Shortest common supersequence
The shortest common supersequence is a problem closely related to the longest common subsequence, which you can use as an external function for this task.
- Task
Given two strings and find the shortest possible sequence which is the shortest common super-sequence of and where both and are a subsequence of Defined as such, is not necessarily unique.
Demonstrate this by printing where
abcbdab
and
bdcaba
C
The C99 code here isn't all that different from Levenstein distance calculation. <lang c>#include <stdio.h>
- include <string.h>
typedef struct link link_t; struct link { int len; char letter; link_t *next; };
// Stores a copy of a SCS of x and y in out. Caller needs to make sure out is long enough. int scs(char *x, char *y, char *out) { int lx = strlen(x), ly = strlen(y); link_t lnk[ly + 1][lx + 1];
for (int i = 0; i < ly; i++) lnk[i][lx] = (link_t) {ly - i, y[i], &lnk[i + 1][lx]};
for (int j = 0; j < lx; j++) lnk[ly][j] = (link_t) {lx - j, x[j], &lnk[ly][j + 1]};
lnk[ly][lx] = (link_t) {0};
for (int i = ly; i--; ) { for (int j = lx; j--; ) { link_t *lp = &lnk[i][j]; if (y[i] == x[j]) { lp->next = &lnk[i+1][j+1]; lp->letter = x[j]; } else if (lnk[i][j+1].len < lnk[i+1][j].len) { lp->next = &lnk[i][j+1]; lp->letter = x[j]; } else { lp->next = &lnk[i+1][j]; lp->letter = y[i]; } lp->len = lp->next->len + 1; } }
for (link_t *lp = &lnk[0][0]; lp; lp = lp->next) *out++ = lp->letter;
return 0; }
int main(void) { char x[] = "abcbdab", y[] = "bdcaba", res[128]; scs(x, y, res); printf("SCS(%s, %s) -> %s\n", x, y, res); return 0; }</lang>
- Output:
SCS(abcbdab, bdcaba) -> abdcabdab
D
<lang d>import std.stdio, std.functional, std.array, std.range;
dstring scs(in dstring x, in dstring y) nothrow @safe {
alias mScs = memoize!scs;
if (x.empty) return y;
if (y.empty) return x;
if (x.front == y.front)
return x.front ~ mScs(x.dropOne, y.dropOne);
if (mScs(x, y.dropOne).length <= mScs(x.dropOne, y).length)
return y.front ~ mScs(x, y.dropOne);
else
return x.front ~ mScs(x.dropOne, y);
}
void main() @safe {
scs("abcbdab", "bdcaba").writeln;
}</lang>
- Output:
abdcabdab
Elixir
uses 'LCS' from here <lang elixir>defmodule SCS do
def scs(u, v) do lcs = LCS.lcs(u, v) |> to_charlist scs(to_charlist(u), to_charlist(v), lcs, []) |> to_string end defp scs(u, v, [], res), do: Enum.reverse(res) ++ u ++ v defp scs([h|ut], [h|vt], [h|lt], res), do: scs(ut, vt, lt, [h|res]) defp scs([h|_]=u, [vh|vt], [h|_]=lcs, res), do: scs(u, vt, lcs, [vh|res]) defp scs([uh|ut], v, lcs, res), do: scs(ut, v, lcs, [uh|res])
end
u = "abcbdab" v = "bdcaba" IO.puts "SCS(#{u}, #{v}) = #{SCS.scs(u, v)}"</lang>
- Output:
SCS(abcbdab, bdcaba) = abdcabdab
Perl
<lang perl>sub lcs { # longest common subsequence
my( $u, $v ) = @_;
return unless length($u) and length($v);
my $longest = ;
for my $first ( 0..length($u)-1 ) {
my $char = substr $u, $first, 1;
my $i = index( $v, $char );
next if -1==$i;
my $next = $char;
$next .= lcs( substr( $u, $first+1), substr( $v, $i+1 ) ) unless $i==length($v)-1;
$longest = $next if length($next) > length($longest);
}
return $longest;
}
sub scs { # shortest common supersequence
my( $u, $v ) = @_;
my @lcs = split //, lcs $u, $v;
my $pat = "(.*)".join("(.*)",@lcs)."(.*)";
my @u = $u =~ /$pat/;
my @v = $v =~ /$pat/;
my $scs = shift(@u).shift(@v);
$scs .= $_.shift(@u).shift(@v) for @lcs;
return $scs;
}
my $u = "abcbdab"; my $v = "bdcaba"; printf "Strings %s %s\n", $u, $v; printf "Longest common subsequence: %s\n", lcs $u, $v; printf "Shortest common supersquence: %s\n", scs $u, $v; </lang>
- Output:
Strings abcbdab bdcaba Longest common subsequence: bcba Shortest common supersquence: abdcabdab
Racket
This program is based on the C implementation, but use memorization instead of dynamic programming. More explanations about the memorization part in http://blog.racket-lang.org/2012/08/dynamic-programming-versus-memoization.html .
