10
\$\begingroup\$

A binary string is a string which contains only characters drawn from 01. A balanced binary string is a binary string which contains exactly as many 0​s as 1​s.

You are given a positive integer n and an arbitrary number of masks, each of which is 2n characters long, and contains only characters drawn from 012. A binary string and a mask match if it is the same length and agrees on the character in every position where the mask doesn't have a 2. E.g. the mask 011022 matches the binary strings 011000, 011001, 011010, 011011.

Given n and the masks as input (separated by newlines), you must output the number of distinct balanced binary strings which match one or more of the masks.

Examples

Input

3
111222
000112
122020
122210
102120

Reasoning

  • The only balanced binary string matching 111222 is 111000.
  • The only balanced binary string matching 000112 is 000111.
  • The balanced binary strings matching 122020 are 111000 (already counted), 110010 and 101010.
  • The balanced binary strings matching 122210 are 110010 (already counted), 101010 (already counted) and 100110.
  • The balanced binary strings matching 102120 are 101100 and 100110 (already counted).

So the output should be

6

Input

10
22222222222222222222

Reasoning

  • There are 20 choose 10 balanced binary strings of length 20.

Output

184756

Winner

The winner will be the one that computes the competition input the fastest, of course treating it the same way as it would any other input. (I use a determined code in order to have a clear winner and avoid cases where different inputs would give different winners. If you think of a better way to find the fastest code, tell me so).

Competition input

http://pastebin.com/2Dg7gbfV

\$\endgroup\$
  • 2
    \$\begingroup\$ Also, I personally highly prefer gist.github.com over pastebin, both for aesthetics and future faults. \$\endgroup\$ – orlp May 13 '15 at 19:40
  • 3
    \$\begingroup\$ @AlbertMasclans I think you should reserve the right to change the input. Otherwise, someone can hardcode the output. \$\endgroup\$ – mbomb007 May 13 '15 at 21:40
  • 1
    \$\begingroup\$ It would be useful if you could post a small example in the question, with the expected result, and an explanation. I might just be slow, but I don't quite get the definition. So for the example, since n=30, we're looking for sequences of 30 0s or 30 1s (with 2 being a wildcard) in any row? Can those sequences overlap? For example, if I find a sequence of 32 1s, does that count as 3 sequences, or as a single sequence? What if I find a sequence of 60 1s (entire row)? Is that 1 sequence, 2 sequences, or 31 sequences? \$\endgroup\$ – Reto Koradi May 14 '15 at 3:37
  • 3
    \$\begingroup\$ So you're asking for the number of unique arrays in this matrix that have equal numbers of 1s and 0s, right? \$\endgroup\$ – ASCIIThenANSI May 14 '15 at 12:00
  • 1
    \$\begingroup\$ Can we have some more test data please? \$\endgroup\$ – alexander-brett May 15 '15 at 0:47
2
\$\begingroup\$

C

If you're not on Linux, or otherwise having trouble compiling, you should probably remove the timing code (clock_gettime).

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

long int binomial(int n, int m) {
  if(m > n/2) {
    m = n - m;
  }
  int i;
  long int result = 1;
  for(i = 0; i < m; i++) {
    result *= n - i;
    result /= i + 1;
  }
  return result;
}

typedef struct isct {
  char *mask;
  int p_len;
  int *p;
} isct;

long int mask_intersect(char *mask1, char *mask2, char *mask_dest, int n) {

  int zero_count = 0;
  int any_count = 0;
  int i;
  for(i = 0; i < n; i++) {
    if(mask1[i] == '2') {
      mask_dest[i] = mask2[i];
    } else if (mask2[i] == '2') {
      mask_dest[i] = mask1[i];
    } else if (mask1[i] == mask2[i]) {
      mask_dest[i] = mask1[i];
    } else {
      return 0;
    }
    if(mask_dest[i] == '2') {
      any_count++;
    } else if (mask_dest[i] == '0') {
      zero_count++;
    }
  }
  if(zero_count > n/2 || any_count + zero_count < n/2) {
    return 0;
  }
  return binomial(any_count, n/2 - zero_count);
}

int main() {

  struct timespec start, end;
  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start);

  int n;
  scanf("%d", &n);
  int nn = 2 * n;

  int m = 0;
  int m_max = 1024;

  char **masks = malloc(m_max * sizeof(char *));

