加密解密的源代码

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				DES算法
				
				
				How to implement the Data Encryption Standard (DES) 
				
				A step by step tutorial 
				Version 1.2 
				
				
				The Data Encryption Standard (DES) algorithm, adopted by the U.S. 
				government in 1977, is a block cipher that transforms 64-bit data blocks 
				under a 56-bit secret key, by means of permutation and substitution. It 
				is officially described in FIPS PUB 46. The DES algorithm is used for 
				many applications within the government and in the private sector. 
				
				This is a tutorial designed to be clear and compact, and to provide a 
				newcomer to the DES with all the necessary information to implement it 
				himself, without having to track down printed works or wade through C 
				source code. I welcome any comments. 
				Matthew Fischer  
				
				
				Here's how to do it, step by step: 
				
				1 Process the key. 
				
				1.1 Get a 64-bit key from the user. (Every 8th bit is considered a 
				parity bit. For a key to have correct parity, each byte should contain 
				an odd number of "1" bits.) 
				
				1.2 Calculate the key schedule. 
				
				1.2.1 Perform the following permutation on the 64-bit key. (The parity 
				bits are discarded, reducing the key to 56 bits. Bit 1 of the permuted 
				block is bit 57 of the original key, bit 2 is bit 49, and so on with bit 
				56 being bit 4 of the original key.) 
				
				Permuted Choice 1 (PC-1) 
				
				57 49 41 33 25 17 9 
				1 58 50 42 34 26 18 
				10 2 59 51 43 35 27 
				19 11 3 60 52 44 36 
				63 55 47 39 31 23 15 
				7 62 54 46 38 30 22 
				14 6 61 53 45 37 29 
				21 13 5 28 20 12 4 
				
				1.2.2 Split the permuted key into two halves. The first 28 bits are 
				called C[0] and the last 28 bits are called D[0]. 
				
				1.2.3 Calculate the 16 subkeys. Start with i = 1. 
				
				1.2.3.1 Perform one or two circular left shifts on both C[i-1] and 
				D[i-1] to get C[i] and D[i], respectively. The number of shifts per 
				iteration are given in the table below. 
				
				Iteration # 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 
				Left Shifts 1 1 2 2 2 2 2 2 1 2 2 2 2 2 2 1 
				
				1.2.3.2 Permute the concatenation C[i]D[i] as indicated below. This 
				will yield K[i], which is 48 bits long. 
				
				Permuted Choice 2 (PC-2) 
				
				14 17 11 24 1 5 
				3 28 15 6 21 10 
				23 19 12 4 26 8 
				16 7 27 20 13 2 
				41 52 31 37 47 55 
				30 40 51 45 33 48 
				44 49 39 56 34 53 
				46 42 50 36 29 32 
				
				1.2.3.3 Loop back to 1.2.3.1 until K[16] has been calculated. 
				
				2 Process a 64-bit data block. 
				
				2.1 Get a 64-bit data block. If the block is shorter than 64 bits, it 
				should be padded as appropriate for the application. 
				
				2.2 Perform the following permutation on the data block. 
				
				Initial Permutation (IP) 
				
				58 50 42 34 26 18 10 2 
				60 52 44 36 28 20 12 4 
				62 54 46 38 30 22 14 6 
				64 56 48 40 32 24 16 8 
				57 49 41 33 25 17 9 1 
				59 51 43 35 27 19 11 3 
				61 53 45 37 29 21 13 5 
				63 55 47 39 31 23 15 7 
				
				2.3 Split the block into two halves. The first 32 bits are called L[0], 
				and the last 32 bits are called R[0]. 
				
				2.4 Apply the 16 subkeys to the data block. Start with i = 1. 
				
				2.4.1 Expand the 32-bit R[i-1] into 48 bits according to the 
				bit-selection function below. 
				
				Expansion (E) 
				
				32 1 2 3 4 5 
				4 5 6 7 8 9 
				8 9 10 11 12 13 
				12 13 14 15 16 17 
				16 17 18 19 20 21 
				20 21 22 23 24 25 
				24 25 26 27 28 29 
				28 29 30 31 32 1 
				
				2.4.2 Exclusive-or E(R[i-1]) with K[i]. 
				
				2.4.3 Break E(R[i-1]) xor K[i] into eight 6-bit blocks. Bits 1-6 are 
				B[1], bits 7-12 are B[2], and so on with bits 43-48 being B[8]. 
				
