des、3des加密算法_数据加密标准（DES）算法

悠悠 2022-12-07 13:16 194阅读 0赞

![158dd3ead192d409d826b784c6643e31.png][]

des、3des加密算法

Data Encryption Standard is a symmetric-key algorithm for the encrypting the data. It comes under block cipher algorithm which follows Feistel structure. Here is the block diagram of Data Encryption Standard.

数据加密标准是用于加密数据的对称密钥算法。 它遵循遵循Feistel结构的分组密码算法。 这是数据加密标准的框图。

![DES Algorithm Block Diagram][]

*Fig1: DES Algorithm Block Diagram \[Image Source: Cryptography and Network Security Principles and Practices 4th Ed by William Stallings\]*

*图1：DES算法框图\[图片来源：William Stallings的密码学和网络安全原理与实践第四版\]*

**Explanation for above diagram:** Each character of plain text converted into binary format. Every time we take 64 bits from that and give as input to DES algorithm, then it processed through 16 rounds and then converted to cipher text.

**上图的说明：**纯文本的每个字符都转换为二进制格式。 每次我们从中获取64位并将其作为DES算法的输入，然后经过16轮处理，然后转换为密文。

**Initial Permutation:** 64 bit plain text goes under initial permutation and then given to round 1. Since initial permutation step receiving 64 bits, it contains an 1×64 matrix which contains numbers from 1 to 64 but in shuffled order. After that, we arrange our original 64 bit text in the order mentioned in that matrix. \[You can see the matrix in below code\]

**初始置换：** 64位纯文本经过初始置换，然后送入第1轮。由于初始置换步骤接收64位，因此它包含一个1×64矩阵，该矩阵包含从1到64的数字，但按随机顺序排列。 之后，我们按照矩阵中提到的顺序排列原始的64位文本。 \[您可以在下面的代码中看到矩阵\]

After initial permutation, 64 bit text passed through 16 rounds. In each round it processed with 48 bit key. That means we need total 16 sub keys, one for each round. See below diagram, it will show what happening in each round of algorithm.

初始排列后，64位文本经过16个回合。 在每个回合中，都使用48位密钥进行处理。 这意味着我们总共需要16个子密钥，每个回合一个。 参见下图，它将显示每轮算法中发生的情况。

![Single Round of DES Algorithm][]

*Fig2: Single Round of DES Algorithm. \[Image Source: Cryptography and Network Security Principles and Practices 4th Ed by William Stallings\]*

*图2：单轮DES算法。* *\[图片来源：William Stallings的密码学和网络安全原理和实践第四版\]*

**Round i:** In each round 64bit text divided into two 32bit parts. Left and Right. You can see in diagram Li-1 and Ri-1. As algorithm says, Right 32bits goes under Expansion Permutation.

**第一轮：**在每一轮中，将64位文本分为两个32位部分。 左和右。 您可以在图L i-1和R i-1中看到。 正如算法所说，Right 32bits在扩展置换下。

**Expansion Permutation:** Right side 32bit part of text given to expansion permutation. It will produce a 48bit text as output. i.e. 16bits added in this step. Some bits below 32 are repeated and arranged in an 1×48 matrix form. We rearrange 32bit text by following the order of that matrix. \[See the matrix in below code\]

**扩展置换：**文本的右侧32位部分用于扩展置换。 它将产生一个48位的文本作为输出。 即在此步骤中添加了16位。 低于32的某些位被重复并以1×48矩阵形式排列。 我们按照矩阵的顺序重新排列32位文本。 \[请参见下面的代码矩阵\]

After expansion permutation we have to XOR the output 48bit with a 48bit sub key. Let see how that 48bit sub key generating from 64bit original key.

