Edit Distance
- 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.