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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.)
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.
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
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]);
}
}
I have a triangulated mesh. Assume it looks like an bumpy surface. I want to be able to find all edges that fall on the surrounding border of the mesh. (forget about inner vertices)
I know I have to find edges that are only connected to one triangle, and collect all these together and that is the answer. But I want to be sure that the vertices of these edges are ordered clockwise around the shape.
I want to do this because I would like to get a polygon line around the outside of mesh.
I hope this is clear enough to understand. In a sense i am trying to "De-Triangulate" the mesh. ha! if there is such a term.
Boundary edges are only referenced by a single triangle in the mesh, so to find them you need to scan through all triangles in the mesh and take the edges with a single reference count. You can do this efficiently (in O(N)) by making use of a hash table.
To convert the edge set to an ordered polygon loop you can use a traversal method:
Pick any unvisited edge segment [v_start,v_next] and add these vertices to the polygon loop.
Find the unvisited edge segment [v_i,v_j] that has either v_i = v_next or v_j = v_next and add the other vertex (the one not equal to v_next) to the polygon loop. Reset v_next as this newly added vertex, mark the edge as visited and continue from 2.
Traversal is done when we get back to v_start.
The traversal will give a polygon loop that could have either clock-wise or counter-clock-wise ordering. A consistent ordering can be established by considering the signed area of the polygon. If the traversal results in the wrong orientation you simply need to reverse the order of the polygon loop vertices.
Well as the saying goes - get it working - then get it working better. I noticed on my above example it assumes all the edges in the edges array do in fact link up in a nice border. This may not be the case in the real world (as I have discovered from my input files i am using!) In fact some of my input files actually have many polygons and all need borders detected. I also wanted to make sure the winding order is correct. So I have fixed that up as well. see below. (Feel I am making headway at last!)
private static List<int> OrganizeEdges(List<int> edges, List<Point> positions)
{
var visited = new Dictionary<int, bool>();
var edgeList = new List<int>();
var resultList = new List<int>();
var nextIndex = -1;
while (resultList.Count < edges.Count)
{
if (nextIndex < 0)
{
for (int i = 0; i < edges.Count; i += 2)
{
if (!visited.ContainsKey(i))
{
nextIndex = edges[i];
break;
}
}
}
for (int i = 0; i < edges.Count; i += 2)
{
if (visited.ContainsKey(i))
continue;
int j = i + 1;
int k = -1;
if (edges[i] == nextIndex)
k = j;
else if (edges[j] == nextIndex)
k = i;
if (k >= 0)
{
var edge = edges[k];
visited[i] = true;
edgeList.Add(nextIndex);
edgeList.Add(edge);
nextIndex = edge;
i = 0;
}
}
// calculate winding order - then add to final result.
var borderPoints = new List<Point>();
edgeList.ForEach(ei => borderPoints.Add(positions[ei]));
var winding = CalculateWindingOrder(borderPoints);
if (winding > 0)
edgeList.Reverse();
resultList.AddRange(edgeList);
edgeList = new List<int>();
nextIndex = -1;
}
return resultList;
}
/// <summary>
/// returns 1 for CW, -1 for CCW, 0 for unknown.
/// </summary>
public static int CalculateWindingOrder(IList<Point> points)
{
// the sign of the 'area' of the polygon is all we are interested in.
var area = CalculateSignedArea(points);
if (area < 0.0)
return 1;
else if (area > 0.0)
return - 1;
return 0; // error condition - not even verts to calculate, non-simple poly, etc.
}
public static double CalculateSignedArea(IList<Point> points)
{
double area = 0.0;
for (int i = 0; i < points.Count; i++)
{
int j = (i + 1) % points.Count;
area += points[i].X * points[j].Y;
area -= points[i].Y * points[j].X;
}
area /= 2.0f;
return area;
}
Traversal Code (not efficient - needs to be tidied up, will get to that at some point) Please Note: I store each segment in the chain as 2 indices - rather than 1 as suggested by Darren. This is purely for my own implementation / rendering needs.
// okay now lets sort the segments so that they make a chain.
var sorted = new List<int>();
var visited = new Dictionary<int, bool>();
var startIndex = edges[0];
var nextIndex = edges[1];
sorted.Add(startIndex);
sorted.Add(nextIndex);
visited[0] = true;
visited[1] = true;
while (nextIndex != startIndex)
{
for (int i = 0; i < edges.Count - 1; i += 2)
{
var j = i + 1;
if (visited.ContainsKey(i) || visited.ContainsKey(j))
continue;
var iIndex = edges[i];
var jIndex = edges[j];
if (iIndex == nextIndex)
{
sorted.Add(nextIndex);
sorted.Add(jIndex);
nextIndex = jIndex;
visited[j] = true;
break;
}
else if (jIndex == nextIndex)
{
sorted.Add(nextIndex);
sorted.Add(iIndex);
nextIndex = iIndex;
visited[i] = true;
break;
}
}
}
return sorted;
The answer to your question depends actually on how triangular mesh is represented in memory. If you use Half-edge data structure, then the algorithm is extremely simple, since everything was already done during Half-edge data structure construction.
