ch05-2018.03.02 ch05-2018.03.02ch05-2018.03.02

A. Levitin “Introduction to the Design & Analysis of Algorithms,” 3rd ed., Ch. 5 ©2012 Pearson
Education, Inc. Upper Saddle River, NJ. All Rights Reserved. 1
Divide-and-Conquer
The most-well known algorithm design strategy:
1. Divide instance of problem into two or more
smaller instances
2. Solve smaller instances recursively
3. Obtain solution to original (larger) instance by
combining these solutions
Distinction between divide and conquer and decrease and conquer is
somewhat vague. Dec/con typically does not solve both subproblems.

Divide-and-Conquer Technique (cont.)
subproblem 2
of size n/2
subproblem 1
of size n/2
a solution to
subproblem 1
a solution to
the original problem
a solution to
subproblem 2
a problem of size n

Divide-and-Conquer Examples
 Sorting: mergesort and quicksort
 Binary tree traversals
 Multiplication of large integers
 Matrix multiplication: Strassen’s algorithm
 Closest-pair and convex-hull algorithms
 Binary search: decrease-by-half (or degenerate divide&conq.)

General Divide-and-Conquer Recurrence
T(n) = aT(n/b) + f (n) where f(n)  (nd), d  0
Master Theorem: If a < bd, T(n)  (nd)
If a = bd, T(n)  (nd log n)
If a > bd, T(n)  (nlog b a )
Note: The same results hold with O instead of .
Examples: T(n) = 4T(n/2) + n  T(n)  ?
T(n) = 4T(n/2) + n2  T(n)  ?
T(n) = 4T(n/2) + n3  T(n)  ?

Mergesort Example
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
3 8 2 9 1 7 4 5
2 3 8 9 1 4 5 7
1 2 3 4 5 7 8 9

Mergesort
 Split array A[0..n-1] in two about equal halves and make
copies of each half in arrays B and C
 Sort arrays B and C recursively
 Merge sorted arrays B and C into array A as follows:
• Repeat the following until no elements remain in one of
the arrays:
– compare the first elements in the remaining
unprocessed portions of the arrays
– copy the smaller of the two into A, while
incrementing the index indicating the unprocessed
portion of that array
• Once all elements in one of the arrays are processed,
copy the remaining unprocessed elements from the other
array into A.

Pseudocode of Mergesort
Thinking about recursive programs: Assume recursive calls work
and ask does the whole program then work?

Pseudocode of Merge

Analysis of Mergesort
 Number of key comparisons – recurrence:
C(n) = …
For merge step:
C’(n) = …

 Number of key comparisons – recurrence:
C(n) = 2C(n/2) + C’(n) = 2C(n/2) + n-1
For merge step – worst case:
C’(n) = n-1
 How to solve?
 Number of calls? Number of active calls?

General Divide-and-Conquer Recurrence
T(n) = aT(n/b) + f (n) where f(n)  (nd), d  0
Master Theorem: If a < bd, T(n)  (nd)
If a = bd, T(n)  (nd log n)
If a > bd, T(n)  (nlog b a )
Using log – Take log b of both sides:
Case 1: logb a < d, T(n)  (nd)
Case 2: logb a = d, T(n)  (nd logb n)
Case 3: logb a > d, T(n)  (nlog b a )

 All cases have same efficiency: Θ(n log n)
 Number of comparisons in the worst case is close to
theoretical minimum for comparison-based sorting:
- Theoretical min: log2 n! ≈ n log2 n - 1.44n
 Space requirement:
• Book’s algorithm?

 All cases have same efficiency: Θ(n log n)
 Number of comparisons in the worst case is close to
theoretical minimum for comparison-based sorting:
- Theoretical min: log2 n! ≈ n log2 n - 1.44n
 Space requirement:
• Book’s algorithm: n + n Sum(1+1/2+1/4+…) = n + n Θ(2)= Θ(3n)
• Can be done Θ(2n) : Don’t copy in sort; just specify bounds, copy
into extra array on merge, then copy back
– Any other space required?
• Can be implemented in place. How …
 Can be implemented without recursion (bottom-up)

Mergesort Example
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
8 3 2 9 7 1 5 4
3 8 2 9 1 7 4 5
2 3 8 9 1 4 5 7
1 2 3 4 5 7 8 9

Quicksort
 Select a pivot (partitioning element) – here, the first element
 Rearrange the list so that all the elements in the first s
positions are smaller than or equal to the pivot and all the
elements in the remaining n-s positions are larger than or
equal to the pivot (see next slide for an algorithm)
 Exchange the pivot with the last element in the first (i.e., )
subarray — the pivot is now in its final position
 Sort the two subarrays recursively
p
A[i]p A[i]p

Quicksort Algorithm
QS (A, l, r)
if l < r then
s = partition (A, l, r)
QS (A, , )
QS (A, , )

Quicksort Algorithm
QS (A, l, r)
if l < r then
s = partition (A, l, r)
QS (A, l , s-1)
QS (A, s+1, r )
Ask later: What if we skip final swap and
use simpler bounds for first recursive call?

