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#	app/src/main/java/de/uni_marburg/powersort/benchmark/Main.java
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app/src/main/java/de/uni_marburg/powersort/MSort/IMPL_M_1.java Normal file
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package de.uni_marburg.powersort.MSort;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.stream.IntStream;
public class IMPL_M_1 <T extends Comparable<? super T>> {
private IMPL_M_1() {
}
/**
* Sorts the given range, using the given workspace array slice
* for temp storage when possible. This method is designed to be
* invoked from public methods (in class Arrays) after performing
* any necessary array bounds checks and expanding parameters into
* the required forms.
*
* @param a the array to be sorted
* @param lo the index of the first element, inclusive, to be sorted
* @param hi the index of the last element, exclusive, to be sorted
* @param c the comparator to use
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
* @since 1.8
*/
protected static int MERGE_COST = 0;
public static void fillWithAscRunsHighToLow(List<Integer> A, int[] runLengths, int runLenFactor) {
//A has a fixed size, but it doesn't have any meaningful values
int n = A.size();
//This ensures that the sum of runLengths multiplied by runLenFactor equals the list size n. If not, an AssertionError is thrown.
assert IntStream.of(runLengths).sum() * runLenFactor == n;
//System.out.println("IntStream Of run length output: "+IntStream.of(runLengths).sum());
//IntStream.of(runLengths).forEach(System.out::println);
for (int i = 0; i < n; i++) {
//putting i in set a, while a is always the last index of n
A.set(i, n - i);
}
int i = 0;
//For each value l in the array runLengths, do the following
// runLengths = {2, 3, 5}, the loop will run three times, with l taking values 2, 3, and 5 respectively.
for (int l : runLengths) {
int L = l * runLenFactor;
List<Integer> sublist = A.subList(i, i + L);
Collections.sort(sublist);
i += L;
}
}
static <T extends Comparable<? super T>> List<T> merge(List<T> run1, List<T> run2) {
List<T> result = new ArrayList<>();
while (!run1.isEmpty() && !run2.isEmpty()) {
//This comparison only works if the lists are sorted
if (run1.get(0).compareTo(run2.get(0)) < 0) {
result.add(run1.remove(0));
} else {
result.add(run2.remove(0));
}
}
/// can be improved by finding out which one is empty and only add the other one
result.addAll(run1);
result.addAll(run2);
return result;
}
static <T extends Comparable<? super T>> void mergeInplace(List<T> a, int i, int m, int j) {
// System.out.printf("Merge(%d, %d, %d)%n", i, m, j);
MERGE_COST += j - i;
List<T> sublist = merge(
new ArrayList<>(a.subList(i, m)),
new ArrayList<>(a.subList(m, j))
);
for (int k = 0; k < sublist.size(); k++) {
a.set(i + k, sublist.get(k));
}
}
static <T extends Comparable<? super T>> int extendRun(List<T> a, int i) {
// if i was the element before end so just return the last element
if (i == a.size() - 1) {
return i + 1;
}
//we're looking at the element next to a[i]
int j = i + 1;
if (a.get(i).compareTo( a.get(j)) <=0) {
while (j < a.size() && a.get(j - 1).compareTo( a.get(j)) <= 0) {
j++;
}
} else {
while (j < a.size() && a.get(j - 1).compareTo(a.get(j)) > 0) {
j++;
}
List<T> sublist = a.subList(i, j);
Collections.reverse(sublist);
}
return j;
}
public static int power(int[] run1, int[] run2, int n) {
int i1 = run1[0], n1 = run1[1];
int i2 = run2[0], n2 = run2[1];
assert i1 >= 0;
assert i2 == i1 + n1;
assert n1 >= 1 && n2 >= 1;
assert i2 + n2 <= n;
double a = ((i1 + n1 / 2.0d) / n);
double b = ((i2 + n2 / 2.0d) / n);
int l = 0;
while (Math.floor(a * Math.pow(2, l)) == Math.floor(b * Math.pow(2, l))) {
l++;
}
return l;
}
public static <T extends Comparable<? super T>> void mergeTopmost2(List<T> a, List<int []> runs) {
assert runs.size() >= 2;
int[] Y = runs.get(runs.size() - 2);
int[] Z = runs.get(runs.size() - 1);
assert Z[0] == Y[0] + Y[1];
