3 ms·

> 1 / (sqrt(2) (sqrt(3) - 1)) I don't mean in any way to demean their research, but I think your exhaustive search could build up formulas in exactly that form

by Pyxl101 11y ago

> 1 / (sqrt(2) (sqrt(3) - 1)) I don't mean in any way to demean their research, but I think your exhaustive search could build up formulas in exactly that form and iterate on them. It starts with a valid pentagon. Then it permutes that pentagon with an evolutionary method by modifying the formula. For example, maybe the lengths it tries are: 1 / 2 1 / sqrt(2) 1 / sqrt(2) * 3 1 / sqrt(2) - 1 1 / sqrt(2) * (3) 1 / sqrt(2) * sqrt(3) It would try both these and many others along the way. 1 / exp(2) ... 2 / sqrt(2) ... Again, not to trivialize, but there are only so formulas made up of a fixed number of terms and operators, and as long as it's easy to check whether a shape is a valid pentagon, and whether it tiles, then I think you could check a considerable number of them. It looks like a number of the pentagon formulas have a few sides with complex lengths, while the others are simple or equal to each other, so you could bias the algorithm to search for those. I'm sure there's way more complexity I'm overlooking, but that's how one might get started.

ntoronto 11y ago
"I'm sure there's way more complexity I'm overlooking, but that's how one might get started." That's how I'd start. There's probably a way to make the search a lot smarter. Here's the part you're overlooking, though: "... there are only so [many] formulas made up of a fixed number of terms and operators..." "So many" = "countably infinite." Paring it down to finitely many would require understanding tantamount to having solved the problem in the first place. [Edit: I can words.]
rtpg 11y ago
This could work for approximations. But if you want the exact value, this might not be super effective, as there are values without closed forms (that is to say, that are not expressible simply with the *-+/ and sqrt/log and other usual operations). So your iteration procedure will probably get to a good approximation, but might not find the "core" expression.
gizmo686 11y ago
In this case, (with the hindsight of knowing the answer), it would probably be easier to enumerate the angles, as they were all rational multiples of pi (with the biggest denominator being 12). Still, that gives a naive 5^144 combinations of angles to test. If you reduce this to proper, reduced fractions this comes down to 5^47 [1]. You can probably improve on this further by requiring that the angles form a (convex) polygon, but I suspect that still leaves you with to many to be practical. [1] http://oeis.org/A005728 http://oeis.org/A005728

by Pyxl101 11y ago

ntoronto 11y ago

"I'm sure there's way more complexity I'm overlooking, but that's how one might get started." That's how I'd start. There's probably a way to make the search a lot smarter. Here's the part you're overlooking, though: "... there are only so [many] formulas made up of a fixed number of terms and operators..." "So many" = "countably infinite." Paring it down to finitely many would require understanding tantamount to having solved the problem in the first place. [Edit: I can words.]

rtpg 11y ago

This could work for approximations. But if you want the exact value, this might not be super effective, as there are values without closed forms (that is to say, that are not expressible simply with the *-+/ and sqrt/log and other usual operations). So your iteration procedure will probably get to a good approximation, but might not find the "core" expression.

gizmo686 11y ago

In this case, (with the hindsight of knowing the answer), it would probably be easier to enumerate the angles, as they were all rational multiples of pi (with the biggest denominator being 12). Still, that gives a naive 5^144 combinations of angles to test. If you reduce this to proper, reduced fractions this comes down to 5^47 [1]. You can probably improve on this further by requiring that the angles form a (convex) polygon, but I suspect that still leaves you with to many to be practical. [1] http://oeis.org/A005728 http://oeis.org/A005728