How to select the first n terms?

Question

3.67/5 (3 votes)

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C++

for(i=0;i<500;i++)
{
    x=valuex[i];
    y=valuey[i];

    A[i][0]=x+y;
    A[i][1]=x*x+3*y;
    A[i][2]=x+2*y*x;
    ……
    A[i][99]=x/2+20*y;
}

In the above code,there are 100 terms:A[i][0]~A[i][99], In fact, I want to use the first n terms of A[i][j],and "n" is the input number by user, how can I realize this efficiently,Thank you for your help.
Best wishes.

Posted 25-Jun-12 16:14pm

Angela2012

Updated 25-Jun-12 19:45pm

v2

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Comments

Sergey Alexandrovich Kryukov 25-Jun-12 23:04pm

What is your criteria for "effectiveness" and what's the problem? What have you done so far?
Besides, I don't know what is the general algorithm to build an expression for right side of assignment to A[i][N] for any N (hidden under '......'). Are you going to write all 100 terms manually, 100 lines of code? Would look very discouraging... :-)
--SA

Angela2012 26-Jun-12 2:32am

Thank you for your help, it's difficult for me to express myself clearly in English, and I learned c++ for only serval months. some problems are too difficult for me. The 100 terms,which stands for zernike polynomials can be generated by writing a function with a input "n"(first n terms ),
http://research.opt.indiana.edu/Library/HVO/ZernikeStringR.m
but the calculation speed will be slow(verified using matlab :0.1s vs 10s): at least three for loops and many if...else...will be used to generate a single item. so I decide to write all 100 terms manually,
because speed is a prerequisite.

Sergey Alexandrovich Kryukov 25-Jun-12 23:30pm

Besides, A[i][0]~A[i][100] would be 101 terms, not 100.

Here is one big problem with your request is: you are trying to get a solution from experts, but only for a part of the problem, not explaining the rest of it and the final goal. A person like myself will consider it as a waste of time, because I would seriously suspect the rest of it is already wrong. Nobody wants to give help which won't be used.

--SA

Angela2012 26-Jun-12 2:55am

Thank you for your advice, It's difficult to explain the whole problem in several sentences, any of your help will be appreciate.

Sandeep Mewara 26-Jun-12 2:40am

It just looks like a Homework question for now. Can you share your thoughts and efforts around it?

Jim Jos 26-Jun-12 3:03am

Could you tell me the n is for i or j?

Angela2012 26-Jun-12 4:32am

n is for j

Jim Jos 26-Jun-12 4:40am

A[i][0] to A[i][99] is a fixed formule? out of hundred you want to use the first n only.. you want to set expressions only upto A[i][n] ? Please clarify..

Angela2012 26-Jun-12 4:47am

yes,A[i][0] to A[i][99] is a fixed formule,but valuex[i] and valuex[i] is different to different i,I want to set expressions only upto A[i][n].

Jim Jos 26-Jun-12 5:05am

Your requirement is done in solution 2? Please give your feedback on solution 2..

2 solutions

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nv3 · Accepted Answer · 2012-06-25T22:20:00

Nice question, although I don't see what the real application is. When I understood your question correctly, you want to compute only the first n terms of

A[i][0]=x+y;
A[i][1]=x*x+3*y;
A[i][2]=x+2*y*x;
……
A[i][99]=x/2+20*y;

thus n being the limit for j. Correct?

There are (at least) two principal approaches of how you can go about that:

(1) Interspersing if statements

while (1)
{
    if (0 >= n) break;
    A[i][0]=x+y;

    if (1 >= n) break;
    A[i][1]=x*x+3*y;

    if (2 >= n) break;
    A[i][2]=x+2*y*x;
     ……

    if (99 >= n) break;
    A[i][99]=x/2+20*y;
    break;
}

That is not elegant, but probably faster than approach number 2:

(2) Placing the arithmetic terms in an array of functions and then looping over that array.

typedef double PolyFunc (double x, double y);

double f0 (double x, double y)
    {return x + y;}

double f1 (double x, double y)
    {return x*x + 3*y;}

...

static PolyFunc* sFunctions[100] = {&f0, &f1, ...};

void ComputeArray (int n)
{
    ...
    for (int j = 0; j < n; ++j)
        A[i][j] = sFunctions[j] (x, y);
}

Hope that helps.

qPCR4vir · Accepted Answer · 2012-06-26T03:10:00

Solution 3

There are many questions about your question...
but...

C#

switch (n)
{
case 99: A[i][99]=x/2+20*y;
...
case 2:   A[i][2]=x+2*y*x;
case 1:   A[i][1]=x*x+3*y;
case 0:   A[i][0]=x+y;
}

will be very fast. Using switch let the compilers make some optimization over the if variant.
A bit confusing bot probably faster:

C#

switch (99-n)
{
case 0:   A[i][99]=x/2+20*y;
...
case 97:   A[i][2]=x+2*y*x;
case 98:   A[i][1]=x*x+3*y;
case 99:   A[i][0]=x+y;

}

Posted 26-Jun-12 3:10am

qPCR4vir

Updated 26-Jun-12 3:42am

v2

Comments

nv3 26-Jun-12 9:53am

Very nice idea! 5.

Angela2012 26-Jun-12 22:42pm

can you explain in detail why "swith(99-n){}" is faster than "switch(n){}",thank you very much.

qPCR4vir 27-Jun-12 7:24am

Actually I don’t know. That is the compiler job. I´m intending to give a tip to the compiler. It will code “switch(n)“ into something like jmp DWORD PTR $LN318@main[edi*4], (edi is our n) that is, a lookup table. (See: http://stackoverflow.com/q/2596320/1458030). I “feel” that with (99-n) this table is simpler or more efficient implemented. But a very gut compiler will figure out by self more than I can. In that case both variant will be equal fast.

qPCR4vir 27-Jun-12 7:28am

Thanks!