A Master Theorem for Discrete Divide and Conquer Recurrences

Transcription

1 A Master Theorem for Discrete Divide and Conquer Recurrences Wojciech Szpankowski Department of Computer Science Purdue University W. Lafayette, IN January 20, 2011 NSF CSoI SODA, 2011 Research supported by NSF Science & Technology Center, and Humboldt Foundation.

2 Outline 1. Divide and Conquer 2. Example: Boncelet s Algorithm 3. Continuous Relaxation of the Recurrence 4. Master Theorem 5. Examples 6. Sketch of Proof.

3 Divide and Conquer Divide and Conquer: A divide and conquer algorithm splits the input into several smaller subproblems, solving each subproblem separately, and then knitting together to solve the original problem. Complexity: A problem of size n is divided into m 2 subproblems of size p j n + δ j and p j n+δ j and each subproblem contributesb j,b j fraction to the final solution; there is a cost a n associated with combining subproblems. Total Cost: The total cost T(n) satisfies the discrete divide and conquer recurrence: T(n) = a n + m b j T ( p j n + δ j ) + m ) b j ( p T j n + δ j (n 2) j=1 j=1 where 0 p j < 1 (e.g., m i=1 p i = 1).

4 Example: Boncelet s Algorithm Boncelet s algorithm is a variable-to-fixed data compression scheme: 1. A variable-to-fixed length encoder partitions a source string over an m- ary alphabet into variable-length phrases. 2. Each phrase belongs to a given dictionary. 3. A dictionary is represented by a complete parsing tree. 4. The dictionary entries correspond to the leaves of the parsing tree.

5 Example: Boncelet s Algorithm Boncelet s algorithm is a variable-to-fixed data compression scheme: 1. A variable-to-fixed length encoder partitions a source string over an m- ary alphabet into variable-length phrases. 2. Each phrase belongs to a given dictionary. 3. A dictionary is represented by a complete parsing tree. 4. The dictionary entries correspond to the leaves of the parsing tree. Example: A binary string (m = 2) with symbol probabilities p 1 and p 2. The expected phrase length d(n) satisfies: d(n) = 1 + p 1 d( p 1 n + δ ) + p 2 d( p 2 n δ )

6 Continuous Relaxation We relax the discrete nature of the recurrence and consider a continuous version: m T(x) = a(x) + b j T(p j x)), x > 1, b j = 0. j=1 Akra and Bazzi (1998) proved that T(x) = Θ ( ( x x s where s 0 is a unique real root of j b jp j s 0 = 1. )) a(u) u s 0 +1du

7 Continuous Relaxation We relax the discrete nature of the recurrence and consider a continuous version: m T(x) = a(x) + b j T(p j x)), x > 1, b j = 0. j=1 Akra and Bazzi (1998) proved that ( T(x) = Θ x s 0 ( x where s 0 is a unique real root of j b jp j s 0 = 1. )) a(u) u s 0 +1du Indeed, by taking Mellin transform of the relaxed recurrence: t(s) = 0 T(x)x s 1 dx we find (for some a(s) and g(s)) t(s) = a(s) + g(s) 1 m j=1 b jp s. i An application of the Wiener-Ikehara theorem leads to T(x) Cx s 0 with C = a( s 0) + g( s 0 ) j b jp s 0 j log(1/p j ).

8 Discrete Divide & Conquer Recurrence by Dirichlet Series For a sequence c(n) define the Dirichlet series as C(s) = n=1 c(n) n s provided it exists for R(s) > σ c for some σ c. Theorem 1 (Perron-Mellin Formula). For all σ > σ c and all x > 0 n<x c(n) + c( x ) [[x Z]] = lim 2 T 1 2πi σ+it σ it C(s) xs s ds. where [[P]] is 1 if P is a true proposition and 0 otherwise.

