cs 3343: analysis of algorithms review for exam 2

CS 3343: Analysis of Algorithms

Review for Exam 2

Exam 2

• Closed book exam

• One cheat sheet allowed (limit to a single page of letter-size paper, double-sided)

• Tuesday, Nov 17, class time + 5 minutes

• Basic calculator (no graphing) is allowed but not necessary

Final grade calculation

• Three components: HQP, Midterm, Final• HQ = HW + Quiz + Participation

: (90% HW + 10% Quiz&Participation)

: One lowest HW dropped

• Midterm = 40% midterm1+ 60% midterm2• Max for each component is 100. • Sort three components (highest, middle, lowest)• Final grade:

– Case 1: HQP is not the highest

55% highest + 30% middle + 15% lowest

– Case 2: HQP is highest

55% middle + 30% HQP + 15% lowest

Materials covered

• Quick Sort• Heap sort, priority queue• Linear time sorting algorithms• Order statistics• Dynamic programming• Greedy algorithm• Graph basics

• Questions will be similar to homework / quizzes– Familiar with the algorithm procedure– Some analysis of time/space complexity– One or two problems on algorithm design

QuickSort

• Quicksort algorithm

• Randimized Quicksort algorithm

• Partition algorithm

• Quicksort/Randomized Quicksort time complexity analysis

Back

Quick sort

Quicksort an n-element array:

1. Divide: Partition the array into two subarrays around a pivot x such that elements in lower subarray x elements in upper subarray.

2. Conquer: Recursively sort the two subarrays.

3. Combine: Trivial.

x x xx ≥ x≥ x

Key: Linear-time partitioning subroutine.

Pseudocode for quicksort

QUICKSORT(A, p, r)if p < r

then q PARTITION(A, p, r) QUICKSORT(A, p, q–1)QUICKSORT(A, q+1, r)

Initial call: QUICKSORT(A, 1, n)

Partition CodePartition(A, p, r) x = A[p]; // pivot is the first element i = p; j = r + 1; while (TRUE) {

repeat i++; until A[i] > x | i >= j; repeat j--; until A[j] < x | j <= i; if (i < j) Swap (A[i], A[j]); else break;

} swap (A[p], A[j]); return j;

i j66 1010 55 88 1313 33 22 1111x = 6

p r

i j66 1010 55 88 1313 33 22 1111

i j66 22 55 88 1313 33 1010 1111

i j66 22 55 88 1313 33 1010 1111

i j66 22 55 33 1313 88 1010 1111

ij66 22 55 33 1313 88 1010 1111

33 22 55 66 1313 88 1010 1111qp r

scan

scan

scan

swap

swap

final swap

Partition example

66 1010 55 88 1111 33 22 1313

33 22 55 66 1111 88 1010 1313

22 33 55 66 88 1010 1111 1313

22 33 55 66 1010 88 1111 1313

22 33 55 66 88 1010 1111 1313

Quick sortexample

Quicksort Runtimes

• Best case runtime Tbest(n) O(n log n)

• Worst case runtime Tworst(n) O(n2)

• Average case runtime Tavg(n) O(n log n)

• Expected runtime of randomized quicksort is O(n log n)

Randomized Partition

• Randomly choose an element as pivot– Every time need to do a partition, throw a die to

decide which element to use as the pivot– Each element has 1/n probability to be selected

Rand-Partition(A, p, r){ d = random(); // draw a random number between 0 and 1 index = p + floor((r-p+1) * d); // p<=index<=q swap(A[p], A[index]); Partition(A, p, r); // now use A[p] as pivot}

Running time of randomized quicksort

• The expected running time is an average of all cases

T(n) =

T(0) + T(n–1) + dn if 0 : n–1 split,T(1) + T(n–2) + dn if 1 : n–2 split,T(n–1) + T(0) + dn if n–1 : 0 split,

nknTkTn

nTn

k

1

0)1()(

1)(

Expectation

nkTn

n

k

1

0)(

2

• Fact:

• Need to Prove: T(n) ≤ c n log (n)

• Assumption: T(k) ≤ ck log (k) for 0 ≤ k ≤ n-1

• Prove by induction

nkTn

nTn

k

1

0)(

2)(

nn

nn

n

cnkk

n

cnT

n

k

8log

2

2log

2)(

221

0

ncn

ncnnT 4

log)(

ncnnT log)( If c ≥ 4

Heaps

• Heap definition

• Heapify

• Buildheap– Procedure and running time (!)

