P = 8 0-RP T = 8 0-BPP T = 8 0-PP T j j j 8 -RP 8 -BPP = 8 -PP k k T j 8 (=)-RP 8 (=)-BPP 8 (=)-PP k k k RP BPP PP k k k 9 (=)-RP T 9 (=)-BPP 9 (=)-PP

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1 Title: Authors: Complexity Classes Characterized by Semi-Random Sources Ryuhei Uehara Aliations: Center for Information Science, Tokyo Woman's Christian University, Suginamiku, Tokyo 67, Japan. Corresponding author: Ryuhei Uehara, Center for Information Science, Tokyo Woman's Christian University, -6- Zempukuji, Suginami-ku, Tokyo 67, Japan. Phone: Fax: Key words: Computational complexity, Semi-random sources, Interactive proof systems. Abstract: The complexity classes PP; BPP and RP are usually dened via probabilistic Turing machines (PTMs) that have access to a perfect random source. A natural question is this: Do these classes change if the PTMs only have access to an arbitrary semi-random source rather than a perfect random source? The notion of semi-randomness are dened by Satha and Vazirani in 98, and this question was rst considered by Vazirani and Vazirani in 985, who proved that RP does not change even if the PTMs only have access to an arbitrary semi-random source. Later, Vazirani proved that BPP does not change, either. In this paper, we show a surprising result that PP collapses to BPP if the PTMs only have access to an arbitrary semi-random source. We also consider the following question: Do the classes PP; BPP and RP change if the PTMs only have access to a specic semi-random source? As answers to this question, we show that RP changes to NP while both BPP and PP change to PSPACE.

2 P = 8 0-RP T = 8 0-BPP T = 8 0-PP T j j j 8 -RP 8 -BPP = 8 -PP k k T j 8 (=)-RP 8 (=)-BPP 8 (=)-PP k k k RP BPP PP k k k 9 (=)-RP T 9 (=)-BPP 9 (=)-PP j 9 -RP 9 -BPP = 9 -PP = PSPACE k NP = 9 0-RP = 9 0-BPP = 9 0-PP Figure : Summary of the results (0 < < ) Introduction Satha and Vazirani introduced, as an extremely general model of an imperfect source of randomness, a \slightly random source" in [SV8], or \semi-random source" in [SV86]. In this paper, we refer to the random source as a \semi-random source." In a semi-random source, the previous bits generated by the source can condition the next bit in an arbitrarily way. Accordingly, the next bit is generated by the ip of a coin whose bias is xed by an adversary who has complete knowledge of the history of the process. The adversary is limited to choosing a bias in [; 0 ] with some positive number 0 < <. We call a semi-random source with parameter a -random source. Gill dened the classes, PP; BPP and RP via PTMs that have access to a perfect random source [Gil77]. What if the PTMs only have access to a semi-random source (rather than a perfect random source)? This question was rst considered by Vazirani and Vazirani [VV85], who dened the class SR p as follows: A language L is in SR p if there exists a PTM M such that () for x L, M accepts with probability greater than for all -random sources and () for x 6 L, M always rejects. Their result is that SR p = RP. Later, Vazirani similarly dened the class SBP P (corresponding to BPP) and prove that SBP P = BPP. In the same manner, we introduce the class 8 -PP (corresponding to PP) as follows: A language L is in 8 -PP if there exists a PTM M such that () for x L, M accepts with probability greater than for all -random sources and () for x 6 L, M accepts with probability less than for all -random sources. We prove that 8 -PP = BPP for 0 < <. Compared with that SR p = RP and SBP P = BPP, this result is very interesting. Recall that in the denition of 8 -PP, the PTM M is required to work for every -random source. This is what the symbol \8" means. This also motivated us to introduce the class 9 -PP. More precisely, we dene the class 9 -PP as follows: A language L is in 9 -PP if there exists a PTM M such that () for x L, M accepts with probability greater than for at least one -random source and () for x 6 L, M accepts with probability less than for all -random sources. The classes 9 -BPP and 9 -RP are dened similarly. We prove that 9 -RP = NP, and 9 -BPP = 9 -PP = PSPACE for 0 < <. These results give new characterizations of NP and PSPACE. We summarize the relations between the classes in Figure. To unify the notations, we use 8 -RP to denote SR p and 8 -BPP to denote SBP P. Shamir showed that interactive proof systems recognize every language in PSPACE [Sha90]. Condon and Ladner investigated interactive proof systems with some restrictions on provers, and showed that some of such systems characterize the classes MA and NP [CL95]. The PTM M which witnesses L(M) 9 -RP can be seen as an interactive proof system with a restriction on the condition of the probability that system accepts. This observation implies that the interactive proof system which accepts with probability either exceeds or 0, characterizes the class NP.

