I need to solve an assignment problem with Google Apps Script. The LinearOptimizationService seemed to be promising so what I tried is to port a corresponding example from the google or-tools .
Unfortunately, I couldn't find out a way how I set up the constraint that each worker is assigned to exactly one task.
The working code (without the constraint) is as follows:
function myFunction() {
// Create the variables
var engine = LinearOptimizationService.createEngine();
const costMatrix = [[7,5,4,8,3]
,[2,5,6,7,1]
,[2,8,9,5,2]
,[2,5,6,7,1]
,[2,8,1,5,9]]
const capacities = [1,1,1,1,1]
// initialize the decision variables
x = new Array(costMatrix.length);
for (let i=0; i<costMatrix.length; i++){
x[i] = new Array(costMatrix[0].length);
for (let j=0; j<costMatrix[0].length; j++){
x[i][j] = engine.addVariable(`x${i}${j}`, 0, 1, LinearOptimizationService.VariableType.INTEGER);
}
}
// Each worker is assigned to at most one task.
for (let i = 0; i < costMatrix.length; ++i) {
constraint = engine.addConstraint(0, 1);
for (let j = 0; j < costMatrix[0].length; ++j) {
constraint.setCoefficient(`x${i}${j}`, 1);
}
}
// Each task is assigned to exactly one worker.
for (let j = 0; j < costMatrix[0].length; ++j) {
constraint = engine.addConstraint(1, 1);
for (let i = 0; i < costMatrix.length; ++i) {
constraint.setCoefficient(`x${i}${j}`, 1);
}
}
for (let i = 0; i < costMatrix.length; ++i) {
for (let j = 0; j < costMatrix[0].length; ++j) {
engine.setObjectiveCoefficient(`x${i}${j}`, costMatrix[i][j]);
}
}
engine.setMinimization()
// Solve the linear program
var solution = engine.solve(30);
if (!solution.isValid()) {
throw 'No solution ' + solution.getStatus();
}
Logger.log('ObjectiveValue: ' + solution.getObjectiveValue());
for (let i=0; i<costMatrix.length; i++){
for (let j=0; j<costMatrix[0].length; j++){
Logger.log(`Value of: x${i}${j} ` + solution.getVariableValue(`x${i}${j}`));
}
}
}
Any help is appreciated.
Thanks
Related
The assignment is the spawn several light and dark feathers according to score points from a quiz. The light feathers symbolize the correct points (light_feather), and the dark feather are the incorrect points (dark_feather) (Each are being tracked). All the feathers are supposed to line up on one line, meaning first light feathers, followed by the dark feathers. I got the quiz dynamics figured out, and the function I have posted here is only for when they press end quiz.
var light_feather:LightFeather = new LightFeather();
var dark_feather:DarkFeather = new DarkFeather();
var good_answers:uint = 0;
var bad_answers:uint = 0;
function avsluttFunc (evt:MouseEvent)
{
var sum_LightFeatherX:Number = 0;
for (var i = 0; i < good_answers; i++) {
addChild(light_feather);
light_feather.x += 12 + (i*16);
light_feather.y = 0;
trace("Lys X-verdi: " + light_feather.x);
sum_LightFeatherX += Number(light_feather.x);
return sum_LightFeatherX;
}
trace(sum_LightFeatherX);
dark_feather.x += sum_LightFeatherX;
for (var j = 1; j <= bad_answers; j++) {
addChild(dark_feather);
dark_feather.x += 12 + (j*16);
dark_feather.y = 0;
trace("Mørk X-verdi: " + dark_feather.x);
}
/*
//Resetter poengsummen
good_answers = 0;
bad_answers = 0;
*/
}
You can do what you are looking for using only one for loop, take a look :
var good_answers:uint = 2;
var bad_answers:uint = 4;
function avsluttFunc(evt:MouseEvent)
{
for (var i:int = 0; i < good_answers + bad_answers; i++) {
var feather:DisplayObject = i < good_answers ? new LightFeather() : new DarkFeather();
feather.x += 12 + i * (feather.width + 1);
feather.y = 0;
addChild(feather);
}
}
This code example will create 4 DarkFeather instances next to 2 LightFeather ones.
Edit :
How to add your objects to an array ?
// feathers array should be accessible for both codes (adding and removing objects)
var feathers:Array = [];
for (var i:int = 0; i < good_answers + bad_answers; i++) {
var feather:DisplayObject = i < good_answers ? new LightFeather() : new DarkFeather();
addChild(feather);
feathers.push(feather);
}
then to remove them from the stage, you can do for example :
for (var i:int = 0; i < feathers.length; i++) {
var feather:DisplayObject = DisplayObject(feathers[i]);
feather.parent.removeChild(feather);
}
Hope that can help.
I want to check for a straight combination in a poker game.
So, I have this array: var tempArr:Array = new Array;
I have this for sorting the array:
for (i = 0; i < 7; i++)
{
tempArr[i] = pValue[i];
}
tempArr.sort( Array.NUMERIC );
pValue is the value of the cards, it's have range from 2 to 14.
So, if I have this Array: tempArray = [2,3,3,4,5,5,6];
How can I check if I have a straight combination in my hand?
