Algorithm: Inverse in place

Algorithm I

Algorithm I (Inverse in place). Replace X[1]X[2]… X[n], a permutation of
{1,2,…,n}, by its inverse. This algorithm is due to Bing-Chao Huang [Inf.
Proc. Letters 12 (1981), 237-238].
I1. [Initialize.] Set m <– n, j <– -1.
I2. [Next element.] Set i <– X[m]. If i < 0, go to step I5 (the element has
already been processed).
I3. [Invert one.] (At this point j < 0 and i = X[m]. If m is not the largest ele-
ment of its cycle, the original permutation had X[-j] = m.) Set X[m] <– j,
j <– -m, m <– i, i <– X[m].
I4. [End of cycle?] If i > 0, go back to I3 (the cycle has not ended); otherwise
set i <– j. (In the latter case, the original permutation had X[-j] = m, and
m is largest in its cycle.)
I5. [Store final value.] Set X[m] <– -i. (Originally X[-i] was equal to m.)
I6. [Loop on m.] Decrease m by 1. If m > 0, go back to I2; otherwise the
algorithm terminates. |

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Data table

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Java program

/**
 * Created with IntelliJ IDEA.
 * User: 1O1O
 * Date: 12/18/13
 * Time: 6:52 PM
 * :)~
 * Inverse in place:ALGORITHMS
 */
public class Main {

    public static void main(String[] args) {
        int[] X = new int[7];
        int m;
        int n=6;
        int i;
        int j;
        X[1]=6;
        X[2]=2;
        X[3]=1;
        X[4]=5;
        X[5]=4;
        X[6]=3;

        /*Print the initial X[i] (1<=i<=n)*/
        System.out.println("The initial X[i] (1<=i<=n) is:");
        for(int k=1; k<=n; k++){
            System.out.println("X["+k+"]="+X[k]);
        }
        System.out.println();

        /*Print the initial permutation*/
        System.out.println("The initial permutation is:");
        System.out.print("| ");
        for(int k=1; k<=n; k++){
            System.out.print((char)(k+96));
            System.out.print(' ');
        }
        System.out.print("|");
        System.out.println();
        System.out.print("| ");
        for(int k=1; k<=n; k++){
            System.out.print((char)(X[k]+96));
            System.out.print(' ');
        }
        System.out.print("|");
        System.out.println();
        System.out.println();

        m=n;                       /*I1*/
        j=-1;

        do{
            i = X[m];                 /*I2*/
            if(i < 0){
            }else {
                do{
                    X[m] = j;               /*I3*/
                    j = -m;
                    m = i;
                    i = X[m];
                }while (i > 0);            /*I4*/
                i = j;
            }
            X[m] = -i;                  /*I5*/
            m--;

        }while (m > 0);

        /*Print the final X[i] (1<=i<=n) after inversed*/
        System.out.println("The final X[i] (1<=i<=n) after inversed is:");
        for(int k=1; k<=n; k++){
            System.out.println("X["+k+"]="+X[k]);
        }
        System.out.println();

        /*Print the final permutation after inversed*/
        System.out.println("The final permutation after inversed is:");
        System.out.print("| ");
        for(int k=1; k<=n; k++){
            System.out.print((char)(k+96));
            System.out.print(' ');
        }
        System.out.print("|");
        System.out.println();
        System.out.print("| ");
        for(int k=1; k<=n; k++){
            System.out.print((char)(X[k]+96));
            System.out.print(' ');
        }
        System.out.print("|");
        System.out.println();
    }
}

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Outputs

The initial X[i] (1<=i<=n) is:
X[1]=6
X[2]=2
X[3]=1
X[4]=5
X[5]=4
X[6]=3

The initial permutation is:
| a b c d e f |
| f b a e d c |

The final X[i] (1<=i<=n) after inversed is:
X[1]=3
X[2]=2
X[3]=6
X[4]=5
X[5]=4
X[6]=1

The final permutation after inversed is:
| a b c d e f |
| c b f e d a |

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Reference

<< The Art of Computer Programming: Fundamental Algorithms >> VOLUME 1, DONALD E. KNUTH