3 Preliminaries We assume a standard Turing machine model. For formal denitions of a deterministic Turing machine (DTM) and a nondeterministic Turing machine (NTM), see [HU79]. A probabilistic Turing machine (PTM) is a Turing machine with distinguished states called coin-tossing states. For formal denitions of a PTM, see [Gil77, BDG88]. Without loss of generality, we assume that an NTM (or PTM) is standardized as follows: It is a polynomial time bounded NTM (or PTM) with exactly two choices (called 0-choice and -choice, respectively) per step. Therefore, its computation tree is a full binary tree whose depth is polynomial in the length of the input. A semi-random source with parameter, or simply a -random source, is dened as follows. Denition [SV8] Let be a number such that 0. A semi-random source with parameter outputs bits X X, such that for all i and for all x ; x ;, Pr[X i = x i j X = x ; ; X i0 = x i0 ] 0 : Denition [VV85] A language L is in 8 -RP if there exists a PTM M such that () for x L, M accepts with probability greater than and () for x 6 L, M always rejects. We note that the class 8 -RP is denoted by SR p in [VV85]. Denition [Vaz86] A language L is in 8 -BPP if there exists a PTM M such that () for x L, M accepts with probability greater than and () for x 6 L, M accepts with probability less than for every -random source. We remind that the class 8 -BPP is denoted by SBP P in [Vaz86]. Denition A language L is in 8 -PP if there exists a PTM M such that () for x L, M accepts with probability greater than and () for x 6 L, M accepts with probability less than for every -random source. We dene the class 9 -RP as follows. Denition 5 A language L is in 9 -RP if there exists a PTM M such that () for x L, M accepts with probability greater than for at least one -random sources and () for x 6 L, M always rejects. Similarly, we dene the classes 9 -BPP and 9 -PP as follows. Denition 6 A language L is in 9 -BPP if there exists a PTM M such that () for x L, M accepts with probability greater than for at least one -random source and () for x 6 L, M accepts with probability less than for every -random source. Denition 7 A language L is in 9 -PP if there exists a PTM M such that () for x L, M accepts with probability greater than for at least one -random source and () for x 6 L, M accepts with probability less than for every -random source. A -random source is a perfect random source. Thus, it is easy to see that RP = 8 (=)-RP = 9 (=)-RP, BPP = 8 (=)-BPP = 9 (=)-BPP, and PP = 8 (=)-PP = 9 (=)-PP. The following proposition follows from the denitions immediately: Proposition For every with 0 < <, the following hold. P = 8 0-RP 8 -RP 8 (=)-RP = RP = 9 (=)-RP 9 -RP 9 0-RP = NP; P = 8 0-BPP 8 -BPP 8 (=)-BPP = BPP = 9 (=)-BPP; 9 0-BPP = NP; P = 8 0-PP 8 -PP 8 (=)-PP = PP = 9 (=)-PP; and 9 0-PP = NP:

4 Proof. For and 0 with 0 0, a 0 -random source is also a -random source. A 0-random source generates with probability between 0 and. Thus for 8 0-RP, 8 0-BPP and 8 0-PP, since every computation for each input must produce same answer, the algorithm must be deterministic. This implies 8 0-RP = 8 0-BPP = 8 0-PP = P. Conversely, for 9 0-RP, 9 0-BPP and 9 0-PP, since it is sucient to accept that at least one computation reaches accepting state, the algorithm is nondeterministic. This implies 9 0-RP = 9 0-BPP = 9 0-PP = NP. Note that for < 0, 9 0 -BPP 9 -BPP and 9 0 -PP 9 -PP do not hold, whereas 9 0 -RP 9 -RP holds. To see this, let M be a PTM such that L(M) 9 0 -BPP and and 0 be two numbers with 0 < 0. It likely holds that L(M) 9 -BPP since the input x with x L(M) satises the condition of the denition of 9 -BPP since a 0 -random source is also a -random source. However, for x 6 L, M must accepts with probability less than for every -random sources. Since the set of -random sources properly contain the set of 0 -random sources when < 0, M may not satisfy the condition. For 0 <, the following results are known. Theorem [VV85] 8 -RP = RP. Theorem [Vaz86] 8 -BPP = BPP. Since the proof of Theorem plays an important role in this paper, we show the outline of the proof. Proof of Theorem [Pap9]. Let L be a language with L BPP, and M 0 be a PTM such that L(M 0 ) = L. Let p(n) be the length of a computation path of M 0 on input of length n. Without loss of generality, we can assume that the number of the accepting path is, by repeating the algorithm enough times, at least p(n) for x L, and at most log p(n)+6 0 p(n) for x 6 L. Let r(n) = d e. A sequence of r(n) bits will be called block. The r(n) possible blocks are denoted by the corresponding binary integers 0; ; ; P r(n) 0. If = ( ; ; ; r(n) ) and = ( ; ; ; r(n) ) r(n) are blocks, then their