Home | History | Annotate | Download | only in examples

Lines Matching refs:left

120       left: number of available bit patterns at length len
126 left: 2..syms - 1, but only the evens (so syms == 8 -> 2, 4, 6)
129 syms == 2 is not saved since that immediately leads to a single code. left
132 left ends at syms-1 since left == syms immediately results in a single code.
133 (left > sym is not allowed since that would result in an incomplete code.)
151 to the num[] array as described above for the (syms, left, len) triplet.
199 len through max inclusive, coding syms symbols, with left bit patterns of
203 local big_t count(int syms, int len, int left)
213 if (syms == left)
217 assert(syms > left && left > 0 && len < max);
220 index = INDEX(syms, left, len);
227 least = (left << 1) - syms;
233 no limit to the code length, this would become: most = left - 1) */
234 most = (((code_t)left << (max - len)) - syms) /
240 got = count(syms - use, len + 1, (left - use) << 1);
255 bit vector to indicate visiting this state. Each (syms,len,left) state
258 local int beenhere(int syms, int len, int left, int mem, int rem)
266 /* point to vector for (syms,left,len), bit in vector for (mem,rem) */
267 index = INDEX(syms, left, len);
317 /* Examine all possible codes from the given node (syms, len, left). Compute
322 local void examine(int syms, int len, int left, int mem, int rem)
329 if (syms == left) {
331 code[len] = left;
334 while (rem < left) {
335 left -= rem;
339 assert(rem == left);
358 if (beenhere(syms, len, left, mem, rem))
363 least = (left << 1) - syms;
369 no limit to the code length, this would become: most = left - 1) */
370 most = (((code_t)left << (max - len)) - syms) /
385 examine(syms - use, len + 1, (left - use) << 1,
399 intermediate code states (syms, left, len). For each completed code,
406 int left; /* number of unused bit patterns at this length */
417 for (left = 2; left < n; left += 2)
421 index = INDEX(n, left, root + 1);
423 examine(n, root + 1, left, 1 << root, 0);
427 if (num[index - 1] && n <= left << 1)
428 examine((n - left) << 1, root + 1, (n - left) << 1,