cse 202: design and analysis of algorithms
TRANSCRIPT
CSE 202: Design and Analysis of Algorithms
Lecture 5
Instructor: Kamalika Chaudhuri
Announcements
• Kamalika’s Office hours today moved to tomorrow,12:15-1:15pm
• Homework 1 due now
• Midterm on Feb 14
Algorithm Design Paradigms
• Exhaustive Search
• Greedy Algorithms: Build a solution incrementally piece by piece
• Divide and Conquer: Divide into parts, solve each part, combine results
• Dynamic Programming: Divide into subtasks, perform subtask by size. Combine smaller subtasks to larger ones
• Hill-climbing: Start with a solution, improve it
Divide and Conquer
• Integer Multiplication
• Closest pair of points on a Plane
• Strassen’s algorithm for matrix multiplication
Integer Multiplication: The Grade School Way
[22] 1 0 1 1 0[5] 1 0 1 --------------------------------------------------- 1 0 1 1 0 0 0 0 0 0 1 0 1 1 0 ---------------------------------------------------[110] 1 1 0 1 1 1 0
How to multiply two n-bit numbers?
1.Create array of n intermediate sums2. Add up the sums
Time per addition = O(n)Total time = O(n2).
Can we do better?
A Simple Divide and Conquer
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Simple Divide and Conquer
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Simple Divide and Conquer
Operations:
1. 4 multiplications of n/2 bit #s2. Shifting by n bits3. 3 additions
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Simple Divide and Conquer
Operations:
1. 4 multiplications of n/2 bit #s2. Shifting by n bits3. 3 additions
Recurrence:T(n) = 4T(n/2) + O(n)
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Simple Divide and Conquer
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
Operations:
1. 4 multiplications of n/2 bit #s2. Shifting by n bits3. 3 additions
Recurrence:T(n) = 4T(n/2) + O(n)T(n) = O(n2)
What is the base case?
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Better Divide and Conquer
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
Need: xLyL, xRyR, (xRyL + xLyR)
Computed by 3 multiplications as:(xRyL + xLyR) = (xL + xR)(yL + yR) - xLyL - xRyR
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Better Divide and Conquer
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
Need: xLyL, xRyR, (xRyL + xLyR)
Computed by 3 multiplications as:(xRyL + xLyR) = (xL + xR)(yL + yR) - xLyL - xRyR
Operations:
1. 3 multiplications of n/2 bit #s2. Shifting by n bits3. 6 additions
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Better Divide and Conquer
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
Need: xLyL, xRyR, (xRyL + xLyR)
Computed by 3 multiplications as:(xRyL + xLyR) = (xL + xR)(yL + yR) - xLyL - xRyR
Operations:
1. 3 multiplications of n/2 bit #s2. Shifting by n bits3. 6 additions
Recurrence:T(n) = 3T(n/2) + O(n)
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
A Better Divide and Conquer
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
Operations:
1. 3 multiplications of n/2 bit #s2. Shifting by n bits3. 6 additions
Recurrence:T(n) = 3T(n/2) + O(n)T(n) = O(n1.59)
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
Need: xLyL, xRyR, (xRyL + xLyR)
Computed by 3 multiplications as:(xRyL + xLyR) = (xL + xR)(yL + yR) - xLyL - xRyR
A Better Divide and Conquer
Operations:
1. 3 multiplications of n/2 bit #s2. Shifting by n bits3. 6 additions
Recurrence:T(n) = 3T(n/2) + O(n)T(n) = O(n1.59)
Best: O(n log n 2O(log * n)) [Furer07])
x
xL xR
=
Problem: How to multiply two n-bit numbers x and y?
y
yL yR
=
y = yL 2n/2 + yR x = xL 2n/2 + xR
Need: xLyL, xRyR, (xRyL + xLyR)
Computed by 3 multiplications as:(xRyL + xLyR) = (xL + xR)(yL + yR) - xLyL - xRyR
xy = (xL 2n/2 + xR)(yL 2n/2 + yR)= xLyL 2n + (xR yL + xLyR) 2n/2 + xRyR
Divide and Conquer
• Integer Multiplication
• Closest pair of points on a Plane
• Strassen’s algorithm for matrix multiplication
Closest Pair of Points on a Plane
Given a set P of n points on the plane, find the two points p and q in P s.t d(p, q) is minimum
p q
Closest Pair of Points on a Plane
Given a set P of n points on the plane, find the two points p and q in P s.t d(p, q) is minimum
Naive solution = O(n2)
p q
What about one dimension?