  while(1) {
    masks[m] = malloc(nn + 1);
    if (scanf("%s", masks[m]) == EOF) {
      break;
    }
    m++;
    if (m == m_max) {
      m_max *= 2;
      masks = realloc(masks, m_max * sizeof(char *));
    }
  }

  int i = 1;
  int i_max = 1024 * 128;

  isct *iscts = malloc(i_max * sizeof(isct));

  iscts[0].mask = malloc(nn);
  iscts[0].p = malloc(m * sizeof(int));

  int j;
  for(j = 0; j < nn; j++) {
    iscts[0].mask[j] = '2';
  }
  for(j = 0; j < m; j++) {
    iscts[0].p[j] = j;
  }
  iscts[0].p_len = m;

  int i_start = 0;
  int i_end = 1;
  int sign = 1;

  long int total = 0;

  int mask_bin_len = 1024 * 1024;
  char* mask_bin = malloc(mask_bin_len);
  int mask_bin_count = 0;

  int p_bin_len = 1024 * 128;
  int* p_bin = malloc(p_bin_len * sizeof(int));
  int p_bin_count = 0;


  while (i_end > i_start) {
    for (j = i_start; j < i_end; j++) {
      if (i + iscts[j].p_len > i_max) {
        i_max *= 2;
        iscts = realloc(iscts, i_max * sizeof(isct));
      }
      isct *isct_orig = iscts + j;
      int x;
      int x_len = 0;
      int i0 = i;
      for (x = 0; x < isct_orig->p_len; x++) {
        if(mask_bin_count + nn > mask_bin_len) {
          mask_bin_len *= 2;
          mask_bin = malloc(mask_bin_len);
          mask_bin_count = 0;
        }
        iscts[i].mask = mask_bin + mask_bin_count;
        mask_bin_count += nn;
        long int count =
            mask_intersect(isct_orig->mask, masks[isct_orig->p[x]], iscts[i].mask, nn);
        if (count > 0) {
          isct_orig->p[x_len] = isct_orig->p[x];
          i++;
          x_len++;
          total += sign * count;
        }
      }
      for (x = 0; x < x_len; x++) {
        int p_len = x_len - x - 1;
        iscts[i0 + x].p_len = p_len;
        if(p_bin_count + p_len > p_bin_len) {
          p_bin_len *= 2;
          p_bin = malloc(p_bin_len * sizeof(int));
          p_bin_count = 0;
        }
        iscts[i0 + x].p = p_bin + p_bin_count;
        p_bin_count += p_len;
        int y;
        for (y = 0; y < p_len; y++) {
          iscts[i0 + x].p[y] = isct_orig->p[x + y + 1];
        }
      }
    }

    sign *= -1;
    i_start = i_end;
    i_end = i;

  }

  printf("%lld\n", total);

  clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &end);

  int seconds = end.tv_sec - start.tv_sec;
  long nanoseconds = end.tv_nsec - start.tv_nsec;
  if(nanoseconds < 0) {
    nanoseconds += 1000000000;
    seconds--;
  }

  printf("%d.%09lds\n", seconds, nanoseconds);
  return 0;
}

Example cases:

robert@unity:~/c/se-mask$ gcc -O3 se-mask.c -lrt -o se-mask
robert@unity:~/c/se-mask$ head testcase-long
30
210211202222222211222112102111220022202222210122222212220210
010222222120210221012002220212102220002222221122222220022212
111022212212022222222220111120022120122121022212211202022010
022121221020201212200211120100202222212222122222102220020212
112200102110212002122122011102201021222222120200211222002220
121102222220221210220212202012110201021201200010222200221002
022220200201222002020110122212211202112011102220212120221111
012220222200211200020022121202212222022012201201210222200212
210211221022122020011220202222010222011101220121102101200122
robert@unity:~/c/se-mask$ ./se-mask < testcase-long
298208861472
0.001615834s
robert@unity:~/c/se-mask$ head testcase-hard
8
0222222222222222
1222222222222222
2022222222222222
2122222222222222
2202222222222222
2212222222222222
2220222222222222
2221222222222222
2222022222222222
robert@unity:~/c/se-mask$ ./se-mask < testcase-hard
12870
3.041261458s
robert@unity:~/c/se-mask$ 

(Times are for an i7-4770K CPU at 4.1 GHz.) Be careful, testcase-hard uses around 3-4 GB of memory.

