Matching Algorithm In Javascript
I'm looking for an implementation of the Blossom algorithm for JavaScript or something similar. I have a set of pairs A - B A - C B - D and I need to pick pairs, assuming that each
Solution 1:
Sure, why not?
/*
Edmonds's maximum matching algorithm
Complexity: O(v^3)
Written by Felipe Lopes de Freitas
Adapted to JavaScript from C++ (http://pastebin.com/NQwxv32y) by גלעד ברקן
*/var MAX = 100,
undef = -2,
empty = -1,
noEdge = 0,
unmatched = 1,
matched = 2,
forward = 0,
reverse = 0;
//Labels are the key to this implementation of the algorithm.functionLabel(){ //An even label in a vertex means there's an alternating path
this.even = undefined; //of even length starting from the root node that ends on the
this.odd = newArray(2); //vertex. To find this path, the backtrace() function is called,
}; //constructing the path by following the content of the labels.//Odd labels are similar, the only difference is that base nodes//of blossoms inside other blossoms may have two. More on this later.functionelem(){ //This is the element of the queue of labels to be analyzed by
this.vertex = undefined;
this.type = undefined; //the augmentMatching() procedure. Each element contains the vertex
}; //where the label is and its type, odd or even.var g = newArray(MAX); //The graph, as an adjacency matrix.for (var i=0; i<MAX; i++){
g[i] = newArray(MAX);
}
//blossom[i] contains the base node of the blossom the vertex ivar blossom = newArray(MAX); //is in. This, together with labels eliminates the need to//contract the graph.//The path arrays are where the backtrace() routine willvar path = newArray(2);
for (var i=0; i<2; i++){
path[i] = newArray(MAX);
}
var endPath = newArray(2); //store the paths it finds. Only two paths need to be//stored. endPath[p] denotes the end of path[p].varmatch = newArray(MAX); //An array of flags. match[i] stores if vertex i is in the matching.//label[i] contains the label assigned to vertex i. It may be undefined,var label = newArray(MAX); //empty (meaning the node is a root) and a node might have even and odd//labels at the same time, which is the case for nonbase nodes of blossomsfor (var i=0; i<MAX; i++){
label[i] = new Label();
}
var queue = newArray(2*MAX); //The queue is necessary for efficiently scanning all labels.var queueFront,queueBack; //A label is enqueued when assigned and dequeued after scanned.for (var i=0; i<2*MAX; i++){
queue[i] = new elem();
}
functioninitGraph(n){
for (var i=0; i<n; i++)
for (var j=0; j<n; j++) g[i][j]=noEdge;
}
functionreadGraph(){
var n = graph.n,
e = graph.e;
//int n,e,a,b;//scanf(" %d %d",&n,&e); //The graph is read and its edges are unmatched by default.
initGraph(n); //Since C++ arrays are 0..n-1 and input 1..n , subtractions for (var i=0; i<e; i++){ //are made for better memory usage.//scanf(" %d %d",&a,&b);var a = graph[i][0],
b = graph[i][1];
if (a!=b)
g[a-1][b-1]=g[b-1][a-1]=unmatched;
}
return n;
}
functioninitAlg(n){ //Initializes all data structures for the augmentMatching()
queueFront=queueBack=0; //function begin. At the start, all labels are undefined,for (var i=0; i<n; i++){ //the queue is empty and a node alone is its own blossom.
blossom[i]=i;
label[i].even=label[i].odd[0]=label[i].odd[1]=undef;
}
}
functionbacktrace (vert, pathNum, stop, parity, direction){
if (vert==stop) return; //pathNum is the number of the path to storeelseif (parity==0){ //vert and parity determine the label to be read.if (direction==reverse){
backtrace(label[vert].even,pathNum,stop,(parity+1)%2,reverse);
path[pathNum][endPath[pathNum]++]=vert;
} //forward means the vertices called first enterelseif (direction==forward){ //the path first, reverse is the opposite.
path[pathNum][endPath[pathNum]++]=vert;
backtrace(label[vert].even,pathNum,stop,(parity+1)%2,forward);
}
}
/*
stop is the stopping condition for the recursion.
Recursion is necessary because of the possible dual odd labels.
having empty at stop means the recursion will only stop after
the whole tree has been climbed. If assigned to a vertex, it'll stop
once it's reached.
*/elseif (parity==1 && label[vert].odd[1]==undef){
if (direction==reverse){
backtrace(label[vert].odd[0],pathNum,stop,(parity+1)%2,reverse);
path[pathNum][endPath[pathNum]++]=vert;
}
elseif (direction==forward){
path[pathNum][endPath[pathNum]++]=vert;
backtrace(label[vert].odd[0],pathNum,stop,(parity+1)%2,forward);
}
}
/*
Dual odd labels are interpreted as follows:
There exists an odd length alternating path starting from the root to this
vertex. To find this path, backtrace from odd[0] to the top of the tree and
from odd[1] to the vertex itself. This, put in the right order, will
constitute said path.
