I want to determine the transitive reduction of this graph:


as of now, I only found the first step of doing this: represent the transitive closure of the graph as an adjacency relation, so this is what I did:


I'm not sure that this is the correct transitive closure of the graph, and I don't know how to move forward in determining its transitive reduction.

  • $\begingroup$ You are missing $(b,e)$. $\endgroup$ – plop Jul 17 '20 at 7:21

If we order the vertices alphabetically, then the adjacency matrix is

$$A=\begin{pmatrix} 0&1&1&1&0\\ 0&0&0&1&0\\ 0&0&0&1&1\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{pmatrix}$$

The adjacency matrix of the transitive closure would be

$$B=\begin{pmatrix} 0&1&1&1&1\\ 0&0&0&1&1\\ 0&0&0&1&1\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{pmatrix}$$

To get the this one you get each row by navigate the graph starting at the node corresponding to that row and mark with a $1$ the columns of the nodes that you manage to visit and with $0$ the rest.


$$AB=\begin{pmatrix} 0&0&0&2&3\\ 0&0&0&0&1\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0 \end{pmatrix}$$

To build the adjacency matrix $C$ of the transitive reduction put a $1$ in position $(i,j)$ if the $(i,j)$ entry of $A$ is non-zero and the $(i,j)$ entry of $AB$ is zero.

So, the adjacency matrix of the transitive reduction should be

$$C=\begin{pmatrix} 0&1&1&0&0\\ 0&0&0&1&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0 \end{pmatrix}$$

So, the transitive reduction looks like this

enter image description here


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