Lecture 5 - BM

Question 1

What does BM stand for?

Answer 1

Boyer Moore (algorithm)

Question 2

How does BM perform compared to KMP and brute force?

Answer 2

It is almost always faster than both.

Question 3

Are all characters in the text checked at BM?

Answer 3

No, most are skipped.

Question 4

In what direction is the string scanned?

Answer 4

right to left (reverse of the other 2 methods)

Question 5

What is used to decide the next comparison?

Answer 5

the text character involved in the mismatch

Question 6

Does BM include any preprocessing?

Answer 6

Yes, we need to record the position of the last occurrence of each character c in the alphabet.

Question 7

While preprocessing what happens if a certain alpha character doesn’t xist in the string?

Answer 7

-1 is entered

Question 8

How big is our BM array?

Answer 8

big enough to fit the whole alphabet and other character, the array is indexed by the ascii value (128)

Question 9

What version of BM will we look at in this course?

Answer 9

Boyer Moore Horspool algorithm , it is the simplified version of BM

Question 10

How big is the ASCII character set?

Answer 10

128 chars

Question 11

What happens upon a mismatch in BM?

Answer 11

We have to slide s along to align last occurance with the mismatched character. If s moves in the incorrect direction we instead move s once position to the right. If occurance doesn’t exist we slide string -> align with the next on the left

Question 12

How many mismatch cases is there in Bm?

Answer 12

3

Question 12

What is case 1 in BM?

Answer 12

There is a mismatch and last occurence exists to the left (yet to check ). In this case:

• slide string to match last occurence with current i (this means i + (m-1) - last occurence index)
• j becomes m-1 (last character of the string as we essentially restart search from right to left)
• sp becomes sp + j - last occurence (the amount pattern has been shifted)

Question 13

What is case 2 in BM?

Answer 13

There is a mismatch and last occurence exists to the right (been checked). In this case:

“we move string along by one place to the right and restart from the end”

• the new value of i is i + (m-1) - (j-1)
• the new value of j is m-1
• sp becomes sp+1 (as shifted by 1)

Question 14

What is case 3 in BM?

Answer 14

When there is a mismatch and the character doesn’t appear.

• i = i + m, shift by the length of the whole string
• j = m-1 (end of string - absolute restart)
• sp = sp + j + 1

Question 15

Where to see all cases?

Answer 15

check slide 182

Question 16

Why can’t last occurrence equal to our char at j?

Answer 16

Since there is a mismatch between char in text and char in string , so this can’t be equal to char at j.

Question 17

What is the complexity of BM?

Answer 17

O(mn)

Question 18

Why is the complexity O(mn)?

Answer 18

As we need at most m char comparisons at every n - (m+1) positions in text (never retract)

Question 19

Is there a better version than O(mn)?

Answer 19

Yes, a linear one O(m+n), but we do not go over this in the course.