Walkthrough: Symmetric LITS 2, 2015 24HPC set

If you’ve struggled with that second symmtric LITS from my 24h set, here’s how to solve it.

Let’s start with the puzzle.



Aside from the basics, there are two techniques involved in here. First, there are some pairs of cells of which at least one needs to be shaded. This information can be mirrored. Below, the dark green dots mark pairs of which one must shaded due to normal LITS rules. These mirror to the light green dots.symlits-2-dots

The second ingredient involves overlaps between opposite areas. For example, looking at the top left and bottom right corner, the areas intersect in a 3×3 square (yellow). Both have the same number of shaded cells in this square, so we can deduce that the two pairs of light green cells have the same number of shaded cells, as well.symlits-2-areas

At this point, give it another try, you’ll probably make it through.


A few cells can be shaded immediately. Also, for parity reasons, the central cell must be unshaded. (The solution has an even number of shaded cells.)


Around the bottom left and top right corner, we find a dot in the bottom two cells of column 3, mirrored to the top of column 7. By intersecting the bottom left and top right areas, we can deduce that the first two cells of row 6  get a dot, mirrored to the end for row 4.


A similar argument chain works with the other pair of corners.


There’s different ways forward now. We can place some more dots, because some of the dots we have need to escape:


Or we deduce that the central right area needs to be a T.


Together, we can make good progress using standard arguments. Without using equal shape rules, we can reach this state.


Now things fall into place nicely.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s