20210424, 19:10  #23 
"Kebbaj Reda"
May 2018
Casablanca, Morocco
2×47 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 20210424 at 19:40 
20210425, 02:18  #24  
Jan 2020
1000_{2} Posts 
[QUOTE=Kebbaj;576761]
Quote:
Last fiddled with by Yusuf on 20210425 at 02:19 

20210425, 04:46  #25  
Oct 2017
11^{2} Posts 
[QUOTE=Yusuf;576792]
Quote:
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". Seems that when the puzzlemaster formulated the real task ("your goal...") he had forgotten the definitions he had made at the beginning of the challenge. "At each spin ... a number is eliminated from the wheel." For the challenge without *, k=number of spins = 7. 

20210504, 22:37  #26 
Jan 2017
2^{3}·3·5 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 = (q1) % 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(nturns+1, n+1))): s.add(calc(n, i, turns)) r = s.symmetric_difference(set(combinations(range(1, n+1), nturns))) if r: break print(n) print(len(r)) print(sorted(r)) 
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