Re: Triangle Numbers Revisited

by tall_man (Parson)
 on Oct 14, 2004 at 18:09 UTC ( #399287=note: print w/replies, xml ) Need Help??

in reply to Triangle Numbers Revisited

There is a theorem by Gauss which shows that finding M as the sum of three triangle numbers is equivalent to finding 8M+3 as the sum of three odd squares. That is, if
```8M + 3 = (2A+1)^2 + (2B+1)^2 + (2C+1)^2
then
M = trinum(A) + trinum(B) + trinum(C)
The computation starts with a bigger number, but the calculations might be easier this way. I'll try it when I get a chance.

Update: Here is code that takes advantage of a few things that are known about Diophantine quadratic problems. For example, it can convert multiples of two into smaller problems. It also screens out a few factors which would make a solution impossible. In many cases, it should be faster than the brute-force triangular number search.

```# Find a triangle-number decomposition by converting to
# a Diophantine squares problem.

use strict;

# These have no quadratic residue for -1.
# If they appear as factors in M an odd number of times
# a solution is impossible.
our @screeners = (3, 7, 11, 19, 23, 31);

# The brute-force search for a quadratic solution.
# We have reduced the number as much as we can before this
# point.
my \$M = shift;
my \$top = int(sqrt(\$M));
my \$last = int(\$top/2) + 1;

my (\$i, \$j, \$rem);
for (\$i = \$top; \$i >= \$last; --\$i) {
\$rem = \$M - \$i*\$i;
\$j = int(sqrt(\$rem));
if (\$j*\$j == \$rem) {
return (\$i,\$j);
}
}
}

# How many times does \$n go into \$M?
sub countpowers {
my (\$M, \$n) = @_;
my \$powercount = 0;
while (\$M % \$n == 0) {
\$M = \$M/\$n;
++\$powercount;
}
return (\$M, \$powercount);
}

# Reduce even numbers before solving by brute force.
# Also, screen out some impossible cases.
# Don't try a full factorization for now.
my \$M = shift;
my \$powercount;
my \$multfactor = 1;

# Screen out odd impossible cases, and factor out pairs of 4k-1 fac
+tors.
foreach my \$num (@screeners) {
(\$M, \$powercount) = countpowers(\$M, \$num);
if (\$powercount % 2 == 1) {
print "Eliminating impossible case with odd-power of 4k-1 fac
+tor \$num\n";
return;
}
# In even cases, half the factors can be multiplied into the res
+ult.
\$powercount = \$powercount/2;
for (my \$i = 0; \$i < \$powercount; \$i++) {
\$multfactor *= \$num;
}
}

# Count powers of 2.  We can handle the case of odd powers of 2.
(\$M, \$powercount) = countpowers(\$M, 2);

if (\$powercount % 2 == 0) {
# Two goes in an even number of times.
if (\$powercount > 0) {
\$multfactor *=  1 << ((\$powercount)/2);
}
return (\$i*\$multfactor, \$j*\$multfactor);
}

# Two goes in an odd number of times.  Use the i+j, i-j trick.
if (\$powercount > 1) {
\$multfactor *=  1 << ((\$powercount - 1)/2);
}
return ((\$i + \$j)*\$multfactor, (\$i - \$j)*\$multfactor);
}

my \$M;
\$M = (\$#ARGV >= 0)? \$ARGV[0] : 987654321;

# Convert from a triangle-number problem to a square-number problem.
my \$M2 = \$M*8 + 3;

# The top-level loop will try odd squares by brute force.
my \$top = int(sqrt(\$M2));
\$top = \$top - 1 if (\$top % 2 == 0);  # start with odd.
my \$last = int(\$top/2) + 1;
my \$k;
for (\$k = \$top; \$k >= \$last; \$k -= 2) {
my \$M3 = \$M2 - \$k*\$k;
if (\$i) {
# Convert solution back to triangle-numbers.
my \$new_i = (\$i - 1)/2;
my \$new_j = (\$j - 1)/2;
my \$new_k = (\$k - 1)/2;
print "Solution: triangle numbers \$new_i, \$new_j, \$new_k\n";
my \$tri_i = (\$new_i+\$new_i*\$new_i)/2;
my \$tri_j = (\$new_j+\$new_j*\$new_j)/2;
my \$tri_k = (\$new_k+\$new_k*\$new_k)/2;
my \$sum = \$tri_i + \$tri_j + \$tri_k;
print "Verifying \$tri_i + \$tri_j + \$tri_k = \$sum = \$M\n";
exit(0);
}
}

Replies are listed 'Best First'.
Re^2: Triangle Numbers Revisited
by tmoertel (Chaplain) on Oct 15, 2004 at 19:09 UTC
Quick question regarding screening out "a few factors which would make a solution impossible": Didn't Gauss prove Fermat's polygonal theorem for triangle numbers, showing that every positive integer can be represented as a sum of at most three triangle numbers? If so, how can there be cases for which a solution is impossible? (Or are you saying that the method you're using only works for some cases?)
I'm talking about an intermediate step. I'm looking for three odd squares that add to the number 8*M+3. I pick the first number, k, by brute force working down from the square root. So then I have to solve:
```  N = 8*M + 3 - k^2
i^2 + j^2 = N
There are choices for k that don't work. I want to eliminate them quickly and move on to the next k in the loop instead of spending time trying all combinations of i and j. Eventually I will find an answer, but that choice of k won't be part of it.

For example, if N is a multiple of an odd power of 3, the quadratic problem can't be solved in integers. So I can eliminate about 1/3 of the possible choices for k.

I hadn't known it, but you're right, Gauss proved that in his diary (which wasn't discovered until he had been dead 50 years).
Re^2: Triangle Numbers Revisited
by Limbic~Region (Chancellor) on Oct 15, 2004 at 22:27 UTC
tall_man,
Wow! It took a bit to make your code with my benchmark, but the results were quite impressive (25_000 random targets between 1 .. 9_999_999
```\$ time ./lr.pl

real    0m21.139s
user    0m20.279s
sys     0m0.040s

\$ time ./tm.pl

real    0m3.935s
user    0m3.585s
sys     0m0.300s

Cheers - L~R

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