I have an ArrayList of colors and their frequency of appearance. My program should calculate a reordering of those items that maximizes the minimum distance between two equal bricks.
For example, given input consisting of 4*brick 1 (x), 3*brick 2 (y), and 5*brick 3 (z), one correct output would be: z y x z x z y x z x y.
My code does not produce good solutions. In particular, sometimes there are 2 equal bricks at the end, which is the worst case.
import java.util.ArrayList;
import java.util.Collections;
public class Calc {
// private ArrayList<Wimpel> w = new ArrayList<Brick>();
private String bKette = "";
public String bestOrder(ArrayList<Brick> w) {
while (!w.isEmpty()) {
if (w.get(0).getFrequency() > 0) {
bChain += w.get(0).getColor() + "|";
Brick brick = new Wimpel(w.get(0).getVariant(), w.get(0).getFrequency() - 1);
w.remove(0);
w.add(brick);
// bestOrder(w);
} else {
w.remove(0);
}
bestOrder(w);
}
return bOrder;
}
public int Solutions(ArrayList<Wimpel> w) {
ArrayList<Brick> tmp = new ArrayList<Brick>(w);
int l = 1;
int counter = (int) w.stream().filter(c -> Collections.max(tmp).getFrequency() == c.getFrequency()).count();
l = (int) (fakultaet(counter) * fakultaet((tmp.size() - counter)));
return l;
}
public static long fakultaet(int n) {
return n == 0 ? 1 : n * fakultaet(n - 1);
}
}
How can make my code choose an optimal order?
We will not perform your exercise for you, but we will give you some advice.
Consider your current approach: it operates by filling the result string by cycling through the bricks, choosing one item from each brick in turn as long as any items remain in that brick. But this approach is certain to fail when one brick contains at least two items more than any other, because then only that brick remains at the end, and all its remaining items have to be inserted one after the other.
That is, the problem is not that your code is buggy per se, but rather that your whole strategy is incorrect for the problem. You need something different.
Now, consider the problem itself. Which items will appear at the shortest distance apart in a correct ordering? Those having the highest frequency, of course. And you can compute that minimum distance based on the frequency and total number of items.
Suppose you arrange these most-constrained items first, at the known best distance.
What's left to do at this point? Well, you potentially have some more bricks with lesser frequency, and some more slots in which to accommodate their items. If you ignore the occupied slots altogether, you can treat this as a smaller version of the same problem you had before.
Related
Let's say we have a 2D-boolean Array as a presentation of a maze, the size of the Array is not fixed and random. The walls are depicted as true:
boolean[][] mazeArray = new boolean[width][height];
The exit of the maze is at a fixed Index. How can I check wether the traversion has reached this certain index or not?
My idea was to create an int[ ] to keep track of the position, it gets updated and overwritten with every step:
int[] location = {1,0};
... But I don't understand why my check in the while-loop doesn't work:
while( location[0] != (maze[0].length-1) && location[1] != (maze[1].length-2) ) {
// traversion with pledge algorithm
}
You're making everything so much harder for yourself. Go easier ways.
Use simpler locations handling
Instead of a location[], simlpy use int destinyX and int destinyY. And as your current position, you should use int positionX and int positionY.
If you'd like the OO-stlye more, or maybe wanna keep the gates open for a solution in 3D or n-D, you could introduce a Location class that has X and Y, and all movement and checks could be handled by that class. Then you'd have Location targetLocation = new Location(x,y); and your current position as Location currentPosition = new Location(x,y);. You then could check with if (currentPosition.equals(targetLocation))...; or in your case while(!currentPosition.equals(targetLocation)) {...}
It seems you have misunderstood the array.length function, or you're using it in an awfully inconvenient way.
At the moment you're blindly shooting at the array lengths of maze[]. This is bad for 2 reasons:
array lengths should not have anything to do with positions inside the array (logical paradox), and
because you could never freely move your destination, it would always stick to the right or bottom outsides of the maze
Use the positioning above, this will clear up that problem.