扩展置换后，我们必须使用48位子密钥对输出48位进行XOR。 让我们看看该48bit子密钥如何从64bit原始密钥生成。

**Permutated Choice 1:** Initially we take a 64 bit key and then apply to permutated choice 1. It contains a 1×56 matrix but with shuffled 1 to 64 numbers except multiples of number 8. i.e. 8, 16, 24, 32, 40, 48, 56, 64 will be discarded. Remaining 64-8 = 56 number will be there in 1×56 matrix. We rearrange key in matrix specified order. \[You can see the matrix in below code\]

**排列选择1：**最初，我们使用64位密钥，然后将其应用于排列选择1。它包含1×56矩阵，但混洗了1至64个数字，但数字8的倍数除外，即8、16、24、32、40， 48、56、64将被丢弃。 剩下的64-8 = 56个数字将出现在1×56矩阵中。 我们以矩阵指定的顺序重新排列键。 \[您可以在下面的代码中看到矩阵\]

**Left Circular Shift:** 56bit key from permutated choice 1 given to left circular shift operation. Here that 56bit key divided into two equal halves of each 28bit. These 28bits shifted depends upon the round number. We already have the data that in each round how many bits circularly we have to shift. You can see this data in shifts array in code.

**左循环移位：**将排列选项1的56位密钥提供给左循环移位操作。 在这里，该56位密钥分为每个28位相等的两半。 这些28位移位取决于轮数。 我们已经有了每轮必须循环移位多少位的数据。 您可以在代码的shifts数组中看到此数据。

**Permutated Choice 2:** Result of Left circular shift 56bit key given to permutated choice 2. This step will produce 48bit sub key. For this it has an 1×48 matrix, in which out of 56, some random 8 bits will be discarded. And remaining 48 will be there. According to this bit positions we have to rearrange the key. You can see this matrix in below code.

**排列选择2：**向排列选择2左移56位密钥的结果。此步骤将产生48位子密钥。 为此，它具有1×48矩阵，其中56个矩阵将被丢弃，其中一些随机的8位将被丢弃。 剩下的48个在那里。 根据该位的位置，我们必须重新排列密钥。 您可以在下面的代码中看到此矩阵。

Now output of permutated choice 2 will be Xor with output of expansion permutation, which results a 48bit one. This 48bit again reduced to 32bit using Substitution boxes \[called S box\].

现在，置换选择2的输出将与扩展置换的输出进行Xor运算，结果为48位。 使用“替换框”(称为“ S框”)将该48位再次减小为32位。

**Substitution boxes \[S box\]:** In DES algorithm we have 8 S boxes. Input for S box is 48bit. And output from S box is 32 bit. The input 48 bit will be divided equally to 8 s boxes from s1, s2, … s8. So each s box will get 48/8= 6 bits as input. This Each S box reduce 6 bits to 4 bits. i.e input for each S box is 6 bits and output is 4 bits. Finally, 8\*4 = 32 bit. Which is final output of S box operation.

**替代盒\[S box\]：**在DES算法中，我们有8个S盒。 S盒的输入为48位。 S盒的输出为32位。 输入的48位将被从s1，s2，...，s8均等地划分为8 s个盒。 因此，每个s盒将获得48/8 = 6位作为输入。 每个S盒将6位减少为4位。 即每个S盒的输入为6位，输出为4位。 最后，8 \* 4 = 32位。 这是S盒操作的最终输出。

Let see how 6bits converted to 4 bits from S box. S box is an 4×16 matrix containing numbers in range 0 to 15. Take example, assume input 6 bits for S box are 011011. In this first and last bit together represents row number. Since maximum number with two bits is 3, S box also contains 0 to 3 rows total of 4. And middle 4 numbers together represent column number. Since maximum number with 4 bits is 15, S box also contains columns 0 to 15 total of 16. So here first and last bit = 01 i.e. row number 1 and middle 4 bits 1101= 13 i.e. column number 13. So for this input the number positioned at row 1 and column 13 will be picked. As mentioned earlier S box only contains number in range 0 to 15. All can be represented in 4 bits. So picked number 4 bits are output for the S box. See the code for all S boxes.

