Oracle America, Inc. v. Google Inc.

Filing 1192

TRIAL BRIEF Google's May 23, 2012 Copyright Liability Trial Brief by Google Inc.. (Attachments: #1 Exhibit A, #2 Exhibit B, #3 Exhibit C, #4 Exhibit D, #5 Exhibit E)(Van Nest, Robert) (Filed on 5/23/2012)

EXHIBIT E

G:\eclair21 - GOOGLE-00-00000525\dalvik\libcore\luni\src\main\java\java\util\Com
mparableTimSort.java
1 /*
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* Copyright (C) 2008 The Android Open Source Project
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*
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* Licensed under the Apache License, Version 2.0 (the "License");
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* you may not use this file except in compliance with the License.
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* You may obtain a copy of the License at
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*
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*
http://www.apache.org/licenses/LICENSE-2.0
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*
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* Unless required by applicable law or agreed to in writing, software
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* distributed under the License is distributed on an "AS IS" BASIS,
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* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or
implied.
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* See the License for the specific language governing permissions and
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* limitations under the License.
15
*/
16
17 package java.util;
18
19 /**
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* This is a near duplicate of {@link TimSort}, modified for use with
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* arrays of objects that implement {@link Comparable}, instead of using
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* explicit comparators.
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*
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* <p>If you are using an optimizing VM, you may find that
ComparableTimSort
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* offers no performance benefit over TimSort in conjunction with a
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* comparator that simply returns {@code
((Comparable)first).compareTo(Second)}.
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* If this is the case, you are better off deleting ComparableTimSort to
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* eliminate the code duplication. (See Arrays.java for details.)
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*/
30 class ComparableTimSort {
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/**
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* This is the minimum sized sequence that will be merged. Shorter
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* sequences will be lengthened by calling binarySort. If the
entire
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* array is less than this length, no merges will be performed.
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*
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* This constant should be a power of two. It was 64 in Tim Peter's
C
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* implementation, but 32 was empirically determined to work better
in
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* this implementation. In the unlikely event that you set this
constant
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* to be a number that's not a power of two, you'll need to change
the
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* {@link #minRunLength} computation.
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*
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* If you decrease this constant, you must change the stackLen
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* computation in the TimSort constructor, or you risk an
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* ArrayOutOfBounds exception. See listsort.txt for a discussion
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* of the minimum stack length required as a function of the length
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* of the array being sorted and the minimum merge sequence length.
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*/
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private static final int MIN_MERGE = 32;
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50
/**
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* The array being sorted.
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*/

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Trial Exhibit 45.2, Page 1 of 17

UNITED STATES DISTRICT COURT
NORTHERN DISTRICT OF CALIFORNIA

TRIAL EXHIBIT
CASE NO. 10-03561 WHA
DATE ENTERED
BY

DEPUTY CLERK

45.2

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private final Object[] a;
/**
* When we get into galloping mode, we stay there until both runs
win less
* often than MIN_GALLOP consecutive times.
*/
private static final int MIN_GALLOP = 7;
/**
* This controls when we get *into* galloping mode. It is
initialized
* to MIN_GALLOP. The mergeLo and mergeHi methods nudge it higher
for
* random data, and lower for highly structured data.
*/
private int minGallop = MIN_GALLOP;
/**
* Maximum initial size of tmp array, which is used for merging.
The array
* can grow to accommodate demand.
*
* Unlike Tim's original C version, we do not allocate this much
storage
* when sorting smaller arrays. This change was required for
performance.
*/
private static final int INITIAL_TMP_STORAGE_LENGTH = 256;
/**
* Temp storage for merges.
*/
private Object[] tmp;
/**
* A stack of pending runs yet to be merged. Run i starts at
* address base[i] and extends for len[i] elements. It's always
* true (so long as the indices are in bounds) that:
*
*
runBase[i] + runLen[i] == runBase[i + 1]
*
* so we could cut the storage for this, but it's a minor amount,
* and keeping all the info explicit simplifies the code.
*/
private int stackSize = 0; // Number of pending runs on stack
private final int[] runBase;
private final int[] runLen;
/**
* Asserts have been placed in if-statements for performace. To
enable them,
* set this field to true and enable them in VM with a command line
flag.
* If you modify this class, please do test the asserts!
*/
private static final boolean DEBUG = false;
/**
* Creates a TimSort instance to maintain the state of an ongoing
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sort.
*
* @param a the array to be sorted
*/
private ComparableTimSort(Object[] a) {
this.a = a;
// Allocate temp storage (which may be increased later if
necessary)
int len = a.length;
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
Object[] newArray = new Object[len < 2 *
INITIAL_TMP_STORAGE_LENGTH ?
len >>> 1 :
INITIAL_TMP_STORAGE_LENGTH];
tmp = newArray;