Suggestion: use a byte[][] or enum[][] for maze
At the moment, you only know if you have a wall at a certain location. If you wanna include other elements, like water, treasure, or the target location, wormholes etc, then you should either:
Use a byte[][] and fill it with values
value 0 could be pathways
value 1 could be walls
value 2 could be the exit
value 3 could be the water etc.
Use constants, like static public final int WATER_CODE = 3;
Or, alternatively, create your own enum:
public enum LocationType {PATH, WALL,EXIT,WATER}
and then have maze be like:
LocationType[][] mazeArray = new LocationType[width][height];
and everything is PATH in the beginning, and you can set up WALLS like this:
mazeArray[x][y] = LocationType.WALL;
or water:
mazeArray[x][y] = LocationType.WATER;
Use class or interface for maze[][]
For the sake of Wormhole or extended functionality, you could also use a class instead of an enum:
abstract class LocationType {}
and then implement certain types, like
class Wall extends LocationType {}
or even
class Wormhole extends LocationType {
public Location leadsTo() { /* */ };
}
and
class Treasure extends LocationType {
public int getAmoundOfGoldCoinsFound() { /* */ };
}
If you implement LocationType as an interface, replace 'extends' by 'implements'
The problem with your code is that you check wrong items in your maze array:
maze[0] is the first "line" of your 2d-array
maze[1] is the second "line" of your 2d-array
Proper way of traversing 2d-array is following (I've replaced your location array with separate x and y variables to better visualize the algorithm).
My algorithm enters the 2d mazeArray line by line and then iterates each line's elements.
public class Maze {
public static void main(String[] args) {
int width = 20;
int height = 20;
boolean[][] mazeArray = new boolean[width][height];
int x = 0;
int y = 0;
while (y < mazeArray.length) {
while (x < mazeArray[y].length) {
System.out.println("Traverse at (" + x + ", " + y + ")");
x += 1;
}
x = 0;
y += 1;
}
}
}
I am trying to build a 4 x 4 sudoku solver by using the genetic algorithm. I have some issues with values converging to local minima. I am using a ranked approach and removing the bottom two ranked answer possibilities and replacing them with a crossover between the two highest ranked answer possibilities. For additional help avoiding local mininma, I am also using mutation. If an answer is not determined within a specific amount of generation, my population is filled with completely new and random state values. However, my algorithm seems to get stuck in local minima. As a fitness function, I am using:
(Total Amount of Open Squares * 7 (possible violations at each square; row, column, and box)) - total Violations
population is an ArrayList of integer arrays in which each array is a possible end state for sudoku based on the input. Fitness is determined for each array in the population.
Would someone be able to assist me in determining why my algorithm converges on local minima or perhaps recommend a technique to use to avoid local minima. Any help is greatly appreciated.
Fitness Function:
public int[] fitnessFunction(ArrayList<int[]> population)
{
int emptySpaces = this.blankData.size();
int maxError = emptySpaces*7;
int[] fitness = new int[populationSize];
for(int i=0; i<population.size();i++)
{
int[] temp = population.get(i);
int value = evaluationFunc(temp);
fitness[i] = maxError - value;
System.out.println("Fitness(i)" + fitness[i]);
}
return fitness;
}
Crossover Function:
public void crossover(ArrayList<int[]> population, int indexWeakest, int indexStrong, int indexSecStrong, int indexSecWeak)
{
int[] tempWeak = new int[16];
int[] tempStrong = new int[16];
int[] tempSecStrong = new int[16];
int[] tempSecWeak = new int[16];
tempStrong = population.get(indexStrong);
tempSecStrong = population.get(indexSecStrong);
tempWeak = population.get(indexWeakest);
tempSecWeak = population.get(indexSecWeak);
population.remove(indexWeakest);
population.remove(indexSecWeak);
int crossoverSite = random.nextInt(14)+1;
for(int i=0;i<tempWeak.length;i++)
{
if(i<crossoverSite)
{
tempWeak[i] = tempStrong[i];
tempSecWeak[i] = tempSecStrong[i];
}
else
{
tempWeak[i] = tempSecStrong[i];
tempSecWeak[i] = tempStrong[i];
}
}
mutation(tempWeak);
mutation(tempSecWeak);
population.add(tempWeak);
population.add(tempSecWeak);
for(int j=0; j<tempWeak.length;j++)
{
System.out.print(tempWeak[j] + ", ");
}
for(int j=0; j<tempWeak.length;j++)
{
System.out.print(tempSecWeak[j] + ", ");
}
}
Mutation Function:
public void mutation(int[] mutate)
{
if(this.blankData.size() > 2)
{
Blank blank = this.blankData.get(0);
int x = blank.getPosition();
