问题
Partition problem is known to be NP-hard. Depending on the particular instance of the problem we can try dynamic programming or some heuristics like differencing (also known as Karmarkar-Karp algorithm).
The latter seems to be very useful for the instances with big numbers (what makes dynamic programming intractable), however not always perfect. What is an efficient way to find a better solution (random, tabu search, other approximations)?
PS: The question has some story behind it. There is a challenge Johnny Goes Shopping available at SPOJ since July 2004. Till now, the challenge has been solved by 1087 users, but only 11 of them scored better than correct Karmarkar-Karp algorithm implementation (with current scoring, Karmarkar-Karp gives 11.796614 points). How to do better? (Answers supported by accepted submission most wanted but please do not reveal your code.)
回答1:
For whatever it's worth, a straightforward, unoptimized Python implementation of the "complete Karmarkar Karp" (CKK) search procedure in [Korf88] -- modified only slightly to bail out of the search after a given time limit (say, 4.95 seconds) and return the best solution found so far -- is sufficient to score 14.204234 on the SPOJ problem, beating the score for Karmarkar-Karp. As of this writing, this is #3 on the rankings (see Edit #2 below)
A slightly more readable presentation of Korf's CKK algorithm can be found in [Mert99].
EDIT #2 - I've implemented Evgeny Kluev's hybrid heuristic of applying Karmarkar-Karp until the list of numbers is below some threshold and then switching over to the exact Horowitz-Sahni subset enumeration method [HS74] (a concise description may be found in [Korf88]). As suspected, my Python implementation required lowering the switchover threshold versus his C++ implementation. With some trial and error, I found that a threshold of 37 was the maximum that allowed my program to finish within the time limit. Yet, even at that lower threshold, I was able to achieve a score of 15.265633, good enough for second place.
I further attempted to incorporate this hybrid KK/HS method into the CKK tree search, basically by using HS as a very aggressive and expensive pruning strategy. In plain CKK, I was unable to find a switchover threshold that even matched the KK/HS method. However, using the ILDS (see below) search strategy for CKK and HS (with a threshold of 25) to prune, I was able to yield a very small gain over the previous score, up to 15.272802. It probably should not be surprising that CKK+ILDS would outperform plain CKK in this context since it would, by design, provide a greater diversity of inputs to the HS phase.
EDIT #1 - I've tried two further refinements to the base CKK algorithm:
"Improved Limited Discrepancy Search" (ILDS) [Korf96] This is an alternative to the natural DFS ordering of paths within the search tree. It has a tendency to explore more diverse solutions earlier on than regular Depth-First Search.
"Speeding up 2-Way Number Partitioning" [Cerq12] This generalizes one of the pruning criteria in CKK from nodes within 4 levels of the leaf nodes to nodes within 5, 6, and 7 levels above leaf nodes.
In my test cases, both of these refinements generally provided noticeable benefits over the original CKK in reducing the number of nodes explored (in the case of the latter) and in arriving at better solutions sooner (in the case of the former). However, within the confines of the SPOJ problem structure, neither of these were sufficient to improve my score.
Given the idiosyncratic nature of this SPOJ problem (i.e.: 5-second time limit and only one specific and undisclosed problem instance), it is hard to give advice on what may actually improve the score*. For example, should we continue to pursue alternate search ordering strategies (e.g.: many of the papers by Wheeler Ruml listed here)? Or should we try incorporating some form of local improvement heuristic to solutions found by CKK in order to help pruning? Or maybe we should abandon CKK-based approaches altogether and try for a dynamic programming approach? How about a PTAS? Without knowing more about the specific shape of the instance used in the SPOJ problem, it's very difficult to guess at what kind of approach would yield the most benefit. Each one has its strengths and weaknesses, depending on the specific properties of a given instance.
*Aside from simply running the same thing faster, say, by implementing in C++ instead of Python.
References
[Cerq12] Cerquides, Jesús, and Pedro Meseguer. "Speeding Up 2-way Number Partitioning." ECAI. 2012, doi:10.3233/978-1-61499-098-7-223
[HS74] Horowitz, Ellis, and Sartaj Sahni. "Computing partitions with applications to the knapsack problem." Journal of the ACM (JACM) 21.2 (1974): 277-292.
[Korf88] Korf, Richard E. (1998), "A complete anytime algorithm for number partitioning", Artificial Intelligence 106 (2): 181–203, doi:10.1016/S0004-3702(98)00086-1,
[Korf96] Korf, Richard E. "Improved limited discrepancy search." AAAI/IAAI, Vol. 1. 1996.