让我们看看如何将6位从S盒转换为4位。 S box是一个4×16矩阵，包含0到15之间的数字。例如，假设S box的输入6位是011011。在此第一位和最后一位共同代表行号。 由于两位的最大数目为3，因此S框还包含0至3行，共4行。中间的4个数字一起代表列号。 由于4位的最大位数为15，所以S框还包含0到15列，共16列。因此，这里的第一位和最后一位= 01，即行号1，中间的4位1101 = 13，即列号13。将选择位于第1行和第13列的数字。 如前所述，S框仅包含0到15范围内的数字。所有这些都可以用4位表示。 因此，将为S盒输出所选的4位数字。 请参阅所有S盒的代码。

**Permutation:** After getting output from all S boxes, we are applying again permutation. Here also a matrix with different arrangements will be there, we have to arrange according to that.

**置换：**从所有S盒获取输出后，我们将再次应用置换。 这里也将有一个具有不同安排的矩阵，我们必须根据此安排。

**Final XOR:** After this permutation, take the left half which initially divided 64bit text to two halves. Do XOR with this permutation output to left 32bit part. This result is new Right part. And Right 32bit part which passed through all permutation will be come as new Left Part. These 2 parts will be the inputs for the second round. Same as keys also, the parts before left shift are next round input keys.

**最终XOR：**完成此排列后，取左半部分，最初将64位文本分为两半。 使用此置换输出到左32位部分进行XOR。 该结果是新的右侧部分。 经过所有排列的右32bit部分将作为新的左部分出现。 这两个部分将作为第二轮的输入。 与键一样，左移之前的部分是下一轮输入键。

All this explanation for a single round for a 62bit plain text. Like this, it passes through total 16 rounds.

所有这些说明仅针对62bit纯文本进行一次回合。 像这样，总共经过16轮。

**32 bit swap:** After completion of 16 rounds, final 64 bits divided into two 32 bit parts and they swap each other.

**32位交换：**完成16回合后，最后的64位分为两个32位部分，它们彼此交换。

**Inverse Initial Permutation:** Here also a matrix will be there, in which bits are just shuffled. No adding or subtracting bits. See the code for this matrix.

**反向初始置换：**这里也将有一个矩阵，其中的各个位都只经过混洗。 没有加或减位。 请参阅此矩阵的代码。

## C语言中的DES算法程序 **(**Program for DES Algorithm in C**)** ##

#include <stdio.h>
     
    int Original_key [64] = { // you can change key if required
    	0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0,
    	0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1,
    	1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0,
    	1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1
    };
     
    int Permutated_Choice1[56] = {
    	  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
    };
     
    int Permutated_Choice2[48] = {
    	  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
    };
     
    int Iintial_Permutation [64] = {
    	  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
    };
     
    int Final_Permutation[] = 
    {
    	  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
    };
     
    int 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
    };
     
    int 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
    };
     
    int S1[4][16] = 
    {
    		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
    };
     
    int S2[4][16] = 
    {
    	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
    };
     
    int S3[4][16] = 
    {
    	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
    };
     
    int S4[4][16] = 
    {
    	 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
    };
     
    int S5[4][16] = 
    {
    	 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
    };
     
    int S6[4][16] = 
    {
    	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
    };
     
    int S7[4][16]= 
    {
    	 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
    };
     
    int S8[4][16]= 
    {
    	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
    };
     
    int shifts_for_each_round[16] = { 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1 };
    int _56bit_key[56];
    int _48bit_key[17][48];
    int text_to_bits[99999], bits_size=0;
    int Left32[17][32], Right32[17][32];
    int EXPtext[48];
    int XORtext[48];
    int X[8][6];
    int X2[32];
    int R[32];
    int chiper_text[64];
    int encrypted_text[64];
     
    int XOR(int a, int b) {
    	return (a ^ b);
    }
     
    void Dec_to_Binary(int n) 
    { 
        int binaryNum[1000]; 
        int i = 0; 
        while (n > 0) { 
            binaryNum[i] = n % 2; 
            n = n / 2; 
            i++; 
        } 
        for (int j = i - 1; j >= 0; j--) {
    			text_to_bits[bits_size++] = binaryNum[j]; 
    	}
    } 
     
    int F1(int i)
    {
    	int r, c, b[6];
    	for (int j = 0; j < 6; j++)
    		b[j] = X[i][j];
     
