Analysis of Array Max Algorithm in Python: Big-O Notation and Induction, Lecture notes of Computer Science

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Operation Counting, Big-O and

(maybe) Induction

array max - python - non-recursive

# Note: Python uses 0 based indexing, so we've switched.

#!/usr/bin/python

def array_max(n, array) :

"""array_max: num * num[pos] -> num

Purpose: Finds the maximum value of the first n elements in

an array

Consumes: a nonnegative number n and an array of numbers

with at least n elements

Produces: a number

Example: array_max(4, [75, 4, 46, 12]) -> 75

m = array[0] # set m to the first element of the array

for i in range (1, n ): # for the rest of the

# elements

m = max(m, array[i]) # set m to the larger of itself

# and element we're checking

return m

What is the slope of the line: T( n ) = 5 + 4 * ( n – 1) A. ¼ B. 4/ C. 2 D. 4 E. 5

#!/usr/bin/python def array_max(n, array): """array_max: num * num[pos] -> num Purpose: Finds the maximum value of the first n elements in an array Consumes: a positive number n and an array of numbers with at least n elements Produces: a number Example: array_max(4, [75, 4, 46, 12]) -> 75 """ if n == 1: # if we are looking for the max of 1

element

return array[0] # return that element else: # otherwise, return the max of the last element

and the result of a recursive call to array_max

return max(array[n-1], array_max(n - 1, array)) array max - python - recursive

array max - python - recursive - op counts

Let A(n) = # of ops in array max(n, array) for any n

def array_max(n, array) : if n == 1: 1 op: comparison return array[0] 1 op: value return else: 0 ops: comparison already done return max(array[n-1], array_max(n - 1, array)) 4 ops: value return, math (max), function call, math (-) + A(n-1), the number of ops for array_max(n - 1, array)

• A(1) = 2
• A(n) = 5 + A(n-1)
• How can we solve for A(n)?

What is the slope of the line: A( n ) = 5 * ( n - 1) + 2 A. ¼ B. 4/ C. 2 D. 4 E. 5

What is the slope of the line: A( n ) = 5 * ( n - 1) + 2 A. ¼ B. 4/ C. 2 D. 4 E. 5  5n – 3 O(n) recursive  4n + 1 O(n) iterative

Big O

• Seamcarve algs

– O (

sqrt (n)

– O(n)

Big O

• Seamcarve algs

– O (

sqrt (n)

– O(n)

Big O

• f = O(g) f,g: Z + ->Z +
• f(n) ≤ c g(n)

Big O

• A(n) = 5 * (n-1) + 2
• Prove A = O(n)
• 5 * (n-1) + 2 ≤ c * n
• 5 * n – 3 ≤ 5 * n // c=
• So, A = O(n)