<lang Racket>#lang racket
(struct link (len letters))
(define (link-add li n letter)
(link (+ n (link-len li))
(cons letter (link-letters li))))
(define (memoize f)
(local ([define table (make-hash)])
(lambda args
(dict-ref! table args (λ () (apply f args))))))
(define scs/list
(memoize
(lambda (x y)
(cond
[(null? x)
(link (length y) y)]
[(null? y)
(link (length x) x)]
[(eq? (car x) (car y))
(link-add (scs/list (cdr x) (cdr y)) 1 (car x))]
[(<= (link-len (scs/list x (cdr y)))
(link-len (scs/list (cdr x) y)))
(link-add (scs/list x (cdr y)) 1 (car y))]
[else
(link-add (scs/list (cdr x) y) 1 (car x))]))))
(define (scs x y)
(list->string (link-letters (scs/list (string->list x) (string->list y)))))
(scs "abcbdab" "bdcaba")</lang>
- Output:
"abdcabdab"
Ruby
uses 'lcs' from here <lang ruby>require 'lcs'
def scs(u, v)
lcs = lcs(u, v)
scs = ""
# Iterate over the characters until LCS processed
ui = vi = li = 0
while li < lcs.size
uc = u[ui]
vc = v[vi]
lc = lcs[li]
if uc==lc
if vc==lc
# Part of the LCS, so consume from all strings
scs << lc
ui += 1
li += 1
else
# char of u = char of LCS, but char of LCS v doesn't so consume just that
scs << vc
end
vi += 1
else
# char of u != char of LCS, so consume just that
scs << uc
ui += 1
end
end
# append remaining characters, which are not in common
scs << u[ui..-1] << v[vi..-1]
end
u = "abcbdab" v = "bdcaba" puts "SCS(#{u}, #{v}) = #{scs(u, v)}"</lang>
- Output:
SCS(abcbdab, bdcaba) = abcbdcaba
Sidef
Uses the lcs function defined here. <lang ruby>func scs(u, v) {
var ls = lcs(u, v).chars
var pat = Regex('(.*)'+ls.join('(.*)')+'(.*)')
u.scan!(pat)
v.scan!(pat)
var ss = (u.shift + v.shift)
ls.each { |c| ss += (c + u.shift + v.shift) }
return ss
}
say scs("abcbdab", "bdcaba")</lang>
- Output:
abdcabdab
Tcl
This example uses either of the lcs implementations from here, assumed renamed to lcs…
<lang tcl>proc scs {u v} {
set lcs [lcs $u $v] set scs ""
# Iterate over the characters until LCS processed
for {set ui [set vi [set li 0]]} {$li<[string length $lcs]} {} {
set uc [string index $u $ui] set vc [string index $v $vi] set lc [string index $lcs $li] if {$uc eq $lc} { if {$vc eq $lc} { # Part of the LCS, so consume from all strings append scs $lc incr ui incr li } else { # char of u = char of LCS, but char of LCS v doesn't so consume just that append scs $vc } incr vi } else { # char of u != char of LCS, so consume just that append scs $uc incr ui }
}
# append remaining characters, which are not in common append scs [string range $u $ui end] [string range $v $vi end] return $scs
}</lang> Demonstrating: <lang tcl>set u "abcbdab" set v "bdcaba" puts "SCS($u,$v) = [scs $u $v]"</lang>
- Output:
SCS(abcbdab,bdcaba) = abdcabdab
zkl
<lang zkl>class Link{ var len,letter,next;
fcn init(l=0,c="",lnk=Void){ len,letter,next=l,c,lnk; }
} fcn scs(x,y,out){
lx,ly:=x.len(),y.len();
lnk:=(ly+1).pump(List,'wrap(_){ (lx+1).pump(List(),Link.create) });
foreach i in (ly){ lnk[i][lx]=Link(ly-i, y[i]) }
foreach j in (lx){ lnk[ly][j]=Link(lx-j, x[j]) }
foreach i,j in ([ly-1..0,-1],[lx-1..0,-1]){
lp:=lnk[i][j];
if (y[i]==x[j]){
lp.next =lnk[i+1][j+1]; lp.letter=x[j];
}else if(lnk[i][j+1].len < lnk[i+1][j].len){
lp.next =lnk[i][j+1]; lp.letter=x[j];
}else{
lp.next =lnk[i+1][j]; lp.letter=y[i];
}
lp.len=lp.next.len + 1;
}
lp:=lnk[0][0]; while(lp){ out.write(lp.letter); lp=lp.next; }
out.close()
}</lang> <lang zkl>scs("abcbdab","bdcaba", Sink(String)).println();</lang>
- Output:
abdcabdab