Start from any boundary half-edge HE_edge* edge0 (it can be found by linear search over all half-edges as the first edge without valid face). Set the current half-edge HE_edge* edge = edge0.
Output the destination edge->vert of the current edge.
The next edge in clockwise order around the shape (and counter-clockwise order around the surrounding "hole") will be edge->next.
Stop when you reach edge0.
To efficiently enumerate the boundary edges in the opposite (counter-clockwise order) the data structure needs to have prev data field, which many existing implementations of Half-edge data structure do provide in addition to next, e.g. MeshLib
This is not a homework question :)
I have a set of rectangles scattered across an image. I want to merge (create a union) of every group of intersected rectangles. If a rectangle doesn't intersect its neighbors, it remains untouched.
The problem is that merged rectangles can intersect rectangles that were previously not in consideration; merged rectangles can also intersect newly-merged rectangles. I want to catch those cases.
So, in my mind, it needs to be iterative (try each rect against every other rect in the set) and recursive (try each merged rect against the set again, including merged rects).
How can I go about this? I'm working in Java, but this is more of an algorithmic question than a language-oriented one.
Thanks!
Edit: added relevant code to better illustrate the poor way in which I'm handling it now.
public static List<BinaryRegion> mergeRegions(List<BinaryRegion> regions)
{
List<BinaryRegion> merged = new ArrayList<BinaryRegion>();
geoModel = new GeometryFactory();
Polygon polys[] = new Polygon[regions.size()];
for (int i = 0; i < regions.size(); i++)
{
Polygon p = convertRectangleToPolygon(regions.get(i)
.getBoundingBox());
polys[i] = p;
}
System.out.println("Converted " + regions.size() + " polys");
for (int i = 0; i < regions.size(); i++)
{
System.out.println("Sending in poly " + i);
ArrayList<Polygon> result = mergePoly(polys[i], polys);
System.out.println("After run, size=" + result.size());
}
return merged;
}
private static ArrayList<Polygon> mergePoly(Polygon p, Polygon[] polys)
{
ArrayList<Polygon> merges = new ArrayList<Polygon>();
for (int i = 0; i < polys.length; i++)
{
if (p.equals(polys[i]))
System.out.println("found the exact match at " + i);
else if (p.intersects(polys[i]))
{
System.out.println("Found intersection at " + i);
System.out.println("Other poly is area "+polys[i].getArea());
Polygon u = (Polygon) p.union(polys[i]);
System.out.println("Merge size="+u.getArea());
merges.add(u);
}
else
merges.add(polys[i]);
}
return merges;
}
Not entirely sure whether this nested iterative approach is really the way to go (especially since I don't see how exactly you're handling the merged regions after your call to mergePoly). Instead of working with one polygon at a time comparing to all other polygons, why don't you keep intermediate steps and rerun the merge until there are no more intersections? Something along the lines of:
private static ArrayList<Polygon> mergePoly(Polygon[] polys)
{
List<Polygon> polygons = new ArrayList<Polygon>(Arrays.asList(polys));
boolean foundIntersection = false;
do
{
foundIntersection = false;
for (int i = 0; i < polygons.size(); i++)
{
Polygon current = polygons.get(i);
for (int j = i + 1; j < polygons.size(); j++)
{
if (current.intersects(polygons.get(j)))
{
foundIntersection = true;
current = (Polygon)current.union(polygons.remove(j--));
System.out.println("Merge size="+u.getArea());
}
}
polygons.set(i, current);
}
} while(foundIntersection);
return polygons;
}
It's been a while since I've worked with Java, but the logic is pretty self-explanatory. You perform two iterations of the polygons. The outside iteration is your "current" polygon, that you will merge all the inner polygons with (removing them from the collection as you go along). After every outer iteration, you'll just set the element at that index with the (possibly) merged polygon, and move on to the next polygon in the series. You'll keep doing this until you no longer get anymore merges. Just keep in mind that this is a horribly naive implementation, and you might be better off breaking it off into "halves" and merging these smaller subsets (think of mergesort).
You can use the sweep line algorithm to find the intersections of rectangles (example1, example2) and the union-find algorithm to effectively merge sets of rectangles.
