eu.interedition.collatex.suffixarray.Skew Maven / Gradle / Ivy
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package eu.interedition.collatex.suffixarray;
import java.util.Arrays;
/**
*
* Straightforward reimplementation of the recursive algorithm given in:
* J. Kärkkäinen and P. Sanders. Simple linear work suffix array construction.
* In Proc. 13th International Conference on Automata, Languages and Programming,
* Springer, 2003
*
*
* This implementation is basically a translation of the C++ version given by Juha
* Kärkkäinen and Peter Sanders.
*
* The implementation of this algorithm makes some assumptions about the input. See
* {@link #buildSuffixArray(int[], int, int)} for details.
*
* @author Michał Nowak (Carrot Search)
* @author Dawid Weiss (Carrot Search)
*/
public final class Skew implements ISuffixArrayBuilder {
/**
* Lexicographic order for pairs.
*/
private final static boolean leq(int a1, int a2, int b1, int b2) {
return (a1 < b1 || (a1 == b1 && a2 <= b2));
}
/**
* Lexicographic order for triples.
*/
private final static boolean leq(int a1, int a2, int a3, int b1, int b2, int b3) {
return (a1 < b1 || (a1 == b1 && leq(a2, a3, b2, b3)));
}
/**
* Stably sort indexes from src[0..n-1] to dst[0..n-1] with values in 0..K from v. A
* constant offset of vi is added to indexes from src.
*/
private final static void radixPass(int[] src, int[] dst, int[] v, int vi,
final int n, final int K, int start, int[] cnt) {
// check counter array's size.
assert cnt.length >= K + 1;
Arrays.fill(cnt, 0, K + 1, 0);
// count occurrences
for (int i = 0; i < n; i++)
cnt[v[start + vi + src[i]]]++;
// exclusive prefix sums
for (int i = 0, sum = 0; i <= K; i++) {
final int t = cnt[i];
cnt[i] = sum;
sum += t;
}
// sort
for (int i = 0; i < n; i++)
dst[cnt[v[start + vi + src[i]]]++] = src[i];
}
/**
* Find the suffix array SA of s[0..n-1] in {1..K}^n. require s[n] = s[n+1] = s[n+2] =
* 0, n >= 2.
*/
static final int[] suffixArray(int[] s, int[] SA, int n, final int K, int start, int[] cnt) {
final int n0 = (n + 2) / 3, n1 = (n + 1) / 3, n2 = n / 3, n02 = n0 + n2;
final int[] s12 = new int[n02 + 3];
s12[n02] = s12[n02 + 1] = s12[n02 + 2] = 0;
final int[] SA12 = new int[n02 + 3];
SA12[n02] = SA12[n02 + 1] = SA12[n02 + 2] = 0;
final int[] s0 = new int[n0];
final int[] SA0 = new int[n0];
/*
* generate positions of mod 1 and mod 2 suffixes the "+(n0-n1)" adds a dummy mod
* 1 suffix if n%3 == 1
*/
for (int i = 0, j = 0; i < n + (n0 - n1); i++)
if ((i % 3) != 0) s12[j++] = i;
// lsb radix sort the mod 1 and mod 2 triples
cnt = ensureSize(cnt, K + 1);
radixPass(s12, SA12, s, +2, n02, K, start, cnt);
radixPass(SA12, s12, s, +1, n02, K, start, cnt);
radixPass(s12, SA12, s, +0, n02, K, start, cnt);
// find lexicographic names of triples
int name = 0, c0 = -1, c1 = -1, c2 = -1;
for (int i = 0; i < n02; i++) {
if (s[start + SA12[i]] != c0 || s[start + SA12[i] + 1] != c1
|| s[start + SA12[i] + 2] != c2) {
name++;
c0 = s[start + SA12[i]];
c1 = s[start + SA12[i] + 1];
c2 = s[start + SA12[i] + 2];
}
if ((SA12[i] % 3) == 1) {
// left half
s12[SA12[i] / 3] = name;
} else {
// right half
s12[SA12[i] / 3 + n0] = name;
}
}
// recurse if names are not yet unique
if (name < n02) {
cnt = suffixArray(s12, SA12, n02, name, start, cnt);
// store unique names in s12 using the suffix array
for (int i = 0; i < n02; i++)
s12[SA12[i]] = i + 1;
} else {
// generate the suffix array of s12 directly
for (int i = 0; i < n02; i++)
SA12[s12[i] - 1] = i;
}
// stably sort the mod 0 suffixes from SA12 by their first character
for (int i = 0, j = 0; i < n02; i++)
if (SA12[i] < n0) s0[j++] = 3 * SA12[i];
radixPass(s0, SA0, s, 0, n0, K, start, cnt);
// merge sorted SA0 suffixes and sorted SA12 suffixes
for (int p = 0, t = n0 - n1, k = 0; k < n; k++) {
// pos of current offset 12 suffix
final int i = (SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2);
// pos of current offset 0 suffix
final int j = SA0[p];
if (SA12[t] < n0 ? leq(s[start + i], s12[SA12[t] + n0], s[start + j],
s12[j / 3]) : leq(s[start + i], s[start + i + 1], s12[SA12[t] - n0 + 1],
s[start + j], s[start + j + 1], s12[j / 3 + n0])) {
// suffix from SA12 is smaller
SA[k] = i;
t++;
if (t == n02) {
// done --- only SA0 suffixes left
for (k++; p < n0; p++, k++)
SA[k] = SA0[p];
}
} else {
SA[k] = j;
p++;
if (p == n0) {
// done --- only SA12 suffixes left
for (k++; t < n02; t++, k++) {
SA[k] = (SA12[t] < n0 ? SA12[t] * 3 + 1 : (SA12[t] - n0) * 3 + 2);
}
}
}
}
return cnt;
}
/**
* Ensure array is large enough or reallocate (no copying).
*/
private static final int[] ensureSize(int[] tab, int length) {
if (tab.length < length) {
tab = null;
tab = new int[length];
}
return tab;
}
/**
* {@inheritDoc}
*
* Additional constraints enforced by Karkkainen-Sanders algorithm:
*
* - non-negative (>0) symbols in the input (because of radix sort)
* input.length >= start + length + 3 (to simplify
* border cases)- length >= 2
*
*
* If the input contains zero or negative values, or has no extra trailing cells,
* adapters can be used in the following way:
*
*
* return new {@link DensePositiveDecorator}(
* new {@link ExtraTrailingCellsDecorator}(
* new {@link Skew}(), 3));
*
*
* @see ExtraTrailingCellsDecorator
* @see DensePositiveDecorator
*/
@Override
public int[] buildSuffixArray(int[] input, int start, int length) {
Tools.assertAlways(input != null, "input must not be null");
Tools.assertAlways(length >= 2, "input length must be >= 2");
Tools.assertAlways(input.length >= start + length + 3, "no extra space after input end");
assert Tools.allPositive(input, start, length);
final int alphabetSize = Tools.max(input, start, length);
final int[] SA = new int[length + 3];
// Preserve the tail of the input (destroyed when constructing the array).
final int[] tail = new int[3];
System.arraycopy(input, start + length, tail, 0, 3);
Arrays.fill(input, start + length, start + length + 3, 0);
suffixArray(input, SA, length, alphabetSize, start, new int[alphabetSize + 2]);
// Reconstruct the input's tail.
System.arraycopy(tail, 0, input, start + length, 3);
return SA;
}
}