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logical deduction – Shaka/shawaka: Trominoes

Once more.. it is me. 🙂

So Let’s go…

With the assistance of marginal regulation talked about in my earlier reply, we will progress.

The apparent place to begin is …

E12-G13 the place an L-tromino is confirmed.

Then if we transfer to the highest proper cell

we see that no triangle suits there. And since K1 additionally needs to be empty, so is L1. It is an I-tromino then! We are able to fill some cells.


Now let’s do that.


If this occurs, H1 and G2 shall be empty cells, however then it causes a tetromino. Thus it reveals that H1 needs to be stuffed.


We are able to discover that J3

is presently surrounded on three sides, due to this fact it should even be an empty cell (keep in mind this logic). It says that the adjoining cell additionally needs to be empty. Therefore we will full one other L-tromino (proven above).

Logic reveals that,

J3 can not lengthen to an I-piece (Trace: think about K5). It should be part of an L-piece. So H4 is cleared.

Additionally we will discover that if M6 is stuffed with a triangle,

we’ll get misplaced in an O-shape, which isn’t a tromino. Thus it’s empty together with the cell beneath it.


Now think about the area I6-L8.

It could possibly comprise an L-tromino, and there are two possibilies. We are able to attempt them out



Now we see the vacancy of

I7, together with its companion H7.

Then we will fill G7.

with a prediction on F6.

After which G8 too.


Now we will full our very first step noting that there’s just one risk for F11. It’s..


Then we will nearly fully fill the correct half of the grid.

I’m not going to clarify it, because it consists easy logical and trial and error strategies.


Then, the underside left half.

We are able to additionally fill this by easy chasing.


Now, this was the half I used to be caught in, till I spotted that,

I could make a BIG L-tromino.


From right here, the remaining is fairly straightforward.

And our finalized grid is..


… stuffed with trominoes!



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