9 Discrete Divide & Conquer Recurrence by Dirichlet Series For a sequence c(n) define the Dirichlet series as C(s) = n=1 c(n) n s provided it exists for R(s) > σ c for some σ c. Theorem 1 (Perron-Mellin Formula). For all σ > σ c and all x > 0 n<x c(n) + c( x ) [[x Z]] = lim 2 T 1 2πi σ+it σ it C(s) xs s ds. where [[P]] is 1 if P is a true proposition and 0 otherwise. Example: Define c(n) = T(n + 2) T(n + 1). Then T(n) = T(2) + lim T 1 2πi c+it c it T(s) (n 3 2 )s s ds for some c > σ T with T(s) = n=1 T(n + 2) T(n + 1) n s. where R(s) > σ T.

10 Assumptions Let a n be a nondecreasing sequence. Define Ã(s) = n=1 a n+2 a n+1 n s which is postulated to exists for R(s) > σ a. Example. Define a n = n σ (logn) α. Then Γ(α + 1) Γ(α + 1) Ã(s) = σ + (s σ) α+1 (s σ) + F(s), α where F(s) is analytic for R(s) > σ 1 and Γ(s) is the gamma function. Define s 0 to be the unique real root of m (b j + b j )p j s = 1. j=1 Other zeros depend on the relation among log(1/p 1 ),...,log(1/p m ).

11 Rationally and Irrationally Related Numbers Definition 1. (i) log(1/p 1 ),...,log(1/p m ) are rationally related if log(1/p 1 ),...,log(1/p m ) are integer multiples of L, that is, log(1/p j ) = n j L, n j Z, (1 j m). (ii) Otherwise log(1/p 1 ),...,log(1/p m ) are irrationally related. Example. If m = 1, then we are always in the rationally related case. For m = 2, if log(1/p 1 )/log(1/p 2 ) = m/n, (m,n integers), then rationally related. Lemma 1. (i) If log(1/p 1 ),...,log(1/p m ) are irrationally related, then s 0 is the only solution on R(s) = s 0. (ii) If log(1/p 1 ),...,log(1/p m ) are rationally related, then there are infinitely many solutions s k = s 0 + 2πik L (k Z) where log(1/p j ) are all integer multiples of L.

12 Evaluation of T(n): A Bird View We estimate T(n) by the Cauchy residue theorem. T(s) = Ã(s)+B(s) 1 m j=1 (b j +b j )ps j, T(n) = 1 2πi c+i T(s) (n 3 2 )s c i s ds

13 Main Master Theorem Theorem 2 (DISCRETE MASTER THEOREM). Let a n = Cn σa (logn) α with min{σ,α} 0. (i) If log(1/p 1 ),...,log(1/p m ) are irrationally related, then T(n) = C 1 + o(1) if σ a 0 and s 0 < 0, C 2 logn + C 2 + o(1) if σ a < s 0 = 0, C 3 (logn) α+1 (1 + +o(1)) if σ a = s 0 = 0 C 4 n s 0 (1 + o(1)) if σ a < s 0 and s 0 > 0, C 5 n s 0(logn) α+1 (1 + o(1)) if σ a = s 0 > 0 and α 1, C 5 n s 0loglogn (1 + o(1)) if σ a = s 0 > 0 and α = 1, C 6 (logn) α (1 + o(1)) if σ a = 0 and s 0 < 0, C 7 n σa (logn) α (1 + o(1)) if σ a > s 0 and σ a > 0. (ii) If log(1/p 1 ),...,log(1/p m ) are rationally related, then T(n) behaves as in the irrationally related case with the following two exceptions: T(n) = { C2 logn + Ψ 2 (logn) + o(1) if σ a < s 0 = 0, Ψ 4 (logn)n s 0 (1 + o(1)) if σ a < s 0 and s 0 > 0, where C 2 is positive and Ψ 2 (t),ψ 4 (t) are periodic functions with period L (with usually countably many discontinuities).

