Edit Distance

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   edit dist.
The first lambda calculus solution of the string edit distance problem follows directly from the mathematical definition. It can be seen that it involves ternary recursion and is therefore exponentially slow in terms of the length of the input strings:
let rec
length = lambda L. if null L then 0 else 1 + length tl L,
min    = lambda x. lambda y. if x < y then x else y,

A =
'A'::'C'::'G'::'T'::nil

,B=
'A'::'G'::'C'::'T'::nil

in let rec

Distance = lambda A. lambda B.
   if      null A then length B
   else if null B then length A
   else
   let As = tl A, Bs = tl B
   in  if hd A = hd B then Distance As Bs
       else 1 + min (Distance As Bs)
               (min (Distance As B)
                    (Distance A Bs))

in Distance A B

{\fB Edit Distance,                                  \fP}
{\fB best case (A=B) O(|A|), worst case exponential. \fP}




The next version of edit distance avoids doing repeated work by storing partial results in an "array" (actually a list of lists) giving the well-known dynamic programming algorithm (DPA). This reduces the time complexity to O(|A|*|B|) where the two strings are A and B.
let rec
  count  = lambda L. lambda B.
           if null B then nil
           else (1 + hd L) :: count tl L tl B,
  last = lambda L. if null tl L then hd L else last tl L,
  min  = lambda x. lambda y. if x < y then x else y,

A = 'a'::'c'::'g'::'t'::'a'::'c'::
    'g'::'t'::'a'::'c'::'g'::'t'::nil  {e.g.}
  
,B = 'a'::'g'::'c'::'t'::'a'::'c'::
     't'::'a'::'c'::'t'::'g'::'t'::nil {e.g.}

in let

Distance = lambda A. lambda B.
  let rec
    Rows = (0 :: count  hd Rows  B)  {the first row }
        :: EachRow A  hd Rows        {the other rows},

    EachRow = lambda A. lambda lastrow.
      if null A then nil
      else
      let rec
        Ach = hd A,

        DoRow = lambda B. lambda NW. lambda W. {NW N}
          if null B then nil                   {W  .}
          else
          let    N  = tl NW
          in let me = if Ach = hd B then hd NW
                      else 1 + min W (min hd N hd NW)
          in me :: DoRow tl B  tl NW  me,

        thisrow = (1 + hd lastrow)
               :: DoRow B lastrow  hd thisrow

      in thisrow :: EachRow tl A  thisrow

  in last (last Rows)

in Distance A B

{\fB Edit Distance, O(|A|*|B|) time and space. \fP}




The final edit distance program reduces the time complexity of O(n*D(A,B)) where the strings are of length ~n, and D(A,B) is the edit distance of A and B.

This program is fast if the strings are similar in which case the edit distance is small. It relies on lazy evaluation or 'call by need' to get this speed up. For a full explanation, see:
L. Allison. Lazy dynamic programming can be eager, Information Processing Letters, 43, pp.207-212, Sept 1992.
let rec
min    = lambda x. lambda y. if x < y then x else y,
length = lambda L. if null L then 0 else 1+length tl L,
last   = lambda L. if null tl L then hd L else last tl L,
index  = lambda n. lambda L.
         if n=1 then hd L else index (n-1) tl L,

acgt = lambda n.
  if n > 0 then 'a'::'c'::'g'::'t'::(acgt (n-4)) else nil,

mutate = lambda L. lambda mutn.
  let rec
    n = length L,
    step = if mutn=0 then 2*n+1 else n/mutn,
    ch = lambda L. lambda st. lambda mtype.
         if null L then nil
         else if st = 0 then
           if mtype=1 or mtype=3 then            {2:1:1}
              'x'::(ch  tl L  step (mtype+1))    {change}
           else if mtype=2 then (ch tl L step 3) {delete}
           else 'y'::(ch L step 1)               {insert}
         else (hd L)::(ch tl L (st-1) mtype)     {copy}
  in ch L (step/2) 1,