mergeInplace(a, Y[0], Z[0], Z[0] + Z[1]);
runs.set(runs.size() - 2, new int[] {Y[0], Y[1] + Z[1], Y[2]});
runs.removeLast();
}
public static <T extends Comparable<? super T>> void powerSort(List<T> a) {
int n = a.size();
int i = 0;
List<int[]> runs = new ArrayList<>();
int j = extendRun(a, i);
runs.add(new int[] {i, j - i, 0});
i = j;
while (i < n) {
j = extendRun(a, i);
int p = power(runs.get(runs.size() - 1), new int[] {i, j - i}, n);
while (p <= runs.getLast()[2]) {
mergeTopmost2(a, runs);
}
runs.add(new int[] {i, j - i, p});
i = j;
}
while (runs.size() >= 2) {
mergeTopmost2(a, runs);
}
}
public static int extendRunIncreasingOnly(List<Integer> a, int i) {
if (i == a.size() - 1) {
return i + 1;
}
int j = i + 1;
while (j < a.size() && a.get(j - 1) <= a.get(j)) {
j++;
}
return j;
}
public static int powerFast(int[] run1, int[] run2, int n) {
int i1 = run1[0], n1 = run1[1];
int i2 = run2[0], n2 = run2[1];
int a = 2 * i1 + n1;
int b = a + n1 + n2;
int l = 0;
while (true) {
l++;
if (a >= n) {
assert b >= a;
a -= n;
b -= n;
} else if (b >= n) {
break;
}
assert a < b && b < n;
a <<= 1;
b <<= 1;
}
return l;
}
//
// ,
// new SortImpl("MSort") {
// @Override
// @SuppressWarnings("unchecked")
// public void sort(Object[] a) {
// // Create a list of type that matches powerSort's requirements
// List<Comparable<Object>> list = new ArrayList<>();
// for (Object obj : a) {
// // Since we know the input will be Integer objects in the benchmark
// list.add((Comparable<Object>) obj);
// }
//
// // Call powerSort with the proper type
// IMPL_M_1.<Comparable<Object>>powerSort(list);
//
// // Copy back to array
// for (int i = 0; i < a.length; i++) {
// a[i] = list.get(i);
// }
// }
// }
}

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/*
* Copyright (c) 2009, 2013, Oracle and/or its affiliates. All rights reserved.
* Copyright 2009 Google Inc. All Rights Reserved.
* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
*
* This code is free software; you can redistribute it and/or modify it
* under the terms of the GNU General Public License version 2 only, as
* published by the Free Software Foundation. Oracle designates this
* particular file as subject to the "Classpath" exception as provided
* by Oracle in the LICENSE file that accompanied this code.
*
* This code is distributed in the hope that it will be useful, but WITHOUT
* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
* version 2 for more details (a copy is included in the LICENSE file that
* accompanied this code).
*
* You should have received a copy of the GNU General Public License version
* 2 along with this work; if not, write to the Free Software Foundation,
* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
*
* Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
* or visit www.oracle.com if you need additional information or have any
* questions.
*/
package de.uni_marburg.powersort.MSort;
import java.util.Comparator;
/**
* A stable, adaptive, iterative mergesort that requires far fewer than
* n lg(n) comparisons when running on partially sorted arrays, while
* offering performance comparable to a traditional mergesort when run
* on random arrays. Like all proper mergesorts, this sort is stable and
* runs O(n log n) time (worst case). In the worst case, this sort requires
* temporary storage space for n/2 object references; in the best case,
* it requires only a small constant amount of space.
*
* This implementation was adapted from Tim Peters's list sort for
* Python, which is described in detail here:
*
* http://svn.python.org/projects/python/trunk/Objects/listsort.txt
*
* Tim's C code may be found here:
*
* http://svn.python.org/projects/python/trunk/Objects/listobject.c
*
* The underlying techniques are described in this paper (and may have
* even earlier origins):
*
* "Optimistic Sorting and Information Theoretic Complexity"
* Peter McIlroy
* SODA (Fourth Annual ACM-SIAM Symposium on Discrete Algorithms),
* pp 467-474, Austin, Texas, 25-27 January 1993.