inner product is dened = i= i i (mod ). Notice that the inner product of two blocks is a bit. Now we construct a PTM M 0 0 simulating M 0. M 0 0 simulates r(n) M 0 s in parallel. Without loss of generality, we can assume that every computation of M 0 has p(n) choices. The jth choice of ith simulation of M 0 is performed as follows; (Simulation of a probabilistic choice:) generate r(n) bits by a -random source in j ; h (i;j) = j i ; select h (i;j) -choice; Notice that j depends only j. In other words, j is used r(n) times of jth choices on the r(n) simulations. At the end of the simulation, M 0 0 accepts if at least half of r(n) simulations accepts, or rejects otherwise. Let T = f( 0 ; ; ; p(n)0 ) j = 0; ; ; r(n) 0g, and B f0; g p(n) be an arbitrary set with jb j p(n). It was shown in [Pap9, the proof of Theorem.] that Pr[ j T \ B j j T j ] < 8 : This imply that M 0 0 accepts with probability greater than for x L, and it accepts with probability less than for x 6 L. Thus L BPP. Notice that M 0 0 works for every -random source with 0 < <. Intuitively, Theorem shows that the PTM can simulate polynomial number of probabilistic choices using polynomial number of bits generated by an arbitrary -random source with high probability. The following lemma due to Luts can be proved using Cherno Bounds [Che5]. P bt P t 0 t i=bt i Lemma [Lut90] Let h(x; y) be a weighted entropy dened by 0x log y 0 ( 0 x) log( 0 y). Then, 0 t i=0 i a i ( 0 a) t0i 0ct for 0 < b < a <, and a i ( 0 a) t0i 0ct for 0 < a < b <, where c = h(b; a) 0 h(b; b). Results for 8 -PP In this section, we prove the following theorem. Theorem 5 8 -PP = BPP for 0 < <.

5 The probability that a PTM with a -random source accepts, depends not only on the number of the accepting paths, but also on the distribution of the accepting paths in the computation tree of the PTM. To consider the distribution, we dene some notations. A -tree is a rooted full binary tree whose every leaf is labeled \accept" or \reject", and every edge is assigned a probability between and 0. A path to a leaf with label \accept" (\reject") is called an accepting (resp. rejecting) path. For a -tree T, we denote by jt j the number of accepting paths of T. Denition 8 For each with 0, a -assignment F to a -tree is a mapping from the set of edges of the -tree to the interval [; 0 ] such that the two edges leaving each internal node are assigned values whose sum is one. The probability that a PTM accepts is completely determined by the PTM and a -assignment. Denition 9 Let T be a -tree, and F be a -assignment to T. The probability of a node of T under F is the product of the values assigned to the edges on the path from root to the node by F. The probability of T under F, denoted by Pr[T j F ], is the sum of the probabilities of the leaves labeled \accept". For a given -tree T and an assignment F to T, F is a maximum assignment if Pr[T j F ] Pr[T j F 0 ] holds for every -assignment F 0 to T. Proposition 6 For every -tree T, there exists a maximum assignment F to T such that F only assigns or 0 to each edge of T. Proof. For a given -tree T, we recursively dene an assignment F as follows: First, we consider the nodes whose sons are leaves. Let v be such a node. If v has at least one son labeled \accept", then assign ( 0 ) to the edge between v and the son, and assign to the other edge incident to v. Otherwise, assign ( 0 ) and to the two edges incident to v, respectively. Secondly, we consider the nodes whose sons are the root of the subtrees already assigned. Let v be such a node, and T 0 and T be such subtrees. If Pr[T 0 j F ] > Pr[T j F ], then assign ( 0 ) to the edge between v and the root of T 0 and assign to the edge between v and the root of T. Otherwise, assign ( 0 ) to the edge between v and the root of T and assign to the edge between v and the root of T 0. A simple induction on the depth of the tree shows that Pr[T j F ] Pr[T j F 0 ] for every -assignment F 0 to T. This proposition shows that to maximize the probability, it is sucient to consider -assignments which only assign the values or 0. For a given -tree T, we denote by F max (T ) the maximum assignment dened in the proof of the proposition. Denition 0 A -tree T is the worst if Pr[T j F max (T) ] Pr[T 0 j F max (T 0 ) ] for every -tree T 0 with jt j = jt 0 j. Proposition 7 Let T be a worst tree. Then, for every -tree T 0 such that Pr[T j F max (T) ] = Pr[T 0 j F max (T 0 ) ], jt j jt 0 j holds. Proof. There exists a worst tree T 00 such that Pr[T 00 j F max (T 00 ) ] Pr[T 0 j F max (T 0 ) ] with j T 00 j = j T 0 j. Here, Pr[T 00 j F max (T 00 ) ] Pr[T j F max (T ) ] holds for worst trees T 00 and T. Thus, jt j jt 00 j = jt 0 j holds. Next, we