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    <p>Building on <a href="https://stackoverflow.com/questions/2936205/number-algorithm/2936248#2936248">@Mark Byers's</a> and <a href="https://stackoverflow.com/questions/2936205/number-algorithm/2936367#2936367">Moron's</a> answers your task can be reformulated as follows:</p> <p>Enumerate all ways to put <code>K</code> zeros into <code>N</code> places (see <a href="http://en.wikipedia.org/wiki/Combination" rel="nofollow noreferrer">combinations with repetition</a> and <a href="http://en.wikipedia.org/wiki/Stars_and_bars_%28probability%29" rel="nofollow noreferrer">Stars and bars</a>).</p> <p>Example: For 15 bits and 1 2 6 1 pattern there are N=5 places (before/after the number and between <code>1</code>s) to put K=2 zeros (number of leading zeros for a flush-right number). Number of ways is <a href="http://en.wikipedia.org/wiki/Binomial_coefficient" rel="nofollow noreferrer">binomial</a>(N + K - 1, K) i.e., binomial(5+2-1, 2) = 15.</p> <p>The key functions in the code below are <a href="http://photon.poly.edu/~hbr/boost/combinations.html#next_combin_counts_desc" rel="nofollow noreferrer"><code>next_combination_counts()</code></a> and <code>comb2number()</code>.</p> <h3>Full program in C</h3> <pre><code>#include &lt;assert.h&gt; #include &lt;stdbool.h&gt; #include &lt;stdio.h&gt; #define SIZE(arr) (sizeof(arr)/sizeof(*(arr))) #define PRInumber "u" typedef unsigned number_t; // swap values pointed to by the pointer static void iter_swap(int* ia, int* ib) { int t = *ia; *ia = *ib; *ib = t; } // see boost::next_combinations_counts() // http://photon.poly.edu/~hbr/boost/combinations.html // http://photon.poly.edu/~hbr/boost/combination.hpp static bool next_combination_counts(int* first, int* last) { /* 0 0 2 0 1 1 0 2 0 1 0 1 1 1 0 2 0 0 */ int* current = last; while (current != first &amp;&amp; *(--current) == 0) { } if (current == first) { if (first != last &amp;&amp; *first != 0) iter_swap(--last, first); return false; } --(*current); iter_swap(--last, current); ++(*(--current)); return true; } // convert combination and pattern to corresponding number // example: comb=[2, 0, 0] pattern=[1,1] =&gt; num=5 (101 binary) static number_t comb2number(int comb[], int comb_size, int pattern[], int pattern_size) { if (pattern_size == 0) return 0; assert(pattern_size &gt; 0); assert(comb_size &gt; pattern_size); // 111 -&gt; 1000 - 1 -&gt; 2**3 - 1 -&gt; (1 &lt;&lt; 3) - 1 // 111 &lt;&lt; 2 -&gt; 11100 number_t num = ((1 &lt;&lt; pattern[pattern_size-1]) - 1) &lt;&lt; comb[pattern_size]; int len = pattern[pattern_size-1] + comb[pattern_size]; for (int i = pattern_size - 1; i--&gt; 0; ) { num += ((1 &lt;&lt; pattern[i]) - 1) &lt;&lt; (comb[i+1] + 1 + len); len += pattern[i] + comb[i+1] + 1; } return num; } // print binary representation of number static void print_binary(number_t number) { if (number &gt; 0) { print_binary(number &gt;&gt; 1); printf("%d", number &amp; 1); } } // print array static void printa(int arr[], int size, const char* suffix) { printf("%s", "{"); for (int i = 0; i &lt; (size - 1); ++i) printf("%d, ", arr[i]); if (size &gt; 0) printf("%d", arr[size - 1]); printf("}%s", suffix); } static void fill0(int* first, int* last) { for ( ; first != last; ++first) *first = 0; } // generate {0,0,...,0,nzeros} combination static void init_comb(int comb[], int comb_size, int nzeros) { fill0(comb, comb + comb_size); comb[comb_size-1] = nzeros; } static int sum(int* first, int* last) { int s = 0; for ( ; first != last; ++first) s += *first; return s; } // calculated max width required to print number (in PRInumber format) static int maxwidth(int comb[], int comb_size, int pattern[], int pattern_size) { int initial_comb[comb_size]; int nzeros = sum(comb, comb + comb_size); init_comb(initial_comb, comb_size, nzeros); return