# Simplest Tiling of the Floor

You should write a program or function which receives a string describing the floor as input and outputs or returns the area of the simplest meta-tiling which could create the given pattern of the floor.

The floor is a part of a square grid. Every square tile is colored either azure or black (represented by a and b in the input).

An example floor:

  aaaa
ababab
aaaaa


A meta-tiling

• is built from an N by M rectangular meta-tile of azure and black squares
• the used meta-tiles are identical up to translation (you cannot rotate or mirror them)
• if the sides of two meta-tiles are connected they should connect along their whole length (i.e. meta-tiles tile the space in a grid-like fashion)

An example meta-tile:

ba
aa


and the meta-tiling created by it:

       .
.
.
babababa
aaaaaaaa
... babababa ...
aaaaaaaa
babababa
aaaaaaaa
.
.
.


This meta-tiling creates the upper shown floor as the left letters show:

       .
.
.
********
***aaaa*
... *ababab* ...
*aaaaa**
********
********
.
.
.


A meta-tiling is simpler than another if the area of its meta-tile is smaller. Our example has an area of 2*2 = 4 which is the smallest possible for the example floor. So the output should be 4 for the example.

## Input

• A string consisting of the characters a b space and newline containing at least one a or b.
• The letters (ab) form one 4-connected (side-by-side connected) shape.
• There will be no unnecessary spaces at the front of the rows i.e. there will be at least one row starting with a or b.
• You can choose of two input format:

• No unnecessary whitespace at the end of rows (as seen in the examples).
• Spaces on the right side of the rows to make all rows the same length as the longest row.
• Trailing newline is optional.

## Output

• A single integer, the area of the smallest possible meta-tile whose tiling contains the input floor.

## Examples

Examples are delimited by dashes. The three parts of an example are input, output and one of the possible smallest meta-tiles.

a

1

a
-----------------
aaaa
aaa
a

1

a
-----------------
aabaab
abaa
aaba

6

aab
aba
-----------------
aabaab
a  a a
aabab

18

aabaab
aaaaaa
aababa
-----------------
ba
aaab

8

baaa
aaab
-----------------
aaaa
ababb
aaaa

10

aaaaa
ababb
-----------------
a aa
ab ba
aba

6

aa
ab
ba
-----------------
aaaa
abab
aaaa

4

aa
ab
-----------------
ba
ba
b

4

ba
ab
-----------------
baa
aba
aab

9

baa
aba
aab
-----------------
aaaa
aabaa
aaaa

6

aaa
aab


This is code golf so the shortest entry wins.

• @Ypnypn Every corner has to touch 3 other corners (except the meta-tiles on the edge of the tiling). I stated it as "if the sides of two meta-tiles are connected they should connect along their whole length". So your given example is illegal. – randomra May 15 '15 at 17:48

## C - 208 bytes

w,q,n,m,r,g,u;t(char*f){for(w=0;f[w++]-10;);for(q=1;;q++)for(n=1;m=q/n,n<=q;++n)if(n*m==q){char t[q];bzero(t,q);r=q;for(g=0;f[g];++g){u=g/w%m*n+g%w%n;r=t[u]+f[g]-195?r:0;if(f[g]&64)t[u]=f[g];}if(r)return r;}}


Equivalent code before golfing:

#include <stdio.h>
#include <strings.h>

int t(char* f) {
int w = 0;
for ( ; f[w++] - 10; );

for (int q = 1; ; q++) {
char t[q];
for (int n = 1; n <= q; ++n) {
int m = q / n;
if (n * m == q) {
bzero(t, q);
int r = q;
for (int g = 0; f[g]; ++g) {
int u = g / w % m * n + g % w % n;
if (t[u] + f[g] == 195) {
r = 0;
}
if (f[g] & 64) {
t[u] = f[g];
}
}
if (r) {
return r;
}
}
}
}
}


The algorithm is fairly brute force, so it should be reasonably obvious how it works based on the code. Here are a few comments anyway:

• Input is expected to have the form with trailing spaces so that all lines have the same length.
• First loop finds the width by looking for first newline character.
• Outer loop is over candidate meta-tile sizes q. Exits with a return when a meta-tile can cover the floor. Note that the loop does not need another exit condition since there is always a solution (worst case is size of input).
• First nested loop and following if enumerates valid meta-tile width/height combinations for size q.
• A character array matching the candidate meta-tile size is zero-initialized.
• Inner loop iterates over all tiles in the floor.
• u is the index in the meta-tile that corresponds to the floor tile.
• If both floor tile and meta-tile tile are a or b and different (sum of a = 97 and b = 98 is 195), there is a mismatch, and the meta-tile size with the attempted dimensions will not work.
• Otherwise, if the floor tile is a or b, the tile color is copied to the candidate meta-tile.
• Returns size of meta-tile when successful match was made, i.e. if the attempted match was not marked as failed.

Test code used:

#include <stdio.h>

extern int t(char* s);

int main()
{
printf("%d\n", t(
"a\n"
));
printf("%d\n", t(
" aaaa\n"
"aaa  \n"
"a    \n"
));
printf("%d\n", t(
"aabaab\n"
"abaa  \n"
"aaba  \n"
));
printf("%d\n", t(
"aabaab\n"
"a  a a\n"
"aabab \n"
));
printf("%d\n", t(
"ba  \n"
"aaab\n"
));
printf("%d\n", t(
" aaaa\n"
"ababb\n"
"aaaa \n"
));
printf("%d\n", t(
" a aa\n"
"ab ba\n"
" aba \n"
));
printf("%d\n", t(
" aaaa\n"
"abab \n"
"aaaa \n"
));
printf("%d\n", t(
"ba \n"
" ba\n"
"  b\n"
));
printf("%d\n", t(
"baa\n"
"aba\n"
"aab\n"
));
printf("%d\n", t(
" aaaa\n"
"aabaa\n"
"aaaa \n"
));
return 0;
}


Output:

1
1
6
18
8
10
6
4
4
9
6