Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Arab Open University - AOU T209 Information and Communication Technologies: People and Interactions Fifth Session 1 Dr. Saatchi, Seyed Mohsen Reference Material This session is based on the following references: – – – – More references: – – 2 Module 5: Security, Book S: Security Module 5: Security, Book N: Numeracy Skills Module 5: Security, Book E: Experiments Module 5: Security, (Text Book) Monograph: Security Techniques in Digital Systems http://www.cacr.math.uwaterloo.ca/hac/ http://en.wikipedia.org/wiki/Cryptography Dr. Saatchi, Seyed Mohsen Topics to be covered in this session Part 2 (Encryption) of Book S – 1. (S.2.3.2) Breaking a code Chapter 2 (Encryption) of Book M (Monograph) – 2. (M.2.2) Working with codes Part 2 (Encryption) of Book S – – 3 (M.2.2.1) Cracking a code (M.2.2.3) Diagraphs (M.2.2.4) Encoding 3. Solve some activities in (S.2.3.2) Breaking a code 4. (S.2.3.3) Encryption using modular multiplication Dr. Saatchi, Seyed Mohsen Continue Part 3 (Modular arithmetic) of Book N – 5. (N.3.3) Modular multiplication 4 (N.3.3.1) Performing multiplication in modular arithmetic (N.3.3.2) The properties of modular multiplication (N.3.3.3) Summary of Section (N.3.3) Dr. Saatchi, Seyed Mohsen Topic 1: (S.2.3.2) Breaking a code Brute force attack: – 5 The method of cracking a code by trying all possible combinations until the correct one is found is known as a brute force attack Dr. Saatchi, Seyed Mohsen Continue Activity 2.7 (exploratory) – How many different arrangements would be possible using the seven letters of the word article? Each letter in the word article appears only once Taking one letter at a time, the first can appear in any of the seven positions; the second in any of the 6 remaining positions; the third in any of the five remaining positions; and so on This gives the total possible number of combinations as: – 6 7 × 6 × 5 × 4 × 3 × 2 × 1= 5040 An expression giving the product of all the integers from 1 to n is known as factorial n and is expressed in mathematical notation as n! Note that the exclamation mark is part of the mathematical expression Dr. Saatchi, Seyed Mohsen Continue Activity 2.8 (self-assessment) – Using a computer that can perform one million calculations per second, calculate how long it would take to try all possible combinations of: (a) 10 different letters – (b) 15 different letters – 15! ÷ 106 = 1 307 674 368 000 ÷ 106 seconds = 1 307 674 seconds = 15.13 days (c) 20 different letters – 7 10! ÷ 106 = 3 628 800 ÷ 106 seconds = 3.6 seconds 20! ÷ 106 = 2.432902008177 × 1018 ÷ 106 seconds = 2.432902008177 × 1012 seconds = 2 432 902 008 177 seconds which is approximately 77 147 years Dr. Saatchi, Seyed Mohsen Continue This account of my attempt to ‘crack a cipher’ illustrates a couple of important points: – – 8 Attempting to break a cipher by brute force attack can be a very lengthy process. Often the content of the message is redundant long before the cipher has been broken. If the enciphered message ‘Attack tomorrow at dawn’ takes a week to break then the sender is using an encryption algorithm with an appropriate level of security for the purpose An alternative, and usually quicker, method of breaking a code is through assumptions derived from a knowledge of the language used in the plaintext message Dr. Saatchi, Seyed Mohsen Topic 2: (M.2.2) Working with codes 9 An encryption system that is intended to obfuscate a plaintext should not be easy to crack It can be disadvantageous to use an encryption procedure that does not have a known method for breaking it since it is then impossible to estimate the effort involved on the part of crackers and impossible to weigh up the costs they face in attempting to break the code Because of the value of knowing how hard it is to crack a code, it has become commonplace to publish the methods of encryption and to encourage their investigation This doctrine is perhaps less relevant for military applications, but is significant for encryption schemes for pervasive, publicly accessible systems Dr. Saatchi, Seyed Mohsen Continue Sub-Topic 2.1: (M.2.2.1) Cracking a code Crackers may know the encryption methods, but initially do not know the key that decrypts or encrypts messages and they do not know what the messages are Example: – – 10 For the Caesar code, there are 26 possible keys. One key, however, leaves the message unaltered (when k = 0) so there are 25 usable keys Sometimes, the first key tried might, by chance, decrypt the message, but on other occasions