Challenge: Add two numbers. In O(n^m) time, where n and m are the two numbers given. Sleeping, or your language equivalent, is not allowed.

Input: Two integers, separated by a space.

Output: The sum of the two integers.

Additional notes: The exact timing doesn't matter, as long as the input 1000 2 takes roughly 1/1000th the time as the input 1000 3


closed as unclear what you're asking by xnor, Luis Mendo, Jonathan Frech, Stewie Griffin, caird coinheringaahing Sep 26 at 7:29

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  • 3
    \$\begingroup\$ O(n^m) time means at most n^m asymptotically. I think you want theta. \$\endgroup\$ – xnor Sep 26 at 6:26
  • 5
    \$\begingroup\$ I presume you want people to use some kind of super slow algorithm, but there's nothing stopping them from doing simple addition followed by an n^m loop of NOPs to pad the runtime. I don't think there's a way around that (we do not allow unobservable requirements) but just a warning that submissions may be less interesting than you envisioned. \$\endgroup\$ – Sanchises Sep 26 at 6:30
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    \$\begingroup\$ Is it OK if a program takes n^m time to complete after printing the sum, or would it have to take the time before printing? \$\endgroup\$ – xnor Sep 26 at 6:35
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    \$\begingroup\$ Why the awkward input requirement. Can I really not just take two arguments? \$\endgroup\$ – Adám Sep 26 at 7:04
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    \$\begingroup\$ Maybe try the sandbox next time to get feedback on your challenge before you post it. And why not try to answer some challenges yourself! Then you get a good feel for what works. \$\endgroup\$ – Sanchises Sep 26 at 9:25

APL (Dyalog Unicode), 4 bytesSBCS

Anonymous tacit infix function:


Try it online!

However, if it really needs to take the two arguments together, separated by a space, we need the following anonymous tacit prefix function:

APL (Dyalog Unicode), 6 bytesSBCS


Try it online!

+ or +/ sum the numbers

 cyclically reshape that to the following length:

* or */ one number raised to the power of the other

 pick the first element of that


Python 3, 34 bytes

lambda n,m:sum(range(n**m))and m+n

Try it online!

Alternatively, as suggested by user202729:

Python 2, 26 bytes

lambda n,m:n**m*[1]and n+m

Try it online!

which also works with Python 3.

  • 1
    \$\begingroup\$ Why sum()? Isn't range(n**m) already \$O(n^m)\$? \$\endgroup\$ – Adám Sep 26 at 8:40
  • \$\begingroup\$ @Adám Yes in Python 2, but not in Python 3 where range doesn't actually create the list: TIO \$\endgroup\$ – xnor Sep 26 at 10:44
  • \$\begingroup\$ What about [1]*? \$\endgroup\$ – user202729 Sep 26 at 11:05
  • \$\begingroup\$ @user202729 : I'd guess you're right. \$\endgroup\$ – Chas Brown Sep 27 at 5:31

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