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 2021-04-24, 19:10 #23 Kebbaj     "Kebbaj Reda" May 2018 Casablanca, Morocco 1428 Posts [QUOTE=Yusuf;] The amount of steps and the amount of spins both equal to 7? No. The amount of steps is = 7 . The amount of spins is a variable. Last fiddled with by Kebbaj on 2021-04-24 at 19:40
2021-04-25, 02:18   #24
Yusuf

Jan 2020

11 Posts

[QUOTE=Kebbaj;576761]
Quote:
 Originally Posted by Yusuf; The amount of steps and the amount of spins both equal to 7? No. The amount of steps is = 7 . The amount of spins is a variable.
But it says "the set is of size n-7", wouldn't that mean that there are 7 spins since there are 7 numbers being taken away from the wheel?

Last fiddled with by Yusuf on 2021-04-25 at 02:19

2021-04-25, 04:46   #25
Dieter

Oct 2017

12410 Posts

[QUOTE=Yusuf;576792]
Quote:
 Originally Posted by Kebbaj But it says "the set is of size n-7", wouldn't that mean that there are 7 spins since there are 7 numbers being taken away from the wheel?
Yes. I'm sure that Kebbaj wanted to say this.

I repeat: Look at #2! That was my first comment to this challenge.

Or look at #8 (uau): "Those refer to different things with "step".

"At each spin ... a number is eliminated from the wheel."

For the challenge without *, k=number of spins = 7.

 2021-05-04, 22:37 #26 uau   Jan 2017 2×3×23 Posts Here's my solution. Just pretty straightforward brute force - enough for the sizes asked for in the question. The program takes the size as input, so calling it with an argument of "7" gives the basic solution, "8" bonus one. Code: #!/usr/bin/python3 from itertools import combinations from math import lcm import sys def calc(n, q, s): a = list(range(1, n+1)) for _ in range(s): w = (q-1) % len(a) a = a[w+1:] + a[:w] a.sort() return tuple(a) turns = int(sys.argv[1]) for n in range(turns, 100): print("Trying", n) s = set() for i in range(lcm(*range(n-turns+1, n+1))): s.add(calc(n, i, turns)) r = s.symmetric_difference(set(combinations(range(1, n+1), n-turns))) if r: break print(n) print(len(r)) print(sorted(r))

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