*
* While the API to this class consists solely of static methods, it is
* (privately) instantiable; a TimSort instance holds the state of an ongoing
* sort, assuming the input array is large enough to warrant the full-blown
* TimSort. Small arrays are sorted in place, using a binary insertion sort.
*
* @author Josh Bloch
*/
public class IMPL_M_2<T>{
/**
* This is the minimum sized sequence that will be merged. Shorter
* sequences will be lengthened by calling binarySort. If the entire
* array is less than this length, no merges will be performed.
*
* This constant should be a power of two. It was 64 in Tim Peter's C
* implementation, but 32 was empirically determined to work better in
* this implementation. In the unlikely event that you set this constant
* to be a number that's not a power of two, you'll need to change the
* {@link #minRunLength} computation.
*
* If you decrease this constant, you must change the stackLen
* computation in the TimSort constructor, or you risk an
* ArrayOutOfBounds exception. See listsort.txt for a discussion
* of the minimum stack length required as a function of the length
* of the array being sorted and the minimum merge sequence length.
*/
private static final int MIN_MERGE = 32;
/**
* The array being sorted.
*/
private final T[] a;
/**
* The comparator for this sort.
*/
private final Comparator<? super T> c;
/**
* 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. A workspace array may optionally be
* provided in constructor, and if so will be used as long as it
* is big enough.
*/
private T[] tmp;
private int tmpBase; // base of tmp array slice
private int tmpLen; // length of tmp array slice
/**
* 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;
/**
* Creates a TimSort instance to maintain the state of an ongoing sort.
*
* @param a the array to be sorted
* @param c the comparator to determine the order of the sort
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
*/
private IMPL_M_2(T[] a, Comparator<? super T> c, T[] work, int workBase, int workLen) {
this.a = a;
this.c = c;
// Allocate temp storage (which may be increased later if necessary)
int len = a.length;
int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ?
len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
if (work == null || workLen < tlen || workBase + tlen > work.length) {
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
T[] newArray = (T[])java.lang.reflect.Array.newInstance
(a.getClass().getComponentType(), tlen);
tmp = newArray;
tmpBase = 0;
tmpLen = tlen;
}
else {
tmp = work;
tmpBase = workBase;
tmpLen = workLen;
}
/*
* 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.
* The maximum value of 49 allows for an array up to length
* Integer.MAX_VALUE-4, if array is filled by the worst case stack size
* increasing scenario. More explanations are given in section 4 of:
* http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf
*/
int stackLen = (len < 120 ? 5 :
len < 1542 ? 10 :
len < 119151 ? 24 : 49);
runBase = new int[stackLen];
runLen = new int[stackLen];
}
/*
* The next method (package private and static) constitutes the
* entire API of this class.
*/
/**
* Sorts the given range, using the given workspace array slice
* for temp storage when possible. This method is designed to be
* invoked from public methods (in class Arrays) after performing
* any necessary array bounds checks and expanding parameters into
* the required forms.
*
* @param a the array to be sorted
* @param lo the index of the first element, inclusive, to be sorted
* @param hi the index of the last element, exclusive, to be sorted
* @param c the comparator to use
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
* @since 1.8
*/
public static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c,
T[] work, int workBase, int workLen) {
assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length;
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) {
int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
binarySort(a, lo, hi, lo + initRunLen, c);
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.
*/
IMPL_M_2<T> ts = new IMPL_M_2<>(a, c, work, workBase, workLen);
int minRun = minRunLength(nRemaining);
do {
// Identify next run
int runLen = countRunAndMakeAscending(a, lo, hi, c);
// 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, c);
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
assert lo == hi;
ts.mergeForceCollapse();
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})
* @param c comparator to used for the sort
*/
@SuppressWarnings("fallthrough")
private static <T> void binarySort(T[] a, int lo, int hi, int start,
Comparator<? super T> c) {
assert lo <= start && start <= hi;
if (start == lo)
start++;
for ( ; start < hi; start++) {
T pivot = a[start];
// Set left (and right) to the index where a[start] (pivot) belongs
int left = lo;
int right = start;
assert left <= right;
/*
* Invariants:
* pivot >= all in [lo, left).
