加速最优路径函数

Question

我有一个脚本，它可以找到从转换成本矩阵中的一个节点到所有其他节点的最佳路径（在此处找到的函数文件夹中的 DPA 函数http://uk.mathworks.com/matlabcentral/fileexchange/39034-finding-optimal-path-on-a-terrain）。

我遇到的问题是这个函数正是我需要的，但我需要它在 16641 x 16641 矩阵 (129*129 x 129*129) 上运行。我目前正在运行它，但正如预期的那样需要很长时间（目前仍在运行 2 小时！）。我将在最后发布整个函数，但我想知道是否有人可以阐明对函数的以下更改是否会使其运行得更快。我相信造成这个过多运行时间的过程是

for fromNode = 1 : m
        for toNode = 1 : m
            aij = P(toNode, fromNode);
            dj = aij + prevStageCostMat(fromNode);
            if dj < stageCostMat(toNode)
                stageCostMat(toNode) = dj;
                predMat(toNode, stage) = fromNode;
            end         
        end % end toNode
    end % end fromNode

其中 m=16641。函数是否必须通过所有节点传播才能给出正确的结果（我想问题是，我是否正确理解了函数在做什么？）或者您认为有什么方法可以加快计算速度？

主功能页面说

% Keep in mind that DPA propagates through all the nodes. Basically DPA
% tries to find optimal path from one nodes to all nodes

这是完整的功能，如果有人能给我任何见解，我将不胜感激（如果您需要更多说明，请告诉我，如果我解释得混乱，请告诉我），谢谢！

function [stageCostMat, predMat, converged] = dpa(P, startNode, endNode, maxIteration)

% This is the function that will perform the DPA.
%
% Assume we have n number of nodes. P matrix is the transition cost matrix
% with dimension of n x n(square matrix). P(toNode, fromNode) shows
% transition cost from fromNode to toNode.
%
% stageCostMat shows the cost at each node for current iteration.
% stageCostMat(c) = current stage cost matrix at node c.
%
% predMat shows parent/predecessor node of each node for every stage.
% predMat(c, s): parent of node c during stage s.
%
% manurung.auralius@gmail.com
% 17.11.2012
% -------------------------------------------------------------------------

% Is the algortihm converged?
converged = 0;

% Cost matrix is a square matrix, m = n
[m, n] = size(P);

% Assume we will probably converge after n stages
stageCostMat = ones(1, m) * inf;

% Initial cost, no initial cost
stageCostMat(startNode) = 0;

% Predecessor matrix to trace back the optimum path, we will record parent of
% each node on each iteration
predMat = zeros(m, maxIteration); 

% Stage-by-stage, we move from start node to terminal node
for stage = 2 : maxIteration    

    % Find connection from any nodes to any nodes, keep the smaller cost
    prevStageCostMat = stageCostMat;
    stageCostMat = ones(1, m) * inf;

    for fromNode = 1 : m
        for toNode = 1 : m
            aij = P(toNode, fromNode);
            dj = aij + prevStageCostMat(fromNode);
            if dj < stageCostMat(toNode)
                stageCostMat(toNode) = dj;
                predMat(toNode, stage) = fromNode;
            end         
        end % end toNode
    end % end fromNode

    % Termination
%     if (stageCostMat == prevStageCostMat)
%         converged = 1;
%         break;
%     end

    if (predMat(endNode, stage) == endNode) || (predMat(endNode, stage-1) > 0) && (predMat(endNode, stage) == predMat(endNode, stage-1))
        converged = 1;
        break;
    end

end

predMat = predMat(:, 1:stage);

Answer 1

更高级一点。 toNode 与 forNode 的当前位置相关，您必须避免 toNode 超出矩阵维度的情况：

for fromNode = 1 : m 
    %relativ to fromNode bc its always on the diagonal
    for toNode = (fromNode-1) : (fromNode+1) 
        %for the first and alse fromNode toNode would be out of the array limits
        if toNode==0 | toNode==m+1
            continue 
        end
        aij = P(toNode, fromNode); 
        dj = aij + prevStageCostMat(fromNode); 
        if dj < stageCostMat(toNode) 
            stageCostMat(toNode) = dj; 
            predMat(toNode, stage) = fromNode; 
        end
    end % end toNode 
end % end fromNode

我建议您在具有旧循环解决方案的数据上尝试此操作，以检查解决方案是否仍然相同。时间应该会大大减少，因为您只计算了 16641 个循环中的 3 个。