Now I have completed the finding 23 sets of x y z values satisfy the condition x^3+y^3=1+z^3 & x
int setsFound = 0;
System.out.println("The first 23 sets ordered by increasing x.");
for (long x = 1; setsFound < 23; x++) {
for (long z = x + 1; z<x*x; z++) {
long y = (long) Math.pow((1 + z*z*z - x*x*x), 1f/3f);
if (x * x * x == 1 + z * z * z - y * y *y && x<y && y<z) {
setsFound++;
System.out.println("X: " + x + ", Y: " + y + ", Z: " + z);
}
}
}
But the code I have is very inefficient, ca开发者_运维技巧n anyone help me to fix this please?
Here is a working code:
int setsFound = 0;
System.out.println("The first 23 sets ordered by increasing x.");
for (long z = 1; setsFound < 23; z++) {
for (long y = z - 1; y > 0; y--) {
long x = (long) Math.pow((1 + z * z * z - y * y * y), 1f/3f);
if(y <= x) break;
if (x * x * x == 1 + z * z * z - y * y *y) {
setsFound++;
System.out.println("X: " + x + ", Y: " + y + ", Z: " + z);
}
}
}
Couple of problems in the old one:
- 1/3 == 0 (because it's integer division) //use 1f/3f
- x and z are swapped - you want z > x, not the other way around
- (long)Math.pow(4*4*4, 1.0/3) == (long)3.9999999999999996 == 3 // use round
If you start the other way, with X < Y < Z by incrementing Y and Z up to a limit, you can gain some efficiencies. Once Z^3 > X^3 + Y^3 + 1, you can skip to the next Y value due to the concavity of the cubic function.
This implementation in C# works pretty fast on a laptop:
UInt64 setsFound = 0;
UInt64 xlim = 10000;
UInt64 ylim = 1000000;
UInt64 zlim = 10000000;
//int ctr = 0;
Console.WriteLine("The first 23 sets ordered by increasing x.");
Parallel.For(1, (long)xlim, new ParallelOptions { MaxDegreeOfParallelism = 4 }, i =>
//for (UInt64 i = 0; i < xlim; i++)
{
UInt64 x = (UInt64)i;
UInt64 xCu = x * x * x;
int zFails = 0;
for (UInt64 y = x + 1; y < ylim; y++)
{
UInt64 yCu = y * y * y;
zFails = 0;
for (UInt64 z = y + 1; z < zlim & zFails < 1; z++)
{
UInt64 zCu = z * z * z;
if (xCu + yCu - zCu == 1)
{
Console.WriteLine(String.Format("{0}: {1}^3 + {2}^3 - {3}^3 = 1", setsFound, x, y, z));
setsFound++;
}
else if (zCu > xCu + yCu - 1)
{
zFails++;
}
}
}
}
);
Obviously you can take out the parallelization. Also, here are the first 19 elements in that set (computer is still running, I'll try to post the last 4 later):
output http://desmond.yfrog.com/Himg640/scaled.php?tn=0&server=640&filename=8qzi.png&xsize=640&ysize=640
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