Blank blank2 = this.blankData.get(1);
int y = blank2.getPosition();
Blank blank3 = this.blankData.get(2);
int z = blank3.getPosition();
int rando = random.nextInt(4) + 1;
if(rando == 2)
{
int rando2 = random.nextInt(4) + 1;
mutate[x] = rando2;
}
if(rando == 3)
{
int rando2 = random.nextInt(4) + 1;
mutate[y] = rando2;
}
if(rando==4)
{
int rando3 = random.nextInt(4) + 1;
mutate[z] = rando3;
}
}
The reason you see rapid convergence is that your methodology for "mating" is not very good. You are always producing two offspring from "mating" of the top two scoring individuals. Imagine what happens when one of the new offspring is the same as your top individual (by chance, no crossover and no mutation, or at least none that have an effect on the fitness). Once this occurs, the top two individuals are identical which eliminates the effectiveness of crossover.
A more typical approach is to replace EVERY individual on every generation. There are lots of possible variations here, but you might do a random choice of two parents weighted fitness.
Regarding population size: I don't know how hard of a problem sudoku is given your genetic representation and fitness function, but I suggest that you think about millions of individuals, not dozens.
If you are working on really hard problems, genetic algorithms are much more effective when you place your population on a 2-D grid and choosing "parents" for each point in the grid from the nearby individuals. You will get local convergence, but each locality will have converged on different solutions; you get a huge amount of variation produced from the borders between the locally-converged areas of the grid.
Another technique you might think about is running to convergence from random populations many times and store the top individual from each run. After you build up a bunch of different local minima genomes, build a new random population from those top individuals.
I think the Sudoku is a permutation problem. therefore i suggest you to use random permutation numbers for initializing population and use the crossover method which Compatible to permutation problems.
I'm trying to write a time efficient algorithm that can detect a group of overlapping circles and make a single circle in the "middle" of the group that will represent that group. The practical application of this is representing GPS locations over a map, put the conversion in to Cartesian co-ordinates is already handled so that's not relevant, the desired effect is that at different zoom levels clusters of close together points just appear as a single circle (that will have the number of points printed in the centre in the final version)
In this example the circles just have a radius of 15 so the distance calculation (Pythagoras) is not being square rooted and compared to 225 for the collision detection. I was trying anything to shave off time, but the problem is this really needs to happen very quickly becasue it's a user facing bit of code that needs to be snappy and good looking.
I've given this a go and I it works with small data sets pretty well. 2 big problems, it takes too long and it can run out of memory if all the points are on top of one another.
The route I've taken is to calculate distance between each point in a first pass, and then take the shortest distance first and start to combine from there, anything that's been combined becomes ineligible for combination on that pass, and the whole list is passed back around to the distance calculations again until nothing changes.
To be honest I think it needs a radical shift in approach and I think it's a little beyond me. I've re factored my code in to one class for ease of posting and generated random points to give an example.
package mergepoints;
import java.awt.Point;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class Merger {
public static void main(String[] args) {
Merger m = new Merger();
m.subProcess(m.createRandomList());
}
private List<Plottable> createRandomList() {
List<Plottable> points = new ArrayList<>();
for (int i = 0; i < 50000; i++) {
Plottable p = new Plottable();
p.location = new Point((int) Math.floor(Math.random() * 1000),
(int) Math.floor(Math.random() * 1000));
points.add(p);
}
return points;
}
private List<Plottable> subProcess(List<Plottable> visible) {
List<PlottableTuple> tuples = new ArrayList<PlottableTuple>();
// create a tuple to store distance and matching objects together,
for (Plottable p : visible) {
PlottableTuple tuple = new PlottableTuple();
tuple.a = p;
tuples.add(tuple);
}
// work out each Plottable relative distance from
// one another and order them by shortest first.