* pivot < all in [right, start).
*/
while (left < right) {
int mid = (left + right) >>> 1;
if (c.compare(pivot, a[mid]) < 0)
right = mid;
else
left = mid + 1;
}
assert left == right;
/*
* 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 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;
}
}
/**
* 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:
*
* 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}.
* @param c the comparator to used for the sort
* @return the length of the run beginning at the specified position in
* the specified array
*/
private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi,
Comparator<? super T> c) {
assert lo < hi;
int runHi = lo + 1;
if (runHi == hi)
return 1;
// Find end of run, and reverse range if descending
if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
runHi++;
reverseRange(a, lo, runHi);
} else { // Ascending
while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
runHi++;
}
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;
}
}
/**
* 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) {
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.
*
* Thanks to Stijn de Gouw, Jurriaan Rot, Frank S. de Boer,
* Richard Bubel and Reiner Hahnle, this is fixed with respect to
* the analysis in "On the Worst-Case Complexity of TimSort" by
* Nicolas Auger, Vincent Jug, Cyril Nicaud, and Carine Pivoteau.
*/
private void mergeCollapse() {
while (stackSize > 1) {
int n = stackSize - 2;
if (n > 0 && runLen[n-1] <= runLen[n] + runLen[n+1] ||
n > 1 && runLen[n-2] <= runLen[n] + runLen[n-1]) {
if (runLen[n - 1] < runLen[n + 1])
n--;
} else if (n < 0 || runLen[n] > runLen[n + 1]) {
break; // Invariant is established
}
mergeAt(n);
}
}
/**
* Merges all runs on the stack until only one remains. This method 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);
}
}
/**
* 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
*/
private void mergeAt(int i) {
assert stackSize >= 2;
assert i >= 0;
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];
assert len1 > 0 && len2 > 0;
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--;
/*
* 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(a[base2], a, base1, len1, 0, c);
assert k >= 0;
base1 += k;
len1 -= k;
if (len1 == 0)
return;
/*
* 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(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
assert len2 >= 0;
if (len2 == 0)
return;
// 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.
* @param c the comparator used to order the range, and to search
* @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 <T> int gallopLeft(T key, T[] a, int base, int len, int hint,
Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len;
int lastOfs = 0;
int ofs = 1;
if (c.compare(key, a[base + hint]) > 0) {
// Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
int maxOfs = len - hint;
while (ofs < maxOfs && c.compare(key, 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 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 && c.compare(key, 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 base
int tmp = lastOfs;
lastOfs = hint - ofs;
ofs = hint - tmp;
}
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);
if (c.compare(key, a[base + m]) > 0)
lastOfs = m + 1; // a[base + m] < key
else
ofs = m; // key <= a[base + m]
}
assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
return ofs;
}
/**
* 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.
* @param c the comparator used to order the range, and to search
* @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
*/
private static <T> int gallopRight(T key, T[] a, int base, int len,
int hint, Comparator<? super T> c) {
assert len > 0 && hint >= 0 && hint < len;
int ofs = 1;
int lastOfs = 0;
if (c.compare(key, a[base + hint]) < 0) {
// Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
int maxOfs = hint + 1;
while (ofs < maxOfs && c.compare(key, 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 && c.compare(key, 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;
}
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);
if (c.compare(key, a[base + m]) < 0)
ofs = m; // key < a[b + m]
else
lastOfs = m + 1; // a[b + m] <= key
}
assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
return ofs;
}
/**
* 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)
*/
private void mergeLo(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy first run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len1);
int cursor1 = tmpBase; // Indexes into tmp array
int cursor2 = base2; // Indexes int a
int dest = base1; // Indexes int a
System.arraycopy(a, base1, tmp, cursor1, len1);
// Move first element of second run and deal with degenerate 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;
}
Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
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
/*
* Do the straightforward thing until (if ever) one run starts
* winning consistently.