consider which -trees can be the worst. From now on, we assume that the two sons of each internal node in a -tree are distinguished as the left and the right sons, respectively. Denition A -tree T is clustered if its rightmost j T j leaves are all labeled \accept". For a given clustered tree T, clearly, the following assignment gives F max (T ): Assign ( 0 ) to every edge incident to a right son, and to every edge incident to a left son. First, we show two lemmas for a clustered tree. Lemma 8 Let T be a clustered tree of depth d. Let t 0 ; t ; ; t k be the integers such that jt j = t k + + t + t 0 with t k > > t > t 0 0, which are uniquely determined by the binary representation of j T j. Then it holds that: Pr[T j F max (T ) ] = kx i=0 i ( 0 ) d0t k0i0i : Proof. For a subtree, its parent is the node whose son is the root of the subtree. From given j T j and a -tree of depth d with no accepting path, we construct a -tree of depth d with jt j accepting paths as follows: 5

6 For k: Let T k be the rightmost subtree of depth t k of the tree with no accepting path. Change all of the label of the leaves of T k from \reject" to \accept". For i (i = k 0 ; k 0 ; ; 0): Let T 0 i+ be the subtree whose parent is as same as T i+. Let T i be the rightmost subtree of depth t i of T 0 i+. (Note that this step works since t i+ > t i.) Change all of the label of the leaves of T i from \reject" to \accept". Since each T i (k i 0) is always taken from rightmost side, we obtain a clustered tree after the construction, and its number of accepting paths is equal to jt j. Thus the constructed tree is the same tree as T. The path from the root of T to the root of T i (k i 0) consists of the path from the root of T to the parent of the root of T k (whose (d 0 t k 0 ) edges are assigned ( 0 )), an edge to the root of T 0 k (which is assigned ), the path to the parent of the root of T k0 (whose (t k 0 t k0 0 ) edges are assigned ( 0 )), an edge to the root of T 0 k0,, and the path to the root of T i. Thus, the probability of the root of T i (k i 0) is given by the product of the probabilities, being equal to i ( 0 ) d0tk0i0i. The constructed clustered tree is a mixture of each T i. Hence the probability of T is given by the sum of every probability of the root of each T i with 0 i k. This implies the lemma. Lemma 9 Let T be a clustered tree of depth d with (s + t) accepting paths with 0 s t. Let T s (resp. T t ) be a clustered tree of depth d 0 with s (resp. t) accepting paths. Let T 0 be a -tree of depth d such that the left son of the root of T 0 is the root of T s, and the right son of the root of T 0 is the root of T t. Then, Pr[T 0 j F max (T 0 ) ] Pr[T j F max (T) ]: Proof. We show the lemma by induction on the depth of the tree. Since it is clear when d = and d =, we assume d >. Let T sl (resp. T tl ) be the subtree rooted at the left son of the root of T s (resp. T t ), and T sr (resp. T tr ) be the subtree rooted at the right son of the root of T s (resp. T t ). We note that Pr[T 0 j F max (T 0 ) ] = Pr[T sl j F max (T 0 ) ] + ( 0 )Pr[T sr j F max (T 0 ) ] + ( 0 )Pr[T tl j F max (T 0 ) ] + ( 0 ) Pr[T tr j F max (T 0 ) ]. Thus the probability of T 0 does not change even if we exchange T sr and T tl. For these four subtrees, four cases arise: Case : jt sl j > 0 and jt tl j = 0. This case is impossible since 0 s t. Case : j T sl j = j T tl j = 0. By exchanging T sr and T tl, we can regard that only T t has accepting paths, where j T tl j = s and j T tr j = t. Thus, by the inductive hypothesis, Pr[T t j F max (T t ) ] Pr[T 00 j F max (T 00 ) ], where T 00 is a clustered tree of depth d 0 with (s + t) accepting paths. Thus the lemma holds. Case : j T sl j > 0 and j T tl j > 0. Since T s and T t are clustered trees, Every path of T sr and T tr is an accepting path. Thus, every path of T t becomes an accepting path by exchanging T sr and T tl. Since j T sl j = s 0 d0 and jt sr j = t 0 d0, by the inductive hypothesis, Pr[T s j F max (T s ) ] Pr[T 000 j F max (T 000 ) ], where T 000 is a clustered tree of depth d 0 with (s + t 0 d0 ) accepting paths. Thus, combining T 000 and T t, we get a clustered tree of depth d with (s + t) accepting paths. This establishes the lemma. Case : jt sl j = 0 and j T tl j > 0. Let T slr (resp. T srr, T tlr ) be the subtree rooted at the right son of the root of T sl (resp. T sr, T tl ), and T sll (resp. T srl, T tll ) be the subtree rooted at the left son of the root of T sl (resp. T sr, T tl ). Here, T slr, T srl and T tll are exchangeable each other, and so are T srr and T tlr. For these four subtrees, four subcases arise: Subcase a: j T srl j = j T tll j = 0. The three edges on the path from the root of T