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/*
* Allocate runs-to-be-merged stack (which cannot be expanded).
The
* stack length requirements are described in listsort.txt. The
C
* version always uses the same stack length (85), but this was
* measured to be too expensive when sorting "mid-sized" arrays
(e.g.,
* 100 elements) in Java. Therefore, we use smaller (but
sufficiently
* large) stack lengths for smaller arrays. The "magic numbers"
in the
* computation below must be changed if MIN_MERGE is decreased.
See
* the MIN_MERGE declaration above for more information.
*/
int stackLen = (len <
120 ? 5 :
len <
1542 ? 10 :
len < 119151 ? 19 : 40);
runBase = new int[stackLen];
runLen = new int[stackLen];

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}
/*
* The next two methods (which are package private and static)
constitute
* the entire API of this class. Each of these methods obeys the
contract
* of the public method with the same signature in java.util.Arrays.
*/
static void sort(Object[] a) {
sort(a, 0, a.length);
}
static void sort(Object[] a, int lo, int hi) {
rangeCheck(a.length, lo, hi);
int nRemaining = hi - lo;
if (nRemaining < 2)
return; // Arrays of size 0 and 1 are always sorted
// If array is small, do a "mini-TimSort" with no merges
if (nRemaining < MIN_MERGE) {
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int initRunLen = countRunAndMakeAscending(a, lo, hi);
binarySort(a, lo, hi, lo + initRunLen);
return;
}
/**
* March over the array once, left to right, finding natural
runs,
* extending short natural runs to minRun elements, and merging
runs
* to maintain stack invariant.
*/
ComparableTimSort ts = new ComparableTimSort(a);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi);

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// If run is short, extend to min(minRun, nRemaining)
if (runLen < minRun) {
int force = nRemaining <= minRun ? nRemaining : minRun;
binarySort(a, lo, lo + force, lo + runLen);
runLen = force;
}
// Push run onto pending-run stack, and maybe merge
ts.pushRun(lo, runLen);
ts.mergeCollapse();
// Advance to find next run
lo += runLen;
nRemaining -= runLen;
} while (nRemaining != 0);
// Merge all remaining runs to complete sort
if (DEBUG) assert lo == hi;
ts.mergeForceCollapse();
if (DEBUG) assert ts.stackSize == 1;
}
/**
* Sorts the specified portion of the specified array using a binary
* insertion sort. This is the best method for sorting small
numbers
* of elements. It requires O(n log n) compares, but O(n^2) data
* movement (worst case).
*
* If the initial part of the specified range is already sorted,
* this method can take advantage of it: the method assumes that the
* elements from index {@code lo}, inclusive, to {@code start},
* exclusive are already sorted.
*
* @param a the array in which a range is to be sorted
* @param lo the index of the first element in the range to be
sorted
* @param hi the index after the last element in the range to be
sorted
* @param start the index of the first element in the range that is
*
not already known to be sorted (@code lo <= start <= hi}
*/
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@SuppressWarnings("fallthrough")
private static void binarySort(Object[] a, int lo, int hi, int
start) {
if (DEBUG) assert lo <= start && start <= hi;
if (start == lo)
start++;
for ( ; start < hi; start++) {
@SuppressWarnings("unchecked")
Comparable<Object> pivot = (Comparable) a[start];
// Set left (and right) to the index where a[start] (pivot)
belongs
int left = lo;
int right = start;
if (DEBUG) assert left <= right;
/*
* Invariants:
*
pivot >= all in [lo, left).
*
pivot < all in [right, start).
*/
while (left < right) {
int mid = (left + right) >>> 1;
if (pivot.compareTo(a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
if (DEBUG) assert left == right;