I've been reading a lot about Markov Decision Processes (using value iteration) lately but I simply can't get my head around them. I've found a lot of resources on the Internet / books, but they all use mathematical formulas that are way too complex for my competencies.
Since this is my first year at college, I've found that the explanations and formulas provided on the web use notions / terms that are way too complicated for me and they assume that the reader knows certain things that I've simply never heard of.
I want to use it on a 2D grid (filled with walls(unattainable), coins(desirable) and enemies that move(which must be avoided at all costs)). The whole goal is to collect all the coins without touching the enemies, and I want to create an AI for the main player using a Markov Decision Process (MDP). Here is how it partially looks like (note that the game-related aspect is not so much of a concern here. I just really want to understand MDPs in general):
From what I understand, a rude simplification of MDPs is that they can create a grid which holds in which direction we need to go (kind of a grid of "arrows" pointing where we need to go, starting at a certain position on the grid) to get to certain goals and avoid certain obstacles. Specific to my situation, that would mean that it allows the player to know in which direction to go to collect the coins and avoid the enemies.
Now, using the MDP terms, it would mean that it creates a collection of states(the grid) which holds certain policies(the action to take -> up, down, right, left) for a certain state(a position on the grid). The policies are determined by the "utility" values of each state, which themselves are calculated by evaluating how much getting there would be beneficial in the short and long term.
Is this correct? Or am I completely on the wrong track?
I'd at least like to know what the variables from the following equation represent in my situation:
(taken from the book "Artificial Intelligence - A Modern Approach" from Russell & Norvig)
I know that s would be a list of all the squares from the grid, a would be a specific action (up / down / right / left), but what about the rest?
How would the reward and utility functions be implemented?
It would be really great if someone knew a simple link which shows pseudo-code to implement a basic version with similarities to my situation in a very slow way, because I don't even know where to start here.
Thank you for your precious time.
(Note: feel free to add / remove tags or tell me in the comments if I should give more details about something or anything like that.)
Yes, the mathematical notation can make it seem much more complicated than it is. Really, it is a very simple idea. I have a implemented a value iteration demo applet that you can play with to get a better idea.
Basically, lets say you have a 2D grid with a robot in it. The robot can try to move North, South, East, West (those are the actions a) but, because its left wheel is slippery, when it tries to move North there is only a .9 probability that it will end up at the square North of it while there is a .1 probability that it will end up at the square West of it (similarly for the other 3 actions). These probabilities are captured by the T() function. Namely, T(s,A,s') will look like:
s A s' T //x=0,y=0 is at the top-left of the screen
x,y North x,y+1 .9 //we do move north
x,y North x-1,y .1 //wheels slipped, so we move West
x,y East x+1,y .9
x,y East x,y-1 .1
x,y South x,y+1 .9
x,y South x-1,y .1
x,y West x-1,y .9
x,y West x,y+1 .1
You then set the Reward to be 0 for all states, but 100 for the goal state, that is, the location you want the robot to get to.
What value-iteration does is its starts by giving a Utility of 100 to the goal state and 0 to all the other states. Then on the first iteration this 100 of utility gets distributed back 1-step from the goal, so all states that can get to the goal state in 1 step (all 4 squares right next to it) will get some utility. Namely, they will get a Utility equal to the probability that from that state we can get to the goal stated. We then continue iterating, at each step we move the utility back 1 more step away from the goal.
In the example above, say you start with R(5,5)= 100 and R(.) = 0 for all other states. So the goal is to get to 5,5.
On the first iteration we set
R(5,6) = gamma * (.9 * 100) + gamma * (.1 * 100)
because on 5,6 if you go North there is a .9 probability of ending up at 5,5, while if you go West there is a .1 probability of ending up at 5,5.
Similarly for (5,4), (4,5), (6,5).
All other states remain with U = 0 after the first iteration of value iteration.
Not a complete answer, but a clarifying remark.
The state is not a single cell. The state contains the information what is in each cell for all concerned cells at once. This means one state element contains the information which cells are solid and which are empty; which ones contain monsters; where are coins; where is the player.
Maybe you could use a map from each cell to its content as state. This does ignore the movement of monsters and player, which are probably very important, too.
The details depend on how you want to model your problem (deciding what belongs to the state and in which form).
Then a policy maps each state to an action like left, right, jump, etc.
First you must understand the problem that is expressed by a MDP before thinking about how algorithms like value iteration work.
I would recommend using Q-learning for your implementation.
Maybe you can use this post I wrote as an inspiration. This is a Q-learning demo with Java source code. This demo is a map with 6 fields and the AI learns where it should go from every state to get to the reward.
Q-learning is a technique for letting the AI learn by itself by giving it reward or punishment.
This example shows the Q-learning used for path finding. A robot learns where it should go from any state.
