Gaussian integer is a complex number in the form \$x+yi\$, where \$x,y\$ are integer and \$i^2=-1\$. The task is to perform such operation for Gaussian integers \$a,b\$, that \$a=q \cdot b+r\$ and \$|r|<|b|\$ (\$q,r\$ are Gaussian integers, \$|z|\$ is defined as \$\sqrt{a^2+b^2}\$ for \$a+bi=z\$).
Need to output only \$r\$.
One of possible approaches is here.

Scoring: the fastest in terms of O-notations algorithm wins, though it can consume more time. By fastest I mean \$O(\left(f(n)\right)^a)\$ is preferred for smaller \$a\$ and \$O(\log^a(\log(n)))\$ will be preferred over \$O(\log^b(n))\$, ...over \$O(n^k)\$, ...over \$O(e^{\alpha n})\$ and so on.
If there's a tie, fastest implementation wins in terms of benchmark (e.g. 100 000 random 32-bit \$b\$ and 64-bit \$a\$ (I mean both real and imaginary part to be 32 or 64 signed integer)) or there can be 2-3 winners.
If unsure about algorithm complexity, leave blank. E.g. I believe Euler's algorithm for integers has \$O(\log(n))\$ time complexity.

Test cases can be obtained from wolframalpha in the form like Mod[151 + 62 I, 2 + I], but in case there are some:

4 mod (3+3*i) = either of 1+3*i, -2-0*i, 4-0*i, 1-3*i
10000 mod 30*i = either 10 or -20
(3140462541696034439-9109410646169016036*i) mod (-81008166-205207311*i) =
    either -75816304-29775984*i or 129391007+51232182*i

\$|b|\$ will not exceed 32 bits, but you don't have to code long arithmetic if your selected language doesn't have built-in one.

Good luck. Of course, WolframLanguage is forbidden. )

  • \$\begingroup\$ It looks to me like the algorithm you link is also log(n) time, which is optimal. You might as well just have the tiebreak be the scoring mechanism. \$\endgroup\$ – xnor Nov 20 at 1:11
  • \$\begingroup\$ I believe there can exist O(1) algorithms when dealing with entire int32 or int64 as a whole. ) \$\endgroup\$ – Alexey Burdin Nov 20 at 1:15
  • 4
    \$\begingroup\$ OK, sure, it's O(1) under the word size model. But formally, any algorithm on a bounded domain of inputs is O(1). I think the whole big-O scoring bit is unnecessary for the question and makes it a chameleon challenge because the "tiebreak" is the real scoring method. \$\endgroup\$ – xnor Nov 20 at 1:23
  • 3
    \$\begingroup\$ Does 32-bit b mean the real and imaginary parts are each 16 bits or each 32 bits? Could you please add some test cases generated this way into the question? Also, it's confusing that a and b are at first used to denote the real and imaginary parts of a complex number, but elsewhere two different complex numbers. \$\endgroup\$ – xnor Nov 20 at 1:26
  • 1
    \$\begingroup\$ Why does Wolfram language have to be forbidden? Are its derivatives? \$\endgroup\$ – my pronoun is monicareinstate Nov 20 at 10:56

Python 2 (PyPy), O(1) operations

def cmod(a1, a2, b1, b2):
	# Compute p = a * conj(b)
	p1 =   a1 * b1 + a2 * b2
	p2 = - a1 * b2 + a2 * b1

	# Compute the norm-squared of b
	b_nsq = b1 ** 2 + b2 ** 2
	b_nsq_half = b_nsq // 2

	# Do a symmetric mod of d1 and d2 by b_nsq
	# That is, into the interval [-b_nsq_half, +b_nsq_half)
	d1 = (p1 + b_nsq_half) % b_nsq - b_nsq_half
	d2 = (p2 + b_nsq_half) % b_nsq - b_nsq_half

	# Compute r = d / b
	r1 = (d1 * b1 - d2 * b2) // b_nsq 
	r2 = (d1 * b2 + d2 * b1) // b_nsq 

	return r1, r2

Try it online!

Uses pure integer arithmetic, with no loops or branching or built-in complex numbers. This should easily complete the benchmark of 100,000 random inputs within a small fraction of a second. Porting this to a faster language would be straightforward and likely improve performance.

One possible low-level optimization is the Karatsuba algorithm trick to do complex multiplication using 3 rather than 4 integer multiplications. But, this might not be worth it for such "small" values.

  • \$\begingroup\$ Sadly the competition ends. ) I should do the test cases... \$\endgroup\$ – Alexey Burdin Nov 20 at 3:06

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