# Solutions Manual for an Introduction to Cryptography with Coding Theory (2nd Edition)

## Wade Trappe, Lawrence C. Washington

The accompanying options handbook to an creation to Cryptography with Coding thought (2nd version) through Wade Trappe, Lawrence C. Washington (Pearson).

174649 38 x Out[257]= 358910 1919 In[258]:= %/.x zero 36599208 one hundred and one In[259]:= Mod Numerator % Out[259]= 69918 38, 358910 , 3876, 9612 ,x Out[258]= PowerMod Denominator % , 1, 984583 , 984583 In[260]:= InterpolatingPolynomial 9581 3876 x Out[260]= 9612 9618 In[261]:= %/.x 3876, 9612 , 23112, 28774 zero seventy eight ,x ,x 9218710 1603 In[262]:= Mod Numerator % Out[262]= 927070 Out[261]= PowerMod Denominator % , 1, 984583 , 984583 23112, 28774 , 432, 178067 In[263]:= InterpolatingPolynomial 149293.

1, zero, 1, 1, zero, 1, 1, zero, 1, zero, zero ,2 In[308]:= Mod s Transpose golay 15 Out[308]= 1, zero, 1, zero, 1, zero, zero, 1, zero, zero, 1, zero ,2 In[309]:= Mod s Transpose golay sixteen Out[309]= zero, zero, zero, 1, zero, zero, zero, zero, zero, zero, zero, zero ,2 we will cease the following on the grounds that this vector has weight 1. we've e16 1 and e4 1. The corrected vector is (1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1) the 1st 12 entries provide the message: (1,1,1,1,1,1,1,1,1,1,1,1) challenge 2. Multiply the obtained vector through the transpose of the parity money.

Chrem([335,1234],[523,4447]); 437040 chrem([-335,1234],[523,4447]); 1234 chrem([335,-1234],[523,4447]); 2324547 chrem([-335,-1234],[523,4447]); 1888741 a hundred > > > > > > > those are the 4 sq. roots. challenge thirteen. 48382&^(83988/4)mod 83987; 60555 60555&^2 mod 83987; 35605 this can be the detrimental of 48382 mod 83987: -48382 mod 83987; 35605 for this reason, we came upon the sq. root of -48382 mod 83987. a hundred and one Chapter 6 - Maple > challenge 1. Encrypt all the messages and spot which one it > >.

Maple challenge 1. (a) when you consider that r is identical in either meassages, the values of okay > are a similar. > half (b) We stick to the process on web page 181. > m1:=809; m2:=22505; s1:=1042; s2:=26272; r:=18357; p:=65539; m1 := 809 m2 := 22505 s1 := 1042 s2 := 26272 r := 18357 p := 65539 > we wish to remedy (s1-s2)k=m1-m2 mod p-1. First, compute a gcd: > gcd(s1-s2,p-1); 6 > because the gcd is 6, there are 6 ideas to the congruence. Divide > the congruence we wish to clear up via 6 to procure ((s1-s2)/6)k = > (m1-m2)/6.

(b) those are exactly the elements of 324 that are 2 and three. eleven. >> powermod(26055,(34807+1)/4,34807) ans = 33573 >> mod(-33573,34807) ans = 1234 12. >> factor(2325781) ans = 523 4447 150 >> powermod(1522756,(523+1)/4,523) ans = 335 >> powermod(1522756,(4447+1)/4,4447) ans = 1234 Now we prepare the 4 mixtures >> crt([335 1234],[523 4447]) ans = 437040 >> crt([-335 1234],[523 4447]) ans = 1234 >> crt([335 -1234],[523 4447]) ans = 2324547 >> crt([-335 -1234],[523 4447]) ans = 1888741.