E (x) = (ax + b) mod m D (x) = a -1 (x - b) mod m For more math formulas, check out our Formula Dossier What 4 concepts are covered in the Affine Cipher Calculator? Each character from the plaintext is always mapped to the same character in the ciphertext as in the Caesar cipher. The basic formula to be used in such a scenario to generate a multiplicative cipher is as follows (Alphabet Number * key)mod (total number of alphabets) The number fetched through output is mapped in the table mentioned above and the corresponding letter is taken as the encrypted letter. Examples for property 4): 24 and 28 are products of primes and prime powers. Each row that contains each integer from 0 to 25 exactly once and therefore yields a unique cipher letter will serve. This is important because if the key shares a factor with the plaintext, it can be easily broken by factoring in the key. the commonly used RSA Cipher is based on the relative slowness of such factoring programs. 7 ((8)= ((23)=23 -22 =4 as 1,3,5,7 are relative prime to 8. Notice, that all we need to find are the different primes, say p1, p2,, pn, as our explicit formula for the number of unique encryptions appears to be: Formula for the number of good keys for any alphabet length M: For an alphabet length M, there are ((M) = M * (1- 1/p1) * (1- 1/p2) ** (1- 1/pn) good keys where each pi is a prime divisor of M. It is really enjoyable to use this simple formula as we just need to find all prime divisors of M and dont have to worry about how often they occur. It is suitable for small-scale applications but not recommended for practical purposes. that 3 and 9 are inverse to each other because of the commutative property of the MOD-multiplication (exhibited by the diagonal as a line of reflection). Instead of performing a transformation before encryption, this implementation allows several alphabets to be specified (see below), thereby accomplishing the same within the encryption process. You can verify this as follows: out of the __ integers that are less than 65, we first cross out all the ___ multiples of __ and then cross out the __ multiples of __ resulting in ______ = 48 good keys. Connect and share knowledge within a single location that is structured and easy to search. } Notice, that property 3) became useless for the calculation process since factors that are relative prime are separated via property 4). Ok, lets continue with the encoding part. What is the inverse of 5 MOD 11? Method 2: Merged: In the alphabet, mod 22 is calculated because the alphabet contains 22 elements. Below is the C++ program that performs the task for us, it just finds all the factors of an entered alphabet length M by testing all the integers less than M for possible factors. Example1: If M=24=3*8=3*23, then ((24) = ((3*23) using property 4) yields = ((3)*((23). 5 7 11 13 17 19 23 25 29 31 35 As an example, lets encode and decode NAT and ANT. It would take quite a long time for a computer to brute-force through a majority of nine million keys. margin-bottom: 16px; Network Security: Multiplicative InverseTopics discussed:1) Explanation on the basics of Multiplicative Inverse for a given number.2) Explanation on the basi. Therefore, no matter how he decides to crack the cipher text, it wont take long. In, this way you can implement Encrypt a plain text and Decrypt a cipher text for Multiplicative cipher in cryptography. Calculate the value of each letter as follows (where a and b are the keys of the password): E (x)= (ax + b) mod m 3. rev2023.5.1.43405. This weirdness is not really weird. Note The advantage with a multiplicative cipher is that it can work with very large keys like 8,953,851. Introduction to Monotonic Stack - Data Structure and Algorithm Tutorials. However, it can be simplified further using the fact that we are considering here alphabets of length M that are powers of a prime p: M=pn for some positive integer n. Thus, our formula simplifies to: u(M) = pn pn/p which simplifies further to = pn - pn-1. I accomplish this. dCode retains ownership of the "Multiplicative Cipher" source code. It converts to the plain letter number 26 so that we now have to encrypt MOD 27. Does the increase of our alphabet length by 1 increase the number