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 letter can only end up once in the output. In the above scenario, a correct result would be

A - C

B - D

because A, B, C, and D all end up in the results. An incorrect result would be

A - B

which would leave C and D out.

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 letter can only end up once in the output. In the above scenario, a correct result would be

A - C

B - D

because A, B, C, and D all end up in the results. An incorrect result would be

A - B

which would leave C and D out.

• javascript
• algorithm
• graph-algorithm

Share Improve this question Follow asked Sep 7, 2014 at 1:09 JeffJeff 36.6k1616 gold badges114114 silver badges234234 bronze badges 9

• join, split, unique, zip by 2 – dandavis Commented Sep 7, 2014 at 1:59
• Calculate the number of repetitions of a character and find the minimum thereby increase output; using a dynamic data structure and removing the processed elements to optimize runtime – user4004613 Commented Sep 7, 2014 at 2:04
• If you end up implementing a general matching algorithm yourself, you might find Nick Harvey's easier to implement than Edmonds's. – David Eisenstat Commented Sep 7, 2014 at 3:10
• 1 Dandavis can you elaborate? – Jeff Commented Sep 7, 2014 at 3:19
• I'll try to generalize this - but do you only have three sets of pairs? Do they all contain a hyphen? – TimSPQR Commented Sep 7, 2014 at 14:48

 |  Show 4 more ments

2 Answers 2

Sorted by: Reset to default 4

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./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.
function Label(){                //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 = new Array(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.

function elem(){               //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 = new Array(MAX);         //The graph, as an adjacency matrix.
for (var i=0; i<MAX; i++){
  g[i] = new Array(MAX);
}
                              //blossom[i] contains the base node of the blossom the vertex i
var blossom = new Array(MAX); //is in. This, together with labels eliminates the need to
                              //contract the graph.

                              //The path arrays are where the backtrace() routine will
var path = new Array(2);
for (var i=0; i<2; i++){
  path[i] = new Array(MAX);
}
var endPath = new Array(2);   //store the paths it finds. Only two paths need to be
                              //stored. endPath[p] denotes the end of path[p].
var match = new Array(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 = new Array(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 blossoms
for (var i=0; i<MAX; i++){
  label[i] = new Label();
}
var queue = new Array(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();
}

function initGraph(n){
     for (var i=0; i<n; i++)
         for (var j=0; j<n; j++) g[i][j]=noEdge;
}

function readGraph(){

    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;
}

function initAlg(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;
     }
}

function backtrace (vert, pathNum, stop, parity, direction){
     if (vert==stop) return;           //pathNum is the number of the path to store
     else if (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 enter
        else if (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.
     */
     else if (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;
        }
        else if (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.
     */
     else if (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;
          }
          else if (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;
          }
     }
}

function enqueue (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.
}

function newBlossom (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 mon 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 this
     for (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.
            }
            else if (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.

            else if (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, except
        if (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);
        }                               
     }
     else if (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);
          }
     }
}

function augmentPath (){           //An augmenting path has been found in the matching
     var 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 simple
             if (g[a][b]==unmatched)   //changing of the matching status.
                g[a][b]=g[b][a]=matched;
             else if (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;
     else if (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;
}

function augmentMatching (n){  //The main analyzing function, with the
     var 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.
                else if (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();
                          return true;
                          /*
                            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.
                          */
                     }
                }
                else if (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);
                }
            }
         }
         else if (nodeLabel==1){ //Similar to above.
            for (var i=0; i<n; i++) if (g[node][i]==matched){
                if (blossom[node]==blossom[i]);
                else if (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();
                          return true;
                     }
                }
                else if (label[i].even==undef && label[i].odd[0]==undef){
                     label[i].even=node;
                     enqueue(i,0);
                }
            }
         }
         /*
           The scanning of this label is plete, 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.
     */
     return false;
}

function findMaximumMatching (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));
}

function main(){
    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);
    }
    return 0;
}

Output:

var graph = [[1,2]
            ,[1,3]
            ,[2,4]];

graph["n"] = 4;
graph["e"] = 3;

main()

1 3
2 4

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 binations are analyzed, the solutions array is presented.

Not very elegant, but seems to work.

JS

var totaldivs = $('div').length;                    //count total divs or objects
    console.log("Total divs = " + totaldivs);

var elements = [];                                  //Read objects from the divs and assign to array elements
for(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 objects
var 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 binations of objects - pair
for(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 = new RegExp("regex","g");

for( var q = 0; q < pair.length; q++)
   {
    for( var r = 0; r < 4; r++)
       {
        var regex = new RegExp( 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);