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I have tried to solve this sequence which, I suppose has been created from the combination of a,b,c,d,e. I've reposted the correct sequence

The sequence:

(a,c), ?, ?, ?, (a,e), ?, ?, (c,d), ?, (b,a)

My logic: We just try to do all the possible combinations :

(A,B) (A,C) (A,D) (A,E)

(B,A) (B,C) (B,D) (B,E)

(C,A) (C,B) (C,D) (C,E)

(D,A) (D,B) (D,C) (D,E)

(E,A) (E,B) (E,C) (E,D)

OR:

(A,A) (A,B) (A,C) (A,D) (A,E)

(B,A) (B,B) (B,C) (B,D) (B,E)

(C,A) (C,B) (C,C) (C,D) (C,E)

(D,A) (D,B) (D,C) (D,D) (D,E)

(E,A) (E,B) (E,C) (E,D) (E,E)

I suppose we should find the criteria with which my friend selected the items made of 2 letters.

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  • $\begingroup$ Are any of (E,A) (E,B) (E,C) (E,D) (E,E) allowed? $\endgroup$ – Weather Vane Apr 21 at 12:58
  • $\begingroup$ @WeatherVane they could be allowed, or atleast i think they are allowed becasue there's letter e $\endgroup$ – ade Apr 21 at 13:03
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My thoughts here are ...

(a,c) (c,e) (e,d) (d,a) (a,e) (e,b) (b,c) (c,d) (d,b) (b,a)

Why this sequence?

The pairs I've listed connect each point in A B C D E to every other point exactly once.

Huh?

Put the letters A to E evenly spaced around a circle.
Each pair of letters in the sequence can be taken as one leg of a journey amongst those 5 points.
Follow the path, first from A to C (the first pair). Continue with the other pairs in order.
The sequence covers all paths between letters, in a continuous route, without retracing any edge.
The result is a fully connected graph for the 5 points, starting and ending at A,
and looks like a 5 pointed star inscribed within a regular pentagon:
enter image description here
For the predefined pairs given initially, I believe my solution is the unique sequence that does this.

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  • $\begingroup$ I'm sorry could your explain your solution to me(I Think I'm stupid),I've understood that there are 5 points from where your start but i don't understand how you selected them and how you "Move" to other groups of letters . thank you for your time @Rubio $\endgroup$ – ade Apr 21 at 20:31
  • $\begingroup$ Nice answer: a circular journey too. $\endgroup$ – Weather Vane Apr 21 at 22:55
  • $\begingroup$ @ade Expanded explanation added to the answer now. $\endgroup$ – Rubio Apr 22 at 4:45
  • $\begingroup$ How did u think a solution like that xD @Rubio $\endgroup$ – ade Apr 22 at 9:30
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Some thoughts:

There are 10 items in the sequence.
There are 10 permutations of the letters a ... e taken in pairs.
There are no repeated pairs used.

So I guess that the 6 missing combinations should be used to fill the 6 gaps
(a,d) (b,c) (b,d) (b,e) (c,e) (d,e)

However the puzzle shows the pair (b,a) is in reverse order.
I was hoping that maybe the pairs in the even-numbered positions are reversed.
But that is not the case.

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