Why is the effective branching factor used for measuring performance of a heuristic function?

1

For search algorithms with heuristic functions, performance of heuristic functions are measured by effective branching factor ${b^*}$ which involves total nodes expanded ${N}$ and depth of the solution ${d}$. I'm not able to find out how different values of ${d}$ affect the performance keeping same ${N}$. Put another way, why not use just the ${N}$ as the performance measure instead of ${b^*}$?

KGhatak

Posted 2019-11-24T11:01:08.023

Reputation: 143

Answers

1

As you found $N$ is the number of nodes that are expanded. The cost of expansion of each node is equal to the number of children of that node. Hence, we use $b^*$ for each node. In other words, the total number of nodes that are involved in the expansion process is $N \times b^*$.

OmG

Posted 2019-11-24T11:01:08.023

Reputation: 1 020

1@ OmG - Why not ${N}$ since you agreed that the cost solely depends on ${N}$? – KGhatak – 2019-11-24T17:03:19.697

@KGhatak because the total number of expanded node is $N \times b^*$. – OmG – 2019-11-24T17:27:17.493

1@ OmG - Not sure if this is the right understanding. Can you provide some reference for it! Just think, we calculated expected ONLY when we don't know for sure. If we know actual N, there is no need to estimate/guess (aka Expected value of) N. What say! – KGhatak – 2019-11-25T04:36:33.777

@KGhatak Unfortunately I don't know which part is confusing for you. In the search, we expand $N$ nodes, and each node has $b^$ children at most. Hence, the complexity is related to $N$ and $b^$. Now, suppose we do not apply $b^$ in the complexity, and another algorithm try to search and expand $N$ nodes to find the solution, but the branching factor of the current algorithm would be $2^{b^}$! Which of them will be a better algorithm in terms of complexity? – OmG – 2019-11-25T10:22:08.267

@ OmG - Let me put it in a different way. Two algorithms, say 1 and 2, try to solve a problem. Their respective values are ${N_1}, {b_1^}, {N_2}$, and ${b_2^}$ such that ${N_1b_1^}={N_2b_2^}$. Which algorithm is efficient given ${N_1>N_2}$. Would you pls explain! – KGhatak – 2019-11-25T18:26:08.593