Thursday, October 22, 2015

Algoritma dan Pemrograman Mencari Transpose Matrik

Algoritma dan Pemrograman Mencari Transpose Matrik

Nama Kelompok :
Andhika Firmanto (D1041141053)
Mahmud Syafi’ie (D1041141001)
Yulizar Pribadi (D1041141057)


Soal :
Buatlah algoritma untuk menentukan transpose matrik dibawah ini, jika diketahui :


Jawaban :
    A. Algoritma Pada Transpose Matrik


[1] [1]  1
[1] [2] ← 2
[1] [3] ← 3
[1] [4] ← 4
[2] [1] ← 5
[2] [2]  6
[2] [3]  7
[2] [4]  8
[3] [1]  1
[3] [2]  3
[3] [3]  5
[3] [4]  7

for  x    1 to 3 do
begin
          for  y    1 to 4 do       
          begin        
                     b [y , x]   a [x , y]
          end
end


for x  1 to 3 do
begin         
          for y  1 to 4 do
                   output   a [x , y]
end

for y  1 to 4 do
begin
           for x  1 to 3 do
                   output  b [y , x]

end




 B. Source Code Program Mencari Transpose Matrik Menggunakan Bahasa C :

#include <stdio.h>
#include <stdlib.h>

void main()
{
           int a[3][4], b[4][3], x, y;

           //Input Matriks A
           printf("Masukan Matriks A\n");
           printf("------------------\n");
           for(x=0;x<3;x++)
           {
                    for(y=0;y<4;y++)
                    {
                              printf("[%x][%x]: ",x+1,y+1);
                              scanf("%x",&a[x][y]);
                    }
           }
           printf("\n\n");

           //Proses transpose
           for (x=0; x<3; x++)
           {
                    for(y=0; y<4; y++)
                    {
                               b[y][x]=a[x][y];
                    }
           }

           //Matriks A
           printf("Matriks A \n");
           printf("------------------\n");
           for(x=0;x<3;x++)
           {
                    for(y=0;y<4;y++)
                    {
                              printf("%3x",a[x][y]);
                    }
                    printf("\n");
           }
           printf("\n\n");

           //Matriks B
           printf("Matriks B \n");
           printf("------------------\n");

           for(y=0;y<4;y++)
           {
                    for(x=0;x<3;x++)
                    {
                              printf("%3x",b[y][x]);
                    }
                    printf("\n");
           }
           printf("\n");
}

   
 C. Hasil Gambar Percobaan :



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