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Sorry. The previously posted image contained a mistake.

Tile D2 was shown with a 180 degree rotation. The image has now been corrected.

This is a straight-forward arrangement puzzle.

You are given a set of tiles showing each a blue and a red arrow pointing in one of 8 directions. Your task is to rearrange these tiles - without rotation or mirroring - such that each blue arrow is pointing to a neighbouring tile containing a red arrow pointing back to the tile. Additionally, all connections need to form a single, continuous cycle, like in the following mini-example:

Example

valid arrangement

Your set contains the following 49 tiles, which must not be rotated or mirrored, just shifted:

CorrectedPuzzle

The grey triangle is just to mark "up" so that you can print those tiles, if you want and don't accidently rotate them.

Question:

Rearrange tiles into a valid 7x7 grid.

This question has at least one solution. Preferably post the solution as image. Alternatively as a grid of tile-IDs using the indexing provided in the tile-set to identify the tiles. (i.e. the first tile is tile (A1) and you use A1 in the position you think it should be in.)

The original post had two more "bonus" questions which have found their answers already. I'm posting them for reference only.

##Question 2:## Take away a single tile of your choosing. Can you rearrange the remaining 48 tiles into either a 6x8 or a 8x6 valid grid? If you can not do this with a single 'loop' you may create a grid which contains multiple closed loops, but each tile must belong to one such loop.

##Question 3:## Can you repeatedly take away a single (freely chosen) tile while managing to rearrange the remaining into *any* rectangular grid? (i.e. find arbitrary solutions for (n x m) grids with 49, 48, (47), 46.... 4 tiles where each 'smaller' set has to be created by taking away a single tile from the next larger set. If the set-number is a prime number, you can skip it (i.e. take away another tile). Again, single closed loops are preferred, but multiple closed loops are allowed.

Both bonus-questions are answered by John Stevens in the answer below.

This puzzle seems to be harder than I thought it would be. To keep things rolling, I'm adding some hints, i.e. some information about specific tiles/positions.

Hint 1

The 4 corner tiles are (without order):

F7, F6, F3 and D7.

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    $\begingroup$ This puzzle seems to be really hands on. [Insert clock pun here.] $\endgroup$
    – Mark N
    Commented Jul 2, 2015 at 15:31
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    $\begingroup$ @BmyGuest: 47 is a prime, it will not be possible to make a rectangle out of 47 tiles. (47x1 isn't really a rectangle that will fit our needs) $\endgroup$
    – CodeNewbie
    Commented Jul 2, 2015 at 16:04
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    $\begingroup$ @MarkN I've already clocked you as someone who makes lots of funny comments! $\endgroup$
    – Rand al'Thor
    Commented Jul 3, 2015 at 9:12
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    $\begingroup$ @MarkN and I didn't have the time to come up with a better pun... $\endgroup$
    – Rand al'Thor
    Commented Jul 3, 2015 at 16:14
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    $\begingroup$ I wrote a program to solve this and it hasn't found a solution. Of course, the usual suspect is a bug, but it did work for a few different examples. Seeing no one came with a solution I thought I'd raise the issue. @BmyGuest - Are you absolutely sure these are the right tiles? Has anybody else tried solving it with a computer? Thanks. $\endgroup$
    – Angkor
    Commented Jul 6, 2015 at 15:06

3 Answers 3

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I mostly used a computer to solve. A random brute force process. I added as much logic as I could determine from the hint and reduced which tiles could be placed in the surrounding cells (3x3 form the corners). I started the process from cell A7 and followed the blue arrow up where I determined cell A6 only had two that could fit, thus reducing the number of random incorrect choices greatly. From there the process followed the blue arrows placing a random tile after filtering out the tiles that logically didn't fit.

