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42 is found to be the sum of three cubes

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Re: 42 is found to be the sum of three cubes

#161
post #152

Earlier quoted context omitted.

I'm not sure I understand your question, but we've posted countless times about this and are happy to answer about specific cases. You can find a lot of information via https://hn.algolia.com/?dateRange=all&page=0&prefix=true&que... and similar searches.

During the submission process where the form say a title is too long you could also be displaying how any transformations will affect it.

Yup, basically this. Another idea would be to add a "please review this title that Hacker News won't let me submit with" button that would get these kinds of things in front of you sooner.

Re: 42 is found to be the sum of three cubes

#162

For context, 42 was the only remaining number below 100 where it wasn’t known if this was possible. The general problem of exactly which numbers are the sum of three cubes is unsolved. https://en.m.wikipedia.org/wiki/Sums_of_three_cubes

Isn't the set of all triples of integers be enumerable, because (inductively) the set of all pairs of integers is enumerable (thanks, Cantor)? Then, if one could enumerate all triples of integers, one could, for each triple, calculate the sum of the cubes. So, integers which are sums of cubes are enumerable. That doesn't mean they're a known recursive set, just recursively enumerable. Am I missing something? Perhaps…

It is not computationally tractable to enumerate all triples of integers up to the size required to find the values for 42.

So certainly any integer N either is or is not the sum of three cubes. Deciding which is the case computationally can be intractable.

Each of the integers in the solution are ~53 bits in length. To enumerate all triples of integers (with plus and minus) requires testing around 2^(53+53+53+1+1+1) = 2^162 items. If you could test one per nanosecond, this is 10^32 years, vastly older than the universe. If you threw 1 trillion such computers at it, it's still 10^23 years, also vastly longer than the age of the universe.

Thus enumeration is simply not possible.

The point of this is they found a solution for 42 after great effort.

Re: 42 is found to be the sum of three cubes

#163
post #143

Earlier quoted context omitted.

Which source?

Sorry, looks like the link has been change to a tweet from http://math.mit.edu/~drew/ I meant the html source of Andrew's dept homepage titled "Life, the Universe, and Everything" at that moment.

Well, pretty close anyway, technically it's only one comment...

Re: 42 is found to be the sum of three cubes

#164

For context, 42 was the only remaining number below 100 where it wasn’t known if this was possible. The general problem of exactly which numbers are the sum of three cubes is unsolved. https://en.m.wikipedia.org/wiki/Sums_of_three_cubes

Isn't the set of all triples of integers be enumerable, because (inductively) the set of all pairs of integers is enumerable (thanks, Cantor)? Then, if one could enumerate all triples of integers, one could, for each triple, calculate the sum of the cubes. So, integers which are sums of cubes are enumerable. That doesn't mean they're a known recursive set, just recursively enumerable. Am I missing something? Perhaps…

Yes, you can use Cantor's diagonal argument to enumerate all triples. No, it does not answer the question whether every integer can be written as the sum of three cubes; it only would if the set of integers was finite.

Re: 42 is found to be the sum of three cubes

#166

For context, 42 was the only remaining number below 100 where it wasn’t known if this was possible. The general problem of exactly which numbers are the sum of three cubes is unsolved. https://en.m.wikipedia.org/wiki/Sums_of_three_cubes

Isn't the set of all triples of integers be enumerable, because (inductively) the set of all pairs of integers is enumerable (thanks, Cantor)? Then, if one could enumerate all triples of integers, one could, for each triple, calculate the sum of the cubes. So, integers which are sums of cubes are enumerable. That doesn't mean they're a known recursive set, just recursively enumerable. Am I missing something? Perhaps…

Many questions about the integers are recursively enumerable (like: what are the counterexamples to Fermat's last theorem?), but this doesn't really reveal anything about any deeper structure. Part of the conjecture is, indeed, that the set of numbers that are a sum of three cubes is recursive, which is given by a simple condition on the residue modulo 9 (see a sibling comment). However, mathematicians don't usually think about recursiveness of sets when they think about whether a problem has been "solved." It's more about whether some structure has been satisfyingly illuminated (maybe, to make something completely up, triples of numbers whose cubes sum to a particular number are found to be in one-to-one correspondence with "tripolar Hopfian unital schemes," and if something is known about them this might be satisfying). Of course, knowing whether or not a set has a recursive description is nice, too.

Here's something I've wondered about. In knot theory, a "knot" is a closed loop of string in 3d space, and if you allow the string to pass through itself it can obviously be unknotted (put into a form where it is a closed loop flat on a table). An unknotting is a sequence of pass-throughs to unknot it, and the unknotting number is the minimal number of pass-throughs among all unknottings. Unknotting sequences are actually recursively enumerable, but it's unknown whether there is a finite-time algorithm to compute unknotting number!

Re: 42 is found to be the sum of three cubes

#168

Earlier quoted context omitted.

Isn't the set of all triples of integers be enumerable, because (inductively) the set of all pairs of integers is enumerable (thanks, Cantor)? Then, if one could enumerate all triples of integers, one could, for each triple, calculate the sum of the cubes. So, integers which are sums of cubes are enumerable. That doesn't mean they're a known recursive set, just recursively enumerable. Am I missing something? Perhaps…

Yes, you can use Cantor's diagonal argument to enumerate all triples. No, it does not answer the question whether every integer can be written as the sum of three cubes; it only would if the set of integers was finite.

Diagonal argument? (That's how you show something is uncountable.)

It's that finite Cartesian products of countable sets are countable https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Se...

Re: 42 is found to be the sum of three cubes

#170
post #168

Earlier quoted context omitted.

Yes, you can use Cantor's diagonal argument to enumerate all triples. No, it does not answer the question whether every integer can be written as the sum of three cubes; it only would if the set of integers was finite.

Diagonal argument? (That's how you show something is uncountable.) It's that finite Cartesian products of countable sets are countable https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Se...

Oh, that's funny, thanks for pointing that out.

I had exactly this in mind (or the proof that the set of rational numbers are countable), but mistakenly thought it was called the 'diagonal argument' because you would get the bijection to the natural numbers by counting diagonally through the table.

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