How to use minimum edit distance to find the distance between two strings with the corresponding alignment using Backtrace?
Minimum Edit Distance with Backtrace for Alignment
Minimum
edit distance is the minimum number of editing operations (insertion, deletion,
substitution) required to convert one string into another. Dynamic programming
is the method used to find the minimum edit distance.
In
the previous posts, we have discussed about the minimum edit distance. Either
the basic version or the Levenshtein distance version is useful to find the
distance. But the major question is how
to align two strings after finding the distance. For example, if I am about
to convert RIVER to SHIVER, I should conclude on the operations that I have to apply
on each character of RIVER. To decide this, we need a corresponding alignment
as per the minimum cost calculated. We can achieve this by using backtrace
pointers in each cell. That is, while we are calculating the minimum cost for each
cell (each subproblem), we also include a pointer in that cell to remember how
did we arrive at the cost for that particular cell. This pointer points to the previous cell which was used in the
calculation of the cost.
In
the example given below, we find the distance between the words DOG and COW
using the basic Minimum Edit Distance algorithm. Also, the Stepbystep process
is shown along with how to use backtrace pointers.
Step 1: Draw the edit
distance matrix. In this, each word is preceded by # symbol which marks the
empty string.



Step 2: Find the
editdistance values using minimum edit distance algorithm to convert # (row side) to #COW (column side) and populate
appropriate cells with the calculated distance.
Number of
operations required to convert;
·
# to # is
0.
·
# to #C is 1.
That is, one insertion (no change with #, insert C)
·
# to #CO is
2. That is, two insertions (no change with #, insert C and O)
·
# to #COW
is 3. That is, three insertions (no change with #, insert C, O and W)



Step 3: Find the
editdistance values using minimum edit distance algorithm to convert #
(column side) to #DOG
(row side) and populate appropriate cells with the calculated distance.
Number of
operations required to convert
·
# to # is 0.
·
# to #D is 1. [one insertion (no change
with #, insert D]
·
# to #DO is 2. [two insertions (no change
#, insert D and O]
·
# to #DOG is 3. [three insertions (no
change #, insert D, O and G]



So far, we have
found the minimum edit distance for 7 subproblems.
[#  # = 0, # 
#C = 1, #  #CO = 2, #  #COW = 3, #  #D = 1, #  #DO = 2, and #  #DOG = 3]


Step 4: From this step
onwards, we try to find the cost for a subproblem by finding the minimum
cost of three subproblems and add 1 with that if the characters intersect at
that cell are different. If the intersecting characters are same, then we add
0 with the diagonal cell value.
[Note: we have used A as the name for this matrix and
included the index numbers for easy understanding]
Use the following
table as a key to find the cost.
If
the row character ≠ column character,
The
cost of the intersecting cell = min(replace, delete, insert).
If
the row character = column character,
The
cost of the intersecting cell = cost of the Replace cell


Subproblem: #D ➔ #C.
Intersecting
cell: A[1,
1]
Same intersecting
characters? NO.
Cost(A[1, 1]) = Min(A[0, 0], A[0, 1], A[1, 0]) + 1
= A[0, 0] + 1
= 0 + 1 = 1
As the cost derived from the cost of cell A[0, 0], a pointer
is included from the current cell A[1, 1] to mark the backtrace path.


Subproblem: #D ➔ #CO.
Intersecting
cell: A[1,
2]
Same intersecting
characters? NO.
Cost(A[1, 2]) = Min(A[0,
1], A[0, 2], A[1, 1]) + 1
= A[1, 1] + 1
= 1 + 1 = 2
As the cost derived from the cost of cell A[1, 1], a pointer
is included from the current cell A[1, 2] to mark the backtrace path.


Justification
for choosing cell A[1, 1] as the previous cell:
We have two cells (A[0,1] and A[1, 1])
with the same minimum cost 1. Which one to choose? We choose A[1, 1]. The
reasons are;
We
are trying to convert ‘D’ to ‘CO’ in this subproblem
We
already used the substitution operation (substitute
character D with C).
So,
the next character ‘O’ in the output can be derived with the insert operation only.
Hence,
we have chosen the cell A[1, 1] which represents insert operation (refer
the key table above) as the
minimum valued cell


Subproblem: #D ➔ #COW.
Intersecting
cell: A[1,
3]
Same intersecting
characters? NO.
Cost(A[1, 3]) = Min(A[0,
2], A[0, 3], A[1, 2]) + 1
= A[1, 2] + 1
= 2 + 1 = 3
As the cost derived from the cost of cell A[1, 2], a pointer
is included from the current cell A[1, 3] to mark the backtrace path.


