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Recurrence Equations

Recurrence Equations. The running time analysis of recursive functions leads naturally to recurrence equations A recurrence equation defines the value of an integer function at input n in terms of values of the function on previous values

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Recurrence Equations

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  1. Recurrence Equations • The running time analysis of recursive functions leads naturally to recurrence equations • A recurrence equation defines the value of an integer function at input n in terms of values of the function on previous values • The specification must also indicate the specific values at small inputs • Example 1: • f(1) = f(2) = 1 (initial conditions) • f(n) = f(n-1) + f(n-2) for all n > 2 • Of course, the above recurrence equation defines the sequence of Fibonacci numbers

  2. Recurrence Equations: Example 2 • Example 2 • Find an exact solution to the following recurrence equation: • Initial condition: c(0) = 0 • Recurrence: c(n) = c(n-1) + n for all n > 0 • To solve by iteration, we repeatedly replace terms with their recurrence expression until the solution is obvious • In principle we should then prove by induction that the solution is valid • c(n) = c(n-1) + n = c(n-2) + n-1 + n = c(n-3) + n-2 + (n-1) + n  = 1 + 2 +  + (n-1) + n = n(n+1)/2

  3. Recurrence Equations: Example 2 • Example 2 • Find an exact solution to the following recurrence equation: • Initial condition: c(0) = 0 • Recurrence: c(n) = c(n-1) + n for all n > 0 • Claim: c(n) = n(n+1)/2 • Proof by induction on n: • Basis (n = 0): c(0) = 0 = 0(0+1)/2 • Inductive Step: Assume that c(k) = k(k+1)/2 and prove c(k+1) = (k+1)((k+1)+1)/2 • c(k+1) = c(k) + k+1 = k(k+1)/2 + (k+1) inductive hypothesis = (k/2 + 1)(k+1) = (k/2+2/2)(k+1) = ((k+2)/2)(k+1) = (k+1)(k+2)/2 = (k+1)((k+1)+1)/2 done

  4. Recurrence Equations: Example 3 • Let c(n) denote the number of times the instruction x = x+1 is executed by the following algorithm: • example(n) { if (n == 1) return for i = 1 to n x = x+ 1 example(n/2) } • Initial condition: c(1) = 0, since x = x+1 is never executed • Recurrence: c(n) = c(n/2) + n for n > 1. • Problem: how to deal with the n/2 • Answer: first prove that c is a nondecreasing function • Then solve for n a power of 2 • Then use the fact that the function is increasing to solve for other values Execution count: n Execution count: c(n/2) // Note: n/2 means integer division, n/2

  5. Recurrence Equations: Example 3 • Initial condition: c(1) = 0 • Recurrence: c(n) = c(n/2) + n for i > 1 • Claim: c(n)  0 for all n  1 The proof is Strong Mathematical Induction on n. The basis case is trivial: c(1) = 0 We must now show that for n > 1, if c(i)  0 for all positive integers i < n, then c(n)  0. But this is also obvious: c(n) = c(n/2) + n  0 + n > 0. Inductive Hypothesis Inductive Conclusion

  6. Recurrence Equations: Example 3 • Initial condition: c(1) = 0 • Recurrence: c(n) = c(n/2) + n for i > 1 • Claim: if 1  i  j, then c(i)  c(j) Let P(n) be the following statement: if 1  i  j  n, then c(i)  c(j) We will prove the P(n) is true for all n  1 by induction on n. Basis step n = 1: immediate, since i = j = n = 1. Inductive Step: assume n > 1 and P(n-1) is true. That is, assumeif 1  i  j  n-1, then c(i)  c(j). We must prove that P(n) is true. If j  n, then j  n-1 and the statement follows by the inductive hypothesis P(n-1). Thus we must show that if 1  i  n, then c(i)  c(n). If i = 1, then c(1) = 0  c(n) for all n  1. Thus assume i  2. Then 1  i/2  n/2  n-1, so c(i/2)  c(n/2) and hence c(i) = c(i/2) + i  c(n/2) + n = c(n).

  7. Recurrence Equations: Example 3 • Initial condition: c(1) = 0 • Recurrence: c(n) = c(n/2) + n • Looking at n = 2k for some k: • c(2k) = c(2k-1) + 2k = c(2k-2) + 2k-1 + 2k … = c(1)+ 21 +  + 2k-1 + 2k = 0 + 21 +  + 2k-1 + 2k = 2k+1 – 2 = 2(2k )– 2 = 2n – 2 • Thus for n a power of 2, c(n) = 2n-2 and hence is (n) • A formal proof of the statement c(n) = 2n-2 for n a power of 2 can be done easily by induction and is assigned as an exercise for the curious.

