Code Golf: Word Search Solver

Question

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Note: This is my first Code Golf challenge/question, so I might not be using the correct format below. I'm not really sure how to tag this particular question, and should this be community wiki? Thanks!

This Code Golf challenge is about solving word searches!

A word search, as defined by Wikipedia, is:

A word search, word find, word seek, word sleuth or mystery word puzzle is a word game that is letters of a word in a grid, that usually has a rectangular or square shape. The objective of this puzzle is to find and mark all the words hidden inside the box. The words may be horizontally, vertically or diagonally. Often a list of the hidden words is provided, but more challenging puzzles may let the player figure them out. Many word search puzzles have a theme to which all the hidden words are related.

The word searches for this challenge will all be rectangular grids with a list of words to find provided. The words can be written vertically, horizontally, or diagonally.

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Input/Output

The user inputs their word search and then inputs a word to be found in their grid. These two inputs are passed to the function that you will be writing. It is up to you how you want to declare and handle these objects.

Using a strategy described below or one of your own, the function finds the specific word in the search and outputs its starting coordinates (simply row number and column number) and ending coordinates. If you find two occurrences of the word, you must output both sets of coordinates. If the word is a palindrome, you may arbitrarily choose one end to be the "start" of the word.

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Example

Input:

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 

Word to find: codegolf

Output:

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

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Strategies

Here are a few strategies you might consider using. It is completely up to you to decide what strategy you want to use; it doesn't have to be in this list.

• Looking for the first letter of the word; on each occurrence, looking at the eight surrounding letters to see whether the next letter of the word is there.
• Same as above, except looking for a part of a word that has two of the same letter side-by-side.
• Counting how often each letter of the alphabet is present in the whole grid, then selecting one of the least-occurring letters from the word you have to find and searching for the letter. On each occurrence of the letter, you look at its eight surrounding letters to see whether the next and previous letters of the word is there.

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

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 4

Perl, 226 char

sub h($){"row ",$%=1+($x=pop)/$W,", column ",1+$x%$W}
@S=map/[^ ]/g,<STDIN>;
B:for(@ARGV){
 for$d(map{$_,-$_}1,$W=@S/$.,$W-1,$W+1){
  A:for$s(0..$#S){
   $t=$s-$d;for(/./g){next A if$S[$t+=$d]ne$_||$t<0}
   print h$s,' --> ',h$t,$/;
   next B
}}}

Character count excludes newlines and indentation, which are included for readability. I assume the matrix of letters is specified with standard input and the target words are command-line arguments.

First line (after the sub h definition) maps all non-space characters into a single array @S. Then iterate over all the target words, over the possible directions that words may appear in ($W is the width of the puzzle), and over all of the letters in the current target word until a match is found.

$ perl wordsearch.pl CODEGOLF RACECAR BYKLHQU <<EOF
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 R A C E 
C A R 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 R A C E C A R R O W G N D 
B C D E J Y E L W X J D F X M 
EOF
row 12, column 8 --> row 5, column 1
row 14, column 3 --> row 14, column 9
row 3, column 12 --> row 9, column 6

Answer 5

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

Answer 6

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).