Topic: Lower Bounds on Randomized Algorithms Date: September 22, 2004 Scribe: Srinath Sridhar

Transcription

1 15-859(M): Randomized Algorithms Lecturer: Anuam Guta Toic: Lower Bounds on Randomized Algorithms Date: Setember 22, 2004 Scribe: Srinath Sridhar 4.1 Introduction In this lecture, we will first consider the roblem of boolean function evaluation and rove that randomized algorithms (in the exected case) erform better than deterministic algorithms. The boolean circuits that we consider, can be naturally extended to obtain a game tree. We will then state Von Neumann s minimax theorem and its extension - Yao s rincile. Finally, we show how Yao s rincile can be used to obtain lower bounds on the exected running time of randomized algorithms. 4.2 Boolean Function Evalutation Figure 4.2.1: A deth 2k AND-OR circuit Boolean functions in general can be formulated using various forms. For the uroses of this lecture we shall focus on functions that have only two kinds of gates: AND and OR. The number of inuts to each gate is restricted to two. The functions are nested by alternating AND and OR gates, so that the circuit corresonding to the function is similar to Figure 4.2. We assume that the circuit forms a full binary tree (each internal node being an AND or OR gate) of height 2k. The number of inuts of the circuit is 2 2k. The root node is an AND gate and all 1

2 the nodes at height 2k 1 are OR gates. Note that, the functions considered are monotone and therefore comlementation is not ermitted and also the wires are not allowed to overla or branch out. The roblem of boolean function evaluation is to comute the result of the function (circuit) while minimizing the number of queries. An algorithm makes a query if it looks at one of the inut values. We make a quick digression to game trees, so that it is easier to comare games with boolean circuits. Game trees are used extensively by comuter rograms to lay games. In a game tree each node corresonds to a osition and each adjacent out-edge a legal move that can be layed from the osition. The root of the game tree is the current osition from which the analysis is being erformed. The even nodes corresond to ositions from which layer 1 (or comuter) icks the move and the odd nodes osition from which layer 2 (oonent) moves. To decide on the best move from the current osition, layer 1 icks the move that maximizes an evaluation function. In doing so he assumes that the oonent lays otimally and therefore icks the move that minimizes the evaluation function at the next ste. The boolean circuit can be naturally viewed as a game tree, where the edges are directed from the root to the leaves. The OR gates corresond to the ositions from which layer 1 moves (MAX nodes) and the AND gates the ositions from which the oonent moves (MIN nodes). The number of rounds of a game tree is defined as the number of moves erformed by a single layer. Therefore a round of the boolean circuit can be thought of as a air of AND followed by an OR gate. In the game tree (boolean circuit) considered above, there are k rounds. It is well known that an AND-OR circuit can be converted to a NOR circuit. Secifically, a circuit where the AND gates and the OR gates alternate can be converted to a NOR circuit, by relacing each gate of the AND-OR circuit with a NOR gate. Consider the the root (AND gate) and the two children (OR gates), assume that the wires of the two children are A, B, C and D. Then, (A B) (C D) = ( (A B) (C D)) = (A NOR B) NOR (C NOR D) Therefore, by induction (on deth) the AND-OR circuit can be converted to a NOR circuit by simle relacement. For the rest of the lecture, we will use an equivalent NOR circuit. We now return back to the question of comuting the value of the circuit for a given set of inuts while minimizing the number of inuts examined (number of queries erformed). Assume that the number of inut values n = 2 2k Exercise The number of inut values examined by any deterministic boolean function evaluation algorithm is n = 2 2k. We make the following imortant observation. Consider any NOR gate, and assume that its inuts are A and B. If a deterministic algorithm comutes that A = 0, then it needs the value of B to comute A NOR B. If a deterministic algorithm comutes that A = 1, then we can assume that it does not comute/examine the value of B. To erform better than the deterministic algorithm, we ick one out of A and B randomly and evaluate/examine its value. If exactly one of A and B 2