  8. Recurrence Equations • Initial condition: c(1) = 0 • Recurrence: c(n) = c(n/2) + n • For n a power of 2, c(n) = 2n-2 • Now, for integers n not equal to a power of 2, let k-1 = log2n • Then k-1  log2n < k and thus 2k-1 n < 2k • Then c(2k-1)  c(n) < c(2k) = 2k+1 – 2 = 4(2k-1)– 2 < 4n • Thus c(n) is O(n) • Since n  3, n/2  n-2  2k – 2 = c(2k-1)  c(n) • Thus c(n) is (n) • It now follows that c(n) is (n) • Whew! What a lot of work. • Don’t worry, there is a theorem on the way to help.

  9. The Master Theorem for Recurrences • Divide and conquer algorithms do the following: • break a problem up into a  2 subproblems of smaller size • recursively solve the subproblems • combine these solutions to solve the intial problem • Often the subproblems all have essentially the same size, say n/b or n/b, where n is the size of the original problem and b  2 is an integer • Such algorithms give rise to recurrences of the formT(n) = aT(n/b) + f(n)where f(n) is the time needed by the combination phase and n/b is understood to be either  n/b or  n/b  • There are also situations where we do not have equality, but rather  or  relations • The Master Theorem provides asymptotic estimates in these situations and, as it happens, the presence of floor or ceiling does not matter.

  10. Example • Mergesort sorts an array of 2 or more elements by recursively sorting the first half, recursively sorting the second half then merging the two sorted subarrays. • The first half of the array has size n/2 and the second half has size n/2. • Moreover, the merge routine for the two subarrays has running time (n) • Thus the recurrence for the running time T(n) of mergesort is T(n) = T(n/2 ) + T(n/2) + n • Since the Master Theorem treats n/2 and n/2 as simply n/2, the recurrence becomesT(n) = 2T(n/2 ) + n

  11. The Master Theorem for Recurrences • Let a, b be integers and k a real number with a  1, b  2 and k  0. • We will use n/b to denote either n/b or n/b • Upper Bound: If T(n)  aT(n/b) + f(n) and f(n) is O(nk), then O(nk) if a < bkT(n) = O(nk lg n) if a = bkO( ) if a > bk

  12. The Master Theorem for Recurrences • Let a, b, k be integers with a  1, b  2 and k  0. • We will use n/b to denote either n/b or n/b • Lower Bound: If T(n)  aT(n/b) + f(n) and f(n) is (nk), then (nk) if a < bkT(n) = (nk lg n) if a = bk ( ) if a > bk

  13. The Master Theorem for Recurrences • Let a, b, k be integers with a  1, b  2 and k  0. • We will use n/b to denote either n/b or n/b • Tight Bound: If T(n) = aT(n/b) + f(n) and f(n) is (nk), then (nk) if a < bkT(n) = (nk lg n) if a = bk ( ) if a > bk

  14. The Master Theorem for Recurrences • Example 1: Mergesort’s recurrence is T(n) = 2T(n/2) + n • In the setting of the Master Theorem, we have a = 2, b = 2 and k = 1 • Since a = bk, the Master Theorem implies Mergesort’s running time is (n1 lg n) = (nlg n)

  15. Example Recurrence Solutions • Example 2: c(n) = c(n/2) + n • The parameters in this case are: a = 1, b = 2, f(n) = n and thus k = 1 • Since a = 1 < bk = 21 = 2, the first case applies. • Since the recurrence is an equality, we conclude that c(n) is (nk) = (n1) = (n)

  16. Example Recurrence Solutions • Example 3: T(n) = 2T(n/2) + n • Parameters: a = 2, b = 2, k = 1 • a = 2 = bk = 21, so second case applies • Since the recurrence is an equality, we conclude that T(n) is (nk lg n) = (n1 lg n) = (n lg n)

  17. Example Recurrence Solutions • Example 4: T(n) = 7T(n/4) + 1 • Parameters: a = 7, b = 4, k = 0 (since 1 = n0) • a = 7 > bk = 40, so third case applies • Since the recurrence is an equality, we conclude thatT(n) is  (n 1.4037 )

  18. Homework • Pages 66-68, problems 14, 15, 17, 32, 33, 35 and 36

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