• Heapsort– Comparison with other sorting algorithms in

terms of running time, stability, and in-place.

• Priority Queue operations

Back

• A heap can be seen as a complete binary tree:

Heaps

16

14 10

8 7 9 3

2 4 1

Perfect binary tree

16 14 10 8 7 9 3 2 4 1

Referencing Heap Elements

• So…Parent(i)

{return i/2;}

Left(i)

{return 2*i;}

right(i)

{return 2*i + 1;}

Heap Operations: Heapify()

Heapify(A, i){ // precondition: subtrees rooted at l and r are heaps

l = Left(i); r = Right(i);if (l <= heap_size(A) && A[l] > A[i])

largest = l;else

largest = i;if (r <= heap_size(A) && A[r] > A[largest])

largest = r;if (largest != i) {

Swap(A, i, largest);Heapify(A, largest);

}} // postcondition: subtree rooted at i is a heap

Among A[l], A[i], A[r],which one is largest?

If violation, fix it.

Heapify() Example

16

4 10

14 7 9 3

2 8 1

16 4 10 14 7 9 3 2 8 1A =

Heapify() Example

16

4 10

14 7 9 3

2 8 1

16 10 14 7 9 3 2 8 1A = 4

Heapify() Example

16

4 10

14 7 9 3

2 8 1

16 10 7 9 3 2 8 1A = 4 14

Heapify() Example

16

14 10

4 7 9 3

2 8 1

16 14 10 7 9 3 2 8 1A = 4

Heapify() Example

16

14 10

4 7 9 3

2 8 1

16 14 10 7 9 3 2 1A = 4 8

Heapify() Example

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 1A = 4

Heapify() Example

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =

BuildHeap()

// given an unsorted array A, make A a heap

BuildHeap(A)

{

heap_size(A) = length(A);

for (i = length[A]/2 downto 1)Heapify(A, i);

}

BuildHeap() Example

• Work through exampleA = {4, 1, 3, 2, 16, 9, 10, 14, 8, 7}

4

1 3

2 16 9 10

14 8 7

4 1 3 2 16 9 10 14 8 7A =

4

1 3

2 16 9 10

14 8 7

4 1 3 2 16 9 10 14 8 7A =

4

1 3

14 16 9 10

2 8 7

4 1 3 14 16 9 10 2 8 7A =

4

1 10

14 16 9 3

2 8 7

4 1 10 14 16 9 3 2 8 7A =

4

16 10

14 1 9 3

2 8 7

4 16 10 14 1 9 3 2 8 7A =

4

16 10

14 7 9 3

2 8 1

4 16 10 14 7 9 3 2 8 1A =

16

4 10

14 7 9 3

2 8 1

16 4 10 14 7 9 3 2 8 1A =

16

14 10

4 7 9 3

2 8 1

16 14 10 4 7 9 3 2 8 1A =

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =

Analyzing BuildHeap(): Tight

• To Heapify() a subtree takes O(h) time where h is the height of the subtree– h = O(lg m), m = # nodes in subtree– The height of most subtrees is small

• Fact: an n-element heap has at most n/2h+1 nodes of height h

• CLR 6.3 uses this fact to prove that BuildHeap() takes O(n) time

n

hh

n

hh

hn

nhnT

22 log

11

log

11 22

)(

Heapsort

Heapsort(A)

{

BuildHeap(A);

for (i = length(A) downto 2)

{

Swap(A[1], A[i]);

heap_size(A) -= 1;

Heapify(A, 1);