it would be necessary to try all 25 possible keys in turn before the message was revealed Dr. Saatchi, Seyed Mohsen Continue – – 11 On average, therefore, it would take 12.5 attempts to find the key Another way of putting this is to say that the probability of cracking the code with one attempt is one in twenty-five or 1/25 = 0.04 Dr. Saatchi, Seyed Mohsen Topic 2.2: (M.2.2.3) Digraphs There are three sources of weakness in the Caesar code: – – – 12 1- The calculations involved in all the encryption, decryption and cracking processes are fairly simple and easily performed, especially with a computer on hand 2- The number of possible keys is small so it would be straightforward to try each of the usable 25 keys until a recognizable message emerged 3- The statistics of the original message are not camouflaged. Therefore, knowledge of the kind of message and its language helps a potential cracker Dr. Saatchi, Seyed Mohsen Continue More Keys the better: – – – – 13 An encryption system that had more potential keys would discourage a cracker from using the brute force approach of trying all the conceivable keys The Caesar code can be extended to provide more keys A simple extension would be to encrypt pairs of letters, which are known as digraphs There are 26 possible first letters in the pair and for each of these there are 26 possible second letters giving a total of 26 × 26, which is often written as 262, combinations and which turns out to be 676 Dr. Saatchi, Seyed Mohsen Continue Some modulo 676 Maths: – – – 14 Now to encrypt a pair of letters with 676 combinations it is possible to work modulo 676 and to draw the codes from Z676 As with modulo 26 addition, modulo 676 addition means adding numbers together as usual, but if the result is 676 or greater, repeatedly subtracting 676 until the result is less than 676 The key is, therefore, a number between 0 and 675, say 637, then encrypting the coded letters ‘DO’ involves adding, using modulo 676 addition, the key to the code for the letter pair, which was calculated to be 92 Dr. Saatchi, Seyed Mohsen Continue 15 The ciphertext is given by: Then, using the properties of modulo 676 addition, Dr. Saatchi, Seyed Mohsen Continue To convert the result to a letter pair, divide it by 26: – 16 53/26 = 2.03846 The whole number part is 2. According to the coding scheme devised in this section, the first letter of the pair is given by 2, which encodes the letter C. The 2 accounts for 2 × 26 = 52 of the original code of 53. The remainder is therefore 53 – 52 = 1. This gives the second letter of the pair as the second letter of the alphabet, B. The encrypted pair is then ‘CB’ Dr. Saatchi, Seyed Mohsen Continue To decrypt the message the pair ‘CB’ is first encoded again which gives, as expected, – 17 2 x 26 + 1 = 53 The decryption key added to the encrypted message should restore the original plaintext message. For the Caesar code and in this case, the decryption key, , is given by the expression: For a key of 637: Therefore, , since Dr. Saatchi, Seyed Mohsen Continue The decrypted message is then given by adding the decryption key to the encrypted message modulo 676: – 18 92 is the encoded version of the original plaintext ‘DO’ An advantage of using digraphs is that there are a larger number of keys to try. The arithmetic too is becoming a little more involved, though not daunting, especially if a computer is available Dr. Saatchi, Seyed Mohsen Sub-Topic 2.3: (M.2.2.4) Encoding 19 Lets look at encrypting more than one or two letters Characters can be grouped together in digraphs or much larger blocks and encrypted in these larger blocks using potentially larger keys Consider a black and white picture of a key as in Figure 1 If black is denoted by a one and white by a zero we can draw out the key on a grid with the numbers highlighting the different areas as shown in Figure 2 (unencrypted) Dr. Saatchi, Seyed Mohsen Continue Figure 1 A picture of a key sampled for use in a computer 20 Dr. Saatchi, Seyed Mohsen Continue 21 Figure 2 The picture of the key encoded in 0s and 1s Dr. Saatchi, Seyed Mohsen Continue The picture is not yet in a form that can be encrypted using the Caesar code. A convention is needed to sweep up the 1s and 0s into groups of 5: – The numbers representing the above codes are (shown in Figure 3): – 22 00000, 00000, 00011, 11111, 10000, 00000, 00000, 00000, 00000, 00000, 00000, 00000, 00000, 00000, 00000 0, 0, 3, 31, 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 Dr. Saatchi, Seyed Mohsen Continue 23 Figure 3 Black and