Given a set P of n points on the line, find the two points p and q in P s.t d(p, q) is minimum
p q
What about one dimension?
Given a set P of n points on the line, find the two points p and q in P s.t d(p, q) is minimum
p q
Property: The closest points are adjacent in sorted order
What about one dimension?
Given a set P of n points on the line, find the two points p and q in P s.t d(p, q) is minimum
Algorithm 11. Sort the points 2. Find a point pi in the sorted set s.t d(pi, pi+1) is minimum
p q
Property: The closest points are adjacent in sorted order
What about one dimension?
Given a set P of n points on the line, find the two points p and q in P s.t d(p, q) is minimum
Algorithm 11. Sort the points 2. Find a point pi in the sorted set s.t d(pi, pi+1) is minimum
p q
Property: The closest points are adjacent in sorted order
Running Time = O(n log n)
Does this work in 2D?
p
q
a
b (a, b) : closest in x-coordinate
(a, q) : closest in y-coordinate
(p, q) : closest
Sorting the points by x or y coordinate, and looking at adjacent pairs does not work!
A Divide and Conquer Approach
Find-Closest-Pair(Q):
1. Find the median x coordinate lx2. L = Points to the left of lx3. R = Rest of the points4. Find-closest-pair(L)5. Find-closest-pair(R)6. Combine the results
How to combine?
lx
RL
Problem Case: p in L, q in R (or vice versa)
A Divide and Conquer Approach
Find-Closest-Pair(Q):
1. Find the median x coordinate lx2. L = Points to the left of lx3. R = Rest of the points4. Find-closest-pair(L)5. Find-closest-pair(R)6. Combine the results
How to combine?
1. Naive combination
2. Can we do better?
Time = O(n2)
lx
RL
A Property of Closest Points
Let: t1 = min x, y in L d(x, y)t2 = min x, y in R d(x, y)t = min (t1, t2)
If d(a, b) < t, where a is in L, b in R, then, a and b lie within distance t of lx
RL
lx
A Property of Closest Points
Let: t1 = min x, y in L d(x, y)t2 = min x, y in R d(x, y)t = min (t1, t2)
If d(a, b) < t, where a is in L, b in R, then, a and b lie within distance t of lx
lx
RL>ta
A Property of Closest Points
Let: t1 = min x, y in L d(x, y)t2 = min x, y in R d(x, y)t = min (t1, t2)
If d(a, b) < t, where a is in L, b in R, then, a and b lie within distance t of lx
lx
RL>t
>t
a
b
A Property of Closest Points
Let: t1 = min x, y in L d(x, y)t2 = min x, y in R d(x, y)t = min (t1, t2)
If d(a, b) < t, where a is in L, b in R, then, a and b lie within distance t of lx
lx
RL>t
>t
a
b
t tWe can safely rule out all points in the grey regions!
Another Property
Let: t1 = min x, y in L d(x, y)t2 = min x, y in R d(x, y)t = min (t1, t2)Sy = all points within distance t of lx in sorted order of y coords
If d(a, b) < t, a in L, b in R, then a and b occur within 15 positions of each other in Sy
lx
t/2
t/2
t
t1L
lx
t2 R
If d(a,b) < t, a in L, b in R, a, b are > 15 positions apart
Fact: Each box in figure can contain at most one point.
then, there are >= 3 rows between a and b
Boxes
Thus, d(a, b) >= 3t/2. Contradiction!
a
b
A Divide and Conquer Approach
Find-Closest-Pair(Q):
1. Find the median x coordinate lx2. L = Points to the left of lx3. R = Rest of the points4. (a1, b1, t1) = Find-closest-pair(L)5. (a2, b2, t2) = Find-closest-pair(R)6. t = min(t1, t2)7. Combine the results
How to combine?
lx
RLS = points in Q
within distance t of Ix
Sy = sort S by y coordsFor p in Sy
Find distance from pto next 15 pointsLet (a, b) be the pair withmin such distance
If d(a, b) < t:Return (a, b, d(a, b))
Else if t1 < t2:Return (a1, b1, t1)
Else:Return (a2, b2, t2)
Property: If a in L, b in R, and if d(a, b) < t, then a and b occur within 15 positions of each other in Sy
A Divide and Conquer Approach
Find-Closest-Pair(Qx, Qy):
1. Find the median x coordinate lx2. L = Points to the left of lx3. R = Rest of the points4. (a1, b1, t1) = Find-closest-pair(Lx, Ly)5. (a2, b2, t2) = Find-closest-pair(Rx, Ry)6. t = min(t1, t2)7. Combine the results
How to combine?
