Day 8—More Divide and Conquer and Master Method for Solving Recurrences
Neil RhodesCSE 101
UC San Diego
MIDTERM April 29, 4-5 PM Center Hall Room 101NOT HSS
5.4 Closest Pair of Points
3
Closest Pair of Points
Closest pair. Given n points in the plane, find a pair with smallest Euclidean distance between them.
Fundamental geometric primitive. Graphics, computer vision, geographic information systems,
molecular modeling, air traffic control. Special case of nearest neighbor, Euclidean MST, Voronoi.
Brute force. Check all pairs of points p and q with Θ(n2) comparisons.
1-D version. O(n log n) easy if points are on a line.
Assumption. No two points have same x coordinate.
to make presentation cleaner
fast closest pair inspired fast algorithms for these problems
4
Closest Pair of Points: First Attempt
Divide. Sub-divide region into 4 quadrants.
L
5
Closest Pair of Points: First Attempt
Divide. Sub-divide region into 4 quadrants.Obstacle. Impossible to ensure n/4 points in each piece.
L
6
Closest Pair of Points
Algorithm. Divide: draw vertical line L so that roughly ½n points on each side.
L
7
Closest Pair of Points
Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. Conquer: find closest pair in each side recursively.
12
21
L
8
Closest Pair of Points
Algorithm. Divide: draw vertical line L so that roughly ½n points on each side. Conquer: find closest pair in each side recursively. Combine: find closest pair with one point in each side. Return best of 3 solutions.
12
218
L
seems like Θ(n2)
9
Closest Pair of Points
Find closest pair with one point in each side, assuming that distance < δ.
12
21
δ = min(12, 21)
L
10
Closest Pair of Points
Find closest pair with one point in each side, assuming that distance < δ. Observation: only need to consider points within δ of line L.
12
21
δ
L
δ = min(12, 21)
11
12
21
1
2
3
45
6
7
δ
Closest Pair of Points
Find closest pair with one point in each side, assuming that distance < δ. Observation: only need to consider points within δ of line L. Sort points in 2δ-strip by their y coordinate.
L
δ = min(12, 21)
12
12
21
1
2
3
45
6
7
δ
Closest Pair of Points
Find closest pair with one point in each side, assuming that distance < δ. Observation: only need to consider points within δ of line L. Sort points in 2δ-strip by their y coordinate. Only check distances of those within 11 positions in sorted list!
L
δ = min(12, 21)
13
Closest Pair of Points
Def. Let si be the point in the 2δ-strip, withthe ith smallest y-coordinate.
Claim. If |i – j| ≥ 12, then the distance betweensi and sj is at least δ.Pf. No two points lie in same ½δ-by-½δ box. Two points at least 2 rows apart
have distance ≥ 2(½δ).
Fact. Still true if we replace 12 with 7.
δ
27
2930
31
28
26
25
δ
½δ
2 rows½δ
½δ
39
i
j
14
Closest Pair Algorithm
Closest-Pair(p1, …, pn) Compute separation line L such that half the points are on one side and half on the other side.
δ1 = Closest-Pair(left half) δ2 = Closest-Pair(right half) δ = min(δ1, δ2)
Delete all points further than δ from separation line L
Sort remaining points by y-coordinate.
Scan points in y-order and compare distance between each point and next 11 neighbors. If any of these distances is less than δ, update δ.
return δ.
O(n log n)
2T(n / 2)
O(n)
O(n log n)
O(n)
15
Closest Pair of Points: Analysis
Running time.
Q. Can we achieve O(n log n)?
A. Yes. Don't sort points in strip from scratch each time. Each recursive returns two lists: all points sorted by y coordinate,
and all points sorted by x coordinate. Sort by merging two pre-sorted lists.
€
T(n) ≤ 2T n /2( ) + O(n) ⇒ T(n) = O(n logn)
€
T(n) ≤ 2T n /2( ) + O(n log n) ⇒ T(n) = O(n log2 n)
16
O(n log n) Closest Pair Algorithm
Closest-Pair(p1, …, pn) Compute separation line L such that half the points are on one side and half on the other side.