of unique encryptions obtained? It would take quite a long time for a . To find the inverse for each good key a, you just need to look back at the 26 by 26 encryption table. Lets add a dot to our alphabet to denote the end of a sentence in the original message. Agree Note the difference in 'D' and 'd': The index value is the same, but the 'd' is. In fact, the cipher E can only be an even cipher letter as only even numbers appear in the E-column. This is very likely in English texts and virtually certain in the German language where on average every 5th letter is an E. Even if an eavesdropper decides to produce all 12 possible plain texts, they can be generated with the help of a computer within a few seconds. On the right we ended up with the explicit formula for ((M) when M consists of one prime power and two primes. To find a multiplicative inverse We need to find a number x such that: If we find the number x such that the equation is true, then x is the inverse of a, and we call it a^-1. Thus, among those numbers that occur twice in the cipher code, 14, 17 and 20, we can eliminate the odd 17. You may see ads that are less relevant to you. We denote 5-1 the inverse of 5. Thus, we now go ahead and practice a bit more computer programming. } 4 Sphero Up to 1 Hour Grades: 5 to 8. However, it turns out to be indispensable when M is not the product of two primes, but say a product of a prime and a prime power. Moreover, you can see that the plain letter V encrypts to the cipher text letter b (=1) when using a=5 as the encoding key. (I.e. Find mod of any numb. 36 modulo 26 = 10 so the letter K would be chosen. This yields the correct plain text: Cipher textanromrjukahhouh013171412179201007714207 0131981819742017178417PLAIN TEXTANTISTHECARRIER As you can see, detecting the most frequent cipher letter is of enormous help in cryptography. First we need to calculate the modular multiplicative inverse of keyA. Now when a=25, we have: 25*25 = 625. Therefore, we just have to add a number in order to get k=111. What are the variants of the Multiplicative cipher. How to pick a symmetric cipher for a given cipher text size? Then we choose a matrix of n x n size, which will be the cipher's key. If a = 1, the Affine cipher is equivalent of a Caesar cipher. In order to create a n x n size matrix, keyphrase length should be square of an integer, i.e., 4, 9, 16. or . 9 Multiplicative cipher encryption|Multiplicative cipher|Multiplicative cipher example|What is multiplicative cipher PLAYFAIR CIPHER WITH EXAMPLE||SUBSTITUTION TECHNIQUE||MATHEMATICS OF. Those are the 8 integers 3, 6, 9, 12, 15, 18, 21, 24. Of course, you dont want to receive any more ambiguous messages. We have to understand why multiplying by a bad key a MOD 26 yields some integers more than once and others not at all. Secondly, we would translate every upper case plain letter into a lower case cipher letter so that we dont reveal information about the beginning of a sentence. As 36=2*2*3*3, the possible keys are basically all numbers not multiples of 2 and/or 3. They seem to not follow any apparent pattern. For the purpose of setting up an explicit formula for ((M), we now try to give the three factors (in parentheses) the same format. Try to explain this, than continue reading! Thus, the number of bad keys can simply be found by dividing the alphabet length M by the only prime divisor p and subtracting 1 from that fraction: M/p - 1. Note: This cipher is closely related to the. Let s be such a reversible function. Our good-key-criterion declares those integers to be good keys that are relative prime to 27. 2) The setwidth command setw() assigns as many spaces as entered in the parentheses for a numerical output in order to have a well-formatted output. 15 Or can we even increase the mere 12 unique encryptions for the Multiplication Cipher by varying the alphabet length? The copy-paste of the page "Multiplicative Cipher" or any of its results, is allowed as long as you cite dCode! Options regulate the case when a letter does not appear in any alphabet: it is not encrypted, but transferred directly to the output. If a single character is encrypted by E(C) = (c * k) % 36 then possible keys k are numbers that are coprime to 36, ie.gcd(k,36)=1.Furthermore it makes not much sense to consider numbers not between 1 and 36, because of the modulo. Moreover, we build the mathematical foundation to understand secure encryption systems such as the RSA encryption. Thus, safer encryptions are necessary. Here, it reduces the number of possible good keys to two. Equivalently stated, 105 divided by 26 leaves a remainder of 1. 