// We may need to do this multiple times for one set so going in own
// method.
// this is the bit that takes ages
setDistances(tuples);
// Sort so that smallest distances are at the top.
// parse the set and combine any pair less than the smallest distance in
// to a combined pin.
// any plottable thats been combine is no longer eligable for combining
// so ignore on this parse.
List<PlottableTuple> sorted = new ArrayList<>(tuples);
Collections.sort(sorted);
Set<Plottable> done = new HashSet<>();
Set<Plottable> mergedSet = new HashSet<>();
for (PlottableTuple pt : sorted) {
if (!done.contains(pt.a) && pt.distance <= 225) {
Plottable merged = combine(pt, done);
done.add(pt.a);
for (PlottableTuple tup : pt.others) {
done.add(tup.a);
}
mergedSet.add(merged);
}
}
// if we haven't processed anything we are done just return visible
// list.
if (done.size() == 0) {
return visible;
} else {
// change the list to represent the new combined plottables and
// repeat the process.
visible.removeAll(done);
visible.addAll(mergedSet);
return subProcess(visible);
}
}
private Plottable combine(PlottableTuple pt, Set<Plottable> done) {
List<Plottable> plottables = new ArrayList<>();
plottables.addAll(pt.a.containingPlottables);
for (PlottableTuple otherTuple : pt.others) {
if (!done.contains(otherTuple.a)) {
plottables.addAll(otherTuple.a.containingPlottables);
}
}
int x = 0;
int y = 0;
for (Plottable p : plottables) {
Point position = p.location;
x += position.x;
y += position.y;
}
x = x / plottables.size();
y = y / plottables.size();
Plottable merged = new Plottable();
merged.containingPlottables.addAll(plottables);
merged.location = new Point(x, y);
return merged;
}
private void setDistances(List<PlottableTuple> tuples) {
System.out.println("pins: " + tuples.size());
int loops = 0;
// Start from the first item and loop through, then repeat but starting
// with the next item.
for (int startIndex = 0; startIndex < tuples.size() - 1; startIndex++) {
// Get the data for the start Plottable
PlottableTuple startTuple = tuples.get(startIndex);
Point startLocation = startTuple.a.location;
for (int i = startIndex + 1; i < tuples.size(); i++) {
loops++;
PlottableTuple compareTuple = tuples.get(i);
double distance = distance(startLocation, compareTuple.a.location);
setDistance(startTuple, compareTuple, distance);
setDistance(compareTuple, startTuple, distance);
}
}
System.out.println("loops " + loops);
}
private void setDistance(PlottableTuple from, PlottableTuple to,
double distance) {
if (distance < from.distance || from.others == null) {
from.distance = distance;
from.others = new HashSet<>();
from.others.add(to);
} else if (distance == from.distance) {
from.others.add(to);
}
}
private double distance(Point a, Point b) {
if (a.equals(b)) {
return 0.0;
}
double result = (((double) a.x - (double) b.x) * ((double) a.x - (double) b.x))
+ (((double) a.y - (double) b.y) * ((double) a.y - (double) b.y));
return result;
}
class PlottableTuple implements Comparable<PlottableTuple> {
public Plottable a;
public Set<PlottableTuple> others;
public double distance;
#Override
public int compareTo(PlottableTuple other) {
return (new Double(distance)).compareTo(other.distance);
}
}
class Plottable {
public Point location;
private Set<Plottable> containingPlottables;
public Plottable(Set<Plottable> plots) {
this.containingPlottables = plots;
}
public Plottable() {
this.containingPlottables = new HashSet<>();
this.containingPlottables.add(this);
}
public Set<Plottable> getContainingPlottables() {
return containingPlottables;
}
}
}
Map all your circles on a 2D grid first. You then only need to compare the circles in a cell with the other circles in that cell and in it's 9 neighbors (you can reduce that to five by using a brick pattern instead of a regular grid).
If you only need to be really approximate, then you can just group all the circles that fall into a cell together. You will probably also want to merge cells that only have a small number of circles together with there neighbors, but this will be fast.
