Previous Level: Tapa-Nurikolor (Level 6)
Next Level: Nurikolor (Level 8)

I saw that the last $2$ levels had a lot of flaws and were not unique. Hence this time I am giving an easy puzzle with uniqueness , with an 8x8 grid and 6 colours.

  • There are colored numbers on the grid, which indicate the number of tiles the group of its color holds.
  • There are tiles with 1 color, which indicate the color of the tile.
  • There are tiles with 2 or more colors, which indicate intersections of colors. All intersections are shown, and these are the only intersections.
  • Grey tiles are not part of any group; they just serve as barriers.
  • The goal is to have every non-grey tile covered by a type of color.
  • 2 by 2 non-grey squares of the same color are illegal.
  • In future levels, there will be multiple numbers of the same color. Their groups must never intersect or be orthogonally adjacent to each other.
  • There will be colored lines in certain places. The same-color group may not cross through the colored lines, although they must border the line.
  • There may be intersections that aren't fully colored. It is also your job to color it.

New :- There are some tiles with two colours which are separated by a horizontal line drawn between them. This means that the tile is fully coloured by either of those 2 colours (you have to find which colour it is coloured with), not by any other colour.

  • $\begingroup$ I am afraid this one is not unique $\endgroup$
    – Retudin
    Oct 13, 2020 at 11:21
  • $\begingroup$ It is, I have checked every-place where unique-ness cannot come, and fixed it. $\endgroup$
    – Anonymous
    Oct 13, 2020 at 11:24
  • $\begingroup$ Do you want my solution with bluorange = blue or orange? $\endgroup$
    – Retudin
    Oct 13, 2020 at 11:26
  • $\begingroup$ I posted the puzzle for solutions right? Of course, post it. $\endgroup$
    – Anonymous
    Oct 13, 2020 at 11:27
  • 2
    $\begingroup$ I think it may not be unique, but... at least it's down to just 2 solutions, so it's kind of semi-unique... (See answer for more explanation) $\endgroup$
    – Player1456
    Oct 14, 2020 at 8:16

1 Answer 1


I will try to solve this, just for once. Here goes.

The puzzle, unsolved. Nurikolor 7

We know that R1C6 is either red or blue. But Blue cannot pass through, as it is blocked by the Red 11 and the Yellow 9. By law of connectivity, the law of isolation, the fact it's neccessary to connect all red and yellow squares, and the 2x2 colored square rule, we can get this far. The red and yellow groups force each other into place (mostly.) Part 1

Now, is R6C7 red, or yellow? It is required that R8C7 be an intersection, which is 2 squares away from R6C7. Red has 1 left compared to Yellow's 3, which, if you subtract 1 to account for the square, is just barely enough to reach it. R6C7 must therefore be Yellow. Green, meanwhile, can be placed in certain spots due to the connectivity rule, and now we're this far. Part 2

(Subtle graphic change) Now let's look at the central squares. Knowing that Red has one space to go, R4C4 cannot be Red (it is 2 squares away), so it must be Green. If R4C5 is Red, Blue is trapped no matter what (with either 2 or 5 squares), so by process of elimination, R5C4 must be Red, which forces all of R4C3, R7C3 and R8C3 to be Green. Using connectivity logic, we get this. Part 3

We now have 3 Oranges, 4 Purples, and 3 Blues left to place. R3C4 cannot be Blue as it would force a 2x2 square with R2C4. So it must be Purple, and two more Blues can be placed. The last square of importance is R2C3, because if it is Purple, no matter where the Oranges are, the result will always break at least one rule, so the wrap-up is done. We can have either... Part 4.1 ...or... Part 4.2 as our final answer. The answer is, as unfortunate as it is, still not unique.

This is my attempt at solving this. =)

Note: I remove the split-color parts if it's guaranteed to be a certain color. I just want it to be more aesthetically pleasing.

  • 1
    $\begingroup$ That is correct, hopefully I reduced the uniqueness to 2 solutions atleast, but the initial logic still remains the same, and that is what I am happy about. $\endgroup$
    – Anonymous
    Oct 14, 2020 at 6:47

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