Forum Discussion
Lookup five characters appearing in any order in a wordlist
Column B = words from the wordlist that contain one or more of the twelve letters
What do you mean one or more?
In the attached from Peter Bartholomew ,I have found All 5 letters matched at the same time not include 1,2,3,4 these 4 scenarios.
About the picture you uploaded,why diptych and forward match?
Are they the expected result?
I am really confused about the expected result?
Can you give more detail explains by an example?
Can you explain
The word list is long... 214,066 words. First, we need to eliminate every word that has adjacent letters from the same side of the square and every word that includes one of the 14 letters of the alphabet that are not included in the puzzle square. That will narrow the list down to 2,000 (give or take 1,000) words.
From the remaining list, the challenge is to find the fewest number of words combined that use all the letters in the puzzle squared. I shoot for two words. That is the real challenge.
Today's letters are
Top of puzzle square - YOD
Right side of puzzle square - WHI
Bottom of puzzle square - ATC
Left side of puzzle square - FRP.
"Forward" contains only letters from the puzzle square (in red), no letters that are on the same side of the square are adjacent to each other, and no letters that aren't in the puzzle, so it meets all the rules.
"Diptych" contains all the rest of the letters in the puzzle square, no letters that are on the same side of the square are adjacent to each other, no letters that aren't in the puzzle, and it starts with "D," the last letter of "Forward," thus meeting all conditions of the puzzle.
Forward and Diptych are one matching pair, but there could be others, and they all must meet the same conditions.
On some days, It might be possible for there to be a word that uses 11 of the letters in the square and meets all other conditions of the puzzle. In that case, a two-letter word could be a match if it starts with the last letter of the first word and uses the twelfth letter in the square.
The second word can re-use letters from the first word as long as it also uses all the remaining letters from the square and meets all the other conditions.
I was mistaken about a one-letter word, because if the first word uses all twelve letters and meets all the other conditions, the puzzle is solved with just the one word. It's rare, though.
Thanks,
Michael
Suppose we have
bsa
olu
te
xit
expected result:
absolute
exit
Or
absolute
exist
Just include like absolute
exist
or must match fully no redundancy letter like absolute
exit ?
The first letter of the second word should be the last letter of the first word?
Only 2 words match whole puzzle square
How about 3?
Re:
, we need to eliminate every word that has adjacent letters from the same side of the squar
Can you provide a sample of this sentence?
eliminate every word that has adjacent letters like
boundless
with adjacent letter ss?
Re:
The word list is long... 214,066 words.
your dictionary has 214,066 words?
But my dictionary on hand only 13,532 words and phrases
Your third line "te" is missing a letter, but we can ignore that for our purposes.
*Yes, "absolute - exist" is okay. There can be redundancy.
*Yes, the first letter of the second word must be the last letter of the first word.
*Yes, you can use all the puzzle letters with three words "bobs - salute - exist," but I don't care to solve for that. I only want to solve for two words.
*Boundless is illegal because of the two ss. In your example letters, boundless would also be illegal because o and u are adjacent, and they appear together on one side of the square "olu."
*If your wordlist is too short, solutions may not be possible. I use the Scrabble word list. It doesn't have all the words that the puzzle maker uses, but it's close enough.
- because o and u are adjacent, and they appear together on one side of the square "olu."
The adjacent letters rule is not as straightforward
why o and u are adjacent？
at last summary：
Have 12 unique letters.
Divided to 2 or 3 groups, each group should be part of an English word regardless of sequence。
The expected result are two English or 3 English words.
Each word in those 2 or 3 words,all letters should be unique. e.g. class boundless are illegal.
And the last letter of the first word should be the first letter of the second word and the first letter of third same as the last of second word. head letter same as former foot letter.