Hoare’s (Two-way) Partition Algorithm

Quicksort Example
5 3 1 9 8 2 4 7

Quicksort Recursion Tree
0 1 2 3 4 5 6 7
5 3 1 9 8 2 4 7
0 .. 7 / s=4
0 .. 3 / s=x
0 .. s-1 / s=?
s+1 .. 3
5 .. 7 / s=y
5 .. y-1 / s=?
y+1 .. 7 / s=?

Analysis of Quicksort
Recurrences:
 Best case:
 Worst case:
 Average case:
Performance:
 Best case:
 Worst case:
 Average case:

Recurrences:
 Best case: T(n) = T(n/2) + Θ (n)
 Worst case: T(n) = T(n-1) + Θ (n)
 Average case:
Performance:
 Best case:
 Worst case:
 Average case:

Recurrences:
 Best case: T(n) = T(n/2) + Θ (n)
 Worst case: T(n) = T(n-1) + Θ (n)
 Average case:
Performance:
 Best case: split in the middle — Θ(n log n)
 Worst case: sorted array! — Θ(n2)
 Average case: random arrays — Θ(n log n)

 Problems:
• Duplicate elements
 Improvements:
• better pivot selection: median of three partitioning
• switch to insertion sort on small sub-arrays
• elimination of recursion – How?
These combine to 20-25% improvement
 Improvements: Dual Pivot
 Method of choice for internal sorting of large files (n ≥ 10000)

Binary Tree Algorithms
Binary tree is a divide-and-conquer ready structure!
Ex. 1: Classic traversals (preorder, inorder, postorder)
Algorithm Inorder(T)
a a
b c b c
d e • • d e
• • • •
Efficiency: Θ(n)

Binary Tree Algorithms
Binary tree is a divide-and-conquer ready structure!
Ex. 1: Classic traversals (preorder, inorder, postorder)
Algorithm Inorder(T)
if T   a a
Inorder(Tleft) b c b c
print(root of T) d e • • d e
Inorder(Tright) • • • •
Efficiency: Θ(n)

Binary Tree Algorithms (cont.)
Ex. 2: Computing the height of a binary tree
T T
L R
h(T) = ??
Efficiency: Θ(n)

Binary Tree Algorithms (cont.)
Ex. 2: Computing the height of a binary tree
T T
L R
h(T) = max{h(TL), h(TR)} + 1 if T   and h() = -1
Efficiency: Θ(n)

Multiplication of Large Integers
Consider the problem of multiplying two (large) n-digit integers
represented by arrays of their digits such as:
A = 12345678901357986429 B = 87654321284820912836
The grade-school algorithm:
a1 a2 … an
b1 b2 … bn
(d10) d11d12 … d1n
(d20) d21d22 … d2n
… … … … … … …
(dn0) dn1dn2 … dnn
Efficiency: ?? one-digit multiplications

Multiplication of Large Integers
Consider the problem of multiplying two (large) n-digit integers
represented by arrays of their digits such as:
A = 12345678901357986429 B = 87654321284820912836
The grade-school algorithm:
a1 a2 … an
b1 b2 … bn
(d10) d11d12 … d1n
(d20) d21d22 … d2n
… … … … … … …
(dn0) dn1dn2 … dnn
Efficiency: n2 one-digit multiplications

First Divide-and-Conquer Algorithm
A small example: A  B where A = 2135 and B = 4014
A = (21·102 + 35), B = (40 ·102 + 14)
So, A  B = (21 ·102 + 35)  (40 ·102 + 14)
= 21  40 ·104 + (21  14 + 35  40) ·102 + 35  14
In general, if A = A1A2 and B = B1B2 (where A and B are n-digit,
and A1, A2, B1, B2 are n / 2-digit numbers),
A  B = A1  B1·10n + (A1  B2 + A2  B1) ·10n/2 + A2  B2
Recurrence for the number of one-digit multiplications M(n):
M(n) = 4M(n/2), M(1) = 1
Solution: M(n) = n2

