1 if n n 0 t n a t n b n k if n n 0
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( 1 ) if n < n 0 , T ( n ) = a T ( n/b ) + ( n k ) if n n 0 . - PowerPoint PPT Presentation

Jan 24, 2023 •329 likes •499 views

The Master Theorem for solving recurrences Theorem 3.1 Let n 0 N , k N 0 and a, b R with a > 0 and b > 1 , and let T : N R satisfy the following recurrence: ( 1 ) if n < n 0 , T ( n ) = a T ( n/b ) + ( n k )

  1. The Master Theorem for solving recurrences Theorem 3.1 Let n 0 ∈ N , k ∈ N 0 and a, b ∈ R with a > 0 and b > 1 , and let T : N → R satisfy the following recurrence: � Θ ( 1 ) if n < n 0 , T ( n ) = a · T ( n/b ) + Θ ( n k ) if n ≥ n 0 . Let c = log b ( a ) ; we call c the critical exponent . Then  Θ ( n c ) if k < c (I) ,   Θ ( n c · lg ( n )) T ( n ) = if k = c (II) ,   Θ ( n k ) if k > c (III) . The n/b in the recurrence may stand for both ⌊ n/b ⌋ and ⌈ n/b ⌉ . More precisely, the theorem holds if we replace a · T ( n/b ) in the recurrence by a 1 · T ( ⌊ n/b ⌋ ) + a 2 · T ( ⌈ n/b ⌉ ) for any a 1 , a 2 ≥ 0 with a 1 + a 2 = a . 1 A&DS Lecture 3 Mary Cryan

  2. The Master Theorem (cont’d) • We don’t have time to prove the Master Theorem in class. You can find the proof in Section 4.4 of [CLRS]. Section 4.4 of [CLR] . Their version of the M.T. is a bit more general than ours. • Homework: To get a feel for the Master Theorem, consider the following examples: T ( n ) = 4T ( n/2 ) + n, 4T ( ⌊ n/2 ⌋ ) + n 2 , T ( n ) = 4T ( n/2 ) + n 3 . T ( n ) = Use unfold-and-sum to answer the first and third of these. We solved the second one by first principles in lecture 2, and it hurt! (mostly because of the ⌊ ⌋ ). 2 A&DS Lecture 3 Mary Cryan

  3. Matrix Multiplication Recall The product of two ( n × n ) -matrices A = ( a ij ) 1 ≤ i,j ≤ n B = ( b ij ) 1 ≤ i,j ≤ n and is the ( n × n ) -matrix C = AB where C = ( c ij ) 1 ≤ i,j ≤ n with entries n � c ij = a ik b kj . k = 1 The Matrix Multiplication Problem Input: ( n × n ) -matrices A and B Output: the ( n × n ) -matrix AB 3 A&DS Lecture 3 Mary Cryan

  4. Matrix Multiplication b 1j b 2j c ij a a a row i = i1 i2 in b nj column j - n multiplications and n additions for each c ij . - there are n 2 different c ij entries. 4 A&DS Lecture 3 Mary Cryan

  5. A straightforward algorithm Algorithm M AT M ULT ( A, B ) 1. n ← number of rows of A 2. for i ← 1 to n do for j ← 1 to n do 3. c ij ← 0 4. for k ← 1 to n do 5. c ij ← c ij + a ik · b kj 6. 7. return C = ( c ij ) 1 ≤ i,j ≤ n Requires Θ ( n 3 ) arithmetic operations (additions and multiplications). 5 A&DS Lecture 3 Mary Cryan

  6. A naive divide-and-conquer algorithm Observe If � � � � A 11 A 12 B 11 B 12 A = B = and A 21 A 22 B 21 B 22 for ( n/2 × n/2 ) -submatrices A ij and B ij then � � A 11 B 11 + A 12 B 21 A 11 B 12 + A 12 B 22 AB = A 21 B 11 + A 22 B 21 A 21 B 12 + A 22 B 22 note: We are assuming n is even. 6 A&DS Lecture 3 Mary Cryan

  7. A naive divide-and-conquer algorithm b A A B 1j B 11 12 11 12 b 2j c ij a a a row i = i1 i2 in A B B A 21 22 22 21 b nj column j Suppose i ≤ n/2 and j ≤ n/2 . Then n/2 n n � � � c ij = a ik b kj = a ik b kj + a ik b kj k = 1 k = 1 k = n/2 + 1 ∈ A 11 B 11 ∈ A 12 B 21 7 A&DS Lecture 3 Mary Cryan

  8. A naive divide-and-conquer algorithm (cont’d) Assume n is a power of 2 . Algorithm D&C-M AT M ULT ( A,B ) 1. n ← number of rows of A 2. if n = 1 then return ( a 11 b 11 ) 3. else Let A ij , B ij (for i,j = 1,2 be ( n/2 × n/2 ) -submatrices such that 4. ! ! A 11 A 12 B 11 B 12 A = and B = A 21 A 22 B 21 B 22 Recursively compute A 11 B 11 , A 12 B 21 , A 11 B 12 , A 12 B 22 , 5. A 21 B 11 , A 22 B 21 , A 21 B 12 , A 22 B 22 Compute C 11 = A 11 B 11 + A 12 B 21 , C 12 = A 11 B 12 + A 12 B 22 , 6. C 21 = A 21 B 11 + A 22 B 21 , C 22 = A 21 B 12 + A 22 B 22 ! C 11 C 12 return 7. C 21 C 22