				2.4.4 Substitute the values found in the S-boxes for all B[j]. Start 
				with j = 1. All values in the S-boxes should be considered 4 bits wide. 
				
				2.4.4.1 Take the 1st and 6th bits of B[j] together as a 2-bit value 
				(call it m) indicating the row in S[j] to look in for the substitution. 
				
				2.4.4.2 Take the 2nd through 5th bits of B[j] together as a 4-bit 
				value (call it n) indicating the column in S[j] to find the substitution. 
				
				2.4.4.3 Replace B[j] with S[j][m][n]. 
				
				Substitution Box 1 (S[1]) 
				
				14 4 13 1 2 15 11 8 3 10 6 12 5 9 0 7 
				0 15 7 4 14 2 13 1 10 6 12 11 9 5 3 8 
				4 1 14 8 13 6 2 11 15 12 9 7 3 10 5 0 
				15 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13 
				
				S[2] 
				
				15 1 8 14 6 11 3 4 9 7 2 13 12 0 5 10 
				3 13 4 7 15 2 8 14 12 0 1 10 6 9 11 5 
				0 14 7 11 10 4 13 1 5 8 12 6 9 3 2 15 
				13 8 10 1 3 15 4 2 11 6 7 12 0 5 14 9 
				
				S[3] 
				
				10 0 9 14 6 3 15 5 1 13 12 7 11 4 2 8 
				13 7 0 9 3 4 6 10 2 8 5 14 12 11 15 1 
				13 6 4 9 8 15 3 0 11 1 2 12 5 10 14 7 
				1 10 13 0 6 9 8 7 4 15 14 3 11 5 2 12 
				
				S[4] 
				
				7 13 14 3 0 6 9 10 1 2 8 5 11 12 4 15 
				13 8 11 5 6 15 0 3 4 7 2 12 1 10 14 9 
				10 6 9 0 12 11 7 13 15 1 3 14 5 2 8 4 
				3 15 0 6 10 1 13 8 9 4 5 11 12 7 2 14 
				
				S[5] 
				
				2 12 4 1 7 10 11 6 8 5 3 15 13 0 14 9 
				14 11 2 12 4 7 13 1 5 0 15 10 3 9 8 6 
				4 2 1 11 10 13 7 8 15 9 12 5 6 3 0 14 
				11 8 12 7 1 14 2 13 6 15 0 9 10 4 5 3 
				
				S[6] 
				
				12 1 10 15 9 2 6 8 0 13 3 4 14 7 5 11 
				10 15 4 2 7 12 9 5 6 1 13 14 0 11 3 8 
				9 14 15 5 2 8 12 3 7 0 4 10 1 13 11 6 
				4 3 2 12 9 5 15 10 11 14 1 7 6 0 8 13 
				
				S[7] 
				
				4 11 2 14 15 0 8 13 3 12 9 7 5 10 6 1 
				13 0 11 7 4 9 1 10 14 3 5 12 2 15 8 6 
				1 4 11 13 12 3 7 14 10 15 6 8 0 5 9 2 
				6 11 13 8 1 4 10 7 9 5 0 15 14 2 3 12 
				
				S[8] 
				
				13 2 8 4 6 15 11 1 10 9 3 14 5 0 12 7 
				1 15 13 8 10 3 7 4 12 5 6 11 0 14 9 2 
				7 11 4 1 9 12 14 2 0 6 10 13 15 3 5 8 
				2 1 14 7 4 10 8 13 15 12 9 0 3 5 6 11 
				
