3. 20 By subtracting a (=101) from the entered plain letter in (pl -'a');. Apr 6, 2013 at 10:46 . A multiplicative cipher is a type of cipher that comes under a monoalphabetic cipher, in which each letter that is present in the plaintext is replaced by a corresponding letter of the ciphertext, according to a fixed multiplication key. 19 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. This online calculator tries to decode substitution cipher without knowing the key. The first character G corresponds to the six. From now on we will use a handy Notation for the set of possible and good keys: 1) All the possible keys for an alphabet length of 26 are clearly all the numbers between 1 and 26, denoted as Z26. Each character in the message is multiplied with this key. First of all, you need to know which one of the 12 good keys was used. Note The advantage with a multiplicative cipher is that it can work with very large keys like 8,953,851. background-color: #620E01; Finally I understand how to calculate the modular multiplicative inverse :) $\endgroup$ - np00. Write to dCode! 11 for the RSA encryption. All symbols to be encrypted must belong to alphabet, Everyone who receives the link will be able to view this calculation, Copyright PlanetCalc Version: 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. Now that we have explored the criteria for unique encryptions and the number of good keys for certain alphabet lengths, it is the nature of Mathematics to generalize the observations and to set up an explicit formula for the number of unique encryptions based on the alphabet length M. For that purpose we have to consider 3 different cases of the alphabet length M 1) If M is a prime number: We observed in the previous section that the prime alphabet length M=29 yields u=28 unique encryptions. To verify this: 262 = 676 =1 MOD 27. Then the if-condition if (ans=='e') is fulfilled so that we enter the encoding part of the program. In order to be able to use the command setw() we have to include the iomanip.h library in #include . For M=31 we have u(31)=30. 17 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. 1 } Certainly none of the cryptosystems we have considered thus far. WAP to find the solutions of equations: a.14x=12mod 18 b.3x+4=6 mod 132. Or can we even increase the mere 12 unique encryptions for the Multiplication Cipher by varying the alphabet length? what are prime divisors of 178247 or of 56272839 ?). (I can not list those here as they depend on the alphabet length M.) We are now able to summarize how to encrypt a message using the multiplication cipher: To encrypt a plain letter P to the cipher letter C using the Multiplication Cipher, we use the encryption function: f : P ( C=(a*P) MOD 26. By using our site, you This process repeats until M is reduced to 1 and therefore less than the smallest factor possible, 2. Lets summarize our discoveries. 23 This is it. The command const is used as a safety feature in C++: both variables are constant and can never be modified in any program. It is suitable for small-scale applications but not recommended for practical purposes. However, there are some additional integers that are not prime (i.e. Cipher textanromrjukahhouha=1ANROMRJUKAHHOUHa=3ANXWEXDYMALLWYLa=5ANTISTHECARRIERa=7ANVCYVFOUABBCOBa=9ANZQKZBIEAVVQIVa=11ANLGULPQIADDGQDa=15ANPUGPLKSAXXUKXa=17ANBKQBZSWAFFKSFa=19ANFYCFVMGAZZYMZa=21ANHSIHTWYAJJSWJa=23ANDEWDXCOAPPECPa=25ANJMOJRGQATTMGT MS Excel as a simple encryption and decryption tool: I created the following table in MS Excel with the CHAR and the MOD function: Cipher textanromrjukahhouhaa-101317141217920100771420739ANXWEXDYMALLWYL521ANTISTHECARRIER715ANVCYVFOUABBCOB93ANZQKZBIEAVVQIV1119ANLGULPQIADDGQD157ANPUGPLKSAXXUKX1723ANBKQBZSWAFFKSF1911ANFYCFVMGAZZYMZ215ANHSIHTWYAJJSWJ2317ANDEWDXCOAPPECP2525ANJMOJRGQATTMGT For example, I created the T in the row a=5 using the Excel-formula =CHAR(65+MOD(E$2*$B4,26)) where the cell E$2 contains 17 and the cell $B4 contains 21 as the decoding key a-1. Boolean algebra of the lattice of subspaces of a vector space? I found a-1 = 2 by simply testing the integers in Z5*={1,2,3,4}. One of the major goals of current Mathematics research is to design faster factoring algorithms as todays are fairly slow. By using this website, you agree with our Cookies Policy. 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. Example1: If M=24=3*8=3*23, then ((24) = ((3*23) using property 4) yields = ((3)*((23). The MOD 26 calculation leaves the 10 unchanged. Combining our three formulas for the number of good keys, we will then be able to develop a general formula for the number of good keys for any given alphabet length M. Lets start with Example1: M=26=p*q=2*13. As an example, lets encode and decode NAT and ANT. Then we perform the reverse operations performed by the encryption algorithm. First we need to calculate the modular multiplicative inverse of keyA. Cryptoanalysis - Cracking the Multiplication Cipher Just like the Cipher Caesar Cipher, the Multiplication is not secure at all. Before considering such encoding techniques, we go ahead and check if the other frequent number, 20, is the cipher E. Checking the E column, we can see that the possible two keys are the bad one a=18 and the good one a=5. Just as 5*1/5 yields 1, 5 * 5-1 shall equal 1 MOD 26. Read about it on wiki, I will provide only example. We factor p1=2 yielding = 2*(2-1)*(3-1)*(5-1) = p1* (p1- 1)*( p2-1)*( p3-1). Examples for property 1): 3 and 5 are two primes. The plain letter c is stored as 103, however, I want the c to equal 2 in compliance with our translation a=0, b=1, c=2, etc. 3.0.4224.0. The encrypted text is the smallest digit of an addition of plaintext and key when both are hexadecimal digits. It is a-1=4 since 3*4 = 12 = 1 MOD 11. Invented by Lester S. Hill in 1929, it was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. We can therefore always find a-1 for a given good key a. Notice that we found the good keys indirectly. 