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According to Norvig in AIMA (Artificial Intelligence: A modern approach), the Depth-first algorithm is not complete (will not always produce a solution) because there are cases when the subtree being descended will be infinite.

On the other hand, the Breadth-first approach is said to be complete if the branching factor is not infinite. But isn't that somewhat the same "thing" as in the case of the subtree being infinite in DFS?

Can't the DFS be said to be complete if the tree's depth is finite? How is then that the BFS is complete and the DFS is not, since the completeness of the BFS relies on the branching factor being finite!

george
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3 Answers3

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A tree can be infinite without having an infinite branching factor. As an example, consider the state tree for Rubik's Cube. Given a configuration of the cube, there is a finite number of moves (18, I believe, since a move consists of picking one of the 9 "planes" and rotating it in one of the two possible directions). However, the tree is infinitely deep, since it is perfectly possible to e.g. only rotate the same plane alternatingly back and forth (never making any progress). In order to prevent a DFS from doing this, one normally caches all the visited states (effectively pruning the state tree) - as you probably know. However, if the state space is too large (or actually infinite), this won't help.

I have not studied AI extensively, but I assume that the rationale for saying that BFS is complete while DFS is not (completeness is, after all, just a term that is defined somewhere) is that infinitely deep trees occur more frequently than trees with infinite branching factors (since having an infinite branching factor means that you have an infinite number of choices, which I believe is not common - games and simulations are usually discrete). Even for finite trees, BFS will normally perform better because DFS is very likely to start out on a wrong path, exploring a large portion of the tree before reaching the goal. Still, as you point out, in a finite tree, DFS will eventually find the solution if it exists.

Aasmund Eldhuset
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  • Yes I thought so myself about the rarity of trees with infinite branching factors in games but it isn't that clearly explained in the book. I thought that the completeness of these algorithms are independent of the context they are applied in, because they are "uninformed". BFS is complete regardless of the fact that "usually game trees do not have infinite branching factors". I most certaintly must be missing something. Thanks for your reply! – george Feb 21 '11 at 17:44
  • I agree with the sentiment that statements about "uninformed" search algorithms should not be conditioned on properties of the graph that cannot be detected by the algorithm. However, I guess that "all graphs have a finite branching factor" is a hidden premise in the book. Uninformed search is described in [these slides](http://goedel.cs.uiowa.edu/classes/145/Fall01/Notes/4u_search.pdf) as "All we know is how to generate new states and recognize a goal state" - it would be reasonable to assume (or even to _define_) that the generator always produces a finite sequence of neighbouring states. – Aasmund Eldhuset Feb 21 '11 at 18:06
  • What is an example of a graph with an infinite branching factor? – ryanwebjackson Sep 29 '22 at 03:45
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    @ryanwebjackson: The [total order](https://en.wikipedia.org/wiki/Total_order) of all natural numbers with the `<=` operator: each natural number is a node, and each number has an edge to itself and to every other number that is greater than it. – Aasmund Eldhuset Sep 30 '22 at 09:46
  • @Aasmund Eldhuset What could that represent in the real world? – ryanwebjackson Sep 30 '22 at 18:44
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DFS can not stuck in cycles (if we have a list of opened and closed states). The algorithm is not complete since it does not find a solution in an infinite space, even though the solution is in depth d which is much lower than infinity.

Imagine a strangely defined state space where each node has same number of successors as following number in Fibonacci sequence. So, it's recursively defined and therefore infinite. We're looking for node 2 (marked green in the graph). If DFS starts with the right branch of tree, it will take infinite number of steps to verify that our node is not there. Therefore it's not complete (it won't finish in reasonable time). BFS would find the solution in 3rd iteration.

infinite space

Rubik's cube state space is finite, it is huge, but finite (human stuck in cycles but DFS won't repeat the same move twice). DFS would find very inefficient way how to solve it, sometimes this kind of solution is infeasible. Usually we consider maximum depth infinite, but our resources (memory) are always finite.

Tombart
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The properties of depth-first search depend strongly on whether the graph-search or tree-search version is used. The graph-search version, which avoids repeated states and redundant paths, is complete in finite state spaces because it will eventually expand every node. The tree-search version, on the other hand, is not complete—for example, in Figure 3.6 the algorithm will follow the Arad–Sibiu–Arad–Sibiu loop foreverenter image description here

Source: AI: a modern approach

baraa yusri
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