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Chapter 9

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Chapter 9

Efficiency of Algorithms

9.3

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

- 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.

- 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.

- 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).

- 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).

- 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)

- 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 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.

- 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)