Microsoft Word Hw#3

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Microsoft Word Hw#3

  1. 1. Computer and Network Security. Name: Khalid Khalil Kamil Matric. No.: G0327887. Solutions to Assignment # 3: 1- Use Euclid’s algorithm to find the greatest common divisor (gcd) of: a. gcd(14,105): Solution: gcd(105,14)=gcd(14, 105 mod 14) = gcd(14, 7) = gcd(7,14 mod 7) = gcd(7,0)=7. b. gcd(180, 1512) Solution: Gcd(1512,180)=gcd(180,1512 mod180)=gcd(180,72)=gcd(72,36)=gcd(36,72mod36) =gcd(36,0)= 36. c. gcd(1001,7655): solution: gcd(7655,1001)=gcd(1001,7655mod1001)=gcd(1001,648)=gcd(648,1001mod648) =gcd(648,353)=gcd(353,648mod353)=gcd(353,295)=gcd(295,353mod295) =gcd(295,58) =gcd(58,295mod58)=gcd(58,5)=gcd(5,58mod5)=gcd(5,3)=gcd(3,5mod3)=gcd(3,2) =gcd(2,3mod2)=gcd(2,1)=gcd(1,2mod1)=gcd(1,0)=1. d. gcd(24140,16762): solution: gcd(24140,16762)=gcd(16762,24140mod16762)=gcd(16762,7378) =gcd(7378,16762mod7378)=gcd(7378,2006)=gcd(2006,7378mod2006) =gcd(2006,1360)=gcd(1360,2006mod1360)=gcd(1360,646)=gcd(646,1360mod646) =gcd(646,68)=gcd(68,646mod68)=gcd(68,34)=gcd(34,68mod34)=gcd(34,0)=34. 2- Find: a. 13 mod 11=2mod11=2. b.875mod9=2mod9=2. c.2594mod48=2mod9=2. d.217mod21: Solution:
  2. 2. 21mod21=2 22mod21=4, 24mod21=(4x4)mod21=16, 28mod21=(16x16)mod21 =256mod21=4, 216mod21=(4x4)mod21=16, 217mod21=[(21mod21)x(216mod21)]mod21=[2x16]mod21=32mod21=11mod21=11. 3-Using Fermat’s theorem, find 3201mod11? Solution: Fermat’s theorem states that: ap-1 ≡1 mod p , provided that, p is a prime number and a is positive integer not divisible by p. 3201mod11=[(310mod11)20 x (31mod11)]mod11 ====== ▼ applying Fermat’s theorem [(1 mod 11)20x (3 mod 11)]mod 11 [1x3] mod 11=3. 4- Perform encryption and decryption using RSA algorithm for the following: a- p=3 q=11 m=5 Solution Key generation: n=p * q=3*11=33. φ(n)=(p-1)(q-1)=2*10=20. Select integer “e” where: gcd(φ(n),e)=1 and 1<e< φ(n) Choose e=3. Calculate d; where d≡e-1mod φ(n) de≡1 mod φ(n) where: d< φ(n) d*3≡1 mod 20 d=7. Public key: KU={3,33}; Private Key: KR={7,33}. Encryption: C=Me mod n = 53mod 33=[(51mod33)(52mod33)]mod33=[5*25]mod33 =26. Decryption: M=Cd mod n=267mod33=[(261mod33)(262mod33)(264mod33)]mod33 =[26*16*(16*16 mod33)]mod33=[26*16*25]mod33=5. b- p=13 q=11 M=9. Key generation:
  3. 3. n=p * q=13*11=143. φ(n)=(p-1)(q-1)=12*10=120. Select integer “e” where: gcd(φ(n),e)=1 and 1<e< φ(n) Choose e=7. Calculate d; where d≡e-1mod φ(n) de≡1 mod φ(n) where: d< φ(n) d*7≡1 mod 120 721≡1mod120 d=721/7=103. Public key: KU={7,143}; Private Key: KR={103,143}. Encryption: C=Me mod n = 97mod 143=[(91mod143)(92mod143)(94mod143)]mod33 =[9*81*(81*81)mod143]mod143=[9*81*126]mod43=48. Decryption: M=Cd mod n=48103mod143 (48mod143)=48. (482mod143)=(48*48)mod143=16. (484mod143)=(16*16)mod143=113. (488mod143)=(113*113)mod143=42. (4816mod143)=(42*42)mod143=48. (4832mod48)=(48*48)mod143=16. (4864mod143)=(16*16mod)143=113. (4896mod143)=(16*113)mod143=92. (48100mod143)=(92*113)mod143=100. (48102mod143)=(100*16)mod143=27. (48103mod143)=(27*48)mod143=9. 5- In a public key system, an intruder intercepts the cipher text C=25, which is destined to a user whose public key is {7,133}. What is the plain text message M? Solution: We have: KU={e,n}={7,133} n=133=20305071110130170191=7*19. p=7, q=19. φ(n) =(7-1)(19-1)=108. de ≡ 1 mod φ(n) 7d ≡ 1 mod108. 7d mod 108=1. Maybe: 7d=108+1? d=109/7=15.57 no Maybe: 7d=108*2+1 d=217/7=31 OK. KR={31,133}. M=Cd mod n =2531 mod 133 251mod133=25. 252mod133=(25*25)mod133=93. 254mod133=(93*93)mod133=4 258mod133=(4*4)mod133=16. 2516mod133=(16*16)mod133=123. 2524mod133=(123*16)mod133=106. 2528mod133=(106*4)mod133=25.
  4. 4. 2530mod133=(25*93)mod133=64. 31 M = 25 mod133=(64*25)mod133=4. M=4.

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