13
$\begingroup$

Since recently Athin showed Circles And Squares, a genre which I invented and am very proud of, I thought "Hey, why not give another one!" But this time, I'm sharing one that I made for both the Puzzler's Club and CTC Discord.

Even though the size might terrify you, it should be approachable, as well as some bits of logic which weren't present in Athin's puzzle. Should be a fun solve!

Rules:

  1. Shade some cells such that all shaded cells form one orthogonally connected area.
  2. Black circles are shaded and white circles are unshaded.
  3. Unshaded area must form squares.
  4. No 2x2 can be fully shaded.

Penpa link: https://tinyurl.com/79j8crjx

enter image description here

$\endgroup$
3
  • $\begingroup$ Do you mean that unshaded regions must be square? $\endgroup$
    – A username
    Mar 15 at 6:12
  • $\begingroup$ Yes, just search for Athin's puzzle to get an example $\endgroup$ Mar 15 at 7:26
  • $\begingroup$ Loving this new puzzle type, and thanks for setting up the penpa, it's way easier than transferring this to grid paper and trying to do it in pencil haha. $\endgroup$ Mar 15 at 17:42

1 Answer 1

12
$\begingroup$

Solution:

enter image description here

For this explanation:

  • Blue = shaded
  • Green = unshaded
  • Yellow = shaded on this move
  • Red = unshaded on this move

On the top row, rows of consecutive white circles must be nxn squares

enter image description here

And must have walls around them:

enter image description here

Likewise for the bottom row:
enter image description here
Certain cells must be unshaded because of the 2x2 rule
enter image description here
And the ones in the center of the top row force a 2x2 square
enter image description here
Likewise, another square is forced
enter image description here
A major breakthrough! These groups must be 3x3 squares because otherwise the squares marked pink would be shaded and disconnected. They cannot be any larger. They must have walls around them.
enter image description here
Some minor deductions help us sort out the area below that
enter image description here
The chunk in the bottom left must remain connected to the rest.
enter image description here
The chunk in the bottom right must connect to the rest, but there are tight restraints on where it can go...
enter image description here
This allows us to fill out the rest of that corner.
enter image description here
A few more deductions...
enter image description here
This square has to be a 2x2.
enter image description here
A few more deductions...
enter image description here
This cell must be shaded to break up the squares to either side.
enter image description here
Which opens up a bunch of stuff.
enter image description here
This square is at least a 2x2.
enter image description here
Because of the pink cell, the two nearby green cells cannot be connected into a 3x3.
enter image description here
This cell can't be connected to the cell to the bottom right, it must be on its own.
enter image description here
Which forces the states of these two.
enter image description here
This group can't be any larger than 2x2 because of the pink cell.
enter image description here
Which forces some more deductions...
enter image description here
Another simple deduction.

enter image description here
The red cell must be empty because otherwise there would be a 2x2 shaded square. It forces the neighbouring square to be 2x2.
enter image description here
Some more simple deductions solve that chunk.
enter image description here
The red cells must be empty because of the 2x2 shaded rule, and can't be connected because of the pink squares.
enter image description here
Certain squares must be filled in.
enter image description here
There are four possibilities for the central two white circles:
enter image description here
enter image description here
enter image description here
enter image description here
But only the last is legal.
enter image description here
These two red cells must form a 2x2 square:
enter image description here
Finishing the square:
enter image description here
A few more deductions and we're done!
enter image description here
Final solution:
enter image description here

Feel free to fix the formatting, spoilers are a pain.

$\endgroup$
1
  • $\begingroup$ Nice job, this is the correct solution! Though not the intended solve path, I'm sure there are much more ways to derive the solution :D Very proud of this genre indeed $\endgroup$ Mar 16 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.