UNIT 4C Iteration: Scalability & Big O. Efficiency

Transcription

1 UNIT 4C Iteratio: Scalability & Big O 1 Efficiecy A computer program should be totally correct, but it should also execute as quickly as possible (time-efficiecy) use memory wisely (storage-efficiecy) How do we compare programs (or algorithms i geeral) with respect to executio time? various computers ru at differet speeds due to differet processors compilers optimize code before executio the same algorithm ca be writte differetly depig o the programmig paradigm 2 1

2 Coutig Operatios We measure time efficiecy by coutig the umber of operatios performed by the algorithm. But what is a operatio? assigmet statemets comparisos retur statemets... 3 Liear Search: Worst Case # let = the legth of list. def search(list, key) idex = 0 1 while idex < list.legth do +1 if list[idex] == key the retur idex idex = idex + 1 retur il 1 Total:

3 Liear Search: Best Case # let = the legth of list. def search(list, key) idex = 0 1 while idex < list.legth do 1 if list[idex] == key the 1 retur idex 1 idex = idex + 1 retur il Total: 4 5 Coutig Operatios How do we kow that each operatio we cout takes the same amout of time? (We do t.) So geerally, we look at the process more abstractly ad cout whatever operatio deps o the amout or size of the data we re processig. For liear search, we would cout the umber of times we compare elemets i the array to the key. 6 3

4 Liear Search: Worst Case Simplified # let = the legth of list. def search(list, key) idex = 0 while idex < list.legth do if list[idex] == key the retur idex idex = idex + 1 retur il Total: 7 Liear Search: Best Case Simplified # let = the legth of list. def search(list, key) idex = 0 while idex < list.legth do if list[idex] == key the 1 retur idex idex = idex + 1 retur il Total: 1 8 4

5 Order of Complexity For very large, we express the umber of operatios as the (time) order of complexity. Order of complexity is ofte expressed usig Big-O otatio: Number of operatios Order of Complexity O() 3+3 O() 2+8 O() Usually does't matter what the costats are... we are oly cocered about the highest power of. 9 O() ( Liear ) Number of Operatios (amout of data) 10 5

6 O() Number of Operatios 30 For a liear algorithm, if you double the amout of data, the amout of work you do doubles 20 (approximately) (amout of data) 11 O(1) ( Costat-Time ) Number of Operatios For a costat-time algorithm, if you double the amout of data, the amout of work you do stays the same = O(1) 1 = O(1) (amout of data) 12 6

7 Liear Search Worst Case: Best Case: O() O(1) Average Case:? 13 Isertio Sort: Worst Case # let = the legth of list. def isort(list) a = list.cloe i = 1 while i!= a.legth do move_left(a, i) -1 i = i + 1 retur a 14 7

8 Isertio Sort: Worst Case # let = the legth of list. def move_left(a, i) x = a.slice!(i) j = i-1 while j >= 0 && a[j] > x do i+1 j = j 1 a.isert(j+1, x) but how log do slice!ad iserttake? 15 move_left(alterate versio) # let = the legth of list. def move_left(a, i) x = a[i] j = i-1 while j >= 0 && a[j] > x do i+1 a[j+1] = a[j] j = j 1 a[j+1] = x 16 8

9 Isertio Sort: Worst Case So the total umber of operatios is + (-1 move_left s) But each move_left performs i+1 operatios, where i varies from 1 to -1: -1 move_left s = Sice = (+1)/2, -1 move_left s = (+1)/2 1 The total umber of operatios is: + (+1)/2 1 = + 2 /2 + /2 1 = 2 /2 + 3/ Order of Complexity Number of operatios 2 O( 2 ) 2 /2 + 3/2-1 O( 2 ) O( 2 ) Order of Complexity Usually does't matter what the costats are... we are oly cocered about the highest power of. 18 9

10 O( 2 ) ( Quadratic ) /2 + 3/2 1 Number of Operatios (amout of data) 19 O( 2 ) Number of Operatios 900 N 2 For a quadratic algorithm, if you double the amout of data, the amout of work you do quadruples (approximately) N (amout of data) 20 10

11 Isertio Sort Worst Case: O( 2 ) Best Case:? We ll compare these algorithms with others soo to see how scalable they really are based o their order of complexities