(xsorted1, ysorted1, δ1) = Closest-Pair(left half) (xsorted2, ysorted2, δ2) = Closest-Pair(right half) δ = min(δ1, δ2)
Merge xsorted1 and xsorted2 into xsorted Merge ysorted1 and ysorted2 into ysorted
Scan points in y-order and compare distance between each point and next 11 neighbors. If any of these distances is less than δ, update δ.
return (xsorted, ysorted, δ).
O(n)
2T(n / 2)
O(n)
O(n)
Master method for solving recurrence relations
Master method idea
A recurrence tree will, in general, have either Time dominated by cost of the leaves
Time dominated by cost of the root
Time evenly distributed across the levels of the tree
18
Time dominated by leaves
19
Tree MethodT (n) = 4T (!n
2")+nn
( n2 )
( n4 ) ( n
4 ) ( n4 ) ( n
4 )
( n2 )
( n4 ) ( n
4 ) ( n4 ) ( n
4 )
( n2 )
( n4 ) ( n
4 ) ( n4 ) ( n
4 )
( n2 )
( n4 ) ( n
4 ) ( n4 ) ( n
4 )
log2 n
n
2n
4n
4log2 n
CSE 202—Day 3 – p.7
Tree MethodT (n) = 4T (!n
2")+n
T (n) = n+2n+4n+ ...+4log2 n
=log2 n
∑i=0
n2i = nlog2 n
∑i=0
2i
= n2logn+1
= n(n+1)
T (n) = Θ(n2)
CSE 202—Day 3 – p.8
Time evenly distributed across levels
20
Tree MethodT (n) = 4T (!n
4")+nn
n4
n16
n16
n16
n16
n4
n16
n16
n16
n16
n4
n16
n16
n16
n16
n4
n16
n16
n16
n16
log4 n
n
n
n
4log4 n
CSE 202—Day 3 – p.12
Tree MethodT (n) = 4T (!n
4")+n
T (n) = n+n+n+ ...+4log4 n
=log2 n
∑i=0
n = nlog2 n
∑i=0
1 = nΘ(logn)
T (n) = Θ(n logn)
CSE 202—Day 3 – p.13
Time dominated by root
21
Tree MethodT (n) = 3T (!n
2")+O(n2)
T (n) = cn2 + cn2(34)
1+ cn2(
34)
2+ ...+ c3log2 n
= cn2(34)
0+ cn2(
34)
1+ ...+ c3log2n(
n2log2 n)
2
= cn2(34)
0+ cn2(
34)
1+ ...+ cn2 3log2n
4log2 n
=log2 n
∑i=0
cn2(34)
i= cn2
log2 n
∑i=0
(34)
i= cn2Θ(1)
T (n) = Θ(n2)CSE 202—Day 3 – p.5
Tree MethodT (n) = 3T (!n
2")+O(n2)
cn2
c( n2 )2
c( n4 )2
c c c
c( n4 )2
c c c
c( n4 )2
c c c
c( n2 )2
c( n4 )2
c c c
c( n4 )2
c c c
c( n4 )2
c c c
c( n2 )2
c( n4 )2
c c c
c( n4 )2
c c c
c( n4 )2
c c c
log2 n
cn2
34 cn2
( 34 )2cn2
c3log2 n
CSE 202—Day 3 – p.4
Limited master method
22
Limited Master MethodRecurrence: T (n) = aT (n/b)+n
n
nb
nb2 ... n
b2
...
nb2 ... n
b2
nb
nb2 ... n
b2
logb n
n
a nb
a2 nb2
alogb n = nlogba
T (n) = n+ab
n+(ab)2n+ ...+alogb n
=logb n
∑i=0
(ab)
in = n
logb n
∑i=0
(ab)
i
Day 3 – p.14
Limited master method
23
Limited Master MethodEvaluate: T (n) = n∑logb n
i=0 (ab)
i
1. ab > 1: Time dominated by leaves. T (n) =
n∑logb ni=0 (a
b)i = n(ab)logb(n+1)−1
ab−1 = O(n(a
b)logb n) =
O(nalogb n
blogb n ) = O(alogb n) = O(nlogb a)
2. ab = 1: Time evenly distributed across all levels.T (n) = n logn = O(n logn).
3. ab < 1: Time dominated by root.
n∑logb ni=0 (a
b)i < n 11− a
b= O(n).