to the root of T srr are assigned, ( 0 ) and ( 0 ), respectively. On the other hand, the three edges on the path from the root of T to the root of T tll are assigned ( 0 ), and, respectively. Thus, even if we exchange T srl and T tll, the probability of T 0 does not increase. Thus the inductive hypothesis for T tl implies the lemma. Subcase b: j T srl j = 0 and j T tll j > 0. First, exchanging T srl and T tll, we get j T srl j > 0, jt srr j > 0 and j T tll j = 0. The inductive hypothesis for T sr shows that the probability of T 0 does not increase even if we replace T sr by a clustered tree with jt sr j accepting paths. If j T srl j = 0 holds, this case is reduced to Subcase a. If j T srl j > 0 holds, then every leaf of T srr is labeled \accept". Since j T tll j = 0 and every leaf of T tlr is labeled \accept", exchanging T srr and T tll and further exchanging T srl and T srr, we see that T 0 is reduced to a clustered tree. Thus, it is sucient to show that these exchanges do not increase the probability. Let p = Pr[T srl j F max (T 0 ) ]. Then the change of the probability is equal to 0( 0 ) + ( 0 ) 0 ( 0 )p + ( 0 ) p = ( 0 )( 0 )(p 0 ), which is less than 0 for 0 < < and 0 < p <. This implies the lemma. Subcase c: jt srl j > 0 and jt tll j = 0. This case is symmetric to Subcase b. 6

7 Subcase d: j T srl j > 0 and j T tll j > 0. Exchanging T slr and T tll, the inductive hypothesis for T s shows that T s can be replaced by a clustered tree with j T s j accepting paths. If j T slr j = 0, then since T srl > 0 and j T tll j = 0 hold, this case is reduced to Subcase b. Otherwise, if jt slr j > 0, then every leaf of T sr is labeled \accept", and jt tll j = 0 holds. Here, exchanging T sr and T tl and further exchanging T slr and T srl, we see that T 0 is reduced to a clustered tree. This implies the lemma. Now, we are ready to show the main lemma in this section. Lemma 0 Every clustered tree is a worst tree. Proof. Let T be a given clustered tree of depth d, and T 0 be any worst tree of depth d with j T j accepting paths. Since T 0 is the worst, it is sucient to show that Pr[T 0 j F max (T 0 ) ] Pr[T j F max (T) ]. Let T 0 l (resp. Tr) 0 be the subtree rooted at the left (resp. right) son of the root of T 0. If either T 0 l or Tr 0 are not the worst, we can improve the probability of T 0 by replacing them. Since this contradicts that T 0 is the worst, both T 0 l and Tr 0 are the worst. By the inductive hypothesis, we can replace T 0 l (resp. T r) 0 by a clustered tree T l (resp. T r ) with j T 0 l j (resp. j T r 0 j ) accepting paths without increasing the probability of T 0. Thus, by Lemma 9, Pr[T 0 j F max (T 0 ) ] Pr[T j F max (T ) ] holds. Remark. A -tree and a -assignment to it correspond to the computation tree of the PTM with a -random source. We consider the PTM M with an arbitrary -random source such that L(M) 8 -PP and the input not in L(M). Then for the -tree T corresponding to the computation tree and every -assignment F to T, Pr[T j F ] < holds. Consequently, Pr[T j F max (T ) ] < holds for the -tree T. Proposition 7 and Lemma 0 show that the maximum number of accepting paths on the computation of M on input not in L(M) is less than the number of accepting paths of a clustered tree T 0 with Pr[T 0 j F max (T 0 ) ]. Finally, we proceed to the proof of the main theorem in this section. Proof of Theorem 5. Theorem shows that 8 -BPP = BPP, and the denition shows that 8 -BPP 8 -PP. Thus BPP 8 -PP holds for 0 <. Thus it is sucient to show that 8 -PP BPP. Let M be a PTM with L(M ) 8 -PP for some, and p(n) be the depth of the computation path of M on input of length n. Fix an x of length n with x 6 L(M ). Let m be a minimum positive integer such that ( 0 )( 0 m ). Since lim m!( 0 )( 0 m ) = 0 >, m must be a constant. Without loss of generality, we assume that p(n) > m +. Here we consider a clustered tree T of depth p(n), with p(n)0 0 p(n)0m0 accepting paths. Here, since p(n)0 0 p(n)0m0 = p(n)0 + p(n)0 + + p(n)0m0, using Lemma 8, Pr[T j F max (T ) ] = 0 ( 0 ) p(n)0(p(n)0)00 + ( 0 ) p(n)0(p(n)0)0 + + m0 ( 0 ) p(n)0(p(n)0m0)0(m0) = ( 0 ) ( m0 ) = ( 0 )( 0 m ). Thus, as mentioned in the remark above, the number of accepting paths of M on input x is less than j T j equal to p(n)0 0 p(n)0m0. Hence if M computes on input x with a perfect random source, M accepts with probability less than or equal to p(n)0 0 p(n)0m0 = p(n) 0. Since m is a constant, by repeating the algorithm enough times, the probability can m+ be improved to a value less than. For inputs in L, the dual arguments work. That is, the dual arguments show that, for a clustered tree T, Pr[T j F min (T) ] Pr[T 0 j F min (T 0 ) ] holds for every -tree T 0 with j T j = j