Thread: Sort 3 letter by ASCII

Hi guys, I am total beginner. I have trouble with my first school work in programming. I have no idea how to sort 3 integers by ASCII ( Lexicographic).
For Example: C A B = A B C.

I cant use cycles, bubble sort and field.

I try some codes, but nothing worked.



Can you give me some help? Thanks.

I think the issue is that he's not allowed to use loops (cycles), which would exclude most sorts, leaving something like a sorting network, which is just a series of if statements and swaps. For 3 elements, 3 if / swap statements are needed. On a side note, 4 elements need 5 if / swap statements, and 10 elements would need 29 if / swap statements. Sorting network is one of the lesser known sorting methods, and only useful for sorting a small and fixed number of elements, sometimes implemented in hardware.

So the idea here is for Jamesss to figure out the 3 if / swap statement needed to sort 3 integers (the ASCII part doesn't really affect how this is done).

I try this but this is probably very bad Doesnt work.

Code:

int main(){
printf("Lexicograpgic:\n");
int  a,b,c;


   scanf("%c",&a,&b,&c);


    {if (a < b)
        {if (a < c)
            {if (b < c)
                printf("\n",&a,&b,&c);
            else
                printf("\n",&a,&c,&b);
            }




        else
            printf("\n",&c,&a,&b);
     }
   else
   {if (b < c)
         {if (a < c)
            printf("\n",&b,&a,&c);
         else
            printf("\n",&b,&c,&a);
   }
      else
         printf("\n",&c,&b,&a);
      }}




    return 0;
}

Originally Posted by Salem

Well your logic looks fine (or at least plausible).

Your use of printf and scanf however needs lots of work.

Eg
scanf("%d %d %d", &a, &b, &c);

and
printf("%d %d %d\n", a, b, c );

I do this but still not work, program write numbers.

Code:

int main(){
printf("Lexicograpgic:\n");
int  a,b,c;


   scanf("%d,%d,%d",&a,&b,&c);


    {if (a < b)
        {if (a < c)
            {if (b < c)
                printf("%d,%d,%d\n",a,b,c);
            else
                printf("%d,%d,%d\n",a,c,b);
            }




        else
            printf("%d,%d,%d\n",c,a,b);
     }
   else
   {if (b < c)
         {if (a < c)
            printf("%d,%d,%d\n",b,a,c);
         else
            printf("%d,%d,%d\n",b,c,a);
   }
      else
         printf("%d,%d,%d\n",c,b,a);
      }}




    return 0;
}

Originally Posted by Salem

I thought you understood.

You have to type in the commas to match your format string.

Code:

$ ./a.out 
Lexicograpgic:
6 2 4
0,0,6

$ ./a.out 
Lexicograpgic:
6,2,4
2,4,6

Ok I have this and it work so thank you, but it work only with numbers and I need letters a-z.

Originally Posted by rcgldr

The logic is ok, just needed to fix scanf and printf lines. This example uses spaces instead of commas for both input and output. Enter characters instead of numbers:

Code:

int main()
{
char a,b,c;
    printf("Lexicograpgic:\n");
    scanf("%c %c %c",&a,&b,&c);
    if (a < b){
        if (a < c){
            if (b < c)
                printf("%c %c %c\n",a,b,c);
            else
                printf("%c %c %c\n",a,c,b);
        } else
            printf("%c %c %c\n",c,a,b);
    } else {
        if (b < c){
            if (a < c)
                printf("%c %c %c\n",b,a,c);
            else
                printf("%c %c %c\n",b,c,a);
        } else
            printf("%c %c %c\n",c,b,a);
    }
    return 0;
}

Example runs with all 6 cases:

Lexicograpgic:
a b c
a b c

Lexicograpgic:
a c b
a b c

Lexicograpgic:
b a c
a b c

Lexicograpgic:
b c a
a b c

Lexicograpgic:
c a b
a b c

Lexicograpgic:
c b a
a b c

Oooo it works! I forgot change int fot char. Thank you mate!

Sorting network version. I'm assuming that no fields means you can't use temp variable, so I used xor swap method:

Code:

int main()
{
char a,b,c;
    printf("Lexicograpgic:\n");
    scanf("%c %c %c",&a,&b,&c);
    if(a > c){a ^= c; c ^= a; a ^= c;}
    if(a > b){a ^= b; b ^= a; a ^= b;}
    if(b > c){b ^= c; c ^= b; b ^= c;}
    printf("%c %c %c\n",a,b,c);
    return 0;
}

For 3 letter case, it's not saving much, but for 4 letter case, only 5 if / swap statements are needed to handle all 24 permutations.

Code:

int main()
{
char a,b,c,d;
    printf("Lexicograpgic:\n");
    scanf("%c %c %c %c",&a,&b,&c,&d);
    if(a > c){a ^= c; c ^= a; a ^= c;}
    if(b > d){b ^= d; d ^= b; b ^= d;}
    if(a > b){a ^= b; b ^= a; a ^= b;}
    if(c > d){c ^= d; d ^= c; c ^= d;}
    if(b > c){b ^= c; c ^= b; b ^= c;}
    printf("%c %c %c %c\n",a,b,c,d);
    return 0;
}

For 10 letter case, 29 if / swap statements would be needed to handle all 3,628,800 permutations.

Originally Posted by rcgldr

Sorting network version. I'm assuming that no fields means you can't use temp variable, so I used xor swap method:

Code:

int main()
{
char a,b,c;
    printf("Lexicograpgic:\n");
    scanf("%c %c %c",&a,&b,&c);
    if(a > c){a ^= c; c ^= a; a ^= c;}
    if(a > b){a ^= b; b ^= a; a ^= b;}
    if(b > c){b ^= c; c ^= b; b ^= c;}
    printf("%c %c %c\n",a,b,c);
    return 0;
}

For 3 letter case, it's not saving much, but for 4 letter case, only 5 if / swap statements are needed to handle all 24 permutations.

Code:

int main()
{
char a,b,c,d;
    printf("Lexicograpgic:\n");
    scanf("%c %c %c %c",&a,&b,&c,&d);
    if(a > c){a ^= c; c ^= a; a ^= c;}
    if(b > d){b ^= d; d ^= b; b ^= d;}
    if(a > b){a ^= b; b ^= a; a ^= b;}
    if(c > d){c ^= d; d ^= c; c ^= d;}
    if(b > c){b ^= c; c ^= b; b ^= c;}
    printf("%c %c %c %c\n",a,b,c,d);
    return 0;
}

For 10 letter case, 29 if / swap statements would be needed to handle all 3,628,800 permutations.

Can I ask you what mean " ^= " ? Your code look better , but this is not clear for me

Code:

 {a ^= c; c ^= a; a ^= c;}

etc.

Using a temp variable t to do swaps for the 3 letter case:

Code:

int main()
{
char a,b,c,t;
    printf("Lexicograpgic:\n");
    scanf("%c %c %c",&a,&b,&c);
    if(a > c){t = a; a = c; c = t;}
    if(a > b){t = a; a = b; b = t;}
    if(b > c){t = b; b = c; c = t;}
    printf("%c %c %c\n",a,b,c);
    return 0;
}

As for the XOR swap, note that x^x == 0.

Code:

    a ^= b;   // a = (a^b)
    b ^= a;   // b = (a^b)^b = a^(b^b) = a^0 = a
    a ^= b;   // a = (a^b)^a = (a^a)^b = 0^b = b