4
$\begingroup$

I'm looking for solution & plausible explanantion which letter comes next in following riddle:

enter image description here

Source (german): https://www.matheboard.de/thread.php?threadid=1459

Update: The correct answer is B. Does anybody know a plausible argument?

$\endgroup$
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio Sep 20 at 5:24
  • $\begingroup$ the correct answer is B but still don't know how to derive it. I updated the question $\endgroup$ – Tim Grosskreutz Sep 21 at 21:04
  • 1
    $\begingroup$ Two answers below put forward answers for B, one with 7 upvotes at this moment - I suggest you look further into those :) $\endgroup$ – Stiv Sep 21 at 23:20
9
$\begingroup$

The answer is B

The secret is Fibonacci series. The edges that connect the letter nodes are either blob ended or flat ended.

We only care about the flat ended edge on each node. So, 'J' has 1 flat ended edge, similarly 'I' has 1 flat ended edge.

The node 'H' has 2 concentric circles around it.

So, we multiply the number of flat ended edges of 'H' by 2 which gives us 2. The node 'G' has 3 flat ended edges. The node 'K' has 5 flat ended edges. The node 'F' has 4 flat ended edges with 2 concentric circles around the node. So, we have 4 times 2, which is 8. The node 'A' has 13 flat ended edges.

If we write the sequence out we have 1,1,2,3,5,8,13. The next one should satisfy 21.

The node that satisfies 21 is 'B'. This is because node 'B' has 3 concentric circles around it and 7 flat ended edges. If we multiply them we get 21. Voila!

$\endgroup$
4
$\begingroup$

My answer is E

Theory:

pointer on J

j, I, K = there is a single line both sides. line between k and j is having a circle in the end toward j side and between j and I, there is a circle at the pointer end, and I side both. so the next letter came up is I. it means away circle is having higher priority.

Pointer on I

I, D, H, J = there are all 3 single lines and away circle is only j side that means and next letter coming is not j that means the letter will not repeat. now between H and D, it moves towards H because there is a circle between d & I towards pointer side on line and there is no circle between I and H. that means no circle letter have high priority than the line where circle is cursor end. also, letter will not repeat

Pointer on letter H

now we already knew that no letter will repeat and priority of line is like away circle line > no circle line > pointer circle line now there is both the line C & G pointing towards H that means the priority is equal but the letter came up in series next is G. that means the letter comes after in alphabetical order is having higher priority.

Pointer on letter G

the pointer moves towards K,

it is clear with a line theory that priority is equal but K comes after B in alphabetical order so pointer moves towards K.

Pointer on letter K

now j and G are both repeated letters we already figure it out that repeated letters will not come up in series again. so pointer moves towards F

Now pointer moves towards A because

if the number of lines is more it has more priority. Now, finally E will come because the number of a line between E & B is same no circle at all. so priority is equal so we will go according to alphabetical order E comes after B in alphabetical order. So answer is E

Rules wise Theory

1) no alphabet will repeat. (first check)
2) the more the number of lines the higher the priority.( second check)
3) the line with circle away from the pointer is having the highest priority, as compared to the line with no circle is higher than the line where circle is toward pointer end. (3rd check)
4) if line priorities are equal we go with alphabet order priority. the letter comes later in alphabetical order is having higher priority. (4th check)

$\endgroup$
  • $\begingroup$ But by this logic it could also be B, C or D... Why choose this letter over any of those? $\endgroup$ – Stiv Aug 30 at 12:29
  • $\begingroup$ @Stiv yes, I later find out that my previous logic was wrong... but this one is 100% correct. $\endgroup$ – Sayed Mohd Ali Aug 30 at 12:31
  • $\begingroup$ You set of rules does apply to the known sequence, but there are so many variables to choose form that many different sets of rules will work. I have no clue what the answer is, but I expect the rule to be much simpler $\endgroup$ – P1storius Sep 2 at 9:25
  • $\begingroup$ @MKBakker if you try you may come up with different rules but I am sure you will need more than one rule to solve this answer. and it is totally depend on the number of lines, type of lines and alphabets. I gave explanation how I able to find out those rules and after that I explained what are 4 rules you need to apply to solve the puzzle. $\endgroup$ – Sayed Mohd Ali Sep 2 at 10:15
  • $\begingroup$ @MKBakker I have other theory maybe you can find out... as you can see in outer circle the letters are in sequence G,H,I,J,K and also in inner circle BCDEF, I think it is just to confuse people instead of lines and alphabetical order... but maybe any intelligent brain can still make theory on it. $\endgroup$ – Sayed Mohd Ali Sep 2 at 10:33
1
$\begingroup$

Is it:

B. Because this is the only option that allows the path to be a Hamiltonian cycle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.