Code Golf: Word Search Solver

Answer 1

AWK - 252 255

NF<2{l=split($1,w,"")}NF>1{for(i=1;i<=NF;)t[x=i++,y=NR]=$i}END{
for(i=1;i<=x;++i)for(j=0;++j<=y;)for(a=0;a<9;++a)if(t[I=i,J=j]==w[1])
for(k=1;k<l;){if(!(T=t[I+=int(a/3)-1,J+=a%3-1])||T!=w[++k])break;if(k==l)
print"row "j (B=", column ")i" --> row "J B I}}

The input should be in the form

....
B L T O O S Q V K R O W G N D 
B C D E J Y E L W X J D F X M 
CODEGOLF

Answer 2

Python - 186 chars

def f(g,W):w=g.find("\n")+1;L=len(W);print" --> ".join("row %s, column %s"%(x/w+1
,x%w+1)for i in range(len(g))for j in(-w-1,-w,-w+1,-1,1,w-1,w,w+1)for x in(i,i+(L
-1)*j)if g[i::j][:L]==W)

testing code:

grid="""A I Y R J J Y T A S V Q T Z E 
X B X G R Z P W V T B K U F O 
E A F L V F J J I A G B A J K 
R E S U R E P U S C Y R S Y K 
F B B Q Y T K O I K H E W G N 
G L W Z F R F H L O R W A R E 
J A O S F U E H Q V L O A Z B 
J F B G I F Q X E E A L W A C 
F W K Z E U U R Z R T N P L D 
F L M P H D F W H F E C G W Z 
B J S V O A O Y D L M S T C R 
B E S J U V T C S O O X P F F 
R J T L C V W R N W L Q U F I 
B L T O O S Q V K R O W G N D 
B C D E J Y E L W X J D F X M """.lower().replace(" ","")
f(grid,"codegolf")
f(grid,"serverfault")
f(grid,"superuser")

This version is 196 chars and accepts the grid without doing any extra tweaking

def f(g,W):w=g.find("\n")+1;L=len(W);print" --> ".join("row %s, column %s"%(x/w+1,x%w/2+1)for i in range(len(g))for j in(-w-2,-w,-w+2,-2,2,w-2,w,w+2)for x in(i,i+(L-1)*j)if g[i::j][:L].lower()==W)

testing code:

grid="""A I Y R J J Y T A S V Q T Z E 
X B X G R Z P W V T B K U F O 
E A F L V F J J I A G B A J K 
R E S U R E P U S C Y R S Y K 
F B B Q Y T K O I K H E W G N 
G L W Z F R F H L O R W A R E 
J A O S F U E H Q V L O A Z B 
J F B G I F Q X E E A L W A C 
F W K Z E U U R Z R T N P L D 
F L M P H D F W H F E C G W Z 
B J S V O A O Y D L M S T C R 
B E S J U V T C S O O X P F F 
R J T L C V W R N W L Q U F I 
B L T O O S Q V K R O W G N D 
B C D E J Y E L W X J D F X M """
f(grid,"codegolf")
f(grid,"serverfault")
f(grid,"superuser")

Answer 3

C++ C, 411 400 388 367 characters

My first attempt ever at a code golf.

#include<stdio.h>
main(){char m[999],w[99];int r,c,l=-1,x=-1,y=-1,i=0,j,k;scanf("%s %d %d"
,w,&r,&c);for(;w[l+1];++l);for(;i<r*c;scanf("%s",&m[i++]));for(;y<2;++y)
for(;x<2;++x)if(x|y)for(i=(y<0)*l;i<r-(y>0)*l;++i)for(j=(x<0)*l;j<c-(x>0
)*l;++j)for(k=0;k<=l&&w[k++]==m[(i+k*y)*c+j+k*x];)k>l?printf(
"row %i, column %i --> row %i, column %i\n",i+1,j+1,i+y*l+1,j+x*l+1):0;}

It compiles without warnings on gcc with no additional flags.

Here's the version with whitespace:

#include<stdio.h>
main() {
    char m[999], w[99];
    int r, c, l = -1, x = -1, y = -1, i = 0, j, k;

    scanf("%s %d %d", w, &r, &c);
    for (; w[l+1]; ++l);
    for (; i < r*c; scanf("%s", &m[i++]));
    for (; y < 2; ++y)
        for (; x < 2; ++x)
            if (x | y)
                for (i = (y<0) * l; i < r - (y>0) * l; ++i)
                    for (j = (x<0) * l; j < c - (x>0) * l; ++j)
                        for (k = 0; k <= l && w[k++] == m[(i+k*y) * c + j+k*x];)
                            k > l ? printf("row %i, column %i --> row %i, column %i\n",
                                    i + 1, j + 1, i + y*l + 1, j + x*l + 1)
                                  : 0;
}

Expected input on stdin looks like:

CODEGOLF 15 15
A I Y R J J Y T A S V Q T Z E
X B X G R Z P W V T B K U F O
...

where 15 15 are the numbers of rows and columns, respectively.

Since this is my first round of golf, and I could find precious little common tricks on the web, I'll list some that I found:

• Try to keep loop bodies to a single statement to save on moustaches (2 chars).
  Example: anything using the comma operator.

