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[Jython-checkins] CVS: jython/org/python/core MergeState.java,NONE,1.1
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From: Finn B. <bc...@us...> - 2002-09-17 16:06:16
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Update of /cvsroot/jython/jython/org/python/core
In directory usw-pr-cvs1:/tmp/cvs-serv12216
Added Files:
MergeState.java
Log Message:
Initial version of Tim Peters stable sort algorithm.
--- NEW FILE: MergeState.java ---
// Copyright 2002 Finn Bock
package org.python.core;
/**
* The MergeState class is a java implementation of the sort created
* Tim Peters and added to CPython2.3.
*
* The algorithm is described in details in the file
* python/dist/src/Objects/listsort.txt in the CPython development CVS.
*/
class MergeState {
/**
* The maximum number of entries in a MergeState's pending-runs stack.
* This is enough to sort arrays of size up to about
* 32 * phi ** MAX_MERGE_PENDING
* where phi ~= 1.618. 85 is ridiculously large enough, good for an
* array with 2**64 elements.
*/
static final int MAX_MERGE_PENDING = 85;
/**
* If a run wins MIN_GALLOP times in a row, we switch to galloping mode,
* and stay there until both runs win less often than MIN_GALLOP
* consecutive times. See listsort.txt for more info.
*/
static final int MIN_GALLOP = 8;
/**
* Initial temp array size
*/
static final int MERGESTATE_TEMP_SIZE = 256;
private PyObject[] a = new PyObject[MERGESTATE_TEMP_SIZE];
private int[] base = new int[MAX_MERGE_PENDING];
private int[] len = new int[MAX_MERGE_PENDING];
private PyObject compare;
private PyObject[] data;
private int size;
private int n;
MergeState(PyObject[] data, int size, PyObject compare) {
this.data = data;
this.compare = compare;
this.size = size;
this.n = 0;
}
public void sort() {
int nremaining = size;
if (nremaining < 2) {
return;
}
int lo = 0;
int hi = nremaining;
int minrun = merge_compute_minrun(nremaining);
boolean[] descending = new boolean[1];
do {
/* Identify next run. */
int n = count_run(lo, hi, descending);
if (descending[0])
reverse_slice(lo, lo + n);
/* If short, extend to min(minrun, nremaining). */
if (n < minrun) {
int force = nremaining < minrun ? nremaining : minrun;
binarysort(lo, lo + force, lo + n);
n = force;
}
/* Push run onto pending-runs stack, and maybe merge. */
//ms.assert_(ms.n < ms.MAX_MERGE_PENDING);
base[this.n] = lo;
len[this.n] = n;
++this.n;
merge_collapse();
/* Advance to find next run */
lo += n;
nremaining -= n;
} while (nremaining != 0);
//assert_(lo == hi);
merge_force_collapse();
//assert_(ms.n == 1);
//assert_(ms.base[0] == 0);
//assert_(ms.len[0] == size);
}
public void getmem(int need) {
if (need <= a.length)
return;
a = new PyObject[need];
}
int count_run(int lo, int hi, boolean[] descending) {
//assert_(lo < hi);
descending[0] = false;
++lo;
if (lo == hi)
return 1;
int n = 2;
if (iflt(data[lo], data[lo-1])) {
descending[0] = true;
for (lo = lo + 1; lo < hi; ++lo, ++n) {
if (! iflt(data[lo], data[lo-1]))
break;
}
} else {
for (lo = lo + 1; lo < hi; ++lo, ++n) {
if (iflt(data[lo], data[lo-1]))
break;
}
}
return n;
}
void merge_lo(int pa, int na, int pb, int nb) {
//debug("merge_lo pa:" + pa + " na:" + na + " pb:" + pb + " nb:" + nb);
int cnt = nb + na;
int origpa = pa;
//dump_data("padata", pa, na);
//dump_data("pbdata", pb, nb);
//assert_(na > 0 && nb > 0 && pa + na == pb);
getmem(na);
System.arraycopy(data, pa, a, 0, na);
int dest = pa;
pa = 0;
data[dest++] = data[pb++];
--nb;
if (nb == 0)
return;
if (na == 1) {
// CopyB;
System.arraycopy(data, pb, data, dest, nb);
data[dest + nb] = a[pa];
return;
}
try {
PyObject compare = this.compare;
for (;;) {
int acount = 0; /* # of time A won in a row */
int bcount = 0; /* # of time B won in a row */
/* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
for (;;) {
boolean k = iflt(data[pb], a[pa]);
if (k) {
data[dest++] = data[pb++];
++bcount;
acount = 0;
--nb;
if (nb == 0)
return;
if (bcount >= MIN_GALLOP)
break;
} else {
data[dest++] = a[pa++];
++acount;
bcount = 0;
--na;
if (na == 1) {
// CopyB;
System.arraycopy(data, pb, data, dest, nb);
data[dest + nb] = a[pa];
na = 0;
return;
}
if (acount >= MIN_GALLOP)
break;
}
}
/* One run is winning so consistently that galloping may
* be a huge win. So try that, and continue galloping until
* (if ever) neither run appears to be winning consistently
* anymore.