snprintf(NULL, 0, "%" PRInumber, comb2number(initial_comb, comb_size, pattern, pattern_size)); } static void process(int comb[], int comb_size, int pattern[], int pattern_size) { // print combination and pattern printa(comb, comb_size, " "); printa(pattern, pattern_size, " "); // print corresponding number for (int i = 0; i &lt; comb[0]; ++i) printf("%s", "0"); number_t number = comb2number(comb, comb_size, pattern, pattern_size); print_binary(number); const int width = maxwidth(comb, comb_size, pattern, pattern_size); printf(" %*" PRInumber "\n", width, number); } // reverse the array static void reverse(int a[], int n) { for (int i = 0, j = n - 1; i &lt; j; ++i, --j) iter_swap(a + i, a + j); } // convert number to pattern // 101101111110100 -&gt; 1, 2, 6, 1 static int number2pattern(number_t num, int pattern[], int nbits, int comb[]) { // SIZE(pattern) &gt;= nbits // SIZE(comb) &gt;= nbits + 1 fill0(pattern, pattern + nbits); fill0(comb, comb + nbits + 1); int i = 0; int pos = 0; for (; i &lt; nbits &amp;&amp; num; ++i) { // skip trailing zeros for ( ; num &amp;&amp; !(num &amp; 1); num &gt;&gt;= 1, ++pos) ++comb[i]; // count number of 1s for ( ; num &amp; 1; num &gt;&gt;=1, ++pos) ++pattern[i]; } assert(i == nbits || pattern[i] == 0); const int pattern_size = i; // skip comb[0] for (int j = 1; j &lt; pattern_size; ++j) --comb[j]; comb[pattern_size] = nbits - pos; reverse(pattern, pattern_size); reverse(comb, pattern_size+1); return pattern_size; } int main(void) { number_t num = 11769; const int nbits = 15; // clear hi bits (required for `comb2number() != num` relation) if (nbits &lt; 8*sizeof(number_t)) num &amp;= ((number_t)1 &lt;&lt; nbits) - 1; else assert(nbits == 8*sizeof(number_t)); // `pattern` defines how 1s are distributed in the number int pattern[nbits]; // `comb` defines how zeros are distributed int comb[nbits+1]; const int pattern_size = number2pattern(num, pattern, nbits, comb); const int comb_size = pattern_size + 1; // check consistency // . find number of leading zeros in a flush-right version int nzeros = nbits; for (int i = 0; i &lt; (pattern_size - 1); ++i) nzeros -= pattern[i] + 1; assert(pattern_size &gt; 0); nzeros -= pattern[pattern_size - 1]; assert(nzeros&gt;=0); // . the same but using combination int nzeros_comb = sum(comb, comb + comb_size); assert(nzeros_comb == nzeros); // enumerate all combinations printf("Combination Pattern Binary Decimal\n"); assert(comb2number(comb, comb_size, pattern, pattern_size) == num); process(comb, comb_size, pattern, pattern_size); // process `num` // . until flush-left number while(next_combination_counts(comb, comb + comb_size)) process(comb, comb_size, pattern, pattern_size); // . until `num` number is encounterd while (comb2number(comb, comb_size, pattern, pattern_size) != num) { process(comb, comb_size, pattern, pattern_size); (void)next_combination_counts(comb, comb + comb_size); } return 0; } </code></pre> <p>Output:</p> <pre><code>Combination Pattern Binary Decimal {1, 0, 0, 1, 0} {1, 2, 6, 1} 010110111111001 11769 {1, 0, 1, 0, 0} {1, 2, 6, 1} 010110011111101 11517 {1, 1, 0, 0, 0} {1, 2, 6, 1} 010011011111101 9981 {2, 0, 0, 0, 0} {1, 2, 6, 1} 001011011111101 5885 {0, 0, 0, 0, 2} {1, 2, 6, 1} 101101111110100 23540 {0, 0, 0, 1, 1} {1, 2, 6, 1} 101101111110010 23538 {0, 0, 0, 2, 0} {1, 2, 6, 1} 101101111110001 23537 {0, 0, 1, 0, 1} {1, 2, 6, 1} 101100111111010 23034 {0, 0, 1, 1, 0} {1, 2, 6, 1} 101100111111001 23033 {0, 0, 2, 0, 0} {1, 2, 6, 1} 101100011111101 22781 {0, 1, 0, 0, 1} {1, 2, 6, 1} 100110111111010 19962 {0, 1, 0, 1, 0} {1, 2, 6, 1} 100110111111001 19961 {0, 1, 1, 0, 0} {1, 2, 6, 1} 100110011111101 19709 {0, 2, 0, 0, 0} {1, 2, 6, 1} 100011011111101 18173 {1, 0, 0, 0, 1} {1, 2, 6, 1} 010110111111010 11770 </code></pre>
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