    	r = b[0] * 2 + b[5];
    	c = 8 * b[1] + 4 * b[2] + 2 * b[3] + b[4];
    	if (i == 0)
    		return S1[r][c];
    	else if (i == 1)
    		return S2[r][c];
    	else if (i == 2)
    		return S3[r][c];
    	else if (i == 3)
    		return S4[r][c];
    	else if (i == 4)
    		return S5[r][c];
    	else if (i == 5)
    		return S6[r][c];
    	else if (i == 6)
    		return S7[r][c];
    	else if (i == 7)
    		return S8[r][c];
    }
     
     
    int PBox(int pos, int bit)
    {
    	int i;
    	for (i = 0; i < 32; i++)
    		if (P[i] == pos + 1)
    			break;
    	R[i] = bit;
    }
     
    int ToBits(int value)
    {
    	int k, j, m;
    	static int i;
    	if (i % 32 == 0)
    		i = 0;
    	for (j = 3; j >= 0; j--) 
    	{
    		m = 1 << j;
    		k = value & m;
    		if (k == 0)
    			X2[3 - j + i] = '0' - 48;
    		else
    			X2[3 - j + i] = '1' - 48;
    	}
    	i = i + 4;
    }
     
    int SBox(int XORtext[])
    {
    	int k = 0;
    	for (int i = 0; i < 8; i++)
    		for (int j = 0; j < 6; j++)
    			X[i][j] = XORtext[k++];
     
    	int value;
    	for (int i = 0; i < 8; i++) 
    	{
    		value = F1(i);
    		ToBits(value);
    	}
    }
     
    void expansion_function(int pos, int bit)
    {
    	for (int i = 0; i < 48; i++)
    		if (E[i] == pos + 1)
    			EXPtext[i] = bit;
    }
     
    void cipher(int Round, int mode)
    {
    	for (int i = 0; i < 32; i++)
    		expansion_function(i, Right32[Round - 1][i]);
     
    	for (int i = 0; i < 48; i++) 
    	{
    		if (mode == 0)
    			XORtext[i] = XOR(EXPtext[i], _48bit_key[Round][i]);
    		else
    			XORtext[i] = XOR(EXPtext[i], _48bit_key[17 - Round][i]);
    	}
     
    	SBox(XORtext);
     
    	for (int i = 0; i < 32; i++)
    		PBox(i, X2[i]);
    	for (int i = 0; i < 32; i++)
    		Right32[Round][i] = XOR(Left32[Round - 1][i], R[i]);
    }
     
    void finalPermutation(int pos, int bit)
    {
    	int i;
    	for (i = 0; i < 64; i++)
    		if (Final_Permutation[i] == pos + 1)
    			break;
    	encrypted_text[i] = bit;
    }
     
    void Encrypt_each_64_bit (int plain_bits [])
    {
    	int IP_result [64] , index=0;
    	for (int i = 0; i < 64; i++) {
    		IP_result [i] = plain_bits[ Iintial_Permutation[i] ];
    	}
    	for (int i = 0; i < 32; i++)
    		Left32[0][i] = IP_result[i];
    	for (int i = 32; i < 64; i++)
    		Right32[0][i - 32] = IP_result[i];
     
    	for (int k = 1; k < 17; k++) 
    	{ // processing through all 16 rounds
    		cipher(k, 0);
     
    		for (int i = 0; i < 32; i++)
    			Left32[k][i] = Right32[k - 1][i]; // right part comes as it is to next round left part
    	}
     
    	for (int i = 0; i < 64; i++) 
    	{ // 32bit swap as well as Final Inverse Permutation
    		if (i < 32)
    			chiper_text[i] = Right32[16][i];
    		else
    			chiper_text[i] = Left32[16][i - 32];
    		finalPermutation(i, chiper_text[i]);
    	}
     
    	for (int i = 0; i < 64; i++)
    		printf("%d", encrypted_text[i]);
    }
     
     
    void convert_Text_to_bits(char *plain_text){
    	for(int i=0;plain_text[i];i++){
    		int asci = plain_text[i];
    		Dec_to_Binary(asci);
    	}
    }
     
