Recently, I got to tackle an old school problem. Only I didn't know it was old school at the time when I started. I suppose that's because I've not been doing hardcore Computer Science type problems in quite some time.

Here's the problem: In TrophyManager, you field a soccer team of 11 players, 10 field players and a keeper. TrExMa calculates a skill for each player on your roster to try to give the user an idea of what the best possible lineup would be. Selecting the keeper is easy. Take the guy with the highest value. Selecting the field players, not so much.

Each player has 14 "visible" attributes. Early players of the game used the instruction manual to determine which attribute is most important for the position the player is in. For example, a Forward must be good at Finishing and Heading, etc. Additionally, each player has a favorite position. When they play "away" from their position, their skill drops. So a player may have a high finishing skill, but since they like to play left back, they're not going to be a very good striker and would have their skill reduced anywhere from 4-40% depending upon a hidden adaptability attribute.

Anyway, how does one figure out the best team? I can set the adaptablity factor from 1-10% and we can then caclulate the skill for every position on the field for every player on the team. Sounds good, right? But how to pick it?

I first looked at the original TrExMa algorithm within Excel. The original app had a drop down to determine the depth of the search. Once I looked at the code it was confirmed. TrExMa used a series of nested loops to build the lineup. If I'm not mistaken it could run from O(n^5) to O(n^10) depending upon the complexity you chose. Wow, no wonder it got very slow. I needed something else. I also realized that the only way this algorithm would guaruntee success would be to run it at O(n^10) setting. Using Javascript, this was simply unacceptable. I couldn't get away with something running for so long.

My first instinct was to use an algorithm similar to Fit First. In this case, find the highest value on the board, and place the player there. cross him out and cross out the position. Do it again until you've filled the positions. Supposedly this runs in O(n log n). Of course, 10 is pretty small for me, as I've only 10 boxes to fill. It's pretty good, but doesn't guaruntee the best match of players to positions. Why? Let's say you've got a player who is adept at a midfield and wing position. And another player who is only good at midfield. If the former is slightly better at midfield, he'll be placed there, and the other player will have to play somewhere else where he won't be as successful.

How do you correct this? I soon realized that this had to be a common problem, I just couldn't figure out what to call it? Eventually I discovered that it's known as an "assignment problem". It's an obvious name, now that I know what it is. The classical use of this problem is to determine the least overall cost of doing multiple things with multiple resources. It's very similar.

Soon I discovered the Hungarian Algorithm. It runs in O(n^4) (potentially O(n^3)). So it's not fast, but it's not insanely slow. It's not exponential, and it runs faster than the Excel app.

So what does the Hungarian Algorithm do? It uses a method of starring and priming to determine when a player is in its ideal place. Essentially, if two players are "best" at a position, it will not accept that assignment until it validates the best organization of those players. It's explained very well in the Wikipedia page.

There are many implementations of the Hungarian Algorithm on the net but none in Javascript. So here's one. I've pulled out some of the specifics for my implementation. You'll want to implement your own loadMatrix and getSolution methods. One thing to note, is that with my problem, we had to reverse the matrix. This is required because the Hungarian Algorithm uses small values to determine the best player (worker) for a position (job). I hope you find this useful.

/* Implementation of the Hungarian Algorithm to determine

* "best" squad. This is a "reverse" implementation.

* References:

* http://en.wikipedia.org/wiki/Hungarian_algorithm

* http://www.ams.jhu.edu/~castello/362/Handouts/hungarian.pdf (Example #2)

* http://www.public.iastate.edu/~ddoty/HungarianAlgorithm.html // Non-square

*/

var Hungarian = window.Hungarian = window.HG = {

/* 2 dimension arrays */

skillMatrix: null,

matrix: null,

stars: null,

/* Single arrays */

rCov: [],

cCov: [],

rows: 0,

cols: 0,

dim: 0,

solutions: 0, // "k"

FORBIDDEN_VALUE: -999999,

/* Rows MUST BE the Formation (Jobs)

* Columns MUST BE the Squad (Workers)

* Therefore, the Rows MUST BE PADDED

*/

hungarianAlgortithm: function(formation, squad) {

HG.init(formation, squad);

// Step 1

HG.matrix = HG.subtractRowMins(HG.matrix);

// Step 2

HG.findZeros(HG.matrix);

var done = false;

while (!done) {

// Step 3

var covCols = HG.coverColumns(HG.matrix);

if (covCols > HG.solutions -1) {

done = true;

}

if (!done) {

// Step 4 (calls Step 5)

done = HG.coverZeros(HG.matrix);

while (done) {

// Step 6

var smallest = HG.findSmallestUncoveredVal(HG.matrix);