}

}

Heapsort Example

• Work through exampleA = {4, 1, 3, 2, 16, 9, 10, 14, 8, 7}

4

1 3

2 16 9 10

14 8 7

4 1 3 2 16 9 10 14 8 7A =

Heapsort Example

• First: build a heap

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =

Heapsort Example

• Swap last and first

1

14 10

8 7 9 3

2 4 16

1 14 10 8 7 9 3 2 4 16A =

Heapsort Example

• Last element sorted

1

14 10

8 7 9 3

2 4 16

1 14 10 8 7 9 3 2 4 16A =

Heapsort Example

• Restore heap on remaining unsorted elements

14

8 10

4 7 9 3

2 1 16 Heapify

14 8 10 4 7 9 3 2 1 16A =

Heapsort Example

• Repeat: swap new last and first

1

8 10

4 7 9 3

2 14 16

1 8 10 4 7 9 3 2 14 16A =

Heapsort Example

• Restore heap

10

8 9

4 7 1 3

2 14 16

10 8 9 4 7 1 3 2 14 16A =

Heapsort Example

• Repeat

9

8 3

4 7 1 2

10 14 16

9 8 3 4 7 1 2 10 14 16A =

Heapsort Example

• Repeat

8

7 3

4 2 1 9

10 14 16

8 7 3 4 2 1 9 10 14 16A =

Heapsort Example

• Repeat

1

2 3

4 7 8 9

10 14 16

1 2 3 4 7 8 9 10 14 16A =

Implementing Priority Queues

HeapMaximum(A){

return A[1];}


HeapExtractMax(A){ if (heap_size[A] < 1) { error; } max = A[1]; A[1] = A[heap_size[A]] heap_size[A] --; Heapify(A, 1); return max;}

HeapExtractMax Example

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =


Swap first and last, then remove last

1

14 10

8 7 9 3

2 4 16

14 10 8 7 9 3 2 4 16A = 1


Heapify

14

8 10

4 7 9 3

2 1

10 7 9 3 2 16A =

16

14 8 4 1


HeapChangeKey(A, i, key){ if (key ≤ A[i]){ // decrease key

A[i] = key; heapify(A, i);

} else { // increase key A[i] = key;

while (i>1 & A[parent(i)]<A[i])swap(A[i], A[parent(i)];

}}

Sift down

Bubble up

HeapChangeKey Example

Increase key

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =


Increase key

16

14 10

15 7 9 3

2 4 1

16 14 10 7 9 3 2 4 1A = 15


Increase key

16

15 10

14 7 9 3

2 4 1

16 10 7 9 3 2 4 1A = 1415

Implementing Priority QueuesHeapInsert(A, key) { heap_size[A] ++; i = heap_size[A];

A[i] = -∞; HeapChangeKey(A, i, key);

}

HeapInsert Example

HeapInsert(A, 17)

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =

HeapInsert Example

HeapInsert(A, 17)

16

14 10

8 7 9 3

2 4 1

16 14 10 8 7 9 3 2 4 1A =

-∞

-∞

-∞ makes it a valid heap

HeapInsert Example

HeapInsert(A, 17)

16

14 10

8 7 9 3

2 4 1

16 10 8 9 3 2 4 1A =

17

1714 7

Now call changeKey

HeapInsert Example

HeapInsert(A, 17)

17

16 10

8 14 9 3

2 4 1

17 10 8 9 3 2 4 1A =

7

716 14

Linear time sorting

• Counting sort– Algorithm and analysis– Conditions that make counting sort linear– Stable property and memory requirement

• Radix sort– Analysis

• Comparison of different sorting algorithms– Running time– Memory requirement– Stable property

Back

Counting sort

for i 1 to kdo C[i] 0

for j 1 to ndo C[A[ j]] C[A[ j]] + 1 ⊳ C[i] = |{key = i}|

for i 2 to kdo C[i] C[i] + C[i–1] ⊳ C[i] = |{key i}|

for j n downto 1do B[C[A[ j]]] A[ j]

C[A[ j]] C[A[ j]] – 1

1.

2.

3.

4.

Initialize

Count

Compute running sum

Re-arrange

Counting sort

A: 44 11 33 44 33

B:

1 2 3 4 5

C: 11 00 22 22

1 2 3 4

C': 11 11 33 55

for i 2 to kdo C[i] C[i] + C[i–1] ⊳ C[i] = |{key i}|

3.

Loop 4: re-arrange

A: 44 11 33 44 33

B: 33

1 2 3 4 5

C: 11 11 33 55

1 2 3 4

C': 11 11 33 55


C[A[ j]] C[A[ j]] – 1

4.

Analysisfor i 1 to k

do C[i] 0

(n)

(k)

(n)

(k)

for j 1 to ndo C[A[ j]] C[A[ j]] + 1

for i 2 to kdo C[i] C[i] + C[i–1]


C[A[ j]] C[A[ j]] – 1

(n + k)

1.

2.

3.

4.

Stable sorting

Counting sort is a stable sort: it preserves the input order among equal elements.

