Quadritaisho: An introduction

Question

_{TL;DR: Nikoli's Tentai Show, but with four different types of symmetry.}

Rules of Quadritaisho:

• Divide the grid into rooms by drawing borders along grid lines.
• Each room contains exactly one symbol (circle, diamond, vertical bar, or horizontal bar).
• Rooms with a vertical bar | are symmetrical along an axis which runs vertically through the bar.
• Rooms with a horizontal bar — are symmetrical along an axis which runs horizontally through the bar.
• Rooms with a circle ● are 180° rotationally symmetrical, with the circle in the central point.
• Rooms with a diamond ◆ are 90° rotationally symmetrical, with the diamond in the central point.

An example of a valid room with each symbol:

The puzzle:

Solve on Penpa+

Answer 1

The complete grid:

Step by step solution:

Start with some basic deductions to draw some edges:

Each area must contain a single symbol (black edges).

Symmetry rules for those edges, and the grid border (green edges)

Now, lets look at cells that can only be reached from a single symbol.

Three cells top-right that must be with the vertical bar below.

Two cells to the left of those that must be with the top-centre diamond.

Two cells at the bottom that must be with the vertical bar bottom-centre.


Applying symmetry rules to those cells (and ensuring those in the top right reach to the vertical bar) gets us to:


Now the marked cell bottom-right must be part of the diamond, and the symmetry rules fully define that region (green).
That defines bottom-centre region (blue).
We can also complete the horizontal-bar region at the top (blue).
And add a couple more edges (red) from symmetry rules.

Now the marked cell in R5C2 can only be part of the remaining horizontal-bar region.
So all the green cells must part of that region. And applying the symmetry rules to the known edges completes that region.
And finally applying symmetry rules to the circle regions (starting with the upper one) completes the grid (blue).

Answer 2

Here's the solution. I'm pretty sure the solution is unique: