This recent puzzle I created was well-received, so here's another one like it, except this time a bit larger and using homophones.

Create a word search with the following conditions:

  1. Use the fifteen words below. They can be spelled up, down, left, right, or any of the 4 diagonal directions.








  1. The size of the word search must be 7 rows by 7 columns. (Bottom right square not used.)

  2. The number of words that pass through each square must fit this pattern:

enter image description here

  1. The accepted answer will clearly and fully explain the flow of logic used from start to finish.
  • 1
    $\begingroup$ How does the bottom-left square have four words passing through it? There's only 3 directions for words to go. $\endgroup$
    – user88
    Commented Jun 12, 2015 at 4:18
  • 2
    $\begingroup$ @JoeZ. Yes, that's quite a hint in itself $\endgroup$
    – JLee
    Commented Jun 12, 2015 at 4:21
  • $\begingroup$ Are you positive that each word only shows up once in the puzzle? $\endgroup$
    – Bailey M
    Commented Jun 12, 2015 at 19:45
  • $\begingroup$ @BaileyM Logically, tea must appear at least twice, right? Tee could appear twice. I don't think there are any others though. $\endgroup$
    – JLee
    Commented Jun 12, 2015 at 19:52
  • $\begingroup$ Ahh, the question I should be asking is: Are tea's letters counted in both tear and team, or just in one of them? $\endgroup$
    – Bailey M
    Commented Jun 12, 2015 at 19:58

2 Answers 2


An old question, bringing back to life..

Here is an answer:


which was basically found by trial-and-error method including a bit of logic.


I will sometimes refer to columns as A,B,C,.. (from left to right), and rows as 1,2,3,.. (from top to bottom), to specify corresponding cells.

The starting point is obviously...

the bottom-left corner. Since four words overlap there, one word should contain another word within it. There are three such words: TEEM, TEAM and TEAR those contain TEE and TEA. The correct word among these cannot lie horizontally or vertically since both A6 and B7 has the value 1. Therefore it must lie along the diagonal. Thus A7 and B6 must be T and E respectively.

As middle cell can be either M or R, the word along the remaining cells of the diagonal should end with those letters i.e. one of those three words aforementioned, or could be a 3-letter word if it does not contain the middle cell; i.e. TEA or TEE. In both cases, E is the second letter (or middle letter) thus F2 should be E.
Starting cells

Those are the letters we can deduce logically.

(This part was already mentioned in Bob's answer. I'm repeating it here just for the completeness.)

From now on, there is no obvious logical path. So we have to take a risk. Guess!

Assume that the middle cell is R. Then the word lying on the diagonal, starting from A7 should be TEAR (C5 cannot be E, otherwise it will give only one word). Considering the rules given by OP, one word is counted only once, therefore the other word along the diagonal should be TEE (it does not share the R).

Since we agreed to count each word only once, we will not use the same word again and again intentionally. Thus the remaining cells should be filled with words those are at least 4-letters long. Keeping that in mind, to fit the density of purple cells on the diagonal, they should be filled as in the following diagram.
failed attempt


there are only two possible ways to get rid of the middle cell (shown in red). But there is only one 4-letter word ending with R and it's already used.

Thus we have to give up on this route. :(

Now consider the following grid.


B6 and D4 have satisfied the condition already, therefore I will call them locked cells. In order to proceed there should be words horizontally and vertically, starting from A7. Take a look at D7. There should be two words through it. Since 3-letter word has been already used, there cannot be words along the dotted lines (horizontal/vertical). If there is a word along the red line, we can't fill the cells beneath it. Hence the other word through D7 should lie along the grey arrow (direction can be changed).

Now let's try to fill the row7.

Starting with the maximum: none of the 6-letter long words (THEYRE, TACKED, TAUGHT) seems to fit, because the letter which comes to the orange cell will be only used once in the grid, violating the criteria of that cell. (TAUGHT, written backward, also does not fit because there is no other word ending with U). And similarly 5-letter long words also do not suit. Thus it should be a 4-letter word. We have to look for the words ending with same letter. The possible candidates are TACT, TAUT, TALE and TARE. We have to choose only one from TAUT or TACT, cause if we choose both, there should be another word starting and ending with T, but we don't have any other word with that property. So I will go with TAUT, TALE and TARE (This is optional).

Now consider C5.

Only one word is left to go through it. We can't choose the horizontal or vertical paths (marked in dotted lines below) because we have to reserve space for the longer words. Therefore that word should lie along the diagonal (line marked in black). Because of 'A', the only possible candidates are TAIL and TEAR. But this is the only chance to fulfill the criterion of A3. Hence it should be TEAR.
more progress


...there must be a word along the black line, since dotted lines are not possible paths. And to fill B3, the word should be read diagonally. Now, keeping in mind that there are much longer words those haven't been used yet, we can fill those cells.


Finally, just by inspection, we can fill the rest of grid



Thanks for reading up to this point!


I have to tell you something.

Did you notice that we could flip or change the position of some words to get other possible answers?

Apart from that, did you notice these two extra words?

Ahhh.... this solution is not correct!

Well, OP stated that he counted one word only once, so according to the given rules this solution is acceptable.


I imposed some constraints to get a unique solution.

1. Every letter in the grid should be used for words at least once.
2. If the same word appears again, you have to count that also, for instance, here you have to count TEA in both TEAM and TEAR.
3. One word can appear more than once.

These constraints do not make the puzzle difficult, but make it easy.

Now let's get back to the original puzzle.

At the beginning, we can deduce three letters in the grid logically as we've done earlier. And there is another hint. Consider the words TEAM, TEEM and TEAR. These are the only words that contains another word in it. So many of those letters are counted twice, meaning: they should lie within the caged areas below.

TEAM and TEEM should lie along the diagonal as they share the ending letter. T of TEAR is counted twice, therefore it must be placed in either of A2 or A4. But if it has to be in A2, we have to fill the 2-letters beneath it with a 2-letter word, which is impossible. Hence we have the starting cells completed as below. (There are two ways for the diagonal to be filled)

There's one more word to be placed in A5-A7. The possible solutions are TEA and TEE only. (I'll go with one solution, hopefully it will not affect the next steps)

Now consider the possible placement of the longest words.


If we allow the diagonal to be a 6-letter word, we can place only one of the other two words.

If we place two of them in the 1st row and last row, there is no space left for the other word.

If we fix one of the 6-letter words in the 7th row, there are three possible ways to fill the the other two words.
row7-fixed-1 row7-fixed-2 row7-fixed-3

In all three cases we will not be able to fill the the grid with remaining 4-letter words. (Note that letters cannot overlap on yellow cells).

Similarly, we can inspect few other cases such as below,

last1 last2 last3

...only to assure that the space is not enough for the remaining words.

Hence we can see,

unfortunately, there's no solution for this puzzle. :(

  • $\begingroup$ Excellently worded, although everything in the addendum feels like it is attempting the impossible. It was never my intention to require a puzzle without any word in there twice, but instead just an answer that fits the numbers. There are multiple similar solutions, making this puzzle not good in my opinion. I'm sorry that this puzzle was so difficult. I will be awarding you 200 bonus pts. $\endgroup$
    – JLee
    Commented Sep 3, 2022 at 18:31
  • $\begingroup$ Thanks @JLee. It's hard to make this type of puzzle, but I like that idea. (Also after reading the bounty description, I saw you have posted exactly 100 questions at the moment : ) ) $\endgroup$
    – ACB
    Commented Sep 4, 2022 at 2:38

About 6% of an answer.

(I've worked out the letters in 3 squares so far. Maybe someone will find this useful while I work out the remaining 94%)

The bottom left corner is marked 4, so one of the words that use that corner must completely overlap another that also uses that corner. Only the beginning of words contain other words so that corner must contain T.

Words that could overlap are:


Overlapping words must extend diagonally from the corner. So square 1,1 is T and square 2,2 is E. The centre square is either M or R. The top right corner must have a diagonal word that starts or ends there (it shares 3 words) so that word must either share the M/R centre square or be 3 letters. There are a number of possibilities for that word but they all require square 6,6 to be E.

The letter frequencies reveal some useful information:

T 18
E 15
A 10
R 5
H 4
I 2
C 2
L 2
U 2
M 2
K 1
D 1
G 1
Y 1

K, D, G and Y appear only once so words must be placed so that these letters don't fall on a square numbered more than 1. Similarly I, C, U, L and M appear twice each, so care must be taken not to place them on squares numbered more than 2.

  • $\begingroup$ Thanks. I have a few more ideas based on the constraints placed on TAUGHT, TACKED and THEYRE but I haven't work out(yet) how these might reveal any more certainty. I've been holding off on reading your original puzzle hoping I can solve this without any extra clues. $\endgroup$
    – Bob
    Commented Jun 12, 2015 at 12:59
  • $\begingroup$ I don't think the original puzzle will give you any clues. There might be times where you need to "trial and error" a bit. I am not 100% sure that it can be 100% logically deduced, but it seems that every time I say that on this site, then someone quickly proves me wrong and shows that it actually can. $\endgroup$
    – JLee
    Commented Jun 12, 2015 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.