This is my contribution to the 'This new puzzle type needs a name' series, started by @Stiv.

I believe I have invented a new type of puzzle...

What is its name?

image of the puzzle

Penpa+ link

A little advice..

Even though you know what to do, this might not be as easy as you think. So, if you don't even find out how to begin, then, oh no baby!

Time for some hints...

Hint 1:

Generally, in the puzzles of this series, @Stiv says, "begin by solving the ...", but, oh no! I can't say that, 'cause you need to solve two puzzles simultaneously!

Hint 2:

Well, the numbers are in red...

  • $\begingroup$ I thought I knew what was going on, then quickly realised I really don't :P $\endgroup$ Sep 14, 2022 at 23:12
  • $\begingroup$ @BeastlyGerbil I'm not surprized that you found it. Who told you were wrong? :) And everyone, please upvote this comment if you think you know what to do. If no one is on the right track, I'll consider adding a hint or two. Well, I can only help you to identity the puzzle, but the deduction is your task. I give you one more hint, look at the bottom row. Edit: This puzzle is a little bit different from the other puzzles of the series. Look at the title. $\endgroup$
    – ACB
    Sep 15, 2022 at 1:09
  • $\begingroup$ Let me know your progress (in comments with spoilers, or even partial answers are welcomed) so I can get an idea. I will add a hint in a few hours. $\endgroup$
    – ACB
    Sep 15, 2022 at 11:32
  • $\begingroup$ Oh I didn't get very far at all, but rot13(Fb V'q nyernql ernyvfrq gur 0u a0 uvag (abg fher vs gur 'onol' ovg vf eryrinag), naq gurer'f boivbhfyl fbzr fbeg bs ababtenz gbb, unira'g jbexrq bhg jung gur fdhnerf naq pvepyrf ner gubhtu be jung 'bireynccrq' zrnaf urer. V gevrq svyyvat n srj pryyf ohg ernyvfrq V qvqa'g xabj ubj gur chmmyrf vagrenpgrq naq jvgubhg gur zvffvat bar V unq ab punapr. V erpxba n fznyyvfu uvag nobhg zrpunavpf/gur zvffvat chmmyr glcr jbhyq or tbbq!) $\endgroup$ Sep 15, 2022 at 14:02
  • $\begingroup$ @BeastlyGerbil Nice! You've got it ! And yeah, it's not important, baby :P. I've given a small clue in my first comment above. Plus, adding a hint now. $\endgroup$
    – ACB
    Sep 15, 2022 at 14:10

1 Answer 1



This is a mashup of nonogram and 0h n0.
Every cell in the grid must be shaded either blue or red. The clues on the outside give the number of cells in each consecutive run of red cells, separated by at least one blue cell. The clues in the grid are always shaded blue, and give the total number of blue cells that can be seen in each of the four cardinal directions before being blocked by a red cell.
A circle in the grid has the same value as the associated circle in that column’s nonogram clue. Likewise, a square in the grid has the same value as the associated square in that row’s nonogram clue.



Name of puzzle:

[Thanks to Stiv for aha/image!]
If we look at each 3*3 block of the solved grid, the blue cells spell out THIS IS AN OHNOGRAM! Puzzle Name Image

Solve path:

We can begin with some nonogram deductions, treating the shapes purely as unknown, non-zero numbers.
Step 1

By 0h n0 rules, the C4 circle must have a value of 2. We can then complete the column.
Step 2

0h n0 rules say that the C5 circle must be at least 2, and nonogram rules say that it can be at most 2, so it is exactly 2. Some nonogram logic follows.
Step 3

0h n0 says the C3 circle must be at least 4, which places one red cell in the column with nonogram. Nonogram then says the R5 square must be 2, so we can place a couple of cells using 0h n0, and another using nonogram. Now 0h n0 says the C3 circle must be 5, and we can finish the column.
Step 4

0h n0 says the C1 circle must be 5, and we can complete the column. We can also trivially say that the R11 square must be 2, which doesn't give any extra cells.
Step 5

Nonogram says the R1 square must be 3, and we can place a few cells from there. Then, 0h n0 says the R2 square must be 2, and nonogram gives even more cells.
Step 6

0h n0 says the bottom C7 circle must be 1, and nonogram places cells. From there, 0h n0 says the top C7 circle must be 4, and we finish the row. Nonogram logic follows.
Step 7

C6 circle and C9 circle can both be shown to be 3 by 0h n0, and C11 circle can be shown as 2 by 0h n0 and nonogram. Nonogram finishes the grid, and we can trivially fill in the rest of the numbers (C8 circle and R7 square both as 2).
Step 8

  • $\begingroup$ Yep, writing up solve path right now, and will re-check the solution $\endgroup$
    – SeptaCube
    Sep 15, 2022 at 18:22
  • $\begingroup$ Oh, rot13(Vf gurer creuncf na npghny svany rkgenpgvba gb trg gur anzr bs gur chmmyr?) $\endgroup$
    – SeptaCube
    Sep 15, 2022 at 18:24
  • 4
    $\begingroup$ Add some lines like this and rot13(vg fcryyf bhg va oybpx yrggref, GUVF VF NA BUABTENZ)! $\endgroup$
    – Stiv
    Sep 15, 2022 at 18:51
  • $\begingroup$ @Stiv or is it rot13(GUVF VF N ABUABTENZ) $\endgroup$
    – caPNCApn
    Sep 15, 2022 at 19:00
  • $\begingroup$ @cap Ooh, good point! Ambiguous, but satisfying either way... :) $\endgroup$
    – Stiv
    Sep 15, 2022 at 19:02

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