*/elseif (parity==1 && label[vert].odd[1]!=undef){
if (direction==reverse){
backtrace(label[vert].odd[0],pathNum,empty,(parity+1)%2,reverse);
backtrace(label[vert].odd[1],pathNum,vert,(parity+1)%2,forward);
path[pathNum][endPath[pathNum]++]=vert;
}
elseif (direction==forward){
backtrace(label[vert].odd[1],pathNum,vert,(parity+1)%2,reverse);
backtrace(label[vert].odd[0],pathNum,empty,(parity+1)%2,forward);
path[pathNum][endPath[pathNum]++]=vert;
}
}
}
functionenqueue (vert, t){
var tmp = new elem(); //Enqueues labels for scanning.
tmp.vertex=vert; //No label that's dequeued during the execution
tmp.type=t; //of augmentMatching() goes back to the queue.
queue[queueBack++]=tmp; //Thus, circular arrays are unnecessary.
}
functionnewBlossom (a, b){ //newBlossom() will be called after the paths are evaluated.var i,base,innerBlossom,innerBase;
for (i=0; path[0][i]==path[1][i]; i++); //Find the lowest common ancestor of a and b
i--; //it will be used to represent the blossom.
base=blossom[path[0][i]]; //Unless it's already contained in another...//In this case, all will be put in the older one.for (var j=i; j<endPath[0]; j++) blossom[path[0][j]]=base;
for (var j=i+1; j<endPath[1]; j++) blossom[path[1][j]]=base; //Set all nodes to thisfor (var p=0; p<2; p++){ //new blossom.for (var j=i+1; j<endPath[p]-1; j++){
if (label[path[p][j]].even==undef){ //Now, new labels will be applied
label[path[p][j]].even=path[p][j+1]; //to indicate the existence of even
enqueue(path[p][j],0); //and odd length paths.
}
elseif (label[path[p][j]].odd[0]==undef && label[path[p][j+1]].even==undef){
label[path[p][j]].odd[0]=path[p][j+1];
enqueue(path[p][j],1); //Labels will only be put if the vertex
} //doesn't have one.elseif (label[path[p][j]].odd[0]==undef && label[path[p][j+1]].even!=undef){
/*
If a vertex doesn't have an odd label, but the next one in the path
has an even label, it means that the current vertex is the base node
of a previous blossom and the next one is contained within it.
The standard labeling procedure will fail in this case. This is fixed
by going to the last node in the path inside this inner blossom and using
it to apply the dual label.
Refer to backtrace() to know how the path will be built.
*/
innerBlossom=blossom[path[p][j]];
innerBase=j;
for (; blossom[j]==innerBlossom && j<endPath[p]-1; j++);
j--;
label[path[p][innerBase]].odd[0]=path[p][j+1];
label[path[p][innerBase]].odd[1]=path[p][j];
enqueue(path[p][innerBase],1);
}
}
}
if (g[a][b]==unmatched){ //All nodes have received labels, exceptif (label[a].odd[0]==undef){ //the ones that called the function in
label[a].odd[0]=b; //the first place. It's possible to
enqueue(a,1); //find out how to label them by
} //analyzing if they're in the matching.if (label[b].odd[0]==undef){
label[b].odd[0]=a;
enqueue(b,1);
}
}
elseif (g[a][b]==matched){
if (label[a].even==undef){
label[a].even=b;
enqueue(a,0);
}
if (label[b].even==undef){
label[b].even=a;
enqueue(b,0);
}
}
}
functionaugmentPath (){ //An augmenting path has been found in the matchingvar a,b; //and is contained in the path arrays.for (var p=0; p<2; p++){
for (var i=0; i<endPath[p]-1; i++){
a=path[p][i]; //Because of labeling, this path is already
b=path[p][i+1]; //lifted and can be augmented by simpleif (g[a][b]==unmatched) //changing of the matching status.
g[a][b]=g[b][a]=matched;
elseif (g[a][b]==matched)
g[a][b]=g[b][a]=unmatched;
}
}
a=path[0][endPath[0]-1];
b=path[1][endPath[1]-1];
if (g[a][b]==unmatched) g[a][b]=g[b][a]=matched;
elseif (g[a][b]==matched) g[a][b]=g[b][a]=unmatched;
//After this, a and b are included in the matching.match[path[0][0]]=match[path[1][0]]=true;
}
functionaugmentMatching (n){ //The main analyzing function, with thevar node,nodeLabel; //goal of finding augmenting paths or
initAlg(n); //concluding that the matching is maximum.for (var i=0; i<n; i++) if (!match[i]){
label[i].even=empty;
enqueue(i,0); //Initialize the queue with the exposed vertices,
} //making them the roots in the forest.while (queueFront<queueBack){
node=queue[queueFront].vertex;
nodeLabel=queue[queueFront].type;
if (nodeLabel==0){
for (var i=0; i<n; i++) if (g[node][i]==unmatched){
if (blossom[node]==blossom[i]);
//Do nothing. Edges inside the same blossom have no meaning.elseif (label[i].even!=undef){
/*
The tree has reached a vertex with a label.