This problem is going to take a reasonable amount of computation no matter how you do it, the question then is: can you do all the computation up-front so that at run-time it's just doing a look-up? I would build a tree-like structure where each layer is all the points that need to be drawn for a given zoom level. It takes more computation up-front, but at run-time you are simply drawing a list of point, fast.
My idea is to decide what the resolution of each zoom level is (ie at zoom level 1 points closer than 15 get merged; at zoom level 2 points closer than 30 get merged), then go through your points making groups of points that are within the 15 of each other and pick a point to represent group that group at the higher zoom. Now you have a 2 layer tree. Then you pass over the second layer grouping all points that are within 30 of each other, and so on all the way up to your highest zoom level. Now save this tree structure to file, and at run-time you can very quickly change zoom levels by simply drawing all points at the appropriate tree level. If you need to add or remove points, that can be done dynamically by figuring out where to attach them to the tree.
There are two downsides to this method that come to mind: 1) it will take a long time to compute the tree, but you only have to do this once, and 2) you'll have to think really carefully about how you build the tree, based on how you want the groupings to be done at higher levels. For example, in the image below the top level may not be the right grouping that you want. Maybe instead building the tree based off the previous layer, you always want to go back to the original points. That said, some loss of precision always happens when you're trying to trade-off for faster run-time.
EDIT
So you have a problem which requires O(n^2) comparisons, you say it has to be done in real-time, can not be pre-computed, and has to be fast. Good luck with that.
Let's analyze the problem a bit; if you do no pre-computation then in order to decide which points can be merged you have to compare every pair of points, that's O(n^2) comparisons. I suggested building a tree before-hand, O(n^2 log n) once, but then runtime is just a lookup, O(1). You could also do something in between where you do some work before and some at run-time, but that's how these problems always go, you have to do a certain amount of computation, you can play games by doing some of it earlier, but at the end of the day you still have to do the computation.
For example, if you're willing to do some pre-computation, you could try keeping two copies of the list of points, one sorted by x-value and one sorted by y-value, then instead of comparing every pair of points, you can do 4 binary searches to find all the points within, say, a 30 unit box of the current point. More complicated so would be slower for a small number of points (say <100), but would reduce the overall complexity to O(n log n), making it faster for large amounts of data.
EDIT 2
If you're worried about multiple points at the same location, then why don't you do a first pass removing the redundant points, then you'll have a smaller "search list"
list searchList = new list()
for pt1 in points :
boolean clean = true
for pt2 in searchList :
if distance(pt1, pt2) < epsilon :
clean = false
break
if clean :
searchList.add(pt1)
// Now you have a smaller list to act on with only 1 point per cluster
// ... I guess this is actually the same as my first suggestion if you make one of these search lists per zoom level. huh.
EDIT 3: Graph Traversal
A totally new approach would be to build a graph out of the points and do some sort of longest-edge-first graph traversal on them. So pick a point, draw it, and traverse its longest edge, draw that point, etc. Repeat this until you come to a point which doesn't have any untraversed edges longer than your zoom resolution. The number of edges per point gives you an easy way to tradeoff speed for correctness. If the number of edges per point was small and constant, say 4, then with a bit of cleverness you could build the graph in O(n) time and also traverse it to draw points in O(n) time. Fast enough to do it on the fly with no pre-computation.
Just a wild guess and something that occurred to me while reading responses from others.
Do a multi-step comparison. Assume your combining distance at the current zoom level is 20 meters. First, subtract (X1 - X2). If This is bigger than 20 meters then you are done, the points are too far. Next, subtract (Y1 - Y2) and do the same thing to reject combining the points.
You could stop here and be happy if you are good with using only horizontal/vertical distances as your metric for combining. Much less math (no squaring or square roots). Pythagoras wouldn't be happy but your users might.
If you really insist on exact answers, do the two subtraction/comparison steps above. If the points are within horizontal and vertical limits, THEN you do the full Pythagoras check with square roots.
Assuming all your points are not highly clustered very close to the combining limit, this should save some CPU cycles.
This is still approximately an O(n^2) technique, but the math should be simpler. If you have the memory, you could store distances between each set of points and then you never have to compute it again. This could take up more memory than you have and also grows at a rate of approximately O(n^2), so be careful.