Solution:

enter image description here

Solution Text Look:

enter image description here

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    $\begingroup$ Congrats. That is the intended solution I have. (So maybe it's the sole solution.) I wouldn't have thought that it is so tricky to find it and require some computer assistance, but one never knows. Well done, Sir. $\endgroup$
    – BmyGuest
    Commented Jul 22, 2015 at 6:51
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    $\begingroup$ Well done! I used another approach; I wrote a genetic algorithm and let „natural” selection do it's thing, but as we know, that takes a lot of time. So here I was, coming to post the solution, just to see that my approach was 11 hours too slow. :D $\endgroup$
    – Cristian Marian
    Commented Jul 22, 2015 at 8:09
  • $\begingroup$ I'm guessing it's the only solution. My program has found the same solution 3 times now. $\endgroup$
    – Crispy
    Commented Jul 22, 2015 at 15:05
  • $\begingroup$ @Crispy, this is the one I found . $\endgroup$
    – Cristian Marian
    Commented Jul 23, 2015 at 8:33
  • $\begingroup$ Well done. Just wanted to say that we were all a bit silly not noticing the mistake: According to @John Stevens's answer, adding all red or blue arrows, in the $x$ or $y$ direction, we should get $0$. But with the original board we get $2$'s and $-2$'s... I actually thought I checked it, but I accidentally added both colours together, so of course this didn't spot the mistake. $\endgroup$
    – Angkor
    Commented Jul 23, 2015 at 12:44
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For Question 2:

Is it not possible to arrange the tiles in a valid configuration if any one is taken away.

Explanation:

If one starts at any square and follows arrows of one color, one will eventually get back to the starting square. I'll just consider the blue arrows from now on, but the same argument also applies to the red arrows.

I'll model the direction of the blue arrow as an ordered pair $(x,y)$, where $x$ is the change in the horizontal coordinate when following the arrow, and $y$ is the change in the vertical coordinate. So, the the move for the A1 tile is $(1,-1)$, the move for the A2 tile is $(1,1)$, the move for the B3 tile is $(0,-1)$, and so on.

In any cycle in a valid configuration, the sum of the moves for all the tiles in that cycle must be $(0,0)$, since if you follow the blue arrows you eventually get back to where you started. Since every tile in a valid configuration is part of a cycle, and no tile is part of more than one cycle, the sum of the moves for all tiles in a valid configuration is $(0,0)$.

Since there is a valid configuration for all 49 tiles, the sum of the moves for all 49 tiles is $(0,0)$. For there to be a valid configuration for 48 tiles, we would need to remove a tile and leave the sum of the moves for the remaining tiles $(0,0)$. This is only possible if there is a tile without a blue arrow. Since all tiles have a blue arrow, there is no valid configuration consisting of any 48 of the 49 tiles in the puzzle.

This result can be used to quickly eliminate possibilities in brute force searches for valid configurations of other sizes: for any subset of tiles, if the sum of the blue moves (or red moves) is not $(0,0)$, then no valid configuration is possible and the subset need not be considered.

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  • $\begingroup$ +1 and surely the right answer to Q2 (and hence Q3 in a way). Thanks. I'll accept primarily the correct Q1 answer though, as that one is the "puzzle" part. $\endgroup$
    – BmyGuest
    Commented Jul 2, 2015 at 19:12
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PRELIMINARIES Q1

A Sennet (rope mat) puzzle can be described as NODAL or PEG

If you think of the pegs arranged along the edges and sides of the board, the continuous one cord sennet or, as the architects call it, Celtic Interlace, will be constructed/ woven on a 2X3 peg; 3X4peg; 4X5 peg board.

The nodes or crossovers can be X or CC back to back. an X needs 4 tiles with 45degree (45,135,225,315) and the available tiles suggest the basic design (before elaboration) is 2X3. This needs six X nodes; but any two X nodes can be replaced by four CC. (If you replace odd numbers of X with 2CCs each the thread will not be continuous/ single.)

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    $\begingroup$ The -1 was from me, but accidental :c(. Can't change until you do an edit though. $\endgroup$
    – BmyGuest
    Commented Jul 3, 2015 at 6:32

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