1. Two equivalent problems
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Two equivalent problems
RMQ on integer arrays  RMQ on ±1 integer arrays
LCA on a Cartesian Tree
Euler tour of the Cartesian tree
RMQ on the array of node-levels
LCA on general trees  RMQ on ±1 integer arrays
Euler tour of the Cartesian tree
RMQ on the array of node-levels

2. Search for k-mismatches
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Search for k-mismatches T
CCGTACGATCAGTACAGTACAGTACTTTTTTAAACCGGAGACTACA P
If O(1)  O(k) time
CCGAACTATC
Problem: Find longest match between P[i,…] and T[j,…]
Data Structure
Concatenate P and T into a string X = T$P
Construct a data structure on X that retrieves FAST the longest match between any pair of suffixes of X
LCA or LCP query

3. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Suffix Tree & LCA 12 11 8 5 2 1 10 9 7 4 6 3 # i
ppi#
ssi
mississippi# p i#
pi# s
ssippi# si 14 $ 13
LCA
Longest match(3,13)
T#P = mississippi#si$

4. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Suffix Array & LCP  RMQ SA
Lcp
Longest match(3,13) 12 11 14 8 5 2 1 10 9 13 7 4 6 3 1 4 2 3
LCP
T#P = mississippi#si
RMQ si
sippi#
sissippi#
ssippi#
ssissippi#
LCP
Surprisingly, also LCA  RMQ

5. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
The RMQ problem
RMQA(i,j) – returns the index of the smallest element in the subarray A[i..j].
A[0]
A[1]
A[2]
A[3]
A[4]
A[5]
A[6]
A[7]
A[8]
A[9] 10 2 12 1 12 13 21 15 14 10
RMQ(2,7) = 3
Trivial solution: Precompute RMQ for every pair of indices.
This takes Q(n2) space, and O(1) query time

6. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
RMQ on a general array
A[0]
A[1]
A[2]
A[3]
A[4]
A[5]
A[6]
A[7]
A[8]
A[9] 10 25 22 34 7 19 9 12 26 16
Cartesian Tree 4 6 2 5 7 1 3 9 8

7. RMQ(i,j) = LCA(i,j) on Cartesian trees
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
RMQ(i,j) = LCA(i,j) on Cartesian trees
A[0]
A[1]
A[2]
A[3]
A[4]
A[5]
A[6]
A[7]
A[8]
A[9] 10 25 22 34 7 19 9 12 26 16 4 6 2 5 7 1 3 9 8

8. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Generic RMQ  LCA  RMQ ±1
LCA(u,v) = shallowest node between u and v during a depth first search traversal of T.
Node at the
lowest level
±1 array 3
Node Level 12 9 3 8 1 3 2 3 1 4 1 7 1 3 5 6 5 3 2 5 1
Euler tour 11 4 10 7 5 6 2 4 7 6
LCAT(4,6) = 3

9. We are left with “RMQ on ±1 array”
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
We are left with “RMQ on ±1 array”
RMQA(i,j) – returns the index of the smallest element in the subarray A[i..j].
A[0]
A[1]
A[2]
A[3]
A[4]
A[5]
A[6]
A[7]
A[8]
A[9] 10 11 12 11 12 13 14 15 14 13
RMQ(2,7) = 3
Recall the trivial solution: Precompute RMQ for every pair of indices. This takes Q(n2) space, and O(1) query time

10. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Sparse Table
Preprocess sub arrays of len 2k, for every k=0,1,…, log n
M(i,j) = index of min value in A[i, i+ 2j -1]
A[0]
A[1]
A[2]
A[3]
A[4]
A[5]
A[6]
A[7]
A[8]
A[9] 10 11 12 11 12 13 14 15 14 13
M(2,0)=2
Total space is O(n log n)
RMQ query ?
M(2,1)=3
M(2,2)=3
M(2,3)=3

11. Querying the Sparse Table
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Querying the Sparse Table i j
2k elements a1
...
...
2k elements
Total space is O(n log n)
RMQ query takes O(1) time

12. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Bucketing
A’[i] = min in the i-th block of A.
B’[i] is the position (index) of that min. A
A’[0]
A’[i]
A’[2n/logn] A’
...
...
...
...
... …
B[0]
B[i]
B[2n/logn]

13. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
Use the Bucketing
A’[0]
A’[i]
A’[2n/logn] A’
...
...
...
...
...
Preprocess A’ for RMQ using SparseTable
Space is (2n/log n) * log (2n/log n) = O(n)
RMQ queries on A’ take O(1) time
Preprocess every block of A’ for border RMQ
Space is O(n), border RMQ take O(1) time.
RMQ(i,j) takes O(1) time, if i,j lie in distinct blocks

14. In-block RMQ over ±1 arrays
Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
In-block RMQ over ±1 arrays
There are normalized blocks
Set Table[BlockEnc, i, j] = RMQ(i,j) X 3 4 5 6 5 4 5 6 5 4 1 3 2 Y
DX = DY +1 -1
Table entries

15. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
LZ-parsing (gzip) # s i 1 12 si 1 p i 3
ssi
mississippi# 2
ppi#
ppi# 1 4
ssippi# #
ppi#
ssippi# 6 3 i#
ppi#
ssippi#
pi# 7 4 11 8 5 2 1 10 9
<m><i><s><si><ssip><pi>
T = mississippi#

16. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
LZ-parsing (gzip)
It is on the path to 6 # s
By maximality check only nodes i 1 12 1 p
Leftmost occ
= 3 < 6 si i 3
ssi
mississippi# 2
ppi#
ppi# 1 4
ssippi#
Leftmost occ
= 3 < 6 #
ppi#
ssippi# 6 3 i#
ppi#
ssippi#
pi# 7 4 11 8 5 2 1 10 9
<ssip>
Longest repeated prefix of T[6,...]
Repeat is on the left of 6
T = mississippi#

17. Prof. Paolo Ferragina, Algoritmi per "Information Retrieval"
LZ-parsing (gzip)
min-leaf
 Leftmost copy # s 3 i 1 12 si 2
Parsing:
Scan T
Visit ST and stop when
min-leaf ≥ current pos 1 p 3 i 3
ssi
mississippi# 4 9 2 2
ppi#
ppi# 1 4
ssippi# #
ppi#
ssippi# 6 3 i#
ppi#
ssippi#
pi# 7 4 11 8 5 2 1 10 9
<m><i><s><si><ssip><pi>
Precompute the min descending leaf
at every node in O(n) time.
T = mississippi#