  9. Analysis of D&C-M AT M ULT T ( n ) is the number of operations done by D&C-M AT M ULT . • Lines 1, 2, 3, 4, 7 require Θ ( 1 ) arithmetic operations • Line 5 requires 8T ( n/2 ) arithmetic operations • Line 6 requires 4 ( n/2 ) 2 = Θ ( n 2 ) arithmetic operations. Remember! Size of matrices is Θ ( n 2 ) , NOT Θ ( n ) We get the recurrence T ( n ) = 8T ( n/2 ) + Θ ( n 2 ) . Since log 2 ( 8 ) = 3 , the Master Theorem yields T ( n ) = Θ ( n 3 ) . (No improvement over M AT M ULT . . . why?) 9 A&DS Lecture 3 Mary Cryan

  10. Strassen’s algorithm (1969) Assume n is a power of 2 . Let � � � � A 11 A 12 B 11 B 12 A = B = . and A 21 A 22 B 21 B 22 We want to compute � � A 11 B 11 + A 12 B 21 A 11 B 12 + A 12 B 22 = AB A 21 B 11 + A 22 B 21 A 21 B 12 + A 22 B 22 � � C 11 C 12 = . C 21 C 22 Strassen’s algorithm uses a trick in applying Divide-and-Conquer. 10 A&DS Lecture 3 Mary Cryan

  11. Strassen’s algorithm (cont’d) Let P 1 = ( A 11 + A 22 )( B 11 + B 22 ) P 2 = ( A 21 + A 22 ) B 11 = A 11 ( B 12 − B 22 ) P 3 ( ∗ ) P 4 = A 22 (− B 11 + B 21 ) P 5 = ( A 11 + A 12 ) B 22 = (− A 11 + A 21 )( B 11 + B 12 ) P 6 P 7 = ( A 12 − A 22 )( B 21 + B 22 ) Then C 11 = P 1 + P 4 − P 5 + P 7 C 12 = P 3 + P 5 C 21 = P 2 + P 4 C 22 = P 1 + P 3 − P 2 + P 6 ( ∗∗ ) 11 A&DS Lecture 3 Mary Cryan

  12. Checking Strassen’s algorithm - C11 We will check the equation for C 11 is correct. Strassen’s algorithm computes C 11 = P1 + P4 − P5 + P7 . We have P1 = ( A11 + A22 )( B11 + B22 ) = A11B11 + A11B22 + A22B11 + A22B22. P4 = A22 (− B11 + B21 ) = A22B21 − A22B11. P5 = ( A11 + A12 ) B22 = A11B22 + A12B22. P7 = ( A12 − A22 )( B21 + B22 ) = A12B21 + A12B22 − A22B21 − A22B22. Then P1 + P4 = A11B11 + A11B22 + A22B22 + A22B21 . Then P1 + P4 − P5 = A11B11 + A22B22 + A22B21 − A12B22 . Then P1 + P4 − P5 + P7 = A11B11 + A12B21 , which is C11 . Class exercise: check other 3 equations. 11-1

  13. Strassen’s algorithm (cont’d) Crucial Observation 7 7 multiplications of ( n/2 × n/2 ) -matrices are needed to Only 7 compute AB . Algorithm S TRASSEN ( A, B ) 1. n ← number of rows of A 2. if n = 1 then return ( a 11 b 11 ) 3. else Determine A ij and B ij for i, j = 1, 2 (as before) 4. Compute P 1 , . . . , P 7 as in ( ∗ ) 5. Compute C 11 , C 12 , C 21 , C 22 as in ( ∗∗ ) 6. � � C 11 C 12 return 7. C 21 C 22 12 A&DS Lecture 3 Mary Cryan

  14. Analysis of Strassen’s algorithm Let T ( n ) be the number of arithmetic operations performed by S TRASSEN . • Lines 1 − 4 and 7 require Θ ( 1 ) arithmetic operations • Line 5 requires 7T ( n/2 ) + Θ ( n 2 ) arithmetic operations • Line 6 requires Θ ( n 2 ) arithmetic operations. remember. We get the recurrence T ( n ) = 7T ( n/2 ) + Θ ( n 2 ) . Since log 2 ( 7 ) ≈ 2.807 > 2 , the Master Theorem yields T ( n ) = Θ ( n log 2 ( 7 ) ) . 13 A&DS Lecture 3 Mary Cryan

  15. Remarks on matrix multiplication • The current best (for asymptotic running time) algorithm is by Coppersmith & Winograd (1987), and has a running time of Θ ( n 2.376 ) . • In practice, the “school” M AT M ULT algorithm tends to outperform Strassen’s algorithm, unless the matrices are huge. • The best known lower bound for matrix multiplication is Ω ( n 2 ) . This is a trivial lower bound (need to look at all entries of each matrix). Amazingly, Ω ( n 2 ) is believed to be “the truth”! Open problem: Can we find a O ( n 2 + o ( 1 ) ) -algorithm for Matrix Multiplication of n × n matrices? 14 A&DS Lecture 3 Mary Cryan

  16. Reading Assignment [CLRS] Section 4.3 “Using the Master method” (pp. 73-75) and Section 28.2 (pp. 735-741). Corresponds to Section 4.3 (pp. 61-63) and Section 31.2 (pp. 739-745) in [CLR]. See Links from course webpage (for history). Problems 1. Exercise 4.3-2, p. 75 of [CLRS]. Ex 4.3-2, page 64 of [CLR]. 2. Exercise 28.2-1, p. 741 of [CLRS]. Ex 31.2-1, page 744 of [CLR]. 3. On page 5, I state that the “school” algorithm M AT M ULT has running time Θ ( n 3 ) . The O ( n 3 ) is fairly easy to see (I think! If not, ask me). Show the Ω ( n 3 ) bound for M AT M ULT . 15 A&DS Lecture 3 Mary Cryan

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