Just for fun... here's another quick visual method:

s1 = 'abcdefghijk1mnOpqrstuvwxyz'
s2 = 'abcdefghijklmn0pqrstuvwxyz'

puts s1, s2
# XOR the binary of each character, will be 0 if the same
s1.size.times {|i| print s1[i]^s2[i].to_i==0?'-':'#'}

Results:
abcdefghijk1mnOpqrstuvwxyz
abcdefghijklmn0pqrstuvwxyz
-----------#--#-----------

(above looks better in a fixed font display)

-Joby