The document discusses different methods for solving recurrence equations: 1. Substitution method involves guessing a solution and using induction to prove it is correct. 2. Iterative method expands the recurrence into a summation and evaluates it. 3. Master method handles recurrences of the form T(n)=aT(n/b)+f(n). It provides 3 cases to bound the solution. 4. Recursion tree method models the recurrence as a tree to track problem sizes and costs at each level. Summing costs across levels provides the total solution.

1. RECURRENCE
EQUATIONS &
ANALYZING
THEM
By:
ALPANA A. INGALE
ROLL NO. 8108

2. RECURRENCES
 A recurrence is an equation or inequality that describes a function in
terms of its value on small inputs.
 The running time of the recursive algorithm can be obtained by a
recurrence.
 To solve the recurrence relation mean to obtain a function defined on
the natural numbers that satisfies the recurrence.
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3. Recurrence Equation Methods –
cont’d
 We solve the recurrence equation by the following methods. That is:
i. Substitution method
ii. Iterative method
iii. Master method
iv. Recurrence or Recursion Tree
 Substitution Method:
1. In this method, first we guess a solution and use mathematical induction to find
the constant and show that the solution works.
2. The substitution method can be used to establish either upper or lower bounds
on a recurrence.
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4. Recurrence Equation Methods –
cont’d
 E.g.: T(n) = 2T(n/2) + n --- (1) is a recurrence relation.
 We guess the solution T(n) = O(nlgn), where g(n) = nlgn is the solution.
 That is: f(n) = T(n) ≤ cnlgn --- (2)
 Then we have to prove that, this solution is true, by using mathematical induction.
 From equation-1, T(n) = 2T(n/2) + n
 T(n) ≤ 2c(n/2) lg(n/2) + n
 T(n) ≤ cn.lg(n/2) + n = cn(lgn – lg2) + n
 T(n) ≤ cnlgn – cn + n.
 So, T(n) ≤ cnlgn – n(c – 1)
 T(n) ≤ cnlgn for c >1
 Now, using mathematical induction, T(n) = T(n-1) + 1.
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5. Recurrence Equation Methods –
cont’d
 For n = 1, if T(1) = 1, then c ≥ 1
 lg1 = 0 => T(1) is not true for n = 1.
 For n = 2, if T(2) = 2T(1) + 2 = 4, then c ≥ 4.
 2lg2 => c ≥ 2.
 Hence, T(n) ≤ cnlgn, c > 1 is true
 For n = 2 => T(2) is true.
 Similarly, T(3), T(4) … is true
 T(k) is true.
 T(k + 1) is true by using Tk.
 T(n ) = T(n – 1) + 1, g(n) = n = O(n)
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6. Recurrence Equation Methods –
cont’d
 T(n) = T(n – 1) + 1
 T(n) = T(n – 2) + 1 + 1
 T(n) = T(n + 3) + 1 + 1 + 1
 Let k = T(n – k) + k.
 Let n – k = 0 (base value).
 n = k
 T₀ + n = n + 1
 Let T₀ = 1 (base case) T(1) = 1, k = n – 1
 T(n) = 1 + n – 1 = n
 So, it is true for n/2 and it is true for n.
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7. Recurrence Equation Methods -
cont’d
 If it is true for n, then it is true for 3.
 If it is true for 3, then it is true for 6 and 7 and if it is true for 6, then it is true for
12 and 13.
 Hence, we conclude that for n ≥ c, where c ≥ 2.
 So, T(n) = O(nlgn) is a solution of T(n) = 2T(n/2) + n.
2. Iterative Method: An iterative method is the method, where the recurrence
relation is solved by considering 3 steps. That is:
1. Step 1: expand the recurrence
2. Step 2: express is as a summation (∑) of terms, dependent only on „n‟ and the initial
condition.
3. Step 3: evaluate the summation (∑).
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8. Recurrence Equation Methods –
cont’d
 T(n) = 3T(n/4) + n
 Expanding the above terms, we get
T(n) = 3(3T(n/16)) + (n/4) + n
T(n) = 3(3(3T(n/164))) + (n/16) + (n/4) + n
T(n) = n + 3(n/4) + 9(n/16) + 2T(T(n/164))
 The recursion stops, when n/4ⁱ ≤ 1, which implies n ≤ 4ⁱ => log₄n = i
 Thus, T(n) = n + 3n/4+ … + 3ⁱ(n/4ⁱ)+ 3log₄n.θ(1)
 T(n) ≤ (n + 3n/4 + 9n/16 + … + 3 log₄n)θ(1)
 T(n) ≤ n + θ(n log₄3) {as 3log₄n = n log₄3}
 T(n) ≤ n(1/(1 – ¾)) + O(n)
 T(n) = 4n + O(n) {as log₄3 < 1} = O(n)
 Hence, T(n) = O(n) (proved).
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k


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)4/3(

9. Recurrence Equation Methods –
cont’d
3. Master Method: The master method is used for solving the following types of
recurrences, T(n) = aT(n/b) + f(n),
where ‘a’ and ‘b’ are constants and a ≥ 1, b > 1.
 In the above recurrence, the problem of size ‘n’ is divided into ‘a’ sub-
problems each of size ‘n/b’.
 Each sub-problem of size ‘n/b’ can be solved recursively in time T(n/b).
 The cost of dividing or splitting the problem and combine the solutions or
result is described by the function f(n).
 Here the size is interpreted as ‘n/b’.
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10. Recurrence Equation Methods –
cont’d
 The T(n) can be bounded asymptotically by the following 3 cases.
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11. Recurrence Equation Methods –
cont’d
4. Recursion Tree Method: Recursion Tree Method is pictorial representation of
an iteration method, which is in the form of a tree, where at each levels,
nodes are expanded.
 It is used to keep track of the size of the remaining arguments in the
recurrence and the non-recursive costs.
 In a recursion tree, each node represents the cost of a single sub-problem.
 We add the cost within each level of the tree to obtain a set of pre-level cost
and then we add all the levels of costs to determine the total cost of all levels
of recursion.
 In general, T(n) = aT(n/b) + f(n)
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12. Recurrence Equation Methods –
cont’d
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13. Recurrence Equation Methods –
cont’d
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