# Connect the colored dots

I've got this:

By connecting the dots, I get this:

Could you do the same for this one:

There is a unique solution.

The rules are: You have to use every square, lines can't cross each other and you can't backtrack to fill the squares. (Every two adjacent squares with the same colour must be directly connected by the coloured line.)

For example, the following image contains a large number of violations of the backtrack rule for the yellow line in the bottom-left part of the image:

• For people that enjoyed this puzzle, this is a screenshot from the Android game "Flow", which has many more instances of this kind of puzzle. – Daniel Wagner Mar 24 '16 at 2:15

The solution is ..............

It's not possible, and here's a proof. (If you see anything wrong, please point it out!) I'm going to be using this grid as a reference:

Consider square A.

A cannot be orange since then J would not be able to be filled.

Therefore, there must be a path WQIAB. (We don't know the color, so I'm calling this the 'mystery path' for now.)

Orange cannot go down because again, J would not be able to be filled.

Orange must be ·JKL in order not to collide with the mystery path,which must be WQIABCDE.

Therefore, orange must go JKL.

Assume our mystery path is not blue. 1

It must continue MNBTb.

Therefore, blue must be ·GHJPVcj·, and cyan must be ·U·.

Orange must be ·JKLRS·.

Our 'mystery path' cannot be green, since that would leave either X or d unreachable.

Assume our mystery path is yellow. (2)

Green has to go through d then; also, it cannot leave open space in e, so it must 'capture' at least one group; the only capturable group is red, but that disconnects the mystery path from the other end of yellow.

So our mystery path is not yellow. (2)

X cannot be green, since yellow would have no way to get out.

Therefore, X is yellow.

Therefore, Y is yellow.

Therefore, W is yellow.

This cuts off the mystery path.

So the mystery path is blue. (1)

QIABCDE must be blue.

JKLMN must be orange.

RSZ must be yellow.

U must be cyan.

What goes through FGHPVcjs? It's not yellow - if it was, then you'd cut off whatever path went through γικ. It's not red - you'd cut off πρ. It's not blue - then blue couldn't go through E. So it must be green.

The right green dot can't connect to the right side of the green "hanging" over the cyan since the left green dot wouldn't be able to connect. Therefore, the right green dot must connect to b.

But then yellow, which we've already determined must go RSZ, has nowhere to go! It's cut off from its other yellow dot.

Therefore, the problem is unsolvable.

• I had found similarly that it was unsolvable, because the bottom left space can be neither yellow nor not-yellow. – Tim C Mar 20 '16 at 4:46
• @TimC and Deusovi. To summarise what you've found so far and what is correct, see hint 1. To reassure you: There is a unique answer and I have the proof in front of me in the form of an image. Unless you specifically ask for hints I won't give any. It took me quite a while to figure the solution out myself. – NZD Mar 20 '16 at 7:44
• @NZD: Can you point out where in my proof I went wrong? – Deusovi Mar 20 '16 at 7:45
• It went wrong in thinking FGHPVcjs can't be yellow – Ivo Beckers Mar 20 '16 at 12:57
• @Ivo: Didn't even consider that γικ could be red. Nicely done! – Deusovi Mar 20 '16 at 12:58