Implementar el Algoritmo Hashing Seguro – 512 ( SHA-512 ) como Paradigma de Programación Funcional

Dada una string S de longitud N , la tarea es encontrar el valor hash SHA-512 de la string S dada .

Ejemplos:

Entrada: S = «GeeksforGeeks»
Salida: acc10c4e0b38617f59e88e49215e2e894afaee5ec948c2af6f44039f03c9fe47a9210e01d5cd926c142bdc9179c2ad30f927a8faf69421ff60a5eaddcf8cb5eaddcf8cb5

Entrada: S = «hola mundo»
Salida:
309ecc489c12d6eb4cc40f50c902f2b4d0ed77ee511a7c7a9bcd3ca86d4cd86f989dd35bc5ff499670da34255b45b0cfd830e81f605dcf7dc5542e93ae9cd76f

Enfoque: siga los pasos a continuación para resolver el problema:

  • Convierte la string dada a la forma binaria .
  • Agregue ‘1’ a la string y luego ‘0’ continuamente hasta que la longitud de la string sea < (N%(1024 – 128)) .
  • Agregue la representación binaria de 128 bits de N en la string S .
  • Encuentra el número de trozos del tamaño de 1024 y guárdalo en una variable, digamos trozos como N/1024 .
  • Divida la string S en 16 partes de 64 caracteres.
  • Amplíe el número de fragmentos a 80 realizando las siguientes operaciones:
    • Itere sobre el rango [16, 80] y luego encuentre 4 valores, digamos WordA, WordB, WordC, WordD como:
      • PalabraA = rotar_derecha(Mensaje[g – 2], 19) ^ rotar_derecha(Mensaje[g – 2], 61) ^ shift_derecha(Mensaje[g – 2], 6) .
      • PalabraB = Mensaje[g – 7] .
      • WordC = rotar_derecha(Mensaje[g – 15], 1) ^ rotar_derecha(Mensaje[g – 15], 8) ^ shift_derecha(Mensaje[g – 15], 7) .
      • PalabraD = Mensaje[g – 16] .
    • Actualice el valor de Message[g] como (WordA + WordB + WordC+ WordD) .
  • Inicialice 8 variables, digamos A , B , C , D , E , F , G , H de tipo 64 bits para almacenar el valor hash final de la string dada S .
  • Atraviese la array Block[] y realice los siguientes pasos:
    • Actualice el valor de A , B , C , D , E , F , G , H usando la función Hash hasta 80 iteraciones girando una por una.
    • Ahora, actualice el valor de A , B , C , D , E , F , G , H mediante la suma de los valores anteriores de A , B , C , D , E , F , G , H y el valor recién actualizado de A , B , C , D , E , F , G , H .
  • Después de completar los pasos anteriores, imprima los valores hexadecimales de A , B , C , D , E , F , G , H para obtener el valor hash de la string dada.

A continuación se muestra la implementación del enfoque anterior:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long int int64;
 
int64 Message[80];
 
// Stores the hexadecimal values for
// calculating hash values
const int64 Constants[80]
    = { 0x428a2f98d728ae22, 0x7137449123ef65cd,
        0xb5c0fbcfec4d3b2f, 0xe9b5dba58189dbbc,
        0x3956c25bf348b538, 0x59f111f1b605d019,
        0x923f82a4af194f9b, 0xab1c5ed5da6d8118,
        0xd807aa98a3030242, 0x12835b0145706fbe,
        0x243185be4ee4b28c, 0x550c7dc3d5ffb4e2,
        0x72be5d74f27b896f, 0x80deb1fe3b1696b1,
        0x9bdc06a725c71235, 0xc19bf174cf692694,
        0xe49b69c19ef14ad2, 0xefbe4786384f25e3,
        0x0fc19dc68b8cd5b5, 0x240ca1cc77ac9c65,
        0x2de92c6f592b0275, 0x4a7484aa6ea6e483,
        0x5cb0a9dcbd41fbd4, 0x76f988da831153b5,
        0x983e5152ee66dfab, 0xa831c66d2db43210,
        0xb00327c898fb213f, 0xbf597fc7beef0ee4,
        0xc6e00bf33da88fc2, 0xd5a79147930aa725,
        0x06ca6351e003826f, 0x142929670a0e6e70,
        0x27b70a8546d22ffc, 0x2e1b21385c26c926,
        0x4d2c6dfc5ac42aed, 0x53380d139d95b3df,
        0x650a73548baf63de, 0x766a0abb3c77b2a8,
        0x81c2c92e47edaee6, 0x92722c851482353b,
        0xa2bfe8a14cf10364, 0xa81a664bbc423001,
        0xc24b8b70d0f89791, 0xc76c51a30654be30,
        0xd192e819d6ef5218, 0xd69906245565a910,
        0xf40e35855771202a, 0x106aa07032bbd1b8,
        0x19a4c116b8d2d0c8, 0x1e376c085141ab53,
        0x2748774cdf8eeb99, 0x34b0bcb5e19b48a8,
        0x391c0cb3c5c95a63, 0x4ed8aa4ae3418acb,
        0x5b9cca4f7763e373, 0x682e6ff3d6b2b8a3,
        0x748f82ee5defb2fc, 0x78a5636f43172f60,
        0x84c87814a1f0ab72, 0x8cc702081a6439ec,
        0x90befffa23631e28, 0xa4506cebde82bde9,
        0xbef9a3f7b2c67915, 0xc67178f2e372532b,
        0xca273eceea26619c, 0xd186b8c721c0c207,
        0xeada7dd6cde0eb1e, 0xf57d4f7fee6ed178,
        0x06f067aa72176fba, 0x0a637dc5a2c898a6,
        0x113f9804bef90dae, 0x1b710b35131c471b,
        0x28db77f523047d84, 0x32caab7b40c72493,
        0x3c9ebe0a15c9bebc, 0x431d67c49c100d4c,
        0x4cc5d4becb3e42b6, 0x597f299cfc657e2a,
        0x5fcb6fab3ad6faec, 0x6c44198c4a475817 };
 