*/
do {
int k = gallop_right(data[pb], a, pa, na, 0);
acount = k;
if (k != 0) {
System.arraycopy(a, pa, data, dest, k);
dest += k;
pa += k;
na -= k;
if (na == 1) {
// CopyB
System.arraycopy(data, pb, data, dest, nb);
data[dest + nb] = a[pa];
na = 0;
return;
}
/* na==0 is impossible now if the comparison
* function is consistent, but we can't assume
* that it is.
*/
if (na == 0)
return;
}
data[dest++] = data[pb++];
--nb;
if (nb == 0)
return;
k = gallop_left(a[pa], data, pb, nb, 0);
bcount = k;
if (k != 0) {
System.arraycopy(data, pb, data, dest, k);
dest += k;
pb += k;
nb -= k;
if (nb == 0)
return;
}
data[dest++] = a[pa++];
--na;
if (na == 1) {
// CopyB;
System.arraycopy(data, pb, data, dest, nb);
data[dest + nb] = a[pa];
na = 0;
return;
}
} while (acount >= MIN_GALLOP || bcount >= MIN_GALLOP);
}
} finally {
if (na != 0)
System.arraycopy(a, pa, data, dest, na);
//dump_data("result", origpa, cnt);
}
}
void merge_hi(int pa, int na, int pb, int nb) {
//debug("merge_hi pa:" + pa + " na:" + na + " pb:" + pb + " nb:" + nb);
int cnt = nb + na;
int origpa = pa;
//dump_data("padata", pa, na);
//dump_data("pbdata", pb, nb);
//assert_(na > 0 && nb > 0 && pa + na == pb);
getmem(nb);
int dest = pb + nb - 1;
int basea = pa;
System.arraycopy(data, pb, a, 0, nb);
pb = nb - 1;
pa += na - 1;
data[dest--] = data[pa--];
--na;
if (na == 0)
return;
if (nb == 1) {
// CopyA;
dest -= na;
pa -= na;
System.arraycopy(data, pa+1, data, dest+1, na);
data[dest] = a[pb];
nb = 0;
return;
}
try {
PyObject compare = this.compare;
for (;;) {
int acount = 0; /* # of time A won in a row */
int bcount = 0; /* # of time B won in a row */
/* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
for (;;) {
boolean k = iflt(a[pb], data[pa]);
if (k) {
data[dest--] = data[pa--];
++acount;
bcount = 0;
--na;
if (na == 0)
return;
if (acount >= MIN_GALLOP)
break;
} else {
data[dest--] = a[pb--];
++bcount;
acount = 0;
--nb;
if (nb == 1) {
// CopyA
dest -= na;
pa -= na;
System.arraycopy(data, pa+1, data, dest+1, na);
data[dest] = a[pb];
nb = 0;
return;
}
if (bcount >= MIN_GALLOP)
break;
}
}
/* One run is winning so consistently that galloping may
* be a huge win. So try that, and continue galloping until
* (if ever) neither run appears to be winning consistently
* anymore.
