Complex nails clear solution for Ulam-Warburton Automaton Hex by StefanPochmann - python coding challenges

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Complex nails solution in Clear category for Ulam-Warburton Automaton Hex by StefanPochmann

from collections import Counter

def automaton(step: int) -> int:
    Z = {0}
    for _ in range(step - 1):
        c = Counter(
            z + d
            for z in Z
            for d in (1, -1, 1j, -1j, 1-1j, 1j-1)
        )
        Z |= {z for z in c if c[z] == 1}
    return len(Z)

March 15, 2023

Comments:

StefanPochmann on March 15, 2023, 1:13 p.m.
If you have a hammer ...

StefanPochmann on March 19, 2023, 4:13 p.m.
Alternatively: ``` from collections import Counter def automaton(step: int) -> int: Z = {0} for _ in range(step - 1): c = Counter( z + round(d.real) + round(d.imag / 3*.5) * 1j for z in Z for i in range(6) for d in [2 * 1j * (i*2/3)] ) Z |= {z for z in c if c[z] == 1} return len(Z) ``` The deltas (what I add to `z`) could be precomputed once, but meh... I like my original solution better anyway.

StefanPochmann on March 20, 2023, 11:21 p.m.
Another alternative: ```python from collections import Counter def automaton(step: int) -> int: Z = {0} for _ in range(step - 1): c = Counter( z + eval(f'{21j(i2/3):.1}') for z in Z for i in range(6) ) Z |= {z for z in c if c[z] == 1} return len(Z) ``` My deltas, as computed by `[eval(f'{21j(i2/3):.1}') for i in range(6)]`, are: `[(2+0j), (1+2j), (-1+2j), (-2+0j), (-1-2j), (1-2j)]`

CDG.Axel on March 21, 2023, 1:54 p.m.
Great solution with counter! Three lines idea... It's like <a href="https://py.checkio.org/mission/ulam-warburton-automaton-hex/publications/CDG.Axel/python-3/3-lines-proc/">my solution </a> without Counter ) <pre class='brush: python'>Z, D = {0}, (1, -1, 1j, -1j, 1-1j, 1j-1) [Z.update({z for z, c in Counter(z+d for z in Z for d in D).items() if c == 1}) for _ in range(step-1)] return len(Z) </pre>

StefanPochmann on March 21, 2023, 2:12 p.m.
Meh, comprehension abuse... no good.

CDG.Axel on March 21, 2023, 2:21 p.m.
Most of code-compacting tricks aren't good). They are in a row with one-letter variables, nested comprehensions, space abusing, etc, etc.

StefanPochmann on March 21, 2023, 2:27 p.m.
Not sure what you mean with space abusing, but one-letter variables can be optimal, and nested comprehensions are fine. Such comprehension abuse on the other hand is never good (except perhaps in code golf, but you weren't doing that). Building a list if you don't want it is just wrong.

CDG.Axel on March 21, 2023, 2:04 p.m.
Another idea: <pre class='brush: python'>from collections import Counter from itertools import product as prod def automaton(step: int) -> int: Z, D, R = {0}, (1, -1, 1j, -1j, 1-1j, 1j-1), range(step - 1) [Z.update({z for z, c in Counter(map(sum, prod(Z, D))).items() if c == 1}) for _ in R] return len(Z) </pre>

List, list and list again

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