I was making ripple effect but failed on uniqueness. That leads me to think of using inequality signs to fix the puzzle. Then, I'm wondering, what if, I make a new one with intentional inequality signs? So, here we are!

Rules (from Nikoli):

  1. The areas divided by bold lines are called "Rooms". Fill in all empty cells with numbers under the following rules.
  2. Each Room contains consecutive numbers starting from 1.
  3. If a number is duplicated in a row or a column, the space between the duplicated numbers must be equal to or larger than the value of the number.
  4. In addition, inequality signs between two cells show which cell has a greater/lesser number.

The puzzle:


Here's penpa link for everyone who prefers to solve it online. I hope you guys can enjoy it!

  • $\begingroup$ For someone unfamiliar with the puzzle type it should be noted that 'Each Room contains consecutive numbers starting from 1' does not mean the numbers have to be placed consecutively. For instance the block of 4 on the top must contain 1, 2, 3, and 4 in some order, but it doesn't have to be '1234' or '4321', it could be '4132'. $\endgroup$ Oct 22, 2021 at 19:01

1 Answer 1


This was a very tricky puzzle, but in the end I believe the solved grid should look like this:

Solved grid

In the step-by-step diagrams that follow I have used three different colours in the small pencil marks:

  • Red in the bottom-right corner indicates an impossible value;
  • Green in the top-right corner indicates the only possible values;
  • Grey in the top-right corner indicates all of the spaces within a shape that may take a particular value of interest.

Large black digits are part of the solution.

Step 1:

Some initial pencil-mark deductions:

Step 1

Predominantly, these are the result of noting that (i) a 1 can never appear at the open end of a '<' sign, and the largest number in any shape can never appear at the closed end, and (ii) a shape of two cells must contain 1 and 2, so immediate inline neighbours cannot contain 2. Furthermore:

- The 4 in column 6 must appear in row 3 or 4;
- R6C3 cannot be 4 or 5, and R6C6 cannot be 5, or it is impossible to place these figures in the two shapes wholly contained by rows 5 and 6;
- No 4 may appear in R3-5C7;
- R1C2 cannot be 3; else, R2C1 must also be 3, and then there is no way to place a 3 in the 2x square in columns 1 and 2.

Step 2:

Assume now for contradiction that R2C6 contains a 2. This soon produces the following state of play by knock-on logical deductions:

Step 2a

(i.e. 2 in R2C6 → 1 in R2C5 → 1 in R4C5 & 2 in R3C5 → 2 in R1C7 & 1 in R1C6 (since cannot be 3 or 4) → 4 in R2C7 & 3 in R3C7 → 4 in R7C7 & 4 in R3C6 & 3 in R4C6)

This means there is now no way to place a 3 in the bottom-right shape, as all its cells are covered by the 3's in R3C7 and R4C6. Contradiction.

Instead, R2C6 contains a 1, and we can make a few further knock-on pencil marks:

Step 2b

R1C6 cannot now be a 3, or there will be nowhere valid for a 3 in the top 1x4 rectangle. Instead, it must contain a 2. This then resolves the 2x2 box below, in the process forcing R1C7 and R2C7 to contain a 3 and 4 between them, which forces R4C7 to be a 5.

Step 2c

Note here that we have also learned some information about the 3's - notably, the top-left corner cannot be a 3, or there is no valid placement for a 3 in the top 1x4 rectangle. This in turn means that no 3 can be in R2C3-4, which forces a 3 to be placed in R4C3-4, which now (as you can see in the diagram above) leaves only one position for a 3 in the left-most 2x2 square: R3C1. Filling this in leads to a few knock-on deductions, including that R1C2 must be a 4 by the inequality, and then R4C1 must also be a 4:

Step 2d

At this point we now realise that R1C3 must be a 2 and R2C7 cannot contain the 4, otherwise there will be no legal space for a 4 in the 2x3 rectangle. Thus, it is a 3. Logical knock-ons end up entirely resolving all 3 of the shapes in the top row:

Step 2e

Step 3:

Now there is only one valid position each for the 2 and 3 in the 2x3 rectangle: R3C4 and R4C3, respectively. Furthermore the bottom-right cell of this shape must contain the 6 - no other number may be placed here. We can now fully resolve the left-most 2x2 box.

Step 3a

There must now be a 3 in R6C2, since no other cell within its shape may contain it. This fixes the position of the 4 within the same shape, which helps fully resolve the 2x3 rectangle (where R3C3 can only be a 5). Further pencil marks can be made:

Step 3b

And now there is only one possible space for the 4 to be placed in the empty shape in rows 5-6: R6C5. We make yet more pencil marks...

Step 3c

Step 4:

Finally, only 2 can now be placed in R5C5. This helps position the 2 in the bottom-right box, and the rest follows!

Solved grid

  • 2
    $\begingroup$ You quite literally beat me to it; was just starting writing up my answer. $\endgroup$
    – Stevo
    Oct 21, 2021 at 22:19
  • 2
    $\begingroup$ Very tricky puzzle indeed - I wasn't even able to make an initial breakthrough $\endgroup$
    – Avi
    Oct 21, 2021 at 23:04
  • $\begingroup$ I think I'm missing something. In the third picture of step 2, why r3c1 cannot be 4? Can you please elaborate it? Thanks! :D $\endgroup$
    – Nusi
    Oct 22, 2021 at 7:25
  • $\begingroup$ @Nusi I think that should now all be fixed! (Unless your eagle eye spots anything else I have foolishly done!) Thanks for pointing that out :) $\endgroup$
    – Stiv
    Oct 22, 2021 at 8:03
  • 1
    $\begingroup$ It should be good right now. Great explanation! :D $\endgroup$
    – Nusi
    Oct 22, 2021 at 15:53

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