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Understanding and Solving Recurrences: Methods and Examples

Recurrences are equations that describe a function in terms of itself, often using smaller inputs. This guide covers the fundamental methods for solving recurrences, including the substitution method, iteration method, and master theorem. Examples illustrate how to apply these methods, such as solving T(n) = 2T(n/2) + n and determining its complexity. We explore how to establish bounds and derive solutions through mathematical induction and techniques like summation. Enhance your understanding of algorithm analysis by mastering recurrence relations.

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Understanding and Solving Recurrences: Methods and Examples

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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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