[Mert99] Mertens, Stephan (1999), A complete anytime algorithm for balanced number partitioning, arXiv:cs/9903011
回答2:
There are many papers describing various advanced algorithms for set partitioning. Here are only two of them:
- "A complete anytime algorithm for number partitioning" by Richard E. Korf.
- "An efficient fully polynomial approximation scheme for the Subset-Sum Problem" by Hans Kellerer et al.
Honestly, I don't know which of them gives more efficient solution. Probably neither of these advanced algorithms are needed to solve that SPOJ problem. Korf's paper is still very useful. Algorithms described there are very simple (to understand and implement). Also he overviews several even simpler algorithms (in section 2). So if you want to know the details of Horowitz-Sahni or Schroeppel-Shamir methods (mentioned below), you can find them in Korf's paper. Also (in section 8) he writes that stochastic approaches do not guarantee good enough solutions. So it is unlikely you get significant improvements with something like hill climbing, simulated annealing, or tabu search.
I tried several simple algorithms and their combinations to solve partitioning problems with size up to 10000, maximum value up to 1014, and time limit 4 sec. They were tested on random uniformly distributed numbers. And optimal solution was found for every problem instance I tried. For some problem instances optimality is guaranteed by algorithm, for others optimality is not 100% guaranteed, but probability of getting sub-optimal solution is very small.
For sizes up to 4 (green area to the left) Karmarkar-Karp algorithm always gives optimal result.
For sizes up to 54 a brute force algorithm is fast enough (red area). There is a choice between Horowitz-Sahni or Schroeppel-Shamir algorithms. I used Horowitz-Sahni because it seems more efficient for given limits. Schroeppel-Shamir uses much less memory (everything fits in L2 cache), so it may be preferable when other CPU cores perform some memory-intensive tasks or to do set partitioning using multiple threads. Or to solve bigger problems with not as strict time limit (where Horowitz-Sahni just runs out of memory).
When size multiplied by sum of all values is less than 5*109 (blue area), dynamic programming approach is applicable. Border between brute force and dynamic programming areas on diagram shows where each algorithm performs better.
Green area to the right is the place where Karmarkar-Karp algorithm gives optimal result with almost 100% probability. Here there are so many perfect partitioning options (with delta 0 or 1) that Karmarkar-Karp algorithm almost certainly finds one of them. It is possible to invent data set where Karmarkar-Karp always gives sub-optimal result. For example {17 13 10 10 10 ...}. If you multiply this to some large number, neither KK nor DP would be able to find optimal solution. Fortunately such data sets are very unlikely in practice. But problem setter could add such data set to make contest more difficult. In this case you can choose some advanced algorithm for better results (but only for grey and right green areas on diagram).
I tried 2 ways to implement Karmarkar-Karp algorithm's priority queue: with max heap and with sorted array. Sorted array option appears to be slightly faster with linear search and significantly faster with binary search.
Yellow area is the place where you can choose between guaranteed optimal result (with DP) or just optimal result with high probability (with Karmarkar-Karp).
Finally, grey area, where neither of simple algorithms by itself gives optimal result. Here we could use Karmarkar-Karp to pre-process data until it is applicable to either Horowitz-Sahni or dynamic programming. In this place there are also many perfect partitioning options, but less than in green area, so Karmarkar-Karp by itself could sometimes miss proper partitioning. Update: As noted by @mhum, it is not necessary to implement dynamic programming algorithm to make things working. Horowitz-Sahni with Karmarkar-Karp pre-processing is enough. But it is essential for Horowitz-Sahni algorithm to work on sizes up to 54 in said time limit to (almost) guarantee optimal partitioning. So C++ or other language with good optimizing compiler and fast computer are preferable.
Here is how I combined Karmarkar-Karp with other algorithms:
template<bool Preprocess = false>
i64 kk(const vector<i64>& values, i64 sum, Log& log)
{
log.name("Karmarkar-Karp");
vector<i64> pq(values.size() * 2);
copy(begin(values), end(values), begin(pq) + values.size());
sort(begin(pq) + values.size(), end(pq));
auto first = end(pq);
auto last = begin(pq) + values.size();
while (first - last > 1)
{
if (Preprocess && first - last <= kHSLimit)
{
hs(last, first, sum, log);
return 0;
}
if (Preprocess && static_cast<double>(first - last) * sum <= kDPLimit)
{
dp(last, first, sum, log);
return 0;
}
const auto diff = *(first - 1) - *(first - 2);
sum -= *(first - 2) * 2;
first -= 2;
const auto place = lower_bound(last, first, diff);
--last;
copy(last + 1, place, last);
*(place - 1) = diff;
}
const auto result = (first - last)? *last: 0;
log(result);
return result;
}
Link to full C++11 implementation. This program only determines difference between partition sums, it does not report the partitions themselves. Warning: if you want to run it on a computer with less than 1 Gb free memory, decrease kHSLimit constant.