Also, you could make a linked list or sorted array of all your points, sorted in order of increasing X or increasing Y. (I don't think you need both, just one). Then walk through the list in sorted order. For each point, check the neighbors out until (X1 - X2) is bigger than your combining distance. and then stop. You don't have to compare each set of points for O(N^2), you only have to compare neighbors that are close in one dimension to quickly prune your large list to a small one. As you move through the list, you only have to compare points that have a bigger X than your current candidate, because you already compared and combined with all previous X values. This gets you closer to the O(n) complexity you want. Of course, you would need to check the Y dimension and fully qualify the points to be combined before you actually do it. Don't just use the X distance to make your combining decision.
I am not familiar with coordinate systems or much of the math dealing with these things at all. What I am trying to do is take a Point (x,y), and find its position in a 1 dimensional array such that it follows this:
(0,2)->0 (1,2)->1 (2,2)->2
(0,1)->4 (1,1)->5 (2,1)->6
(0,0)->8 (1,0)->9 (2,0)->10
where the arrows are showing what value the coordinates should map to. Notice that an index is skipped after each row. I'm think it'll end up being a fairly trivial solution, but I can't find any questions similar to this and I haven't had any luck coming up with ideas myself. I do know the width and height of the 2 dimensional array. Thank you for any help!
My question is perhaps ambiguous or using the wrong terminology, my apologies.
I know that the coordinate (0,0) will be the bottom left position. I also know that the top left coordinate should be placed at index 0. Each new row skips an index by 1. The size of the coordinate system varies, but I know the number of rows and number of columns.
First step, flip the values upside down, keep points in tact:
(0,2)->8 (1,2)->9 (2,2)->10
(0,1)->4 (1,1)->5 (2,1)->6
(0,0)->0 (1,0)->1 (2,0)->2
You'll notice that y affects the output by a factor of 4 and x by a factor of 1.
Thus we get a very simple 4y + x.
Now to get back to the original, you'll notice the transformation is (x,y) <- (x,2-y) (that is, if we transform each point above with this transformation, we get the original required mapping).
So, substituting it into the equation, we get (2-y)*4 + x.
Now this is specific to 3x3, but I'm sure you'll be able to generalize it by replacing 2 and 4 by variables.
If you want to reduce the dimension and avoid overlapping you need a space-filling-curve, for example a morton curve. Your example looks like a peano curve because it's a 3x3 matrix. These curves is difficult to calculate but have some nice things. But if you just look for self-avoiding curves you can create your own? Read here: http://www.fractalcurves.com/Root4Square.html.
I was beaten to the formula, here is the bruteforce using a Map.
public class MapPointToIndex {
private Map<Point, Integer> map;
private int index, rowcount;
public MapPointToIndex(int rows, int columns) {
map = new HashMap<Point, Integer>();
for (int i = rows - 1; i >= 0; i--) {
index += rowcount;
for (int j = 0; j < columns; j++) {
Point p = new Point(j, i);
map.put(p, index);
index++;
}
rowcount = 1;
}
}
public int getIndex(Point point){
return map.get(point);
}
public static void main(String[] args) {
MapPointToIndex one = new MapPointToIndex(3, 3);
System.out.println(one.map);
}
}
Out:
{java.awt.Point[x=0,y=0]=8, java.awt.Point[x=2,y=2]=2, java.awt.Point[x=1,y=2]=1, java.awt.Point[x=2,y=1]=6, java.awt.Point[x=1,y=1]=5, java.awt.Point[x=2,y=0]=10, java.awt.Point[x=0,y=2]=0, java.awt.Point[x=1,y=0]=9, java.awt.Point[x=0,y=1]=4}
I was wondering if I could get some advice on increasing the overall efficiency of a program that implements a genetic algorithm. Yes this is an assignment question, but I have already completed the assignment on my own and am simply looking for a way to get it to perform better
Problem Description
My program at the moment reads a given chain made of the types of constituents, h or p. (For example: hphpphhphpphphhpphph) For each H and P it generated a random move (Up, Down, Left, Right) and adds the move to an arrayList contained in the "Chromosome" Object. At the start the program is generating 19 moves for 10,000 Chromosomes
SecureRandom sec = new SecureRandom();
byte[] sbuf = sec.generateSeed(8);
ByteBuffer bb = ByteBuffer.wrap(sbuf);