3 is 1, then the randomized algorithm has a 0.5 robability of comuting/examining the wire with value 1 first. This leads us to the following algorithm. Assume that the tree used by the recursive function is T, rooted at r. Further assume that the two children of r are L and R. Eval-Rand ( Tree T ) 1. ick x {L, R} uniformly at random 2. a := Eval-Rand ( subtree rooted at x ) 3. if a = 1 then 4. else (a) return 0 (a) Eval-Rand ( subtree rooted at y {L, R}\x ) Theorem The exected number of queries erformed by function Eval-Rand: cost(eval-rand) 3 k = n log 4 3 Proof: By induction on the number of rounds of the circuit. Let X k be the random variable that denotes the number of queries erformed by the algorithm on a circuit with k rounds. We want to comute E[X k ]. Let o denote the outut of the circuit. Assume that the values carried by the two wires to r are i 1 1 and i1 2 and the values carried by the wires to the left gate at deth one are i2 1 and i 2 2 and that of the right gate being i2 3 and i2 4. We break the analysis of the exected cost into two cases. The two cases are shown in Figure 4.2. Case 1: Outut o = 1. In this case, though it might be necessary to examine both inuts of deth 1, randomization hels imrove the cost at deth 2. We have, (o = 1) = (i 1 1 = 0 and i 2 1 = 0) (i 1 1 = 0) = (i 2 1 = 1 or i 2 2 = 1), (i 1 2 = 0) = (i 2 3 = 1 or i 2 4 = 1) In this case, both the recursive calls Eval-Rand ( Tree rooted at L ) and Eval-Rand ( Tree rooted at R ) are erformed. Consider either of the two calls. The child x is chosen at random and evaluated with cost E[X k 1 ]. The value a = 0 with robability 0.5 and the function has to erform an additional recursive call with cost E[X k 1 ]. The value a = 1 with robability 0.5 in which case we incur no additional cost besides E[X k 1 ] used to comute the value of a. Therefore in total, E[X k o = 1] = 2E[X k 1 ]Pr[x = 0] + E[X k 1 ]Pr[x = 1] E[X k 1 ] + 0.5E[X k 1 ] < 3E[X k 1 ] 3

4 Figure 4.2.2: Two cases in the analysis of the number of queries erformed by Eval-Rand Case 2: Outut o = 0. Using the same notations, we have, (o = 0) = (i 1 1 = 1 or i 1 2 = 1) In the worst case, exactly one of i 1 i = 1 and w.l.o.g assume i1 1 = 0 and i1 2 = 1. i 1 1 = 0 = (i 2 1 = 1 or i 2 2 = 1) and i 1 2 = 1 = (i 2 3 = 0 and i 2 4 = 0) Once again, w.l.o.g assume i 2 1 = 1 and i2 2 = 0. In the function call Eval-Rand( T ), x = L and therefore a = 0 with robability 0.5. In this case, the entire of right subtree has to be queried with cost 2E[X k 1 ]. Furthermore, in the function call Eval-Rand(Tree rooted at L), a = 1 with robability 0.5 in which case the exected number of queries is E[X k 1 ]; a = 0 with robability 0.5 in which case the exected number of queries is E[X k 1 ]. Now consider the function call Eval-Rand(T ), x = R and therefore a = 1 with robability 0.5, and here we incur a cost of 2E[X k 1 ] to comute the value of a. Using the notation, a 1, a l and a r to denote the values of a in the function calls Eval-Rand(T ), Eval-Rand(L) and Eval-Rand(R) resectively we have, E[X k o = 0] E[X k a 1 = 0]Pr[a 1 = 0] + E[X k a 1 = 1]Pr[a 1 = 1] = 0.5(4E[X k 1 ]Pr[a l = 0] + 3E[X k 1 ]Pr[a l = 1]) + 0.5(2E[X k 1 ]) = 0.5( )E[X k 1 ] < 3E[X k 1 ] The roof follows induction (on the number of rounds) with a trivial base case. 4

5 Moves H T H 1-1 T -1 1 Figure 4.3.3: Payoff Matrix for matching ennies game. Row Player: Roberta and Column Player: Charles 4.3 Games In this section we will focus on two layer zero sum games. Zero sum games have the roerty that one layer s loss is exactly the other layer s gain. A standard examle is the game stone, aer and scissors. We will assume that the two layers are Roberta and Charles. The ayoff (rules) of a game is usually exressed in a ayoff matrix M, where the rows corresond to the ossible moves for Roberta and the columns the ossible moves for Charles. An entry M i,j in the matrix corresonds to the ayoff awarded to Roberta if she lays i and Charles lays j. In general, if the game suorts n moves for Roberta and m for Charles, the ayoff matrix is of size n m. We will consider the game of matching ennies. Roberta and Charles simultaneously show one side of their enny. If the dislayed sides match then Roberta wins one dollar and if the sides don t match then Charles wins a dollar. The ayoff matrix for the game is shown in Figure If Roberta icks move i, then she is guaranteed a ayoff min j M i,j. The lower bounds for the gain of Roberta ˆV R and Charles ˆV C is given by ˆV R = max i min M i,j j and ˆV C = min j max M i,j. i In the game of matching ennies, ˆV R = 1 and ˆV C = 1. Claim For any ayoff matrix ˆV R ˆV C Proof: Let ˆV R = M i,k and ˆV C = M l,j. We have for all m, ˆV R M i,m, and secifically, ˆV R M i,j. Similarly, for all m, ˆV C M m,j, and secifically ˆV C M i,j and therefore ˆV R ˆV C. In games where V = ˆV R = ˆV C, the game is said to have a solution and V is called the value of the game. In such games, the two layers can advertise what they are going to lay without any harm. Definition A ure strategy is defined as a deterministic choice to lay in a game. Definition A mixed strategy is defined by a robability distribution on deterministic strategies. 5