The robot starts at a random place, it keeps memory of the score while it explores the area, whenever it reaches the goal, we repeat with a new random start. After enough repetitions the score values will be stationary (convergence).
In this example the action outcome is deterministic (transition probability is 1) and the action selection is random. The score values are calculated by the Q-learning algorithm Q(s,a).
The image shows the states (A,B,C,D,E,F), possible actions from the states and the reward given.
Result Q*(s,a)
Policy Π*(s)
Qlearning.java
import java.text.DecimalFormat;
import java.util.Random;
/**
* #author Kunuk Nykjaer
*/
public class Qlearning {
final DecimalFormat df = new DecimalFormat("#.##");
// path finding
final double alpha = 0.1;
final double gamma = 0.9;
// states A,B,C,D,E,F
// e.g. from A we can go to B or D
// from C we can only go to C
// C is goal state, reward 100 when B->C or F->C
//
// _______
// |A|B|C|
// |_____|
// |D|E|F|
// |_____|
//
final int stateA = 0;
final int stateB = 1;
final int stateC = 2;
final int stateD = 3;
final int stateE = 4;
final int stateF = 5;
final int statesCount = 6;
final int[] states = new int[]{stateA,stateB,stateC,stateD,stateE,stateF};
// http://en.wikipedia.org/wiki/Q-learning
// http://people.revoledu.com/kardi/tutorial/ReinforcementLearning/Q-Learning.htm
// Q(s,a)= Q(s,a) + alpha * (R(s,a) + gamma * Max(next state, all actions) - Q(s,a))
int[][] R = new int[statesCount][statesCount]; // reward lookup
double[][] Q = new double[statesCount][statesCount]; // Q learning
int[] actionsFromA = new int[] { stateB, stateD };
int[] actionsFromB = new int[] { stateA, stateC, stateE };
int[] actionsFromC = new int[] { stateC };
int[] actionsFromD = new int[] { stateA, stateE };
int[] actionsFromE = new int[] { stateB, stateD, stateF };
int[] actionsFromF = new int[] { stateC, stateE };
int[][] actions = new int[][] { actionsFromA, actionsFromB, actionsFromC,
actionsFromD, actionsFromE, actionsFromF };
String[] stateNames = new String[] { "A", "B", "C", "D", "E", "F" };
public Qlearning() {
init();
}
public void init() {
R[stateB][stateC] = 100; // from b to c
R[stateF][stateC] = 100; // from f to c
}
public static void main(String[] args) {
long BEGIN = System.currentTimeMillis();
Qlearning obj = new Qlearning();
obj.run();
obj.printResult();
obj.showPolicy();
long END = System.currentTimeMillis();
System.out.println("Time: " + (END - BEGIN) / 1000.0 + " sec.");
}
void run() {
/*
1. Set parameter , and environment reward matrix R
2. Initialize matrix Q as zero matrix
3. For each episode: Select random initial state
Do while not reach goal state o
Select one among all possible actions for the current state o
Using this possible action, consider to go to the next state o
Get maximum Q value of this next state based on all possible actions o
Compute o Set the next state as the current state
*/
// For each episode
Random rand = new Random();
for (int i = 0; i < 1000; i++) { // train episodes
// Select random initial state
int state = rand.nextInt(statesCount);
while (state != stateC) // goal state
{
// Select one among all possible actions for the current state
int[] actionsFromState = actions[state];
// Selection strategy is random in this example
int index = rand.nextInt(actionsFromState.length);
int action = actionsFromState[index];
// Action outcome is set to deterministic in this example
// Transition probability is 1
int nextState = action; // data structure
// Using this possible action, consider to go to the next state
double q = Q(state, action);
double maxQ = maxQ(nextState);
int r = R(state, action);
double value = q + alpha * (r + gamma * maxQ - q);
setQ(state, action, value);
// Set the next state as the current state
state = nextState;
}
}
}
double maxQ(int s) {
int[] actionsFromState = actions[s];
double maxValue = Double.MIN_VALUE;
for (int i = 0; i < actionsFromState.length; i++) {
int nextState = actionsFromState[i];
double value = Q[s][nextState];
if (value > maxValue)
maxValue = value;
}
return maxValue;
}
// get policy from state
int policy(int state) {
int[] actionsFromState = actions[state];
double maxValue = Double.MIN_VALUE;
int policyGotoState = state; // default goto self if not found
for (int i = 0; i < actionsFromState.length; i++) {
int nextState = actionsFromState[i];
double value = Q[state][nextState];
if (value > maxValue) {
maxValue = value;
policyGotoState = nextState;
}
}
return policyGotoState;
}
double Q(int s, int a) {
return Q[s][a];
}
void setQ(int s, int a, double value) {
Q[s][a] = value;
}
int R(int s, int a) {
return R[s][a];
}
void printResult() {
System.out.println("Print result");
for (int i = 0; i < Q.length; i++) {
System.out.print("out from " + stateNames[i] + ": ");
for (int j = 0; j < Q[i].length; j++) {
System.out.print(df.format(Q[i][j]) + " ");
}
System.out.println();
}
}
// policy is maxQ(states)
void showPolicy() {
System.out.println("\nshowPolicy");
for (int i = 0; i < states.length; i++) {
int from = states[i];
int to = policy(from);
System.out.println("from "+stateNames[from]+" goto "+stateNames[to]);
}
}
}
Print result
out from A: 0 90 0 72,9 0 0
out from B: 81 0 100 0 81 0
out from C: 0 0 0 0 0 0
out from D: 81 0 0 0 81 0
out from E: 0 90 0 72,9 0 90
out from F: 0 0 100 0 81 0
showPolicy
from a goto B
from b goto C
from c goto C
from d goto A
from e goto B
from f goto C
Time: 0.025 sec.
