Question

Apply poker test to the following sequence with level of significance equation.
equation 1001101000010000101111011
equation 01110100101101100100110. 

09 Jan 2026
Answer :
Word Count : 661
We are asked to apply the poker test to the given binary sequence with a significance level (\alpha = 0.05). Let’s solve it step by step manually. The poker test is a statistical test for randomness based on counting patterns in blocks of digits, usually of length 4 (like poker hands: all different, pairs, three of a kind, etc.). Step 1: Combine the sequence The full sequence is: ``` 100110100001000010111101101110100101101100100110 ``` Count the total number of bits: * First line: 25 bits * Second line: 23 bits * Total: 48 bits We will divide into blocks of 4 bits (common for the poker test). [ \text{Number of 4-bit blocks} = \frac{48}{4} = 12 \text{ blocks} ] Step 2: List all 4-bit blocks 1. 1001 2. 1010 3. 0001 4. 0000 5. 1011 6. 1101 7. 1011 8. 1010 9. 0101 10. 1011 11. 0010 12. 0110 Step 3: Classify each block Poker test categories (for 4-bit blocks): 1. All
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Related Questions

Alice wants to use the ElGamal digital signature scheme with public parameters equation, equation, secret value equation and equation. She wants to sign the message equation and send it to Bob. She chooses equation as the secret value. Explain the procedure that Alice will use for computing the signature of the message. What information will she send Bob? 

 

Let equation. Find the product of x2 + 2x + 1 + (f(x)) and x2 + 3x - 1 + (f(x)) using the algorithm in page 23, block 1. You should show all the steps as in example 11, page 22, block 1. 

 

Find the inverse of 13 (mod 51) using extended euclidean algorithm.

In this exercise, we introduce you to Hill cipher. In this cipher, we convert our message to numbers, just as in affine cipher. However, instead of encrypting character by character, we encrypt pairs of characters by multiplying them with an invertible matrix with co-efficients in Z26.

 

Here is an example: Suppose we want to ENCRYPT "ALLISWELL". Since we require the plaintext to have even number of characters, we pad the message with the character ‘X’. We break up the message into pairs of characters AL, LI, SW, EL and LX. We convert each pair of characters into a pair elements in Z26 as follows:

 

equation\underline{\overline{\begin{aligned}AL&\quad(0,11)\\LI&\quad(11,8)\\SW&\quad(18,22)\\EL&\quad(4,11)\\LX&\quad(11,23)\end{aligned}}}">

 

Next, we choose an invertible equation matrix with coefficients in equation, for example, equation.
This matrix has determinant equation and equation is a unit in equation with inverse equation. We write each pair of elements in equation as a column vector and multiply it by A:
equation
We then convert each pair of numbers to a pair of characters and write them down. In this example, we get the cipher text "LSPFYGXUEN" corresponding to the plain text "ALLWELL". To decrypt, we convert pairs of characters to pairs of numbers and multiply by equation and we have
equation
Decrypt the text "TWDXHUJLUENN" which was encrypted using the Hill’s cipher with the matrix equation as the encryption matrix.

) Let equation. We represent the field equation by equation. Let us write equation. The table of values is given below:
 

 


i
 

  


γi
 

 Vector  


i
 

  


γi
 

 Vector
0  


1
 

  


(0,0,0,1)
 

 8  


γ2+1
 

  


(0,1,0,1)
 
1  


γ
 

  


(0,0,1,0)
 

 9  


γ3+γ
 

  


(1,0,1,0)
 
2  


γ2
 

  


(0,1,0,0)
 

 10  


γ2+γ+1
 

  


(0,1,1,1)
 
3  


γ3
 

  


(1,0,0,0)
 

 11  


γ3+γ2+γ
 

  


(1,1,1,0)
 
4  


γ+1
 

  


(0,0,1,1)
 

 12  


γ3+γ2+γ+1
 

  


(1,1,1,1)
 
5  


γ2+γ
 

  


(0,1,1,0)
 

 13  


γ3+γ2+1
 

  


(1,1,0,1)
 
6  


γ3+γ2
 

  


(1,1,0,0)
 

 14  


γ3+1
 

  


(1,0,0,1)
 
7  


γ3+γ+1
 

  


(1,0,1,1)
 

      

 

 i) Prepare logarithm and antilogarithm tables as given in page 23 of block 1. 
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Find the keys for the second round.
6) a) Considering the bytes 10001001 and 10101010 as elements of the field equation, where g(X) is the polynomial X8 + X4 + X3 + X + 1, find their product and quotient. 

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