T 0 j. The number of accepting paths on M is greater than the number of accepting paths of a clustered tree T of depth p(n) such that Pr[T j F min (T ) ], and so on. Thus, L BPP. Results for 9 -RP, 9 -BPP, and 9 -PP In this section, we show the result for the classes 9 -RP, 9 -BPP, and 9 -PP. Theorem 9 -RP = NP for 0 < <. Proof. Since 9 -RP NP is shown in Proposition, it is sucient to show that NP 9 -RP. Let M be an NTM with L(M ) NP, and p(n) be the length of M 's computation on input of length n. Let q(n) = d0 log(8(p(n)+)) p e: (0)) We note that q(n) > 0, since log( p ( 0 )) < 0 when 0 < <. We construct a PTM M0 which simulates M, with a specic -random source. M 0 simulates M straightforwardly if M is not in a nondeterministic state. Otherwise, M 0 simulates as follows: When M nondeterministically selects 0-choice (resp. -choice), M 0 assigns ( 0 ) to the probability that the outcome of a coin tossing is 0 (resp. ); tosses a coin q(n) times; and selects i-choice, where i is the majority of the outcomes. log( 7

8 It is clear that M 0 simulates M in time polynomial in n, and M 0 reject x for x 6 L. We consider the probability that P M 0 accepts x for x L. M 0 fails to simulate one nondeterministic choice of M with probability equal to q(n) 0 q(n) P i=0 i q(n)0i ( 0 ) i. By Lemma, since 0 < < ( 0 ) <, we have q(n) 0 q(n) i=0 i q(n)0i ( 0 p ) i q(n) log( (0)) 0 log(8(p(n)+)) =. Thus the probability that 8(p(n)+) M 0 successes in simulating at most p(n) nondeterministic choices of M is P p(n) 0 q(n)0i ( 0 ) i 0 q(n) i=0 0 q(n) i p(n) 8(p(n)+) : n Here, e 0p < 0 n+ p holds for 0 < p < and any positive integer n. Thus, Pr[M 0 accepts x for x L ] p(n) 0 8(p(n)+) > e 0 8 >. Consequently, L 9 -RP. Notice that M 0 works for a specic -random source with 0 < <. Intuitively, Theorem shows that the PTM can simulate polynomial number of nondeterministic choices using polynomial number of bits generated by a specic -random source with high probability. To prove that 9 -BPP = 9 -PP = PSPACE for 0 < <, we need several denitions and lemmas. Denition [Pap9] A probabilistic alternating Turing machine (PATM) is an alternating polynomial time Turing machine M, every computation of which on input x of length n have equal length p(n) for some polynomial p. The computation alternates between states in two disjoint sets, which we call K + and K max. Consider a conguration C in a computation of the PATM M. The acceptance count of conguration C, denoted by AC(C), is dened as follows: If the state of C is an accepting state, then AC(C) = ; if the state of C is a rejecting state, then AC(C) = 0; if the state of C is in K +, then AC(C) = AC(C 0 ) + AC(C ), where C 0 and C are the two successor congurations; and if the state of C is in K max, then AC(C) = maxfac(c 0 ); AC(C )g, where C 0 and C are the two successor congurations. Denition [Pap9] A language L is in ABPP if there exists a PATM M, whose initial conguration is denoted by C I, such that () for x L, AC(C I ) p(n) and () for x 6 L, AC(C I ) p(n), where n is the length of the inputs. Intuitively, a state in K + is a probabilistic state, and a state in K max is a nondeterministic state. For ABPP, the following lemma holds: Lemma [Pap9] ABPP = PSPACE. Now, we consider how to simulate a PTM with a specic -random source by a DTM with polynomial space. When can be represented in constant space (e.g., = ), it is clear that 9 -PP PSPACE. We claim that 9 -PP PSPACE even if is arbitrary. Since it is not essential in this article, the proof of the claim is shown in appendix A. Now we are ready to show the theorem. Theorem 9 -BPP = 9 -PP = PSPACE for 0 < <. Proof. Using Lemma, it is sucient to show that ABPP 9 -BPP for 0 < <, since 9 -BPP 9 -PP holds. Let M be a PATM such that L(M ) ABPP, and p(n) be the length of the computation of M on input x of length n. Without loss of generality, we assume that AC(C I ) 6 6 p(n) for x L, and AC(C I ) 6 p(n) for x 6 L, where C I denotes the acceptance count of the initial conguration of M. On the computation of M, a state pair contains a nondeterministic state and the next probabilistic state. That is, a computation of M contains p(n) state pairs. Here, the PTM M 0 0 constructed in the proof of Theorem simulates probabilistic choices with an arbitrary - random source. On the other hand, the PTM M 0 constructed in the proof of Theorem simulates nondeterministic choices with a specic -random source. By combining M 0 0 and M, 0 we construct a PTM M 0 which simulates M with a specic -random source. Let q(n) = d0 log(8(p(n)+)) log p(n)+6 p e, and r(n) = d log( (0)) 0 e. A block and an inner product are dened in the same way as those in the proof of Theorem for r(n). M 0 simulates r(n) M s in parallel to simulate probabilistic choices. The jth state pair of ith simulation of an M is performed by the following: (Simulation of a nondeterministic choice:) 8

