Continuing our look at some toplogical methods, today we’ll see the evasiveness conjecture in decision tree complexity. In the next lecture, we’ll see how we can sometimes analyze the complexity using fixed point theorems, and their generalizations (like the Hopf index formula), following the work of Kahn, Saks, and Sturtevant. These two lectures are co-blogged with Elisa Celis, with a lot of input from Lovasz’s lecture notes.
Decision tree complexity and evasiveness
Consider a boolean function on n bits. We define the decision tree complexity of f as follows. Given an unknown input , you are allowed to ask about the values of various bits of x, e.g. . Your goal is to compute using as few questions as possible, and your questions can be adaptive, depending on answers to previous questions. The complexity of such a strategy is the maximum number of questions asked for any . The decision tree complexity, written , is the minimum complexity of any strategy that computes f. (There are many other interesting models of decision complexity, see e.g. this survey.)
Clearly , because we can trivially query all the bits of x, and then output f(x). A function f is called evasive if this upper bound is met, i.e. . As an example, consider the parity function , where is addition modulo 2. Clearly is evasive because after bits of x are asked about, the setting of the final bit determines the value of f.
For a more general example, consider any such that is odd. In this case, for every , exactly one of or has the same property that the number of inputs resulting in a 1 is odd. (These two functions are the natural restriction of f to functions on n-1 bits, which results from fixing the value of the ith bit.) Thus an adversary could keep answering questions “?” so that the restricted function retains this property. Since the number of inputs yielding a 1 is always odd, the restricted function always takes both possible values, implying that f is evasive–the advesary ensures that the value cannot be determined until all n possible questions are asked.
For an example of a non-evasive property, think of a point as speciying a directed graph on vertices, where there is exactly one directed edges connecting every pair of vertices, and x specifies the direction of this edge (this is called a tournament). Thinking of the vertices as players, a directed edge from u to v means that u defeats v. Now if the digraph specified by x has one vertex that defeats everyone else. What is ?
Well, first we can conduct a single elimination tournament, where vertex 1 plays vertex 2, and the winner players vertex 3, and the winner of that players vertex 4, etc. At the end, there is only one remaining vertex that remains undefeated. Now asking more questions, we can determine whether indeeds defeats everyone else. The total number of questions was , hence , implying that f is not evasive.
Montone graph properties and the evasiveness conjecture
Let . In general, we can encode an arbitrary undirected N-vertex graph as an element . A function is called a graph property if relabeling the vertices of doesn’t affect the value of . The function f is monotone if the value of the function can never change from 1 to 0 when flipping one of the input bits from 0 to 1. In the setting of graph properties, this corresponds to those which are maintained under addition of edges to the graph, e.g. “is G connected?” or “does G have a k-clique?”
Evasiveness Conjecture (Aanderaa-Karp-Rosenberg): Every non-trivial monotone graph property is evasive.
Here, non-trivial means that , where and denote the all-zeros and all-ones strings, respectively.
For example, consider the example “is G connected?” The adversary is simple: When asked about a possible edge , she answers NO unless this answer would imply that the graph is disconnected. In other words, she answers NO unless she has answered NO already for all edges across a cut except for , in which case she has to answer YES.
Now, suppose there is a strategy which figures out the connectivity of without asking a question about some edge . In this case, the conclusion must be that because the adversary always maintains that by answering everything in the future YES, she could force the graph to be connected. In this case, the edges answered YES have to form a spanning tree of G (otherwise by answering all unasked questions NO, the graph would become disconnected). Consider a path P from i to j in this YES spanning tree. Let be the edge of P which was asked about last. Clearly the adversary answered YES for , but this contradicts the advesary’s strategy. Since has not been asked yet, the adversary is safe to answer NO for , and still later by answering YES on , she could force the graph to be connected. Thus no such strategy exists, and connectivity is evasive.