## sneaky bug

Hello. I wrote a simple program that helps me simplify radicals:

Code:
#include <iostream>
#include <math.h>

typedef struct
{
unsigned int x, y;
}pair;

bool breakcube(unsigned int radi, unsigned short inx, pair& pr)
{
unsigned int cb = 0;
for (unsigned int i = 2; i < radi; i++)
{
cb=i;
for (unsigned short j = 1; j < inx; j++)
cb*=i;
for (unsigned int j = 1; j < radi; j++)
{
{
pr.x = cb;
pr.y = j;
return true;

}
}
}
return false;
}
int main()
{
unsigned int   cb   = 0;
unsigned short inx  = 0;
char     buff[64];
while(true)
{
std::cout << "\nEnter radicand ('quit' to terminate): ";
std::cin >> buff;
if (!strcmp(buff, "quit"))
return 0;
std::cout << "\nEnter index: ";
std::cin >> inx;
if ((double)pow(radi, 1.0/inx) == cb) // force the comparison
{
std::cout << "\nPerfect " << inx << " : ";
for (unsigned short i = 0; i < inx; i++)
{
std::cout << cb;
if (i!=inx-1)
std::cout << " * ";
}
std::cout << " = " << pow(cb, inx);
continue;
}
pair pr;
std::cout << "\n" << radi << " can be broken by: " << pr.x << " * " << pr.y;
else
std::cout << "\n" << radi << " cannot be broken";
}
return 0;
}
The problem happens when you try to break a perfect cube. The funny bit is that it works for some cases, and it doesn't work for others. For example, if you input 8 with an index of 3, the program works as expected, and says that it has found a perfect square. However if you enter 64 (4*4*4) it fails to recognize the perfect cube. I have ran this through the debugger, everything should work as expected...
 Recognitions: Homework Help Science Advisor I'm having a bit of trouble following the code, but you can check if a number is a cube with Code: bool isCube(int n, int& root) { root = (int)(0.5 + pow(n, 1.0/3)); // Nearest integer to the cube root return root * root * root == n; } and you can check for divisibility by perfect squares with Code: void simplify (int radicand, int& factor) { factor = 1; // Check for factors of 2; replace operations with shifts for better speed. while (radicand%4 == 0) { radicand /= 4; factor *= 2; } // Check for odd factors int limit = (int)(sqrt(radicand) + .0001); // Add small constant to avoid rounding the wrong way on perfect squares for (int n = 3; n <= limit; n += 2) { if (radicand % (n*n) == 0) { radicand /= n * n; factor *= n; limit = (int)(sqrt(radicand) + .0001); } } } Of course directly factoring the number would probably be faster, because then you could stop work on 33812897387 when you hit 18,670 = isqrt(348586571) instead of 183,882 = isqrt(33812897387). A quick change to the above code could make that improvement. Of course using a list of primes could speed things along as well.
 Recognitions: Gold Member A quick look over it makes me think you're having rounding/casting issues. Code: cb = (unsigned int)pow(radi, 1.0/inx); if ((double)pow(radi, 1.0/inx) == cb) // force the comparison If you cast a floating point value to an int, it doesn't round to the nearest number, it simply throws away everything after the decimal. This, coupled with the tendency for floating point values to be "close" (say, 3.99999999999 or 4.000000001 instead of 4.0), you're going to run into issues when doing the comparison the way that you are. Try looking at the round() function, or maybe rethink how you're doing your check. Hope that helps.

## sneaky bug

Thanks for the replies guys. CRGreathouse, the program helps break down radicals of any index, so explicitly checking for a cube root is not going to work here. The problem is, I have tried stepping through the code using a debugger, string outputs, rounding, pretty much everything I can think off, everything should work as expected, but it doesn't. Bill_B, if it was some rounding problem, wouldn't it present itself for other indexes too? For simplicity's sakes, here's the problem demonstrated in python:

Code:
#! /usr/bin/python

if fR == int(fR):
print str(iRad) + " is a cube"
else:
print str(iRad) + " is not a cube"

Recognitions:
Gold Member
 if it was some rounding problem, wouldn't it present itself for other indexes too?
It does. Try 4^6, for example. I'm sure there are others.