A: 44 11 33 44 33

B: 11 33 33 44 44

Why this is important?What other algorithms have this property?

Radix sort

• Similar to sorting the address books

• Treat each digit as a key

• Start from the least significant bit

198099109123518183599340199540380128115295384700101594539614696382408360201039258538614386507628681328936

Most significant Least significant

Time complexity

• Sort each of the d digits by counting sort• Total cost: d (n + k)

– k = 10– Total cost: Θ(dn)

• Partition the d digits into groups of 3– Total cost: (n+103)d/3

• We work with binaries rather than decimals– Partition a binary number into groups of r bits– Total cost: (n+2r)d/r– Choose r = log n– Total cost: dn / log n– Compare with dn log n

• Catch: faster than quicksort only when n is very large

Order statistics

• Select vs Randomize Select– Note: select is different from selection sort– Algorithm– Worst-case vs average-case (expected)

running time analysis

• Worst-case Linear-time Select– Algorithm and analysis– Comparison between sorting and select

algorithmsBack

Randomized select algorithm

RAND-SELECT(A, p, q, i) ⊳ i th smallest of A[ p . . q] if p = q & i > 1 then error!r RAND-PARTITION(A, p, q)k r – p + 1 ⊳ k = rank(A[r])if i = k then return A[ r]if i < k

then return RAND-SELECT( A, p, r – 1, i )else return RAND-SELECT( A, r + 1, q, i – k )

A[r] A[r] A[r] A[r]rp q

k

77 1010 55 88 1111 33 22 1313

33 22 55 77 1111 88 1010 1313

1010

1010 88 1111 1313

88 1010

Complete example: select the 6th smallest element.

i = 6

k = 4

i = 6 – 4 = 2

k = 3

i = 2 < k

k = 2

i = 2 = k

Note: here we always used first element as pivot to do the partition (instead of rand-partition).

Running time of randomized select

• For upper bound, assume ith element always falls in larger side of partition

• The expected running time is an average of all cases

T(n) ≤

T(max(0, n–1)) + n if 0 : n–1 split,T(max(1, n–2)) + n if 1 : n–2 split,T(max(n–1, 0)) + n if n–1 : 0 split,

nknkTn

nTn

k

1

0)1,max(

1)(

Expectation

Substitution method

Assume: T(k) ≤ ck for all k < n

nkTn

nknkTn

nTn

nk

n

k

1

2

1

0)(

2)1,max(

1)(

nknc

nkTn

nTn

nk

n

nk

1

2

1

2

2)(

2)(

cncn

ncnncn

nn

nc

nT )4

(4

38

32)(

2

if c ≥ 4Therefore, T(n) = O(n)

Want to show T(n) = O(n). So need to prove T(n) ≤ cn for n > n0

Worst-case linear-time select

if i = k then return xelseif i < k

then recursively SELECT the i th smallest element in the

lower partelse recursively SELECT the (i–

k)th smallest element in the upper part

SELECT(i, n)1. Divide the n elements into groups of 5. Find

the median of each 5-element group by rote.2. Recursively SELECT the median x of the n/5

group medians to be the pivot.3. Partition around the pivot x. Let k = rank(x).4.

Same as RAND-SELECT

Developing the recurrence

if i = k then return xelseif i < k

then recursively SELECT the i th smallest element in the

lower partelse recursively SELECT the (i–

k)th smallest element in the upper part

SELECT(i, n)1. Divide the n elements into groups of 5. Find

the median of each 5-element group by rote.2. Recursively SELECT the median x of the n/5

group medians to be the pivot.3. Partition around the pivot x. Let k = rank(x).4.

T(n)

(n)

T(n/5)

(n)

T(7n/10+3)

nnTnTnT

3

107

51

)(

Solving the recurrence

if c ≥ 20 and n ≥ 60cn

ncncn

ncn

ncncn

nncncnT

)20/(

20/19

4/35

)3107()5()(

Assumption: T(k) ck for all k < n

if n ≥ 60

Dynamic programming

• Proofs of optimal substructure property• Defining recurrence • Solve DP problem without recursive calls• Example DP problems:

– Shortest path problem– LCS– Restaurant location– Knapsack– Scheduling– Other similar problems

Back

Elements of dynamic programming

• Optimal sub-structures– Optimal solutions to the original problem

contains optimal solutions to sub-problems

• Overlapping sub-problems– Some sub-problems appear in many solutions

Two steps to dynamic programming

• Formulate the solution as a recurrence relation of solutions to subproblems.

