Algorithm 2:
|
Indel pair redundancy check Algorithm |
Input:
|
Two candidate redundant indels A and B’s information |
|
same type T (either insertion or deletion), T
A = T
B,, same length L
A = L
B,
|
|
allele information S
A,
S
B, position information P
A,
P
B, where P
A < = P
B,
|
|
reference genome sequence S. |
Output:
|
A pair indels A, B are redundant or not: Redundancy |
1 |
Set Redundancy = False; |
2 |
Phase 1: template substring formation |
3 |
Form template substring for insertion type S
I or for deletion type S
D separately |
4 |
S
I
= Substring in reference genome with P
B - P
A; |
5 |
S
D = Substring in reference genome with P
B - P
A + L
A; |
6 |
Phase 2: variant substring formation for insertion type |
7 |
if
T
A = T
B = Insertion then
|
8 |
Insert S
A in front of template substring S
I to form variant substring V
A for indel A; |
9 |
Append S
B at the end of template substring S
I to form variant substring V
B for indel B; |
10 |
if
V
A = V
B
then
|
11 |
Redundancy found: Redundancy = True; |
12 |
else
|
13 |
No Redundancy |
14 |
Phase 3: variant substring formation for deletion type |
15 |
if
T
A = T
B = Deletion then
|
16 |
Cut S
A in front of template string S
D, form variant substring V
A for indel A; |
17 |
Cut S
B at the end of template string S
D, form variant substring V
B for indel B; |
18 |
if
V
A = V
B
then
|
19 |
Redundancy found: Redundancy = True; |
20 |
else
|
21 |
No Redundancy |
22 |
return Redundancy |