Chapter 9
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Chapter 9. Efficiency of Algorithms. 9.3. Efficiency of Algorithms. Efficiency of Algorithms. Two aspects of algorithm efficiency are important: 1. the amount of time required for execution 2. amount of memory space needed when it runs.

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Chapter 9

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Chapter 9

Chapter 9

Efficiency of Algorithms


Chapter 9

9.3

Efficiency of Algorithms


Efficiency of algorithms

Efficiency of Algorithms

  • Two aspects of algorithm efficiency are important:

    • 1. the amount of time required for execution

    • 2. amount of memory space needed when it runs.

  • Algorithms are usually analyzed in terms of their best case, worst case, and average.

  • How can the time efficiency of an algorithm be calculated?

    • Have to factor in the input size to the algorithm

    • Nature of the input data


Efficiency of algorithms1

Efficiency of Algorithms

  • Because the time efficiency can be influenced by many physical parameters of a computing device, i.e. processor speed, memory size, multi-core, etc., a method of analysis must be used that is not a factor of the processing platform.

  • Evaluation of algorithms can be based on the number of elementary operations required by the algorithm.

    • Elementary operations are addition, subtraction, multiplication, division and comparison.

    • All elementary ops are viewed as taking 1 time unit on any system.


Example

Example

  • Consider algorithms A & B designed to accomplish the same task. For input size of n

    • A requires 10n to 20n elementary operations.

    • B requires 2n2 to 4n2 elementary operations.

  • Which algorithm is more efficient?

    • for n ≤ 10, 2n2 < 20n, and hence, B is better

    • for n > 10, 20n < 2n2, and in this case A is better

  • The answer is dependent on the size of the input. For a small n B is better, but for a larger inputs A wins. It is important to understand the constraints on the order.


Definition

Definition

  • Let A be an algorithm

    • Suppose the number of elementary ops performed when A is executed for an input of size n depends on n alone and not on the nature of the input data; say it equals f(n). If f(n)is Θ(g(n)) then, A is of order g(n).

    • Suppose the number of elementary operations performed when A is executed for an input of size n depends on the nature of the input data as well as on n.

      • Let b(n) be the minimum number of elementary operations required to execute A for all possible input sets of size n. If b(n) is Θ(g(n)), we say A has a best case order of g(n).

      • Let w(n) be the maximum number of elementary operations required to execute A for all possible input sets of size n. If w(n) is Θ(g(n)), we say A has a worst case order of g(n).


Time comparisons of algorithm orders

Time Comparisons of Algorithm Orders


Example1

Example

  • Consider the following algorithm segmentp =0, x=2for i = 2 to np = (p + i) * xnext I

  • Compute the actual number of elementary ops?

    • 1 multi, 1 add for each iteration.

    • 2 ops per iteration.

    • num of iteration = n – 2 + 1 = n-1.

    • num of elementary ops = 2(n-1)

  • Find an order from among the set of power functions

    • by theorem on polynomial orders, 2n – 2 is Θ(n)

    • and thus, segment is Θ(n).


Example2

Example

  • Consider a segment with a nested loop (loop inside of a loop)s=0for i=1 to n for j= 1 to is=s + j * (i-j+1) next jnext i

  • Compute the number of elementary ops.

    • 2 adds, 1 multi, and 1 minus (4 ops) for each iteration

    • inside loop (j) iterates i=1 1, i=2 2, i=3 3 … i=nn times

    • inside loop: 1 + 2 + 3 + … + n = n(n+1)/2

    • num of ops = 4 * n(n+1)/2 = 2*n(n+1) = 2n2 + 2n

  • Find the order among the set of power functions

    • 2n(n+1) = 2n2 + 2n is Θ(n2),

    • hence, segment is Θ(n2)


Example3

Example

  • Consider the segment where the floor function is used.for i = ⎣n/2⎦ to n a = n – Inext I

  • Compute the actual number of subtractions performed.

    • 1 subtraction for each iteration.

    • loop iterates n - ⎣n/2⎦ + 1 times.

      • n is even: ⎣n/2⎦ = n/2

      • n – n/2 + 1 = (2n – n + 2)/2 = (n+2)/2

      • n is odd: ⎣n/2⎦ = (n-1)/2

      • n – (n-1)/2 + 1 = [2n – (n-1) + 2]/2 = (n+3)/2

  • Find an order for this segment

    • (n+2)/2 is Θ(n) and hence, segment is Θ(n)


Sequential search

Sequential Search

  • Sequential search occurs when every element in the list is compared to a particular x until either a match is found or the end of the list.


Example4

Example

  • Find the best and worst case orders for sequential search.

    • Best case: the best case occurs when the first element of the list is the item that is being pursued (searched for). The search takes 1 comparison. Θ(1)

    • Worst case: the element being searched for is the last element or not in the last (these two situations are equal). The search takes n comparisons. Θ(n)


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