7
$\begingroup$

Well, thanks to Len, I managed to unlock my phone again, but now everyone knows my pattern, so I need to choose a new one.

For brevity, I won't include the explanation of how an Android pattern lock screen works again, but you still need to follow those rules (see previous puzzle).

Here's the rules for my new pattern:

  1. I still want to use all nine dots
  2. I'm still obsessed with mirroring, so it must still have reflective symmetry
  3. I don't like odd numbers, so the pattern's resulting line must have exactly two "ends"*

What should my new pattern be?

* by this I mean that visually, the final line formed by the pattern has exactly two dots which have a single line connecting to them. So, for example the pattern 2 1 3 has two line endings (at 1 & 3), but 5 2 1 3 has three (at 1, 3 & 5). All solutions found for the previous puzzle have three line endings.

$\endgroup$
0

1 Answer 1

8
$\begingroup$

Woah... I think I got it after a whole lot of trying around. Here is my train of thought:

  1. Since we have two open Ends and reflective symmetry, both ends must be mirrored.
  2. Since we can never directly go to an already visited point, the last point must be an open end - so one of the two ends.
  3. The two end segments (the last part, direct connection) cannot be directly connected, because we cannot leave the path and come back to a point.

Valid endpoints: 1-4 3-6 and 1-4 8-9

A . . .  B . . .              . . .   . . .   . . .   . . .
  | . |    | . .    Invalid:  . . .   .\./.   .\. .   . | .
  . . .    . ._.              ._._.   . . .   . .\.   . | .

Trying to connect the two valid positions for endpoints I realized I couldn't start at one of them, because I cannot leave the path and come back on the path - and without coming back on the path, it is no longer symmetric. So one of the endpoints had to be a hidden double-line (like 2-1-3) and the other the real endpoint.

So my path had to start on the mirroring point of this hidden path, since the mirrored path without the end-points would end there and has to be symmetric.

So the start-points for Figure A was 4 or 6 and for B was 4 or 8.

Some more trying around gave me this:

. ._. 4-5-6-3-2-8-9-7-1 |_|_| |_|_.

$\endgroup$
1
  • 1
    $\begingroup$ :) Awesome logic to narrow it down. This is indeed the pattern I was looking for. Well done! $\endgroup$
    – Alconja
    Feb 12, 2015 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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