CSE 202—Day 3 – p.15
Limited master method
Examples
24
Limited Master MethodExamples:
• T (n) = 3T (!n2")+n: a
b = 32 > 1.
T (n) = O(nlog23) = O(n1.6)• T (n) = 4T (!n
4")+n: ab = 1. T (n) = O(n logn).
• T (n) = 4T (!n5")+n. a
b < 1. T (n) = O(n).
CSE 202—Day 3 – p.16
Limited Master MethodExamples:
• T (n) = 3T (!n2")+n: a
b = 32 > 1.
T (n) = O(nlog23) = O(n1.6)• T (n) = 4T (!n
4")+n: ab = 1. T (n) = O(n logn).
• T (n) = 4T (!n5")+n. a
b < 1. T (n) = O(n).
CSE 202—Day 3 – p.16
Limited Master MethodExamples:
• T (n) = 3T (!n2")+n: a
b = 32 > 1.
T (n) = O(nlog23) = O(n1.6)• T (n) = 4T (!n
4")+n: ab = 1. T (n) = O(n logn).
• T (n) = 4T (!n5")+n. a
b < 1. T (n) = O(n).
CSE 202—Day 3 – p.16
Master Method
Recurrence:
Case 1: f(n) polynomially smaller than (Lots of tiny subproblems)
25
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb ) Day 3 – p.17
Master Method
Recurrence:
Case 2: (All levels about the same)
26
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 2: f (n) = Θ(nlogb a) (All levels about the same)
T (n) = logb nΘ(nlogb a)CSE 202—Day 3 – p.18
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 2: f (n) = Θ(nlogb a) (All levels about the same)
T (n) = logb nΘ(nlogb a)CSE 202—Day 3 – p.18
Master Method
Recurrence:
Case 3: f(n) polynomially larger than (Big step is most expensive)
27
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 1: f (n) polynomially smaller than nlogb a (Lotsof tiny suproblems)
f (n) = O(nlogb a−ε)
T (n) = Θ(alogb n) = Θ(nlogb a) = Θ(nlogalogb )
CSE 202—Day 3 – p.17
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 3: f (n) polynomially larger than nlogb a (Big stepis most expensive)
f (n) = Ω(nlogb a+ε)T (n) = Θ( f (n))
CSE 202—Day 3 – p.19
Master MethodRecurrence: T (n) = aT (n/b)+ f (n)
f (n)
f ( nb )
f ( nb2 ) ... f ( n
b2 )
...
f ( nb2 ) ... f ( n
b2 )
f ( nb )
f ( nb2 ) ... f ( n
b2 )
logb n
f (n)
a f ( nb )
a2 f ( nb2 )
alogb n = nlogba
Case 3: f (n) polynomially larger than nlogb a (Big stepis most expensive)
f (n) = Ω(nlogb a+ε)T (n) = Θ( f (n))
Day 3 – p.19
Master Method
Additional requirement: For case 3, also need regularity: there exists c > 1 such that for
sufficiently large n: f(n) ≥ c•af(n/b). That is, each lower level is at least a constant fraction smaller than the higher level. Actually, this regularity implies polynomially larger, so case 3 need meet only this requirement.
Polynomials or are regular.
28
Master Method• For case 3, also need regularity: there exists c < 1
such that for sufficiently large n: c f (n) ≥ a f (nb).
That is, each lower level costs less than theprevious higher level. Actually, this regularityimplies polynomially larger, so case 3 need meetonly this requirement.Polynomials or f (n) = cnk log j n for c,k, j ≥ 0are regular.
• Book shows how floors and ceilings within call toT can be ignored in order analysis
T (n) = T ("nb#), S(n) = T ($n
b%), R(n) = T (
nb)
T (n) = Θ(S(n)) = Θ(R(n))Day 3 – p.20
Master Method
Examples
29
Master MethodExamples:
• T (n) = 3T (n2)+O(n2): f (n) = Ω(nlog23+.2).
T (n) = O(n2)• T (n) = 3T (n
2)+n1.4: f (n) = O(nlog2 3−.1).T (n) = O(nlog2 3).
• T (n) = 16T (n4)+10n2 +3n+5:
f (n) = Θ(nlog4 16). T (n) = O(n2 logn).