A = acgt 100     {e.g.}
,B = mutate A 4  {e.g.}


in let

Distance = lambda A. lambda B.
 let rec
  MainDiag = OneDiag A B  hd Uppers  (-1 :: hd Lowers),
  Uppers   = EachDiag A B (MainDiag::Uppers), {upper diags}
  Lowers   = EachDiag B A (MainDiag::Lowers), {lower diags}

  OneDiag = lambda A. lambda B.
            lambda diagAbove. lambda diagBelow.
   let rec
    DoDiag= lambda A. lambda B. lambda NW. lambda N. lambda W.
       if null A  or  null B then nil
       else                                   { NW N  }
       let me = if hd A = hd B then NW        { W  me }
{fast}          else 1+if hd W < NW then hd W else min hd N NW
{slow}        { else 1+min NW (min hd N  hd W) }

       in me::DoDiag  tl A  tl B  me  tl N  tl W, {along diag}
                          {hope these ^^^^  ^^^^not evaluated}

    thisdiag = (1+hd diagBelow)
           :: DoDiag A B  hd thisdiag  diagAbove  tl diagBelow

   in thisdiag,

  EachDiag =  lambda A. lambda B. lambda Diags.
    if null B then nil
    else (OneDiag A tl B hd tl tl Diags hd Diags) {one diag &}
       :: EachDiag A tl B tl Diags                {the others}

 in let LAB = (length A) - (length B)
 in last if      LAB=0 then MainDiag
         else if LAB > 0 then index   LAB  Lowers
         else   {LAB < 0}     index (-LAB) Uppers

in Distance A B

{\fB Edit-Distance, diagonal orientation. \fP}





For the record, the last algorithm in Haskell

module Edit_IPL_V43_N4 (d) where
-- compute the edit distance of sequences a and b.
d a b =
 let
   -- diagonal from the top-left element
   mainDiag = oneDiag  a b (head uppers) ( -1 : (head lowers))

   -- diagonals above the mainDiag
   uppers   = eachDiag a b (mainDiag : uppers)

   -- diagonals below the mainDiag
   lowers   = eachDiag b a (mainDiag : lowers)  -- !

   oneDiag a b diagAbove diagBelow =  -- \   \   \
    let                               --  \   \   \
      doDiag [] b nw n w = []         --   \  nw   n
      doDiag a [] nw n w = []         --    \   \
      doDiag (a:as) (b:bs) nw n w =   --      w   me
       let me = if a==b then nw       -- dynamic programming DPA
                else 1+min3 (head w) nw (head n)
       in  me : (doDiag as bs me (tail n) (tail w))

      firstelt = 1+(head diagBelow)
      thisdiag =
        firstelt:(doDiag a b firstelt diagAbove (tail diagBelow))

    in thisdiag

   min3 x y z =
   -- see L. Allison, Lazy Dynamic-Programming can be Eager
   --     Inf. Proc. Letters 43(4) pp207-212, Sept' 1992
     if x < y then x else min y z   -- makes it O(|a|*D(a,b))
     -- min x (min y z)             -- makes it O(|a|*|b|)
   -- the fast one does not always evaluate all three values.

   eachDiag a [] diags = []
   eachDiag a (b:bs) (lastDiag:diags) =
    let nextDiag = head(tail diags)
    in  (oneDiag a bs nextDiag lastDiag):(eachDiag a bs diags)

   -- which is the diagonal containing the bottom R.H. elt?
   lab = (length a) - (length b)

 in last( if      lab == 0 then mainDiag
          else if lab > 0 then lowers !! ( lab-1)
          else                 uppers !! (-lab-1) )

 -- module under Gnu 'copyleft' GPL General Public Licence.

See: L. Allison. Lazy dynamic programming can be eager, Information Processing Letters, 43, pp.207-212, Sept 1992.

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↑ © L. Allison, www.allisons.org/ll/   (or as otherwise indicated).
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