WKES3311:ANALISIS ALGORITMA

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WKES3311ANALISIS ALGORITMA
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WHICH ALGORITHM?

• The higher growth rate?
• OR
• The lower growth rate?

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ASYMPTOTIC ANALYSIS

• Refers to the study of an algorithm as the input
  size gets big or reaches a limit
• It has its limitation

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ALGORITHM BEHAVIOUR

• Terms describing the running time equation for an
  algorithm

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UPPER BOUND O(f(n)) BIG OH

• The highest growth rate that an algorithm can
  have
• To make statement we say

Worst / best/average
f(n)
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UPPER BOUND PRECISE STATEMENT

• For T(n) a non-negatively valued function, T(n)
  is in set if there exist two positive
  constants c and n0 such that for all n gt
  n0

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EXAMPLE 1

• Consider the sequential search algorithm for
  finding a specific value in an array of integers

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EXAMPLE 2

• For a particular algorithm, T(n) c1n2 c2n in
  the average case where c1 and c2 are positive
  numbers

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LOWER BOUND ?(g(n)) BIG OMEGA DEFINITION

• For T(n) a non-negatively valued function, T(n)
  is in set ?(g(n)) if there exist two
• OR
• T(n) is in the set ?(g(n)) if there exists a
  positive constant c such that

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EXAMPLE 1

• Assume T(n)c1n2 c2n for c1 and c2 gt 0.
• So, T(n) ? c1n2 for c c1 and n01.

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EXAMPLE 2
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?(h(n)) THE BIG THETA

• When the upper and lower bound for an algorithm
  are the same within a constant factor we indicate
  this by using ?(big-Theta) notation.

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EXAMPLE 1

• Sequential Search Algorithm
• So we say that this algorithm is both in O(n) and
  ?(n) OR JUST ?(n)

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SIMPLIFYING RULES

• If f(n) is in
  then f(n) is in O(h(n))
• If f(n) for any
  constant kgt0, then f(n) is in O(g(n)
• If f1(n) is in is in
  O(g2(n)),
  O(max(g1(n),g2(n)))
• If f1(n) is in O(g1(n)) and f2(n) is in
• Rule also holds for ? and ? notations

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CALCULATING THE RUNNING TIME

• Example 1
• Since the assignment statement takes constant
  time, it is ?(1)

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EXAMPLE 2

• sum 0
• for (i1iltni)
• sum n

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EXAMPLE 3

• Asignment (sum0)
• Second loop
• First loop
• c3j(sum)
• Therefore its c3

• sum 0
• for (j1jltnj) //First for loop
• for (i1iltji)// is a double loop
• sum
• for (k0kltnk) // Second for loop
• Akk

• By rule(3),

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EXAMPLE 4

• sum10
• for (k1kltnk2) // Do log n times
• for(j1jltnj) // Do n times
• sum
• sum20
• For(k1kltnk22) // Do log n times
• for(j1jltkj) // Do k times
• sum2

• simply

• simply

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EXAMPLE 5

• Long fact (int n) //Compute n!
• //To fit n! in a long variable, require nlt12
• Assert((ngt0)(nlt12)), Input out of range)
• if (nlt1) return 1
• return n fact(n-1)

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RECURRENCE RELATIONSHIP T(n)T(n-1) 1

• T(n) T(n-1)1
• (T(n-2)1)1, how can we write this in terms of
  summation?
• T(n-3)3 , therefore we conclude
• T(n) T(n-(n-1))

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PROOF BY INDUCTION

• Theorem1 T(n) T(n-1) 1 T(1)0 has
  closed-form solution T(n)n-1
• Proof To prove the base case we observe

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SPACE BOUNDS

• The
  to measure space are similar to those used to
• Concepts of can also
  be applied with

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HOMEWORKS

• What are the space requirements for a two
  dimensional array?
• What are the space requirements for n Boolean
  flags?