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Now keeping k=2 fixed we take all possible combinations of i,j. If you think for yourself you will see that for k=2 i=3, j=1 in this combination we will get another new edge, the rest of the graph will remain as before.
Now we think of k=1 as the middle node. If you want to change your mind, you can put the new edges in the book yourself before looking down.
This time only one edge will be added. If i=4, j=2 we can go from 4 to 1 to node 2 at 23 cost. Then the graph will look like this:
Now let’s take node number 3 as the middle node. Try it again yourself and then see the section below.
Now some interesting things happen. 4 to 1 used to cost 20. Now keeping 3 between 4 to 1 will cost 8+7=15. Notice we didn’t have the edge at 3 to 1 at the very beginning. But while updating we put the edge, now going directly from 3 to 1, in the original graph actually going 3->2->1.
Similarly the path of 23 cost of 4 to 2 will be updated to 13.
For k=4 there will be no more updates because no node can go to 4.
Now we can tell the cost to go from any node to any node by looking at the matrix. Having infinity means there is no path to that node.
After building the adjacency matrix from the input then the task is very simple:
In the code we run the loop for k from 1 even though in the example we took 2 earlier. It can actually be done in any order, even if you take 1 first, the algorithm will work.
In this way, we only got the cost of the path, how to get the path?
Suppose we have a matrix next[][]. Now by next[i][j] we mean the next node to go to from i to j. Then next[i][j] = j for all i,j at the very beginning. Because there is no “middle node” in the beginning and the shortest path is still not finished.
Now when we update matrix[i][j] in the loop it means we go through a node k in between. Note that we may not go directly from i to k by edges in the original graph, we just know that there is a node k between nodes i,j that we need to go to before going to j. The next node on the path from i to k is stored in next[i][k]! Then next[i][j] = next[i][k].
In the findpath function we keep j fixed and traverse the next array until we reach j. Then we got the path!
Transitive Closure:
Let our adjacency matrix look like this:
matrix[i][j] = 1 if there is a direct edge from i to j
matrix[i][j] = 0 if no edge
1
2
matrix[i][j] = 1 if there is a direct edge from i to j
matrix[i][j] = 0 if no edge
Now we want to create a matrix that can tell if one or more edges from i to j can be used. If we wanted, we could find the shortest path by replacing o with infinity as above. But in this case “OR” and “AND” operation can be done faster. Now the update condition will be like this:
This means matrix[i][j] will contain 1 only when either “matrix[i][j] contains 1” or “matrix[i][k] and matrix[k][j] both contain 1. That means either go directly or use a node k in between.
Complexity:
3 nested loops running over the number of nodes, time complexity is O(n^3). Space complexity is O(n^2) as 2D matrix is used.
Some questions:
1. Would the algorithm work if the k loop was placed inside the i,j loop?
2. Will Floyd Warshall work if there is a negative cost on the graph?
Related problems:
Page Hopping
05-2 Rendezvous
Minimum Transport Cost
Asterix and Obelix
Many more problems
Happy Coding!
Matrix chain multiplication
We will now look at another classic dynamic programming problem called matrix chain multiplication. It is important to learn because many types of problems can be solved using its concepts. Before reading this article you should have an understanding of dynamic programming. Apart from this, you must have an understanding of matrix.
I’m sure you all know the terms of matrix multiplication, but I also want to remind you. Suppose we have two matrices A1,A2
and their dimensions are m∗n and p∗q
. So a few properties are important to us:
A matrix can be multiplied by two if and only if n=p
is That means the number of columns of the first matrix should be equal to the number of rows of the second matrix.
If the previous condition holds then after multiplication we have A3
Get the matrix whose dimension is m∗q
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.
While doing matrix multiplication we need to multiply and add some numbers which we can call scalar multiplication. We need to multiply the total scalar to the total m∗n∗q
or m∗p∗q
times
You can check the terms by multiplying them in two matrix notebooks. Another thing to remember is that A1∗A2
And A2∗A1
Not one, i.e. we cannot do matrix multiplication by breaking the order.
When this text refers to multiplication or simply “multiplication” it refers to the numbers that must be multiplied inside when doing matrix multiplication.
[Notice 23 April 2020: Started a new series on Dynamic Programming. A new version of the article is available here.
So if we have 3 matrices A1,A2,A3 having dimension m∗n
,n∗p, p∗q then A4=A1∗A2∗A3 matrix will have dimension m∗q. This is understood from the 2nd property above. Now we can do this matrix multiplication A1*A2*A3 in two ways, with brackets they can be shown as: (A1∗A2)∗A3,A1∗(A2∗A3). That is, we can multiply A1∗A2 with A3, or A1 with A2∗A3. But the order must be fixed, (A1∗A3)∗A2
It may not be valid.
You understand that multiplication can be done by placing brackets in many ways by keeping the order correct. But how the bracket is placed is very important. Let’s say A1,A2,A3
Its dimensions are 10×100, 100×5, 5×50.
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