I know this is a fairly old post, but i came across it when looking for MDP related questions, I did want to note (for folks coming in here) a few more comments about when you stated what "s" and "a" were.
I think for a you are absolutely correct it's your list of [up,down,left,right].
However for s it's really the location in the grid and s' is the location you can go to.
What that means is that you pick a state, and then you pick a particular s' and go through all the actions that can take you to that sprime, which you use to figure out those values. (pick a max out of those). Finally you go for the next s' and do the same thing, when you've exhausted all the s' values then you find the max of what you just finished searching on.
Suppose you picked a grid cell in the corner, you'd only have 2 states you could possibly move to (assuming bottom left corner), depending on how you choose to "name" your states, we could in this case assume a state is an x,y coordinate, so your current state s is 1,1 and your s' (or s prime) list is x+1,y and x,y+1 (no diagonal in this example) (The Summation part that goes over all s')
Also you don't have it listed in your equation, but the max is of a or the action that gives you the max, so first you pick the s' that gives you the max and then within that you pick the action (at least this is my understanding of the algorithm).
So if you had
x,y+1 left = 10
x,y+1 right = 5
x+1,y left = 3
x+1,y right 2
You'll pick x,y+1 as your s', but then you'll need to pick an action that is maximized which is in this case left for x,y+1. I'm not sure if there is a subtle difference between just finding the maximum number and finding the state then the maximum number though so maybe someone someday can clarify that for me.
If your movements are deterministic (meaning if you say go forward, you go forward with 100% certainty), then it's pretty easy you have one action, However if they are non deterministic, you have a say 80% certainty then you should consider the other actions which could get you there. This is the context of the slippery wheel that Jose mentioned above.
I don't want to detract what others have said, but just to give some additional information.
I am interested in ways to improve or come up with algorithms that are able to solve the Travelling salesman problem for about n = 100 to 200 cities.
The wikipedia link I gave lists various optimizations, but it does so at a pretty high level, and I don't know how to go about actually implementing them in code.
There are industrial strength solvers out there, such as Concorde, but those are way too complex for what I want, and the classic solutions that flood the searches for TSP all present randomized algorithms or the classic backtracking or dynamic programming algorithms that only work for about 20 cities.
So, does anyone know how to implement a simple (by simple I mean that an implementation doesn't take more than 100-200 lines of code) TSP solver that works in reasonable time (a few seconds) for at least 100 cities? I am only interested in exact solutions.
You may assume that the input will be randomly generated, so I don't care for inputs that are aimed specifically at breaking a certain algorithm.
200 lines and no libraries is a tough constraint. The advanced solvers use branch and bound with the Held–Karp relaxation, and I'm not sure if even the most basic version of that would fit into 200 normal lines. Nevertheless, here's an outline.
Held Karp
One way to write TSP as an integer program is as follows (Dantzig, Fulkerson, Johnson). For all edges e, constant we denotes the length of edge e, and variable xe is 1 if edge e is on the tour and 0 otherwise. For all subsets S of vertices, ∂(S) denotes the edges connecting a vertex in S with a vertex not in S.
minimize sumedges e we xe
subject to
1. for all vertices v, sumedges e in ∂({v}) xe = 2
2. for all nonempty proper subsets S of vertices, sumedges e in ∂(S) xe ≥ 2
3. for all edges e in E, xe in {0, 1}
Condition 1 ensures that the set of edges is a collection of tours. Condition 2 ensures that there's only one. (Otherwise, let S be the set of vertices visited by one of the tours.) The Held–Karp relaxation is obtained by making this change.
