CSE 202: Design and Analysis of Algorithms
Transcript of CSE 202: Design and Analysis of Algorithms
![Page 1: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/1.jpg)
CSE 202: Design and Analysis of Algorithms
Lecture 5
Instructor: Kamalika Chaudhuri
![Page 2: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/2.jpg)
Announcements
• Kamalika’s Office hours today moved to tomorrow,12:15-1:15pm
• Homework 1 due now
• Midterm on Feb 14
![Page 3: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/3.jpg)
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
![Page 4: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/4.jpg)
Divide and Conquer
• Integer Multiplication
• Closest pair of points on a Plane
• Strassen’s algorithm for matrix multiplication
![Page 5: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/5.jpg)
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?
![Page 6: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/6.jpg)
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
![Page 7: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/7.jpg)
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
![Page 8: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/8.jpg)
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
![Page 9: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/9.jpg)
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
![Page 10: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/10.jpg)
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
![Page 11: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/11.jpg)
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
![Page 12: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/12.jpg)
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
![Page 13: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/13.jpg)
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
![Page 14: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/14.jpg)
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
![Page 15: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/15.jpg)
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
![Page 16: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/16.jpg)
Divide and Conquer
• Integer Multiplication
• Closest pair of points on a Plane
• Strassen’s algorithm for matrix multiplication
![Page 17: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/17.jpg)
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
![Page 18: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/18.jpg)
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
![Page 19: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/19.jpg)
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
![Page 20: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/20.jpg)
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
![Page 21: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/21.jpg)
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
![Page 22: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/22.jpg)
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)
![Page 23: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/23.jpg)
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!
![Page 24: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/24.jpg)
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)
![Page 25: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/25.jpg)
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
![Page 26: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/26.jpg)
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
![Page 27: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/27.jpg)
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
![Page 28: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/28.jpg)
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
![Page 29: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/29.jpg)
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!
![Page 30: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/30.jpg)
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
![Page 31: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/31.jpg)
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
![Page 32: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/32.jpg)
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
![Page 33: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/33.jpg)
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
![Page 34: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/34.jpg)
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?
![Page 35: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/35.jpg)
Divide and Conquer
• Integer Multiplication
• Closest pair of points on a Plane
• Strassen’s algorithm for matrix multiplication
![Page 36: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/36.jpg)
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
![Page 37: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/37.jpg)
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
![Page 38: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/38.jpg)
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
![Page 39: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/39.jpg)
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
![Page 40: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/40.jpg)
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
![Page 41: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/41.jpg)
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
![Page 42: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/42.jpg)
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)
![Page 43: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/43.jpg)
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)
![Page 44: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/44.jpg)
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)
![Page 45: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/45.jpg)
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
![Page 46: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/46.jpg)
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!)
![Page 47: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/47.jpg)
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)
![Page 48: CSE 202: Design and Analysis of Algorithms](https://reader033.fdocuments.us/reader033/viewer/2022051521/586910be1a28ab65328b9675/html5/thumbnails/48.jpg)
Dynamic Programming
• String Reconstruction
• ...