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I need to find any one of all the possible square quadruples that sum to a given number. The number can be as large as $10^{12}$. I have tried a $o(n)$ complexity algorithm but that is not sufficient to calculate within the time limit.

Cettt
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bunny
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    "Within the time limit"...? Is this a contest of some sort? If so, is asking for help like this acceptable under the rules? – Andrew D. Hwang Jan 10 '17 at 13:47
  • It depends on what you want. If you want to find all the 4-square representations of an integer N, then you may not get the efficiency. If you want just one representation, then that can be done very efficiently. To give you an idea, run this online 4-square calculator and increase the number N and see what happens. http://www.mathcelebrity.com/foursquare.php?num=+178&pl=Show+Lagrange+Four-Square+Notation – user25406 Jan 10 '17 at 18:56
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    This may be related to the current CodeChef challenge. See here: If you can follow the links given there, you may be able to check how close a match this is. A bound such as $10^{12}$ is often a tell-tale sign. If you agree with my assessment, upvote this comment, and vote to close the question. – Jyrki Lahtonen Jan 10 '17 at 22:04
  • I removed two comments for now. The comments were undoubtedly posted in good faith, but there is a chance that they amount to aiding somebody win a monetary prize by cheating. – Jyrki Lahtonen Jan 10 '17 at 22:08
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    @Jyrki I know you were having trouble on the other thread viewing the challenge; I don't know if you figured it out or not, but here is a (phone screenshot!) challenge snippet; indeed $10^{12}$ is present. – pjs36 Jan 11 '17 at 03:13
  • Possible duplicate of Express Integer as Sum of Four Squares – bunny Jan 11 '17 at 06:45

1 Answers1

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For some weird reason, I cannot make comments due to low reputation, so I'm forced to write this comment as answer.

You need to make clear if you are searching integer solutions. Otherwise, your solution is trivial $n=x^2+x^2+x^2+x^2 \to x=\sqrt{\frac{n}{4}}$

Anyways, if you are searching integer solutions, should all be different numbers, or can they be repeated?

Is 0 allowed as solution?

If only integers larger than 0 without repetition are allowed, the worst algorithm cannot be larger than $o(\sqrt{n})$

Do you need a fast algorithm on average, or for every run? A greedy algorithm running in parallel would reduce your average time.

This is a greedy algorithm:

$n=a^2+b^2+c^2+d^2$

Assume $a\geq b\geq c\geq d \geq 1$

The largest possible values for a, b, c, d are:

$a= truncate(\sqrt{n-3})=\lfloor \sqrt{n-3} \rfloor$

$b=\lfloor \sqrt{n-a^2-2} \rfloor$

$c=\lfloor \sqrt{n-a^2-b^2-1} \rfloor$

$d=\lfloor \sqrt{n-a^2-b^2-c^2} \rfloor$

It is either the solution, or you get a remaining r:

$r=n-(a^2+b^2+c^2+d^2)$

If r>0 then you have up to 4 integer errors $e_i \in \mathbb{Z}$

$n=(a+e_a)^2+(b+e_b)^2+(c+e_c)^2+(d+e_d)^2$

$r=e_a^2+e_b^2+e_c^2+e_d^2+2.(a.e_a+b.e_b+c.e_c+d.e_d)$

But because n is large, $e_a<<e_b<<e_c<<e_d$, with a high probability of $e_a=0$, $e_b=0$, so assume $e_a=e_b=e_c=0$ and solve for $r=e_d^2+2.(d.e_d)$

if $e_d$ is not an integer, solve for

$e_a=e_b=0$

$r=e_c^2+e_d^2+2.(c.e_c+d.e_d)$

$c+e_c \geq d+e_d$ (because c>d by definition)

$e_c<0$

$e_d>0$

At this point, you will most probably have found a solution. Otherwise assume that only $e_a=0$, and solve the other errors.

In the worst case, $e_a \neq 0$, but at least you know that $e_a<0$, and $a+e_a \geq b+e_b \geq c+e_c \geq d+e_d$

cezudidu
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