3. for all edges e in E, xe in {0, 1}
3. for all edges e in E, 0 ≤ xe ≤ 1
Held–Karp is a linear program but it has an exponential number of constraints. One way to solve it is to introduce Lagrange multipliers and then do subgradient optimization. That boils down to a loop that computes a minimum spanning tree and then updates some vectors, but the details are sort of involved. Besides "Held–Karp" and "subgradient (descent|optimization)", "1-tree" is another useful search term.
(A slower alternative is to write an LP solver and introduce subtour constraints as they are violated by previous optima. This means writing an LP solver and a min-cut procedure, which is also more code, but it might extend better to more exotic TSP constraints.)
Branch and bound
By "partial solution", I mean an partial assignment of variables to 0 or 1, where an edge assigned 1 is definitely in the tour, and an edge assigned 0 is definitely out. Evaluating Held–Karp with these side constraints gives a lower bound on the optimum tour that respects the decisions already made (an extension).
Branch and bound maintains a set of partial solutions, at least one of which extends to an optimal solution. The pseudocode for one variant, depth-first search with best-first backtracking is as follows.
let h be an empty minheap of partial solutions, ordered by Held–Karp value
let bestsolsofar = null
let cursol be the partial solution with no variables assigned
loop
while cursol is not a complete solution and cursol's H–K value is at least as good as the value of bestsolsofar
choose a branching variable v
let sol0 be cursol union {v -> 0}
let sol1 be cursol union {v -> 1}
evaluate sol0 and sol1
let cursol be the better of the two; put the other in h
end while
if cursol is better than bestsolsofar then
let bestsolsofar = cursol
delete all heap nodes worse than cursol
end if
if h is empty then stop; we've found the optimal solution
pop the minimum element of h and store it in cursol
end loop
The idea of branch and bound is that there's a search tree of partial solutions. The point of solving Held–Karp is that the value of the LP is at most the length OPT of the optimal tour but also conjectured to be at least 3/4 OPT (in practice, usually closer to OPT).
The one detail in the pseudocode I've left out is how to choose the branching variable. The goal is usually to make the "hard" decisions first, so fixing a variable whose value is already near 0 or 1 is probably not wise. One option is to choose the closest to 0.5, but there are many, many others.
EDIT
Java implementation. 198 nonblank, noncomment lines. I forgot that 1-trees don't work with assigning variables to 1, so I branch by finding a vertex whose 1-tree has degree >2 and delete each edge in turn. This program accepts TSPLIB instances in EUC_2D format, e.g., eil51.tsp and eil76.tsp and eil101.tsp and lin105.tsp from http://www2.iwr.uni-heidelberg.de/groups/comopt/software/TSPLIB95/tsp/.
// simple exact TSP solver based on branch-and-bound/Held--Karp
import java.io.*;
import java.util.*;
import java.util.regex.*;
public class TSP {
// number of cities
private int n;
// city locations
private double[] x;
private double[] y;
// cost matrix
private double[][] cost;
// matrix of adjusted costs
private double[][] costWithPi;
Node bestNode = new Node();
public static void main(String[] args) throws IOException {
// read the input in TSPLIB format
// assume TYPE: TSP, EDGE_WEIGHT_TYPE: EUC_2D
// no error checking
TSP tsp = new TSP();
tsp.readInput(new InputStreamReader(System.in));
tsp.solve();
}
public void readInput(Reader r) throws IOException {
BufferedReader in = new BufferedReader(r);
Pattern specification = Pattern.compile("\\s*([A-Z_]+)\\s*(:\\s*([0-9]+))?\\s*");
Pattern data = Pattern.compile("\\s*([0-9]+)\\s+([-+.0-9Ee]+)\\s+([-+.0-9Ee]+)\\s*");
String line;
while ((line = in.readLine()) != null) {
Matcher m = specification.matcher(line);
if (!m.matches()) continue;
String keyword = m.group(1);
if (keyword.equals("DIMENSION")) {
n = Integer.parseInt(m.group(3));
cost = new double[n][n];
} else if (keyword.equals("NODE_COORD_SECTION")) {
x = new double[n];
y = new double[n];
for (int k = 0; k < n; k++) {
line = in.readLine();
m = data.matcher(line);
m.matches();
int i = Integer.parseInt(m.group(1)) - 1;
x[i] = Double.parseDouble(m.group(2));
y[i] = Double.parseDouble(m.group(3));
}
// TSPLIB distances are rounded to the nearest integer to avoid the sum of square roots problem
