Member Avatar for FraidaL

This is a minor issue that I'm having in a larger program. The program is for factoring a second degree polynomial. The function below is supposed to find the factors, which is does, but let's say a=6, b=1, c=-12 so a*c=-72 and I want the two numbers that will add to 1. Instead of 9 and -8, this code gives me -9 and 8. I can't figure out why. If anyone could help me with this I would really appreciate it.

void findFactors(int a, int b, int c, int&f1, int&f2)
{
    int f;
    int i=0;
    f=-int(sqrt(abs(a*c)));
    while(f<=int(sqrt(abs(a*c))))
    {
        if(f==0)
            f++;
        if((a*c)%f==0)
        {
            f1=(a*c)/f, f2=f;
            i++;
        }

        f++;
    }

    if(f!=-int(sqrt(abs(a*c))))
    {
        f1==0, f2==0;
    }
}
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  • Latest Post Latest Post by histrungalot

Dani AI

Generated

Quick diagnosis and a safe fix.

The symptom you showed (first finding 9 and -8, then being overwritten by -9 and 8) is caused by two things: the loop keeps scanning every divisor pair and keeps updating f1/f2 instead of stopping when the matching pair is found, and the fallback at the end uses the comparison operator (==) instead of assignment so it never resets the outputs. correctly suggested an early exit and pointed out the == mistake. Below is a clearer, more robust approach that (1) only iterates positive divisors (faster), (2) checks the signed complement pairings, (3) uses 64-bit arithmetic to avoid overflow, and (4) returns success/failure explicitly.

// returns true and sets f1,f2 when two integers exist with f1*f2 == a*c and f1+f2 == b
bool findFactorsInt(int a, int b, int c, int &f1, int &f2) {
    using ll = long long;
    ll prod = static_cast<ll>(a) * static_cast<ll>(c);
    if (prod == 0) { f1 = 0; f2 = b; return true; } // handle ac == 0 quickly
    ll absprod = std::llabs(prod);
    ll limit = static_cast<ll>(std::sqrt((long double)absprod));
    for (ll d = 1; d <= limit; ++d) {
        if (absprod % d != 0) continue;
        ll e = prod / d;                // signed complement, d > 0
        if (d + e == b) { f1 = static_cast<int>(e); f2 = static_cast<int>(d); return true; }
        if (-d - e == b) { f1 = static_cast<int>(-e); f2 = static_cast<int>(-d); return true; }
    }
    f1 = f2 = 0;
    return false;
}

Troubleshooting tips

  • Use = for assignment. f1==0 is a comparison and does nothing.
  • Return immediately when you find the matching pair instead of letting later iterations overwrite it.
  • Use long long for a*c if a and c can be large.
  • After you get m and n (the two numbers), split the middle term (ax^2 + mx + nx + c) and factor by grouping to get the final binomials.

This keeps the search small and predictable and avoids the “wrap around” effect you observed.

Member Avatar for histrungalot

Its wraping around.

$ ./a.out
i(0) f1(9) f2(-8)    <------ Should have stopped here
i(1) f1(12) f2(-6)
i(2) f1(18) f2(-4)
i(3) f1(24) f2(-3)
i(4) f1(36) f2(-2)
i(5) f1(72) f2(-1)
i(6) f1(-72) f2(1)
i(7) f1(-36) f2(2)
i(8) f1(-24) f2(3)
i(9) f1(-18) f2(4)
i(10) f1(-12) f2(6)
i(11) f1(-9) f2(8)  <------ When all of the way to here
-9 8
$
Try adding a break if f1+f2 equal b. I tested it a little and looks like it works
$ ./a.out
i(0) f1(9) f2(-8)
9 -8
$
Code
void findFactors(int a, int b, int c, int&f1, int&f2)
{
    int f;
    int i=0;
    f=-int(sqrt(abs(a*c)));
    while(f<=int(sqrt(abs(a*c))))
    {
        if(f==0)
            f++;
        if((a*c)%f==0)
        {
            f1=(a*c)/f, f2=f;
            //-----------------------------------
            // Debug
            //cout << "i(" << i << ") f1(" << f1 << ") f2(" <<f2 << ")" << endl;
            //-----------------------------------
            // Break if all is well
            if ( f1+f2 == b) {
               break;
            }
            i++;
        }

        f++;
    }
    //--------------------------------------
    // Was't sure what this is for?
    if(f!=-int(sqrt(abs(a*c))))
    {
        //----------------------------------
        // Why are you using '==' here? 
        // The values of f1 and f2 will be
        // unchanged.
        f1==0, f2==0;
    }
}
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