HG.matrix = HG.uncoverSmallest(smallest, HG.matrix);

done = HG.coverZeros(HG.matrix);

}

}

}

return HG.getSolution(formation, squad)

},

init: function(formation, squad) {

HG.cols = squad.players.length;

HG.rows = formation.length;

HG.dim = Math.max(HG.rows, HG.cols);

HG.solutions = HG.dim;

HG.skillMatrix = HG.initMatrix(HG.rows, HG.cols);

HG.matrix = HG.initMatrix(HG.dim, HG.dim);

HG.stars = HG.initMatrix(HG.dim, HG.dim);

HG.matrix = HG.loadMatrix(squad, formation, HG.matrix, true);

HG.skillMatrix = HG.loadMatrix(squad, formation, HG.skillMatrix, false);

HG.rCov = new Array(HG.dim);

HG.cCov = new Array(HG.dim);

HG.initArray(HG.cCov, 0); // Zero it out

HG.initArray(HG.rCov, 0);

},

initMatrix: function(sizeX, sizeY) {

var matrix = new Array(sizeX);

for (var i=0; i<sizeX; i++) {

matrix[i] = new Array(sizeY);

HG.initArray(matrix[i], 0);

}

return matrix;

},

// Takes an array of positions as a formation.

// Takes a squad which contains an array of players

loadMatrix: function(squad, formation, matrix, reverse) {

matrix = loadYourMatrix(squad, formation, matrix); // I've removed my implementation here. Far too much stuff

if (reverse) {

// This reverses the matrix. We need to to create a cost based solution.

matrix = HG.reverseMatrix(HG.findMaxValue(matrix), matrix);

}

return matrix;

},

findMaxValue: function(matrix) {

var max = 0.0;

for (var i = 0; i < matrix.length; i ++) {

for (var j = 0; j < matrix[i].length; j++) {

if (matrix[i][j] > max) {

max = matrix[i][j];

}

}

}

return Number(max);

},

reverseMatrix: function(max, matrix) {

for (var i = 0; i < matrix.length; i ++) {

for (var j = 0; j < matrix[i].length; j++) {

matrix[i][j] = (Number(max) - Number(matrix[i][j])).toFixed(0);

}

}

return matrix;

},

subtractRowMins: function(matrix) {

for (var i = 0; i < matrix.length; i ++) {

var min = Number.MAX_VALUE;

for (var j = 0; j < matrix[i].length; j++) {

if (matrix[i][j] < min) {

min = Number(matrix[i][j]);

}

}

for (var k = 0; k < matrix[i].length; k++) {

matrix[i][k] = matrix[i][k] - min;

}

}

return matrix;

},

subtractColMins: function(matrix) {

for (var j = 0; j < matrix[0].length; j ++) {

var min = Number.MAX_VALUE;

for (var i = 0; i < matrix.length; i++) {

if (matrix[i][j] < min) {

min = Number(matrix[i][j]);

}

}

for (var k = 0; k < matrix[0].length; k++) {

matrix[k][j] = matrix[k][j] - min;

}

}

return matrix;

},

findZeros: function(matrix) {

for (var i = 0; i < matrix.length; i++) {

for (var j = 0; j < matrix[i].length; j++) {

if (matrix[i][j] == 0) {

if (HG.rCov[i] == 0 && HG.cCov[j] == 0) {

HG.stars[i][j] = 1;

HG.cCov[j] = 1;

HG.rCov[i] = 1;

}

}

}

}

// Clear Covers

HG.initArray(HG.cCov,0);

HG.initArray(HG.rCov,0);

},

initArray: function(theArray, initVal) {

for (var i = 0; i < theArray.length; i++) {

theArray[i] = Number(initVal);

}

},

coverColumns: function(matrix) {

var count = 0;

for (var i=0; i < matrix.length; i++) {

for (var j=0; j < matrix[i].length; j++) {

if (HG.stars[i][j] == 1) {

HG.cCov[j] = 1;

}

}

}

for (var k=0; k < HG.cCov.length; k++) {

count = Number(HG.cCov[k]) + Number(count);

}

return count;

},

/**

* step 4

* Cover all the uncovered zeros one by one until no more

* cover the row and uncover the column

*/

coverZeros: function(matrix) {

var retVal = true;

var zero = HG.findUncoveredZero(matrix); // Returns a Coords object..

while (zero.row > -1 && retVal == true) {

HG.stars[zero.row][zero.col] = 2 //Prime it

var starCol = HG.foundStarInRow(zero.row, matrix);

if (starCol > -1) {

HG.rCov[zero.row] = 1;

HG.cCov[starCol] = 0;

} else {

HG.starZeroInRow(zero); // Step 5

retVal = false;

