Question

Question 4 (10 marks) When analysing the complexity of algorithms, there are three main approaches: worst case, best case and average case. As an example, consider measuring the complexity of list-merging by counting the number of comparisons used As a test example, assume the following A1: There are two ordered lists, each of length 4, say A2: Neither list contains repeats, so a! < a2 < аз < a4 and bl <b2 < b3 < b4 A3: The lists are disjoint: no item occurs in both lists. Assumptions A2 and A3 together mean that we have eight items A, B,..., H such that Furthemore, since the number of comparisons used to merge a pair La, Lb will always be the same as the number required to merge the pair Lb, La we will make the further assumption A4: Item H is in list Lb; so b4- H. Then a best case would be La -[A, B, C, D, Lb [E, F, G, H] which requires only 4 comparisons. Let us abbreviate this case to ABCD, EFGH Using a similar abbreviation, a worst case would be ACEG, BDFH, requiring 7 compar sons What about an average case? This could be anything, but to calculate an average number of comparisons we will need to know the distribution of possible inputs. So we make a final simplifying assumption: A5: Any pair of lists La,Lb satisfying A1-A4 is equally likely as input to be merged (a) Calculate the expected value (average) of the number of comparisons required to merge pairs of lists satisfying A1-A5 Hint: First try the corresponding calculation for pairs of lists of length 3. You should find the answer is 4.5. You could get this by writing out all possible pairs of lists (b) Now try the corresponding problem for lists of length 5, but not by listing all possibilities. Rather, first use bionmial coefficients to express the probabilities of ısing each of 9,8,7,6 and 5 comparisons Hint: You will have b5 = J; the value of a5 is crucial (c) [bonus marks Find (and justify) a formula for the average number of comparisons required to merge two sorted lists of length n. The more compact your formula, the better

Answer 1

ans:'-

a) consider following table the list size is 3

S.NO	CASE	COMPARISONS REQUIRED
1	ABC,DEF	3
2	ABD,CEF	4
3	ACD,BEF	4
4	BCD,AEF	4
5	ABE,CDF	5
6	ACE,BDF	5
7	ADE,BCF	5
8	BCE,ADF	5
9	BDE,ACF	5
10	CDE,ABF	5

SO EXPECTED VALUE IS = 1*3+4*3+6*5/10 = 4.5

NOW go for n = 4

i`m showing you a way by which you can calculate easily

if n = 4 then comparisons will be 4,5,6 and 7

when comparison will be 4

then all the elements of the first list will be smaller than all the elements of the second list

so only one combination is possible i.e, ABCD,EFGH

when comparison will be 5

then we have following fixed formula

- - - E, -F G H

it means first 3 characters of first list and only the first character of the second list can be changed

so from 4 we can choose any 3

so 4_C3 = 4

when comparison will be 6

then we have following fixed formula

- - -F, - -G H

it means the first 3 place of the first list and the first 2 elements of the second list can be changed

so from remaining 5 characters we have to find combination of 3

so 5_C3 = 10 ways are there

- - - G, - - - H

so we have 3 positions can be fixed with the remaining 6 characters

6_C3 = 20

so expected value = 1*4 + 4* 5 + 10*6 + 20*7 /35 = 6.4

b) now for n =5

we can directly write

for comparison 5 = 1

for comparison 6 = 5_C1 = 5

for comparison 7 = 6_C2 = 15

for comparison 8 = 7_{C3 = 35}

for comparison 9 = 8_C4 = 70

expected value = 1*5 + 5*6 + 15*7 + 35*8 + 70*9 /126 = 8.33

c)now we can simply write that as formula

NOW

FOR n items in a list

E(comparison) =

$\sum_{i=0}^{n-1}^{n-1+ i}\left C_{i}^{X(n+i)} \right )/\sum_{i=0}^{n-1}^{n-1+i}C_{i}$

use this formula when n = 3

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