*/
do {
assert len1 > 1 && len2 > 0;
if (c.compare(a[cursor2], 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 {
assert len1 > 1 && len2 > 0;
count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
if (count1 != 0) {
System.arraycopy(tmp, cursor1, a, dest, count1);
dest += count1;
cursor1 += count1;
len1 -= count1;
if (len1 <= 1) // len1 == 1 || len1 == 0
break outer;
}
a[dest++] = a[cursor2++];
if (--len2 == 0)
break outer;
count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
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
if (len1 == 1) {
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 {
assert len2 == 0;
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)
*/
private void mergeHi(int base1, int len1, int base2, int len2) {
assert len1 > 0 && len2 > 0 && base1 + len1 == base2;
// Copy second run into temp array
T[] a = this.a; // For performance
T[] tmp = ensureCapacity(len2);
int tmpBase = this.tmpBase;
System.arraycopy(a, base2, tmp, tmpBase, len2);
int cursor1 = base1 + len1 - 1; // Indexes into a
int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array
int dest = base2 + len2 - 1; // Indexes into a
// Move last element of first run and deal with degenerate cases
a[dest--] = a[cursor1--];
if (--len1 == 0) {
System.arraycopy(tmp, tmpBase, 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;
}
Comparator<? super T> c = this.c; // Use local variable for performance
int minGallop = this.minGallop; // " " " " "
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
/*
* Do the straightforward thing until (if ever) one run
* appears to win consistently.
*/
do {
assert len1 > 0 && len2 > 1;
if (c.compare(tmp[cursor2], 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)
* neither run appears to be winning consistently anymore.
*/
do {
assert len1 > 0 && len2 > 1;
count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
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;
count2 = len2 - gallopLeft(a[cursor1], tmp, tmpBase, len2, len2 - 1, c);
if (count2 != 0) {
dest -= count2;
cursor2 -= count2;
len2 -= count2;
System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
if (len2 <= 1) // len2 == 1 || len2 == 0
break outer;
}
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
if (len2 == 1) {
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 {
assert len1 == 0;
assert len2 > 0;
System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
}
}
/**
* 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 T[] ensureCapacity(int minCapacity) {
if (tmpLen < minCapacity) {
// Compute smallest power of 2 > minCapacity
int newSize = -1 >>> Integer.numberOfLeadingZeros(minCapacity);
newSize++;
if (newSize < 0) // Not bloody likely!
newSize = minCapacity;
else
newSize = Math.min(newSize, a.length >>> 1);
@SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
T[] newArray = (T[])java.lang.reflect.Array.newInstance
(a.getClass().getComponentType(), newSize);
tmp = newArray;
tmpLen = newSize;
tmpBase = 0;
}
return tmp;
}
}

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app/src/main/java/de/uni_marburg/powersort/benchmark/Main.java Normal file
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app/src/main/java/de/uni_marburg/powersort/msort/IMPL_M_1.java
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package de.uni_marburg.powersort.msort;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.stream.IntStream;
public class IMPL_M_1 {
private IMPL_M_1() {}
/**
* Sorts the given range, using the given workspace array slice
* for temp storage when possible. This method is designed to be
* invoked from public methods (in class Arrays) after performing
* any necessary array bounds checks and expanding parameters into
* the required forms.
*
* @param a the array to be sorted
* @param lo the index of the first element, inclusive, to be sorted
* @param hi the index of the last element, exclusive, to be sorted
* @param c the comparator to use
* @param work a workspace array (slice)
* @param workBase origin of usable space in work array