This is pretty much an implementation of inclusion-exclusion method blutorange came up with, but done so that it will handle intersections of any depth. The code as written is spending a lot of time on memory allocation, and will get even faster once I optimize the memory management.

I shaved off around 25% on testcase-hard, but the performance on the original (testcase-long) is pretty much unchanged, since not much memory allocation is going on there. I'm going to tune a bit more before I call it: I think I might be able to get a 25%-50% improvement on testcase-long too.

Mathematica

Once I noticed this was a #SAT problem, I realized I could use Mathematica's built-in SatisfiabilityCount:

AbsoluteTiming[
 (* download test case *)
 input = Map[FromDigits, 
   Characters[
    Rest[StringSplit[
      Import["http://pastebin.com/raw.php?i=2Dg7gbfV", 
       "Text"]]]], {2}]; n = Length[First[input]];
 (* create boolean function *)
 bool = BooleanCountingFunction[{n/2}, n] @@ Array[x, n] && 
   Or @@ Table[
     And @@ MapIndexed[# == 2 || Xor[# == 1, x[First[#2]]] &, i], {i, 
      input}];
 (* count instances *)
 SatisfiabilityCount[bool, Array[x, n]]
]

Output:

{1.296944, 298208861472}

That's 298,208,861,472 masks in 1.3 seconds (i7-3517U @ 1.9 GHz), including the time spent downloading the test case from pastebin.

\$\endgroup\$
  • \$\begingroup\$ So I've rewritten this in C... unfortunately it's too fast for me to use gperftools on it! I'll find some harder test cases before I post tomorrow. \$\endgroup\$ – 2012rcampion May 16 '15 at 4:46
  • \$\begingroup\$ testcase-hard can be completed very quickly if your code looks for masks that can be combined. If your code does this, then delete every other line (so just the /^2*02*$/ cases remain). I don't think that case can be optimized. \$\endgroup\$ – 2012rcampion May 16 '15 at 20:41
4
\$\begingroup\$

ruby, pretty fast, but it depends upon the input

Now speed-up by a factor of 2~2.5 by switching from strings to integers.

Usage:

cat <input> | ruby this.script.rb

Eg.

mad_gaksha@madlab ~/tmp $ ruby c50138.rb < c50138.inp2
number of matches: 298208861472
took 0.05726237 s

The number of matches for a single mask a readily calculated by the binomial coefficient. So for example 122020 needs 3 2s filled, 1 0 and 2 1. Thus there are nCr(3,2)=nCr(3,1)=3!/(2!*1!)=3 different binary strings matching this mask.

An intersection between n masks m_1, m_2, ... m_n is a mask q, such that a binary string s matches q only iff it matches all masks m_i.

If we take two masks m_1 and m_2, its intersection is easily computed. Just set m_1[i]=m_2[i] if m_1[i]==2. The intersection between 122020 and 111222 is 111020:

122020 (matched by 3 strings, 111000 110010 101010)
111222 (matched by 1 string, 111000)
111020 (matched by 1 string, 111000)

The two individual masks are matched by 3+1=4 strings, the interesection mask is matched by one string, thus there are 3+1-1=3 unique strings matching one or both masks.

I'll call N(m_1,m_2,...) the number of strings matched all m_i. Applying the same logic as above, we can compute the number of unique strings matched by at least one mask, given by the inclusion exclusion principle and see below as well, like this:

N(m_1) + N(m_2) + ... + N(m_n) - N(m_1,m_2) - ... - N(m_n-1,m_n) + N(m_1,m_2,m_3) + N(m_1,m_2,m_4) + ... N(m_n-2,m_n-1,m_n) - N(m_1,m_2,m_3,m_4) -+ ...

There are many, many, many combinations of taking, say 30 masks out of 200.

So this solution makes the assumption that not many high-order intersections of the input masks exist, ie. most n-tuples of n>2 masks will not have any common matches.

Use the code here, the code at ideone may be out-dated.

I added a function remove_duplicates that can be used to pre-process the input and delete masks m_i such that all strings that match it also match another mask m_j.,For the current input, this actually takes longer as there are no such masks (or not many), so the function isn't applied to the data yet in the code below.

