You can change your choice at any time on our, Modular Multiplicative Inverse Calculator. Options regulate the case when a letter does not appear in any alphabet: it is not encrypted, but transferred directly to the output. div#home { This corresponds to the K. If "GEHEIMNIS" would be completely encoded by this procedure, the ciphertext would be: "SMVMYKNYC". We also turn the plaintext into digraphs (or trigraphs) and each of these into a column vector. For now, lets focus on the lower case letters. Reminder : dCode is free to use. Code Or can we even increase the mere 12 unique encryptions for the Multiplication Cipher by varying the alphabet length? Coincidence? If you are able to invent a fast factoring algorithm, you will not have to worry about a future job. Note the difference in 'D' and 'd': The index value is the same, but the 'd' is. Since any plain letter fulfills the condition in while(cl!='~') The loop is reentered and the next cipher letter is displayed in cout << cl; We can then end this while loop by entering ~ and then choose to either encode, decode or exit the program. The inverse function returns the n-th character for a number n in L. To n, the length of the list L is added or subtracted as often as necessary until the index lies in the list. b ^ ^ ^ 8 ^ a G n n n n n R R R f h h h h h h $ u R ` R R R n n n n 7 R j n n n f R f k \ ^ % n n `d P ^ v$ .$ r % T 0 G $ r 2 % n n n n Chapter 2 Multiplicative Cipher In this chapter we will study the Multiplicative Cipher. an idea ? Similarly, the multiples of a=7 will translate an F (=5) into an 0 (=a) because 7 does so. We can see in the table that an A will always translate into 0 (=a) since the product of any such key a with 0 (=A) yields 0. What 1 formula is used for the Affine Cipher Calculator? will translate the H (=7) into a (=0), because 5*7 = 35 = 0 MOD 35. 23 Modular inverse of a matrix. We saw that an alphabet length of M=26 produces 12 unique encryptions, since the even numbers as multiples of 2 and the 13 are the 13 bad keys. The key should be kept secret and only shared with authorized parties. for M=29 we have u(29)=28. In our example, after subtracting 101 from the plain letter c we get the desired 2 that is now multiplied by a=5 yielding 10. } Multiplicative encryption uses a key $ k $ (an integer) and an alphabet. 28 equals 2*2*7 so that all the keys that are multiples of 2 or 7 do not and all non-multiples of 2 or 7 do produce a unique encryption: Z28* = {1, 3, 5, 9, 11, 13, 15, 17, 19, 23, 25, 27} allowing only 12 different unique encryptions. To find the multiplicative inverse of a real number, simply divide 1 by that number. Which ones are those? 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. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. If we dont want to give an eavesdropper any additional information about our secret message, we would firstly either not use such characters at all or we would expand our alphabet length and encode them just like the other plain letters. How does the j decode to the H, and the u decode to the E? Thus, dividing is performed slightly different: instead of dividing by 5 or multiplying by 1/5, we first write 5-1 (instead of 1/5) where 5-1 now equals an integer and multiply both sides by that integer 5-1. They seem to not follow any apparent pattern. The key should be changed frequently to prevent cryptographic attacks. 19 The number fetched through output is mapped in the table mentioned above and the corresponding letter is taken as the encrypted letter. We can therefore always find a-1 for a given good key a. In order to have a modular multiplicative inverse, determinant and modulo (length of the alphabet) should be coprime integers, refer to Modular Multiplicative Inverse Calculator. Aha, there is 105 = 21*5 so that 21*5 = 1 MOD 26. color: #ffffff; But the modular multiplicative inverse is a different thing, that's why you can see our inverse modulo calculator below. Learn more. That are those that dont have a common divisor with 26. Example3: Now, it is your turn. Activity Code: 65665. It would take quite a long time for a . The explanation of cipher, which is below the calculator, assumes an elementary knowledge of matrices. It is not difficult to understand that the length of such numbers requires the usage of computers. Lets investigate this in the following section. Then we perform the reverse operations performed by the encryption algorithm. A summary of our explorations for the number of good keys shows: 1) u(p) = p - 1, if M is prime M=p. Apr 6, 2013 