The parity of this label indicates that an odd length
alternating path has been found. If this path is between
roots, we have an augmenting path, else there's an
alternating cycle, a blossom.
*/
endPath[0]=endPath[1]=0;
backtrace(node,0,empty,0,reverse);
backtrace(i,1,empty,0,reverse);
//Call the backtracing function to find out.if (path[0][0]==path[1][0]) newBlossom(node,i);
/*
If the same root node is reached, a blossom was found.
Start the labelling procedure to create pseudo-contraction.
*/else {
augmentPath();
returntrue;
/*
If the roots are different, we have an augmenting path.
Improve the matching by augmenting this path.
Now some labels might make no sense, stop the function,
returning that it was successful in improving.
*/
}
}
elseif (label[i].even==undef && label[i].odd[0]==undef){
//If an unseen vertex is found, report the existing path//by labeling it accordingly.
label[i].odd[0]=node;
enqueue(i,1);
}
}
}
elseif (nodeLabel==1){ //Similar to above.for (var i=0; i<n; i++) if (g[node][i]==matched){
if (blossom[node]==blossom[i]);
elseif (label[i].odd[0]!=undef){
endPath[0]=endPath[1]=0;
backtrace(node,0,empty,1,reverse);
backtrace(i,1,empty,1,reverse);
if (path[0][0]==path[1][0]) newBlossom(node,i);
else {
augmentPath();
returntrue;
}
}
elseif (label[i].even==undef && label[i].odd[0]==undef){
label[i].even=node;
enqueue(i,0);
}
}
}
/*
The scanning of this label is complete, dequeue it and
keep going to the next one.
*/
queueFront++;
}
/*
If the function reaches this point, the queue is empty, all
labels have been scanned. The algorithm couldn't find an augmenting
path. Therefore, it concludes the matching is maximum.
*/returnfalse;
}
functionfindMaximumMatching (n){
//Initialize it with the empty matching.for (var i=0; i<n; i++) match[i]=false;
//Run augmentMatching(), it'll keep improving the matching.//Eventually, it will no longer find a path and break the loop,//at this point, the current matching is maximum.while (augmentMatching(n));
}
functionmain(){
var n;
n=readGraph();
findMaximumMatching(n);
for (var i=0; i<n; i++){
for (var j=i+1; j<n; j++) if (g[i][j]==matched)
console.log(i+1,j+1);
}
return0;
}
Output:
var graph = [[1,2]
,[1,3]
,[2,4]];
graph["n"] = 4;
graph["e"] = 3;
main()
1324Solution 2:
Just for fun - FIDDLE.
It reads the contents of the divs, then creates an array of all possible permutations, "pairs" (3 'variables' taken 2 at a time).
Then the individual letters that make up the contents of the divs are parsed out.
Then each letter is matched to the pairs to see how many times it occurs - if it is just one time for each pair, that particular 'pair' is added to the solution array.
After all of the combinations are analyzed, the solutions array is presented.
Not very elegant, but seems to work.
JS
var totaldivs = $('div').length; //count total divs or objectsconsole.log("Total divs = " + totaldivs);
var elements = []; //Read objects from the divs and assign to array elementsfor(var i = 0; i < totaldivs; i++ )
{
elements[i] = $( 'div:eq(' + i + ')' ).text();
elements[i] = elements[i].replace(/\s+/g, '');
console.log(elements[i]);
}
var objects = []; //make array of all individual objectsvar loopvar = 0;
for(var n = 0; n < totaldivs; n++)
{
poshyphen = elements[n].indexOf('-');
objects[loopvar] = elements[n].substring(0, poshyphen);
objects[loopvar+1] = elements[n].substring(poshyphen+1, elements[n].length);
loopvar = loopvar + 2;
}
$('.putmehere2').html(objects);
var pair = [];
var pairindex = 0; //make and array of all combinations of objects - pairfor(var r = 0; r < totaldivs; r++)
{
for(var s = 0; s <totaldivs; s++)
{
if(elements[r] != elements[s])
{
pair[pairindex] = elements[r] + '-' + elements[s];
pairindex++;
}
}
}
$('.putmehere').html(pair);
var solution = [];
var sol = 0;
var count = [];
var reg = newRegExp("regex","g");
for( var q = 0; q < pair.length; q++)
{
for( var r = 0; r < 4; r++)
{
var regex = newRegExp( objects[r], 'g' );
count[r] = (pair[q].match( regex )||[]).length;
console.log("Pair= " + pair[q] + " Objects = " + objects[r] + " Count= " + count[r]);
}
if( count[0] == 1 && count[1] == 1 && count[2] == 1 && count[3] == 1 )
{
solution.push( pair[q]);
}
}
$('.putmehere2').html(solution);
Post a Comment for "Matching Algorithm In Javascript"