minimiztion the distance travel

hi everyone

i have to write a code such that the time reaching the two points is minimum where different different routes are available to reach the point

for example :

A=[ 0 2 100 2 10 100 100;

2 0 3 100 100 100 100;

100 3 0 100 100 5 2;

100 100 100 0 4 3 100;

10 100 100 2 0 2 1;

100 100 5 3 2 0 4;

100 100 2 100 1 4 0];

the above matrix denotes the time to reach from one point to other (for example: from point(1) to point(5) time will 10 sec) but when the time is 100 (point(1) and point(3)) that way has to b skipped and we have to find another route (ex: to reach point 1 fro 3 the new route can be (3-->2-->1)or(3-->6-->4-->1)) the another route must have minimum time travel.

can we use optimization technique??

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/graph.shortestpath.html

https://www.mathworks.com/help/matlab/ref/graph.shortestpathtree.html

If you need to calculate the shortest path from each point to each other point then you will need to call one of the above multiple times.

Answer 1

[Bruno Luong]

Checkout function distances

Warning : Your adjacent matrix is not symmetric !

>> A

A =

     0     2   100     2    10   100   100
     2     0     3   100   100   100   100
   100     3     0   100   100     5     2
   100   100   100     0     4     3   100
    10   100   100     2     0     2     1
   100   100     5     3     2     0     4
   100   100     2   100     1     4     0

>> A==A'

ans =

  7×7 logical array

   1   1   1   0   1   1   1
   1   1   1   1   1   1   1
   1   1   1   1   1   1   1
   0   1   1   1   0   1   1
   1   1   1   0   1   1   1
   1   1   1   1   1   1   1
   1   1   1   1   1   1   1

Apply on symmetrized A

G = graph((A+A')/2);
distances(G)

Result:

ans =

     0     2     5    11     8    10     7
     2     0     3     9     6     8     5
     5     3     0     6     3     5     2
    11     9     6     0     3     3     4
     8     6     3     3     0     2     1
    10     8     5     3     2     0     3
     7     5     2     4     1     3     0