20190802, 11:16  #1 
Aug 2002
8451_{10} Posts 
August 2019

20190803, 19:55  #2 
"Kebbaj Reda"
May 2018
Casablanca, Morocco
2×7^{2} Posts 
Hello everybody.
Understanding the question is always the first difficulty of challange. B> C Restriction. Are we talking about variable or alphabetic chain to sort? 
20190804, 05:19  #3 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2·17·293 Posts 
That kind of means "ordered pair", or "totally ordered set", i.e. all the permutations in which B is after C are not accepted. Or viceversa, WLOG. You can read the operator as "after" or "before", or "ascents" or "descents", "B is after (or before) C", etc, or even completely eliminate the operator. I learned (most of) my (related) math without it, and only occasionally seen it in combinatorics or so, I mean you can just write "ABCD" instead of "A<B<C<D", and it is still clear in the context of total ordered sets. The dilemma (my dilemma) of "after" versus "before" comes from the fact that many authors use one or the other, with the meaning defined locally in their own texts. I had two professors in the uni and one of them used to write "A<B<C" and the other "A>B>C" for the same thing, the ordered triplet (A, B, C), or ABC, which was an eternal resource of fun and confusion. If you think deeper, both notations make sense, haha...
Last fiddled with by LaurV on 20190804 at 05:24 
20190805, 12:44  #4 
Just call me Henry
"David"
Sep 2007
Liverpool (GMT/BST)
2^{5}×11×17 Posts 
I am struggling to come up with a small example that requires 3 permutations never mind 4 or more (4 may be easier as 2*2=4). Is this puzzle solvable with pen and paper?

20190805, 17:25  #5 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·19·59 Posts 
I read this puzzle again everyday and everyday I understand the challenge a bit more than the day before.
At the current rate I should completely understand it in a couple of months, Max. 
20190805, 18:11  #6  
"Kebbaj Reda"
May 2018
Casablanca, Morocco
2·7^{2} Posts 
Quote:
In two months you will understand Rachid. I'm joking Rachid. I do not understand too !! We're getting old. 

20190806, 07:56  #7  
Oct 2017
174_{8} Posts 
Quote:
There are 9! possibilities to write abcdefghi ... ihgfedcba. We can interprete the first permutation as a<b<c...<i. Here is c<g and in the last permutation is g<c. Counting only the permutations fulfilling B>C, C>D, and E>G (b has to be right of c and so on) we have only 30,240 permutations. There are 36 pairs ab ac ... ai bc bd...bi...hi. In our exemple we have given relations between b,c and c,d and b,d and e,g. The other 32 relations aren‘t fixed, f.e. a<b and a>b is possible. The permutations A<D<C<B<F<G<E<H<I and I<H<G<E<F<D<C<B<A are two of the 30240 permutations fulfilling the fixed conditions. Furthermore they contain („cover“) all 2*32 undefined relations, f.e. a<b (the first) and a>b (the second). We have to search for conditions (restrictions) for which it isn‘t possible to find three permutations covering the fixed and the undetermined pairs. They have to be free of contradictions (feasible). This sentence has been additioned (the challenge has been changed at least one time). I suppose that someone has sent a<b and b<c and c<a to the puzzlemaster. 

20190806, 08:31  #8  
"Kebbaj Reda"
May 2018
Casablanca, Morocco
62_{16} Posts 
Quote:
Your explanations are very clear. Thank you dieter, "The weight carried by a group is a feather". 

20190806, 10:46  #9 
Aug 2019
Marseille
5 Posts 
I still quite don't get it as from what I understand it shouldn't be an easy problem. Maybe I'm missing something here as I can't see why the simple solution to enforce a specific permutation shouldn't work here. For example
the restrictions A<B, B<C, C<D, D<E, E<F, F<G, G<H, H<I are fulfilled by just one permutation A<B<C<D<E<F<G<H<I 
20190806, 14:07  #10  
Oct 2017
2^{2}·31 Posts 
Quote:


20190806, 14:43  #11 
Aug 2019
Marseille
5 Posts 
ok, so "no three permutations" means "more than three permutations"? (sorry, I'm not a native speaker and the problem is very strangely worded for me)

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