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Constructing a string pattern

Richard Mateosian
Posted 2 years ago

I am working on a helper function to use with the New York Times online puzzle called Spelling Bee. I have been going in circles for a while, but I can't seem to get the syntax right. Here is what I want to do.

Input: A string of English letters. Exact length doesn't matter. Assume 7. For example "wdleano"

Output: A variable containing a string pattern, such as I would obtain by explicitly typing 

z := "w" | "d" | "l" | "e" | "a" | "n" | "o"

Then I can construct a selection function such as 

StringMatchQ[#, z ..] &

to apply to a list of words (such as Wordlist[]).

Conceptually, I want to Riffle Characters[the input word] with the | operator, but I can't get the quotes to appear around the individual characters. I tried 

Style[Characters["wdleano"], ShowStringCharacters -> True]

but then I get something with a head of Style rather than List, so I can't feed it to Riffle. And I'm not sure how to handle the | operator in Riffle.

I'm sure I'm overlooking something simple, but I'm new to the Wolfram language. Please help.

Thanks. ...RM
POSTED BY: Richard Mateosian
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Rohit Namjoshi
Rohit Namjoshi
Posted 2 years ago

It is a good practice to separate input/output from computation. That permits more flexibility e.g. suppose you want to perform the computation using a string read from a file or some other source rather than user input, or write an automated test case. So I would modify the function to take letters as an argument

ClearAll@beeCandidates;
beeCandidates[letters_String] := Module[{pattern, firstCharacter},
  pattern = Alternatives @@ Characters[letters];
  firstCharacter = StringTake[letters, 1];
  Select[WordList[], StringMatchQ[#, pattern ..] && StringContainsQ[#, firstCharacter] && StringLength[#] > 3 &]]

beeCandidates["fetbind"]
POSTED BY: Rohit Namjoshi

For anyone interested, here is the final version. It's my first Wolfram Language function, so I'm sure it can be improved.

bee[] := Module[ {letters, z, k},

  letters = InputString[];

  z = Alternatives @@ Characters[letters];

  k = StringTake[letters, 1];

  Select[WordList[],     StringMatchQ[#, z ..] && StringContainsQ[#, k] && 

 StringLength[#] > 3 &]

    ]

Here is what the output looks like if the input string is fetbind:

In:= bee[]

Out = {"beef", "befit", "benefit", "biff", "defend", "define", "defined", \ "definite", "deft", "diffident", "edified", "effendi", "effete", \ "feed", "feint", "fend", "fete", "fetid", "fief", "fiend", "fife", \ "fifteen", "find", "fine", "finite", "fitted", "identified", \ "indefinite", "infinite", "tiff"}
POSTED BY: Richard Mateosian
Eric Rimbey
Eric Rimbey
Posted 2 years ago 
Select[
  WordList[], 
  And[StringMatchQ[#, z ..], StringContainsQ[#, "w"], StringLength[#] > 3] &]
POSTED BY: Eric Rimbey

Hi, Eric -

Your suggestion seems to work. Here is how I am using it:

In: z = Alternatives @@ Characters["wdleano"]

Out: "w" | "d" | "l" | "e" | "a" | "n" | "o"

In: Select [Select[Select[WordList[], StringMatchQ[#, z ..] &],    StringMatchQ[#, ___ ~~ "w" ~~ ___] &], StringLength[#] > 3 &]

Out: {"allow", "anew", "awed", "dawdle", "dawn", "deadwood", "dowel", \ "down", "download", "dwell", "endow", "endowed", "lawn", "lowland", \ "newel", "owned", "waddle", "wade", "wale", "wall", "wallow", "wand", \ "wane", "weal", "wean", "weaned", "wedded", "weed", "weld", "well", \ "wend", "woad", "wold", "wood", "wooded", "wooden", "woodland", \ "wool", "woolen"}

I constructed the second StringMatch selector manually, but I think it's a small step to automate that.

Thanks. ...RM
POSTED BY: Richard Mateosian
Eric Rimbey
Eric Rimbey
Posted 2 years ago

The expression

"w" | "d" | "l" | "e" | "a" | "n" | "o"

is shorthand for

Alternatives["w", "d", "l", "e", "a", "n", "o"]

Here is a way to get there:

input = "wdleano";
chars = Characters[input];(* {"w", "d", "l", "e", "a", "n", "o"} *)
pattern = Alternatives @@ chars;(* "w" | "d" | "l" | "e" | "a" | "n" | "o" *)

But at this point I didn't understand how you wanted to apply this, because this pattern will match individual characters, but not longer words. So, I'd need more context to help with the next step.
POSTED BY: Eric Rimbey
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