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 2019-08-02, 11:16 #1 Xyzzy     "Mike" Aug 2002 1E1E16 Posts August 2019
 2019-08-03, 19:55 #2 Kebbaj     "Kebbaj Reda" May 2018 Casablanca, Morocco 5·11 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?
 2019-08-04, 05:19 #3 LaurV Romulan Interpreter     Jun 2011 Thailand 223716 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 "AB>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 2019-08-04 at 05:24
 2019-08-05, 12:44 #4 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 2×2,861 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?
 2019-08-05, 17:25 #5 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 2×13×73 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.
2019-08-05, 18:11   #6
Kebbaj

"Kebbaj Reda"
May 2018
Casablanca, Morocco

3716 Posts

Quote:
 Originally Posted by a1call 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.

In two months you will understand Rachid.
I'm joking Rachid.
I do not understand too !!
We're getting old.

2019-08-06, 07:56   #7
Dieter

Oct 2017

1318 Posts

Quote:
 Originally Posted by Kebbaj In two months you will understand Rachid. I'm joking Rachid. I do not understand too !! We're getting old.
Sometimes it‘s helpful to hear the same in other words:
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.

2019-08-06, 08:31   #8
Kebbaj

"Kebbaj Reda"
May 2018
Casablanca, Morocco

5×11 Posts

Quote:
 Originally Posted by Dieter Sometimes it‘s helpful to hear the same in other words: There are 9! possibilities to write abcdefghi ... ihgfedcba. We can interprete the first permutation as aC, 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. ab is possible. The permutations Ab (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

Thank you dieter,

"The weight carried by a group is a feather".

 2019-08-06, 10:46 #9 Bolero   Aug 2019 Marseille 48 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
2019-08-06, 14:07   #10
Dieter

Oct 2017

89 Posts

Quote:
 Originally Posted by Bolero 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
... and so it's no solution, because we are searching for restrictions for which the pairs can NOT be covered by one or two or three permutations.

2019-08-06, 14:43   #11
Bolero

Aug 2019
Marseille

1002 Posts

Quote:
 Originally Posted by Dieter ... and so it's no solution, because we are searching for restrictions for which the pairs can NOT be covered by one or two or three permutations.
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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