Justification
for choosing cell A[1, 2] as the previous cell:
We have two cells (A[0, 2] and A[1, 2])
with the same minimum cost 1. Which one to choose? We choose A[1, 2]. The
reasons are;
We
are trying to convert ‘D’ to ‘COW’ in this subproblem
We
already used the substitution operation
(substitute character D with C).
So,
the next characters ‘O’ and ‘W’ in the output can be derived with the insert operation only.
Hence, we have chosen the cell A[1, 2] which represents insert
operation (refer the key
table above) as the minimum valued cell.


Subproblem: #DO ➔ #C.
Intersecting
cell: A[2,
1]
Same intersecting
characters? NO.
Cost(A[2, 1]) = Min(A[1, 0], A[1, 1], A[2, 0]) + 1
= A[1, 1] + 1
= 1 + 1 = 2
As the cost derived from the cost of cell A[1, 1], a pointer
is included from the current cell A[2, 1] to mark the backtrace path.


Justification
for choosing cell A[1, 1] as the previous cell:
We have two cells (A[1, 0] and A[1, 1])
with the same minimum cost 1. Which one to choose? We choose A[1, 1]. The
reasons are;
We
are trying to convert ‘DO’ to ‘C’ in this subproblem
We
already used the substitution operation
(substitute character D with C).
So,
the next character ‘O’ has to be removed in the output. Hence, a delete operation used.
Hence,
we have chosen the cell A[1, 1] which represents delete operation (refer
the key table above) as the
minimum valued cell.


Subproblem: #DO ➔ #CO.
Intersecting
cell: A[2, 2]
Same
intersecting characters? YES
Cost(A[2, 2]) = Min(A[1, 1], A[1, 2], A[2, 1]) + 0
= A[1, 1] + 0
= 1 + 0 = 1
As the cost derived from the cost of cell A[1, 1], a pointer
is included from the current cell A[2, 2] to mark the backtrace path.


Subproblem: #DO ➔ #COW.
Intersecting
cell: A[2,
3]
Same intersecting
characters? NO.
Cost(A[2, 3]) = Min(A[1, 2], A[1, 3], A[2, 2]) + 1
= A[2, 2] + 1
= 1 + 1 = 2
As the cost derived from the cost of cell A[2, 2], a pointer
is included from the current cell A[2, 3] to mark the backtrace path.


Subproblem: #DOG ➔ #C.
Intersecting
cell: A[3,
1]
Same intersecting
characters? NO.
Cost(A[3, 1]) = Min(A[2, 0], A[2, 1], A[3, 0]) + 1
= A[2, 1] + 1
= 2 + 1 = 3


Justification
for choosing cell A[3, 1] as the previous cell:
We have two cells (A[2, 0] and A[2, 1])
with the same minimum cost 1. Which one to choose? We choose A[2, 1]. The
reasons are;
We
are trying to convert ‘DOG’ to ‘C’ in this subproblem
We
already used the substitution operation
(substitute character D with C) and delete
operation (deleted a character ‘O’). [refer subproblem #DO Ã
#C]
So,
the next character ‘G’ has to be removed in the output as well. Hence, a delete operation used.
Hence,
we have chosen the cell A[2, 1] which represents delete operation (refer
the key table above) as the
minimum valued cell.


Subproblem: #DOG ➔ #CO.
Intersecting
cell: A[3,
2]
Same intersecting
characters? NO
Cost(A[3, 2]) = Min(A[2, 1], A[2, 2], A[3, 1]) + 1
= A[2, 2] + 1
= 1 + 1 = 2


Subproblem: #DOG ➔ #COW.
Intersecting
cell: A[3,
3]
Same intersecting
characters? NO.
Cost(A[3, 3]) = Min(A[2, 2], A[2, 3], A[3, 2]) + 1
= A[2, 2] + 1
= 1 + 1 = 2

Now,
to learn the alignment, we need to traverse back from the last cell (the
minimum edit distance between DOG and COW) using the pointer that starts at
that cell. Refer the image below;
The
cost at cell A[3, 3] was derived from cell A[2, 2] by incrementing the cost at
A[2, 2] due to the substitution operation.
The
cost at cell A[2, 2] was derived from cell A[1, 1]. The cost of A[1, 1] is used
as it is because of No change. (the
column character = row character in the intersecting cell).
The
cost at cell A[1, 1] was derived from cell A[0, 0] by incrementing the cost at
A[0, 0] due to the substitution operation.
Hence,
the result of Minimum Edit Distance with alignment is;
D

O

G







C

O

W

Substitution

No change

Substitution

No comments:
Post a comment