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I've got this:
enter image description here

By connecting the dots, I get this:
enter image description here

Could you do the same for this one:
enter image description here

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:
enter image description here

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  • $\begingroup$ 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. $\endgroup$
    – Daniel Wagner
    Mar 24, 2016 at 2:15
  • $\begingroup$ Flow is also available on iPhone! $\endgroup$
    – ribs2spare
    May 18, 2022 at 13:27
  • $\begingroup$ They don't list it as a rule, but every "Flow Free" puzzle I've encountered thus far (whether you pay for the extras or not) can be elegantly accomplished WITHOUT any "tight turn arounds". In other words, All loop backwards have at least one cell between them. In a Flow Free puzzle, the zig-zag yellow section would not be needed. $\endgroup$
    – alife
    Nov 27, 2022 at 2:50

3 Answers 3

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

enter image description here

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  • $\begingroup$ Yes, and notice that the unwritten rule of "no tight turnarrounds" is obeyed beautifully here. No zigzag nonsense. Every time a path attempts to reverse the direction, there is always an empty square between each leg of it. $\endgroup$
    – alife
    Nov 27, 2022 at 2:53
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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:

enter image description here

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.

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  • $\begingroup$ I had found similarly that it was unsolvable, because the bottom left space can be neither yellow nor not-yellow. $\endgroup$
    – Tim C
    Mar 20, 2016 at 4:46
  • $\begingroup$ @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. $\endgroup$
    – NZD
    Mar 20, 2016 at 7:44
  • $\begingroup$ @NZD: Can you point out where in my proof I went wrong? $\endgroup$
    – Deusovi ♦
    Mar 20, 2016 at 7:45
  • $\begingroup$ It went wrong in thinking FGHPVcjs can't be yellow $\endgroup$
    – Ivo
    Mar 20, 2016 at 12:57
  • $\begingroup$ @Ivo: Didn't even consider that γικ could be red. Nicely done! $\endgroup$
    – Deusovi ♦
    Mar 20, 2016 at 12:58
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I’m not sure but I took a crack at it with a screen shot. is this a fair answer?

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    $\begingroup$ This violates the rule that you can't backtrack. Your lines for the orange and yellow dots in particular backtrack several times. $\endgroup$
    – F1Krazy
    May 17, 2022 at 21:52
  • $\begingroup$ @F1Krazy but what exactly "backtracking" means? His solution seems fine to me in this kind of puzzle (all cells are used, no diagonal jumps, no crossings) $\endgroup$
    – Vitalii Vasylenko
    Sep 28, 2022 at 22:29
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    $\begingroup$ @VitaliiVasylenko The question says clearly: "You can't backtrack to fill the squares. Every two adjacent squares with the same colour must be directly connected by the coloured line." This is not the case in Aldo's solution. Ordinarily, yes, it would be fine, but in the case of this specific puzzle, it violates that extra constraint. $\endgroup$
    – F1Krazy
    Sep 29, 2022 at 6:46
  • $\begingroup$ @F1Krazy, that's not what the "no backtracking" means. The no backtracking rule the OP is referencing has to do with not walking backwards within the same path to then branch out again elsewhere. This means that every line path always has a "from" and a "to". If there were "backtracking", a path might suddenly end without a dot, the user would backup along that same pathway, and branch out elsewhere. One of the seemingly unwritten rules that Flow does obey is an avoidance of tight turn arounds. There's always a gap of at least one cell when a path changes a direction backwards. $\endgroup$
    – alife
    Nov 27, 2022 at 2:57

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