Logic problem to solve

Obvious things I already figured out:
1.If you can prove that it's impossible to fill every square, you prove that it's impossible to do what's asked.
2.Since you have to do a loop, the starting point technically doesn't matter. Although it doesn't prove anything, it might make it easier to determine whether it's possible to do what's asked.
3.From testing multiple times, it seems impossible to do the loop. Haven't clearly figured out why yet.
4.Since the starting point doesn't matter, proving that going on every square exactly once isn't possible by starting off one of the squares in a grid of 3x4 in any corner, you prove is for the rest of the grid since you can just rotate/mirror it to prove the rotated/symmetrical cases.

File: grid 2.jpg (131KB, 723x521px) Image search: [Google]

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More specific facts:
1.If you arrive at the one of the extremities of the red paths, you have to go through the entire red path, because otherwise you end up with an impossible case.
2.If you land into one of the points of the blue loop, you have to complete the loop otherwise you end up with a failed attempt again. You also have to make sure that when going into the blue loop, you have a way out.
3.I'm unsure about this, but it probably doesn't matter whether you first fill out the squares that go through the blue/red paths before going through the "undetermined" path. This idea comes from the fact that since it's a loop, you have to go through the blue/red paths at some point anyway, so since the starting point is free to choose, choosing to complete the paths first or last doesn't matter. It's also probable that the blue/red paths need to be followed one right after the other, since some of the red paths extremities correspond to the entry/exit points of the blue loop.

File: grid 3.jpg (76KB, 734x530px) Image search: [Google]

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So, my current strategy is to prove that for any of the upper right squares with a pink corner marker lead to a case where you fail.
The squares with a pink marker indicate every square that is not one the blue/red paths from previous picture.
So, for every of these squares, with the blue/red paths logic in mind, I might be able to prove that you always end up in a failed attempt.
At this point it's still backtracking though so I need to figure out further logical aspects to be able to justify myself.

>>8806499
Forgot to mention that
5.of course, if you find a single working example, you're done, no need to justify hoe you found the working path.

Color the grid black and white like a chess board.

If you start on a black square, the next square you land on will be white (and vice versa), and if you want to finish on that black square, the penultimate square must be white.

Will it be white?

>>8806592
>tries to solve problem for 1 hour because I think it works
>end up finding it doesn't work, try 1-2 more hours with 6-7 GIMP layers to find why it doesn't work
>make thread on sci
>instant solution
...I'm glad I made this thread, even though I feel ridiculously stupid. It's probably a recurrent thing here though.
Well thanks anon, at least I can go back to getting some sleep and studying stuff.

>>8806478
https://en.m.wikipedia.org/wiki/Seven_Bridges_of_Königsberg

>>8806636
Oh right, I read about this method somewhere a year ago or so, it's nice to look it up again.
I thought it'd be inefficient for this case but actually once you found any point that branches into an odd amount of paths it's good, no need to check every square.
I still prefer the chess-like solution though, since it stands by itself in logic.

>>8806478
>go to every grid square exactly once, and end up back where you started
Since this is a logic problem, the answer is in the (trick) question; if you end up where you started, you've visited that square twice.
"NO".

>>8806636
>>8806680
Seven Bridges of Königsberg is about finding an Eulerian path while the Knight Tour is about finding a Hamiltonian path.
Those are two very different problems.

>There are 35 squares.
>You need 35 moves for a complete closed tour
>If you start from a white square then you will be on a black square after an odd number of moves
>It's reversed if you start from a black square

File: 1491425822135.jpg (70KB, 726x522px) Image search: [Google]

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Solved it.

>>8806871
>end up back where you started
you can't get from your 34 back to A in a knight's move, and even if you could you still failed - see
>>8806720

>>8806478
Hint: Given (x,y), you can transform the point by matrices like (1, 0; 0, 2) and so on.

On some N x M (where N may or may not = M) boards, where N x M is even, it MAY be possible that, after the penultimate move, the knight could next return to its original square. (This has been proved for N=M=8).
For any N x M = odd number, the knight's penultimate move is to the same colour square as its origin square, and therefore cannot reach its origin square on the next move. See 'knights move black-white-black arguments above.
ie in >>8806871 suppose squares 14, 6, *34*, 26, 10 are black. Squares 14, 22, 16, and *A* are therefore also black.
And even if you land on the origin square, that's the second time you've been there, which loses as stipulated by the rules.