Rash thoughts about .NET, C#, F# and Dynamics NAV.

In the first part of this series I showed a naïve algorithm for the Damerau-Levenshtein distance which needs O(m*n) space. In the last post I improved the algorithm to use only O(m+n) space. This time I will show a more functional implementation which uses only immutable F#-Lists and works still in O(m+n) space. This version doesn’t need any mutable data.

/// Calcs the damerau levenshtein distance.    
let calcDL (a:'a array) (b: 'a array) =       
  let n = a.Length + 1
  let m = b.Length + 1
  
  let processCell i j act l1 l2 ll1 =
    let cost = 
      if a.[i-1] = b.[j-1] then 0 else 1
    let deletion = l2 + 1
    let insertion = act + 1
    let substitution = l1 + cost
    let min1 =  
      deletion 
      |> min insertion 
      |> min substitution

    if i > 1 && j > 1 &&
      a.[i-1] = b.[j-2] && a.[i-2] = b.[j-1] then
        min min1 <| ll1 + cost
    else
      min1
  
  let processLine i lastL lastLastL =
    let processNext (actL,lastL,lastLastL) j =
      match actL with 
        | act::actRest -> 
          match lastL with
            | l1::l2::lastRest -> 
              if i > 1 && j > 1 then
                match lastLastL with
                  | ll1::lastLastRest -> 
                    (processCell i j act l1 l2 ll1 :: actL,
                     l2::lastRest,
                     lastLastRest)
                  | _ -> failwith "can't be"
              else
                (processCell i j act l1 l2 0 :: actL,
                 l2::lastRest,
                 lastLastL)                 
            | _ -> failwith "can't be"
        | [] -> failwith "can't be"
      
    let (act,last,lastLast) =
      [1..b.Length]
        |> List.fold_left processNext ([i],lastL,lastLastL)
    act |> List.rev
    
  let (lastLine,lastLastLine) =               
    [1..a.Length]
      |> List.fold_left
          (fun (lastL,lastLastL) i -> 
             (processLine i lastL lastLastL,lastL))
          ([0..m-1],[])
              
  lastLine.[b.Length]    
 
let damerauLevenshtein(a:'a array) (b:'a array) =
  if a.Length > b.Length then
    calcDL a b
  else
    calcDL b a

I admit the code is still a little messy but it works fine. Maybe I will find the time to cleanup a bit and post a final version.

Last time I showed a naïve implementation of the Damerau-Levenshtein-Distance in F# that needs O(m*n) space. This is really bad if we want to compute the edit distance of large sequences (e.g. DNA sequences). If we look at the algorithm we can easily see that only the last two lines of the (n*m)-matrix are used. This observation leads to a improvement where we compute the distance with only 3 additional arrays of size min(n,m).

/// Calcs the damerau levenshtein distance.    
let calcDL (a:'a array) (b: 'a array) =       
  let n = a.Length + 1
  let m = b.Length + 1
  let lastLine = ref (Array.init m (fun i -> i))
  let lastLastLine = ref (Array.create m 0)
  let actLine = ref (Array.create m 0)
    
  for i in [1..a.Length] do
    (!actLine).[0] <- i      
    for j in [1..b.Length] do          
      let cost = 
        if a.[i-1] = b.[j-1] then 0 else 1
      let deletion = (!lastLine).[j] + 1
      let insertion = (!actLine).[j-1] + 1
      let substitution = (!lastLine).[j-1] + cost
      (!actLine).[j] <- 
        deletion 
        |> min insertion 
        |> min substitution

      if i > 1 && j > 1 then
        if a.[i-1] = b.[j-2] && a.[i-2] = b.[j-1] then
          let transposition = (!lastLastLine).[j-2] + cost  
          (!actLine).[j] <- min (!actLine).[j] transposition
    
    // swap lines
    let temp = !lastLastLine
    lastLastLine := !lastLine
    lastLine := !actLine
    actLine := temp
            
  (!lastLine).[b.Length]

 
let damerauLevenshtein(a:'a array) (b:'a array) =
  if a.Length > b.Length then
    calcDL a b
  else
    calcDL b a

This version of the algorithm needs only O(n+m) space but is not really "functional" style. I will show a more "F#-stylish" version in part III.