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Recurrences

Recurrences. Solving Recurrences. A recurrence is an equation or inequality that describes a function in terms of itself by using smaller inputs The expression: is a recurrence. Solving Recurrences. Examples: T ( n ) = 2 T ( n /2) + ( n ) T ( n ) = ( n lg n )

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Recurrences

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  1. Recurrences

  2. Solving Recurrences • A recurrence is an equation or inequality that describes a function in terms of itself by using smaller inputs • The expression: • is a recurrence.

  3. Solving Recurrences • Examples: • T(n) = 2 T(n/2) + (n) • T(n) = (n lg n) • T(n) = 2T(n/2) + n • T(n) = (n lg n) • T(n) = 2(T(n/2)+ 17) + n • T(n) = (n lg n) • Three methods for solving recurrences • Substitution method • Iteration method • Master method

  4. Recurrence Examples

  5. Substitution Method • The substitution method • “making a good guess method” • Guess the form of the answer, then • use induction to find the constants and show that solution works • Our goal: show that T(n) = 2T(n/2) + n = O(n lg n)

  6. Substitution MethodT(n) = 2T(n/2) + n = O(n lg n) • Thus, we need to show that T(n)  c n lg n with an appropriate choice of c • Inductive hypothesis: assume T(n/2)  c (n/2) lg (n/2) • Substitute back into recurrence to show thatT(n)  c n lg n follows, when c  1 • T(n) = 2T(n/2) + n  2(c (n/2) lg (n/2)) + n = cn lg(n/2) + n = cn lg n – cn lg 2 + n = cn lg n – cn + n  cn lg n for c  1 = O(n lg n) for c  1

  7. Substitution Method • Consider • Simplify it by letting m = lg n n = 2m • Rename S(m) = T(2m) • S(m) = 2 S(m/2) + m = O(m lg m) • Changing back from S(m) to T(n), we obtain • T(n) = T(2m) = S(m) = O(m lg m) = O(lg n lg lg n)

  8. Iteration Method • Iteration method: • Expand the recurrence k times • Work some algebra to express as a summation • Evaluate the summation

  9. T(n) = c + T(n-1) = c + c + T(n-2) = 2c + T(n-2) = 2c + c + T(n-3) = 3c + T(n-3) … kc + T(n-k) = ck + T(n-k) • So far for nk we have • T(n) = ck + T(n-k) • To stop the recursion, we should have • n - k = 0  k = n • T(n) = cn + T(0) = cn • Thus in general T(n) = O(n)

  10. T(n) = n + T(n-1) = n + n-1 + T(n-2) = n + n-1 + n-2 + T(n-3) = n + n-1 + n-2 + n-3 + T(n-4) = … = n + n-1 + n-2 + n-3 + … + (n-k+1) + T(n-k) = for nk • To stop the recursion, we should have n - k = 0  k = n

  11. T(n) = 2 T(n/2) + c 1 = 2(2 T(n/2/2) + c) + c 2 = 22 T(n/22) + 2c + c = 22(2 T(n/22/2) + c) + (22-1)c 3 = 23 T(n/23) + 4c + 3c = 23 T(n/23) + (23-1)c = 23(2 T(n/23/2) + c) + 7c 4 = 24 T(n/24) + (24-1)c = … = 2kT(n/2k) + (2k - 1)ck

  12. So far for nk we have • T(n) = 2k T(n/2k) + (2k - 1)c • To stop the recursion, we should have • n/2k = 1  k = lg n • T(n) = 2lgnT(n/2lgn) + (2lgn - 1)c = nT(n/n) + (n - 1)c = nT(1) + (n-1)c = nc + (n-1)c = nc + nc – c = 2cn – c=cn – c/2 cn = O(n) for all n ½

  13. T(n) = aT(n/b) + cn 1 = a(aT(n/b/b) + cn/b) + cn 2 = a2T(n/b2) + cna/b + cn = a2T(n/b2) + cn(a/b + 1) = a2(aT(n/b2/b) + cn/b2) + cn(a/b + 1) 3 = a3T(n/b3) + cn(a2/b2) + cn(a/b + 1) = a3T(n/b3) + cn(a2/b2 + a/b + 1) = … = akT(n/bk) + cn(ak-1/bk-1 + ak-2/bk-2 + … + a2/b2 + a/b + 1) k

  14. So we have • T(n) = akT(n/bk) + cn(ak-1/bk-1 + ... + a2/b2 +a/b+1) • To stop the recursion, we should have • n/bk = 1  n = bk  k = logbn • T(n) = akT(1) + cn(ak-1/bk-1 +...+ a2/b2 + a/b+1) = akc + cn(ak-1/bk-1 + ... + a2/b2 + a/b + 1) = cak + cn(ak-1/bk-1 + ... + a2/b2 + a/b + 1) = cnak/bk + cn(ak-1/bk-1 + ... +a2/b2 +a/b+1) = cn(ak/bk+ ... + a2/b2 + a/b + 1)

  15. So with k = logbn • T(n) = cn(ak/bk+ ... + a2/b2 + a/b + 1) • What if a = b? • T(n) = cn(1+ ... + 1+ 1 + 1) // k+1 times = cn(k + 1) = cn(logbn + 1) = (n logb n)

  16. So with k = logbn • T(n) = cn(ak/bk+ ... + a2/b2 + a/b + 1) • What if ab? • Recall that • (xk + xk-1 + … + x + 1) = (xk+1 -1)/(x-1) • So: • T(n) = cn · (1) = (n)

  17. So with k = logbn • T(n) = cn(ak/bk+ ... + a2/b2 + a/b + 1) • What if a  b? • T(n) = cn · (ak / bk) = cn · (alogbn / blogbn) = cn· (alogbn / n) recall logarithm fact: alogbn = nlogba = cn· (nlogba / n) = (cn· nlogba / n) = (nlogba)

  18. So…

  19. The Master Theorem • Given: a divide and conquer algorithm • An algorithm that divides the problem of size n into a subproblems, each of size n/b • Let the cost of each stage (i.e., the work to divide the problem + combine solved subproblems) be described by the function f(n) • Then, the Master Theorem gives us a cookbook for the algorithm’s running time:

  20. The Master Theorem • if T(n) = aT(n/b) + f(n) where a≥ 1 & b > 1 • then

  21. Using The Master MethodCase 1 • T(n) = 9T(n/3) + n • a = 9, b = 3, f(n) = n • nlogb a = nlog39 = n2 • since f(n) = O(nlog39 - ) = O(n2-0.5) = O(n1.5) • where  = 0.5 • case 1 applies: • Thus the solution is • T(n) = (n2)

  22. Using The Master Method Case 2 • T(n) = T(2n/3) + 1 • a = 1, b = 3/2, f(n) = 1 • nlogba = nlog3/21 = n0 = 1 • since f(n) = (nlogba) = (1) • case 2 applies: • Thus the solution is • T(n) = (lg n)

  23. Using The Master Method Case 3 • T(n) = 3T(n/4) + n lg n • a = 3, b = 4, f(n) = n lg n • nlogba = nlog43 = n0.793 = n0.8 • Since f(n) = W(nlog43+) = W(n0.8+0.2) = W(n) • where   0.2, and for sufficiently large n, • a.f(n/b) = 3(n/4) lg(n/4)  (3/4) n lg n for c = 3/4 • case 3 applies: • Thus the solution is • T(n) = (n lg n)

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