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/*
* The invariants still hold: pivot >= all in [lo, left) and
* pivot < all in [left, start), so pivot belongs at left.
Note
* that if there are elements equal to pivot, left points to
the
* first slot after them -- that's why this sort is stable.
* Slide elements over to make room to make room for pivot.
*/
int n = start - left; // The number of elements to move
// Switch is just an optimization for arraycopy in default
case
switch(n) {
case 2: a[left + 2] = a[left + 1];
case 1: a[left + 1] = a[left];
break;
default: System.arraycopy(a, left, a, left + 1, n);
}
a[left] = pivot;

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}
}
/**
* Returns the length of the run beginning at the specified position
in
* the specified array and reverses the run if it is descending
(ensuring
* that the run will always be ascending when the method returns).
*
* A run is the longest ascending sequence with:
*
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*
a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
*
* or the longest descending sequence with:
*
*
a[lo] > a[lo + 1] > a[lo + 2] > ...
*
* For its intended use in a stable mergesort, the strictness of the
* definition of "descending" is needed so that the call can safely
* reverse a descending sequence without violating stability.
*
* @param a the array in which a run is to be counted and possibly
reversed
* @param lo index of the first element in the run
* @param hi index after the last element that may be contained in
the run.
It is required that @code{lo < hi}.
* @return the length of the run beginning at the specified
position in
*
the specified array
*/
@SuppressWarnings("unchecked")
private static int countRunAndMakeAscending(Object[] a, int lo, int
hi) {
if (DEBUG) assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
// Find end of run, and reverse range if descending
if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { //
Descending
while(runHi < hi && ((Comparable)
a[runHi]).compareTo(a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else {
// Ascending
while (runHi < hi && ((Comparable)
a[runHi]).compareTo(a[runHi - 1]) >= 0)
runHi++;
}

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return runHi - lo;
}
/**
* Reverse the specified range of the specified array.
*
* @param a the array in which a range is to be reversed
* @param lo the index of the first element in the range to be
reversed
* @param hi the index after the last element in the range to be
reversed
*/
private static void reverseRange(Object[] a, int lo, int hi) {
hi--;
while (lo < hi) {
Object t = a[lo];
a[lo++] = a[hi];
a[hi--] = t;
}
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}
/**
* Returns the minimum acceptable run length for an array of the
specified
* length. Natural runs shorter than this will be extended with
* {@link #binarySort}.
*
* Roughly speaking, the computation is:
*
* If n < MIN_MERGE, return n (it's too small to bother with fancy
stuff).
* Else if n is an exact power of 2, return MIN_MERGE/2.
* Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that
n/k
*
is close to, but strictly less than, an exact power of 2.
*
* For the rationale, see listsort.txt.
*
* @param n the length of the array to be sorted
* @return the length of the minimum run to be merged
*/
private static int minRunLength(int n) {
if (DEBUG) assert n >= 0;
int r = 0;
// Becomes 1 if any 1 bits are shifted off
while (n >= MIN_MERGE) {
r |= (n & 1);
n >>= 1;
}
return n + r;
}
/**
* Pushes the specified run onto the pending-run stack.
*
* @param runBase index of the first element in the run
* @param runLen the number of elements in the run
*/
private void pushRun(int runBase, int runLen) {
this.runBase[stackSize] = runBase;
this.runLen[stackSize] = runLen;
stackSize++;
}
/**
* Examines the stack of runs waiting to be merged and merges
adjacent runs
* until the stack invariants are reestablished:
*
*
1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
*
2. runLen[i - 2] > runLen[i - 1]
*
* This method is called each time a new run is pushed onto the
stack,
* so the invariants are guaranteed to hold for i < stackSize upon
* entry to the method.
*/
private void mergeCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
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if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1]) {
if (runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
} else if (runLen[n] <= runLen[n + 1]) {
mergeAt(n);
} else {
break; // Invariant is established
}
}
}
/**
* Merges all runs on the stack until only one remains.
is
* called once, to complete the sort.
*/
private void mergeForceCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n - 1] < runLen[n + 1])
n--;
mergeAt(n);
}
}