• Specify an order to solve the subproblems so you always have what you need.

Optimal subpaths

• Claim: if a path startgoal is optimal, any sub-path, startx, or xgoal, or xy, where x, y is on the optimal path, is also the shortest.

• Proof by contradiction– If the subpath between x and y is not the shortest, we can

replace it with the shorter one, which will reduce the total length of the new path => the optimal path from start to goal is not the shortest => contradiction!

– Hence, the subpath xy must be the shortest among all paths from x to y

start goalx ya

bc

b’

a + b + c is shortest

b’ < b

a + b’ + c < a + b + c

Dynamic programming illustration3 9 1 2

3 2 5 2

2 4 2 3

3 6 3 3

1 2 3 2

5 3 3 3 3

2 3 3 9 3

6 2 3 7 4

4 6 3 1 3

3 12 13 15

6 8 13 15

9 11 13 16

11 14 17 20

17 17 18 20

0

5

7

13

17

S

G

F(i-1, j) + dist(i-1, j, i, j) F(i, j) = min

F(i, j-1) + dist(i, j-1, i, j)

Trace back

3 9 1 2

3 2 5 2

2 4 2 3

3 6 3 3

1 2 3 2

5 3 3 3 3

2 3 3 9 3

6 2 3 7 4

4 6 3 1 3

3 12 13 15

6 8 13 15

9 11 13 16

11 14 17 20

17 17 18 20

0

5

7

13

17

Longest Common Subsequence

• Given two sequences x[1 . . m] and y[1 . . n], find a longest subsequence common to them both.

x: A B C B D A B

y: B D C A B A

“a” not “the”

BCBA = LCS(x, y)

functional notation, but not a function

Optimal substructure

• Notice that the LCS problem has optimal substructure: parts of the final solution are solutions of subproblems.– If z = LCS(x, y), then any prefix of z is an LCS of a prefix of

x and a prefix of y.

• Subproblems: “find LCS of pairs of prefixes of x and y”

x

y

m

nz

i

j

Finding length of LCS

• Let c[i, j] be the length of LCS(x[1..i], y[1..j])=> c[m, n] is the length of LCS(x, y)

• If x[m] = y[n]c[m, n] = c[m-1, n-1] + 1

• If x[m] != y[n]c[m, n] = max { c[m-1, n], c[m, n-1] }

x

y

m

n

DP Algorithm

• Key: find out the correct order to solve the sub-problems• Total number of sub-problems: m * n

c[i, j] =c[i–1, j–1] + 1 if x[i] = y[j],max{c[i–1, j], c[i, j–1]} otherwise.

C(i, j)

0

m

0 n

i

j

LCS Example (0)j 0 1 2 3 4 5

0

1

2

3

4

i

X[i]

A

B

C

B

Y[j] BB ACD

X = ABCB; m = |X| = 4Y = BDCAB; n = |Y| = 5Allocate array c[5,6]

ABCBBDCAB

LCS Example (1)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0

for i = 1 to m c[i,0] = 0 for j = 1 to n c[0,j] = 0

ABCBBDCAB

X[i]

Y[j]

LCS Example (2)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0

if ( Xi == Yj ) c[i,j] = c[i-1,j-1] + 1 else c[i,j] = max( c[i-1,j], c[i,j-1] )

0

ABCBBDCAB

X[i]

Y[j]

LCS Example (3)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


0 0 0

ABCBBDCAB

X[i]

Y[j]

LCS Example (4)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


0 0 0 1

ABCBBDCAB

X[i]

Y[j]

LCS Example (5)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


000 1 1

ABCBBDCAB

X[i]

Y[j]

LCS Example (6)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


0 0 10 1

1

ABCBBDCAB

X[i]

Y[j]

LCS Example (7)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


1000 1

1 1 11

ABCBBDCAB

X[i]

Y[j]

LCS Example (8)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


1000 1

1 1 1 1 2

ABCBBDCAB

X[i]

Y[j]

LCS Example (9)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0

if ( Xi == Yj )c[i,j] = c[i-1,j-1] + 1

else c[i,j] = max( c[i-1,j], c[i,j-1] )

1000 1

21 1 11

1 1

ABCBBDCAB

X[i]

Y[j]

