Mystery of the One Word Wordsearch

Question

How have you been? Please, come in, have a seat. Would you like a drink? Wine? Soda? No? Yes, I have orange juice... Ok, enjoy!

Anyway, I know how you love wordsearches, so I made one exclusively for you! I spent hours on it! I'm such a good friend! Oh, but it's special you see: It's in an 8x8 grid, and of course, the words can be left, right, up, up-right... any compass direction, you see. But the gimmick is that there's only one word you need to find: It's "BEE"! It only appears once! Oh, and to make it more devious, there are only two distinct letters in the entire wordsearch! "B" and "E", of course... Here, let me show it to you. Isn't it beautiful? I worked so ha-

splat

OH MY GOD

WHAT HAVE YOU DONE

D-Did you just SPILL YOUR ORANGE JUICE ON MY WONDERFUL WORDSEARCH??? ;-; HOW COULD YOU?!?

I spent so much time making this wordsearch... Now you'll never find the word and enjoy my beautiful wordsearch... :'(

(Or can you?)


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Beautifully crafted puzzle +1 – Mordechai Apr 09 '17 at 13:38 — Mordechai, Apr 09 '17 at 13:38

boboquack · Accepted Answer · 2017-11-10T06:25:57.630

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 Es, 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 Bs, 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 Bs 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 Bs, 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 Zs to be Es in the following diagrams, and chase the Es 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 Es because if one of them were a B, the other two would be E and we would have two BEEs:

. . . 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 BEEs.

. . . 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 BEEs:

. . . 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 Bs, we would have multiple Bs:

. . 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 Es, we would have two BEEs:

. 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 Es otherwise we would have multiple BEEs:

. 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 Es 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 Es for the same reason. Finally, we have W also being E for the same reason again. But then there are NO BEEs 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 BEEs. 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 BEEs:

. . . . 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 Es. Now any square marked by a # must be an E because otherwise we have multiple BEEs:

. . . . 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 again:

. . . . 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 again:

. . . . 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 BEEs 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!

A fantastic job well done! After deducing the "diamond" of Es, the reasoning I used to determine that the + isn't E was a little more abstract. I asserted that if the BEE's B was anywhere else, it won't stop the spread of E's into becoming the cute grid pattern, where we will either have multiple BEEs or none. Regardless, bravo on your absolutely concrete solution! — greenturtle3141, Apr 09 '17 at 13:51
Why such a short answer, did you run out of !><pre>s? Seriously, this answer deserves a bounty of recognition once eligible. — humn, Apr 10 '17 at 00:05
@humn Oh no, there's plenty of space for more! I've only used 10886 characters out of the available 30000! :D — boboquack, Apr 10 '17 at 00:07

Edward H · Answer 2 · 2022-02-26T18:58:11.233

Oh darn, is there any way I could get another cup of that, perhaps? ...No? Ok fine, here we go then...

Let JUICE-EE a diamond

If a BEE is formed anywhere (shaded gray in image below), then an EE elsewhere must not see a B, so they extend to a line of Es (arrows in image below). Several cells aren't a B because the Es meet up to create a secondary BEE.

Parity is for sHOR NETcessary

Consider any non-self-intersecting loop of _E_ segments, examplified below. All shaded cells have to be the same, since otherwise whenever some two shaded cells are distinct, at least two BEEs will form along the whole loop.

Shade the puzzle's cells as below.

We focus on the blue loop of _E_ segments. For contradiction assume all blue cells are E, then the green cells and the single purple E join to form a loop of _E_ segments, so all green cells must be E. Then the red and the yellow cells each form a _E_ loop and they're all E. More and more _E_ loops form until the whole grid is full of Es, which mustn't happen. Therefore every blue cell is a B.

Re-shade the cells as below.

Focus on the green _E_ loop. For contradiction assume all green cells are E. Then the red cells form various _E_ loops so all red cells are E, and the yellow cells form various _E_ loops so all yellow cells are E. Next the red loops extend to two of the pink cells, and the yellow loops extend to two of the orange cells, which are all E. Due to the pink and orange, the purple cells join the green cells and are all E. Finally all shaded cells in the image are E. The remaining white cells must all be B to prevent multiple BEEs, and there will be no BEE left on the grid. A contradiction is reached, so all green cells are B.

We atTANG E RINEconstruction of the BEE

I've completed your word search! Can I get a refill of my orange juice, now? Still no? D'oh...

Nautilus · Answer 3 · 2017-04-11T13:33:12.907

0

More of an intuitive solution than a systematical one, but I'll try my hand at this:

If we paint the puzzle like a chessboard, all the initially-shown Es except for the one at the bottom will be on a square of the same color. Apart from one exception, these rules must apply:

1. If a B neighbours an E, the next letter must be B.
2. If there are two Es next to each other, all the orthogonal or diagonal line they belong to must consist of Es.

Starting from the rightmost E and going diagonally to exploit the fact that all Es but one would be at identically-colored squares, and then making the same kind of movement for the E at the bottom, it can give us this:

B B B B E B B B
B B B E B E B B
E B E B B B E B
B E B B B B B E
E B E B B B E B
B B B E B E B B
B B B B E B B B
B B B E E E B B

It can be fixed with a small change:

B B B B E B B B
B B B E B E B B
E B E B B B E B
B E B B B B B E
E B E B B B E B
B B B E B E B B
B B B B E B B B
E E E E E E E E - Only the squares from e3 to e1 read as BEE.

edited Apr 11 '17 at 13:33

answered Apr 11 '17 at 12:54

Nautilus

6,534
10
32

How does your answer add to the identical ones already given? You should always look at existing answers before providing one of your own, to ensure you are not just adding a duplicate. This is not the first time you've been told this; please stop. – Rubio Apr 11 '17 at 13:52
2

It's NOT a duplicate. I used a different reasoning and gave a (slightly) different answer, so how the heck is it identical? Just because you hold a grudge against me doesn't mean every answer I give deserves being downvoted to death. – Nautilus Apr 11 '17 at 13:55
For the record, if someone downvoted because he/she thought it's too intuitive and not a "good enough" answer, I could somewhat understand it, but I can't stand false accusations like yours. – Nautilus Apr 11 '17 at 14:02
2

Wouldn't your two rules only apply if the location of BEE is already known? This is a start to an intuitive solution, but as it stands it's a little unconvincing because while you found a valid way to place BEE, I don't think you have proven that it can't be anywhere else, which imo is 1/2 - 2/3 of the puzzle. – greenturtle3141 Apr 11 '17 at 14:18


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Mystery of the One Word Wordsearch

3 Answers3


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