# Score a game of Go

Scoring a Go game is a task that is not all too easy. In the past there have been several debates about how to design rules to cover all the strange corner cases that may occur. Luckily, in this task you don't have to do complicated stuff like life and death or seki detection. In this task, you have to implement a program that scores a game according to the Tromp-Taylor rules without Komi.
The scoring procedure is pretty simple:

a point P, not colored C, is said to reach C, if there is a path of (vertically or horizontally) adjacent points of P's color from P to a point of color C.
A player's score is the number of points of her color, plus the number of empty points that reach only her color.

For example, consider the following board. X, O and - denote black, white and uncoloured intersections:

- - - X - O - - -
- - - X - O - - -
- - - X - O - - -
- - - X O - - O -
X X X O - O O - -
- - - X O - - O O
- - - X - O - - -
- - - X - O - X -
- - - - - O - - -


Applying the scoring rule yields the following result. x, o and - represent uncoloured intersections that are counted as black, white and nobody's points.

x x x X - O o o o
x x x X - O o o o
x x x X - O o o o
x x x X O o o O o
X X X O o O O o o
- - - X O - - O O
- - - X - O - - -
- - - X - O - X -
- - - - - O - - -


According to the diagram, black has 23 points, white has 29 points of territory. Thus, your program should print W+6 for this board.

I hope it is clear enough this way.

## Input and output

The input is a string that contains exactly of the characters X, O, - where n is not known at compile time. Your program should ignore all other characters in the input stream. Behavior is undefined if there is no integer n such that the amount of XO- characters equals . You may assume that n is in [0, 255].

The sequence of characters is to be interpreted as a Go-board of n rows and columns. The output is the absolute value of the difference of the total amount of points of white and black in decimal representation. If white has more points, it is prefixed by W+, if black has more points it is prefixed by B+. In the case that both players have an equal amount of points, the output is Jigo.

Input is to be read in an implementation defined manner. Input may not be part of the source code.

## Winning conditions

This is code-golf. Usual code-golf conventions apply. The submission with the least amount of characters in its source wins. Only programs that fully implement the specification may win.

## Test cases

Input:

- - - X - O - - -
- - - X - O - - -
- - - X - O - - -
- - - X O - - O -
X X X O - O O - -
- - - X O - - O O
- - - X - O - - -
- - - X - O - X -
- - - - - O - - -


Output: W+6

Input:

Xavier is insane -- says Oliver


Output: Jigo

Inpout:

Code-Golf


Output: Jigo

Input:

-XXXXXXX-XOOOOOOOXXO-OXXXOXXXOX--XOXXOOX
-
XOOXXOX--XOXXXOXXXO-OXXOOOOOOOX-XXXXXXX-


Output: B+21

Input:

- - X O O O O X X - - - - - - X O O -
- X X O X O X X O X X X X X X - X O -
- X O O X X X - O O O X O O X X X O -
- X O O O X X O O O O O O X X X O - -
- - X X O X - X X X X O O O O O O O -
- - X O O X X X - X X X O O O X X O -
- - X O - O X O X O O O O O X X X O -
- X O O - O O O X X X X X O O X O - -
- X X X O - - - O X O X X X O X O - -
X O O O O - - O - O O O O X X X O O -
X X O - - - O - - O O X X - - X X O O
X O O O - - O - O O X - - - - X O O X
- X X X O O X O O X X - - - - X X X X
X - X X X O X X O O X - - X X O X O O
X X O O X O X O X X - - - X O O O O -
X O - O X X X O X - - - - - X O - - -
O O - O X O O O O X X - X X X X O - -
O O - O O O X O X X - - X - X X O - -
- - O - - O X X X - - - - X O O O - -


Output: B+6

More testcases will come soon.

## reference implementation

I have created a reference implementation written in ANSI C. This implementation reads input from the standard input and writes output to the standard output.