This method

/**
* Merges the two runs at stack indices i and i+1. Run i must be
* the penultimate or antepenultimate run on the stack. In other
words,
* i must be equal to stackSize-2 or stackSize-3.
*
* @param i stack index of the first of the two runs to merge
*/
@SuppressWarnings("unchecked")
private void mergeAt(int i) {
if (DEBUG) assert stackSize >= 2;
if (DEBUG) assert i >= 0;
if (DEBUG) assert i == stackSize - 2 || i == stackSize - 3;
int base1 = runBase[i];
int len1 = runLen[i];
int base2 = runBase[i + 1];
int len2 = runLen[i + 1];
if (DEBUG) assert len1 > 0 && len2 > 0;
if (DEBUG) assert base1 + len1 == base2;
/*
* 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.
*/
runLen[i] = len1 + len2;
if (i == stackSize - 3) {
runBase[i + 1] = runBase[i + 2];
runLen[i + 1] = runLen[i + 2];
}
stackSize--;
/*
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* Find where the first element of run2 goes in run1. Prior
elements
* in run1 can be ignored (because they're already in place).
*/
int k = gallopRight((Comparable<Object>) a[base2], a, base1,
len1, 0);
if (DEBUG) assert k >= 0;
base1 += k;
len1 -= k;
if (len1 == 0)
return;

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/*
* Find where the last element of run1 goes in run2. Subsequent
elements
* in run2 can be ignored (because they're already in place).
*/
len2 = gallopLeft((Comparable<Object>) a[base1 + len1 - 1], a,
base2, len2, len2 - 1);
if (DEBUG) assert len2 >= 0;
if (len2 == 0)
return;

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// Merge remaining runs, using tmp array with min(len1, len2)
elements
if (len1 <= len2)
mergeLo(base1, len1, base2, len2);
else
mergeHi(base1, len1, base2, len2);
}
/**
* Locates the position at which to insert the specified key into
the
* specified sorted range; if the range contains an element equal to
key,
* returns the index of the leftmost equal element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint <
n.
*
The closer hint is to the result, the faster this method will
run.
* @return the int k, 0 <= k <= n such that a[b + k - 1] < key <=
a[b + k],
*
pretending that a[b - 1] is minus infinity and a[b + n] is
infinity.
*
In other words, key belongs at index b + k; or in other words,
*
the first k elements of a should precede key, and the last n k
*
should follow it.
*/
private static int gallopLeft(Comparable<Object> key, Object[] a,
int base, int len, int hint) {
if (DEBUG) assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
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int ofs = 1;
if (key.compareTo(a[base + hint]) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <=
a[base+hint+ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) >
0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0)
// int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;

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// Make offsets relative to base
lastOfs += hint;
ofs += hint;
} else { // key <= a[base + hint]
// Gallop left until a[base+hint-ofs] < key <=
a[base+hint-lastOfs]
final int maxOfs = hint + 1;
while (ofs < maxOfs && key.compareTo(a[base + hint - ofs])
<= 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0)
// int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;

