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view pyPRADA_1.2/tools/bwa-0.5.7-mh/is.c @ 0:acc2ca1a3ba4
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author | siyuan |
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date | Thu, 20 Feb 2014 00:44:58 -0500 |
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/* * sais.c for sais-lite * Copyright (c) 2008 Yuta Mori All Rights Reserved. * * Permission is hereby granted, free of charge, to any person * obtaining a copy of this software and associated documentation * files (the "Software"), to deal in the Software without * restriction, including without limitation the rights to use, * copy, modify, merge, publish, distribute, sublicense, and/or sell * copies of the Software, and to permit persons to whom the * Software is furnished to do so, subject to the following * conditions: * * The above copyright notice and this permission notice shall be * included in all copies or substantial portions of the Software. * * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES * OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND * NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT * HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, * WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING * FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR * OTHER DEALINGS IN THE SOFTWARE. */ #include <stdlib.h> typedef unsigned char ubyte_t; #define chr(i) (cs == sizeof(int) ? ((const int *)T)[i]:((const unsigned char *)T)[i]) /* find the start or end of each bucket */ static void getCounts(const unsigned char *T, int *C, int n, int k, int cs) { int i; for (i = 0; i < k; ++i) C[i] = 0; for (i = 0; i < n; ++i) ++C[chr(i)]; } static void getBuckets(const int *C, int *B, int k, int end) { int i, sum = 0; if (end) { for (i = 0; i < k; ++i) { sum += C[i]; B[i] = sum; } } else { for (i = 0; i < k; ++i) { sum += C[i]; B[i] = sum - C[i]; } } } /* compute SA */ static void induceSA(const unsigned char *T, int *SA, int *C, int *B, int n, int k, int cs) { int *b, i, j; int c0, c1; /* compute SAl */ if (C == B) getCounts(T, C, n, k, cs); getBuckets(C, B, k, 0); /* find starts of buckets */ j = n - 1; b = SA + B[c1 = chr(j)]; *b++ = ((0 < j) && (chr(j - 1) < c1)) ? ~j : j; for (i = 0; i < n; ++i) { j = SA[i], SA[i] = ~j; if (0 < j) { --j; if ((c0 = chr(j)) != c1) { B[c1] = b - SA; b = SA + B[c1 = c0]; } *b++ = ((0 < j) && (chr(j - 1) < c1)) ? ~j : j; } } /* compute SAs */ if (C == B) getCounts(T, C, n, k, cs); getBuckets(C, B, k, 1); /* find ends of buckets */ for (i = n - 1, b = SA + B[c1 = 0]; 0 <= i; --i) { if (0 < (j = SA[i])) { --j; if ((c0 = chr(j)) != c1) { B[c1] = b - SA; b = SA + B[c1 = c0]; } *--b = ((j == 0) || (chr(j - 1) > c1)) ? ~j : j; } else SA[i] = ~j; } } /* * find the suffix array SA of T[0..n-1] in {0..k-1}^n use a working * space (excluding T and SA) of at most 2n+O(1) for a constant alphabet */ static int sais_main(const unsigned char *T, int *SA, int fs, int n, int k, int cs) { int *C, *B, *RA; int i, j, c, m, p, q, plen, qlen, name; int c0, c1; int diff; /* stage 1: reduce the problem by at least 1/2 sort all the * S-substrings */ if (k <= fs) { C = SA + n; B = (k <= (fs - k)) ? C + k : C; } else if ((C = B = (int *) malloc(k * sizeof(int))) == NULL) return -2; getCounts(T, C, n, k, cs); getBuckets(C, B, k, 1); /* find ends of buckets */ for (i = 0; i < n; ++i) SA[i] = 0; for (i = n - 2, c = 0, c1 = chr(n - 1); 0 <= i; --i, c1 = c0) { if ((c0 = chr(i)) < (c1 + c)) c = 1; else if (c != 0) SA[--B[c1]] = i + 1, c = 0; } induceSA(T, SA, C, B, n, k, cs); if (fs < k) free(C); /* compact all the sorted substrings into the first m items of SA * 2*m must be not larger than n (proveable) */ for (i = 0, m = 0; i < n; ++i) { p = SA[i]; if ((0 < p) && (chr(p - 1) > (c0 = chr(p)))) { for (j = p + 1; (j < n) && (c0 == (c1 = chr(j))); ++j); if ((j < n) && (c0 < c1)) SA[m++] = p; } } for (i = m; i < n; ++i) SA[i] = 0; /* init the name array buffer */ /* store the length of all substrings */ for (i = n - 2, j = n, c = 0, c1 = chr(n - 1); 0 <= i; --i, c1 = c0) { if ((c0 = chr(i)) < (c1 + c)) c = 1; else if (c != 0) { SA[m + ((i + 1) >> 1)] = j - i - 1; j = i + 1; c = 0; } } /* find the lexicographic names of all substrings */ for (i = 0, name = 0, q = n, qlen = 0; i < m; ++i) { p = SA[i], plen = SA[m + (p >> 1)], diff = 1; if (plen == qlen) { for (j = 0; (j < plen) && (chr(p + j) == chr(q + j)); j++); if (j == plen) diff = 0; } if (diff != 0) ++name, q = p, qlen = plen; SA[m + (p >> 1)] = name; } /* stage 2: solve the reduced problem recurse if names are not yet * unique */ if (name < m) { RA = SA + n + fs - m; for (i = n - 1, j = m - 1; m <= i; --i) { if (SA[i] != 0) RA[j--] = SA[i] - 1; } if (sais_main((unsigned char *) RA, SA, fs + n - m * 2, m, name, sizeof(int)) != 0) return -2; for (i = n - 2, j = m - 1, c = 0, c1 = chr(n - 1); 0 <= i; --i, c1 = c0) { if ((c0 = chr(i)) < (c1 + c)) c = 1; else if (c != 0) RA[j--] = i + 1, c = 0; /* get p1 */ } for (i = 0; i < m; ++i) SA[i] = RA[SA[i]]; /* get index */ } /* stage 3: induce the result for the original problem */ if (k <= fs) { C = SA + n; B = (k <= (fs - k)) ? C + k : C; } else if ((C = B = (int *) malloc(k * sizeof(int))) == NULL) return -2; /* put all left-most S characters into their buckets */ getCounts(T, C, n, k, cs); getBuckets(C, B, k, 1); /* find ends of buckets */ for (i = m; i < n; ++i) SA[i] = 0; /* init SA[m..n-1] */ for (i = m - 1; 0 <= i; --i) { j = SA[i], SA[i] = 0; SA[--B[chr(j)]] = j; } induceSA(T, SA, C, B, n, k, cs); if (fs < k) free(C); return 0; } /** * Constructs the suffix array of a given string. * @param T[0..n-1] The input string. * @param SA[0..n] The output array of suffixes. * @param n The length of the given string. * @return 0 if no error occurred */ int is_sa(const ubyte_t *T, int *SA, int n) { if ((T == NULL) || (SA == NULL) || (n < 0)) return -1; SA[0] = n; if (n <= 1) { if (n == 1) SA[1] = 0; return 0; } return sais_main(T, SA+1, 0, n, 256, 1); } /** * Constructs the burrows-wheeler transformed string of a given string. * @param T[0..n-1] The input string. * @param n The length of the given string. * @return The primary index if no error occurred, -1 or -2 otherwise. */ int is_bwt(ubyte_t *T, int n) { int *SA, i, primary = 0; SA = (int*)calloc(n+1, sizeof(int)); is_sa(T, SA, n); for (i = 0; i <= n; ++i) { if (SA[i] == 0) primary = i; else SA[i] = T[SA[i] - 1]; } for (i = 0; i < primary; ++i) T[i] = SA[i]; for (; i < n; ++i) T[i] = SA[i + 1]; free(SA); return primary; }