回答3:
EDIT Here's a implementation that starts with Karmarkar-Karp differencing then tries to optimize the resulting partitions.
The only optimizations that time allows are giving 1 from one partition to the other and swapping 1 for 1 between both partitions.
My implementation of Karmarkar-Karp at the beginning must be inaccurate since the resulting score with just Karmarkar-Karp is 2.711483 not 11.796614 points cited by OP. The score goes to 7.718049 when the optimizations are used.
SPOILER WARNING C# submission code follows
using System;
using System.Collections.Generic;
using System.Linq;
public class Test
{
// some comparer's to lazily avoid using a proper max-heap implementation
public class Index0 : IComparer<long[]>
{
public int Compare(long[] x, long[] y)
{
if(x[0] == y[0]) return 0;
return x[0] < y[0] ? -1 : 1;
}
public static Index0 Inst = new Index0();
}
public class Index1 : IComparer<long[]>
{
public int Compare(long[] x, long[] y)
{
if(x[1] == y[1]) return 0;
return x[1] < y[1] ? -1 : 1;
}
}
public static void Main()
{
// load the data
var start = DateTime.Now;
var list = new List<long[]>();
int size = int.Parse(Console.ReadLine());
for(int i=1; i<=size; i++) {
var tuple = new long[]{ long.Parse(Console.ReadLine()), i };
list.Add(tuple);
}
list.Sort((x, y) => { if(x[0] == y[0]) return 0; return x[0] < y[0] ? -1 : 1; });
// Karmarkar-Karp differences
List<long[]> diffs = new List<long[]>();
while(list.Count > 1) {
// get max
var b = list[list.Count - 1];
list.RemoveAt(list.Count - 1);
// get max
var a = list[list.Count - 1];
list.RemoveAt(list.Count - 1);
// (b - a)
var diff = b[0] - a[0];
var tuple = new long[]{ diff, -1 };
diffs.Add(new long[] { a[0], b[0], diff, a[1], b[1] });
// insert (b - a) back in
var fnd = list.BinarySearch(tuple, new Index0());
list.Insert(fnd < 0 ? ~fnd : fnd, tuple);
}
var approx = list[0];
list.Clear();
// setup paritions
var listA = new List<long[]>();
var listB = new List<long[]>();
long sumA = 0;
long sumB = 0;
// Karmarkar-Karp rebuild partitions from differences
bool toggle = false;
for(int i=diffs.Count-1; i>=0; i--) {
var inB = listB.BinarySearch(new long[]{diffs[i][2]}, Index0.Inst);
var inA = listA.BinarySearch(new long[]{diffs[i][2]}, Index0.Inst);
if(inB >= 0 && inA >= 0) {
toggle = !toggle;
}
if(toggle == false) {
if(inB >= 0) {
listB.RemoveAt(inB);
}else if(inA >= 0) {
listA.RemoveAt(inA);
}
var tb = new long[]{diffs[i][1], diffs[i][4]};
var ta = new long[]{diffs[i][0], diffs[i][3]};
var fb = listB.BinarySearch(tb, Index0.Inst);
var fa = listA.BinarySearch(ta, Index0.Inst);
listB.Insert(fb < 0 ? ~fb : fb, tb);
listA.Insert(fa < 0 ? ~fa : fa, ta);
} else {
if(inA >= 0) {
listA.RemoveAt(inA);
}else if(inB >= 0) {
listB.RemoveAt(inB);
}
var tb = new long[]{diffs[i][1], diffs[i][4]};
var ta = new long[]{diffs[i][0], diffs[i][3]};
var fb = listA.BinarySearch(tb, Index0.Inst);
var fa = listB.BinarySearch(ta, Index0.Inst);
listA.Insert(fb < 0 ? ~fb : fb, tb);
listB.Insert(fa < 0 ? ~fa : fa, ta);
}
}
listA.ForEach(a => sumA += a[0]);
listB.ForEach(b => sumB += b[0]);
// optimize our partitions with give/take 1 or swap 1 for 1
bool change = false;
while(DateTime.Now.Subtract(start).TotalSeconds < 4.8) {
