Converting from ternary to decimal

[kgehrma]

I need a function to change a ternary into a decimal. I already made one that goes from binary to decimal, but i am a little stumped on the logic of the ternary function

here is my binary one

int change_bin(char *bin)   
{
  int  b , k, n;
  int  len, sum = 0; 
 
  len = strlen(bin) - 1;
  for(k = 0; k <= len; k++) 
  {
      b = 1;
      n = (bin[k] - '0'); 
      b = b<<(len-k);
      sum = sum + n * b;
  }
  return(sum);
}

[anduril462]

Read up on positional notation, then apply similiar techniques as you did in your binary. Note that in your binary example << serves as multiplying by a power of two. E.g. << 1 is multiplying by 2^1, << 3 is like multiplying by 2^3.

[kgehrma]

i understand what you are saying i am just a little lost on how to get my ideas down on code.

[anduril462]

I'm not going to write the code for you. Give it a shot, and lets see what you come up with. As a hint, it will look very similar to change_bin. Remember, the << is multiplication by powers of 2 (binary is base 2), so for base 3, you will need to adjust that line to make b a power of 3 instead.

[kgehrma]

I understand what your saying, if i include <math.h> can't i do the pow(x,y). Or is that not at all what you are trying to say.

int change_bin(char *bin)   
{
  int  b , k, n;
  int  len, sum = 0; 
 
  len = strlen(bin) - 1;
  for(k = 0; k <= len; k++) 
  {
      b = 1;
      n = (bin[k] - '0'); 
      b = b<<(len-k);
      // so the above line might look something like b = pow(b,(len-k));
      sum = sum + n * b;
  }
  return(sum);
}

[anduril462]

Well, that's one way to do it. My thought was more like starting b at 1 outside the loop, then multiplying it each time through the loop. A little more efficient.

[kgehrma]

soo something like this?

int change_bin(char *bin)   
{
  int  b , k, n;
  int  len, sum = 0; 
 
  len = strlen(bin) - 1;
  b = 1;
  for(k = 0; k <= len; k++) 
  {
      n = (bin[k] - '0'); 
      b = b<<(len-k);
      sum = sum + n * b;
      b++;
  }
  return(sum);
}

[nonoob]

I think b goes up too rapidly. I thought you wanted base 3. You want b to be 3 to-the-exponent (len-k)

[kgehrma]

I think b goes up too rapidly.

i am sorry, but this math is just confusing the hell out of me, i might take a cell phone picture of this piece of paper i have been brain storming on just to show haha.

i understand c, and i understand trinary but God help me if i have to combine the two.

[nonoob]

It's easier if you start at the other end. The "units" digit.
for (k = len; k >=0; k--)
Then you could initialize b to 1 outside the loop.
In each iteration you do b = b * 3; b will go 1, 3, 9, 27, 81...
Get rid of the b++ thing.

[kgehrma]

It's easier if you start at the other end. The "units" digit.
for (k = len; k >=0; k--)
Then you could initialize b to 1 outside the loop.
In each iteration you do b = b * 3; b will go 1, 3, 9, 27, 81...
Get rid of the b++ thing.

So something like this?

b = 1;

for(k = 0; k <= len; k++) 
  {
   
      b = b * 3;
      n = (bin[k] - '0'); 
      b = b<<(len-k);
      sum = sum + n * b;
  
  }

[nonoob]

Except your loop has to go from the other end like I said. And you have to get rid of
b = b<<(len-k);

[kgehrma]

so this is my function:
the input: 22121002010111021
yeilds:370008669
it should be: 123336223? right?

int tri2dec(char *bin)
{
  int  b , k, n;
  int  len, sum = 0; 
 
  len = strlen(bin) - 1;
  b = 1;
  for(k = len; k >=0; k--) 
  {   
      b = b * 3;
      n = (bin[k] - '0'); 
      sum = sum + n * b;
 
  }
  return(sum);
}

[anduril462]

The rightmost digit needs to be processed with b = 1 (it's the 1's column). The next digit to the left is the 3's column, then the 9's, etc. Move your b = b * 3 down to the bottom of the loop.

[kgehrma]

The rightmost digit needs to be processed with b = 1 (it's the 1's column). The next digit to the left is the 3's column, then the 9's, etc. Move your b = b * 3 down to the bottom of the loop.

dear lord... i finally got it, thanks so much