at 10:46 . This inverse modulo calculator calculates the modular multiplicative inverse of a given integer a modulo m. First of all, there is a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x, and it is not the same as modular multiplicative inverse. Certainly, it might be a double encoded message that has to be decoded twice, possibly using two different keys or even two different ciphers. Alternatively, the non-alphabet letters in the key and the plain text can also be filtered out to increase the security. Therefore, Formula for the number of good keys if M is a prime power: If M = pn , the number of good keys is u(M) = pn - pn-1. Before we conclude this section with the highlight of creating a sole formula for ((M) from these four properties, we will consider 2 examples for each of the 4 properties of Eulers (-function. Example2: M=81=34 has again 3 as the only prime divisor and thus b = 81/3 1 = 34/3 1 = 33 1 = 26 bad keys. It has to be placed after the cout command as in: cout << setw(2) << j*factor. All symbols to be encrypted must belong to alphabet, Everyone who receives the link will be able to view this calculation, Copyright PlanetCalc Version: Therefore, we just need to divide 27 by the only prime divisor 3 and subtract 1 at the end to find the number of bad keys: 8 = 27/3 1. Therefore, no matter how he decides to crack the cipher text, it wont take long. It was encoded MOD 26. Now, lets look at examples for MOD arithmetic: Example2: The inverse of a=3 is a-1 = 2 MOD 5 because a * a-1 = 3*2 = 6 = 1 MOD 5. 565), Improving the copy in the close modal and post notices - 2023 edition, New blog post from our CEO Prashanth: Community is the future of AI, Link between Cipher suites and certificate key. (I.e. The trick is now that if we enter more than one letter all but the first entered letter are buffered (which means temporarily stored in the computers RAM) until read in in cin >> pl; inside the while loop. Furthermore it makes not much sense to consider numbers not between 1 and 36, because of the modulo. The modular multiplicative inverse of a modulo m can be found with the Extended Euclidean algorithm. Examples are: 4 and 5 are relatively prime because gcd(4,5)=1. Firstly I have no idea how they derived this formula, but I think I have a general idea. Now, lets come to the highlight of this section: I will show you in a few steps how to compute ((M) for any M from one equation instead of combining the four properties? For the same reason, an alphabet length of M=31 produces u=30 unique encryptions. 16 color: #ffffff; This allows us to force results to belong to the same alphabet. (Definition). ((24) = ((23 *3) = ((23)*((3) = (23-22)*(3-1) = 4*2 = 8 as 1,5,7,11,13,17,19,23 are relative prime to 24. Note: This cipher is closely related to the. Subsequently, ( is computed by property 1) if such factors are primes or by property 2) if they are prime powers. Each character from the plaintext is always mapped to the same character in the ciphertext as in the Caesar cipher. A little computer program turns out to be again very valuable as the number of good keys can be easily determined by first finding all prime factors of M to then use the above explicit formula. 18 Example: D = 3, so $ 3 \times 17 \mod 26 \equiv 25 $ and the letter at rank 25 is Z. This is also the case when the letter is in the key. if the letter e (the most frequent letter in the English language) occurs 20 times in the plain text its replacement letter will appear 20 times in the cipher text. This eventually enables us to calculate the number of integers that are relative prime to these primes and prime powers. Well, I leave all the entered non-letters such as ! How would anyone ever break even this basic, amateurish cipher/encryption scheme? The multiplicative cipher has little interest, but it is often used for learning computer science and ciphers. Combining this fact with the fact that each key a possesses a decoding key a-1, the set of the good keys forms a commutative group with the unit element 1. Which number would that be? When doing so we will discover very important mathematical encryption tools such as Eulers (-function, Eulers and Lagranges Theorem and study further examples of groups, rings and fields. Why is that? Lab 2. For an alphabet length of 26 this corresponds to 12 keys: 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23 and 25. I will couple the Multiplication Cipher with the Caesar Cipher (which produces 26 unique encryptions) to obtain a super encryption that will allow 12*26=312 possible unique encryptions. Multiplicative encryption uses a key k k (an integer) and an alphabet. I want to show you an example where we used it already. 