S = points in Q within distance t of Ix
Sy = sort S by y coordsFor p in Sy
Find distance from pto next 15 pointsLet (a, b) be the pair withmin such distance
If d(a, b) < t:Return (a, b, d(a, b))
Else if t1 < t2:Return (a1, b1, t1)
Else:Return (a2, b2, t2)
Property: If a in L, b in R, and if d(a, b) < t, then a and b occur within 15 positions of each other in Sy
Implementing Divide + Combine:1. Maintain sorted lists of the input points:
Px = sorted by x, Py = sorted by y2. Computing Lx,Ly,Rx, Ry from Qx, Qy: O(|Q|) time2. Computing Sy from Qy : O(|S|) time3. Combination procedure: O(|Sy|) time
A Divide and Conquer Approach
Find-Closest-Pair(Qx, Qy):
1. Find the median x coordinate lx2. L = Points to the left of lx3. R = Rest of the points4. (a1, b1, t1) = Find-closest-pair(Lx, Ly)5. (a2, b2, t2) = Find-closest-pair(Rx, Ry)6. t = min(t1, t2)7. Combine the results
How to combine?
S = points in Q within distance t of Ix
Sy = sort S by y coordsFor p in Sy
Find distance from pto next 15 pointsLet (a, b) be the pair withmin such distance
If d(a, b) < t:Return (a, b, d(a, b))
Else if t1 < t2:Return (a1, b1, t1)
Else:Return (a2, b2, t2)
Property: If a in L, b in R, and if d(a, b) < t, then a and b occur within 15 positions of each other in Sy
Implementing Divide + Combine:
T(n) = 2 T(n/2) + cnT(n) = (n log n)Θ
Summary: Closest Pair of Points
Given a set P of n points on the plane, find the two points p and q in P s.t d(p, q) is minimum
Find-Closest-Pair(Qx,Qy):
1. Find the median x coordinate lx2. L = Points to the left of lx3. R = Rest of the points4. (a1, b1, t1) = Find-closest-pair(Lx, Ly)5. (a2, b2, t2) = Find-closest-pair(Rx, Ry)6. t = min(t1, t2)7. Let Sy = points within distance t of Ix sorted by y 8. For each p in Sy
Find distance from p to next 15 points in Sy
Let (a, b) be the closest such pair9. Report the closest pair out of:
(a, b), (a1, b1), (a2, b2)
Recurrence:
T(n) = 2T(n/2) + cnT(n) = O(n log n)
What is a base case?
Divide and Conquer
• Integer Multiplication
• Closest pair of points on a Plane
• Strassen’s algorithm for matrix multiplication
Matrix Multiplication
Problem: Given two n x n matrices X and Y, compute Z = XY
Zij = Xik YkjΣ
X =i
j
(i,j)
X Y Z
Naive Method: O(n3) time
A Simple Divide and Conquer
Problem: Given two n x n matrices X and Y, compute Z = XY
X =A B
C DY =
E F
G H
Z =AE + BG AF + BH
CE + DG CF + DH
Algorithm 1: 1. Compute AE, BG, CE, DG, AF, BH, CF, DH2. Compute AE + BG, CE + DG, AF + BH, CF + DH
Operations: 8 n/2 x n/2 matrix multiplications, 4 additions
Recurrence: T(n) = 8 T(n/2) + O(n2) T(n) = O(n3)
A..H: n/2 x n/2
Strassen’s Algorithm
Problem: Given two n x n matrices X and Y, compute Z = XY
X =A B
C DY =
E F
G H
Z =P5 + P4 - P2 + P6 P1 + P2
P3 + P4 P1 + P5 - P3 - P7
P1 = A(F - H)P2 = (A + B)H
P3 = (C + D)EP4 = D(G - E)
P5 = (A + D)(E + H)P6 = (B - D)(G + H)
P7 = (A - C)(E + F)
Operations: 7 n/2 x n/2 matrix multiplications, O(1) additions
Recurrence: T(n) = 7 T(n/2) + O(n2) T(n) = O(n2.81)
A..H: n/2 x n/2
Strassen’s Algorithm
Problem: Given two n x n matrices X and Y, compute Z = XY
X =A B
C DY =
E F