change = false;
// give one from A to B
for(int i=0; i<listA.Count; i++) {
var a = listA[i];
if(Math.Abs(sumA - sumB) > Math.Abs((sumA - a[0]) - (sumB + a[0]))) {
var fb = listB.BinarySearch(a, Index0.Inst);
listB.Insert(fb < 0 ? ~fb : fb, a);
listA.RemoveAt(i);
i--;
sumA -= a[0];
sumB += a[0];
change = true;
} else {break;}
}
// give one from B to A
for(int i=0; i<listB.Count; i++) {
var b = listB[i];
if(Math.Abs(sumA - sumB) > Math.Abs((sumA + b[0]) - (sumB - b[0]))) {
var fa = listA.BinarySearch(b, Index0.Inst);
listA.Insert(fa < 0 ? ~fa : fa, b);
listB.RemoveAt(i);
i--;
sumA += b[0];
sumB -= b[0];
change = true;
} else {break;}
}
// swap 1 for 1
for(int i=0; i<listA.Count; i++) {
var a = listA[i];
for(int j=0; j<listB.Count; j++) {
var b = listB[j];
if(Math.Abs(sumA - sumB) > Math.Abs((sumA - a[0] + b[0]) - (sumB -b[0] + a[0]))) {
listA.RemoveAt(i);
listB.RemoveAt(j);
var fa = listA.BinarySearch(b, Index0.Inst);
var fb = listB.BinarySearch(a, Index0.Inst);
listA.Insert(fa < 0 ? ~fa : fa, b);
listB.Insert(fb < 0 ? ~fb : fb, a);
sumA = sumA - a[0] + b[0];
sumB = sumB - b[0] + a[0];
change = true;
break;
}
}
}
//
if(change == false) { break; }
}
/*
// further optimization with 2 for 1 swaps
while(DateTime.Now.Subtract(start).TotalSeconds < 4.8) {
change = false;
// trade 2 for 1
for(int i=0; i<listA.Count >> 1; i++) {
var a1 = listA[i];
var a2 = listA[listA.Count - 1 - i];
for(int j=0; j<listB.Count; j++) {
var b = listB[j];
if(Math.Abs(sumA - sumB) > Math.Abs((sumA - a1[0] - a2[0] + b[0]) - (sumB - b[0] + a1[0] + a2[0]))) {
listA.RemoveAt(listA.Count - 1 - i);
listA.RemoveAt(i);
listB.RemoveAt(j);
var fa = listA.BinarySearch(b, Index0.Inst);
var fb1 = listB.BinarySearch(a1, Index0.Inst);
var fb2 = listB.BinarySearch(a2, Index0.Inst);
listA.Insert(fa < 0 ? ~fa : fa, b);
listB.Insert(fb1 < 0 ? ~fb1 : fb1, a1);
listB.Insert(fb2 < 0 ? ~fb2 : fb2, a2);
sumA = sumA - a1[0] - a2[0] + b[0];
sumB = sumB - b[0] + a1[0] + a2[0];
change = true;
break;
}
}
}
//
if(DateTime.Now.Subtract(start).TotalSeconds > 4.8) { break; }
// trade 2 for 1
for(int i=0; i<listB.Count >> 1; i++) {
var b1 = listB[i];
var b2 = listB[listB.Count - 1 - i];
for(int j=0; j<listA.Count; j++) {
var a = listA[j];
if(Math.Abs(sumA - sumB) > Math.Abs((sumA - a[0] + b1[0] + b2[0]) - (sumB - b1[0] - b2[0] + a[0]))) {
listB.RemoveAt(listB.Count - 1 - i);
listB.RemoveAt(i);
listA.RemoveAt(j);
var fa1 = listA.BinarySearch(b1, Index0.Inst);
var fa2 = listA.BinarySearch(b2, Index0.Inst);
var fb = listB.BinarySearch(a, Index0.Inst);
listA.Insert(fa1 < 0 ? ~fa1 : fa1, b1);
listA.Insert(fa2 < 0 ? ~fa2 : fa2, b2);
listB.Insert(fb < 0 ? ~fb : fb, a);
sumA = sumA - a[0] + b1[0] + b2[0];
sumB = sumB - b1[0] - b2[0] + a[0];
change = true;
break;
}
}
}
//
if(change == false) { break; }
}
*/
// output the correct ordered values
listA.Sort(new Index1());
foreach(var t in listA) {
Console.WriteLine(t[1]);
}
// DEBUG/TESTING
//Console.WriteLine(approx[0]);
//foreach(var t in listA) Console.Write(": " + t[0] + "," + t[1]);
//Console.WriteLine();
//foreach(var t in listB) Console.Write(": " + t[0] + "," + t[1]);
}
}
来源：https://stackoverflow.com/questions/32354215/better-results-in-set-partition-than-by-differencing