古典密码加密解密之多表代换

张泽旭

发布于 2018-12-10 16:41:02

1.3K0

发布于 2018-12-10 16:41:02

多表代换密码首先将明文M 分为由n 个字母组成的分组, , … ，对每个分组的加密为 ≡ + ( ), = , , … 其中，（A,B）是密钥，A 是 × 的可逆矩阵，满足gcd(|A|,N)=1,( |A|是A 的行列式)， = (, , … ), = (, , … ), = (, , … ),对密文的解密为 ≡ −( − )( ), = , , …

java实现多表代换加密

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
public class Mangboot {
    private static Map<Integer, String> maps = new HashMap<Integer,String>();
    public static int getKey(Map map,String value){
        int key = 0;
        Iterator it  = map.entrySet().iterator();
        while(it.hasNext()){
            Map.Entry entry = (Entry) it.next();
            Object obj = entry.getValue();
            if(obj!=null&&obj.equals(value)){
                key=(int)entry.getKey();
            }
        }
        return key;
    }
    /**
     * 对字符和数字进行一一映射
     */
    public static void fuzhi(){
        String str = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        for(int i=0;i<str.length();i++){
            maps.put(i, str.substring(i,i+1));
        }
        
    }
    /**
     * 加密的主要算法
     * @param a  输入的A矩阵（密钥）
     * @param d  需要加密的密文转换后的矩阵
     * @param c  保存加密后的结果
     * @param b  矩阵B（密钥）
     * @param n  矩阵的维数
     */
    public static void addmatrixMultiply(int[][]a,int [][]d,int [][]c,int b[][],int n){
        for(int i=0;i<n;i++)
            for(int j = 0;j <1;j++){
                int sum = a[i][0]*d[0][j];
                for(int k = 1;k<n;k++)
                    sum +=a[i][k]*d[k][j];
                c[i][j] = sum+b[i][j];
            }
    }
         public static void main(String[] args) {
        fuzhi();
        Scanner scanner = new Scanner(System.in);
        System.out.println("请输入你要加密的明文:");
        String str = scanner.nextLine();
        System.out.println("请输入密钥");
        System.out.println("请输入A矩阵的维数:");
        int n = scanner.nextInt();
        System.out.println("开始创建A矩阵:");
        int a[][] = new int[n][n];
        for(int i =0;i<n;i++){
            for(int j= 0;j<n;j++){
                a[i][j] = scanner.nextInt();
            }
        }
        System.out.println("开始创建B矩阵:");
        int b[][] = new int[n][1];
        int c[][] = new int[n][1];
        int d[][] = new int[n][1];
        for(int j= 0;j<n;j++){
            b[j][0] = scanner.nextInt();
        }
        System.out.println("获得的密文为:");
        for(int i = 0;i<str.length()/n;i++){
            String sub = str.substring(i*n, (i+1)*n);
            for(int j= 0;j<n;j++){
                d[j][0] = getKey(maps, sub.substring(j, j+1));
            }
            addmatrixMultiply(a, d, c,b, n);
            for(int j= 0;j<n;j++){
                System.out.print(maps.get(c[j][0]%26));
            }    
        }        
    }
}

java实现多表代换解密

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
public class Mangbootjiemi {
    private static Map<Integer, String> maps = new HashMap<Integer, String>();
    public static int getKey(Map map, String value) {
        int key = 0;
        Iterator it = map.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry entry = (Entry) it.next();
            Object obj = entry.getValue();
            if (obj != null && obj.equals(value)) {
                key = (int) entry.getKey();
            }
        }
        return key;
    }
    public static void fuzhi() {
        String str = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        for (int i = 0; i < str.length(); i++) {
            maps.put(i, str.substring(i, i + 1));
        }
    }
    public static void addmatrixMultiply(int[][] a, int[][] d, int[][] c, int b[][], int n) {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < 1; j++) {
                int sum = a[i][0] * d[0][j];
                for (int k = 1; k < n; k++)
                    sum += a[i][k] * d[k][j];
                c[i][j] = sum;
            }
    }
    public static void main(String[] args) {
        fuzhi();
        Scanner scanner = new Scanner(System.in);
        System.out.println("请输入你要解密的密文:");
        String str = scanner.nextLine();
        System.out.println("请输入密钥");
        System.out.println("请输入A-1矩阵的维数:");
        int n = scanner.nextInt();
        System.out.println("开始创建A-1矩阵:");
        int a[][] = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                a[i][j] = scanner.nextInt();
            }
        }
        System.out.println("开始创建B矩阵:");
        int b[][] = new int[n][1];
        int c[][] = new int[n][1];
        int d[][] = new int[n][1];
        for (int j = 0; j < n; j++) {
            b[j][0] = scanner.nextInt();
        }
        System.out.println("获得的明文为:");
        for (int i = 0; i < str.length() / n; i++) {
            String sub = str.substring(i * n, (i + 1) * n);
            for (int j = 0; j < n; j++) {
                d[j][0] = getKey(maps, sub.substring(j, j + 1)) - b[j][0];
            }
            addmatrixMultiply(a, d, c, b, n);
            for (int j = 0; j < n; j++) {
            
                if (c[j][0]%26 < 0) {
                    System.out.print(   maps.get(( c[j][0]% 26) + 26));
                    
                }else
                
                System.out.print(maps.get(c[j][0] % 26));
            }
        }
    }
}

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原始发表：2017年10月13日，如有侵权请联系 cloudcommunity@tencent.com 删除

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