Second Divide-and-Conquer Algorithm
A  B = A1  B1·10n + (A1  B2 + A2  B1) ·10n/2 + A2  B2
The idea is to decrease the number of multiplications from 4 to 3:
(A1 + A2 )  (B1 + B2 ) = A1  B1 + (A1  B2 + A2  B1) + A2  B2,
I.e., (A1  B2 + A2  B1) = (A1 + A2 )  (B1 + B2 ) - A1  B1 - A2  B2,
which requires only 3 multiplications at the expense of (4-1=3)
extra additions and subtractions (2 adds and 2 subs – 1 add)
Recurrence for the number of multiplications M(n):
M(n) = 3M(n/2), M(1) = 1
Solution: M(n) = 3log 2n = nlog 23 ≈ n1.585

Example of Large-Integer Multiplication
2135  4014

Strassen’s Matrix Multiplication
Strassen observed [1969] that the product of two matrices can
be computed as follows:
C00 C01 A00 A01 B00 B01
= *
C10 C11 A10 A11 B10 B11
M1 + M4 - M5 + M7 M3 + M5
=
M2 + M4 M1 + M3 - M2 + M6
 Example: M2 + M4 = (A10 + A11)  B00 + A11  (B10 - B00)
 = A10B00 + A11B10

Formulas for Strassen’s Algorithm
M1 = (A00 + A11)  (B00 + B11)
M2 = (A10 + A11)  B00
M3 = A00  (B01 - B11)
M4 = A11  (B10 - B00)
M5 = (A00 + A01)  B11
M6 = (A10 - A00)  (B00 + B01)
M7 = (A01 - A11)  (B10 + B11)

Analysis of Strassen’s Algorithm
If n is not a power of 2, matrices can be padded with zeros.
Number of multiplications:
M(n) = 7M(n/2), M(1) = 1
Solution: M(n) = 7log 2n = nlog 27 ≈ n2.807 vs. n3 of brute-force alg.
Algorithms with better asymptotic efficiency are known but they
are even more complex.

Closest-Pair Problem by Divide-and-Conquer
Step 1 Divide the points given into two subsets Pl and Pr by a
vertical line x = m so that half the points lie to the left or on
the line and half the points lie to the right or on the line.
x = m
d l
d
r
d d

Closest-Pair Driver
 CP_Driver(A)
 Pre => A(1..n), n > 1, is an array of (X, Y), pairs
 Post => Returns distance between closest pair in A
 P(1 .. n) := Copy of A, sorted by X
 Q(1 .. n) := Copy of A, sorted by Y
 return CP(P, Q)

Closest-Pair Problem by Divide-and-Conquer
 CP(P, Q) – Pre=> P (1..n) and Q (1..n) have same points, P is x-sorted, Q y-sorted
• If n <= 3 then return CPBF(P)
• Copy P(1 .. n/2) into PL(1 .. n/2)
• Distribute-copy the same n/2 points from Q into QL (anti-merge)
• Copy P(n/2+1 .. n) into PR(1 .. n/2)
• Distribute-copy the same n/2 points from Q into QR !! Careful
• dL := CP(PL, QL); dR := CP(PR, QR)
• d := min(dL, dR); dsq := d ** 2
• midx = P(n/2).x
• strip(1 .. num) => strip(i) := Q(k) iff | Q(k).x – midx | <= d
• For i in 1 .. Num – 1 loop
• k := i + 1
• while k <= num and (strip(i).y – strip(k).y)**2 < dsq loop
• dsq := min (dsq, (strip(i).x – strip(k).x)**2 + (strip(i).y – strip(k).y)**2)
• k := k + 1
• return sqrt(dsq)

CP Notes
 CP(P, Q) – Pre=>P and Q have same points, P x sorted, Q y sorted y, P(1..n), n> 1
• Copy P(1 .. n/2) into PL(1 .. n/2) and same n/2 points from Q into QL
• Copy P(n/2+1 .. n) into PR(1 .. n/2)and same n/2 points from Q into QR
• -- Divide Q based on Q(i).x < P(n/2).x. if Q(k).x = P(n/2).x must send to correct side
• -- N.B. P and Q are sorted. PL, PR, QL, QR are distributed, not resorted.
• d := min(dL, dR) ; dsq := d** 2 -- For efficiency
• midx = P(n/2).x -- x value of middle point
• -- Copy points of Q within distance d of middle into strip. Num such points.
• For i in 1 .. Num – 1 loop k := i + 1
• k := k + 1
• -- CP can add sorted P index to points of Q, simplifying distribution!