}

if (retVal == true) {

zero = HG.findUncoveredZero(matrix);

}

}

return retVal;

},

findUncoveredZero: function(matrix) {

var coords = new HgCoords();

for (var i=0; i< matrix.length; i++) {

for (var j=0; j < matrix[i].length; j++) {

if (matrix[i][j] == 0 && HG.rCov[i] == 0 && HG.cCov[j] == 0) {

coords.row = i;

coords.col = j;

j = matrix[i].length;

i = matrix.length - 1;

}

}

}

return coords;

},

foundStarInRow: function(zeroRow, matrix) {

var retVal = -1;

for (var j = 0; j < matrix[zeroRow].length; j++) {

if (HG.stars[zeroRow][j] == 1) {

retVal = j;

j = matrix[zeroRow].length;

}

}

return retVal;

},

/**

* step 5

* augmenting path algorithm

* go back to step 3

*/

starZeroInRow: function(zero) { // Takes a Coords Object

TrU.log("Step 5: Uncovered Zero:" + zero.row + "," + zero.col, TrU.DEBUG );

var done = false;

var count = 0;

var path = HG.initMatrix(HG.dim*2, 2);

path[count][0] = zero.row;

path[count][1] = zero.col;

while (!done) {

var row = HG.findStarInCol(path[count][1]);

if (row > -1) {

count++;

path[count][0] = row;

path[count][1] = path[count - 1][1];

} else {

done = true;

}

if (!done) {

var col = HG.findPrimeInRow(path[count][0]);

count++;

path[count][0] = path[count - 1][0];

path[count][1] = col;

}

}

HG.convertPath(path, count);

// Clear Covers

HG.initArray(HG.cCov,0);

HG.initArray(HG.rCov,0);

HG.erasePrimes();

},

findStarInCol: function(col) {

var retVal = -1;

for (var i = 0; i < HG.stars.length; i++) {

if (HG.stars[i][col] == 1) {

retVal = i;

i = HG.stars.length;

}

}

return retVal;

},

findPrimeInRow: function(row) {

var retVal = -1;

for (var j=0; j< HG.stars[row].length; j++) {

if (HG.stars[row][j] == 2) {

retVal = j;

j = HG.stars[row].length;

}

}

return retVal;

},

/* Should convert all primes to stars and reset all stars.

* Count is needed to be sure we look at all items in the path

*/

convertPath: function(path, count) {

HG.logMatrix(path, "Step 5: Converting Path. Count = " + count);

for (var i=0; i < count+1; i++) {

var x = path[i][0];

var y = path[i][1];

if (HG.stars[x][y] == 1) {

HG.stars[x][y] = 0;

} else if (HG.stars[x][y] == 2) {

HG.stars[x][y] = 1;

}

}

},

erasePrimes: function() {

for (var i=0; i<HG.stars.length; i++) {

for (var j=0; j < HG.stars[i].length; j++){

if (HG.stars[i][j] == 2) {

HG.stars[i][j] = 0;

}

}

}

},

findSmallestUncoveredVal: function(matrix) {

var min = Number.MAX_VALUE;

for (var i = 0; i < matrix.length; i++) {

for (var j = 0; j < matrix[i].length; j++) {

if (HG.rCov[i] == 0 && HG.cCov[j] == 0) {

if (min > matrix[i][j]) {

min = matrix[i][j];

}

}

}

}

return min;

},

/**

* step 6

* modify the matrix

* if the row is covered, add the smallest value

* if the column is not covered, subtract the smallest value

*/

uncoverSmallest: function(smallest, matrix) {

TrU.log("Uncover Smallest: "+ smallest);

HG.logMatrix(matrix, "B4 Smallest uncovered");

for (var i = 0; i < matrix.length; i++) {

for (var j = 0; j < matrix[i].length; j++) {

if (HG.rCov[i] == 1) {

matrix[i][j] += smallest;

}

if (HG.cCov[j] == 0) {

matrix[i][j] -= smallest;

}

}

}

HG.logMatrix(matrix, "Smallest uncovered");

return matrix;

},

getSolution: function(formation, squad) {

var total = 0;

var lineup = [];

// Changed from length of stars, since we must ignore some rows due to padding.

for (var i = 0; i < HG.rows; i++) {

for (var j = 0; j < HG.cols; j++) {

if (HG.stars[i][j] == 1) {

/* the player (worker) at index j is the best player

* for poisition (job) at index i in your initial arrays.

*/

lineup.push(getThePlayer(i,j));

}

}

}

return lineup;

}

}

function HgCoords() {

this.row = -1;

this.col = -1;

}

## 3 comments:

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Great Job. But a bit silly.

Great stuff, thanks!

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