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} = _______ _____ __________ ____ ________ ___ ________ ______ ___ ______ _____.
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Related Questions

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

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.

 Apply the frequency test, serial test and autocorrelation test to the following sequence at level of significance equation:

 011001110000110010011100. 

 Find a recurrence that generates the sequence 110110110110110. 

 

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. 

 Alice wants to use the Digital Signature algorithm for signing messages. She chooses equation, equation, equation and equation. Alice wants to sign the message equation. She chooses the secret value equation. Explain the procedure that Alice will use for computing the signature. What information will she send Bob? 

Decrypt each of the following cipher texts:
    i) Text: "CBBGYAEBBFZCFEPXYAEBB", encrypted with affine cipher with key (7,2). 
    ii) Text:"KSTYZKESLNZUV", encrypted with Vigenère cipher with key "RESULT". 
   b) Another version of the columnar transposition cipher is the cipher using a key word. In this cipher, we encrypt as follows: Given a key word, we remove all the duplicate characters in the key word. For example, if the key word is ‘SECRET’, we remove the second ‘E’ and use ‘SECRT’ as the key word. To encrypt, we form a table as follows: In the first row, we write down the key word. In the following rows, we write the plaintext. Suppose we want to encrypt the text ‘ATTACKATDAWN’. We make a table as follows:

S E C R T
A T T A C
K A T D A
W N X X X

 

Then we read off the columns in alphabetical order. We first read the column under ‘C’, followed by the columns under ‘E’, ‘R’, ‘S’ and ‘T’. We get the cipher text TTX TAN ADX AKW CAX. To decrypt, we reverse the process. Note that, since we know the length of the keyword, we can find the length of the columns by dividing the length of the message by the length of the keyword.

 

Given the ciphertext ‘HNDWUEOESSRORUTXLARFASUXTINOOGFNEGASTORX’ and the key word ‘LANCE’, find the plaintext.

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