| From the introduction, each row and each column of the 6x6 Word Square
contains the characters PLANE and a blank cell in some order, so that no
character or space can be repeated in a row or a column. By clue 7, the
1st letter in both rows 3 and 4 is A; there are two possible arrangements
given this information: 1) A is in row 3, column 1 and row 4, column 2,
with a space in row 4, column 1; or 2) A is in row 3, column 2 and row 4,
column 1, with a space in row 3, column 1. Using both these arrangements,
we will recover more of the letters in the Word Square. By clue 1, we have
L then P down column 1. If L were in row 5 and P in row 6, clue 9 conflicts.
So, L is in column 1, row 1 and P in column 1, row 2. Then N is in column 1,
row 6 (clue 4) and E in column 1, row 5. By clue 11, P is 1st down column 4.
P cannot be in row 2 of column 4, with space in row 1, since we already have
a P in row 2; so P is in column 4, row 1. E is the 2nd letter down column 2
(3), so in arrangement 2) E would be in column 2, row 2. Since E isn't the
2nd letter in row 3 (12), E would also have to be in row 2 of arrangement 1).
In both cases, there would be a letter rather than a space in column 2, row 1.
Since the P and L are already used in row 1 and the A and E are already in
column 2, N must be in column 2, row 1. By clue 2, A is the 4th letter down
column 4 and must be in row 5. E is the 5th letter in row 1 (8) and is thus
in column 5 or 6; by clue 10, E in column 5 can't be in row 1; E is in
column 6, row 1. Then by clue 13, in column 6 space is in row 5 and A in
row 6. By clue 10, P is followed immediately by E down column 5; P is in
row 3 or 5. However, if P is in row 3 and E in row 4 of column 5, the P
is also in row 4 of column 4 (5)--no, since there is already a P in column 4.
So, P is in column 5, row 5 and E in column 5, row 6. In row 5, N must be
in column 3 and L in column 2. Since A is the 4th letter down column 4 (2),
column 4, row 6 must contain a letter--L is the only one that doesn't
appear in column 4 or row 6. We now will now further test arrangements
1) and 2). Trying 1), with A in clue 7 in row 3, column 1 and row 4,
column 2, with a space in row 4, column 1, by clue 5 in row 4 P would be
in column 3 and E in column 4. By elimination, E in row 3 would be in
column 3. Since E isn't the 2nd letter across row 3 (12), column 2 of
row 3 would have P in it, with the column 2 space in row 6. However, P
in row 6 would be in column 3, where we would already have a P. Therefore,
arrangement 1) fails. So, by clue 7, A is in row 3, column 2 and row 4,
column 1, with a space in row 3, column 1. By clue 12, E isn't the 2nd
letter across row 3; E must be in column 4 of row 3 and by elimination
in column 3, row 4--preceded by P in column 2, row 4 (5). The column 2
space is in row 6, with P in row 6 in column 3. By elimination, the P
in column 6 is in row 3; then in row 3, L must be column 2 and N in
column 5. By clue 6, L is 4th in row 4 and must be in column 5. Then
L in column 6 is in row 2 with N in row 4. The row 4 space is in
column 4, with column 4, row 2 being N. By clue 14, space in column 3
is in row 1, with A in row 2. Finally, in column 5, A is in row 1 and
space in row 2. In sum, Word Square 7 is as follows:
| | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | L | N | | P | A | E |
| 2 | P | E | A | N | | L |
| 3 | | A | L | E | N | P |
| 4 | A | P | E | | L | N |
| 5 | E | L | N | A | P | |
| 6 | N | | P | L | E | A |
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