12
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This is an entry for Fortnightly Topic Challenge #44: Introduce a new grid deduction genre to the community

I had an idea for what I think is a new grid deduction puzzle which has some aspects of Statue Park and some from Skyscrapers, so Office Park seems a good temporary name. It may well have appeared before, but I don't recall ever seeing anything like it. The closest I can find on PSE is a 3-D Statue Park by jafe. This one isn't too hard, befitting an introduction.

Like Statue Park, there is a set of shapes to place on a grid, but these shapes are three-dimensional polycubes...see the link for pictures of the 8 tetracubes (4 cubes, of course). Blocks must be placed without cantilevers, so there cannot be "air" underneath a hanging cube. Blocks must be placed so that no two blocks are touching, not even diagonally, and such that the empty squares form one orthogonally connected region.

Clues are given in a manner similar to Skyscrapers, where a number outside of the grid indicates the number of blocks that can be seen when looking along the adjacent row/column; the usual rules of impaired vision from Skyscrapers apply. Note that seeing two different levels of the same block only counts once.

For this puzzle, the shapes to be placed are the 7 "free" tetracubes, that is the ones unique up to reflection and rotation. More details on the pieces, including possible layouts, are given below. I hope you enjoy!

Grid

Solver Helps

Pieces

I
  1111 or 4

L
  31 or 112 or 111
                 1

O
  22 or 11
        11

T
  121 or  1
         111

V (this is the one that has left- and right-handed versions)
  12 or 11
  1     2

W
  12
   1

S
  11  or  11
   11    11

Text Version

   -----------------
 1 | | | | | | | | | 
   -----------------
   | | | | | | | | | 4
   -----------------
 2 | | | | | | | | | 
   -----------------
   | | | | | | | | | 2
   -----------------
   | | | | | | | | | 
   -----------------
   | | | | | | | | | 
   -----------------
   | | | | | | | | | 
   -----------------
   | | | | | | | | |
   ----------------- 
      3   3   1
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I think the crucial breakthrough is

There is a 4 in row 2, which mean it must see 4 different blocks with heights from 1-4 each. Since there is only 1 block with height 4(I) and 1 block with height 3(L), those must belong to that row.

Following that, we look at column 2. That column sees 3 blocks, so it must see blocks of 3 different heights. The remaining blocks are all of height 2, so the I must be in that column. Since the I must be in column 2 and these blocks cannot even touch diagonally, then the block that must provide the height of 2 in row 2 must be the O. (all the rest of the blocks will touch either the height of 3 block or height of 1 block diagonally). For similar reasons, the block providing the height of 1 in row 2 must be the T block.

So, just from this, we can get up till here:

OP_1

Next, if both the L and O blocks extend downwards, then we cannot satisfy row 1. So, L must extend downwards and O upwards to satsify constraints of row 1 and row 3.

OP_2

Now, another deduction: There are 3 pieces left but each of the '3' columns see 2 pieces of their own. This means the S piece is the first piece they see and it extends across both columns, using with its length of 2 in between the columns. Using grey colour to mark the areas that cannot be occupied by a block, we get:

OP_3

Now, it is easy to see that the cell in R5C4 must be of height 2. That means that is a V block and we can fill that in.

OP_4

For the final step, we must satisfy the constraints of both the 2nd column and 4th rows, so there is only one orientation for the W block and we are done.

OP_5

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  • 1
    $\begingroup$ Nailed it! Exactly the solution path I expected. Hope you enjoyed! $\endgroup$ – Jeremy Dover Nov 30 '20 at 16:48

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