How to find the weight of the heaviest path of an undirected, unweighted graph (M(d) of the graph), using adjacency matrix

Question

In the image A-L is the longest path, but L-M is the heaviest The heaviest path of the graph is the path with the most edges connected to it, for this current case the heaviest path is L-M with 19 edges: L-K, K-J, J-I, I-H, H-G, G-F, F-E, F-M, M-N, M-O, M-P, M-Q, M-R, M-S, M-T, M-U, M-V, M-W, M-X. I am trying to create a program that will check that the inputted matrix represents a tree graph, and if it does, that returns the number of edges on the heaviest path of the graph. In other words it will return the M(d) of the graph. For this current case the output will be the number of edges connected to L-M path which is 19.

But I can not find a way to do that. Here I created a function to find the row with the least 1s to set it as a root, but don't know how to proceed.

let matrix = [
  [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],


//find row with least 1s
function findRowWithLeastOnes(matrix_data) {
  let counter = 0;
  let row = 0;
  for (let i = 0; i < matrix_data.length; i++) {
    for (let j = 0; j < matrix_data[i].length; j++) {
      if (matrix_data[i][j] === 1) {
        counter++;
      }
    }
    if (counter < matrix_data.length) {
      row = i;
      console.log(row);
      return row;
    }
  }
}
findRowWithLeastOnes(matrix);

Answer 1

We can use the following algorithm:

• Turn the matrix into an adjacency list so to avoid scanning zeroes all the time, and to have easy access to the degree of a node, which contributes to the weight.

• Choose any node to start with (e.g. 0) and find the heaviest path from it. We can use a depth-first search for that (for instance, using recursion). From all paths that are found, keep the heaviest. To avoid an edge being traversed back and forth, remember what the previous node was in the traversal: it should not be revisited. If there was a previous node, we should avoid counting the traversed edge twice in the sum of weights.

• Take the last node in that path: this is the node that is the furthest away from our starting node in terms of weight. This node will be an end-point of the heaviest path we are looking for.

• Use the same algorithm a second time to find the heaviest path from that node. This is the path we need.

If your question is also about determining whether a graph is a tree, you would essentially need to check there are no cycles (a depth first traversal can be used again) and that from one node all others can be reached (again depth first can serve). Maybe you'd also want to check that the graph is undirected, i.e. that the matrix is mirrored along the main diagonal.

function isUndirected(matrix) {
    if (matrix.length !== matrix[0]?.length) return false; // must be square
    for (let i = 0; i < matrix.length; i++) {
        for (let j = i; j < matrix.length; j++) {
            if (matrix[i][j] !== matrix[j][i]) return false;
        }
    }
    return true;
}

function isTree(matrix) {
    const visited = Array(matrix.length).fill(false);

    function dfs(node, previous=-1) {
        if (visited[node]) return false; // cycle
        visited[node] = true;
        const neighbors = matrix[node];
        for (let neighbor = 0; neighbor < neighbors.length; neighbor++) {
            if (neighbor != previous && neighbors[neighbor] && !dfs(neighbor, node)) return false;
        }
        return true;
    }
    // When no cycles were found and all nodes were reachable from node 0 => tree!
    return dfs(0) && !visited.includes(false);
}

function findHeaviestPathFrom(adj, node, previous=-1) {
    const neighbors = adj[node];
    let heaviest = {
        nodes: [],
        weight: 0
    };
    for (let neighbor of neighbors) {
        if (neighbor != previous) {
            const path = findHeaviestPathFrom(adj, neighbor, node);
            if (path.weight > heaviest.weight) heaviest = path;
        }
    }
    heaviest.nodes.push(node);
    heaviest.weight += neighbors.length - (previous !== -1)    
    return heaviest;
}

function findHeaviestPath(matrix) {
    // Turn the matrix into an adjacency list
    let adj = matrix.map(row => row.map((isEdge, i) => isEdge ? i : -1).filter(i => i > -1));
    // Find the node that has the heaviest connection to node 0
    let remote = findHeaviestPathFrom(adj, 0).nodes[0];
    // Find heaviest path starting in that node
    return findHeaviestPathFrom(adj, remote);
}

const matrix = [
  [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
  [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
];

if (!isUndirected(matrix)) console.log("The graph is directed");
else if (!isTree(matrix)) console.log("The graph is not a tree");
else console.log("Heaviest path: ", JSON.stringify(findHeaviestPath(matrix)));