Draw four separate loops in the grid, one blue, one green, one grey and one red. Loops connect centres of cells horizontally and vertically and cannot cross over themselves or share any cell with each other. If a coloured cell is visited by the loop, the loop must be the same colour as the cell. Each circle in the grid must be visited by one of the loops. When passing through a white circle, the loop must go straight through the circle, and then immediately make a turn in the next cell on at least one side. When passing through a black circle, the loop must make turn in the circled cell and continue straight on both sides.

Solve on Penpa+ (no answer check)

Empty grid

  • 1
    $\begingroup$ There is no restriction about how the coloured loops are allowed to interact with each other except for the coloured cells must be on their respective coloured loops. Correct? $\endgroup$
    – LeppyR64
    Commented Apr 5 at 16:16
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    $\begingroup$ Guess it's better to just spell out all the rules every time. $\endgroup$
    – Jafe
    Commented Apr 5 at 16:20
  • $\begingroup$ Thanks I guess my confusion was around the term "separate". It's more clear now. $\endgroup$
    – LeppyR64
    Commented Apr 5 at 16:28
  • $\begingroup$ @PDT Thanks, that's really nice to hear! This took a few more tries to make than I like, but I'm really happy with the result. $\endgroup$
    – Jafe
    Commented Apr 6 at 7:49
  • $\begingroup$ I feel the strong urge to downvote this for putting Go stones in the center of fields instead of the intersections, but I won't ;-) Nice puzzle. $\endgroup$
    – quarague
    Commented Apr 6 at 8:49

2 Answers 2


Whilst a solution image has been posted already, a solution path has not. Since this is a pretty tricky puzzle with some interesting deductions, I figure it is worth explaining to other potential solvers how exactly to see this puzzle through from start to finish, ultimately ending up with this solved state:

Final solved state

Step 1:

First, some obvious deductions by the basic rules of masyu:

Step 1a

Now consider the cells known to be in the grey loop.

If these connect to each other somehow by passing in between the pairs of red and blue loop cells, this leaves only three possible path sections around this loop for the red and blue loops - but we need four for those loops to both enter and exit this top-left region.

Step 1b

Thus the grey loop must pass outside the red and blue cells and occupy spaces around the edge of the board, forcing some of the red and blue loops...

Step 1c

Step 2:

For red to escape from the top-left corner without cutting across the grey loop, it must pass through R5C3-4. This requires the white circle in R4C2 to be part of the red loop.

Step 2a

It is also clear at this stage that the two green squares cannot be part of a loop that goes outside any of the others.

Step 2b

In fact, inspection reveals that it is required for each loop to circle the green loop fully (one part of the loop going round the leftmost green square, the other part going around the rightmost green square - no coloured loop can exit the top left corner of the grid with both parts passing by the same green 'end'), which in turn means that we are dealing with four 'concentric' loops where the green must be entirely contained inside the blue loop, which is inside the red loop, which is inside the grey. We can fill in more loop sections for each colour...

Step 2c

...and in fact, we can make several further logical deductions on the left-hand side concerning where the grey and red loops must pass, forcing more of the loops. Plus in that top-right corner we need to extend the green loop 2 spaces from the black circle in each direction, which has several other knock-on deductions.

Step 2d

Step 3:

We can keep extending these loop segments in the top-left and top-right, ultimately joining the two grey loop segments.

Step 3a

And we can now conclude that the green loop's black circle in R9C6 must have a segment leading downwards rather than up, as that route is now blocked. This enables us to resolve the top-left quadrant entirely...

Step 3b

...and we can also keep extending those segments in the top-right quadrant downwards, leading us to realise that the black circle in R10C15 must be part of the blue loop.

Step 3c

Step 4:

Focus now on the black circle in R14C4.

This must be part of the blue loop. If it were grey, then attempting to resolve the white circle above it with red would result in the red loop making a turn afterwards that blocks off blue.

Step 4a

Now there is only one way for the blue loop to squeeze its way out of the bottom-left corner whilst satisfying the white circle currently sandwiched between it and the green loop.

Step 4b

There now remains only one way to resolve the green loop fully:

Step 4c

Plus the white circle in R11C12 must be passed through vertically (to avoid closing off the blue loop too early).

Step 4d

Step 5:

Almost there now.

The white circle in R15C15 must be passed through horizontally, which forces the red and grey loops to completeness:

Step 5a

And now all that remains is to complete the blue loop, and we are done!

Final solved state

  • $\begingroup$ Excellent. I didn't have time to turn my lunch-break solve into a full solution. You have clearly put significant effort into your answer. $\endgroup$ Commented Apr 6 at 13:23
  • $\begingroup$ Thanks @DanielMathias - it was definitely a tricky write-up! $\endgroup$
    – Stiv
    Commented Apr 6 at 13:55
  • $\begingroup$ Monstrous this is what I wanted to do but I don’t have a way with words to explain things like you obviously do. $\endgroup$
    – PDT
    Commented Apr 6 at 16:55

Working mostly with numberlink logic and correcting errors on the fly, I have found this solution:

enter image description here

  • $\begingroup$ I solved this too yesterday but you pipped me, however there are a few pretty obvious deductions at the start you could add and deductions based on where the grey loop cannot be (there are bottle necks created) and the green (there is only one way to connect the green) loop has to be that dictates the nature of the other loops. $\endgroup$
    – PDT
    Commented Apr 6 at 3:17
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    $\begingroup$ There are also other deductions but they are so abstract that I cba to write them out, I hate providing step by step guides to complex grid deduction puzzles. $\endgroup$
    – PDT
    Commented Apr 6 at 3:26
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    $\begingroup$ @PDT On PSE, the best kind of answer to a grid-deduction puzzle would demonstrate to a solver exactly how they can go about solving it themselves, so explaining the logical steps they would need to take in the process is a key part in helping them understand it (explaining why we generally ask for the step-by-step explanation in answers). $\endgroup$
    – Stiv
    Commented Apr 6 at 12:27
  • $\begingroup$ @ Stiv I know that and I wasn’t denying it. $\endgroup$
    – PDT
    Commented Apr 6 at 13:26
  • $\begingroup$ @PDT if you want a tip on how best to do it, it is much much easier if you just write the answer as you are solving, that way you can explain what you just did and take screenshots as you go without having to go back and try and remember every step. Saves a lot of time overall too! $\endgroup$ Commented Apr 7 at 18:36

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