				2.4.4.4 Loop back to 2.4.4.1 until all 8 blocks have been replaced. 
				
				2.4.5 Permute the concatenation of B[1] through B[8] as indicated below. 
				
				Permutation P 
				
				16 7 20 21 
				29 12 28 17 
				1 15 23 26 
				5 18 31 10 
				2 8 24 14 
				32 27 3 9 
				19 13 30 6 
				22 11 4 25 
				
				2.4.6 Exclusive-or the resulting value with L[i-1]. Thus, all together, 
				your R[i] = L[i-1] xor P(S[1](B[1])...S[8](B[8])), where B[j] is a 6-bit 
				block of E(R[i-1]) xor K[i]. (The function for R[i] is written as, R[i] = 
				L[i-1] xor f(R[i-1], K[i]).) 
				
				2.4.7 L[i] = R[i-1]. 
				
				2.4.8 Loop back to 2.4.1 until K[16] has been applied. 
				
				2.5 Perform the following permutation on the block R[16]L[16]. 
				
				Final Permutation (IP**-1) 
				
				40 8 48 16 56 24 64 32 
				39 7 47 15 55 23 63 31 
				38 6 46 14 54 22 62 30 
				37 5 45 13 53 21 61 29 
				36 4 44 12 52 20 60 28 
				35 3 43 11 51 19 59 27 
				34 2 42 10 50 18 58 26 
				33 1 41 9 49 17 57 25 
				
				
				This has been a description of how to use the DES algorithm to encrypt 
				one 64-bit block. To decrypt, use the same process, but just use the keys 
				K[i] in reverse order. That is, instead of applying K[1] for the first 
				iteration, apply K[16], and then K[15] for the second, on down to K[1]. 
				
				Summaries: 
				
				Key schedule: 
				C[0]D[0] = PC1(key) 
				for 1 				C[i] = LS[i](C[i-1]) 
				D[i] = LS[i](D[i-1]) 
				K[i] = PC2(C[i]D[i]) 
				
				Encipherment: 
				L[0]R[0] = IP(plain block) 
				for 1 				L[i] = R[i-1] 
				R[i] = L[i-1] xor f(R[i-1], K[i]) 
				cipher block = FP(R[16]L[16]) 
				
				Decipherment: 
				R[16]L[16] = IP(cipher block) 
				for 1 				R[i-1] = L[i] 
				L[i-1] = R[i] xor f(L[i], K[i]) 
				plain block = FP(L[0]R[0]) 
				
				
				To encrypt or decrypt more than 64 bits there are four official modes 
				(defined in FIPS PUB 81). One is to go through the above-described 
				process for each block in succession. This is called Electronic Codebook 
				(ECB) mode. A stronger method is to exclusive-or each plaintext block 
				with the preceding ciphertext block prior to encryption. (The first 
				block is exclusive-or'ed with a secret 64-bit initialization vector 
				(IV).) This is called Cipher Block Chaining (CBC) mode. The other two 
				modes are Output Feedback (OFB) and Cipher Feedback (CFB). 
				
				When it comes to padding the data block, there are several options. One 
				is to simply append zeros. Two suggested by FIPS PUB 81 are, if the data 
				is binary data, fill up the block with bits that are the opposite of the 
				last bit of data, or, if the data is ASCII data, fill up the block with 
				random bytes and put the ASCII character for the number of pad bytes in 
				the last byte of the block. Another technique is to pad the block with 
				random bytes and in the last 3 bits store the original number of data bytes. 
				
				The DES algorithm can also be used to calculate checksums up to 64 bits 
				long (see FIPS PUB 113). If the number of data bits to be checksummed is 
				not a multiple of 64, the last data block should be padded with zeros. If 
				the data is ASCII data, the first bit of each byte should be set to 0. 
				The data is then encrypted in CBC mode with IV = 0. The leftmost n bits 
				(where 16 				block are an n-bit checksum 
							

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