24 In order to simplify the representation of the alphabets, the following abbreviation has been introduced: The minus sign in the following letter 1-letter 2 is extended to all the letters between the two flanking letters. They are very special primes as they must consist of 100 digits or more. Thus, x indeed is the modular multiplicative inverse of a modulo m. Everyone who receives the link will be able to view this calculation, Copyright PlanetCalc Version: The alphabet function sL returns the smallest index at which it occurs to a letter that is present in L. The index of the first character can be configured. This calculator uses Hill cipher to encrypt/decrypt a block of text. To show this, let's look at this equation: This is a linear diophantine equation with two unknowns; refer to Linear Diophantine Equations Solver. 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. This modulo calculator performs arithmetic operations modulo p over a given math expression. It is not difficult to find the encoded E in English documents as every 8th letter on average is an E (about 13%), it is therefore by far the most frequent letter. The 14 as the possible cipher E then tells him to test the keys a=10 and a=23. 6*3=18. Simply: Z26* = {1,3,5,7,9,11,15,17,19,21,23,25}. Simply: Z26 = {0,1,2,3,, 24,25}. The first time the loop passes the line cout << cl; the translated plain letter pl that was read in as cin >> pl; before the while loop is output as its cipher letter cl. The index of coincidence is unchanged from plain text. WAP to implement Additive cipher(key=20), Multiplicative cipher(key=15)and affine cipher(key=15,20). Here is a non-calculator way to understand why 25 is inverse to itself: Since 25 = -1 MOD 26, it follows 25 * 25 = (-1) * (-1) = 1 MOD 26. Therefore, the set of all encoding keys must equal the set of all decoding keys. Here is the C++ Code for the encryption and decryption of the multiplication cipher: //Multiplication Cipher using the good key a=5 //Author: Nils Hahnfeld, 9/22/99 #include #include void main() { char cl,pl,ans; int a=5, ainverse=21; //as a-1*a=21*5=105=1 MOD 26 clrscr(); do { cout << "Multiplication Cipher: (e)ncode or (d)ecode or (~) to exit:" ; cin >> ans; if (ans=='e') { cout<< "Enter plain text: "<< endl; cin >> pl; while(pl!='~') { if ((pl>='a') && (pl<='z')) cl='a' + (a*(pl -'a'))%26; else if ((pl>='A') && (pl<='Z')) cl='A' + (a*(pl -'A'))%26; else cl=pl; cout << cl; cin >> pl; } } else if (ans=='d') { cout << "Enter cipher text: " << endl; cin >> cl; while(cl!='~') { if ((cl>='a') && (cl<='z')) pl='a' + (ainverse*(cl -'a'))%26; else if ((cl>='A') && (cl<='Z')) pl='A' + (ainverse*(cl -'A'))%26; else pl=cl; cout << pl; cin >> cl; } } } while(ans!='~'); } Programmers Remarks: Can you understand the code yourself? Can you? Cite as source (bibliography): The following table shows the numbers relative prime to M for the first 21 integers. //Author: Nils Hahnfeld 10/15/99 //Factoring program #include #include #include void main() { int M, factor ; clrscr(); do { cout << "Enter the integer that you want to factor or 0 to exit: M="; cin >> M; factor=2; while(factor <= M) { if (M%factor==0) //check all integers less than M as factors { cout << factor << endl; M/=factor; factor=1; } factor++; } }while(M!=0); } Programmers remarks: Starting with 2, this program checks the integers from 2 to M-1 as potential factors of M in if (M%factor==0). background-color: #620E01; Instead of adding a number as we did in the Caesar Cipher, we will now multiply each plain letter by an integer a, our secret encoding key. It would take quite a long time for a computer to brute-force through a majority of nine million keys. 343 and 14 are not relative prime since gcd(343,14)=7. Step 3: Lets see how decryption can be done using the above formula: Ciphertext = QCCSWJUPQCCSW and multiplication inverse key = 15, Ciphertext: Q > 16 Decryption: (16*15) mod 26 Plaintext: 6 > G, Ciphertext: C > 2 Decryption: (2*15) mod 26 Plaintext: 4 > E, Ciphertext: S > 18 Decryption: (18*15) mod 26 Plaintext: 10 > K, Ciphertext: W > 22 Decryption: (22*15) mod 26 Plaintext: 18 > S, Ciphertext: J > 9 Decryption: (9*15) mod 26 Plaintext: 5 > F, Ciphertext: U > 20 Decryption: (20*15) mod 26 Plaintext: 14 > O, Ciphertext: P > 15 Decryption: (15*15) mod 26 Plaintext: 17 > R, After decryption the plain text = GEEKSFORGEEKS. Extracting arguments from a list of function calls. the number of unique encryptions u are dependent on the chosen alphabet length M. Since u can be expressed as a formula that involves M, namely u=M-1, we say that u is a function of M and write u(M)=M-1. As a small example we consider Vigenre with the following two alphabets: In both cases, both the plaintext and the key should both consist of the text "0123456789abcdABCD". 