14 Extensions and Remarks 1. We can handle any a n sequence with Dirichlet series Ã(s): Ã(s) = g 0 (s) ( ) β0 log 1 s σa (s σ a ) α 0 + ( ) βj J log 1 s σa g j (s) (s σ a ) α j j=1 + F(s), F(s) is analytic, g 0 (σ a ) 0, β j non-negative integers, and α 0 real. Then the solutiont(n) of the divide and conquer recurrence has the form: T(n) C n σ (logn) α (loglogn) β or T(n) Ψ(logn)n s 0 σ = max{σ,s 0 }), depending whether logp 1,...logp m are irrationally or rationally related. 2. The periodic function Ψ(t) has the following building blocks λ t n 1 λ B n t logn L λ 1 +1 where λ > 1 andb n is such that n 1 B nλ (logn)/l converges absolutely. This function is discontinuous at t = {logn/l}, where{x} = x x denotes the fractional part of a real number x.

15 Examples Example 1. Irrationally Related; Case 4: T(n) = 2T( n/2 ) + 3T( n/6 ) + nlogn Here σ a = 1 since a n = nlogn. The equation 2 2 s s = 1 has the (real) solution s 0 = > 1, and finally log(1/2)/log(1/6) are irrationally related. Thus by our Master Theorem Case 4 for some constant C > 0 T(n) Cn s 0

16 Examples Example 2. Irrationally Related; Case 6: Consider the recurrence T(n) = 2T( n/2 ) + 8 n2 T( 3n/4 ) + 9 logn. Here σ a = s 0 = 2, and we deal with irrationally related case. Furthermore, Ã(s) = slog 1 s 2 + G(s) for G(s) analytic for R(s) > 1. By Master Theorem Case 6 T(n) Cn 2 loglogn. Example 3. Rationally Related (m = 1); Case 3: Next consider T(n) = T( n/2 ) + logn. Here σ a = s 0 = 0, and we have rational case (m = 1). Since Ã(s) = 1 s + G(s) we conclude T(n) C(logn) 2.

17 Examples Example 4: Karatsuba algorithm: Rationally Related (m = 1): T(n) = 3T( n/2 ) + n Here, s 0 = (log3)/(log2) = and s 0 > σ a = 1. Thus for some periodic function Ψ(t). T(n) = Ψ(logn)n log3 log2 (1 + o(1))

18 Examples Example 5. Rationally Related (m = 1). The recurrence T(n) = 1 2 T( n/2 ) + 1 n is not covered by our Master Theorem but our methodology still works. Here σ a = s 0 = 1 < 0. It follows that T(n) = C logn n + Ψ(logn) ( ) 1 + o n n for a periodic function Ψ(t). Example 6: Mergesort. Rationally Related. The mergesort recurrences are T(n) = T( n/2 ) + T( n/2 ) + n 1, Y (n) = Y( n/2 ) + Y ( n/2 ) + n/2. Here σ a = s 0 = 1 and we deal with the rationally related case. By our Master Theorem (cf. Flajolet & Golin, 1994) T(n) = Y (n) = 1 nlogn + nψ(logn) + o(n), log2 1 nlogn + nψ(logn) + o(n). 2log2

19 Boncelet s Algorithm Revisited Let a sequence X be generated by a memoryless source over alphabet A of size m with symbol probabilities p i, i A. Using the Boncelet s parsing tree, we parse X into phrases {v 1,...v n } of length l(v 1 ),..., l(v n ) with phrase probabilities P(v 1 ),...,P(v n ). Phrase Length and its Probability Generating Function: Let D n denote the phrase length and define the probability generating function as C(n,y) = E[y Dn ] It satisfies the following discrete divide and conquer recurrence: C(n,y) = y m p i C([p i n + δ i ],y) i=1 The expected phrase length d(n) = E[D n ] = C (n,1) satisfies the following discrete divide and conquer recurrence: d(n) = 1 + m p i d([p i n + δ i ]) i=1 with d(0) = = d(m 1) = 0.