/* http://codegolf.stackexchange.com/q/6693/134
* reference implementation
* by user FUZxxl
*/

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

#define MAXBOARD 256

/* bit 0x01: black colour
* bit 0x02: white colour
* bit 0x04: there is a stone on the intersection
*/

enum colour {
UNCOLOURED    = 0x0,
BLACK_REACHED = 0x1,
WHITE_REACHED = 0x2,
BOTH_REACHED  = 0x3,
HAS_STONE     = 0x4,
BLACK         = 0x5,
WHITE         = 0x6
};

static enum colour board[MAXBOARD * MAXBOARD] = { 0 };

static int bsize = 0;

static void fill_board(void);
static void show_score(void);

int main()
{
fill_board();
show_score();
return EXIT_SUCCESS;
}

{
int n = 0;
int invalue;

while ((invalue = getchar()) != EOF) {
switch (invalue) {
case '-': board[n++] = UNCOLOURED; break;
case 'X': board[n++] = BLACK; break;
case 'O': board[n++] = WHITE; break;
}
}

while (bsize * bsize < n) bsize++;

/* your program may exhibit undefined behaviour if this is true */
if (bsize * bsize != n) exit(EXIT_FAILURE);
}

static void fill_board(void)
{
int i,j;
int changes;
enum colour here, top, bottom, left, right, accum;

do {
changes = 0;

for (i = 0; i < bsize; ++i) {
for (j = 0; j < bsize; ++j) {

here   = board[i * bsize + j];
if (here >= BOTH_REACHED) continue;

top    = i == 0 ? UNCOLOURED : board[(i - 1) * bsize + j];
left   = j == 0 ? UNCOLOURED : board[i * bsize + j - 1];
bottom = i == bsize-1 ? UNCOLOURED : board[(i + 1) * bsize + j];
right  = j == bsize-1 ? UNCOLOURED : board[i * bsize + j + 1];

accum = here | top | bottom | left | right;
accum &= ~HAS_STONE;

changes |= board[i * bsize + j] != accum;

board[i * bsize + j] = accum;

}
}

} while (changes);
}

static void show_score(void) {
int w = 0, b = 0, n;

for (n = 0; n < bsize*bsize; ++n) switch (board[n] & ~HAS_STONE) {
case BLACK_REACHED: ++b; break;
case WHITE_REACHED: ++w; break;
}

if (b != w)
printf("%s%i\n",b>w?"B+":"W+",abs(b-w));
else
printf("Jigo\n");
}

• Presumably you mean the last output to be W+7? – dmckee --- ex-moderator kitten Jul 20 '12 at 23:42
• No... How do you come to this conclusion? – FUZxxl Jul 21 '12 at 14:59
• Er...I assumed that S+ was a typo (because you earlier listed possible output as either W+, B+, or Jigo) and I looked at my keyboard and saw the S is near W... Or do you use Dvorak? – dmckee --- ex-moderator kitten Jul 21 '12 at 15:07
• @dmckee I suppose the "S" comes from the german "Schwarz" instead of "Black". – Howard Jul 21 '12 at 15:12
• Oh... You're right. Sorry for that – FUZxxl Jul 21 '12 at 16:00