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// Make offsets relative to base
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
}
if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[base+lastOfs] < key <= a[base+ofs], so key belongs
somewhere
* to the right of lastOfs but no farther right than ofs. Do a
binary
* search, with invariant a[base + lastOfs - 1] < key <= a[base
+ ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);

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if (key.compareTo(a[base + m]) > 0)
lastOfs = m + 1; // a[base + m] < key
else
ofs = m;
// key <= a[base + m]
}
if (DEBUG) assert lastOfs == ofs;
key <= a[base + ofs]
return ofs;

// so a[base + ofs - 1] <

}
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/**
* Like gallopLeft, except that if the range contains an element
equal to
* key, gallopRight returns the index after the rightmost equal
element.
*
* @param key the key whose insertion point to search for
* @param a the array in which to search
* @param base the index of the first element in the range
* @param len the length of the range; must be > 0
* @param hint the index at which to begin the search, 0 <= hint <
n.
*
The closer hint is to the result, the faster this method will
run.
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key <
a[b + k]
*/
private static int gallopRight(Comparable<Object> key, Object[] a,
int base, int len, int hint) {
if (DEBUG) assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int lastOfs = 0;
if (key.compareTo(a[base + hint]) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <
0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0)
// int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to b
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
} else { // a[b + hint] <= key
// Gallop right until a[b+hint + lastOfs] <= key < a[b+hint
+ ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && key.compareTo(a[base + hint + ofs])
>= 0) {
lastOfs = ofs;
ofs = (ofs << 1) + 1;
if (ofs <= 0)
// int overflow
ofs = maxOfs;
}
if (ofs > maxOfs)
ofs = maxOfs;
// Make offsets relative to b
lastOfs += hint;
ofs += hint;
}
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G:\eclair21 - GOOGLE-00-00000525\dalvik\libcore\luni\src\main\java\java\util\Com
mparableTimSort.java
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if (DEBUG) assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;
/*
* Now a[b + lastOfs] <= key < a[b + ofs], so key belongs
somewhere to
* the right of lastOfs but no farther right than ofs. Do a
binary
* search, with invariant a[b + lastOfs - 1] <= key < a[b +
ofs].
*/
lastOfs++;
while (lastOfs < ofs) {
int m = lastOfs + ((ofs - lastOfs) >>> 1);

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if (key.compareTo(a[base + m]) < 0)
ofs = m;
// key < a[b + m]
else
lastOfs = m + 1; // a[b + m] <= key
}
if (DEBUG) assert lastOfs == ofs;
< a[b + ofs]
return ofs;

// so a[b + ofs - 1] <= key

}
/**
* Merges two adjacent runs in place, in a stable fashion. The
first
* element of the first run must be greater than the first element
of the
* second run (a[base1] > a[base2]), and the last element of the
first run
* (a[base1 + len1-1]) must be greater than all elements of the
second run.
*
* For performance, this method should be called only when len1 <=
len2;
* its twin, mergeHi should be called if len1 >= len2. (Either
method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
*
(must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
@SuppressWarnings("unchecked")
private void mergeLo(int base1, int len1, int base2, int len2) {
if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
Object[] a = this.a; // For performance
Object[] tmp = ensureCapacity(len1);
System.arraycopy(a, base1, tmp, 0, len1);
int cursor1 = 0;
int cursor2 = base2;
int dest = base1;

// Indexes into tmp array
// Indexes int a
// Indexes int a

// Move first element of second run and deal with degenerate
Page 12 of 17

Trial Exhibit 45.2, Page 12 of 17

G:\eclair21 - GOOGLE-00-00000525\dalvik\libcore\luni\src\main\java\java\util\Com
mparableTimSort.java

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cases
a[dest++] = a[cursor2++];
if (--len2 == 0) {
System.arraycopy(tmp, cursor1, a, dest, len1);
return;
}
if (len1 == 1) {
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end
of merge
return;
}
int minGallop = this.minGallop; // Use local variable for
performance
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run
won
int count2 = 0; // Number of times in a row that second run
won