G H
Z =P5 + P4 - P2 + P6 P1 + P2
P3 + P4 P1 + P5 - P3 - P7
P1 = A(F - H)P2 = (A + B)H
P3 = (C + D)EP4 = D(G - E)
P5 = (A + D)(E + H)P6 = (B - D)(G + H)
P7 = (A - C)(E + F)
Operations: 7 n/2 x n/2 matrix multiplications, O(1) additionsRecurrence: T(n) = 7 T(n/2) + O(n2) T(n) = O(n2.81)
A..H: n/2 x n/2
Algorithm Design Paradigms
• Exhaustive Search
• Greedy Algorithms: Build a solution incrementally piece by piece
• Divide and Conquer: Divide into parts, solve each part, combine results
• Dynamic Programming: Divide into subtasks, perform subtask by size. Combine smaller subtasks to larger ones
• Hill-climbing: Start with a solution, improve it
function Fib1(n)if n = 1 return 1if n = 2 return 1return Fib1(n-1) + Fib1(n-2)
Dynamic Programming (DP): A Simple Example
Problem: Compute the n-th Fibonacci number
Recursive Solution Running Time:
T(n) = T(n-1) + T(n-2) + 1
function Fib1(n)if n = 1 return 1if n = 2 return 1return Fib1(n-1) + Fib1(n-2)
Dynamic Programming (DP): A Simple Example
Problem: Compute the n-th Fibonacci number
Recursive Solution Running Time:
T(n) = T(n-1) + T(n-2) + 1
Running time: O(cn)
T(n) = O(cn)
function Fib1(n)if n = 1 return 1if n = 2 return 1return Fib1(n-1) + Fib1(n-2)
Dynamic Programming (DP): A Simple Example
function Fib2(n)Create an array fib[1..n]fib[1] = 1fib[2] = 1for i = 3 to n: fib[i] = fib[i-1] + fib[i-2]return fib[n]
Problem: Compute the n-th Fibonacci number
Recursive Solution
Dynamic Programming Solution
Running Time:
T(n) = T(n-1) + T(n-2) + 1
Running time: O(cn)
T(n) = O(cn)
function Fib1(n)if n = 1 return 1if n = 2 return 1return Fib1(n-1) + Fib1(n-2)
Dynamic Programming (DP): A Simple Example
function Fib2(n)Create an array fib[1..n]fib[1] = 1fib[2] = 1for i = 3 to n: fib[i] = fib[i-1] + fib[i-2]return fib[n]
Problem: Compute the n-th Fibonacci number
Recursive Solution
Dynamic Programming Solution
Running Time:
T(n) = T(n-1) + T(n-2) + 1
Running time: O(cn)
T(n) = O(n)
Running Time:
Running time: O(n)
T(n) = O(cn)
function Fib1(n)if n = 1 return 1if n = 2 return 1return Fib1(n-1) + Fib1(n-2)
Why does DP do better?
function Fib2(n)Create an array fib[1..n]fib[1] = 1fib[2] = 1for i = 3 to n: fib[i] = fib[i-1] + fib[i-2]return fib[n]
Problem: Compute the n-th Fibonacci number
Recursive Solution
Dynamic Programming Solution
Running time: O(cn)
Running time: O(n)
F(5)
F(4)
F(2)F(3)
F(2) F(1)
F(3)
F(1)F(2)
Recursion Tree
Dynamic Programming
Main Steps:
1. Divide the problem into subtasks
2. Define the subtasks recursively (express larger subtasks in terms of smaller ones)
3. Find the right order for solving the subtasks (but do not solve them recursively!)
Dynamic Programming
Main Steps:
1. Divide the problem into subtasks: compute fib[i]
2. Define the subtasks recursively (express larger subtasks in terms of smaller ones)
3. Find the right order for solving the subtasks (i = 1,..,n)
function Fib2(n)Create an array fib[1..n]fib[1] = 1fib[2] = 1for i = 3 to n: fib[i] = fib[i-1] + fib[i-2]return fib[n]
Dynamic Programming Solution
Running time: O(n)
Dynamic Programming
• String Reconstruction
• ...