Algorithm Analysis: Running Time Big O

1. Algorithm Analysis: Running Time Big O

2. Introduction • An algorithm analysis of a program is a step by step procedure for accomplishing that program • In order to learn about an algorithm, we need to analyze it • This means we need to study the specification of the algorithm and draw conclusion about the implementation of that algorithm (the program) will perform in general

3. The issues that should be considered in analyzing an algorithm are: • The running time of a program as a function of its inputs • The total or maximum memory space needed for program data • The total size of the program code • Whether the program correctly computes the desired result • The complexity of the program. For example, how easy it is to read, understand, and modify the program • The robustness of the program. For example, how well does it deal with unexpected or erroneous inputs

4. In this course, we consider the running time of the algorithm. • The main factors that effect the running time are the algorithm itself, input data, the computer system, etc. • The performance of a computer is determined by • The hardware • The programming language used and • The operating system • To calculate the running time of a general C++ program, we first need to define a few rules. • In our rules , we are going to assume that the effect of hardware and software systems used in the machines are independent of the running time of our C++ program

5. Rule 1: • The time required to fetch an integer from memory is a constant t(fetch), • The time required to store an integer in memory is also a constant t(store) • For example the running time of • x = y • is: • t(fetch) + t(store) • because we need to fetch y from memory store it into x • Similarly the running time of • x = 1 • is also t(fetch) + t(store) because typically any constant is stored in the memory before it is fetched.

6. Rule 2 • The time required to perform elementary operations on integers, such as addition t(+), subtraction t(-), multiplication t(*), division t(/), and comparison t(cmp), are all constants. • For Example the running time of • y= x+1 • is: • 2t(fetch) + t(store) + t (+) • because you • need to fetch x and 1: 2*t(fetch) • then add them together: t(+) • and place the result into y: t(store)

7. Rule 3: • The time required to call a function is a constant, t(call) • And the time required to return a function is a constant, t(return) • Rule 4: • The time required to pass an integer argument to a function or procedure is the same as the time required to store an integer in memory, t(store)

8. For example the running time of • y = f(x) • is: • t(fetch) + 2t(store) + t(call) + t(f(x)) • Because you need • To fetch the value of x: t (fetch) • Pass x to the function and store it into parameter: t (store) • Call the function f(x): t (call) • Run the function: t (f(x)) • Store the returned result into y: t (store)

9. Rule 5: • The time required for the address calculation implied by an array subscripting operation like a[i] is a constant, t([ ]). • This time does not include the time to compute the subscript expression, nor does it include the time to access (fetch or store) the array element • For example, the running time of • y = a[i] • is: • 3t(fetch) + t([ ]) + t(store) • Because you need • To fetch the value of i: t(fetch) • To fetch the value of a: t(fetch) • To find the address of a[i]: t([ ]) • To fetch the value of a[i]: t(fetch) • To store the value of a[i] into y: t(store)

10. Rule 6: • The time required to calculate a fixed amount of storage from the heap using operator new is a constant, t(new) • This time does not include any time required for initialization of the storage (calling a constructor). • Similarly, the time required to return a fixed amount of storage to the heap using operator delete is a constant, t(delete). • This time does not include any time spent cleaning up the storage before it is returned to the heap (calling destructor) • For example the running time of • int* ptr = new int; • is: • t(new) + t(store) • Because you need • To allocate a memory: t(new) • And to store its address into ptr: t(store) • For example the running time of • delete ptr; • is: • t(fetch ) + t(delete) • Because you need • To fetch the address from ptr : t(fetch) • And delete specifies location: t(delete)

11. int x Sum (int n) • { • int result =0; • for (int i=1; i<=n; ++i) • result += i; • return result • }

12. int func (int a[ ], int n, int x) • { • int result = a[n]; • for (int i=n-1; i>=0; --i) • result =result *x + a[i]; • return result • }

