Computing GCDs over the Gaussian Integers: Difference between revisions
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I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee. |
I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee or proof of correctness. One can sketchily argue something like this however: |
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''Claim'': <math>\mathbb{Z}[i]</math> is a Euclidean domain, and hence a PID, and hence a UFD.<br /> |
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Consider the norm function <math>\nu(a + bi) = a^2 + b^2</math>. We write the standard norm on the complex plane as <math>||a+bi||_{\mathbb{C}} = \sqrt{a^2 + b^2}</math>. We show that for all <math>x \neq 0</math> and <math>y</math> we have, <math> y = qx + r </math> where <math>\nu(r) = 0</math> or <math>\nu(r) < \nu(x)</math>. Consider <math>x</math> and <math>y</math> as points on the complex plane. Since <math>x \neq 0</math> we have that <math>y/x</math> is another point in the complex plane. Consider <math>\mathbb{Z}[i]</math> as the points with integer coordinates in <math>\mathbb{C}</math>. For any point <math>\zeta \in \mathbb{C}</math> we can find a <math>\zeta' \in \mathbb{Z}[i]</math> such that <math>||\zeta - \zeta'||_{\mathbb{C}} \leq \sqrt{2}/2</math> by taking the <math>\zeta'</math> to be the nearest lattice point to <math>\zeta</math>. Since any point on the plane is contained in a square whose vertices are lattice points and whose side lengths are one, there must be a lattice point nearer than half the diagonal of the square. We then take <math>z</math> to be the nearest lattice point to <math>y/x</math>. We then have <math>y = zx + (y - zx)</math>. We compute the norm of <math>\nu(y - zx)</math>. We have: <math> \sqrt{\nu(y - zx)} = ||y - zx||_{\mathbb{C}} = ||x||_{\mathbb{C}} \cdot ||y/x - z||_{\mathbb{C}} \leq ||x||_{\mathbb{C}} \frac{\sqrt{2}}{2} < ||x||_{\mathbb{C}} = \sqrt{\nu(x)} </math> It follows that <math>\nu(y - zy) < \nu(x)</math>. We obtain that <math>\mathbb{Z}[i]</math> is a Euclidean domain. Thus <math>\mathbb{Z}[i]</math> is a PID and hence a UFD by results given in class. |
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<pre> |
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#!/usr/bin/perl |
#!/usr/bin/perl |
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use Math::Complex; |
use Math::Complex; |
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&gcd($z1, $z2); |
&gcd($z1, $z2); |
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</pre> |
Latest revision as of 14:56, 26 November 2011
I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee or proof of correctness. One can sketchily argue something like this however:
Claim: is a Euclidean domain, and hence a PID, and hence a UFD.
Consider the norm function . We write the standard norm on the complex plane as . We show that for all and we have, where or . Consider and as points on the complex plane. Since we have that is another point in the complex plane. Consider as the points with integer coordinates in . For any point we can find a such that by taking the to be the nearest lattice point to . Since any point on the plane is contained in a square whose vertices are lattice points and whose side lengths are one, there must be a lattice point nearer than half the diagonal of the square. We then take to be the nearest lattice point to . We then have . We compute the norm of . We have: It follows that . We obtain that is a Euclidean domain. Thus is a PID and hence a UFD by results given in class.
#!/usr/bin/perl use Math::Complex; ## A Quick hack for computing GCDs of Gaussian integers. $z2 = 857 + i; $z1 = 255; sub gcd { # the Euclidean algorithm my $x = $_[0]; my $y = $_[1]; if ($x * $y == 0) { print "Done!\n"; } else { $q = &approx($x/$y); $r = $x - $q*$y; print "($x) = ($q)($y) + ($r)\n"; &gcd($y,$r); } } sub approx { # find the nearest Gaussian integer to a point on the complex plane my $z = $_[0]; my $x = int(Re($z)); my $y = int(Im($z)); if (abs($z - (($x+1) + i*$y) ) < 1/sqrt(2)) { return ($x+1) + i*$y; } elsif (abs($z - (($x) + i*($y+1)) ) < 1/sqrt(2)) { return $x + i*($y+1); } elsif (abs($z - (($x+1) + i*($y+1)) ) < 1/sqrt(2)) { return ($x+1) + i*($y+1); } else { return $x + i*$y; } } &gcd($z1, $z2);