    void key56to48(int round, int pos, int bit)
    {
    	int i;
    	for (i = 0; i < 56; i++)
    		if (Permutated_Choice2[i] == pos + 1)
    			break;
    	_48bit_key[round][i] = bit;
    }
     
    int key64to56(int pos, int bit)
    {
    	int i;
    	for (i = 0; i < 56; i++)
    		if (Permutated_Choice1[i] == pos + 1)
    			break;
    	_56bit_key[i] = bit;
    }
     
    void key64to48(int key[])
    {
    	int k, backup[17][2];
    	int CD[17][56];
    	int C[17][28], D[17][28];
     
    	for (int i = 0; i < 64; i++)
    		key64to56(i, key[i]);
     
    	for (int i = 0; i < 56; i++)
    		if (i < 28)
    			C[0][i] = _56bit_key[i];
    		else
    			D[0][i - 28] = _56bit_key[i];
     
    	for (int x = 1; x < 17; x++) 
    	{
    		int shift = shifts_for_each_round[x - 1];
     
    		for (int i = 0; i < shift; i++)
    			backup[x - 1][i] = C[x - 1][i];
    		for (int i = 0; i < (28 - shift); i++)
    			C[x][i] = C[x - 1][i + shift];
    		k = 0;
    		for (int i = 28 - shift; i < 28; i++)
    			C[x][i] = backup[x - 1][k++];
     
    		for (int i = 0; i < shift; i++)
    			backup[x - 1][i] = D[x - 1][i];
    		for (int i = 0; i < (28 - shift); i++)
    			D[x][i] = D[x - 1][i + shift];
    		k = 0;
    		for (int i = 28 - shift; i < 28; i++)
    			D[x][i] = backup[x - 1][k++];
    	}
     
    	for (int j = 0; j < 17; j++) 
    	{
    		for (int i = 0; i < 28; i++)
    			CD[j][i] = C[j][i];
    		for (int i = 28; i < 56; i++)
    			CD[j][i] = D[j][i - 28];
    	}
     
    	for (int j = 1; j < 17; j++)
    		for (int i = 0; i < 56; i++)
    			key56to48(j, i, CD[j][i]);
    }
     
    int main(){
    	char plain_text[] = "tomarrow we wiil be declaring war";
    	convert_Text_to_bits(plain_text);
    	key64to48(Original_key); // it creates all keys for all rounds
    	int _64bit_sets = bits_size/64;
    	printf("Decrypted output is\n");
    	for(int i=0;i<= _64bit_sets ;i++) {
    		Encrypt_each_64_bit (text_to_bits + 64*i);
    	}
    	return 0;
    }

**Output**

**输出量**

*Decrypted output is* *0000111001101001001100011010111010010110111010111111111000010111001011111011111101010011011101011011000000111011100100000010110101000101011000011001000000101000001010011110101001011000111010011001110010110011011110110001101110000000001000001001000110111010*

*解密后的输出是* *0000111001101001001100011010111010010110111010111111111111000010111001011111011101010011011101011011000000111011100100000010110101000101011000011000000001000001010000010100111101010010110001110100110011100101100101100110110101101101110110000000*

> 翻译自: [https://www.thecrazyprogrammer.com/2019/01/data-encryption-standard-des-algorithm.html][https_www.thecrazyprogrammer.com_2019_01_data-encryption-standard-des-algorithm.html]

des、3des加密算法

[158dd3ead192d409d826b784c6643e31.png]: /images/20221123/7ac31615e27341a384c1f19f68939a12.png
[DES Algorithm Block Diagram]: /images/20221123/e5debd0c9d344b51bd4e28432ecfdf85.png
[Single Round of DES Algorithm]: /images/20221123/7750052fa36b4e1db85e3c3c624c9b08.png
[https_www.thecrazyprogrammer.com_2019_01_data-encryption-standard-des-algorithm.html]: https://www.thecrazyprogrammer.com/2019/01/data-encryption-standard-des-algorithm.html