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/*
* Do the straightforward thing until (if ever) one run
starts
* winning consistently.
*/
do {
if (DEBUG) assert len1 > 1 && len2 > 0;
if (((Comparable) a[cursor2]).compareTo(tmp[cursor1]) <
0) {
a[dest++] = a[cursor2++];
count2++;
count1 = 0;
if (--len2 == 0)
break outer;
} else {
a[dest++] = tmp[cursor1++];
count1++;
count2 = 0;
if (--len1 == 1)
break outer;
}
} while ((count1 | count2) < minGallop);
/*
* 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 {
if (DEBUG) assert len1 > 1 && len2 > 0;
count1 = gallopRight((Comparable) a[cursor2], tmp,
cursor1, len1, 0);
if (count1 != 0) {
System.arraycopy(tmp, cursor1, a, dest, count1);
dest += count1;
cursor1 += count1;
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Trial Exhibit 45.2, Page 13 of 17

G:\eclair21 - GOOGLE-00-00000525\dalvik\libcore\luni\src\main\java\java\util\Com
mparableTimSort.java
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len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
}
a[dest++] = a[cursor2++];
if (--len2 == 0)
break outer;
count2 = gallopLeft((Comparable) tmp[cursor1], a,
cursor2, len2, 0);
if (count2 != 0) {
System.arraycopy(a, cursor2, a, dest, count2);
dest += count2;
cursor2 += count2;
len2 -= count2;
if (len2 == 0)
break outer;
}
a[dest++] = tmp[cursor1++];
if (--len1 == 1)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back
to field

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if (len1 == 1) {
if (DEBUG) assert len2 > 0;
System.arraycopy(a, cursor2, a, dest, len2);
a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end
of merge
} else if (len1 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
if (DEBUG) assert len2 == 0;
if (DEBUG) assert len1 > 1;
System.arraycopy(tmp, cursor1, a, dest, len1);
}
}
/**
* Like mergeLo, except that this method should be called only if
* len1 >= len2; mergeLo should be called if len1 <= len2. (Either
method
* may be called if len1 == len2.)
*
* @param base1 index of first element in first run to be merged
* @param len1 length of first run to be merged (must be > 0)
* @param base2 index of first element in second run to be merged
*
(must be aBase + aLen)
* @param len2 length of second run to be merged (must be > 0)
*/
@SuppressWarnings("unchecked")
private void mergeHi(int base1, int len1, int base2, int len2) {
if (DEBUG) assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
Page 14 of 17

Trial Exhibit 45.2, Page 14 of 17

G:\eclair21 - GOOGLE-00-00000525\dalvik\libcore\luni\src\main\java\java\util\Com
mparableTimSort.java
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// Copy second run into temp array
Object[] a = this.a; // For performance
Object[] tmp = ensureCapacity(len2);
System.arraycopy(a, base2, tmp, 0, len2);
int cursor1 = base1 + len1 - 1;
int cursor2 = len2 - 1;
int dest = base2 + len2 - 1;

// Indexes into a
// Indexes into tmp array
// Indexes into a

// Move last element of first run and deal with degenerate cases
a[dest--] = a[cursor1--];
if (--len1 == 0) {
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
return;
}
if (len2 == 1) {
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2];
return;
}
int minGallop = this.minGallop; // Use local variable for
performance
outer:
while (true) {
int count1 = 0; // Number of times in a row that first run
won
int count2 = 0; // Number of times in a row that second run
won

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/*
* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
do {
if (DEBUG) assert len1 > 0 && len2 > 1;
if (((Comparable) tmp[cursor2]).compareTo(a[cursor1]) <
0) {
a[dest--] = a[cursor1--];
count1++;
count2 = 0;
if (--len1 == 0)
break outer;
} else {
a[dest--] = tmp[cursor2--];
count2++;
count1 = 0;
if (--len2 == 1)
break outer;
}
} while ((count1 | count2) < minGallop);
/*
* One run is winning so consistently that galloping may be
a
* huge win. So try that, and continue galloping until (if
ever)
Page 15 of 17