13. Using constant times such as t(fetch), t(store), t(delete), t(new), t(+), …, ect makes our running time accurate • However, in order to make life simple, we can consider the approximate running time of any constant to be the same time as t(1). • For example, the running time of • y = x + 1 • is 3 because it includes two “fetches” and one “store” in which all are constants • For a loop there are two cases: • If we know the exact number of iterations, the running time becomes constant t(1) • If we do not know the exact number of iterations, the running time becomes t(n) where n is the number of iterations

14. n n n i i i+1 S S S 4 4 3 i=0 i=0 i=0 • int Geometric (int x, int n) • { • int sum = 0; • for (int i=0; i<=n; ++i) • { • int prod = 1; • for (int j=0; j<i; ++j) • prod = prod * x; • sum = sum + prod; • } • return result • }

15. Asymptotic Notation • Suppose the running time of two algorithms A and B are TA(n) and TB(n), respectively where n is the size of the problem • How can we determine TA(n) is better than TB(n)? • One way to do that is if we know the size n ahead of time for some n=no. Then we may say that algorithm A is performing better than algorithm B for n= no • But this is a special case for n=no. How about n = n1, or n=n2? Is A better than B for other cases too? • Unfortunately, this is not an easy answer. We cannot expect that the size of n to be known ahead of time. But we may be able to say that under certain situations TA(n) is better than TB(n) for all n >= n1

16. To understand the running times of the algorithms we need to make some definitions: • Definition: • Consider a function f(n) which is non-negative for all integers n>=0. We say that “f(n) is big oh of g(n)” which we write (f(n) is O(g(n)) if there exists an integer no and a constant c > 0 such that for all integers n >=0, f(n) <=c g(n) • Example: • Show that f(n) = 8n + 128 is O(n2) • 8n + 128 <= c n2 (lets set c = 1) • 0 <= cn2 -8n -128 • 0 <= (n-16) (n+8) • Thus we can say that for constant c =1 and no >= 16 f(n) is O(n2)

17. f(n)=n2 f(n)=4n2 f(n)=2n2 400 f(n)=8n+128 200 10 5 15 20

18. The names of common big O expressions

19. Conventions for writing Big Oh Expression • Certain conventions have evolved which concern how big oh expression normally written: • First, it is common practice when writing big oh expression to drop all but the most significant items. Thus instead of O(n2 + nlogn + n) we simply write O(n2) • Second, it is common practice to drop constant coefficients. Thus, instead of O(3n2), we write O(n2). As a special case of this rule, if the function is a constant, instead of, say O(1024), we simply write O(1)

20. int func (int a[ ], int n, int x) • { • int result = a[n]; • for (int i=n-1; i>=0; --i) • result =result *x + a[i]; • return result • } The total running time is: O(16n + 14) = O(max(16n, 14)) = O(16n) = O(n)

21. int PrefixSums (int a[ ], int n) • { • for (int j=n-1; i>=0; --j) • { • int sum = 0; • for (int i=0; i<=j; ++i) • sum = sum + a[i]; • a[j] = sum; • } • return result • }

22. Sorting

23. The efficiency of data handling can often be substantially increased if the data are sorted • For example, it is practically impossible to find a name in the telephone directory if the items are not sorted • In order to sort a set of item such as numbers or words, two properties must be considered • The number of comparisons required to arrange the data • The number of data movement

24. Depending on the sorting algorithm, the exact number of comparisons or exact number of movements may not always be easy to determine • Therefore, the number of comparisons and movements are approximated with big-O notations • Some sorting algorithm may do more movement of data than comparison of data • It is up to the programmer to decide which algorithm is more appropriate for specific set of data • For example, if only small keys are compared such as integers or characters, then comparison are relatively fast and inexpensive • But if complex and big objects should be compared, then comparison can be quite costly

25. If on the other hand, the data items moved are large, and the movement is relatively done more, then movement stands out as determining factor rather than comparison • Further, a simple method may only be 20% less efficient than a more elaborated algorithm • If sorting is used in a program once in a while and only for small set of data, then using more complicated algorithm may not be desirable • However, if size of data set is large, 20% can make significant difference and should not be ignored • Lets look at different sorting algorithms now