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
double dx = x[i] - x[j];
double dy = y[i] - y[j];
cost[i][j] = Math.rint(Math.sqrt(dx * dx + dy * dy));
}
}
}
}
}
public void solve() {
bestNode.lowerBound = Double.MAX_VALUE;
Node currentNode = new Node();
currentNode.excluded = new boolean[n][n];
costWithPi = new double[n][n];
computeHeldKarp(currentNode);
PriorityQueue<Node> pq = new PriorityQueue<Node>(11, new NodeComparator());
do {
do {
boolean isTour = true;
int i = -1;
for (int j = 0; j < n; j++) {
if (currentNode.degree[j] > 2 && (i < 0 || currentNode.degree[j] < currentNode.degree[i])) i = j;
}
if (i < 0) {
if (currentNode.lowerBound < bestNode.lowerBound) {
bestNode = currentNode;
System.err.printf("%.0f", bestNode.lowerBound);
}
break;
}
System.err.printf(".");
PriorityQueue<Node> children = new PriorityQueue<Node>(11, new NodeComparator());
children.add(exclude(currentNode, i, currentNode.parent[i]));
for (int j = 0; j < n; j++) {
if (currentNode.parent[j] == i) children.add(exclude(currentNode, i, j));
}
currentNode = children.poll();
pq.addAll(children);
} while (currentNode.lowerBound < bestNode.lowerBound);
System.err.printf("%n");
currentNode = pq.poll();
} while (currentNode != null && currentNode.lowerBound < bestNode.lowerBound);
// output suitable for gnuplot
// set style data vector
System.out.printf("# %.0f%n", bestNode.lowerBound);
int j = 0;
do {
int i = bestNode.parent[j];
System.out.printf("%f\t%f\t%f\t%f%n", x[j], y[j], x[i] - x[j], y[i] - y[j]);
j = i;
} while (j != 0);
}
private Node exclude(Node node, int i, int j) {
Node child = new Node();
child.excluded = node.excluded.clone();
child.excluded[i] = node.excluded[i].clone();
child.excluded[j] = node.excluded[j].clone();
child.excluded[i][j] = true;
child.excluded[j][i] = true;
computeHeldKarp(child);
return child;
}
private void computeHeldKarp(Node node) {
node.pi = new double[n];
node.lowerBound = Double.MIN_VALUE;
node.degree = new int[n];
node.parent = new int[n];
double lambda = 0.1;
while (lambda > 1e-06) {
double previousLowerBound = node.lowerBound;
computeOneTree(node);
if (!(node.lowerBound < bestNode.lowerBound)) return;
if (!(node.lowerBound < previousLowerBound)) lambda *= 0.9;
int denom = 0;
for (int i = 1; i < n; i++) {
int d = node.degree[i] - 2;
denom += d * d;
}
if (denom == 0) return;
double t = lambda * node.lowerBound / denom;
for (int i = 1; i < n; i++) node.pi[i] += t * (node.degree[i] - 2);
}
}
private void computeOneTree(Node node) {
// compute adjusted costs
node.lowerBound = 0.0;
Arrays.fill(node.degree, 0);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) costWithPi[i][j] = node.excluded[i][j] ? Double.MAX_VALUE : cost[i][j] + node.pi[i] + node.pi[j];
}
int firstNeighbor;
int secondNeighbor;
// find the two cheapest edges from 0
if (costWithPi[0][2] < costWithPi[0][1]) {
firstNeighbor = 2;
secondNeighbor = 1;
} else {
firstNeighbor = 1;
secondNeighbor = 2;
}
for (int j = 3; j < n; j++) {
if (costWithPi[0][j] < costWithPi[0][secondNeighbor]) {
if (costWithPi[0][j] < costWithPi[0][firstNeighbor]) {
secondNeighbor = firstNeighbor;
firstNeighbor = j;
} else {
secondNeighbor = j;
}
}
}
addEdge(node, 0, firstNeighbor);
Arrays.fill(node.parent, firstNeighbor);
node.parent[firstNeighbor] = 0;
// compute the minimum spanning tree on nodes 1..n-1
double[] minCost = costWithPi[firstNeighbor].clone();
for (int k = 2; k < n; k++) {
int i;
for (i = 1; i < n; i++) {
if (node.degree[i] == 0) break;
}
for (int j = i + 1; j < n; j++) {
if (node.degree[j] == 0 && minCost[j] < minCost[i]) i = j;
}
addEdge(node, node.parent[i], i);
for (int j = 1; j < n; j++) {
if (node.degree[j] == 0 && costWithPi[i][j] < minCost[j]) {
minCost[j] = costWithPi[i][j];
node.parent[j] = i;
}
}
}
addEdge(node, 0, secondNeighbor);
node.parent[0] = secondNeighbor;
node.lowerBound = Math.rint(node.lowerBound);
}
private void addEdge(Node node, int i, int j) {
double q = node.lowerBound;
node.lowerBound += costWithPi[i][j];
node.degree[i]++;
node.degree[j]++;
}
}
class Node {
public boolean[][] excluded;
// Held--Karp solution
public double[] pi;
public double lowerBound;
public int[] degree;
public int[] parent;
}
class NodeComparator implements Comparator<Node> {
public int compare(Node a, Node b) {
return Double.compare(a.lowerBound, b.lowerBound);
}
}
If your graph satisfy the triangle inequality and you want a guarantee of 3/2 within the optimum I suggest the christofides algorithm. I've wrote an implementation in php at phpclasses.org.