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G:\eclair21 - GOOGLE-00-00000525\dalvik\libcore\luni\src\main\java\java\util\Com
mparableTimSort.java
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* neither run appears to be winning consistently anymore.
*/
do {
if (DEBUG) assert len1 > 0 && len2 > 1;
count1 = len1 - gallopRight((Comparable) tmp[cursor2],
a, base1, len1, len1 - 1);
if (count1 != 0) {
dest -= count1;
cursor1 -= count1;
len1 -= count1;
System.arraycopy(a, cursor1 + 1, a, dest + 1,
count1);
if (len1 == 0)
break outer;
}
a[dest--] = tmp[cursor2--];
if (--len2 == 1)
break outer;

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count2 = len2 - gallopLeft((Comparable) a[cursor1], tmp,
0, len2, len2 - 1);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
System.arraycopy(tmp, cursor2 + 1, a, dest + 1,
count2);
if (len2 <= 1)
break outer; // len2 == 1 || len2 == 0
}
a[dest--] = a[cursor1--];
if (--len1 == 0)
break outer;
minGallop--;
} while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
if (minGallop < 0)
minGallop = 0;
minGallop += 2; // Penalize for leaving gallop mode
} // End of "outer" loop
this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back
to field

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if (len2 == 1) {
if (DEBUG) assert len1 > 0;
dest -= len1;
cursor1 -= len1;
System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
a[dest] = tmp[cursor2]; // Move first elt of run2 to front
of merge
} else if (len2 == 0) {
throw new IllegalArgumentException(
"Comparison method violates its general contract!");
} else {
if (DEBUG) assert len1 == 0;
if (DEBUG) assert len2 > 0;
System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
}
}
/**
Page 16 of 17

Trial Exhibit 45.2, Page 16 of 17

G:\eclair21 - GOOGLE-00-00000525\dalvik\libcore\luni\src\main\java\java\util\Com
mparableTimSort.java
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891 }

* Ensures that the external array tmp has at least the specified
* number of elements, increasing its size if necessary. The size
* increases exponentially to ensure amortized linear time
complexity.
*
* @param minCapacity the minimum required capacity of the tmp array
* @return tmp, whether or not it grew
*/
private Object[] ensureCapacity(int minCapacity) {
if (tmp.length < minCapacity) {
// Compute smallest power of 2 > minCapacity
int newSize = minCapacity;
newSize |= newSize >> 1;
newSize |= newSize >> 2;
newSize |= newSize >> 4;
newSize |= newSize >> 8;
newSize |= newSize >> 16;
newSize++;
if (newSize < 0) // Not bloody likely!
newSize = minCapacity;
else
newSize = Math.min(newSize, a.length >>> 1);
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
Object[] newArray = new Object[newSize];
tmp = newArray;
}
return tmp;
}
/**
* Checks that fromIndex and toIndex are in range, and throws an
* appropriate exception if they aren't.
*
* @param arrayLen the length of the array
* @param fromIndex the index of the first element of the range
* @param toIndex the index after the last element of the range
* @throws IllegalArgumentException if fromIndex > toIndex
* @throws ArrayIndexOutOfBoundsException if fromIndex < 0
*
or toIndex > arrayLen
*/
private static void rangeCheck(int arrayLen, int fromIndex, int
toIndex) {
if (fromIndex > toIndex)
throw new IllegalArgumentException("fromIndex(" + fromIndex
+
") > toIndex(" + toIndex+")");
if (fromIndex < 0)
throw new ArrayIndexOutOfBoundsException(fromIndex);
if (toIndex > arrayLen)
throw new ArrayIndexOutOfBoundsException(toIndex);
}

Page 17 of 17

Trial Exhibit 45.2, Page 17 of 17