9 () when M nondeterministically selects l-choice (l = 0 or ), assign ( 0 ) to the probability that the outcome of a coin tossing is l; and () select k-choice, where k is the majority of the outcomes of q(n) coin tossing. (Simulation of a probabilistic choice:) () generate r(n) bits by a -random source in j ; () select h (i;j) -choice, where h (i;j) = j i; At the end of the simulation, M 0 accepts if at least half of the r(n) simulations accepts, or rejects otherwise. When x L, as in the proof of Theorem, M 0 successes in simulating p(n) nondeterministic choices with probability greater than e 0=8. In the case, as in the proof of Theorem, the majority of M s rejects with probability less than. Thus 8 M 0 accepts with probability greater than e 0=8 7 8 >. On the other hand, when x 6 L, as in the proof of Theorem, the majority of M s accepts with probability less than for every nondeterministic choices. 8 Consequently, M 0 rejects with probability greater than. Thus L 9 -BPP. In the proof, notice that we can't replace a simulation of a probabilistic choice by a fair coin tossing. Although fair coin tossing works for x L, it does not work for x 6 L, since a probabilistic choice must be simulated for every -random source. 5 Concluding Remarks The outline of the proof of Lemma is the following: Babai introduced \Arthur-Merlin games", and the class AM(poly) dened by the games [Bab85]. An Arthur-Merlin game directly corresponds to the computation of a PATM; an Arthur's (resp. Merlin's) turn corresponds to a state in K + (resp. K max ). Thus it is easy to see that ABPP = AM(poly): On the other hand, Goldwasser, Micali, and Racko introduced Interactive Proof Systems and the class IP dened by the systems [GMR85], and Goldwasser and Sipser showed that IP = AM(poly) [GS86]. Moreover, Shamir showed that PSPACE = IP [Sha90]. Thus ABPP = AM(poly) = IP = PSPACE: An Arthur-Merlin games directly corresponds to a language in ABPP, and we have shown that the language is also in 9 -BPP. The class APP is dened as follows. Denition [Pap9] A language L is in APP if there exists a PATM M, whose initial conguration is denoted by C I, such that () for x L, AC(C I ) > p(n) and () for x 6 L, AC(C I ) < p(n), where n is the length of the inputs. We note that in the same way, a \game against Nature" introduced by Papadimitriou [Pap8] directly corresponds to a language in APP, and we can show that the language is also in 9 -PP. As noted in section, 9 -PP PSPACE = ABPP holds. In fact, we can show directly that 9 -BPP ABPP, using the gadget in Figure. Although the gadget in the gure only works for =, Knuth and Yao showed that for an arbitrarily rational number, we can construct such a gadget [KY76]. On the other hand, we have shown that ABPP 9 -BPP indirectly. In this sense, roughly speaking, 9 -PP looks like lower than ABPP. Arthur-Merlin games and games against Nature have alternations. In other words, they are represented by Turing machines which have probabilistic states and nondeterministic states, or by quantied Boolean expressions which have \random" quantiers and existential quantiers (e.g., see SSAT in [Pap9]). Alternations disappear if we use the semi-random sources. For instance, we can dene a \-random" quantiers and construct a kind of satisability problem, being PSPACE-complete, and has only \-random" quantiers. In the two points noticed above, the result that 9 -BPP = 9 -PP = PSPACE is an extension of the results in a series of the research for interactive proof systems. Condon and Ladner investigated interactive proof systems with some restrictions on provers, and showed that some of such systems characterize the classes MA and NP [CL95]. We dene an interactive proof system with a restriction on the condition of the probability that system accepts. Denition 5 A language L is is restricted-am(poly) if there exists a PATM M, whose initial condition is denoted by C I, such that () for x L, AC(C I ) p(n) and () for x 6 L, AC(C I ) = 0, where n is the length of the inputs. 9

