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Statue Park: Five

Five keys, five doors, seven guards, and only seven questionsFirst to Five PointsFive SlitherlinksFive Letter Cryptex CodeThe Ludicrous Loop: over a thousand cells of circular logic!Five Hats and Three LogiciansFive balls weighingStatue Park (Loop)Diabolical Deceptions: A 333rd Birthday Tribute to J.S. BachModified Intersection Puzzle

2

$begingroup$

This is a Statue Park puzzle (originally constructed for the 2019 24-Hour Puzzle Championship, as part of a Tarot card themed set -- no prizes for guessing which rank this puzzle was).

Rules of Statue Park:

• Shade some cells of the grid to form the given set of pieces. Pieces may be rotated or reflected.


• Pieces cannot be orthogonally adjacent (though they can touch at a corner).


• All unshaded cells must be (orthogonally) connected.


• Any cells with black circles must be shaded; any cells with white circles must be unshaded.

logical-deduction grid-deduction

share|improve this question

asked 51 mins ago

Deusovi♦Deusovi

64.6k6223281

$endgroup$

add a comment | 

2

$begingroup$

This is a Statue Park puzzle (originally constructed for the 2019 24-Hour Puzzle Championship, as part of a Tarot card themed set -- no prizes for guessing which rank this puzzle was).

Rules of Statue Park:

• Shade some cells of the grid to form the given set of pieces. Pieces may be rotated or reflected.


• Pieces cannot be orthogonally adjacent (though they can touch at a corner).


• All unshaded cells must be (orthogonally) connected.


• Any cells with black circles must be shaded; any cells with white circles must be unshaded.

logical-deduction grid-deduction

share|improve this question

asked 51 mins ago

Deusovi♦Deusovi

64.6k6223281

$endgroup$

add a comment | 

$begingroup$

This is a Statue Park puzzle (originally constructed for the 2019 24-Hour Puzzle Championship, as part of a Tarot card themed set -- no prizes for guessing which rank this puzzle was).

Rules of Statue Park:

• Shade some cells of the grid to form the given set of pieces. Pieces may be rotated or reflected.


• Pieces cannot be orthogonally adjacent (though they can touch at a corner).


• All unshaded cells must be (orthogonally) connected.


• Any cells with black circles must be shaded; any cells with white circles must be unshaded.

logical-deduction grid-deduction

share|improve this question

asked 51 mins ago

Deusovi♦Deusovi

64.6k6223281

$endgroup$

share|improve this question

asked 51 mins ago

Deusovi♦Deusovi

64.6k6223281

share|improve this question

asked 51 mins ago

Deusovi♦Deusovi

64.6k6223281

asked 51 mins ago

Deusovi♦Deusovi

64.6k6223281

asked 51 mins ago

Deusovi♦Deusovi

64.6k6223281

1 Answer
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2

$begingroup$

I believe this does the trick:

Rough line of reasoning:

The whole width of the puzzle is just enough to fit the four letters upright (putting any of them sideways won't work). The problem is the last 2 x 1 piece. This must go in the gap between two letters somehow.


Observe that for the white squares to stay contiguous, the I must be the second letter. It can't fit in the third space, and if it were in the first or fourth then it would close off a set of white squares.


With the I in place, we conjecture that the concave shape of the I's sides will permit the fitting of the 2 x 1 piece, combined with the empty space in the bottom half of the F. A quick thought confirms that the 2 x 1 piece cannot possibly fit between any other two letters. After trying to unsuccessfully fit the F in the leftmost spot such that there is room for the 2 x 1 piece, we can deduce that the F must be in the third place, to the immediate right of the I. We can find the correct placement of the F and the 2 x 1 piece from there. The rest is trivial.

share|improve this answer

answered 27 mins ago

greenturtle3141greenturtle3141

5,69012154

$endgroup$

add a comment | 

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2

$begingroup$

I believe this does the trick:

Rough line of reasoning:

The whole width of the puzzle is just enough to fit the four letters upright (putting any of them sideways won't work). The problem is the last 2 x 1 piece. This must go in the gap between two letters somehow.


Observe that for the white squares to stay contiguous, the I must be the second letter. It can't fit in the third space, and if it were in the first or fourth then it would close off a set of white squares.


With the I in place, we conjecture that the concave shape of the I's sides will permit the fitting of the 2 x 1 piece, combined with the empty space in the bottom half of the F. A quick thought confirms that the 2 x 1 piece cannot possibly fit between any other two letters. After trying to unsuccessfully fit the F in the leftmost spot such that there is room for the 2 x 1 piece, we can deduce that the F must be in the third place, to the immediate right of the I. We can find the correct placement of the F and the 2 x 1 piece from there. The rest is trivial.

share|improve this answer

answered 27 mins ago

greenturtle3141greenturtle3141

5,69012154

$endgroup$

add a comment | 

2

$begingroup$

I believe this does the trick:

Rough line of reasoning:

The whole width of the puzzle is just enough to fit the four letters upright (putting any of them sideways won't work). The problem is the last 2 x 1 piece. This must go in the gap between two letters somehow.


Observe that for the white squares to stay contiguous, the I must be the second letter. It can't fit in the third space, and if it were in the first or fourth then it would close off a set of white squares.


With the I in place, we conjecture that the concave shape of the I's sides will permit the fitting of the 2 x 1 piece, combined with the empty space in the bottom half of the F. A quick thought confirms that the 2 x 1 piece cannot possibly fit between any other two letters. After trying to unsuccessfully fit the F in the leftmost spot such that there is room for the 2 x 1 piece, we can deduce that the F must be in the third place, to the immediate right of the I. We can find the correct placement of the F and the 2 x 1 piece from there. The rest is trivial.

share|improve this answer

answered 27 mins ago

greenturtle3141greenturtle3141

5,69012154

$endgroup$

add a comment | 

$begingroup$

I believe this does the trick:

Rough line of reasoning:

The whole width of the puzzle is just enough to fit the four letters upright (putting any of them sideways won't work). The problem is the last 2 x 1 piece. This must go in the gap between two letters somehow.


Observe that for the white squares to stay contiguous, the I must be the second letter. It can't fit in the third space, and if it were in the first or fourth then it would close off a set of white squares.


With the I in place, we conjecture that the concave shape of the I's sides will permit the fitting of the 2 x 1 piece, combined with the empty space in the bottom half of the F. A quick thought confirms that the 2 x 1 piece cannot possibly fit between any other two letters. After trying to unsuccessfully fit the F in the leftmost spot such that there is room for the 2 x 1 piece, we can deduce that the F must be in the third place, to the immediate right of the I. We can find the correct placement of the F and the 2 x 1 piece from there. The rest is trivial.

share|improve this answer

answered 27 mins ago

greenturtle3141greenturtle3141

5,69012154

$endgroup$

share|improve this answer

answered 27 mins ago

greenturtle3141greenturtle3141

5,69012154

answered 27 mins ago

greenturtle3141greenturtle3141

5,69012154

answered 27 mins ago

greenturtle3141greenturtle3141

5,69012154