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/* Original source code:
* G. Nong, S. Zhang and W. H. Chan, Two Efficient Algorithms for Linear Time Suffix Array Construction, IEEE Transactions on Computers, To Appear
* http://www.cs.sysu.edu.cn/nong/index.files/Two%20Efficient%20Algorithms%20for%20Linear%20Suffix%20Array%20Construction.pdf
*/
import "bit-vector.jsx";
class OArray
{
var offset : int;
var array : int[];
function constructor (array : int[])
{
this.array = array;
this.offset = 0;
}
function constructor (array : int[], offset : int)
{
this.array = array;
this.offset = offset;
}
function get (index : int) : int
{
return this.array[index + this.offset];
}
function set (index : int, value : int) : void
{
this.array[index + this.offset] = value;
}
function isS (index : int) : boolean
{
return this.array[index + this.offset] < this.array[index + this.offset + 1];
}
function compare (index1 : int, index2 : int) : boolean
{
return this.array[index1 + this.offset] == this.array[index2 + this.offset];
}
}
class SAIS
{
static function _isLMS (t : BitVector, i : int) : boolean
{
return i > 0 && t.get(i) && !t.get(i - 1);
}
// find the start or end of each bucket
static function _getBuckets(s : OArray, bkt : int[], n : int, K : int, end : boolean) : void
{
var sum = 0;
for (var i = 0; i <= K; i++)
{
bkt[i] = 0; // clear all buckets
}
for (var i = 0; i < n; i++)
{
bkt[s.get(i)]++; // compute the size of each bucket
}
for (var i = 0; i <= K; i++)
{
sum += bkt[i];
bkt[i] = end ? sum : sum - bkt[i];
}
}
// compute SAl
static function _induceSAl(t : BitVector, SA : int[], s : OArray, bkt : int[], n : int, K : int, end : boolean) : void
{
SAIS._getBuckets(s, bkt, n, K, end); // find starts of buckets
for (var i = 0; i < n; i++)
{
var j = SA[i] - 1;
if (j >= 0 && !t.get(j))
{
SA[bkt[s.get(j)]++] = j;
}
}
}
// compute SAs
static function _induceSAs(t : BitVector, SA : int[], s : OArray, bkt : int[], n : int, K : int, end : boolean) : void
{
SAIS._getBuckets(s, bkt, n, K, end); // find ends of buckets
for (var i = n - 1; i >= 0; i--)
{
var j = SA[i] - 1;
if (j >=0 && t.get(j))
{
SA[--bkt[s.get(j)]] = j;
}
}
}
// find the suffix array SA of s[0..n-1] in {1..K}^n
// require s[n-1]=0 (the sentinel!), n>=2
// use a working space (excluding s and SA) of at most 2.25n+O(1) for a constant alphabet
static function make(source : string) : int[]
{
var charCodes = [] : int[];
charCodes.length = source.length;
var maxCode = 0;
for (var i = 0; i < source.length; i++)
{
var code = source.charCodeAt(i);
charCodes[i] = code;
maxCode = (code > maxCode) ? code : maxCode;
}
var SA = [] : int[];
SA.length = source.length;
var s = new OArray(charCodes);
SAIS._make(s, SA, source.length, maxCode);
return SA;
}
static function _make(s : OArray, SA : int[], n : int, K : int) : void
{
// Classify the type of each character
var t = new BitVector();
t.set(n - 2, false);
t.set(n - 1, true); // the sentinel must be in s1, important!!!
for (var i = n - 3; i >= 0; i--)
{
t.set(i, (s.isS(i) || (s.compare(i, i + 1) && t.get(i + 1))));
}
// stage 1: reduce the problem by at least 1/2
// sort all the S-substrings
var bkt = [] : int[];
bkt.length = K + 1;
SAIS._getBuckets(s, bkt, n, K, true); // find ends of buckets
for (var i = 0; i < n; i++)
{
SA[i] = -1;
}
for (var i = 1; i < n; i++)
{
if (SAIS._isLMS(t, i))
{
SA[--bkt[s.get(i)]] = i;
}
}
SAIS._induceSAl(t, SA, s, bkt, n, K, false);
SAIS._induceSAs(t, SA, s, bkt, n, K, true);
// compact all the sorted substrings into the first n1 items of SA
// 2*n1 must be not larger than n (proveable)
var n1 = 0;
for (var i = 0; i < n; i++)
{
if (SAIS._isLMS(t, SA[i]))
{
SA[n1++] = SA[i];
}
}
// find the lexicographic names of all substrings
for (var i = n1; i < n; i++)
{
SA[i]=-1; // init the name array buffer
}
var name = 0;
var prev = -1;
for (i = 0; i < n1; i++)
{
var pos = SA[i];
var diff = false;
for (var d = 0; d < n; d++)
{
if (prev == -1 || !s.compare(pos + d, prev + d) || t.get(pos + d) != t.get(prev + d))
{
diff = true;
break;
}
else if (d > 0 && (SAIS._isLMS(t, pos+d) || SAIS._isLMS(t, prev + d)))
{
break;
}
}
if (diff)
{
name++;
prev = pos;
}
pos = (pos % 2 == 0) ? pos / 2 : (pos - 1) / 2;
SA[n1 + pos] = name - 1;
}
for (var i = n - 1, j = n - 1; i >= n1; i--)
{
if (SA[i] >= 0)
{
SA[j--] = SA[i];
}
}
// stage 2: solve the reduced problem
// recurse if names are not yet unique
var SA1 = SA;
var s1 = new OArray(SA, n - n1);
if (name < n1)
{
SAIS._make(s1, SA1, n1, name - 1);
}
else
{
// generate the suffix array of s1 directly
for (i = 0; i < n1; i++)
{
SA1[s1.get(i)] = i;
}
}
// stage 3: induce the result for the original problem
bkt = [] : int[];
bkt.length = K + 1;
// put all left-most S characters into their buckets
SAIS._getBuckets(s, bkt, n, K, true); // find ends of buckets
for (i = 1, j = 0; i < n; i++)
{
if (SAIS._isLMS(t, i))
{
s1.set(j++, i); // get p1
}
}
for (i = 0; i < n1; i++)
{
SA1[i] = s1.get(SA1[i]); // get index in s
}
for (i = n1; i < n; i++)
{
SA[i] = -1; // init SA[n1..n-1]
}
for (i = n1 - 1; i >= 0; i--)
{
j = SA[i];
SA[i] = -1;
SA[--bkt[s.get(j)]] = j;
}
SAIS._induceSAl(t, SA, s, bkt, n, K, false);
SAIS._induceSAs(t, SA, s, bkt, n, K, true);
}
}
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