Code:

# factorial table
FAC = [1]
def gen_fac(n)
  n.times do |i|
    FAC << FAC[i]*(i+1)
  end
end

# generates a mask such that it is matched by each string that matches m and n
def diff_mask(m,n)
  (0..m.size-1).map do |i|
    c1 = m[i]
    c2 = n[i]
    c1^c2==1 ? break : c1&c2
  end
end

# counts the number of possible balanced strings matching the mask
def count_mask(m)
  n = m.size/2
  c0 = n-m.count(0)
  c1 = n-m.count(1)
  if c0<0 || c1<0
    0
  else
    FAC[c0+c1]/(FAC[c0]*FAC[c1])
  end
end

# removes masks contained in another
def remove_duplicates(m)
  m.each do |x|
    s = x.join
    m.delete_if do |y|
      r = /\A#{s.gsub(?3,?.)}\Z/
      (!x.equal?(y) && y =~ r) ? true : false
    end
  end
end

#intersection masks of cn masks from m.size masks
def mask_diff_combinations(m,n=1,s=m.size,diff1=[3]*m[0].size,j=-1,&b)
  (j+1..s-1).each do |i|
    diff2 = diff_mask(diff1,m[i])
    if diff2
      mask_diff_combinations(m,n+1,s,diff2,i,&b) if n<s
      yield diff2,n
    end
  end
end

# counts the number of balanced strings matched by at least one mask
def count_n_masks(m)
  sum = 0
  mask_diff_combinations(m) do |mask,i|
    sum += i%2==1 ? count_mask(mask) : -count_mask(mask)
  end
  sum
end

time = Time.now

# parse input
d = STDIN.each_line.map do |line|
  line.chomp.strip.gsub('2','3')
end
d.delete_if(&:empty?)
d.shift
d.map!{|x|x.chars.map(&:to_i)}

# generate factorial table
gen_fac([d.size,d[0].size].max+1)

# count masks
puts "number of matches: #{count_n_masks(d)}"
puts "took #{Time.now-time} s"

This is called the inclusion exclusion principle, but before somebody had pointed me to it I had my own proof, so here it goes. Doing something yourself feels great though.

Let us consider the case of 2 masks, call then 0 and 1, first. We take every balanced binary string and classify it according to which mask(s) it matches. c0 is the number of those that match only mask 0, c1 the nunber of those that match only 1, c01 those that match mask 0 and 1.

Let s0 be the number sum of the number of matches for each mask (they may overlap). Let s1 be the sum of the number of matches for each pair (2-combination) of masks. Let s_i be the sum of the number of matches for each (i+1) combination of masks. The number of matches of n-masks is the number of binary strings matching all masks.

If there are n masks, the desired output is the sum of all c's, ie. c = c0+...+cn+c01+c02+...+c(n-2)(n-1)+c012+...+c(n-3)(n-2)(n-1)+...+c0123...(n-2)(n-1). What the program computes is the alternating sum of all s's, ie. s = s_0-s_1+s_2-+...+-s_(n-1). We wish to proof that s==c.

n=1 is obvious. Consider n=2. Counting all matches of mask 0 gives c0+c01 (the number of strings matching only 0 + those matching both 0 and 1), counting all matches of 1 gives c1+c02. We can illustrate this as follows:

0: c0 c01
1: c1 c10

By definition, s0 = c0 + c1 + c12. s1 is the sum number of matches of each 2-combination of [0,1], ie. all uniqye c_ijs. Keep in mind that c01=c10.

s0 = c0 + c1 + 2 c01
s1 = c01
s = s0 - s1 = c0 + c1 + c01 = c

Thus s=c for n=2.

Now consider n=3.

0  : c0 + c01 + c02 + c012
1  : c1 + c01 + c12 + c012
2  : c2 + c12 + c02 + c012
01 : c01 + c012
02 : c02 + c012
12 : c12 + c012
012: c012

s0 = c0 + c1 + c2 + 2 (c01+c02+c03) + 3 c012
s1 = c01 + c02 + c12 + 3 c012
s2 = c012

s0 = c__0 + 2 c__1 + 3 c__2
s1 =          c__1 + 3 c__2
s2 =                   c__2

s = s0 - s1 + s2 = ... = c0 + c1 + c2 + c01 + c02 + c03 + c012 = c__0 + c__1 + c__2 = c

Thus s=c for n=3. c__i represents the of all cs with (i+1) indices, eg c__1 = c01 for n=2 and c__1 = c01 + c02 + c12 for n==3.