As of 2013, It is possible to solve for 100 cities using only the exact formulation in Cplex. Add degree equations for each vertex, but include subtour-avoiding constraints only as they appear. Most of them are not necessary. Cplex has an example on this.
You should be able to solve for 100 cities. You will have to iterate every time a new subtour is found. I ran an example here and in a couple of minutes and 100 iterations later I got my results.
I took Held-Karp algorithm from concorde library and 25 cities are solved in 0.15 seconds. This performance is perfectly good for me! You can extract the code (writen in ANSI C) of held-karp from concorde library: http://www.math.uwaterloo.ca/tsp/concorde/downloads/downloads.htm. If the download has the extension gz, it should be tgz. You might need to rename it. Then you should make little ajustments to port in in VC++. First take the file heldkarp h and c (rename it cpp) and other about 5 files, make adjustments and it should work calling CCheldkarp_small(...) with edgelen: euclid_ceiling_edgelen.
TSP is an NP-hard problem. (As far as we know) there is no algorithm for NP-hard problems which runs in polynomial time, so you ask for something that doesn't exist.
It's either fast enough to finish in a reasonable time and then it's not exact, or exact but won't finish in your lifetime for 100 cities.
To give a dumb answer: me too. Everyone is interrested in such algorithm, but as others already stated: I does not (yet?) exist. Esp your combination of exact, 200 nodes, few seconds runtime and just 200 lines of code is impossible. You already know that is it NP hard and if you got the slightest impression of asymptotic behaviour you should know that there is no way of achieving this (except you prove that NP=P, and even that I would say thats not possible). Even the exact commercial solvers need for such instances far more than some seconds and as you can imagine they have far more than 200 lines of code (even when you just consider their kernels).
EDIT: The wiki algorithms are the "usual suspects" of the field: Linear Programming and branch-and-bound. Their solutions for the instances with thousands of nodes took Years to solve (they just did it with very very much CPUs parallel, so they can do it faster). Some even use for the branch-and-bound problem specific knowledge for the bounding, so they are no general approaches.
Branch and bound just enumerates all possible paths (e.g. with backtracking) and applies once it has a solution this for to stop a started recursion when it can prove that the result is not better than the already found solution (e.g. if you just visited 2 of your cities and the path is already longer than a found 200 city tour. You can discard all tours that start with that 2 city combination). Here you can invest very much problem specific knowledge in the function that tells you, that the path is not going to be better than the already found solution. The better it is, the less paths you have to look at, the faster is your algorithm.
Linear Programming is an optimization method so solve linear inequality problems. It works in polynomial time (simplex just practically, but that doesnt matter here), but the solution is real. When you have the additional constraint that the solution must be integer, it gets NP-complete. For small instances it is possible, e.g. one method to solve it, then look which variable of the solution violates the integer part and add addition inequalities to change it (this is called cutting-plane, the name cames from the fact that the inequalities define (higher-dimensional) plane, the solution space is a polytop and by adding additional inequalities you cut something with a plane from the polytop). The topic is very complex and even a general simple simplex is hard to understand when you dont want dive deep into the math. There are several good books about, one of the betters is from Chvatal, Linear Programming, but there are several more.
I have a theory, but I've never had the time to pursue it:
The TSP is a bounding problem (single shape where all points lie on the perimeter) where the optimal solution is that solution that has the shortest perimeter.
There are plenty of simple ways to get all the points that lie on a minimum bounding perimeter (imagine a large elastic band stretched around a bunch of nails in a large board.)
My theory is that if you start pushing in on the elastic band so that the length of band increases by the same amount between adjacent points on the perimeter, and each segment remains in the shape of an eliptical arc, the stretched elastic will cross points on the optimal path before crossing points on non-optimal paths. See this page on mathopenref.com on drawing ellipses--particularly steps 5 and 6. Points on the bounding perimeter can be viewed as focal points of the ellipse (F1, F2) in the images below.
What I don't know is if the "bubble stretching" process needs to be reset after each new point is added, or if the existing "bubbles" continue to grow and each new point on the perimeter causes only the localized "bubble" to turn into two line segments. I'll leave that for you to figure out.