// Function to convert a binary string
// to hexa-decimal value
string gethex(string bin)
{
    if (bin == "0000")
        return "0";
    if (bin == "0001")
        return "1";
    if (bin == "0010")
        return "2";
    if (bin == "0011")
        return "3";
    if (bin == "0100")
        return "4";
    if (bin == "0101")
        return "5";
    if (bin == "0110")
        return "6";
    if (bin == "0111")
        return "7";
    if (bin == "1000")
        return "8";
    if (bin == "1001")
        return "9";
    if (bin == "1010")
        return "a";
    if (bin == "1011")
        return "b";
    if (bin == "1100")
        return "c";
    if (bin == "1101")
        return "d";
    if (bin == "1110")
        return "e";
    if (bin == "1111")
        return "f";
}
 
// Function to convert a decimal value
// to hexa decimal value
string decimaltohex(int64 deci)
{
    // Stores the value as string
    string EQBIN = bitset<64>(deci).to_string();
 
    // Stores the equivalent hexa decimal
    string hexstring = "";
    string temp;
 
    // Traverse the string EQBIN
    for (unsigned int i = 0;
         i < EQBIN.length(); i += 4) {
        temp = EQBIN.substr(i, 4);
        hexstring += gethex(temp);
    }
 
    // Return the hexstring
    return hexstring;
}
 
// Function to convert a binary
// string to decimal value
int64 BintoDec(string bin)
{
 
    int64 value = bitset<64>(bin)
                      .to_ullong();
    return value;
}
 
// Function to right rotate x by n bits
int64 rotate_right(int64 x, int n)
{
    return (x >> n) | (x << (64 - n));
}
 
// Function to right shift x by n bits
int64 shift_right(int64 x, int n)
{
    return (x >> n);
}
 
// Function to divide the string
// into chunks
void separator(string getBlock)
{
    // Stores the size of chunks
    int chunknum = 0;
 
    // Traverse the string S
    for (unsigned int i = 0;
         i < getBlock.length();
         i += 64, ++chunknum) {
 
        // Update the Message[chunknum]
        Message[chunknum]
            = BintoDec(getBlock.substr(i, 64));
    }
 
    // Iterate over the range [16, 80]
    for (int g = 16; g < 80; ++g) {
 
        // Find the WordA
        int64 WordA = rotate_right(Message[g - 2], 19)
                      ^ rotate_right(Message[g - 2], 61)
                      ^ shift_right(Message[g - 2], 6);
 
        // Find the WordB
        int64 WordB = Message[g - 7];
 
        // Find the WordC
        int64 WordC = rotate_right(Message[g - 15], 1)
                      ^ rotate_right(Message[g - 15], 8)
                      ^ shift_right(Message[g - 15], 7);
 
        // Find the WordD
        int64 WordD = Message[g - 16];
 
        // Find the resultant code
        int64 T = WordA + WordB + WordC + WordD;
 
        // Return the resultant Hash Code
        Message[g] = T;
    }
}
 
// Function to find the major of a, b, c
int64 maj(int64 a, int64 b, int64 c)
{
    return (a & b) ^ (b & c) ^ (c & a);
}
 
// Function to find the ch value of a,
// b, and c
int64 Ch(int64 e, int64 f, int64 g)
{
    return (e & f) ^ (~e & g);
}
 
// Function to find the Bitwise XOR with
// the right rotate over 14, 18, and 41
int64 sigmaE(int64 e)
{
    // Return the resultant value
    return rotate_right(e, 14)
           ^ rotate_right(e, 18)
           ^ rotate_right(e, 41);
}
 
// Function to find the Bitwise XOR with
// the right rotate over 28, 34, and 39
int64 sigmaA(int64 a)
{
 
    // Return the resultant value
    return rotate_right(a, 28)
           ^ rotate_right(a, 34)
           ^ rotate_right(a, 39);
}
 