26. Insertion Sort • Start with first two element of the array, data[0], and data[1] • If they are out of order then an interchange takes place • Next data[2] is considered and placed into its proper position • If data[2] is smaller than data[0], it is placed before data[0] by shifting down data[0] and data[1] by one position • Otherwise, if data[2] is between data[0] and data[1], we just need to shift down data [1] and place data[2] in the second position • Otherwise, data[2] remain as where it is in the array • Next data[3] is considered and the same process repeats • And so on

27. Algorithm and code for insertion sort InsertionSort(data[], n) for (i=1, i<n, i++) move all elements data[j] greater than data[i] by one position; place data[i] in its proper position;

28. 5 2 5 2 2 2 5 5 5 2 3 5 5 3 5 3 3 3 8 8 8 8 8 8 1 1 1 1 1 1 Example of Insertion Sort Put tmp=2 in position 1 tmp = 2 Moving 5 down Put tmp=3 in position 2 tmp = 3 Moving 5 down

29. 2 2 2 1 2 2 2 2 2 3 2 3 3 3 3 3 5 5 3 3 3 5 5 5 8 5 5 8 5 5 8 8 1 1 1 8 8 8 8 8 Since 5 is less than 8 no shifting is required tmp = 8 Put tmp=1 in position 1 Moving 5 down Moving 2 down Moving 3 down Moving 8 down tmp=1

30. Advantage of insertion sort: • If the data are already sorted, they remain sorted and basically no movement is not necessary • Disadvantage of insertion sort: • An item that is already in its right place may have to be moved temporary in one iteration and be moved back into its original place • Complexity of Insertion Sort: • Best case: This happens when the data are already sorted. It takes O(n) to go through the elements • Worst case: This happens when the data are in reverse order, then for the ith item (i-1) movement is necessary • Total movement = 1 + 2 + .. . +(n-1) = n(n-1)/2 which is O(n2) • The average case is approximately half of the worst case which is still O(n2)

31. Selection Sort • Select the minimum in the array and swap it with the first element • Then select the second minimum in the array and swap it with the second element • And so on until everything is sorted

32. Algorithm and code for selection sort • SelectionSort(data[ ],n) • for (i=0; i<n-1; i++) • Select the smallest element among data[i] … data[n-1]; • Swap it with data[i]

33. 1 5 1 1 2 2 2 2 3 3 3 3 8 8 8 8 5 5 5 1 Example of Selection Sort The first minimum is searched in the entire array which is 1 Swap 1 with the first position The second minimum is 2 Swap it with the second position

34. 1 1 1 1 2 2 2 2 3 3 3 3 5 8 8 8 5 8 5 5 The third minimum is 3 Swap 1 with the third position The fourth minimum is 5 Swap it with the forth position

35. Complexity of Selection Sort • The number of comparison and/or movements is the same in each case (best case, average case and worst case) • The number of comparison is equal to • Total = (n-1) + (n-2) + (n-3) + …. + 1 • = n(n-1)/2 • which is O(n2)

36. Bubble Sort • Start from the bottom and move the required elements up (i.e. bubble the elements up) • Two adjacent elements are interchanged if they are found to be out of order with respect to each other • First data[n-1] and data[n-2] are compared and swapped if they are not in order • Then data[n-2] and data[n-3] are swapped if they are not in order • And so on

37. Algorithm and code for bubble sort • BubbleSort(data[ ],n) • for (i=0; i<n-1; i++) • for (j=n-1; j>i; --j) • swap elements in position j and j-1 if they are out of order

38. 1 5 1 1 5 5 1 1 5 2 5 2 5 2 5 2 5 1 swap no swap swap swap swap 1 3 2 3 5 2 2 2 2 3 3 3 3 3 1 3 3 8 8 8 8 8 8 8 8 8 1 swap no swap Example of Bubble Sort Iteration 1: Start from the last element up to the first element and bubble the smaller elements up Iteration 2: Start from the last element up to second element and bubble the smaller elements up