CP Assignment: Performance Measurement
 CP(P, Q) – Pre=>P and Q have same points, P x sorted, Q y sorted y, P(1..n), n> 1
• -- Count number of calls here, using a GLOBAL VARIABLE
• Copy P(1 .. n/2) into PL(1 .. n/2) and same n/2 points from Q into QL
• Copy P(n/2+1 .. n) into PR(1 .. n/2)and same n/2 points from Q into QR
• d := min(dL, dR) ; dsq := d** 2
• midx = P(n/2).x
• For i in 1 .. Num – 1 loop k := i + 1
• -- Count number of distance calculations here, using a global
• k := k + 1

Closest Pair by Divide-and-Conquer (cont.)
Step 2 Find recursively the closest pairs for the left and right
subsets.
Step 3 Set d = min{dl, dr}
We can limit our attention to the points in the symmetric
vertical strip S of width 2d as possible closest pair. (The
points are stored and processed in increasing order of
their y coordinates.)
Step 4 Scan the points in the vertical strip S from the lowest up.
For every point p(x,y) in the strip, inspect points in
in the strip that may be closer to p than d. There can be
no more than 5 such points following p on the strip list!

Closest Pair by Divide-and-Conquer (cont.)
How many points could be in each strip? n/2
Brute force would look at all n/2 x n/2 pairs
Why can we restrict ourselves to calculating the distance
from each point to the next 5 points
Because only the next 5 points could be within distance d of the
current point. Why?
Consider a square of size d: No two points in it can be closer than
d. How many points can it contain. No more than 4 (since 5
won’t fit).
Thus a rectangle of size dx2d can only contain 8 points (or 6 if
duplicates are not allowed). So from a point, only check the
next 7 (or 5) points.

Efficiency of the Closest-Pair Algorithm
Running time of the algorithm is described by
T(n) = 2T(n/2) + M(n), where M(n)  O(n)
By the Master Theorem (with a = 2, b = 2, d = 1)
T(n)  O(n log n)

Quickhull Algorithm
Convex hull: smallest convex set that includes given points
 Assume points are sorted by x-coordinate values
 Identify extreme points P1 and P2 (leftmost and rightmost)
 Compute upper hull recursively:
• find point Pmax that is farthest away from line P1P2
• compute the upper hull of the points to the left of line P1Pmax
• compute the upper hull of the points to the left of line PmaxP2
 Compute lower hull in a similar manner
P1
P2
Pmax

Efficiency of Quickhull Algorithm
 Finding point farthest away from line P1P2 can be done in
linear time
 Time efficiency:
• Assume nu and nl points in upper and lower hulls
• Performance: T(n) = T(?) + T(?) + Θ(?)
• Worst case: T(n) = T(?) + T(?) + Θ(?)
• Expected case: T(n) = T(?) + T(?) + Θ(?)

linear time
• Performance: T(n) = T(nu) + T(nl) + Θ(n)
• Worst case: T(n) = T(1) + T(n-1) + Θ n) = ?
• Expected case: T(n) = T(n/2) + T(n/2) + Θ(n) = ?

linear time
• Worst case: T(n) = T(1) + T(n-1) + Θ(n) = Θ(n^2)
• Expected case: T(n) = T(n/2) + T(n/2) + Θ(n) = Θ(n lg n)
Can we do better?

 Finding point farthest from line P1P2 can be done in Θ(n)
• Worst case: T(n) = T(1) + T(n-1) + Θ(n) = Θ(n^2)
• Expected case: T(n) = T(n/2) + T(n/2) + Θ(n) = Θ(n lg n)
Can we do better? T(n) = T(n/4) + T(n/4) + Θ(n) = Θ(n)

 Finding point farthest from line P1P2 can be done in Θ(n)
• worst case: Θ(n2) (as quicksort)
• average case: Θ(n lg n) or Θ(n)
(assuming reasonable distribution of points)
 If points are not initially sorted by x-coordinate value, this
can be accomplished in O(n log n) time
 Several O(n log n) algorithms for convex hull are known
 Field is called Computational Geometry

ch05-2018.03.02 ch05-2018.03.02ch05-2018.03.02

More Related Content

Similar to ch05-2018.03.02 ch05-2018.03.02ch05-2018.03.02

More from Shanmuganathan C

Recently uploaded

ch05-2018.03.02 ch05-2018.03.02ch05-2018.03.02