6 We will now focus on mixed strategies. Let i be the robability that Roberto icks move i and let q j be the robability that Charles icks move j. Therefore, Exected ayoff = n m i q j M i,j = T Mq, i=1 j=1 where vectors and q contain the robabilities of selecting moves. Similar to the revious definition for ure strategies, we can obtain lower bounds on the ayoffs for Roberta and Charles as ˆV R = max min T Mq q and ˆV C = min q max T Mq Note that the min and max are comuted over all ossible distributions and q over deterministic choices. Theorem (Von Neumann s Minimax Theorem [Ne28]) For any two erson zero sum game with game matrix M, max min T Mq = min max T Mq q q Note that in the above theorem, if Roberta announces the robability distribution that she is using, Charles icks a single move (corresonding to the maximum term in vector T M) and his strategy becomes deterministic. Therefore an equivalent definition is ˆV R = max min j T Me j, where e j is a vector with 0 in all ositions excet j which is set to 1. This equivalent definition of ˆV R leads to Loomis theorem. Theorem (Loomis Theorem [Lo46]) For any two layer zero sum game secified by game matrix M, max min T Me j = min max e T j i Mq i 4.4 Yao s Princile We can use the above ideas from game theory to lower bound the running time of randomized algorithms. This is the only known technique currently used to establish lower bounds. Consider any randomized algorithm that uses a finite number of random bits. Fixing the values of the random bits secifies a deterministic algorithm. The values for the random bits can themselves be used to label each such deterministic algorithm. Informally a randomized algorithm can be viewed as secifying a robability distribution over several deterministic algorithms. Consider a matrix M similar to the game matrix. For any randomized algorithm, the rows of M corresond to different inuts and the columns corresond to different deterministic algorithms 6

7 (obtained by fixing the random bits of a randomized algorithm). The ayoff M i,j of the matrix corresonds to the running time of algorithm A j on inut I i. Let vector denote the robability distribution on the inuts and vector q the robability distribution on the algorithms and e j the (0, 1)-vector with 1 in exactly osition j. Using the same ideas from the revious section, we have, M i,j = C(A j, I i ), E[C(A j, I )] = T Me j, E[C(A q, I i )] = e T i Mq Using we have, max min E[C(I, A q )] = min q q max E[C(I, A q )] Droing the max term on the left side and the min term on the right side from the above equation gives us Yao s rincile. Proosition (Yao s Princile [Ya77]) For all distributions over I and q over A, min E[C(I, A)] max E[C(I, A q)] A A I I We will now show a simle examle and illustrate how Yao s rincile can be alied to establish a lower bound on randomized algorithms. The find-bill roblem can be stated as follows. There are n boxes and exactly one box contains a dollar bill, and the rest of the boxes are emty. A robe is defined as oening a box to see if it contains the dollar bill. The objective is to locate the box containing the dollar bill while minimizing the number of robes erformed. Consider the following randomized algorithm: 1. select x {H, T } uniformly at random 2. if x = H then 3. else (a) robe boxes in order from 1 through n and sto if bill is located (a) robe boxes in order from n through 1 and sto if bill is located The exected number of robes made by the algorithm is (n+1)/2. Since, if the dollar bill is in the i th box, then with robability 0.5, i robes are made and with robability 0.5, (n i + 1) robes are needed. Lemma A lower bound on the exected number of robes required by any randomized algorithm to solve the find-bill roblem is (n + 1)/2. 7

8 Proof: To use Yao s rincile we need to secify a distribution on the inut. In this case, we assume that the bill is located in any one of the n boxes uniformly at random. We only consider deterministic algorithms that does not robe the same box twice. By symmetry we can assume that the robe order for the deterministic algorithm is 1 through n. We therefore have, min E[C(A, I )] = A A n i=1 i n = 1 n n i=1 max I I E[C(I, A q)], i = n by using Yao s rincile. Therefore any randomized algorithm A q requires at least (n + 1)/2 robes in exectation. References [Lo46] L. H. Loomis. On a theorem of von Neumann Proceedings of the National Academy of Sciences of the U.S.A., 32: , [Ne28] J. von Neumann. Zur Theorie der Gesellschaftssiele. Mathematische Annalen, 100: , [Sn85] M. Snir. Lower bounds on robabilistic linear decision trees. Theoretical Comuter Science, 38:69-82, [Ya77] A. C-C. Yao. Probabilistic comutations: Towards a unified measure of comlexity. Proceedings of the 17th Annual Symosium on Foundations of Comuter Science, ,