { new[]{0,1,2}, new[]{3,4,5}, new[]{6,7,8}, new[]{0,3,6}, new[]{1,4,7}, new[]{2,5,8}, new[]{0,4,8}, new[]{2,4,6} };
    foreach(var a in i)
    {
        var n = b[a[0]] * b[a[1]] * b[a[2]];
        if(n==1) return 1;
        if(n==8) return 2;
    }
    return 0;
}
于 2010-02-12T17:19:13.397 に答える
0

C, 113 characters

f(int*b){char*s="012345678036147258048264\0";int r=0;while(!r&&*s){int q=r=3;while(q--)r&=b[*s++-'0'];}return r;}

I think it works? My first code golf, be gentle.

Every 3 digits encodes 3 cells that need to match. The inner while checks a triad. The outer while checks all 8.

于 2010-02-12T20:38:14.503 に答える
0

Java, 155 chars

After much toil on my first code golf, I was able to pare down the function to 155 chars (Curse you array brackets!). With some math tricks I was able to generalize the three-cell-check for horizontals, verticals, and diagonals. Also, I independently discovered what I see Eric Pi also noted, about testing triplet equivalence with bitwise ands. My method:

int i=-1,j,w=0;int[]a={0,0,2,0,9,3,3,1,3,1,1,1,1,3,2,4};while(++i<4)for(j=a[i];j<a[i+4];j+=a[i+8])if((g[j]&g[j+a[i+12]]&g[j+2*a[i+12]])>0)w=g[j];return w;

Also, I made a class to generate all valid boards for testing (not quite as simple as it sounds). For those of you interested in trying to best 155 in Java, here's my testing class:

public class TicTacToe
{
public static void main(String[] args)
{
    int[][] boards = generateBoards();

    for(int i = 0; i < boards.length; ++i)
    {
        int winner = getWinner(boards[i]);

        System.out.println(winner + "  " + boards[i][0] + " " + boards[i][1] + " " + boards[i][2]);
        System.out.println(        "   " + boards[i][3] + " " + boards[i][4] + " " + boards[i][5]);
        System.out.println(        "   " + boards[i][6] + " " + boards[i][7] + " " + boards[i][8]);
        System.out.println();
    }
}

public static int getWinner(int[] g)
{
    int i=-1,j,w=0;int[]a={0,0,2,0,9,3,3,1,3,1,1,1,1,3,2,4};while(++i<4)for(j=a[i];j<a[i+4];j+=a[i+8])if((g[j]&g[j+a[i+12]]&g[j+2*a[i+12]])>0)w=g[j];return w;
}

public static boolean isValid(int[] board)
{
    // O = 0 : X = 1
    int counts[] = new int[2];

    // Count the number of Xs and Os
    for(int i = 0; i < 9; ++i)
        if(board[i] > 0)
            ++counts[board[i] - 1];

    // Make sure the counts make sense. If not return "no"
    if(!(counts[1] == counts[0] || counts[1] == counts[0] + 1))
        return false;

    // Now we're going to total the number of horizontal/vertical wins
    int wins[] = new int[2];

    // Check rows
    if(board[0] != 0 && board[0] == board[1] && board[1] == board[2]) ++wins[board[0] - 1];
    if(board[3] != 0 && board[3] == board[4] && board[4] == board[5]) ++wins[board[3] - 1];
    if(board[6] != 0 && board[6] == board[7] && board[7] == board[8]) ++wins[board[6] - 1];

    // Check columns
    if(board[0] != 0 && board[0] == board[3] && board[3] == board[6]) ++wins[board[0] - 1];
    if(board[1] != 0 && board[1] == board[4] && board[4] == board[7]) ++wins[board[1] - 1];
    if(board[2] != 0 && board[2] == board[5] && board[5] == board[8]) ++wins[board[2] - 1];

    // Make sure the win counts make sense
    if(wins[0] > 1 && wins[1] > 1)
        return false;

    // Hmmmm... I guess it's a valid board
    return true;
}

public static int[][] generateBoards()
{
    int boardSize = 9;
    int permutationCount = (int)Math.pow(4, 9);
    int[][] boards = new int[permutationCount][boardSize];
    int actualIndex = 0;

    for(int i = 0; i < permutationCount; ++i)
    {
        boolean isUnique = true;

        for(int j = 0; j < boardSize; ++j)
        {
            int x = (i >>> j) & 3;

            if(x == 3)
                isUnique = false;

            boards[actualIndex][j] = x;
        }

        if(isUnique && isValid(boards[actualIndex]))
            ++actualIndex;
    }

    return Arrays.copyOf(boards, actualIndex);
}
}

Not bad I suppose for simple java without any exotic function calls. Enjoy!

于 2010-04-02T07:23:38.853 に答える