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- Let B be the set of all binary strings of length 8. {“00000000”, “00000001”, … “11111110”, “11111111”} How many elements of B have no adjacent digits with the same value? How many elements of B are palindromes? How many elements of B contain two or more 0s?Consider the code C={00010, 01100,01011,11010,01010}. What is the minimum distance of C? For what k is the code C k-error detecting? Write the set of error patterns that C can detect?For what values of c isle(1.2.3)|= 1? A. + B. - V14 C.tV14 D. + 41
- Given two prime numbers, P1 and P2, your task is to reach from P1 to P2 by changing only one digit at a time and also the number you get after changing a digit must also be a prime. For example suppose that P1 is 1033 and P2 is 8179 (both primes), we can reach from P1 to P2 in following way:- 1033-> 1733-> 3733-> 3739-> 3779-> 8779-> 8179. The first digit must be non-zero that is you can not go from 4-digit to 3-digit numbers. Note that there may be multiple way to get to destination but you have to find the way in minimum number of steps. For example the minimum number of steps to get from P1 to P2 in following examples are as follows 1033 8179 ANWSER:- 6 1373 8017 ANWSER:- 7 1033 1033 ANWSER:- 0 Hint: Apply BFSQ8/F(A,B,C,D)= [m(0,2,5,8,10,13)+[d(1,7,12,15) G(A,B,C,D)= [m(6,7,11,12)+[d(0,2,3,8,13,15) Chose the correct answer None of them F= A'C'D'+ BD+ A'BD+ ABC', G= ABC'D'+ AB'CD +CD+ A'CD' F= BD+ A'D', G= AC'D'+ CD+ ABCD+ A'C F= A'C'D+ ABD+ B'D', G= A'BC+ ABC+ AC'D' F= A'B'D'+ AB'D'+ BC'D, G= A'CD'+ BCD+ ACD+ AC'D'Project 1 dealt with single symbol Huffman Coding. Project 2 deals with Extended Huffman Codes. For instance, for a source emitting two symbols A and B, the second order extension involves coding messages AA, AB, BA and BB (22 in number). The third order extension involves messages such as AAA, AAB, etc. (23 in number). The probabilities of such strings are computed by multiplying the individual probabilities. For this project, use the Matlab code you have developed in Project 1 to perform third, fourth and fifth order extensions of a source message. 1. Choose an alphabet a set of at least six (6) symbols with assigned probabilities. 2. Compute the third, fourth and fifth order extension probabilities. 3. Using the built-in algorithm, derive the Huffman Code for each extension. 4. Compute the following quantities: (i) Average length of the codeword; (ii) The code efficiency; (iii) The Compression Ratio; (iv) Speed of computation.
- Q3/Full the following blanks (15)10 =( )Gray=( )BCD(5321)=( )EX-3 O O None of them O O (15)10 (1000)Gray-(0001 0110 )BCD(5321)=(10010 )EX-3 O (15)10 = (1000) Gray=( 1001 0011 )BCD(5321)=(10010 )EX-3 (15)10 (1111)Gray-(0001 1001 )BCD(5321)=(10010 )EX-3 (15)10 (1111)Gray-(0001 1000 )BCD(5321)=(10010 )EX-3لتكن المصفوفة [x=[1,4,3,2;8,7,9,0;4,6,7,3 [Y=[1,6,7;5,6,4;7,8,9;0,3,4 [W=[1,3,4,5;7,6,8,9;0,9,1,2;4,5,2,3 اكتب كلا من الايعازات التالية : كون المصفوفة الفرعية من الصفوفةx حيث ان z= 6. 8 9. 9. 2.Substitution Cipher is an old technique that substitutes one thing for another. For example, alphabet messages can be ciphered by applying the following substitution. plaintext: abcdefghijklmnopqrstuvwxyz | | V Vciphertext: mnbvcxzasdfghjklpoiuytrewq However, this type of ciphering produces patterns since the same plaintext strings produce the same cipher text. Polyalphabetic Cipher, on the other hand mitigates this problem as plaintext characters are not always replaced with the same cipher text character. The replacement of each character dependson a key as well as on the position of the character in the text; for example, by using a formula such as: for (int i=0; i < length of P; i++) C[i] = P[i] + K + (i mod 3) In essence, if the key, K, is choses as K=10, then 10 is added to characters in position 0, 3, 6, …; and 11 is added for those in…
- Question4/ Combine the following two chromosomes using the PMX method to form two new children (cut points) (r1=2) and (r2=6) Suppose indicators start from zero Chromosome 1 6 4 5 7 8 9 2 3 1 Chromosome 2 1 2 3 4 5 6 7 9 m∞0 8True or false: =taut ((Vx)ƒx = gx \ (fx = y → gx = z)) → ((\x) fx = gx → (fx = y \ gx = z)) ? Select one: True O Falsedef mystery (1st); for idx in range(1, len(1st)); tmp = 1st[idx) idx2 = idx while 1dx2 > 9 and 1st[idx2-11 tmp: 1st[idx2] 1st[10/2 - 11 1dx2 = 1dx2 - 1 1st[idx2] = tmp print(1st) a. If we call this function as follows: mystery(lt) where ist 15, 2, 8, 11, what is printed out t clear about what is printed out, don't make me try to figure it out). b. What does this function do? c. What is the complexity of this function? Oni Ora, On³ Ologinil, Onioginil? Explain your reasoning