# (3 of 11: Akari) What is Pyramid Cult's Favorite Car?

Dear PSE users and moderators,
I’m new here in PSE, but I really need your help. There was this person who gave me a black envelope consisting 10+1 pages of puzzles, and also a scribble saying: “Find our favorites and you will be accepted to join our ‘pyramid cult’. Feel free to ask for help from your beloved friends on PSE. They will surely guide you into all the truth.” I’m also a newbie on grid puzzles, so, could you please give me any hint to solve these? It’s getting harder and harder later on..
- athin

Jump to the first page: #1 Numberlink | Previous page: #2 Moon-or-Sun | Next page: #4 Hitori

Rules:

1. Place light bulbs (O) according to the following rules.
2. Light bulbs may be placed in any of the white triangles, the number in the triangle shows how many light bulbs are next to it, adjacent to its side.
3. Each light bulb illuminates from bulb to black triangle or outer frame in its 3-directions (parallel with one of the board sides).
4. Every white triangle must be illuminated and a light bulb can not illuminate another light bulb.

Special thanks to chaotic_iak for testing this puzzle series!

• “I’m new here in PSE” says someone who’s been here almost 3 years and just passed 12345 rep. Hah! :)
– Rubio
Commented Jul 29, 2019 at 12:52

Finished grid:

The car:

The letter in cells with lamps in them spell FERRARI.

Reasoning

I'm going use red for lamps, yellow for illuminated cells and grey for not-yet-illuminated cells that cannot have a lamp in them.

To start off, we notice that the cell marked with orange has to be a lamp. If it wasn't, completing the 2 would illuminate two sides of the 2 below it (circled) and leave no room for two lamps.

To complete the 2, we have to illuminate one side of the 2 below it. This means that the cell below the corner 2 has to be a lamp, because only one of the other two cells next to the 2 can be a lamp (the other being illuminated from above).

In the central triangle, the bottom cell cannot be a lamp because this would force the cell next to the central-left 1 to be a lamp. This in turn would illuminate two sides of the 2 above it (circled), leaving no room for two lamps around it.

Since the central triangle's bottom cell must be empty, we now know where the two lamps of the 2 next to it go. Fill those in.

From here, the 1 next to the 2 we just filled only has one option. This leaves only one option for the 1's in the bottom left corner as well.

The 2 in the bottom right corner is forced now, and filling that in solves the 2 above it.

The 1 above the central triangle only has one option as well. Also, add lamps to the bottom centre and the centralmost cell because they can't be illuminated from anywhere else.

The circled cell cannot be a lamp, because it would make it impossible to complete the 2 above it.

This cell can only be illuminated from one place now.

The circled cell cannot be a lamp, because that would block off the only cells that can illuminate the dark cell on the right.

Only one way to complete the grid now.

• Nicely done! :D Commented Jul 29, 2019 at 14:23

My solution:

Step 1:

The cyan cell would prevent both red cells from being lamps - a contradiction. Hence the first four lamps can be positioned.

Step 2:

If the cyan cell was a lamp, both red cells can't be lamps, again a contradiction. So both '1' triangles adjacent to it can be fixed to lamps.

Step 3:

Again, if the cyan cell was a lamp, both red cells can't be lamps. So the '1' triangle adjacent to it can be fixed to a lamp. The central region needs illuminating, and so a lamp must be placed in the centre. The associated '2' triangle can now be completed.

Step 4:

The yellow cell needs illuminating, and itself is the only cell that can do this. This in turn forces both of the green cells to be lamps.

Step 5:

Again, the cyan cell forces a contradiction. The red cell to the right of the '2' triangle still needs illuminating, and this can only be done by itself.

Step 6:

All the cells left to light up are in yellow. The lower two force a lamp, then the remaining '2' triangle is forced, which forces the final cell.