*/
do {
int k = gallop_right(a[pb], data, basea, na, na-1);
acount = k = na - k;
if (k != 0) {
dest -= k;
pa -= k;
System.arraycopy(data, pa+1, data, dest+1, k);
na -= k;
if (na == 0)
return;
}
data[dest--] = a[pb--];
--nb;
if (nb == 1) {
// CopyA
dest -= na;
pa -= na;
System.arraycopy(data, pa+1, data, dest+1, na);
data[dest] = a[pb];
nb = 0;
return;
}
k = gallop_left(data[pa], a, 0, nb, nb-1);
bcount = k = nb - k;
if (k != 0) {
dest -= k;
pb -= k;
System.arraycopy(a, pb+1, data, dest+1, k);
nb -= k;
if (nb == 1) {
// CopyA
dest -= na;
pa -= na;
System.arraycopy(data, pa+1, data, dest+1, na);
data[dest] = a[pb];
nb = 0;
return;
}
/* nb==0 is impossible now if the comparison
* function is consistent, but we can't assume
* that it is.
*/
if (nb == 0)
return;
}
data[dest--] = data[pa--];
--na;
if (na == 0)
return;
} while (acount >= MIN_GALLOP || bcount >= MIN_GALLOP);
}
} finally {
if (nb != 0)
System.arraycopy(a, 0, data, dest-(nb-1), nb);
//dump_data("result", origpa, cnt);
}
}
/*
Locate the proper position of key in a sorted vector; if the vector contains
an element equal to key, return the position immediately to the left of
the leftmost equal element. [gallop_right() does the same except returns
the position to the right of the rightmost equal element (if any).]
"a" is a sorted vector with n elements, starting at a[0]. n must be > 0.
"hint" is an index at which to begin the search, 0 <= hint < n. The closer
hint is to the final result, the faster this runs.
The return value is the int k in 0..n such that
a[k-1] < key <= a[k]
pretending that *(a-1) is minus infinity and a[n] is plus infinity. IOW,
key belongs at index k; or, IOW, the first k elements of a should precede
key, and the last n-k should follow key.
Returns -1 on error. See listsort.txt for info on the method.
*/
private int gallop_left(PyObject key, PyObject[] data, int a, int n,
int hint)
{
//assert_(n > 0 && hint >= 0 && hint < n);
a += hint;
int ofs = 1;
int lastofs = 0;
if (iflt(data[a], key)) {
/* a[hint] < key -- gallop right, until
* a[hint + lastofs] < key <= a[hint + ofs]
*/
int maxofs = n - hint; // data[a + n - 1] is highest
while (ofs < maxofs) {
if (iflt(data[a + ofs], key)) {
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxofs;
} else {
// key < data[a + hint + ofs]
break;
}
}
if (ofs > maxofs)
ofs = maxofs;
// Translate back to offsets relative to a.
lastofs += hint;
ofs += hint;
} else {
/* key <= a[hint] -- gallop left, until
* a[hint - ofs] < key <= a[hint - lastofs]
*/
int maxofs = hint + 1; // data[a] is lowest
while (ofs < maxofs) {
if (iflt(data[a - ofs], key))
break;
// key <= data[a + hint - ofs]
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxofs;
}
if (ofs > maxofs)
ofs = maxofs;
// Translate back to offsets relative to a.
int k = lastofs;
lastofs = hint - ofs;
ofs = hint - k;
}
a -= hint;
//assert_(-1 <= lastofs && lastofs < ofs && ofs <= n);
/* Now a[lastofs] < key <= a[ofs], so key belongs somewhere to the
* right of lastofs but no farther right than ofs. Do a binary
* search, with invariant a[lastofs-1] < key <= a[ofs].
*/
++lastofs;
while (lastofs < ofs) {
int m = lastofs + ((ofs - lastofs) >> 1);
if (iflt(data[a + m], key))
lastofs = m+1; // data[a + m] < key
else
ofs = m; // key <= data[a + m]
}
//assert_(lastofs == ofs); // so data[a + ofs -1] < key <= data[a+ofs]
return ofs;
}
/*
* Exactly like gallop_left(), except that if key already exists in a[0:n],
* finds the position immediately to the right of the rightmost equal value.
*
* The return value is the int k in 0..n such that
* a[k-1] <= key < a[k]
* or -1 if error.
*
* The code duplication is massive, but this is enough different given that
* we're sticking to "<" comparisons that it's much harder to follow if
* written as one routine with yet another "left or right?" flag.
*/
private int gallop_right(PyObject key, PyObject[] data, int a, int n,
int hint)
{
//assert_(n > 0 && hint >= 0 && hint < n);
a += hint;
int lastofs = 0;
int ofs = 1;
if (iflt(key, data[a])) {
/* key < a[hint] -- gallop left, until
* a[hint - ofs] <= key < a[hint - lastofs]
*/
int maxofs = hint + 1; /* data[a] is lowest */
while (ofs < maxofs) {
if (iflt(key, data[a - ofs])) {
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) // int overflow
ofs = maxofs;
} else {
/* a[hint - ofs] <= key */
break;
}
}
if (ofs > maxofs)
ofs = maxofs;
/* Translate back to positive offsets relative to &a[0]. */
int k = lastofs;
lastofs = hint - ofs;
ofs = hint - k;
} else {
/* a[hint] <= key -- gallop right, until
* a[hint + lastofs] <= key < a[hint + ofs]
*/
int maxofs = n - hint; /* data[a + n - 1] is highest */
while (ofs < maxofs) {
if (iflt(key, data[a + ofs]))
break;
/* a[hint + ofs] <= key */
lastofs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0) /* int overflow */
ofs = maxofs;
}
if (ofs > maxofs)
ofs = maxofs;
/* Translate back to offsets relative to &a[0]. */
lastofs += hint;
ofs += hint;
}
a -= hint;
//assert_(-1 <= lastofs && lastofs < ofs && ofs <= n);
/* Now a[lastofs] <= key < a[ofs], so key belongs somewhere to the
* right of lastofs but no farther right than ofs. Do a binary
* search, with invariant a[lastofs-1] <= key < a[ofs].
*/
++lastofs;
while (lastofs < ofs) {
int m = lastofs + ((ofs - lastofs) >> 1);
if (iflt(key, data[a + m]))
ofs = m; // key < data[a + m]
else
lastofs = m+1; // data[a + m] <= key
}
//assert_(lastofs == ofs); // so data[a + ofs -1] <= key < data[a+ofs]
return ofs;
}
void merge_at(int i) {
//assert_(n >= 2);
//assert_(i >= 0);
//assert_(i == n - 2 || i == n - 3);
int pa = base[i];
int pb = base[i+1];
int na = len[i];
int nb = len[i+1];
//assert_(na > 0 && nb > 0);
//assert_(pa + na == pb);
// Record the length of the combined runs; if i is the 3rd-last
// run now, also slide over the last run (which isn't involved
// in this merge). The current run i+1 goes away in any case.
if (i == n - 3) {
len[i+1] = len[i+2];
base[i+1] = base[i+2];
}
len[i] = na + nb;
--n;
// Where does b start in a? Elements in a before that can be
// ignored (already in place).
int k = gallop_right(data[pb], data, pa, na, 0);
pa += k;
na -= k;
if (na == 0)
return;
// Where does a end in b? Elements in b after that can be
// ignored (already in place).
nb = gallop_left(data[pa + na - 1], data, pb, nb, nb-1);
if (nb == 0)
return;
// Merge what remains of the runs, using a temp array with
// min(na, nb) elements.
if (na <= nb)
merge_lo(pa, na, pb, nb);
else
merge_hi(pa, na, pb, nb);
}
/* Examine the stack of runs waiting to be merged, merging adjacent runs
* until the stack invariants are re-established:
*
* 1. len[-3] > len[-2] + len[-1]
* 2. len[-2] > len[-1]
*
* See listsort.txt for more info.