* @param workLen usable size of work array
* @since 1.8
*/
protected static int MERGE_COST = 0;
// Example usage
// int[] runs = new int[] {5, 3, 3, 14, 1, 2}; // example from slides
// //runs = new int[]{9, 16, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7};
//
// ArrayList<Integer> a = new ArrayList<>(IntStream.range(0, Arrays.stream(runs).sum()).boxed().collect(Collectors.toList()));
//
// System.out.println();
// fillWithAscRunsHighToLow(a, runs);
// MERGE_COST = 0;
// System.out.println("Sorting with Powersort:");
// powersort(a, this::extendRunIncreasingOnly);
// System.out.println("Merge cost: " + MERGE_COST);
// runs = [5,3,3,14,1,2];
// runs = [9,16,7,7,7,7,7,7,7,7,7,7];
//
// a = list(range(sum(runs)));
// fill_with_asc_runs_high_to_low(a, runs);
// MERGE_COST = 0;
// System.out.println("Sorting with Powersort:");
// powersort(a, extendRunIncreasingOnly);
// System.out.println("Merge cost: " + MERGE_COST);
public static void fillWithAscRunsHighToLow(List<Integer> A, int[] runLengths, int runLenFactor) {
int n = A.size();
assert IntStream.of(runLengths).sum() * runLenFactor == n;
for (int i = 0; i < n; i++) {
A.set(i, n - i);
}
int i = 0;
for (int l : runLengths) {
int L = l * runLenFactor;
List<Integer> sublist = A.subList(i, i + L);
Collections.sort(sublist);
i += L;
}
}
private static List<Integer> merge(List<Integer> run1, List<Integer> run2) {
List<Integer> result = new ArrayList<>();
while (!run1.isEmpty() && !run2.isEmpty()) {
if (run1.get(0) < run2.get(0)) {
result.add(run1.remove(0));
} else {
result.add(run2.remove(0));
}
}
result.addAll(run1);
result.addAll(run2);
return result;
}
private static void mergeInplace(List<Integer> a, int i, int m, int j) {
System.out.printf("Merge(%d, %d, %d)%n", i, m, j);
MERGE_COST += j - i;
List<Integer> sublist = merge(
new ArrayList<>(a.subList(i, m)),
new ArrayList<>(a.subList(m, j))
);
for (int k = 0; k < sublist.size(); k++) {
a.set(i + k, sublist.get(k));
}
}
static int extendRun(List<Integer> a, int i) {
if (i == a.size() - 1) {
return i + 1;
}
int j = i + 1;
if (a.get(i) <= a.get(j)) {
while (j < a.size() && a.get(j - 1) <= a.get(j)) {
j++;
}
} else {
while (j < a.size() && a.get(j - 1) > a.get(j)) {
j++;
}
List<Integer> sublist = a.subList(i, j);
Collections.reverse(sublist);
}
return j;
}
private static int extendRunIncreasingOnly(List<Integer> a, int i) {
if (i == a.size() - 1) {
return i + 1;
}
int j = i + 1;
while (j < a.size() && a.get(j - 1) <= a.get(j)) {
j++;
}
return j;
}
public static int power(int[] run1, int[] run2, int n) {
int i1 = run1[0], n1 = run1[1];
int i2 = run2[0], n2 = run2[1];
assert i1 >= 0;
assert i2 == i1 + n1;
assert n1 >= 1 && n2 >= 1;
assert i2 + n2 <= n;
double a = ((i1 + n1 / 2.0d) / n);
double b = ((i2 + n2 / 2.0d) / n);
int l = 0;
while (Math.floor(a * Math.pow(2, l)) == Math.floor(b * Math.pow(2, l))) {
l++;
}
return l;
}
public static int powerFast(int[] run1, int[] run2, int n) {
int i1 = run1[0], n1 = run1[1];
int i2 = run2[0], n2 = run2[1];
int a = 2 * i1 + n1;
int b = a + n1 + n2;
int l = 0;
while (true) {
l++;
if (a >= n) {
assert b >= a;
a -= n;
b -= n;
} else if (b >= n) {
break;
}
assert a < b && b < n;
a <<= 1;
b <<= 1;
}
return l;
}
public static void mergeTopmost2(List<Integer> a, List<int[]> runs) {
assert runs.size() >= 2;
int[] Y = runs.get(runs.size() - 2);
int[] Z = runs.get(runs.size() - 1);
assert Z[0] == Y[0] + Y[1];
mergeInplace(a, Y[0], Z[0], Z[0] + Z[1]);
runs.set(runs.size() - 2, new int[] {Y[0], Y[1] + Z[1], Y[2]});
runs.remove(runs.size() - 1);
}
public static void powerSort(List<Integer> a) {
int n = a.size();
int i = 0;
List<int[]> runs = new ArrayList<>();
int j = extendRun(a, i);
runs.add(new int[] {i, j - i, 0});
i = j;
while (i < n) {
j = extendRun(a, i);
int p = power(runs.get(runs.size() - 1), new int[] {i, j - i}, n);
while (p <= runs.get(runs.size() - 1)[2]) {
mergeTopmost2(a, runs);
}
runs.add(new int[] {i, j - i, p});
i = j;
}
while (runs.size() >= 2) {
mergeTopmost2(a, runs);
}
}
/* """Fills the given array A with ascending runs of the given list of run
lengths.