// Function to generate the hash code
void Func(int64 a, int64 b, int64 c,
          int64& d, int64 e, int64 f,
          int64 g, int64& h, int K)
{
    // Find the Hash Code
    int64 T1 = h + Ch(e, f, g) + sigmaE(e) + Message[K]
               + Constants[K];
    int64 T2 = sigmaA(a) + maj(a, b, c);
 
    d = d + T1;
    h = T1 + T2;
}
 
// Function to convert the hash value
// of a given string
string SHA512(string myString)
{
    // Stores the 8 blocks of size 64
    int64 A = 0x6a09e667f3bcc908;
    int64 B = 0xbb67ae8584caa73b;
    int64 C = 0x3c6ef372fe94f82b;
    int64 D = 0xa54ff53a5f1d36f1;
    int64 E = 0x510e527fade682d1;
    int64 F = 0x9b05688c2b3e6c1f;
    int64 G = 0x1f83d9abfb41bd6b;
    int64 H = 0x5be0cd19137e2179;
 
    int64 AA, BB, CC, DD, EE, FF, GG, HH;
 
    stringstream fixedstream;
 
    // Traverse the string S
    for (int i = 0;
         i < myString.size(); ++i) {
 
        // Add the character to stream
        fixedstream << bitset<8>(myString[i]);
    }
 
    // Stores string of size 1024
    string s1024;
 
    // Stores the string in the
    // fixedstream
    s1024 = fixedstream.str();
 
    // Stores the length of string
    int orilen = s1024.length();
    int tobeadded;
 
    // Find modded string length
    int modded = s1024.length() % 1024;
 
    // If 1024-128 is greater than modded
    if (1024 - modded >= 128) {
        tobeadded = 1024 - modded;
    }
 
    // Else if 1024-128 is less than modded
    else if (1024 - modded < 128) {
        tobeadded = 2048 - modded;
    }
 
    // Append 1 to string
    s1024 += "1";
 
    // Append tobeadded-129 zeros
    // in the string
    for (int y = 0; y < tobeadded - 129; y++) {
        s1024 += "0";
    }
 
    // Stores the binary representation
    // of string length
    string lengthbits
        = std::bitset<128>(orilen).to_string();
 
    // Append the lengthbits to string
    s1024 += lengthbits;
 
    // Find the count of chunks of
    // size 1024 each
    int blocksnumber = s1024.length() / 1024;
 
    // Stores the numbering of chunks
    int chunknum = 0;
 
    // Stores hash value of each blocks
    string Blocks[blocksnumber];
 
    // Traverse the string s1024
    for (int i = 0; i < s1024.length();
         i += 1024, ++chunknum) {
        Blocks[chunknum] = s1024.substr(i, 1024);
    }
 
    // Traverse tha array Blocks[]
    for (int letsgo = 0;
         letsgo < blocksnumber;
         ++letsgo) {
 
        // Divide the current string
        // into 80 blocks size 16 each
        separator(Blocks[letsgo]);
 
        AA = A;
        BB = B;
        CC = C;
        DD = D;
        EE = E;
        FF = F;
        GG = G;
        HH = H;
 
        int count = 0;
 
        // Find hash values
        for (int i = 0; i < 10; i++) {
 
            // Find the Hash Values
 
            Func(A, B, C, D, E, F, G, H, count);
            count++;
            Func(H, A, B, C, D, E, F, G, count);
            count++;
            Func(G, H, A, B, C, D, E, F, count);
            count++;
            Func(F, G, H, A, B, C, D, E, count);
            count++;
            Func(E, F, G, H, A, B, C, D, count);
            count++;
            Func(D, E, F, G, H, A, B, C, count);
            count++;
            Func(C, D, E, F, G, H, A, B, count);
            count++;
            Func(B, C, D, E, F, G, H, A, count);
            count++;
        }
 
        // Update the value of A, B, C,
        // D, E, F, G, H
 
        A += AA;
        B += BB;
        C += CC;
        D += DD;
        E += EE;
        F += FF;
        G += GG;
        H += HH;
    }
 
    stringstream output;
 
    // Print the hexadecimal value of
    // strings as the resultant SHA-512
    output << decimaltohex(A);
    output << decimaltohex(B);
    output << decimaltohex(C);
    output << decimaltohex(D);
    output << decimaltohex(E);
    output << decimaltohex(F);
    output << decimaltohex(G);
    output << decimaltohex(H);
 
    // Return the string
    return output.str();
}
 
// Driver Code
int main()
{
    // Input
    string S = "GeeksForGeeks";
 
    // Function Call
    cout << S << ": " << SHA512(S);
 
    return 0;
}
Producción: 

GeeksForGeeks: 0acc10c4e0b38617f59e88e49215e2e894afaee5ec948c2af6f44039f03c9fe47a9210e01d5cd926c142bdc9179c2ad30f927a8faf69421ff60a5eaddcf8cb9c

 

Complejidad temporal: O(N)
Espacio auxiliar: O(1)

Publicación traducida automáticamente

Artículo escrito por therain0605 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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