white data grouped into five bit groups and encoded (still unencrypted) Dr. Saatchi, Seyed Mohsen Continue 24 Each number corresponding to a group of five bits can be encrypted separately and the entire picture encrypted as the string of numbers taken row after row Figure 4 shows how the data encrypted using modulo 32 addition would appear if the encrypted data were treated as a picture when the key is 22 and is the same for each block of data Dr. Saatchi, Seyed Mohsen Continue Figure 4 The encrypted image reproduced as a picture 25 Dr. Saatchi, Seyed Mohsen Continue 26 The image is obfuscated but the original image still shows through. It needs to make the obfuscation more thorough Figure 5 shows the image encrypted in blocks of five bits using modulo 32 addition with a key selected at random for each successive block. The encryption appears to be very successful since the original image has been obscured completely. Of course, to decrypt the encrypted image, the corresponding random sequence of decryption keys must also be sent to the recipient, presumably in secret Dr. Saatchi, Seyed Mohsen Continue Figure 5 The image encrypted using a random succession of keys 27 Dr. Saatchi, Seyed Mohsen Continue 28 End of section 2.2 from the Monograph study material Dr. Saatchi, Seyed Mohsen Topic 3: Solve some activities in (S.2.3.2) Breaking a code Activity 2.12 (self-assessment) – – Using a Caesar code and a digraph coding similar to the one described by Monk, what would be: (a) the ciphertext for the letter pair M E using an encryption key of 576? 29 M is the 13th letter and E the 5th letter of the alphabet, so they will be coded as 12 and 4 respectively. The numerical code for the letter pair is: 12 × 26 + 4 = 316 c p+K mod 676 316+576 mod 676 892 mod 676 216 mod 676 216 = 8 × 26 + 8, A code of 8 is given to the letter in the 9th position in the alphabet.So using a key of 576 the ciphertext for the letter pair ME is II Dr. Saatchi, Seyed Mohsen Continue – (b) the plaintext for the ciphertext 376 which was encoded with a key of 149? 30 K+ 676 mod 676, So 676 - K mod 676 676-149 mod 676 527 mod 676 p c+ mod 676 376+527 mod 676 903 mod 676 227 mod 676, So the coded plaintext letter pair is 227 26 divides into 227 8 times leaving a remainder of 19 So the plaintext letters have been coded as 8 and 19 respectively These codes represent the letters in the 9th and 20th positions of the alphabet = IT Dr. Saatchi, Seyed Mohsen Continue Activity 2.13 (self-assessment) – – How many possible keys would there be if we encrypted the plaintext using: (a) trigraphs? – (b) four-letter groupings? 31 A trigraph (a group of three letters) would result in 263 = 17 576 keys A four-letter grouping would result in 264 = 456 976 keys Dr. Saatchi, Seyed Mohsen Continue Activity 2.14 (self-assessment) – What is the probability of cracking the Caesar cipher in one attempt when a three-letter key is used? 32 Between 1 and around 263 = 17 576 attempts would be needed, giving a probability of 1 / 17576 = 0.0000569 of cracking it in one attempt Dr. Saatchi, Seyed Mohsen Topic 4: (S.2.3.3) Encryption using modular multiplication Book N: Numeracy – 33 Students should work through all of Section 3.3 ‘Modular multiplication’ in Book N. A summary follows Dr. Saatchi, Seyed Mohsen Topic 5: (N.3.3) Modular multiplication Sub-Topic 5.1: (N.3.3.1) Performing multiplication in modular arithmetic Multiplication in modular arithmetic is carried out in a similar manner to addition Example: Calculate 6 × 7 mod 8 – First multiply the two integers in the conventional way: – Then divide the result by 8: – – 42 ÷ 8 = 5, remainder 2 Express the answer as the remainder (or residue). In mathematical language I can express this as: 34 6 × 7 = 42 6×7 2 mod 8 Dr. Saatchi, Seyed Mohsen Sub-Topic 5.2: (N.3.3.2) The properties of modular multiplication 35 Figure 6 shows multiplication tables for modulos 4, 5, 6 and 7 To multiply two numbers together, find the first number in the top horizontal number line and the second number in the left vertical number line The result of modular multiplication is in the matrix where the column and row of the numbers intersect Dr. Saatchi, Seyed Mohsen Continue Figure 6 Multiplication tables for modulos 4, 5, 6 and 7 36 Dr. Saatchi, Seyed Mohsen Continue Example: Find 3 × 4 mod 7, Use the modulo 7 table (in Figure 7) – – 1- The property of closure – 37 Read from the 3 in the top horizontal number line and right from the 4 in the left vertical number line The result (5) is shown in the grid where the column and row