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. Can we increase the number of unique encryptions by further extending our alphabet? Try to understand as much as possible first, then continue reading. Generally: An alphabet of length M has the keys: ZM = {0,1,2,3,, M-2,M-1} 2) Now, the good keys are the ones that are relative prime to 26 as listed above and are denoted as Z26*. Now, how do you decrypt the above message? How many multiples of 3 will not produce a unique encryption? For example, if we have the number 7, the multiplicative inverse, or reciprocal, would be 1/7 because when you multiply 7 and 1/7 together, you get 1. I leave the translation from an upper case plain letter to a lower case cipher letter as an easy exercise for you. I will complete the first ones and leave the second ones for you as exercises. For the fraction a/b, the multiplicative inverse is b/a. The grey rows show what would be expected for the order, and the red one shows what your text gives for the order: If we use a value which is not co-prime, such as 2, we will not get unique characters for the mapping: Bib: @misc{asecuritysite_99257, title = {Multiplication Cipher}, year={2023}, organization = {Asecuritysite.com}, author = {Buchanan, William J}, url = {https://asecuritysite.com/coding/mult}, note={Accessed: May 02, 2023}, howpublished={\url{https://asecuritysite.com/coding/mult}} }. As you can see on the wiki, decryption function for affine cipher for the following encrytption function: E (input) = a*input + b mod m is defined as: D (enc) = a^-1 * (enc - b) mod m The only possible problem here can be computation of a^-1, which is modular multiplicative inverse. Remember that the first 3 ciphers are meant to familiarize you with basic encryption systems. Since, for the standard alphabet, there are 12 numbers less than 26 which are . 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. Step 4: So, once the calculation part is done now you can easily encrypt your given plain text. 4 color: #ffffff; Also, each B and each M turn into 2 (=c) since 2*1 = 2 MOD 26 and 2*14 = 28 = 2 MOD 26. The basic formula to be used in such a scenario to generate a multiplicative cipher is as follows . This shows that when using an encoding key that is one less than the alphabet length M, namely a = M-1, then the decoding key must also equal M-1, a-1 = M-1. Our ultimate goal is not to develop a formula for the number of bad keys but rather for the number of good keys. In affine cipher each letter in an alphabet is mapped to its numeric equivalent, encrypted using a simple mathematical function, and converted back to a letter. It is easy to implement and easy to understand, and it does not require any large amount of computational power. You should have realized that decoding means to undo the original multiplication. Example4: If M=39=3*13=p*q, then the formula yields u(39) = (3-1)*(13-1) = 2*12 = 24. the commonly used RSA Cipher is based on the relative slowness of such factoring programs. In fact, all the upper case letters on Excel are 65 numbers higher than those we are using, the lower case letters on Excel are 97 numbers above ours (i.e. Therefore, a simple prime check program would be sufficient to find the divisors p of M. We then set up the factors of the form (1- 1/p), multiply them and eventually multiply that answer by M. Example1: Say M=180, then a prime check program yields the prime factors 2,3 and 5, so that ((180) = 180 * (1-1/2) * (1-1/3) * (1-1/5) = 180 * (1/2) * (2/3) * (4/5) = 90 * (2/3) * (4/5) = 60 * (4/5) = 48 Example2: Say M=360, since 360=2*180 the prime factors are again 2,3 and 5, so that ((360) = 360 * (1-1/2) * (1-1/3) * (1-1/5) = 360 * (1/2) * (2/3) * (4/5) = 180 * (2/3) * (4/5) = 120 * (4/5) = 96 Example3 is for you: Say M=90, since 90=____ the prime factors are _______, so that ((90) = 90 * (1-1/__) * (1-1/__) * (1-1/__) = 90 * ____________________ = _______________ = _______________ = ___ Of course, I could have computed the answers in the above examples right away but I wanted to give you the chance of brushing up on your skills to multiply fractions.
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