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ASSIGNMENT 6

FOUNDATION OF ALGORITHMS

CHINMAY KULKARNI

(ck1166)

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1.  Given 2 graphs G1 and G2 !e need "# pr#ve "ha" $GraphIs#%#rphis%& '#r $ G1

G2& !hich checks !he"her "he 2 graphs are is#%#rphs #' each #"her is N pr#*e%+

,-pp#se "here is an a*g#ri"h% !hich veri.es "ha" G1 and G2 '#**#! is#%#rphis% #r

n#"+ I" can e d#ne / checking !he"her "he graph G1 and G2 are per%-"a"i#ns #'

each #"her+ I' /es "hen re"-rn "he '#**#!ing per%-"a"i#n and i' n# "hen graphs G 1

and G2 are n#" is#%#rphic+

0nce !e ge" "he per%-"a"i#n !e app*/ "ha" per%-"a"i#n "# #ne #' "he graph and

check !he"her "ha" graph is iden"ica* "# #"her graph+

 # ge" "he per%-"a"i#n "he a*g#ri"h% re-ires 0(%2) !here % is n-%er #' ver"ices

and "# check !he"her "he per%-"a"i#n app*ied "# a graph genera"es an iden"ica*

graph "# #"her graph has 0(%3n) "i%e c#%p*e4i"/+

0vera** "he a*g#ri"h%s c#%p*e4i"/ is 0(%2) hence i" e*#ngs "# N+

2. pr#ve N ≠coNP, then P≠NP

 We can follow a contrapositive way to prove this equation,

 If P=NP then, NP≠coNP,

 Suppose X   ∈ NP, then X   ∈  P and since P is closed under complement that’s why X’∈  P and

hence

X   ∈ conP. (1)

For X  ∈

 coNP, X  ' ∈

 P and since P is closed under complement that’s why X  ∈

 P and therefore

X   ∈  NP (2).

From (1) and (2),

We can say that NP=coNP.

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3.

5hen !e c#nver" "his circ-i" "# ##*ean e4pressi#n !e ge"7

(89 ⋁  (81  ⋀ 82 ⋀   ¬ 89))   ⋀   (89 ⋀  (81  ⋁ 82))   ⋀   (81  ⋀ 82 ⋀

¬ 89)

And !e app*/ a** "he c#%ina"i#ns #' va*-es "# 81 82 89 !e a*!a/s ge" a va*-e #' :7

81  82  89  0-"p-"

  : : : :

  : : 1 :

: 1 : :

  : 1 1 :

  1 : : :

  1 : 1 :

  1 1 : :

  1 1 1 :

Hence !e can sa/ "ha" i" is -nsa"is.a*e+

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4. A ;is

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5e need "# pr#ve "his 'ac"

$A"&   ∈  s-se" s-% and AF   ∈  ARII0N !here s-% #' n-%ers in AF (2pD2")

I' "here e4is"s a s-se" A in , !h#se s-% is " "hen hence "he re%aining n-%ers in

, !i** have s-%

(p D ") "here'#re "here is par"i"i#n in AF s-ch "ha" #"h have s-% as (pD")+

And n#! i' "here is par"i"i#n in AF s-ch "ha" #"h c#n"ain s-% pD" "hen #ne #' "he

par"i"i#n !i** c#n"ain pD2" and re%#ving "his give -s "he se" #' n-%ers !h# have

s-% as "+

Hence !e have red-ced ,-se" s-% pr#*e% "# ar"i"i#n pr#*e% and as !e kn#!

"ha" ,-se" s-% is N c#%p*e"e "here'#re !e can sa/ "ha" ARII0N I, N

c#%p*e"e+

6. a. pse-d#c#de '#r rec-rsive :D1 knapsack pr#*e%

  de' knapsacker#0neRec-rsive ("arge"5" a*-es 5eigh"s n)7

  "arge"5"he "arge" !eigh" "# e achived !i"h %a4i%-%

c#rresp#nding va*-es

  a*-es va*-es ass#cia"ed !i"h n i"e%s

  5eigh"s !eigh"s ass#cia"ed !i"h va*-es in n+

  i' n: #r "arge"5":7 ase case+  re"-rn :

  i' !eigh"nD1O& "arge"5"7 "his %eans "ha" "arge" !eigh" "#

achieve is *ess "han "he

c-rren" !eigh" in "he !eigh" arra/+ hen !e !i** search '#r #"her

!eigh"s in arra/

  Re"-rn knapsacker#0neRec-rsive ("arge"5" a*-es 5eigh"s nD

1)

  E*se7 re"-rn "he %a4 va*-e e"!een i' n"h !eigh" is inc*-ded #r i' i" is

n#" inc*-ded+

  Re"-rn %a4(va*-esnD1O3 knapsacker#0neRec-rsive ("arge"5"D

5eigh"snD1O a*-es 5eigh"s nD1) ) knapsacker#0neRec-rsive ("arge"5" a*-es

5eigh"s nD1))

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6.c. he c#%p*e4i"/ #' d/na%ic pr#gra%%ing a*g#ri"h% '#r :D1 knapsack pr#*e% is

0(n("arge"5)) !here n is n-%er #' ch#ices #' !eigh"s and va*-es+

6.d. N0 ≠NP, because we cannot generalize solution of 1 NP complete solution to all the NP

problems.

 Because, there are large number of NP Problems of different kinds ,we can reduce all Np Problems to

Np complete problems . The NP complete problems can be verified in polynomial time but cannot be

solved in polynomial time.