intersect Looking at Figure 6 and Figure 7, we can see that every operation produces a result that is a member of the set, so the tables indicate that modular multiplication has the property of closure Dr. Saatchi, Seyed Mohsen Continue 38 Figure 7 Multiplication tables for modulos 4 to 11, with identity highlighted Continue Dr. Saatchi, Seyed Mohsen 2- Identity Property (Activity 28 (exploratory)) – Remember that the identity is defined as e in the general equation: – – It is that element of the set which when combined mathematically with any other element of the set will leave that element unaltered Here we can replace the symbol º (the symbol used to denote any mathematical operation) with the multiplication symbol × giving: 39 a º e = a mod n or e º a = a mod n a×e a mod n or e × a Dr. Saatchi, Seyed Mohsen a mod n Continue – What the question is asking is: is there an element within each set which, if multiplied by a, leaves a unaltered? Yes there is 40 If any element of the set is multiplied by 1 it remains unaltered Thus 1 is the identity in modular multiplication You can check this for yourself from the tables (in Figure 6 and Figure 7) by locating 1 in the top horizontal number line and looking down its column Every number in it should be identical to the number appearing in the left-hand vertical number column of the same row Dr. Saatchi, Seyed Mohsen Continue 3- The multiplicative inverse: – – 41 You should recall from Section N.3.2.2 that under modular addition every element in the set could be combined with another to return a result equal to the identity of the set Figure 7 shows multiplication tables for modulos 4 to 11 with the identity highlighted Dr. Saatchi, Seyed Mohsen Continue Activity 29 (exploratory) – – 42 Study the tables in Figure 7. Does every element in every set have a multiplicative inverse? That is, for any element of the set is there another element which, when combined by multiplication, returns a result equal to the identity of the set under multiplication? No. The identity of the set under multiplication is 1 and not every element in every set can be multiplied with another element to produce the result 1 mod n Dr. Saatchi, Seyed Mohsen Continue Activity 30 (exploratory) – – – 43 Can you see any pattern emerging in the tables of Figure 7? Look closely at those elements that do produce a multiplicative inverse. What is their relationship with the modulus? In all the tables the only elements that have a multiplicative inverse are those that share a highest common factor of 1 with the modulus. (In effect this means that the only factor they share with the modulus is 1) When the element and the modulus share a highest common factor that is greater than 1, there is no multiplicative inverse Dr. Saatchi, Seyed Mohsen Continue Concept of Coprime: – – 44 Two or more numbers whose highest common factor is 1 are said to be coprime. (Often the expression relatively prime is used as an alternative to coprime but in this module I will stick with the term coprime) Of course, when the modulus itself is a prime number then it will be coprime with all the members of the group, since, by definition, a prime number has no factors other than 1 and itself Dr. Saatchi, Seyed Mohsen Continue Significance of this multiplicative inverse property in cryptography: – – – Unlike modular addition, modular multiplication doesn’t always yield a single solution to the congruence a º x b mod n Sometimes there is a single solution, sometimes there is more than one solution, and sometimes there is no solution at all One important point to note is that if the modulus is prime then there will always be one single solution 45 Where a number is coprime with the modulus, multiplication by any other number in the group produces a single solution. Thus when the modulus itself is prime (so that all the elements in the set are coprime with it) every element produces a single solution Dr. Saatchi, Seyed Mohsen Continue Activity 31 (self-assessment) – How many solutions exist for the value of x in the congruence 4 × x 2 mod n 46 (a) when n = 5? (b) when n = 6? (c) when n = 7? (d) when n = 8? (e) when n = 9? (f) when n = 10? (g) when n = 11? 1 2 1 0 1 2 1 Dr. Saatchi, Seyed Mohsen Continue Example: – Imagine I need to send you secretly the number 4 and that we have already agreed between us an encryption method whereby I will work in modulo 9 and encrypt any number that I send you by multiplying it by 3: – 3 mod 9 I send you the encrypted number 3. Would you be able to decrypt my message? 