*/
void merge_collapse() {
while (this.n > 1) {
int n = this.n - 2;
if (n > 0 && len[n-1] <= len[n] + len[n+1]) {
if (len[n-1] < len[n+1])
--n;
merge_at(n);
} else if (len[n] <= len[n+1]) {
merge_at(n);
} else {
break;
}
}
}
/* Regardless of invariants, merge all runs on the stack until only one
* remains. This is used at the end of the mergesort.
*
* Returns 0 on success, -1 on error.
*/
void merge_force_collapse() {
while (this.n > 1) {
int n = this.n - 2;
if (n > 0 && len[n-1] < len[n+1])
--n;
merge_at(n);
}
}
/* Compute a good value for the minimum run length; natural runs shorter
* than this are boosted artificially via binary insertion.
*
* If n < 64, return n (it's too small to bother with fancy stuff).
* Else if n is an exact power of 2, return 32.
* Else return an int k, 32 <= k <= 64, such that n/k is close to, but
* strictly less than, an exact power of 2.
*
* See listsort.txt for more info.
*/
int merge_compute_minrun(int n) {
int r = 0; // becomes 1 if any 1 bits are shifted off
//assert_(n >= 0);
while (n >= 64) {
r |= n & 1;
n >>= 1;
}
return n + r;
}
void assert_(boolean expr) {
if (!expr)
throw new RuntimeException("assert");
}
private boolean iflt(PyObject x, PyObject y) {
//assert_(x != null);
//assert_(y != null);
if (compare == null) {
/* NOTE: we rely on the fact here that the sorting algorithm
only ever checks whether k<0, i.e., whether x<y. So we
invoke the rich comparison function with _lt ('<'), and
return -1 when it returns true and 0 when it returns
false. */
return x._lt(y).__nonzero__();
}
PyObject ret = compare.__call__(x, y);
if (ret instanceof PyInteger) {
int v = ((PyInteger)ret).getValue();
return v < 0;
}
throw Py.TypeError("comparision function must return int");
}
void reverse_slice(int lo, int hi) {
--hi;
while (lo < hi) {
PyObject t = data[lo];
data[lo] = data[hi];
data[hi] = t;
++lo;
--hi;
}
}
/*
* binarysort is the best method for sorting small arrays: it does
* few compares, but can do data movement quadratic in the number of
* elements.
* [lo, hi) is a contiguous slice of a list, and is sorted via
* binary insertion. This sort is stable.
* On entry, must have lo <= start <= hi, and that [lo, start) is already
* sorted (pass start == lo if you don't know!).
* If islt() complains return -1, else 0.
* Even in case of error, the output slice will be some permutation of
* the input (nothing is lost or duplicated).
*/
void binarysort(int lo, int hi, int start) {
//debug("binarysort: lo:" + lo + " hi:" + hi + " start:" + start);
int p;
//assert_(lo <= start && start <= hi);
/* assert [lo, start) is sorted */
if (lo == start)
++start;
for (; start < hi; ++start) {
/* set l to where *start belongs */
int l = lo;
int r = start;
PyObject pivot = data[r];
// Invariants:
// pivot >= all in [lo, l).
// pivot < all in [r, start).
// The second is vacuously true at the start.
//assert_(l < r);
do {
p = l + ((r - l) >> 1);
if (iflt(pivot, data[p]))
r = p;
else
l = p+1;
} while (l < r);
//assert_(l == r);
// The invariants still hold, so pivot >= all in [lo, l) and
// pivot < all in [l, start), so pivot belongs at l. Note
// that if there are elements equal to pivot, l points to the
// first slot after them -- that's why this sort is stable.
// Slide over to make room.
for (p = start; p > l; --p)
data[p] = data[p - 1];
data[l] = pivot;
}
//dump_data("binsort", lo, hi - lo);
}
private void dump_data(String txt, int lo, int n) {
/*
System.out.print(txt + ":");
for (int i = 0; i < n; i++)
System.out.print(data[lo + i] + " ");
System.out.println();
*/
}
private void debug(String str) {
//System.out.println(str);
}
}
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