Number Slope™1 is an original2 grid deduction puzzle similar to Sudoku. It is solved on an n-by-n grid, tiled with n n-ominoes. The rules are:

  • Each number from 1 to n must occur once and only once in each row, column, and n-omino.
  • Any row or column of adjacent cells within each n-omino must have numbers which are either increasing or decreasing along the row or column (these are called slopes).

That second rule may be a bit confusing, so here is a diagram showing the slopes in an 11-omino and a valid filling:


Here's another diagram, this time giving an example of the solving process for a 3x3 puzzle. The process is quite similar to solving a Sudoku, but note that in step 3 we can deduce how to fill the vertical tromino, since the numbers have to form a slope.


Got that? Good. Now go solve this puzzle. Good luck!


1Or, to give it a more credible (read: Japanese) name, Sukobai, as Google Translate tells me.
2At least, I believe it is. Tell me in the comments if it isn't.

  • $\begingroup$ I'm pretty sure these exist somewhere, but I don't remember what they're called. $\endgroup$
    – dcfyj
    Jan 9, 2017 at 13:34
  • $\begingroup$ I've seen "irregular" / jigsaw Sudokus before (it's easy to find them online too). But the "slope" concept is new to me. $\endgroup$ Jan 9, 2017 at 17:43

1 Answer 1


What a fun puzzle!

Green means a note.
Blue means an important deduction.
Red means can be deduced by shallow sudoku logic, or shallow slope logic (I.e. If you have a 1 and 3, a 2 goes in between.)

Start off by focusing on 1s and 8s, because they're limited in where they can go (they have to be in a corner). In particular, that top middle area let's us deduce a 1.

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Consider where 1 can be in row 3.

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Consider where 2 can be in column 5.

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Consider where 8 can be in column 5.

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In the middle bottom area, 8 can only fit in row 7. Therefore, in the bottom right area, 8 cannot be in row 7. We get 8 has to be in column 7.

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Since 8s are in corners, we can deduce those 8s, and chase 8s all around the board. This also lets us sort out that 1/8 thing at the top, and then chase 1s around the board.

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Now that we've placed all 1s, 2s behave very similarly with needing to be in corners, or right next to ones. We can deduce the two blue twos, then chase.

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Again, now that all 8s are placed, 7s behave similarly. We deduce the blue 7 in the bottom right, the red 7 at the top, then the blue 7 at the top, then chase.

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Now most of our 7s are in, 6s behave similarly. Importantly, looking at the top middle area, we can use the only possible positions of 6 to deduce a 6 at the top right, then chase.

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Now deduce the blue 4 (it can't be 3 or 5, and it has to be between 2 and 6), and it will give you the five right above it. That's enough to sudoku chase the rest of the puzzle easily.

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  • 1
    $\begingroup$ You're too fast. I was almost there :) $\endgroup$
    – Rubio
    Jan 9, 2017 at 10:38
  • 7
    $\begingroup$ What can I say? I love a good grid-deduction puzzle :) $\endgroup$ Jan 9, 2017 at 10:39
  • $\begingroup$ This is simply brilliant (and fast of course). $\endgroup$
    – ABcDexter
    Jan 9, 2017 at 11:14
  • $\begingroup$ Nice solution! In the "twos" section you can have only the left one blue. The second blue comes straight from the two reds. $\endgroup$
    – pajonk
    Jan 9, 2017 at 11:19
  • 1
    $\begingroup$ Ah, fair point. I simply took these screenshots as I solved the puzzle, and I noticed the top right blue first. It didn't lead to any sudoku chase, so I then found the bottom left one and I think I just left them both blue. $\endgroup$ Jan 9, 2017 at 11:30

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