More precisely, the array is first filled n, n-1, ..., 1
and then for i=0..l-1 segments of runLengths.get(i) * runLenFactor
are sorted ascending.
The sum of all lengths in runLengths times runLenFactor should be equal to the
length of A.
"""*/
/* static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c,
T[] work, int workBase, int workLen) {
assert c != null && a != null && lo >= 0 && lo <= hi && hi <= a.length;
}*/
/*
public static final int MIN_MERGE=24;
public int mergeCost=0;
private final T []sortedArray;
public PowerSort(T[] sortedArray) {
super();
this.sortedArray = sortedArray;
}
ArrayList<Integer> run1 = new ArrayList<>();
ArrayList<Integer> run2 = new ArrayList<>();
private AbstractList<Integer> merge(ArrayList <Integer> run1, ArrayList<Integer> run2) {
ArrayList<Integer> result = new ArrayList<>();
while(run1.size() > 0 && run2.size() >0) {
if (run1.getFirst()<run2.getFirst()){
result.add(run1.getFirst());
run1.removeFirst();
}else {
result.add(run2.getFirst());
run2.removeFirst();
}
result.addAll(run1);
result.addAll(run2);
}
return result;
}
public void mergeInplace(int[] a, int[] i, int[] m, int[] j) {
//System.out.println("merge(" + i + "," + m + "," + j + ")");
System.out.printf("Merge(%d, %d, %d)%n", i, m, j);
// this.mergeCost += j - i;
for(int s =0; s < i.length && s< j.length ; s++) {
// int[] leftSubarray = copyOfRange(a, i, m);
// int[] rightSubarray = copyOfRange(a, m, j);
// int[] mergedSubarray = merge(leftSubarray, rightSubarray);
// System.arraycopy(mergedSubarray, 0, a, i, mergedSubarray.length);
//// mergeCost += j[s] - i[s];
// System.arraycopy(merge(Arrays.copyOfRange(a, i, m), Arrays.copyOfRange(a, m, j)), 0, a, i, j - i);
// a[i:j]=merge(a[i:m],a[m:j]);
}
}
public void power(int run1,int run2, int n) {
int i = run1;
int n1 = run1;
int j = run2;
int n2 = run2;
int a=(i + n1/2) / n;
int b=(j + n2/2) / n;
int l =0;
//while( Math.floor(a * 2**1)){
// Math.floor(b * );
// }
}
public void sorting(final int[] Array, final int left, final int right) {
}*/
}

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app/src/test/java/de/uni_marburg/powersort/MSort/PowerSortT.java Normal file
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package de.uni_marburg.powersort.MSort;
import static de.uni_marburg.powersort.MSort.IMPL_M_1.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class PowerSortT {
public static void main(String[] args) {
testFillWithAscRunsHighToLow();
testMerge();
testMergeInplace();
testExtendRun();
testPower();
testPowerFast();
testMergeTopmost2();
testPowerSort();
}
// Test for fillWithAscRunsHighToLow
public static void testFillWithAscRunsHighToLow() {
List<Integer> A = new ArrayList<>(Collections.nCopies(10, 0));
int[] runLengths = {2, 3, 5};
int runLenFactor = 1;
fillWithAscRunsHighToLow(A, runLengths, runLenFactor);
System.out.println("Test fillWithAscRunsHighToLow: " + A);
}
// Test for merge
public static void testMerge() {
List<Integer> run1 = new ArrayList<>(Arrays.asList(1, 4, 6));
List<Integer> run2 = new ArrayList<>(Arrays.asList(2, 3, 5));
List<Integer> result = merge(run1, run2);
System.out.println("Test merge: " + result);
}
// Test for mergeInplace
public static void testMergeInplace() {
List<Integer> A = new ArrayList<>(Arrays.asList(1, 4, 6, 2, 3, 5));
mergeInplace(A, 0, 3, 6);
System.out.println("Test mergeInplace: " + A);
}
// Test for extendRun
public static void testExtendRun() {
List<Integer> A = new ArrayList<>(Arrays.asList(1, 2, 3, 6, 5, 4));
int endIndex = extendRun(A, 0);
System.out.println("Test extendRun (from 0): " + endIndex);
System.out.println("Modified List: " + A);
}
