19
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This is a puzzle that I've worked on-and-off for quite a few years, but not being a competent coder, I haven't worked out a definitive solution. I suspect the 1st part is impossible but haven't proved it and the subsequent iterations are enough of a challenge that I thought I'd include it all here.

Create a 5x5 grid of letters containing the English words for each of the 1st 18 counting numbers (i.e. 'one, two, three, ..., seventeen, eighteen) to be found in the style of Boggle. I'm fairly sure (but not positive) that this cannot be done - however, the numbers 'one' through 'seventeen' can be done as can 1-18 except 'five' or 1-18 except 'fifteen'.

So the challenge is either to be the first to find 1-18 or to conclusively prove that it's impossible. If it's the latter then post your solutions to 1-17, 1-18 less 5, 1-18 less 15, etc.

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  • $\begingroup$ Can we rotate the cubes to use different letters if needed? $\endgroup$ – Nautilus Dec 23 '19 at 20:38
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    $\begingroup$ Can the same cube be used more than once in the same word, by your rules? $\endgroup$ – Stiv Dec 24 '19 at 0:04
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    $\begingroup$ You must use one set of cubes (i.e. Boggle dice) and per Boggle rules the same cube cannot be used more than once in each word. E.g. therefore to make the word 'seventeen' you need at least 4 of your cubes to show 'E' and 2 to show 'N' (as well as a 'V' and an 'S'). $\endgroup$ – Channel_Swimmer Dec 24 '19 at 12:42
20
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Update

Here is one of two solutions I have found for numbers ONE to NINETEEN:

enter image description here

I used fresh approach which was less computationally intensive.


I have found a solution to the puzzle with numbers ONE to EIGHTEEN:


enter image description here

At first I tried the logical solution but then went by the 'computer-puzzle' tag.
It's obviously impossible to generate and check every combination of letters.
The first thing was to ignore FOUR SIX SEVEN EIGHT as they are part of the teens.
I noticed two distinct groups of numbers with letters in common:
ELEVEN TWELVE SEVENTEEN FIVE
THIRTEEN FOURTEEN ONE THREE TEN
and I hoped that these could be bridged by the TWO, and other TEENs fitting in.
So I permuted one number at a time, restricting with letters to be shared.
Otherwise the list of perms would grow too big to manage.
However, my best solution was short by three numbers, not just one number as asked.

So I reviewed the strategy.
I started by permuting the part-word TEEN without any rotations or reflections.
Then joined the FOURTEEN to the T followed by THIRTEEN joined to the R and then
the SIXTEEN, SEVENTEEN by joining SIX, SEVEN to a TEEN where they would fit.
I then decided to join TWO to the exisiting O and perhaps rewind to here if it failed.
By now I had 3 billion combinations in a disk file - smaller than the first effort.

Now I added TWELVE forcing it to use the existing T, W and V which reduced the perms.
Presumably some of the numbers must now come 'for free' and I removed any perms that
did not contain ONE, THREE, TEN and ELEVEN and then added NINE where it would fit.
This left me with 72 million sets to approach the last stage of the puzzle.
I solved two of the partial solutions requested and went for the full lot, hey presto!
Actually there were three non-symmetric solutions.

To be clear, here are the individual numbers for the one shown:

enter image description here

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    $\begingroup$ Great job! I stared at your solution for ages going "Aha, but where is NINETEEN!" ...and then I reacquainted myself with the rules... D'oh! +1 $\endgroup$ – Stiv Dec 31 '19 at 18:39
  • $\begingroup$ This is so close to getting NINETEEN too. Are you sure that's not in either of your other solutions? $\endgroup$ – nickgard Jan 2 at 16:42
  • $\begingroup$ @nickgard I wondered at the puzzle condition which stated that a solution was known with EIGHT but not EIGHTEEN. Yet here's the same scenario with NINE but not NINETEEN, with more than one TEEN available. My approach wasn't exhaustive — I had to place reasonable limits on the perms — so there are probably more than 3 solutions, and possibly one with the NINETEEN. I started to make another run with different conditions after Stiv's comment but I have spent enough time on this puzzle :) $\endgroup$ – Weather Vane Jan 2 at 18:17
  • $\begingroup$ ... with only a Y introduced there could also be TWENTY and as ONE...FIVE are already there it's tempting to think it could go to 25 in 25 tiles. The minimum letter count requirement for that is 25, but at least two I's are needed to cover the perms (although only one is needed in any one word), so it's definitely not possible. $\endgroup$ – Weather Vane Jan 2 at 18:32
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    $\begingroup$ @Stiv I had to come back to this puzzle and make 19 to obtain a satisfactory closure ); $\endgroup$ – Weather Vane Jan 7 at 22:02
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This is a very interesting puzzle. I am not sure how to approach it rigorously, other than applying the deductions already mentioned, therefore I did a lot of trial and error. I have not come up with a solution yet, so I will keep trying, but I thought I'd share my progress as there have been no new developments on this problem for some time. Here is my solution then to words 1-17 minus fifteen:

ELEOF
SVNWU
FIETR
NEXIF
RHTHG

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  • $\begingroup$ It looks as if EIGHTEEN is there but sadly the E in the centre was already used when you reach the last E in TEEN. My two best solutions so far lack (8 17 18) and (1 9 14). $\endgroup$ – Weather Vane Dec 29 '19 at 16:42

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