Day 3 – p.21
Master MethodExamples:
• T (n) = 3T (n2)+O(n2): f (n) = Ω(nlog23+.2).
T (n) = O(n2)• T (n) = 3T (n
2)+n1.4: f (n) = O(nlog2 3−.1).T (n) = O(nlog2 3).
• T (n) = 16T (n4)+10n2 +3n+5:
f (n) = Θ(nlog4 16). T (n) = O(n2 logn).
Day 3 – p.21
Master MethodExamples:
• T (n) = 3T (n2)+O(n2): f (n) = Ω(nlog23+.2).
T (n) = O(n2)• T (n) = 3T (n
2)+n1.4: f (n) = O(nlog2 3−.1).T (n) = O(nlog2 3).
• T (n) = 16T (n4)+10n2 +3n+5:
f (n) = Θ(nlog4 16). T (n) = O(n2 logn).
Day 3 – p.21
5.5 Integer Multiplication
31
Integer Arithmetic
Add. Given two n-digit integers a and b, compute a + b. O(n) bit operations.
Multiply. Given two n-digit integers a and b, compute a × b. Brute force solution: Θ(n2) bit operations.
1
1
0
0
0
1
1
1
0
0
1
1
1
1
0
0
1
1
1
1
0
1
0
1
00000000
01010101
01010101
01010101
01010101
01010101
00000000
0100000000001011
1
0
1
1
1
1
1
0
0
*
1
011 1
110 1+
010 1
111
010 1
011 1
100 0
10111
Add
Multiply
32
To multiply two n-digit integers: Multiply four ½n-digit integers. Add two ½n-digit integers, and shift to obtain result.
Divide-and-Conquer Multiplication: Warmup
€
T(n) = 4T n /2( )recursive calls1 2 4 3 4
+ Θ(n)add, shift1 2 3
⇒ T(n) =Θ(n2 )
€
x = 2n / 2 ⋅ x1 + x0
y = 2n / 2 ⋅ y1 + y0
xy = 2n / 2 ⋅ x1 + x0( ) 2n / 2 ⋅ y1 + y0( ) = 2n ⋅ x1y1 + 2n / 2 ⋅ x1y0 + x0 y1( ) + x0 y0
assumes n is a power of 2Master method: logba = log24 = 2
f(n) = O(n2-1). Case 1:
33
To multiply two n-digit integers: Add two ½n digit integers. Multiply three ½n-digit integers. Add, subtract, and shift ½n-digit integers to obtain result.
Theorem. [Karatsuba-Ofman, 1962] Can multiply two n-digit integers in O(n1.585) bit operations.
Karatsuba Multiplication
€
x = 2n / 2 ⋅ x1 + x0
y = 2n / 2 ⋅ y1 + y0
xy = 2n ⋅ x1y1 + 2n / 2 ⋅ x1y0 + x0 y1( ) + x0 y0
= 2n ⋅ x1y1 + 2n / 2 ⋅ (x1 + x0 ) (y1 + y0 ) − x1y1 − x0 y0( ) + x0 y0
€
T(n) ≤ T n /2 ( ) + T n /2 ( ) + T 1+ n /2 ( )recursive calls
1 2 4 4 4 4 4 4 4 3 4 4 4 4 4 4 4 + Θ(n)
add, subtract, shift1 2 4 3 4
⇒ T(n) = O(n log 2 3 ) = O(n1.585 )
A B CA C
Karatsuba: Master Method
34
€
T(n) =0 if n =1
3T(n /2) + n otherwise
logba = log23 = (roughly) 1.585
f(n) = n
f(n) = O(n1.585 - .1) (polynomially smaller)
T(n) = θ(n1.585)
35
Karatsuba: Recursion Tree
€
T(n) =0 if n =1
3T(n /2) + n otherwise
n
3(n/2)
9(n/4)
3k (n / 2k)
3 lg n (2)
. . .
. . .
T(n)
T(n/2)
T(n/4) T(n/4)
T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2)
T(n / 2k)
T(n/4)
T(n/2)
T(n/4) T(n/4)T(n/4)
T(n/2)
T(n/4) T(n/4)T(n/4)
. . .
. . .
€
T(n) = n 32( )k
k=0
log2 n
∑ = 32( )1+ log2 n
−132 −1
= 3nlog2 3 − 2