## GolfScript (105 bytes)

{'-XO'?}/]-1-.{2*3%}%{.,:N),{.*N=}?/{{[{.2$+1={+.}*}*]}%zip}N*[]*.1--,\}2*-.{.0>'W+B+'2/=\abs}{;'Jigo'}if  Flood-fill adapted from this earlier answer of mine. The solution flood-fills one copy of the original board with X, and another with O. Thus empty cells which are reachable by both colours are scored for both, but cancel in the subtraction. • Fair enough. You might win this round. – FUZxxl Feb 19 '15 at 20:55 # C (438434413382364336322298294292 290 characters) #define I b[d*g+e a;b[65536];c;d;e;f;g;main(){for(;d=getchar()+1;f++)b[f]=d-80?d-89?d-46&& f--:5:6,g+=g*g<f;while(!c--)for(d=g;d--;)for(e=g;e--;)I]<3?a=3&(I]|!!d*I -g]|!!e*I-1]|(d<g-1)*I+g]|(e<g-1)*I+1]),c&=I]==a,I]=a:0;while(f--)c+=b[f ]%2-b[f]/2%2;printf(c?"%c+%i":"Jigo",c>0?66:87,abs(c));}  All newlines except the first one inserted for enhanced legibility. A commented and slightly more legible version can be found here. This answer is essentially the reference solution but with all that useless stuff (such as types [who needs something different from int anyway?] and standards compliance [return value of main? please!]) ## Corrections and improvements ### 438 → 434 Dropped explicit initialization of variables after I convinced myself that they are automatically initialized to 0 according to standard. ### 434 → 413 Removed case statement: If an uncoloured intersection is reachable from both black and white, we can count it as one point for both to simplify the program. Switch of logical branches to avoid negation. ### 413 → 382 Assign d to getchar()+1 to save one pair of parenthesis. Under the assumption that b is initialized to zeroes, reorder case statements, discarding all break statements. (a>b?c:0) is longer than (a>b)*c. (d+1)*g+e is longer than d*g+e+g. ### 382 → 364 Improved looping, no newlines in the output, shorter output routines. ### 364 → 336 Got rid of that switch statement. (Thanks, Howard!), track difference of points for two characters. Negate c for one character. four characters in the big or clause. ### 336 → 323 Replacing & by % allows removal of braces for four characters. Fused the square-root with the input loop for nine or so characters (yeah!), removed an if for one char. ### 323 → 298 Introduced the macro H to replace the often occuring and bulky b[d*g+e] construct. ### 298 → 294 Change a&=~4 to a&=3 as we only every observe the lowest three bytes of a. Also changed to loop body from ((a=I])<3)?a|=... to I]<3?a=I]|... which is two characters shorter. Also, introduce h instead of reusing c, which is one character shorter. ### 294 → 292 Eliminate h variable. If we test c with !c-- instead of !c++, c equals 0 at the end of the flood-fill loop and can thus be used for the purpose h was used before (i.e. score keeping). ### 292 → 290 Replace the construct d-46?f--:0 with d-46&&f-- which is shorter by a character and combine the two assignments to a in the inner loop. • You may replace the switch statement with something like {b[f]=d-80?d-89?d-46?f--:0:5:6;f++;} to save several characters. – Howard Jul 26 '12 at 13:42 • @Howard: Yeah. That worked really nice! Thank you – FUZxxl Jul 26 '12 at 21:06 • "For enhanced legibility" — as if. – tomsmeding Sep 22 '14 at 19:18 • @tomsmeding Well, scrolling one line is much less legible. Also, a commented version is linked to. – FUZxxl Sep 22 '14 at 20:39 • @FUZxxl That was meant jokingly. :) Also, you're right in saying it's better than on 1 line. – tomsmeding Sep 22 '14 at 20:41 ## J (140136131129119117 116 characters) After ramping up my J skills, I can finally provide a submission in J. It's a little bit long though. exit echo>(*{'Jigo';('B+',":);'W+',":@|)+/,-/1 2=/(]OR(0=[)*[:OR((,.|.)0,i:1)|.!.0])^:_~($~,~@%:@#)3-.~'-XO'i:1!:1]3


The algorithm implemented by this submission is very similar to the reference implementation but different in the way occupied fields are handled.

Here is the solution split up into more parts for easier understandability. The golfed solution is slightly different from that, but the difference isn't very large.

input =. 3 -.~ '-XO' i: 1!:1 ] 3
board =. ($~ ,~@%:@#) input NB. shift up, down, left, right rotm =. (,. |.) 0 , i: 1 fill =. ] OR (0 = [) * [: OR rotm |.!.0 ] filledboard =. fill^:_~ board score =. +/ , -/ 1 2 =/ filledboard echo > (* { 'Jigo' ; ('B+' , ":) ; ('W+', ":@|)) score exit 0  ### GolfScript, 190 characters {"XO-"?)},:v,.),\{\.*=}+,~:s.*:0\,{s%!\2*+}/:r;88{0v@{=\2*+}+/}:%~79%1${{:<.r|r^2*|<2/r|r^|<2s?:h/|<h*|1$|1$^2?(&}*}:a~@@a\;.2$|2${^2*)2base{+}*}:m~@2$|@m-.{"BW"1/1$0>="+"@abs}{;"Jigo"}if


The script became much longer than I thought in the beginning. Pass any input on STDIN and the output will then be printed when the program terminates.

# Ruby (314)

could be made shorter with some more time:

q={?-=>0,?X=>5,?O=>6};b=[];\$<.chars{|c|(t=q[c])&&b<<t}
z=Math.sqrt b.size
loop{c=b.each_with_index.map{|h,i|
next h if h>2
x=i%z
y=i/z
u=y<1?0:b[i-z]
l=x<1?0:b[i-1]
d=y>z-2?0:b[i+z]
r=x>z-2?0:b[i+1]
~4&(h|u|d|l|r)}
break if c==b
b=c}
b.map!{|h|h&~4}
s=b.count(1)-b.count(2)
puts s==0?"Jigo":s>0?"B+#{s}":"W+#{-s}"
`