Random numberGen = new Random(bb.getLong());
int numberMoves = chromosoneData.length();
moveList = new ArrayList(numberMoves);
for (int a = 0; a < numberMoves; a++) {
int randomMove = numberGen.nextInt(4);
char typeChro = chromosoneData.charAt(a);
if (randomMove == 0) {
moveList.add(Move.Down);
} else if (randomMove == 1) {
moveList.add(Move.Up);
} else if (randomMove == 2) {
moveList.add(Move.Left);
} else if (randomMove == 3) {
moveList.add(Move.Right);
}
}
After this comes the selection of chromosomes from the Population to crossover. My crossover function selections the first chromosome at random from the fittest 20% of the population and the other at random from outside of the top 20%. The chosen chromosomes are then crossed and a mutation function is called. I believe the area in which I am taking the biggest hit is calculating the fitness of each Chromosome. Currently my fitness function creates a 2d Array to act as a grid, places the moves in order from the move list generated by the function shown above, and then loops through the array to do the fitness calculation. (I.E. found and H at location [2,1] is Cord [1,1] [3,1] [2,0] or [2,2] also an H and if an H is found it just increments the count of bonds found)
After the calculation is complete the least fit chromosome is removed from my population and the new one is added and then the array list of chromosomes is sorted. Rinse and repeat until target solution is found
If you guys want to see more of my code to prove I actually did the work before asking for help just let me know (dont want to post to much so other students cant just copy pasta my stuff)
As suggested in the comments I have ran the profiler on my application (have never used it before, only a first year CS student) and my initial guess on where i am having issues was somewhat incorrect. It seems from what the profiler is telling me is that the big hotspots are:
When comparing the new chromosome to the others in the population to determine its position. I am doing this by implementing Comparable:
public int compareTo(Chromosome other) {
if(this.fitness >= other.fitness)
return 1;
if(this.fitness ==other.fitness )
return 0;
else
return -1;
}
The other area of issue described is in my actual evolution function, consuming about 40% of the CPU time. A codesample from said method below
double topPercentile = highestValue;
topPercentile = topPercentile * .20;
topPercentile = Math.ceil(topPercentile);
randomOne = numberGen.nextInt((int) topPercentile);
//Lower Bount for random two so it comes from outside of top 20%
int randomTwo = numberGen.nextInt(highestValue - (int) topPercentile);
randomTwo = randomTwo + 25;
//System.out.println("Selecting First: " + randomOne + " Selecting Second: " + randomTwo);
Chromosome firstChrom = (Chromosome) populationList.get(randomOne);
Chromosome secondChrom = (Chromosome) populationList.get(randomTwo);
//System.out.println("Selected 2 Chromosones Crossing Over");
Chromosome resultantChromosome = firstChrom.crossOver(secondChrom);
populationList.add(resultantChromosome);
Collections.sort(populationList);
populationList.remove(highestValue);
Chromosome bestResult = (Chromosome) populationList.get(0);
The other main preformance hit is the inital population seeding which is performed by the first code sample in the post
I believe the area in which I am taking the biggest hit is calculating the fitness of each Chromosome
If you are not sure then I assume you have not run a profiler on the program yet.
If you want to improve the performance, profiling is the first thing you should do.
Instead of repeatedly sorting your population, use a collection that maintains its contents already sorted. (e.g. TreeSet)
If your fitness measure is consistent across generations (i.e. not dependent on other members of the population) then I hope at least that you are storing that in the Chromosome object so you only calculate it once for each member of the population. With that in place you'd only be calculating fitness on the newly generated/assembled chromosome each iteration. Without more information on how fitness if calculated it's difficult to be able to offer any optimisations in that area.
Your random number generator seed doesn't need to be cryptographically strong.
Random numberGen = new Random();
A minor speedup when seeding your population is to remove all the testing and branching:
static Move[] moves = {Move.Down, Move.Up, Move.Left, Move.Right};
...
moveList.add(moves[randomMove]);