1 00:00:02,300 --> 00:00:09,110 Let's take a look at our essay first the algorithm is not a new development. 2 00:00:09,140 --> 00:00:13,910 It was created back in 1977. 3 00:00:13,940 --> 00:00:16,490 It's not the first asymmetric algorithm either. 4 00:00:18,240 --> 00:00:23,370 The first application of the asymmetric cryptography scheme was Diffie and Hellmann secure key exchange 5 00:00:23,370 --> 00:00:28,530 protocol the Diffie Hellman key exchange is still implemented today. 6 00:00:28,830 --> 00:00:36,640 For example in public key infrastructure in sec this solution solve the problem with exchanging confidential 7 00:00:36,640 --> 00:00:38,300 data through an untrusted channel. 8 00:00:38,320 --> 00:00:42,530 What keeping the confidentiality of the data. 9 00:00:42,720 --> 00:00:44,470 The point was not to send the key. 10 00:00:44,760 --> 00:00:53,450 Which should not be done but to challenge a random message instead and receive a response both parties 11 00:00:53,450 --> 00:01:00,580 make computations based on the challenge response messages a first user picks a key and generates a 12 00:01:00,580 --> 00:01:01,440 challenge. 13 00:01:02,390 --> 00:01:05,800 The second user picks a key receives the challenge. 14 00:01:05,960 --> 00:01:09,640 Compute something based on it generates and sends a response. 15 00:01:10,680 --> 00:01:16,530 Both users assume that they have provided the same key the result of the random cogeneration was the 16 00:01:16,530 --> 00:01:19,240 same combination of digits. 17 00:01:19,260 --> 00:01:20,700 Let's go back to rsa 18 00:01:24,030 --> 00:01:27,570 the algorithm was granted a patent that expired in 2000. 19 00:01:29,340 --> 00:01:37,480 To be able to use a before that time a license fee had to be paid the patent rights were given to RSA 20 00:01:37,480 --> 00:01:40,650 data security on the expiry. 21 00:01:40,790 --> 00:01:46,580 The RSA algorithm has been implemented in a number of solutions and software. 22 00:01:46,660 --> 00:01:53,240 The basis for the operation here is key generation the encryption and decryption steps are relatively 23 00:01:53,240 --> 00:01:55,970 uncomplicated. 24 00:01:56,030 --> 00:02:04,140 Remember that the key the ciphertext and the plaintext will be you can represent any string as a number. 25 00:02:05,110 --> 00:02:11,320 This is how computers operate. 26 00:02:11,320 --> 00:02:15,040 Let's consider the key generating mechanism. 27 00:02:15,160 --> 00:02:24,690 You have to pick two large primes the banks have to be large enough to be represented using 1024 2048 28 00:02:25,050 --> 00:02:29,340 or preferably four thousand ninety six bits. 29 00:02:29,440 --> 00:02:41,580 In this presentation we'll pick the standard prime numbers 7 and of computer product it's 77. 30 00:02:41,710 --> 00:02:47,110 The next step is to find an exponent that satisfies an equation where the greatest common divisor of 31 00:02:47,110 --> 00:02:53,930 the exponent and the product of the two primes minus one will be 1 as in the following formula. 32 00:02:55,450 --> 00:02:57,760 This equation has many solutions. 33 00:02:59,350 --> 00:03:03,900 In practice the smallest found is usually chosen. 34 00:03:03,950 --> 00:03:09,270 The value has no bearing on the security of the ciphertext. 35 00:03:09,490 --> 00:03:15,310 The smaller it is the faster the encryption will be for the two primes will picked. 36 00:03:15,340 --> 00:03:16,000 You can choose. 37 00:03:16,000 --> 00:03:19,210 E equals 37. 38 00:03:19,460 --> 00:03:22,690 And this way you've found your public key. 39 00:03:22,820 --> 00:03:29,900 The public key is made from the values and the 77 and 37. 40 00:03:30,010 --> 00:03:35,140 Now a private key that corresponds to the public key has to be generated. 41 00:03:35,180 --> 00:03:40,330 The private key will be generated based on the following equation. 42 00:03:40,360 --> 00:03:46,300 You have to multiply B and D in such a way that you find a product of multiplying P minus one times 43 00:03:46,330 --> 00:03:49,500 Q minus one as a Madieu operation parameter 44 00:03:52,330 --> 00:04:01,150 for the values we calculated earlier D'Ken B 13 the value D will from this point be your private key. 45 00:04:01,180 --> 00:04:09,120 It amounts to 13. 46 00:04:09,210 --> 00:04:13,750 Why is this all working from what we officially know. 47 00:04:13,930 --> 00:04:22,840 Finding the bodies and cube based on the value N is computationally complex as it turns out. 48 00:04:22,960 --> 00:04:26,820 What we're able to calculate the Kedy by having the product. 49 00:04:27,100 --> 00:04:34,810 If we don't know the values p and q we must resort to checking all possible solutions to the equation. 50 00:04:34,830 --> 00:04:38,430 This is presumed infeasible because there are large prime numbers 51 00:04:41,580 --> 00:04:44,820 encryption itself is easy. 52 00:04:44,880 --> 00:04:50,720 It involves assuming that the plaintext has a value that is a bit smaller than n. 53 00:04:50,800 --> 00:04:54,740 Next you perform a modular exponentiation. 54 00:04:54,930 --> 00:05:03,100 The ciphertext you get is the plaintext raised the e power must do low n the values e n n are parts 55 00:05:03,100 --> 00:05:07,170 of your public key. 56 00:05:07,310 --> 00:05:11,000 If the plaintext was to the ciphertext will be 51 57 00:05:14,310 --> 00:05:24,340 decryption as inverting this process to decode the ciphertext razy to the DB power model N. 58 00:05:24,480 --> 00:05:26,580 In this way the result will recover to.