10 e C C C CC 9 -PP or e CWe e e CW e C C C CC - u u PATM A A AU A Q AA A AA AU AU + QQQQs AU u nondeterministic state probabilistic state Figure : gadget for with = It is easy to see that restricted-am(poly) NP holds. We can show that 9 -RP restricted-am(poly), using the gadget shown in Figure. This implies the following corollary: Corollary restricted-am(poly) = NP. Acknowledgments I would like to thank Seinosuke Toda, Zhi-Zhong Chen, and Misao Nagayama for many helpful discussions and suggestions. References [Bab85] L. Babai. Trading Group Theory for Randomness. In Proc. 7th ACM Symp. on the Theory of Computing, pages {9. ACM, 985. [BDG88] J.L. Balcazar, J. Daz, and J. Gabarro. Structural Complexity I. Springer-Verlag, 988. [Che5] [CL95] H. Cherno. A Measure of Asymptotic Eciency for Tests of a Hypothesis Based on the Sum of Observations. Ann. of Math. Statist., :9{509, 95. A. Condon and R. Ladner. Interactive Proof Systems with Polynomially Bounded Strategies. Journal of Computer and System Science, 50:506{58, 995. [Gil77] J. Gill. Computational Complexity of Probabilistic Turing Machines. SIAM J. Comput., 6():675{695, Dec [GMR85] S. Goldwasser, S. Micali, and C. Racko. The Knowledge Complexity of Interactive Proof-Systems. In Proc. 7th ACM Symp. on the Theory of Computing, pages 9{0. ACM, 985. [GS86] [HU79] [KY76] S. Goldwasser and M. Sipser. Private Coins versus Public Coins in Interactive Proof Systems. In Proc. 8th ACM Symp. on the Theory of Computing, pages 59{68. ACM, 986. J.E. Hopcroft and J.D. Ullman. Introduction to Automata Theory, Languages, and Computation. Addison- Wesley Publishing Company, 979. D.E. Knuth and A.C. Yao. The complexity of nonuniform random number generation. In J.F.Traub, editor, Algorithms and Complexity, New Directions and Results, pages 57{8. Academic, New York,

11 [Lut90] J.H. Lutz. Pseudorandom Sources for BPP. Journal of Computer and System Science, :07{0, 990. [Pap8] C.H. Papadimitriou. Games Against Nature. In Proc. th Symp. on Foundations of Computer Science, pages 6{50. IEEE, 98. [Pap9] C.H. Papadimitriou. Computational Complexity. Addison-Wesley Publishing Company, 99. [Sha90] [SV8] A. Shamir. IP=PSPACE. In Proc. st Symp. on Foundations of Computer Science, pages {5. IEEE, 990. M.S. Satha and U.V. Vazirani. Generating Quasi-random Sequences from Slightly random Sources. In Proc. 5th Symp. on Foundations of Computer Science, pages {0. IEEE, 98. [SV86] M.S. Satha and U.V. Vazirani. Generating Quasi-random Sequences from Semi-random Sounces. Journal of Computer and System Science, :75{87, 986. [Vaz86] U. Vazirani. Randomness, Adversaries and Computation. PhD thesis, University of California, Berkeley, 986. [VV85] U.V. Vazirani and V.V. Vazirani. Random Polynomial Time is Equal to Slightly-random Polynomial Time. In Proc. 6th Symp. on Foundations of Computer Science, pages 7{8. IEEE, 985. A Proof for 9 -PP PSPACE We show the following lemma to deal with arbitrary. Lemma 5 Let L be a language with L 9 -PP for some. Then there exists a number 0 such that; L 9 0 -PP and 0 can be represented in polynomial space for the input length. Proof. Let L be a language with L 9 -PP for some, and M be a PTM such that L(M ) = L. Let p be the length of a computation of M. (We write only p, depending on the input length, for short.) Let d = p. p+ p(0) p0 Since d can be represented in polynomial space for the input length, there exists some 0 with j 0 0 j such that 0 can be represented in polynomial space for the input length. Without loss of generality, we assume that 0 >. We simulate M in polynomial space by substituting 0 for. Then the error of the probability of a leaf by substituting is equal to i ( 0 ) p0i 0 0i ( 0 0 ) p0i for some i with 0 i p. If the sum of every error of the probability of the leaf is less than half of the probability of any leaf, L(M ) is also in 9 0 -PP. Here, the minimum probability of a leaf is equal to p. On the other hand, an error of the probability of a leaf is at most maxf 0p 0 p ; ( 0 ) p 0 ( 0 0 ) p g. Two cases arise. Case : 0p 0 p < ( 0 ) p 0 ( 0 0 ) p. Since M has p leaves, the sum of every error is at most Using Taylor series, we get p j i ( 0 ) p0i 0 0i ( 0 0 ) p0i j < p (( 0 ) p 0 ( 0 0 ) p ) : p (( 0 ) p 0 ( 0 0 ) p ) < p pd( 0 ) p0 = p : Case : 0p 0 p < ( 0 ) p 0 ( 0 0 ) p. The sum of every error is at most p j i ( 0 ) p0i 0 0i ( 0 0 ) p0i j < p ( 0p 0 p ) < p pd p0 = p0 p 0 < p : In each case, the sum of every error of the probability of any computation tree is less than a half of the probability of any leaf. This implies the lemma. We show the main lemma in this section: Lemma 6 For arbitrary with 0 < <, 9 -PP PSPACE: Proof. Let L be a language with L 9 -PP for some. Let M 5 be a PTM, such that L = L(M 5 ). Let 0 be a number satisfying the conditions in the proof of Lemma 5. We construct an NTM M 0 5, which accepts L as follows;

12 (i) nondeterministically compute 0 ; (ii) simulate all computations of M 5, and counts up its probability by using 0 instead of ; and (iv) accept if the probability is greater than ; or reject otherwise. Clearly, M 0 5 uses at most polynomial space for the input length, and L = L(M 0 5). Thus L PSPACE.