// Test for power
public static void testPower() {
int[] run1 = {0, 3};
int[] run2 = {3, 3};
int n = 6;
int powerValue = power(run1, run2, n);
System.out.println("Test power: " + powerValue);
}
// Test for powerFast
public static void testPowerFast() {
int[] run1 = {0, 3};
int[] run2 = {3, 3};
int n = 6;
int powerFastValue = powerFast(run1, run2, n);
System.out.println("Test powerFast: " + powerFastValue);
}
// Test for mergeTopmost2
public static void testMergeTopmost2() {
List<Integer> A = new ArrayList<>(Arrays.asList(1, 3, 5, 2, 4, 6));
List<int[]> runs = new ArrayList<>();
runs.add(new int[]{0, 3, 1});
runs.add(new int[]{3, 3, 1});
mergeTopmost2(A, runs);
System.out.println("Test mergeTopmost2: " + A);
}
// Test for powerSort
public static void testPowerSort() {
List<Integer> A = new ArrayList<>(Arrays.asList(10, 9, 8, 7, 6, 5, 4, 3, 2, 1));
powerSort(A);
System.out.println("Test powerSort: " + A);
}
}

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app/src/test/java/de/uni_marburg/powersort/MSort/PowerSortTest.java Normal file
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package de.uni_marburg.powersort.MSort;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import org.junit.jupiter.api.Test;
import static de.uni_marburg.powersort.MSort.IMPL_M_1.MERGE_COST;
import static de.uni_marburg.powersort.MSort.IMPL_M_1.fillWithAscRunsHighToLow;
import static de.uni_marburg.powersort.MSort.IMPL_M_1.powerSort;
class PowerSortTest {
@Test
public void testWithOneInputList() {
// List<Integer> list = new ArrayList<>(List.of(5, 2, 8, 12, 1, 6));
// extendRun(list, 0);
//System.out.println(list);
// example from slides he wrote this
int[] runs = {5, 3, 3, 14, 1, 2};
runs = new int[]{9, 16, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7};
List<Integer> a = new ArrayList<>(IntStream.range(0, Arrays.stream(runs).sum()).boxed().collect(Collectors.toList()));
System.out.println();
fillWithAscRunsHighToLow(a, runs, 1);
MERGE_COST = 0;
System.out.println("Sorting with Powersort:");
powerSort(a);
System.out.println("Merge cost: " + MERGE_COST);
}
@Test
public void testWithFinnInputList() {
List<Integer> numbers = List.of(new Integer[] {24, 25, 26, 27, 28, 21, 22, 23, 18, 19, 20, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 3, 1, 2});
powerSort(numbers);
System.out.println("Result: ");
System.out.println(new ArrayList<>(List.of(numbers)));
}
}

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app/src/test/java/de/uni_marburg/powersort/msort/PowerSortTest.java
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package de.uni_marburg.powersort.msort;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.IntStream;
import org.junit.jupiter.api.Test;
import static de.uni_marburg.powersort.msort.IMPL_M_1.MERGE_COST;
import static de.uni_marburg.powersort.msort.IMPL_M_1.extendRun;
import static de.uni_marburg.powersort.msort.IMPL_M_1.fillWithAscRunsHighToLow;
import static de.uni_marburg.powersort.msort.IMPL_M_1.powerSort;
class PowerSortTest {
@Test
public void testWithOneInputList() {
List<Integer> list = new ArrayList<>(List.of(5, 2, 8, 12, 1, 6));
extendRun(list, 0);
System.out.println(list);
// example from slides he wrote this
int[] runs = {5, 3, 3, 14, 1, 2};
// runs = new int[]{9, 16, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7};
List<Integer> a = new ArrayList<>(IntStream.range(0, Arrays.stream(runs).sum()).boxed().collect(Collectors.toList()));
System.out.println();
fillWithAscRunsHighToLow(a, runs, 1);
MERGE_COST = 0;
System.out.println("Sorting with Powersort:");
powerSort(a);
System.out.println("Merge cost: " + MERGE_COST);
}
}