8 replies to this topic

[VITUSH]

Hi Iam trying to write this in c++:
http://latex.codecog...e j}^{n}x_i-x_j
So far Ive got this:

double ljmi[max]; 

	int i=0;

	int j;

	int p = 0;

	for(j=0;j<k, i<k;j++)       //k <=> n in this case

	{

		if(j=i && j-1<0)

		{

			j = j+1;

			do

			{

			ljmi[p] *= (x[i]-x[j]);

			}

			while(j=k);

			{

			p+1;

			i+1;

			}

		}

		else if(j=i && j-1>0)

		{

			j = j-1;

			do

			{

			ljmi[p] *= (x[i]-x[j]);

			}

			while(j=k);

			p+1;

			i+1;

		}

		else

		{

			do

			{

			ljmi[p] *= (x[i]-x[j]);

			}

			while(j=k);

			p+1;

			i+1;

		}

	}

(I used conditions instead i!=j because it gave me bad result)
This should fill me an array ljmi[max].
Now when I do:

cout << arraysum(ljmi, k);

or

cout << ljmi[a]; //"a" is any number between 0 and k-1

Nothing happens, there is no value printed and I cant exit window with ENTER.
Yeah its a mess but I dont have much experience with cycles - where is mistake and is there a way to make it Small and Stupid?
thx4re

[WingedPanther]

You need an inner and outer loop. The outer loop contains the inner loop and an increment of sum. The inner loop accumulates a product as long as its variable isn't the same value as the outer variable.

[VITUSH]

thanks for your reply. I have this:

for(int i=0;i<k;i++)

	{

		for(int j=0;j<k;j++)

		{

			product[i] *= x[i] - x[j];

		}

	}

but i dont know how to make a condition that i != j. If I put it into inner loop:

for(int j=0;i!=j,j<k;j++)

Iam getting exactly the same result for:

for(int i = 0;i<k;i++)

      cout << product[i] << endl;

Which is this:

0

-0

0

-0

.

.

.

Could you please be more specific about that condition? Thanks

[rocketboy9000]

The comma operator in C simply discards its first argument.
You want to put a if statement inside the for loops.

[Flying Dutchman]

By if you probably mean logical expression, such as and &&, or ||, not !, complement ~, xor ^.

[WingedPanther]

for(int i=0;i<k;i++)
{
  product = 1; 
  for(int j=0;j<k;j++)
  {
    if (i!=j) product *= x[i] - x[j];
  }
  sum += product;
}

[VITUSH]

WingedPanther said:

for(int i=0;i<k;i++)

{

  product = 1; 

  for(int j=0;j<k;j++)

  {

    if (i!=j) product *= x[i] - x[j];

  }

  sum += product;

}

Thanks for the code, but I had something similar to this and its giving me bad result (same behavior as my code). For example:

for input k=3; x[3-1] = {1, 2, 3}; it gives me: 6 thats not correct:
(1-2)*(1-3)+(2-1)*(2-3)+(3-1)*(3-1) != 6 but 3
and for input k=3; x[3-1] = {4, 5, 6}; it gives me 9 also not correct:
(4-5)*(4-6)+(5-4)*(5-6)+(6-4)*(6-5) != 9 but also 3

[rocketboy9000]

k=3; x[3-1] = {1, 2, 3};

That code above is wrong. You declare x to be 3-1 = 2 numbers, and then give it 3 numbers.

[VITUSH]

rocketboy9000 said:

k=3; x[3-1] = {1, 2, 3};

That code above is wrong. You declare x to be 3-1 = 2 numbers, and then give it 3 numbers.

that's not code man. That's input...
=> for given x[0], x[1], x[2]