47 4×3 If you look at the table in Figure 8 you will see that the result 3 arises from the congruence 4 × 3 mod 9 and the congruence 7× 3 mod 9. You couldn’t be sure whether my original number had been 4 or 7 Dr. Saatchi, Seyed Mohsen Continue Figure 8 Multiplication table for modulo 9 48 Dr. Saatchi, Seyed Mohsen Continue – If instead we had agreed to use one of the elements that produces a single solution for multiplication in modulo 9 this ambiguity would not have arisen. Let’s say we had agreed on 7 as the multiplier: – 49 4×7 1 mod 9 Since the result of 1 has no duplicates for the solution of x in the congruence x×7 1 mod 9 you could be confident that your decryption recovers my original value Dr. Saatchi, Seyed Mohsen Continue To take this a step further, how would you have decrypted this message? – – – – 50 You could have tested every element in modulo 9, multiplying it with 7 until you found one that produced the result 1 Or instead you could have found the multiplicative inverse of 7 and multiplied 1 by this to recover my original value. (Remember that successive multiplication of a number by a multiplicative inverse pair will leave the number unaltered.) The multiplicative inverse of 7 mod 9 is 4 (which you can check in the table of Figure 8 if you need to reassure yourself of this) You would then evaluate the congruence 1 × 4 4 mod 9 to recover the original value of 4 Dr. Saatchi, Seyed Mohsen Continue Activity 32 (exploratory) – – – 51 In Figure 8 the multiplication table for modulo 9 is reproduced. Do any of the elements in the set produce a single solution when multiplied by each of the other set elements? Yes. When I look at the columns under the numbers 1, 2, 4, 5, 7 and 8 in the modulo 9 multiplication table, I find that there are no duplicates, so these six elements do produce a single solution when multiplied by any other member of the set The interesting characteristic of these six elements is that they are all coprime with the modulus, which means they are also elements which have a multiplicative inverse. The elements 3 and 6 share a common factor (3) with the modulus. These elements all show duplicates in the column Dr. Saatchi, Seyed Mohsen Continue Activity 33 (self-assessment) – Using multiplicative inverses find the solution for x in each of the following: (a) x × 3 – 9 mod 10 The multiplicative inverse of 7 mod 10 is 3, so the value of x can be found from the congruence 9×3 7 mod 10. So x = 7 (c) x × 9 – 52 The multiplicative inverse of 3 mod 10 is 7, so the value of x can be found from the congruence 5×7 5 mod 10. So x = 5 (b) x × 7 – 5 mod 10 6 mod 10 The multiplicative inverse of 9 mod 10 is 9, so the value of x can be found from the congruence 6×9 4 mod 10. So x = 4 Dr. Saatchi, Seyed Mohsen Continue 4- The associative property – Modular multiplication is associative. In other words: a × (b × c) mod n 5- The commutative property – There is one more property to check before we leave modular multiplication – the commutative property. You saw in Section 3.2.2 that in conventional arithmetic multiplication is commutative. A simple example can be used to investigate whether or not modular multiplication is commutative: – 53 (a × b) × c 5×7 7×5 3 mod 8 3 mod 8 So, 5 × 7 mod 8 7 × 5 mod 8 Dr. Saatchi, Seyed Mohsen Continue – This demonstrates that modular multiplication is commutative. I can state this generally as: 54 a × b mod n b × a mod n Dr. Saatchi, Seyed Mohsen Sub-Topic 5.3: (N.3.3.3) Summary of Section (N.3.3) 1- The group has one element, the identity e such that – – – 2- Within a group, any element (which we will call a) that is coprime with the modulus n has a multiplicative inverse (which we will call ) such that – 55 a × e a or e× a a For modular multiplication the identity e is 1 In modular multiplication the result of a × 0 mod n is always 0 a× e mod n 1 mod n 3- There is no single solution for x to the equation a × x b mod n unless a is coprime with the modulus Dr. Saatchi, Seyed Mohsen Continue 4- When the modulus itself is prime, it is coprime with all elements of the group 5- When an element (which we will call x) of a group is successively multiplied by a multiplicative inverse pair (a and ) the result is x: x×a× 56 x mod n 6- Modular multiplication is associative 7- Modular multiplication is commutative Dr. Saatchi, Seyed Mohsen Topic 6: Preparation for next session – – Continue reading about Module 5 Try to study hard for Quiz3 in next session – 57 From 1st session to 4th session The due date of TMA04 is Apr. 16 Dr. Saatchi, Seyed Mohsen