1 The nonogram is underconstrained under normal nonogram rules. Instead, the solver must use the crossword conventions given in the puzzle (reproduced below) to arrive at the unique solution that satisfies them. 1. The grid must have 180-degree rotational symmetry 2. Every white square must be a part of an across entry and a down entry 3. Every entry in the grid must be at least three letters long Rules 2 and 3 together imply that every white square must be a part of a row and a column of 3 or more white squares (3-square rule). A detailed solve path is given in the following pages. For the solution to the resulting crossword, go to the last page. Nonogram Crossword Samuel Yeom
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1
The nonogram is underconstrained under normal nonogram rules.
Instead, the solver must use the crossword conventions given in the
puzzle (reproduced below) to arrive at the unique solution that satisfies
them.
1. The grid must have 180-degree rotational symmetry
2. Every white square must be a part of an across entry and a
down entry
3. Every entry in the grid must be at least three letters long
Rules 2 and 3 together imply that every white square must be a part of
a row and a column of 3 or more white squares (3-square rule).
A detailed solve path is given in the following pages. For the solution to
the resulting crossword, go to the last page.
Nonogram CrosswordSamuel Yeom
2
2 4 1 3
1 1 1 1 1 1 1
2 1 1 2 3 1 1 1 1 4 2 2 1 1 2
1 2
1 1
1 1
2
1
4
1 2
4 4 X
2 1
4
1
2
1 1
1 1
2 1
By rotational symmetry, the middle row must be palindromic. Since
there is an even number (2) of groups of black squares in the middle
row, the center square must be white. This creates two spaces of 7
squares, each of which must fit a group of 4 black. Therefore, the
middles of these spaces must be black.
3
2 4 1 3
1 1 1 1 1 1 1
2 1 1 2 3 1 1 1 1 4 2 2 1 1 2
1 2
1 1
1 1
2
1
4
1 2
4 4 X X X X X X X
2 1
4
1
2
1 1
1 1
2 1
The red squares are either both black or both white by rotational
symmetry. If they were both black, the center square would violate the
3-square rule. Therefore, they must be both white, and then there is
only one way to complete the middle row while satisfying the 3-square
rule.
4
2 4 1 3
1 1 1 1 1 1 1
2 1 1 2 3 1 1 1 1 4 2 2 1 1 2
1 2 X X X X X X X X
1 1 X X X X X X X X
1 1 X X X X X X X X
2 X X X X X X X X
1 X X X X X X X X
4 X X X X X X X X
1 2 X X X X
4 4 X X X X X X X
2 1 X X X X
4 X X X X X X X X
1 X X X X X X X X
2 X X X X X X X X
1 1 X X X X X X X X
1 1 X X X X X X X X
2 1 X X X X X X X X
We can mark much of the first four columns and the last four columns
as white based on the column clues.
5
2 4 1 3
1 1 1 1 1 1 1
2 1 1 2 3 1 1 1 1 4 2 2 1 1 2
1 2 X X X X X X X X
1 1 X X X X X X X X
1 1 X X X X X X X X
2 X X X X X X X X
1 X X X X X X X X
4 X X X X X X X X
1 2 X X X X X X
4 4 X X X X X X X
2 1 X X X X X X
4 X X X X X X X X
1 X X X X X X X X
2 X X X X X X X X
1 1 X X X X X X X X
1 1 X X X X X X X X
2 1 X X X X X X X X
The red square must be white because of the 2 in its row clue. We can
then fill in the 1st column and, by the 3-square rule in the 7th row, the
4th column as well. The right side of the grid is also filled in by
rotational symmetry. (From now on, rotational symmetry will not be
explicitly mentioned.)
6
2 4 1 3
1 1 1 1 1 1 1
2 1 1 2 3 1 1 1 1 4 2 2 1 1 2
1 2 X X X X X X X X
1 1 X X X X X X X X
1 1 X X X X X X X X
2 X X X X X X X X X
1 X X X X X X X X X
4 X X X X X X X X X
1 2 X X X X X X X X X X X X
4 4 X X X X X X X
2 1 X X X X X X X X X X X X
4 X X X X X X X X X
1 X X X X X X X X X
2 X X X X X X X X X
1 1 X X X X X X X X
1 1 X X X X X X X X
2 1 X X X X X X X X
We complete the 9th row, filling in the red square as black. If the red
square is part of the 3 in its column clue, there is not enough room to
fit the 2 and 1 in the upper half of the column while leaving 3 spaces
between every group of black squares (3-square rule). Therefore, it
must be part of the 1, and the group of 3 black squares must be at the
bottom due to the 3-square rule.
7
2 4 1 3
1 1 1 1 1 1 1
2 1 1 2 3 1 1 1 1 4 2 2 1 1 2
1 2 X X X X X X X X X X X X
1 1 X X X X X X X X X X X X X
1 1 X X X X X X X X X X X X X
2 X X X X X X X X X
1 X X X X X X X X X X
4 X X X X X X X X X X
1 2 X X X X X X X X X X X X
4 4 X X X X X X X
2 1 X X X X X X X X X X X X
4 X X X X X X X X X X
1 X X X X X X X X X X
2 X X X X X X X X X
1 1 X X X X X X X X X X X X X
1 1 X X X X X X X X X X X X X
2 1 X X X X X X X X X X X X
Now there is only one way to fit the group of 4 squares in the 6th
column while satisfying the 3-square rule. We can then complete the
first three rows easily.
8
2 4 1 3
1 1 1 1 1 1 1
2 1 1 2 3 1 1 1 1 4 2 2 1 1 2
1 2 X X X X X X X X X X X X
1 1 X X X X X X X X X X X X X
1 1 X X X X X X X X X X X X X
2 X X X X X X X X X X X X X
1 X X X X X X X X X X X X X X
4 X X X X X X X X X X X
1 2 X X X X X X X X X X X X
4 4 X X X X X X X
2 1 X X X X X X X X X X X X
4 X X X X X X X X X X X
1 X X X X X X X X X X X X X X
2 X X X X X X X X X X X X X
1 1 X X X X X X X X X X X X X
1 1 X X X X X X X X X X X X X
2 1 X X X X X X X X X X X X
The group of 4 black squares in the 6th row is forced, and the rest of
the grid follows.
9
Once the nonogram is correctly solved, the solver is given the clues to
the crossword. The nonogram grid is mostly consistent with the clues,
but it is black in a few places where letters should go. Reading the
letters in these places, we get AMOEBA as the answer.
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