Find the G-spot

Logic Level 3

B2, C2 or D2 B4, C4 or D4 B1, C1 or D1 B3, C3 or D3

S Broekhuis by S Broekhuis , 50, Netherlands

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

S Broekhuis
Jan 25, 2021

There are 9 ways to fill the grid with only one where M and S are adjacent: In that solution we find G to be in cell D3 so the answer is "B3, C3 or D3".

1 Helpful 0 Interesting 1 Brilliant 0 Confused
Saya Suka
Jan 26, 2021

Let's name the puzzle pieces F, J, O & S (by their positions at the corner of each piece) for easier reference.

We're told that M on O-piece and S on S-piece are adjacent, including diagonally. M is at the top of the piece and S is at the corner of North and East legs. O has a height of 3 and S 2, so they cannot be stacked together on top of the other on a 4 by 4 board with a combined height of 3 + 2 = 5 > 4. With the S-piece's East leg being a hindrance, S can only be on M's right in 2 ways relative to each other, but wherever the combo pieces are on the board, considering that the other pieces with only South & short West legs (of only 1 box) both, the top of M is either outside of the board, or inaccessible in 2 ways : either the space is physically blocked by R of S-piece from the East side (to match the West legs of other pieces) or the open path from the right is far too deep compared to the shortness of the West legs. If it isn't the top of M that's unreachable, then it just means that another blocked space would be created the same way, only that it's on the opposite direction, at the bottom of O.

Going back to the original compositions of O-S pieces, the combined height is 3 ~ 4 depending on how we arrange them, but the width is constant at 3. The East leg of S-piece, T can only be on column 3 or 4 on row B or C, but we know that the other pieces have appendage on their left sides, so they can only be on the right side that may actually allow some room for those legs. As T must occupy the middle of the 3rd column, J-piece with the longer South leg cannot find any space to squeeze in there (3rd column) and has to monopolize most of the 4th. That would push F-piece to the middle column by J-piece's J-K-L on its right and O-S combo's M-N-O on its left.

Now that we know F-G must be on the 3rd column, we may conclude that the last option must be the answer as the only one with any coordinates with column 3 in it.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Basically, the reasoning came from this sequence below :

1) M & S are adjacent but S cannot be on top of M for a total combined height of 3 (from M's) + 2 (from S's) = 5 that's outside the possible range of 4 × 4 board

2) therefore, S must occupy the same column as M's P for M-S pieces total combined width of 3

3) since either M's N or S's T (or both) will occupy at least one of the middle rows B & C, H can only be placed in an inward direction compared to its long southern leg, so that JKL leg must take the easternmost column 4

4) because JKL from H-piece already occupied 3 out of 4 grids within column 4 the same way that MNO from M-piece and RS-P from M-S combo piece both did within columns 1 & 2 respectively, FG with a height of 2 can never fit inside of any of the three rows mentioned before (neither 4 nor 1 nor 2), so the only possibility left is for G to be on column 3.

Saya Suka - 1 month, 1 week ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...