For n=4, a pattern starts to emerge:

0:   c0 + c01 + c02 + c03 + c012 + c013 + c023 + c0123
1:   c1 + c01 + c12 + c13 + c102 + c103 + c123 + c0123
2:   c2 + c02 + c12 + c23 + c201 + c203 + c213 + c0123
3:   c3 + c03 + c13 + c23 + c301 + c302 + c312 + c0123

01:  c01 + c012 + c013 + c0123
02:  c02 + c012 + c023 + c0123
03:  c03 + c013 + c023 + c0123
12:  c11 + c012 + c123 + c0123
13:  c13 + c013 + c123 + c0123
23:  c23 + c023 + c123 + c0123

012:  c012 + c0123
013:  c013 + c0123
023:  c023 + c0123
123:  c123 + c0123

0123: c0123

s0 = c__0 + 2 c__1 + 3 c__2 + 4 c__3
s1 =          c__1 + 3 c__2 + 6 c__3
s2 =                   c__2 + 4 c__3
s3 =                            c__3

s = s0 - s1 + s2 - s3 = c__0 + c__1 + c__2 + c__3 = c

Thus s==c for n=4.

In general, we get binomial coefficients like this (↓ is i, → is j):

   0  1  2  3  4  5  6  .  .  .

0  1  2  3  4  5  6  7  .  .  .
1     1  3  6  10 15 21 .  .  .
2        1  4  10 20 35 .  .  .
3           1  5  15 35 .  .  .
4              1  6  21 .  .  .
5                 1  7  .  .  .
6                    1  .  .  . 
.                       .
.                          .
.                             .

To see this, consider that for some i and j, there are:

  • x = ncr(n,i+1): combinations C for the intersection of (i+1) mask out of n
  • y = ncr(n-i-1,j-i): for each combination C above, there are y different combinations for the intersection of (j+2) masks out of those containing C
  • z = ncr(n,j+1): different combinations for the intersection of (j+1) masks out of n

As that may sound confusing, here's the defintion applied to an example. For i=1, j=2, n=4, it looks like this (cf. above):

01:  c01 + c012 + c013 + c0123
02:  c02 + c012 + c023 + c0123
03:  c03 + c013 + c023 + c0123
12:  c11 + c012 + c123 + c0123
13:  c13 + c013 + c123 + c0123
23:  c23 + c023 + c123 + c0123

So here x=6 (01, 02, 03, 12, 13, 23), y=2 (two c's with three indices for each combination), z=4 (c012, c013, c023, c123).

In total, there are x*y coefficients c with (j+1) indices, and there are z different ones, so each occurs x*y/z times, which we call the coefficient k_ij. By simple algebra, we get k_ij = ncr(n,i+1) ncr(n-i-1,j-i) / ncr(n,j+1) = ncr(j+1,i+1).

So the index is given by k_ij = nCr(j+1,i+1) If you recall all the defintions, all we need to show is that the alternating sum of each column gives 1.

The alternating sum s0 - s1 + s2 - s3 +- ... +- s(n-1) can thus be expressed as:

s_j = c__j * ∑[(-1)^(i+j) k_ij] for i=0..n-1
     = c__j * ∑[(-1)^(i+j) nCr(j+1,i+1)] for i=0..n-1
     = c__j * ∑[(-1)^(i+j) nCr(j+1,i)]{i=0..n} - (-1)^0 nCr(j+1,0)
     = (-1)^j c__j

s   = ∑[(-1)^j  s_j] for j = 0..n-1
    = ∑[(-1)^j (-1)^j c__j)] for j=0..n-1
    = ∑[c__j] for j=0..n-1
    = c

Thus s=c for all n=1,2,3,...

