Well, I believe I've found your beautiful BEE
! Sorry for being so clumsy!
. . . . . . . .
. . . E . . . .
. . . . . . E .
. . . . . . . E
. . E . . . E .
/-\
. . . .|.|. . .
| |
. . . .|.|. . .
| |
. . . .|E|. . .
\-/
(going down)
Explanation:
First, the X
can't be a B
because of the following chain of deductions where all the numbers are E
:
. . . . . . . .
. . . E 7 X . .
. . . . 6 . E .
. . . . 5 . . E
. . E . 4 . E .
. . . . 3 1 . .
. . . . 2 . . .
. . . . E . . .
Then the X
here can't be a B
for almost the same reason:
. . . . X . . .
. . . E 7 E . .
. . . . 6 . E .
. . . . 5 . . E
. . E . 4 . E .
. . . . 3 1 . .
. . . . 2 . . .
. . . . E . . .
The X here can't be a B
because of the !
- it can't be a B
because of !41
but neither can it be an E
because of X!4
:
. . . . E . . .
. . . E . E . .
. . X . . . E .
. . . ! . . . E
. . E . 4 . E .
. . . . 3 1 . .
. . . . 2 . . .
. . . . E . . .
The X
can't be a B
here because of the !
- if the !
is a B
then !31
is a BEE
, but if not XE!
is a BEE
:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. X . . . . . E
. . E . . . E .
. . . ! 3 1 . .
. . . . 2 . . .
. . . . E . . .
Now, if X
was a B
here, we'd have this, and the same argument applies to the space marked 1
:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E . . . E .
. . . X 3 1 . .
. . . . 2 . . .
. . . . E . . .
Obviously the space marked X
can't be B
here:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E . . . E .
. . . E + E . .
. . . . X . . .
. . . . E . . .
Now, if the spot +
in the above diagram were not to be a B
, then either:
the 6th row is all E
s, or the BEE
is on the 6th row
So case bashing assuming Y
is a B
, we have:
- Here and in its mirror image across the
EEX
line, 1
through 6
are all B
s, and then we have a BEE
at 6EX
:
. . . . E . . .
. . 4 E 3 E 2 .
. . E . . . E .
. E 5 . . . 1 E
. . E . . . E .
. . 6 E X E Y .
. . . . E . . .
. . . . E . . .
- Here and in its mirror image across the
EEX!
line, the 1
is a B
, but then we have a BEE
at either 1!E
or !XE
:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . 1 . E
. . E . ! . E .
. . E E X E E Y
. . . . E . . .
. . . . E . . .
- Finally, here,
1
has to be an E
, so Z
has to be a B
, so one of ZE!
and !EE
is a B
. . . . E . . .
. . . E . E . .
. . E . . . E .
Z E ! . . . . E
1 . E . . . E .
Y E E E X E E E
. . . . E . . .
. . . . E . . .
- So then we must have the following:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E . . . E .
E E E E E E E E
. . . . E . . .
. . . . E . . .
- From here, we need to check the fifth line. Assume that the line has a
B
. Then:
- If the
X
is a B
, the 1
is a B
, leading to a BEE
going up from the 1
:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E . X . E .
E E E E E E E E
. . . . E . 1 .
. . . . E . . .
- If the
X
is a B
and the Y
is an E
, we can chase B
s around with numbers to get 2EY
as a BEE
. The same applies with the mirror image around the EEEE
line (we can ignore the stray E
)
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E Y E X E .
E E E E E E E E
. . . 2 E 1 . .
. . . . E . . .
- If both the
X
and the Y
are B
s, in the following series of diagrams the Z
shows a place that can't be an E
because the !
would result in a contradiction of YE!
and !EZ
:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E X E Y E .
E E E E E E E E
. . . ! E Z . .
. . . . E . . .
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E X E Y E .
E E E E E E E E
. . . Z E ! . .
. . . . E . . .
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E X E Y E Z
E E E E E E E E
. . . . E ! . .
. . . . E . . .
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. Z E X E Y E .
E E E E E E E E
. . . ! E . . .
. . . . E . . .
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. ! E X E Y E .
E E E E E E E E
. Z . . E . . .
. . . . E . . .
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E X E Y E !
E E E E E E E E
. . . . E . . Z
. . . . E . . .
- So we've got quite a few squares that must now be
B
:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
- So now we can consider the
Z
s to be E
s in the following diagrams, and chase the E
s around with numbers to get another B
:
. . . 3 E 2 . .
. . . E . E . .