When I execute the program with an input of 64,3 - after this line -

Code:
 		cb = (unsigned int)pow(radi, 1.0/inx);
cb equals 3, not 4. (due to the truncation by the cast to int)

So when this line is executed -

Code:
  	if ((double)pow(radi, 1.0/inx) == cb) // force the comparison
you're comparing 4 == 3, so it doesn't see it as a cube.

I still stand by my original answer.

Recognitions:
Homework Help
 Quote by computerex Thanks for the replies guys. CRGreathouse, the program helps break down radicals of any index, so explicitly checking for a cube root is not going to work here.
That's what the rest of my post was for. The natural translation for nth roots instead of square roots:

Code:
int simplify (int radicand, int idx) {
int factor = 1;

// Check for factors of 2; replace operations with shifts for better speed.
while (radicand%pow(2, idx) == 0) {
factor *= 2;
}

// Check for odd factors
int limit = (int)(pow(radicand, 1.0/idx) + .0001); // Add small constant to avoid rounding the wrong way on perfect powers
for (int n = 3; n <= limit; n += 2) {
while (radicand % pow(n, idx) == 0) {
factor *= n;
limit = (int)(pow(radicand, 1.0/idx) + .0001);
}
}
return factor;
}
This assumes you have an integer power function. If not, code it:
Code:
int pow(int a, int b) {
int c = 1;
while (b > 1) {
if (b&1)
c *= a;
a *= a;
b >>= 1;
}
return c * a;
}
It's probably worthwhile, as before, to pull out any factors you find rather than looking only for divisibility by prime powers of large enough exponent. This way you stop the computation sooner and don't compute as many idx-th powers.
 Thanks for that. I am not accomplished at mathematical algorithms, so don't expect any good code coming from me. :)
 Here's the final program, in working condition. It might not be the fastest, but it helps nonetheless. CRGreathouse, your suggestions were great, but a bit over my head buddy. Bill_B, thank you as well. EDIT: Nope, false alarm. :( It certainly is better, it now recognizes 64 as a perfect cube, but it doesn't recognize 125...I am gloomy now. Code: #include #include typedef struct { unsigned int x, y; }pair; bool breakcube(unsigned int radi, unsigned short inx, pair& pr) { unsigned int cb = 0; for (unsigned int i = 2; i < radi; i++) { cb=i; for (unsigned short j = 1; j < inx; j++) cb*=i; for (unsigned int j = 1; j < radi; j++) { if (cb*j==radi) { pr.x = cb; pr.y = j; return true; } } } return false; } bool isPerfect(int rad, int idx) { double rt = pow(rad, 1.0/idx); rt = round(rt); if (pow(rt, idx)==rad) return true; return false; } int main() { unsigned int radi = 0; double cb = 0; unsigned short inx = 0; char buff[64]; while(true) { std::cout << "\nEnter radicand ('quit' to terminate): "; std::cin >> buff; if (!strcmp(buff, "quit")) return 0; radi=(unsigned int)atof(buff); std::cout << "\nEnter index: "; std::cin >> inx; cb = pow(radi, 1.0/inx); if (isPerfect(radi, inx)) { std::cout << "\nPerfect " << inx << " : "; for (unsigned short i = 0; i < inx; i++) { std::cout << cb; if (i!=inx-1) std::cout << " * "; } std::cout << " = " << pow(cb, inx); continue; } pair pr; if (breakcube(radi, inx, pr)) std::cout << "\n" << radi << " can be broken by: " << pr.x << " * " << pr.y; else std::cout << "\n" << radi << " cannot be broken"; } return 0; }

Recognitions:
Homework Help
 Quote by computerex Here's the final program, in working condition. It might not be the fastest, but it helps nonetheless. CRGreathouse, your suggestions were great, but a bit over my head buddy.
Your problem is almost certainly with rounding. At this point

Code:
bool isPerfect(int rad, int idx)
{
rt = round(rt);
return true;
return false;
}
output rad, rt, and pow(rt, idx) and see how they look. My best guess is that pow(rad, idx) is wrong since it's a floating-point number which may not match rad exactly; my second guess is that rt is rounded wrong.
 Ah! This seems to fix the problem! Code: bool isPerfect(int rad, int idx) { double rt = pow(rad, 1.0/idx); rt = round(rt); if (round(pow(rt, idx))==rad) return true; return false; } Thanks! Can someone confirm to make sure?

Recognitions:
Homework Help
 Quote by computerex Thanks! Can someone confirm to make sure?
It looks good, and tests good for all 32-bit values with index up to 10 million.