25 We know already that: ((60) = ((22*3*5) = (22-21)*(3-1)*(5-1)((M) = ((p12* p2* p3) = (p12- p11)*( p2-1)*( p3-1). It thus gives a great example that we are only guaranteed to solve this equation for numbers that form a group with respect to multiplication MOD 26. We factor p1=2 yielding = 2*(2-1)*(3-1)*(5-1) = p1* (p1- 1)*( p2-1)*( p3-1). We first found the bad keys as the multiples of the prime divisors of the alphabet length M. Consequently, the good keys are the remaining integers less than M. Again a perfect task for a computer, especially when we have to find the prime divisors of bigger integers. Generally: The good keys are those as that are relative prime to M and are denoted as ZM*. v l X X X How to recognize a Multiplicative ciphertext? Example3: Doing arithmetic MOD 7, the inverse of a=3 is a-1 = 5 because a * a-1 = 3*5 = 15 = 1 MOD 7. 3) If the alphabet length M is a product of two prime numbers p and q The last case we have to study is when M is a product of two primes. color: #ffffff; For a given alphabet, there are only a few possible keys. 19 This is important because even if a key is secure when it is first chosen, it can become less secure over time as technology and methods for breaking encryption increase. In this video u will learn how to encrypt the message using multiplicative cipher technique.Plain text to cipher text.Calculator tricks. When a letter occurs in several alphabets, the first of these alphabets is used. Multiplicative Cipher In a Multiplicative cipher, each character of the alphabet is assigned a value (starting at a zero index [A=0, B=1, etc]) and a coprime key to the length of the alphabet is chosen. Thus they have the following restrictions: Online calculator: Modular Multiplicative Inverse Calculator - PLANETCALC Since the number of unique encryptions u is a function of the alphabet length M, we may write in function notation: u(M) to denote the number of unique encryptions (which equals the number of good keys) as a function of M. I.e. Since one can be divided without remainder only by one, the equation above has the solution only if . For M=31 we have u(31)=30. You can verify this as follows: out of the 38 (=p*q-1) integers that are less than 39, we first cross out all the 12 (=13-1) multiples of 3 {3,6,9,12,15,18,21,24,27,30,33,36} and then cross out the 2 (=3-1) multiples of 13 {13,26} resulting in 38 12 2 = 24 good keys. Cryptography Tutorial - Multiplication Cipher : Decode It - TI89 The x values are the ones that we can choose independently, here the length of the alphabet M. Each y-value is dependent on the choice of x, i.e. 8 23 Example2: For M=9=32 we have u(9) = 32 - 31 = 9 3 = 6 which are the 6 good keys a=1,2,4,5,7,8. However, when using MOD arithmetic and solving 23=5*P MOD 26, we dont deal with fractions but only integers. unchanged so that you can detect the format of the original message easier. He decodes all the other cipher letters by finding their corresponding number in the 23rd row (see above) and then goes up that column to find the original plain letter. } For a check: the same eight integers 1,5,7,11,13,17,19,23 are relative prime to 30 and are thus the good keys for M=30. color: #ffffff; The ultimate trick to yet produce the same format is factoring: from each parentheses we factor the first integer (which is a divisor of M) and obtain: ((60) = 22*(1 -1/2) * 3*(1 -1/3) * 5 * (1 -1/5)((M) = p12 * (1 -1/ p1) * p2*(1 -1/ p2) * p3 * (1 -1/ p3) = 22*3*5*(1 -1/2)*(1 -1/3)*(1 -1/5) = p12* p2* p3*(1 -1/ p1)*(1 -1/ p2) * (1 -1/ p3) = 60*(1 -1/2)*(1 -1/3)*(1 -1/5) = M * (1 -1/ p1) * (1 -1/ p2) * (1 -1/ p3). In fact, any character is stored as a number: i.e. For classical methods, the alphabet often consists only of the uppercase letters (A-Z). This brute force approach will work fast enough for integers M that have 10 digits or less.