R.E. Boss wrote:
>
> What happens after
>
> 'levenshtein' Ldist 'malenstein'
>
> or
>
> 'levenshtein' Ldist1 'malenstein'
>
> ?
> [...]
>
I paste the answer from my terminal:
----------------------------
$ ~/j602/bin/jconsole
Ldist1=:>.&#`((>:@$:}:)~<.(>:@$:}:)<....@=&{:+$:&}:)@.(0<*&#)
'levenshtein' Ldist1 'malenstein'
4
----------------------------
Since in your other post the claim was that the correct answer is 5,
I suppose we should figure out "where the truth lies" (coincidentally
I watched movie with this title earlier today).
Your program gives a diagram that is a 5-edit solution:
> 'levenshtein' Levdist 'malenstein'
> 5 NB. is the right answer
>
> p
> +---+-+---+-+--+-+-+-+
> |l |e|ven|s|ht|e|i|n|
> +---+-+---+-+--+-+-+-+
> |mal|e|n |s|t |e|i|n|
> +---+-+---+-+--+-+-+-+
>
But there is also a 4 edits solution:
1. insert h: malenstein -> malenshtein
2. substitute l with v: -> mavenshtein
3. substitute a with e: -> mevenshtein
4. substitute m with l: -> levenshtein
This solution alone of course does not prove the optimality of 4.
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