• Use a conditional expression (?:) instead of an if when you can (1 char).
  Example: instead of if(x)printf("");, write x?printf(""):0;. This works because printf returns int.

• Use the fact that booleans are 0 or 1 when used as an int (? chars).
  Example: instead of (y>0?r-l:r), write r-(y>0)*l. The savings here are in the parentheses, which were necessary in this situation.

• while loops are never shorter than for loops, and longer most of the time (>= 0 chars).
  Example: while(x)y; is just as long as for(;x;y); but with for you get the loop body to play with as well, without paying for the extra ;.

• Put your loop increments into the last expression that uses the loop counter (1 char).
  Example: instead of for(;x<b;++x)printf("%i",x);, write for(;x<b;)printf("%i",x++);.

• Leave off main's return type (4 chars).

• Leave off main's return statement (9 chars).

• Use & and | instead of && and || whenever possible (1 char).
  Example: instead of if(x||y), write if(x|y). This only works if you don't depend on the short-circuit behaviour of && and ||.

• Inline the strlen function and remove #include<string.h> (11 chars).

If you don't care about compiler warnings:

• Do not include any headers (many chars).

Answer 4

Python, 318 strokes.

from itertools import*
f=lambda x,y,i,j,n:(x-i+1,y-j+1)*(not n)or all((2*i+j,x+1,y+1,x<c,y<d))and a[x][y*2]==n[0]and f(x+i,y+j,i,j,n[1:])
w=raw_input
a=list(iter(w,''))
c=len(a)
d=len(a[0])/2
r=range
z=r(-1,2)
s='row %d, column %d'
for x in product(r(c),r(d),z,z,[w()]):
 if f(*x):print s%(x[0]+1,x[1]+1),'-->',s%f(*x)

---

Sample input:

A I Y R J J Y T A S V Q T Z E 
X C O D E G O L F B K U F O Z
E A F L V F J J I A G B A J K 
R E S U R E P U S C Y R S Y K 
F B B Q Y T K O I K H E W G N 
G L W Z F R F H L O R W A R E 
J A O S F U E H Q V L O A Z B 
J F B G I F Q X E E A L W A C 
F W K Z E U U R Z R T N P L D 
F L M P H D F W H F E C G W Z 
B J S V O A O Y D L M S T C R 
B E S J U V T C S O O X P F F 
R J T L C V W R N W L Q U F I 
B L T O O S Q V K R O W G N D 
B C D E J Y E L W X J D F X M 

CODEGOLF

Output:

$ python wordsearch < wordsearchinput.txt
row 2, column 2 --> row 2, column 9
row 12, column 8 --> row 5, column 1

Released under the "edit-the-answer-instead-of-commenting-on-how-I-should-fix-this" license.

Answer 5

Perl - 243

(Not including linebreaks, all of which are optional)

Mostly de-golfed code can be found here.

$/="";@w=split//,pop;($b=<>)=~y/ //d;$W=1+index$b,"\n";
for(1,2){for(0,$W-2..$W){$p=join"."x$_,@w;
push@m,[$-[0],$+[0]-1]while$b=~/$p/sg}
@w=reverse@w;@m=map[reverse@$_],@m}
$t="row %d, column %d";
printf"$t --> $t\n",map{1+$_/$W,1+$_%$W}@$_ for@m;

Usage: run as a script. Either wordsearch.pl boardfile searchword to read from a file, or wordsearch.pl searchword to read the board from STDIN (hey, pop is two characters shorter than shift!)

The idea is that we read the board into a string (throwing away the extra spaces between letters), and also make note of how wide it is (by counting characters until the first newline). Then we do a series of regex matches to find the word, using the s regex modifier to allow a match to span lines. Given a board that's 4 letters wide (so each line is 5 chars including newline), and the search word "CAR", we can make the patterns /CAR/ to match forwards, /C...A...R/ to match going southwest, /C....A....R/ to match going downwards, and /C.....A.....R/ to match going southeast. For each match, we record the starting and ending positions of the match.

Then we reverse the search word ("CAR" -> "RAC") and do it again, which lets us match the word going left, up, northwest, and northeast. Of course, we want to make sure that the start/end positions are also swapped. To save code, I swap the match positions twice; the matches for the forward word are swapped both times (so they come out un-swapped), while the matches for the backward word are only swapped once, so they will come out swapped.

Finally, it's just a matter of taking the match offsets into the board string and turning them into row/column pairs, and formatting the output how the problem wants it.

Answer 6

Python: 491 - 353 characters

Still quite some room for improvement I guess, all comments welcome.

Strategy: finding all occurences of the first letter, and then trying to fetch the word in all directions.

import sys;r=range(-1,2);k=''.join;q="row %s, column %s"
a=[l[:-1:2]for l in sys.stdin]
b=sys.argv[1];c=len(a[0])
for x,y in[(x/c,x%c)for x,h in enumerate(k(map(k,a)))]:
 for i,j in[(i,j)for i in r for j in r if i or j<>0]:
  if k([a[x+z*i][y+z*j]for z in range(len(b))if 0<=x+z*i<c and 0<=y+z*j<len(a)])==b:print q%(x+1,y+1)+" --> "+q%(x+z*i+1,y+z*j+1)

Example usage:

cat input.txt | python test.py CODEGOLF

Example output:

row 12, column 8 --> row 5, column 1