20 Main Results for Boncelet s Algorithm Theorem 3. Consider anm-ary memoryless source with probabilitiesp i > 0 and the entropy rate H = m i=1 p ilog(1/p i ). (i) If log(1/p 1 ),...log(1/p m ) are irrationally related, then where d(n) = 1 H logn α H + o(1), α = E (0) H H 2 2H, H 2 = m i=1 p ilog 2 p i, and E (0) is the derivative at s = 0 of a Dirichlet series E(s) arises from the discrete nature of the recurrence. (ii) If log(1/p 1 ),...log(1/p m ) are rationally related, then d(n) = 1 H logn α + Ψ(logn) H + O(n η ) for some η > 0, where Ψ(t) is a periodic function of bounded variation that has usually an infinite number of discontinuities.

21 Redundancy of the Boncelet s Algorithm Corollary 1. Let R n denote the redundancy of the Boncelet code: R n = logn E[D n ] H = logn d(n) H. (i) If log(1/p 1 ),...log(1/p m ) are irrationally related, then R n = Hα ( ) 1 logn + o. logn (ii) If log(1/p 1 ),...log(1/p m ) are rationally related, then ( ) Hα + Ψ(logn) 1 R n = + o. logn logn Tunstall Code Redundancy: R T n = H logn ( logh H 2 2H ) ( ) 1 + o logn for irrational case; in the rational case there is aperiodic function. Example. Consider p = 1/3 and q = 2/3. Then one computes α = E (0) H H 2 2H while for the Tunstall code logh H 2 2H

22 Limiting Distribution for the Phrase length Theorem 4. Consider a memoryless source generating a sequence of length n parsed by the Boncelet algorithm. If (p 1,...,p m ) is not the uniform distribution, then the phrase length D n satisfies the central limit law, that is, D n 1 H logn (H2 ) N(0,1), logn H 3 1 H where N(0,1) denotes the standard normal distribution, and VarD n E[D n ] = logn H + O(1), H 1 ) logn 3 H ( H2 for n.

23 Sketch of Proof 1. From the recurrence we have T(s) m = Ã(s) + b j j=1 n=1 T ( p j (n + 2) + δ j ) T ( p j (n + 1) + δ j ) n s. But defining k + 2 δj n = for some integer k, we arrive at p j 2 n=1 T ( p j (n + 2) + δ j ) T ( p j (n + 1) + δ j ) n s = G j (s)+ k=1 T(k + 2) T(k + 1) ( k+2 δj ) s. p 2 j for an explicit (and simple) analytic function G j (s).

24 Sketch of Proof Continuation 2. We now compare the last sum to p s j T(s) and obtain k=1 T(k + 2) T(k + 1) ( k+2 δj ) s = p 2 j k=1 T(k + 2) T(k + 1) (k/p j ) s = p s j T(s) E j (s), where E j (s) = (T(k + 2) T(k + 1)) 1 (k/p j ) 1 ( s k+2 δj p 2 j k=1 ) s. 3. Defining E(s) = m b j E j (s) and G(s) = j=1 m b j G j (s) j=1 we finally obtain our final formula T(s) = Ã(s) + G(s) E(s) 1 m j=1 b. j p s j

25 Sketch of Proof Binary Boncelet s Algorithm 1. Let s 0 (y) be the real zero of y(p s+1 + q s+1 ) = 1, q = 1 p. 2. Define C(s,y) = n=1 which from the basic recurrence becomes C(n + 2,y) C(n + 1,y) n s. C(s,y) = (y 1) E(s,y) 1 y(p s+1 + q s+1 ), where E(s,y) converges for R(s) > s 0 (y) 1 and satisfies E(0,y) = 0 and E(s,1) = By Mellin-Perron formula and residue theorem we can prove that C(n,y) = (1 + O(y 1))n s 0 (y) (1 + o(1)) where s 0 (y) = y 1 ( H + H2 2H 1 ) (y 1) 2 + O((y 1) 3 ). 3 H

26 That s It THANK YOU