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*. In order to increase the probability of this, the alphabet is expanded, so its length becomes the prime integer. The multiplicative cipher is a special case of the Affine cipher where B is 0. The only disadvantage is that the minus sign itself has to be written as "---", so as not to be confused as a range operator. Encrypted text: The quick brown fox jumps over the lazy dog. For classical methods, the alphabet often consists only of the uppercase letters (A-Z). Or are they possibly the primes between 1 and 25? 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 modular multiplicative inverse of a modulo m can be found with the Extended Euclidean algorithm. Counter examples are: 45 and 18 are not relative prime since gcd(45,18)=9 and not 1. Finally, I have to add the usual 65 = A (why?) To do so, we have to look at the encryption equation C=a*P MOD 26 and solve it for the desired plain text letter P. In order to solve an equation like 23=5*P for P using the rational numbers, we would divide by 5 or multiply by 1/5 to obtain the real solution P=23/5. a feedback ? The following steps take place: In the example, an overflow has occurred in the third letter, so that modulo |L| = 4 is calculated. Wonderful, that is all we need to solve our encryption function C= a*P MOD 26 for the plain letter P in order to then decode the encrypted message: Multiplying both sides of our encryption equation the equation yields a-1*C = a-1*(a*P) (1) = (a-1*a)*P (2) = 1*P (3) = P MOD 26 (4) Remark: Solving this equation required the 4 group properties: the existence of an inverse and the closure in (1), the associative property in (2), the inverse property in (3) and the unit element property in (4). (Attacks). Divide the letters of the message into groups of two or three. To do so, I distinguish between upper and lower case letters since they are encoded slightly different. You could also define a to be a different good key. ((21)=________________________ as 1,2,4,5,8,10,11,13,16,17,19,20 are relative prime to 21. Moreover, multiplying any two good keys yields again a good key. The key should be changed frequently to prevent cryptographic attacks. Say, we want to encrypt the plain letter C=67. What would you do? Lets simply test all possible keys of the multiplication ciphers MOD 26: PLAIN LETTER 0000000000000000000000000 a ABCDEFGHIJKLMNOPQRSTUVWXYZ00000000000000000000000000010123456789101112131415161718192021222324252024681012141618202224024681012141618202224303691215182124147101316192225258111417202340481216202426101418220481216202426101418225051015202549141924381318232712172216111621606121824410162228142006121824410162228142070714212916234111825613201815223101724512198081624614224122021018081624614224122021018909181101921120312214132251423615247162581710010204142481821222616010204142481821222616110112271831425102161721324920516112238194151201224102282061841621401224102282061841621413013013013013013013013013013013013013013140142164186208221024120142164186208221024121501541982312116520924132176211025143187221116016622122188241442010016622122188241442010170178251672415623145221342112320112191011891801810220124221462416801810220124221462416819019125241710322158120136251811423169221147200201482221610424181260201482221610424181262102116116122171272231813832419149425201510522022181410622420161284022181410622420161284230232017141185225221916131074124211815129632402422201816141210864202422201816141210864225025242322212019181716151413121110987654321 We learned already that the key a=2 (as can be seen in the 3rd row) does not produce a unique encryption. The solution shows the work for the Standard Algorithm. M23456789101112131415161718192021( (M)12242648121041268816618812 Similar to our notation, the properties of Eulers (-function that computes the number of integers that are relatively prime to M and wrote similarly to our notation: Eulers (-function: 1) ((p) = p-1 for a prime p. 2) ((pn) = pn - pn-1 for a prime power pn. 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. The handling of non-alphabet characters (convert, skip, ) can be set in the options - but this is not a function of the actual encryption process itself. An easier way to determine the decoding key a-1 Decoding a message turns out to be really easy once we know the decoding key a-1. Sphero Up to 1 Hour Grades: 5 to 8. We denote 5-1 the inverse of 5. Firstly I have no idea how they derived this formula, but I think I have a general idea. >def unshift (key, ch): offset = ord (ch) - ASC_A return chr ( ( (key [0] * (offset + key [1])) % WIDTH) + ASC_A) Note The advantage with a multiplicative cipher is that it can work with very large keys like 8,953,851. This principle of finding the number of bad keys holds true for any alphabet length that is a prime power: There are M/p multiples of p less or equal to M, and therefore M/p - 1 many less than M. And we are only interested in those integers less than M since we are calculating MOD M which involves the integers 0 to M-1. Say M=26=2*13=n*m. Since n and m are two distinct primes, they certainly are relative prime, so that the condition for property 4) is fulfilled. For example, take the list L = "ABCD", whose length is 4. For each character of the plain message, apply the following calculation: ($ 26 $ being the number of letters in the alphabet).

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