Set a bucket array to save if you got a card in hand
var t:Array = [];
//t[2] = 1;mean you have 2
// t[3] = 0;mean you don't have 3
//first initialize t
for(var i:int = 0; i < 15; i++)
{
t[i] = 0;
}
//then set the values from tempArray
for (var j:int = 0; j < tempArray.length; j++)
{
t[tempArray[j]] = 1;
}
//if you have a straight combination in your hand
//you will get a k that t[k] & t[k-1]& t[k-2] & t[k-3] & t[k-4] == 1
var existed:boolean = false;//the flag save if you got a straight combination
for (var k:int = t.length - 1; k >= 4; k--)
{
if (t[k] == 0)
{
continue;
}
if ((t[k] & t[k-1] & t[k-2] & t[k-3] & t[k-4]) == 1)
{
existed = true;
break;
}
}
I have the following scenario:
if (event.status == AMFResultEvent.SUCCESS) {
var lev1:uint = 0;
var lev2:uint = 0;
var lev3:uint = 0;
var lev4:uint = 0;
var lev5:uint = 0;
var lev6:uint = 0;
for (var i:int = 0; i < event.result.length; i++) {
if (mainLevel == "1") {
lev1++;
}
if (mainLevel == "2") {
lev2++;
}
if (mainLevel == "3") {
lev3++;
}
if (mainLevel == "4") {
lev4++;
}
if (mainLevel == "5") {
lev5++;
}
if (mainLevel == "6") {
lev6++;
}
}
for (var j:int = 1; j < 7; j++) {
_row = new StatisticsRow(event.result[j], this);
_rowsPlace.addChild(_row);
_row.y = (_row.height +1) * j;
_row.codeLevel.htmlText = j; // works as it should
// need to access variables lev1 - lev6, called by something like "lev"+j here:
_row.amount.htmlText =
}
// traces correct amounts of mainLevels from the i loop:
trace ("level 1: " + lev1);
trace ("level 2: " + lev2);
trace ("level 3: " + lev3);
trace ("level 4: " + lev4);
trace ("level 5: " + lev5);
trace ("level 6: " + lev6);
}
I'm missing something obvious here, as the ["lev"]+j doen't work. How can I dynamically acces the lev1 - lev6 in the j-loop? As the code comment at the bottoms shows, this traces as expected.
Thanks in advance!
You can access them with brackets, string concatenation, and the this keyword. Here's an example of how you would use bracket notation in a loop:
for (var i:int = 0; i <= 6; i++) {
var currLev = this["lev"+i];
// do stuff to currLev
}
Thanks for answering guys!
I had a lousy approach to my problem anyway, and should have used an array right away:
var mainLevels:Array = new Array();
for (var n:int = 1; n < 7; n++) {
mainLevels[n] = 0;
}
if (event.status == AMFResultEvent.SUCCESS) {
for (var i:int = 0; i < event.result.length; i++) {
var data = event.result[i];
var correctCode:String = data["correct"];
var mainLevelFound:uint = uint(correctCode.substr(0, 1));
for (var k:int = 1; k < 7; k++) {
if (k == mainLevelFound) {
mainLevels[k]++;
}
}
}
for (var j:int = 1; j < 7; j++) {
_row = new StatisticsRow(event.result[j], this);
_rowsPlace.addChild(_row);
_row.y = (_row.height +1) * j;
_row.codeLevel.htmlText = j;
// Now this works as a reference to mainLevels[*] created above!
_row.amount.htmlText = mainLevels[j];
}
Thanks again for your effort :)
i have code
dict_a[box_1_a] = text_1_a;
dict_a[box_2_a] = text_2_a;
dict_a[box_3_a] = text_3_a;
dict_a[box_4_a] = text_4_a;
dict_a[box_5_a] = text_5_a;
dict_a[box_6_a] = text_6_a;
how to summarize the code looks like this
for (var i:int = 1; i <= 6; i++)
{
dict_a[box_(i)_a] = text_(i)_a;
}
Thanks before
If boxes and texts are class members, correct syntax should be
for (var i:int = 1; i <= 6; i++)
{
dict_a[this["box_" + i + "_a"]] = this["text_" + i + "_a"];
}
To do this, you should first save your box_1_a and text objects in an array, so you can iterate over them via:
for(int i=0; i<=5; i++) {
dict_a[box_a[i]] = text_a[i];
}
I don't know if the syntax is correct for AS/Flash, but that's how it will basically work.
I am adding two matrices (or possibly many) in ActionScript 3.0.
Now my problem is how can I add indexes in array that is something like this?
array1[1,2,3,4] + array2[2,4,5,6] = answer[3,6,8,10]
This function adds up all the arrays that are passed to it:
function sumOfArrays(...args):Array
{
var sum:Array = [];
var arrays:Array = [];
var longestArrayLength:uint = 0;
for (var i:int = 0, n:int = args.length; i < n; i++)
{
if (args[i] is Array)
{
arrays.push(args[i]);
longestArrayLength = args[i].length > longestArrayLength ? args[i].length : longestArrayLength;
}
}
for (var j:int = 0; j < longestArrayLength; j++)
{
sum[j] = 0;
for (i = 0; i < n; i++)
{
sum[j] += isNaN(arrays[i][j]) ? 0 : arrays[i][j];
}
}
return sum;
}
It can be used like this:
var sum:Array = sumOfArrays(array1, array2);
That's not possible. Arrays only allows access via one index. You'd have to write a method on your own for this. But be aware of the fact, that null would be referenced on answer at, 0, 1, 2, 4, 5 and so on.