\$\endgroup\$
  • 1
    \$\begingroup\$ I am not sure if you are aware, but the method you are applying is en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle, so you don't have to prove it, if this was what you are trying to do. Also while not needed for the test cases you could eliminate from the groups masks that are completely included in another mask in the group. For example in TC5: 0011 < 2211, 0022 < 0222. I think that this makes the groups no larger than 2*n, though it is still too big in the worst case. \$\endgroup\$ – nutki May 15 '15 at 8:16
  • \$\begingroup\$ @nutki I hadn't been aware of this, so thanks for the link. Occasionally, proving and thinking about something for oneself is still a nice exercise, though:). As for your suggestion, yes, it had occured to me to do that, but I don't think I'm going to implement it unless a test case is added that requires it to get the result in a reasonable amount of time. \$\endgroup\$ – blutorange May 15 '15 at 9:24
  • \$\begingroup\$ @blutorange did u think to use decision tree ? \$\endgroup\$ – Abr001am May 15 '15 at 16:24
  • \$\begingroup\$ I think you mean intersection (matches both masks), not union (matches one or the other mask). \$\endgroup\$ – 2012rcampion May 15 '15 at 16:37
  • \$\begingroup\$ @2012rcampion As in unifying two masks the term union makes sense to me and I xan define that way, but you're right, in the interest of mutual understanding I've chabged. @Agawa001 Can you be more specific? Also, if you've got some good idea to make this faster, feel free to use any ideas from this answer for your program/answer. For now, it's fast enough for the large test case, and if made multi-threaded it should take <0.1s, which is below any meaningful measurement/comparison, so for harder test cases are needed. \$\endgroup\$ – blutorange May 15 '15 at 17:05
1
\$\begingroup\$

C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <gsl/gsl_combination.h>

int main (int argc, char *argv[]) {

    printf ("reading\n");
    char buffer[100];
    gets(buffer);
    char n = atoi(buffer);

    char *masks[1000];
    masks[0] = malloc(2 * n * sizeof(char));
    char c,nrows,j,biggestzerorun,biggestonerun,currentzerorun,currentonerun = 0;

    while ((c = getchar()) && c != EOF) {
        if (c == '\n') {
            nrows++;
            if (currentonerun > biggestonerun) {
                biggestonerun = currentonerun;
            }
            if (currentzerorun > biggestzerorun) {
                biggestzerorun = currentzerorun;
            }
            j=currentonerun=currentzerorun=0;
            masks[nrows] = malloc(2 * n * sizeof(char));
        } else if (c == '0') {
            masks[nrows][j++] = 1;
            currentzerorun++;
            if (currentonerun > biggestonerun) {
                biggestonerun = currentonerun;
            }
            currentonerun=0;
        } else if (c == '1') {
            masks[nrows][j++] = 2;
            currentonerun++;
            if (currentzerorun > biggestzerorun) {
                biggestzerorun = currentzerorun;
            }
            currentonerun=0;
        } else if (c == '2') {
            masks[nrows][j++] = 3;
            currentonerun++;
            currentzerorun++;
        }
    }
    if (currentonerun > biggestonerun) {
        biggestonerun = currentonerun;
    }
    if (currentzerorun > biggestzerorun) {
        biggestzerorun = currentzerorun;
    }

    printf("preparing combinations\n");

    int nmatches=0;

    gsl_combination *combination = gsl_combination_calloc(2*n, n);

    printf("entering loop:\n");

    do {
        char vector[2*n];
        char currentindex, previousindex;
        currentonerun = 0;
        memset(vector, 1, 2*n);


        // gsl_combination_fprintf (stdout, combination, "%u ");
        // printf(": ");

        for (char k=0; k<n; k++) {
            previousindex = currentindex;
            currentindex = gsl_combination_get(combination, k);
            if (k>0) {
                if (currentindex - previousindex == 1) {
                    currentonerun++;
                    if (currentonerun > biggestonerun) {
                        goto NEXT;
                    }
                } else {
                    currentonerun=0;
                    if (currentindex - previousindex > biggestzerorun) {
                        goto NEXT;
                    }
                }
            }
            vector[currentindex] = 2;
        }

        for (char k=0; k<=nrows; k++) {
            char ismatch = 1;
            for (char l=0; l<2*n; l++) {
                if (!(vector[l] & masks[k][l])) {
                    ismatch = 0;
                    break;
                }
            }
            if (ismatch) {
                nmatches++;
                break;
            }
        }

        NEXT: 1;

    } while (
        gsl_combination_next(combination) == GSL_SUCCESS
    );

    printf ("RESULT: %i\n", nmatches);

    gsl_combination_free(combination);
    for (; nrows>=0; nrows--) {
        free(masks[nrows]);
    }
}

Good luck getting the big input to run on this - it'll probably take all night to get through approx. 60^30 permutations! Maybe an intermediate sized dataset might be a good idea?

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.