Factorising quadratic trinomials where the coefficient of X^2 is not 1
Author
Ok, i'm factorising quadratic trinomials, and i want to find a really fast way to do it. (i would also like to know how to do it without using guess and check.)
What i want to know, is what is the fastest way to do this?
From,
Nice coder
[Edited by - Nice Coder on July 23, 2005 10:41:21 PM]
Author
Private Sub Form_Load()Dim t As IntegerDim u As Integertim = TimerOpen "C:\Q.txt" For Input As #1Open "C:\A.txt" For Output As #2Do Until EOF(1)Input #1, tInput #1, uPrint #2, "=X^2 " & IIf(t >= 0, "+ ", "- ") & IIf(Abs(t) <> 1, Abs(t), "") & "X " & IIf(u >= 0, "+ ", "- ") & Abs(u)lim = 10DoFor x = -lim To limFor y = -lim To lims = x + yp = x * yDoEventsIf s = t And p = u Then Print #2, "=(X " & IIf(x >= 0, "+ ", "- ") & Abs(x) & ")(X " & IIf(y >= 0, "+ ", "- ") & Abs(y) & ")" GoTo nextnum:End IfNext yNext xlim = lim * 2Loopnextnum:Print #2, ""LoopClose #1Print #2, "Time required: " & Timer - timClose #2EndEnd SubThats my code so far.
Its now currently at 0.0032552083 seconds per node. (Which is fast, but we can do better!)
From,
Nice coder
Your timing method includes time spent on outputting the diagnostic messages. I/O is really slow stuff.
Author
Hmm, yes.
But is there anything faster? (ie. for the algorithm?)
Also, if the coefficient of X^2 is not 1, how would i factorise them?
From,
NIce coder
[Edited by - Nice Coder on July 23, 2005 10:13:08 PM]
But is there anything faster? (ie. for the algorithm?)
Also, if the coefficient of X^2 is not 1, how would i factorise them?
From,
NIce coder
[Edited by - Nice Coder on July 23, 2005 10:13:08 PM]
Quote:Original post by Nice Coder
Hmm, yes.
But is there anything faster? (ie. for the algorithm?)
Actually use the quadratic formula perhaps? (which would be a hell of a lot more general, too)
Quote:Also, if the coefficient of X^2 is not 1, how would i factorise them?
Just divide through by said coefficient first (and list it as a separate constant factor in the result); a constant factor will not affect where the roots are.
No promises that I didn't screw any of this up.
The math:
(a*x+f)*(x+e)=a*x*x+b*x+c
e=b/a/2+sqrt(b*b-(4*a*c))/a/2
f=c/e
A program:
Sample run:
a*x^2+b*x+c
a=2
b=13
c=15
(2*x+3)*(x+5)
[Edited by - smart_idiot on July 24, 2005 3:58:52 AM]
The math:
(a*x+f)*(x+e)=a*x*x+b*x+c
e=b/a/2+sqrt(b*b-(4*a*c))/a/2
f=c/e
A program:
#include <iostream>#include <complex>#include <string>#include <sstream>std::string comp(const std::complex<double> &var) { std::stringstream stream; if(var.real() && var.imag()) { stream << "(" << var.real(); if(var.imag() >= 0) stream << '+' << var.imag() << "i)"; else stream << var.imag() << "i)"; } else if(var.imag()) { if(var.imag() >= 0) stream << var.imag() << "i"; else stream << var.imag() << "i"; } else if(var.real()) stream << var.real(); return stream.str(); }std::string mult(const std::string &a, const std::string &b) { if(a.empty()) return a; else if(a == "1") return b; else return a + "*" + b; }std::string add(const std::string &a, const std::string &b) { if(a.empty()) { if(b.empty()) return "0"; else return b; } else if(b.empty()) return a; else if(b[0] == '-') return a+"-"+(b.c_str()+1); else return a + "+" + b; }int main() { std::complex<double> a, b, c; std::cout << "a*x^2+b*x+c\na="; std::cin >> a; std::cout << "b="; std::cin >> b; std::cout << "c="; std::cin >> c; std::complex<double> e=(0.5*b/a)+(0.5*std::sqrt(b*b-(4.0*a*c))/a), f=c/e; // (a*x+f)*(x+e) std::cout << "("<<add(mult(comp(a),"x"),comp(f))<<")*("<<add("x",comp(e))<<")" << std::endl; }Sample run:
a*x^2+b*x+c
a=2
b=13
c=15
(2*x+3)*(x+5)
[Edited by - smart_idiot on July 24, 2005 3:58:52 AM]
Author
Many thanks to you smart_idiot!
(i would rate++ you, but i've already rated you fully.)
Um, smart idiot, when i try
a = -2
b = -13
c = -15
It spits back an answer of
(-2X + 1.5)(X - 10)
Yet, for X = 17:
-2 * x ^ 2 - (13 * x) - 15 = -814
(-2 * x + 1.5) * (x - 10) = 227.5
From,
Nice coder
(i would rate++ you, but i've already rated you fully.)
Um, smart idiot, when i try
a = -2
b = -13
c = -15
It spits back an answer of
(-2X + 1.5)(X - 10)
Yet, for X = 17:
-2 * x ^ 2 - (13 * x) - 15 = -814
(-2 * x + 1.5) * (x - 10) = 227.5
From,
Nice coder
I kind of goofed up my math so double check what I wrote. I think it's correct, now. Hey, I had a disclaimer. [wink]
Author
ok, it works now.
Should it be producing whole numbers? (ie. shouldn't factorisation require all of the numbers to remain integers?)
From,
Nice coder
Should it be producing whole numbers? (ie. shouldn't factorisation require all of the numbers to remain integers?)
From,
Nice coder
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