39. 1 1 1 1 2 2 2 2 swap 3 5 3 5 5 3 3 5 8 8 8 8 no swap no swap Example of Bubble Sort Iteration 3: Start from the last element up to third element and bubble the smaller elements up Iteration 4: Start from the last element up to fourth element and bubble the smaller elements up

40. Complexity of Bubble Sort • The number of comparison and/or movements is the same in each case (best case, average case and worst case) • The number of comparison is equal to • Total = (n-1) + (n-2) + (n-3) + …. + 1 • = n(n-1)/2 • which is O(n2)

41. Comparing the bubble sort with insertion and selection sorts we can say that: • For the average case, bubble sort makes approximately twice as many comparisons and the same number of moves as insertion sort • Bubble sort also, on average, makes as many comparison as selection sort and n times more moves than selection sort • Between theses three types of sorts “Selection Sort” is generally better algorithm because if array is already sorted running time only takes O(n) which is relatively faster than other algorithms

42. Quick Sort • This is known to be the best sorting method. • In this scheme: • One of the elements in the array is chosen as pivot • Then the array is divided into sub-arrays • The elements smaller than the pivot goes into one sub-array • The elements bigger than the pivot goes into another sub-array • The pivot goes in the middle of these two sub-arrays • Then each sub-array is partitioned the same way as the original array and process repeats recursively

43. Algorithm of quick sort • QuickSort(array[ ]) • if length (array) > 1 • choose a pivot; // partition array into array1 and array2 • while there are elements left in array • include elements either in array1 // if element <= pivot • or in array2 // if element >= pivot • QuickSort(array1); • QuickSort(array2); • Complexity of quick sort • The best case is when the arrays are always partitioned equally • For the best case, the running time is O(nlogn) • The running time for the average case is also O(nlogn) • The worst case happens if pivot is always either the smallest element in the array or largest number in the array. • In the worst case, the running time moves toward O(n2)

44. By example Select pivot Partition 43 92 81 75 0 31 26 13 57 65 0 13 92 43 81 26 75 31 57 Example of Quick Sort 65 65

45. Recursively apply quicksort to both partitions Result will ultimately be a sorted array 0 13 65 92 43 81 26 75 31 57 0 13 26 31 43 57 65 75 81 92

46. Radix Sort • Radix refers to the base of the number. For example radix for decimal numbers is 10 or for hex numbers is 16 or for English alphabets is 26. • Radix sort has been called the bin sort in the past • The name bin sort comes from mechanical devices that were used to sort keypunched cards • Cards would be directed into bins and returned to the deck in a new order and then redirected into bins again • For integer data, the repeated passes of a radix sort focus on the ones place value, then on the tens place value, then on the thousands place value, etc • For character based data, focus would be placed on the right-most character, then the second most right-character, etc

47. Algorithm and Code for Radix Sort Assuming the numbers to be sorted are all decimal integers • RadixSort(array[ ]) • for (d = 1; d <= the position of the leftmost digit of longest number; i+=) • distribute all numbers among piles 0 through 9 according to he dth digit • Put all integers on one list

48. Example of Radix Sort Assume the data are: 459 254 472 534 649 239 432 654 477 Radix sort will arrange the values into 10 bins based upon the ones place value 0 1 2 472 432 3 4 254 534 654 5 6 7 477 8 9 459 649 239

49. The sublists are collected and made into one large bin (in order given) 472 432 254 534 654 477 459 649 239 Then Radix sort will arrange the values into 10 bins based upon the tens place value 0 1 2 3 432 534 239 4 649 5 254 654 459 6 7 472 477 8 9

50. The sublists are collected and made into one large bin (in order given) 432 534 239 649 254 654 459 472 477 Radix sort will arrange the values into 10 bins based upon the hundreds place value (done!) 0 1 2 239 254 3 4 432 459 472 477 5 534 6 649 654 7 8 9 • The sublists are collected and the numbers are sorted 239 254 432 459 472 477 534 649 654