. 5 E 4 . 1 E Z
. E . . . . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
. . . 2 E 1 . .
. . . E . E . .
. 4 E 3 . Z E B
. E . . . . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
. . . 1 E Z . .
. . . E . E . .
. 3 E 2 . B E B
. E . . . . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
. . . Z E B . .
. . . E . E . .
. 2 E 1 . B E B
. E . . . . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
. . . B E B . .
. . . E . E . .
. 1 E Z . B E B
. E . . . . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
. . . B E B . .
. . . E . E . .
. Z E B . B E B
. E . . . . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
- Now the
XYZ
must all be E
s because if one of them were a B
, the other two would be E
and we would have two BEE
s:
. . . B E B . .
. . . E . E . .
. B E B . B E B
. E X Y Z . . E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
. . . B E B . .
. . . E . E . .
. B E B . B E B
. E E E X Y Z E
. B E B E B E B
E E E E E E E E
. B . B E B . B
. . . . E . . .
- If any of
XYZ
were a B
, we would have multiple BEE
s.
. . . B E B . .
. . . E . E . .
. B E B X B E B
. E E E E E E E
. B E B E B E B
E E E E E E E E
. B Y B E B Z B
. . . . E . . .
- If any of
XYZ
were a B
, all three would have to be, and there would be three BEE
s:
. . . B E B . .
. . X E Y E Z .
. B E B E B E B
. E E E E E E E
. B E B E B E B
E E E E E E E E
. B E B E B E B
. . . . E . . .
- If any of
WXYZ
were B
s, we would have multiple B
s:
. . X B E B Y .
. W E E E E E Z
. B E B E B E B
. E E E E E E E
. B E B E B E B
E E E E E E E E
. B E B E B E B
. . . . E . . .
- If
X
or Y
were E
s, we would have two BEE
s:
. X E B E B E Y
. E E E E E E E
. B E B E B E B
. E E E E E E E
. B E B E B E B
E E E E E E E E
. B E B E B E B
. . . . E . . .
WXYZ
must all be E
s otherwise we would have multiple BEE
s:
. B E B E B E B
. E E E E E E E
W B E B E B E B
. E E E E E E E
X B E B E B E B
E E E E E E E E
. B E B E B E B
. . . Y E Z . .
VWXYZ
must be E
s for the same reason:
. B E B E B E B
V E E E E E E E
E B E B E B E B
W E E E E E E E
E B E B E B E B
E E E E E E E E
X B E B E B E B
. . Y E E E Z .
- Now
XYZ
must be E
s for the same reason. Finally, we have W
also being E
for the same reason again. But then there are NO BEE
s in the grid:
X B E B E B E B
E E E E E E E E
E B E B E B E B
E E E E E E E E
E B E B E B E B
E E E E E E E E
E B E B E B E B
W Y E E E E E Z
- Phew! So now we know this must be the case (where we are assuming the lowercase
e
is not a B
):
. . . . E . . .
. . . E . E . .
. . E . . . E
. E . . . . . E
. . E E E E E .
E E E E e E E E
. . . . E . . .
. . . . E . . .
- Now if the
X
was a B
, the Y
must be a B
leading to multiple BEE
s. The same argument applies reflected in the EEEE
line (ignoring the stray E
):
. . . . E . . .
. . . E . E . .
. . E . . . E
. E . . . . . E
. X E E E E E .
E E E E E E E E
. . . Y E . . .
. . . . E . . .
- Finally, the
X
cannot be an E
otherwise we would have two BEE
s:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
X E E E E E E E
E E E E E E E E
. . . . E . . .
. . . . E . . .
- So our second assumption was wrong, and the fifth row must then consist of all
E
s. Now any square marked by a #
must be an E
because otherwise we have multiple BEE
s:
. . . . E . . .
. . . E . E . .
. . E . . . E .
# E # # # # # E
E E E E E E E E
E E E E E E E E
# # # # E # # #
. . . . E . . .
. . . . E . . .
. . . E . E . .
# # E # # # E #
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
# # # # E # # #
. . . . E . . .
# # # E # E # #
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
- And one last time... well, obviously there aren't any
BEE
s in here:
# # # # E # # #
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
E E E E E E E E
So then our initial assumption must be wrong, and we have this, with a BEE
already in there:
. . . . E . . .
. . . E . E . .
. . E . . . E .
. E . . . . . E
. . E . . . E .
. . . E B E . .
. . . . E . . .
. . . . E . . .
So there's the BEE
!