LCS Example (10)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


1000 1

1 21 11

1 1 2

ABCBBDCAB

X[i]

Y[j]

LCS Example (11)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0

if ( Xi == Yj )c[i,j] = c[i-1,j-1] + 1


1000 1

1 21 1

1 1 2

1

22

ABCBBDCAB

X[i]

Y[j]

LCS Example (12)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


1000 1

1 21 1

1 1 2

1

22

1

ABCBBDCAB

X[i]

Y[j]

LCS Example (13)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0

if ( Xi == Yj )c[i,j] = c[i-1,j-1] + 1


1000 1

1 21 1

1 1 2

1

22

1 1 2 2

ABCBBDCAB

X[i]

Y[j]

LCS Example (14)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

B

BB ACD

0

0

00000

0

0

0


1000 1

1 21 1

1 1 2

1

22

1 1 2 2 3

ABCBBDCAB

X[i]

Y[j]

LCS Algorithm Running Time

• LCS algorithm calculates the values of each entry of the array c[m,n]

• So what is the running time?

O(m*n)

since each c[i,j] is calculated in constant time, and there are m*n elements in the array

How to find actual LCS

• The algorithm just found the length of LCS, but not LCS itself.• How to find the actual LCS?• For each c[i,j] we know how it was acquired:

• A match happens only when the first equation is taken

• So we can start from c[m,n] and go backwards, remember x[i] whenever c[i,j] = c[i-1, j-1]+1.

2

2 3

2 For example, here c[i,j] = c[i-1,j-1] +1 = 2+1=3

otherwise]),1[],1,[max(

],[][ if1]1,1[],[

jicjic

jyixjicjic

Finding LCSj 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

BB ACD

0

0

00000

0

0

0

1000 1

1 21 1

1 1 2

1

22

1 1 2 2 3B

X[i]

Y[j]

Time for trace back: O(m+n).

Finding LCS (2)j 0 1 2 3 4 5

0

1

2

3

4

i

A

B

C

BB ACD

0

0

00000

0

0

0

1000 1

1 21 1

1 1 2

1

22

1 1 2 2 3B

B C BLCS (reversed order):

LCS (straight order): B C B (this string turned out to be a palindrome)

X[i]

Y[j]

Restaurant location problem

• You work in the fast food business• Your company plans to open up new restaurants in

Texas along I-35

• Towns along the highway called t1, t2, …, tn

• Restaurants at ti has estimated annual profit pi

• No two restaurants can be located within 10 miles of each other due to some regulation

• Your boss wants to maximize the total profit• You want a big bonus

10 mile

A DP algorithm

• Suppose you’ve already found the optimal solution

• It will either include tn or not include tn

• Case 1: tn not included in optimal solution

– Best solution same as best solution for t1 , …, tn-1

• Case 2: tn included in optimal solution

– Best solution is pn + best solution for t1 , …, tj , where j < n is the largest index so that dist(tj, tn) ≥ 10

Recurrence formulation

• Let S(i) be the total profit of the optimal solution when the first i towns are considered (not necessarily selected)– S(n) is the optimal solution to the complete problem

S(n-1)

S(j) + pn j < n & dist (tj, tn) ≥ 10S(n) = max

S(i-1)

S(j) + pi j < i & dist (tj, ti) ≥ 10S(i) = max

Generalize

Number of sub-problems: n. Boundary condition: S(0) = 0.

Dependency: ii-1jS

Example

• Natural greedy 1: 6 + 3 + 4 + 12 = 25• Natural greedy 2: 12 + 9 + 3 = 24

5 2 2 6 6 63 10 7

6 7 9 8 3 3 2 4 12 5

Distance (mi)

Profit (100k)

6 7 9 9 10 12 12 14 26 26S(i)

S(i-1)

S(j) + pi j < i & dist (tj, ti) ≥ 10S(i) = max

100

07 3 4 12

dummy

Optimal: 26

Complexity

• Time: (nk), where k is the maximum number of towns that are within 10 miles to the left of any town– In the worst case, (n2)

– Can be improved to (n) with some preprocessing tricks

• Memory: Θ(n)

Knapsack problem

Three versions:

0-1 knapsack problem: take each item or leave it

Fractional knapsack problem: items are divisible

Unbounded knapsack problem: unlimited supplies of each item.

Which one is easiest to solve?

•Each item has a value and a weight•Objective: maximize value•Constraint: knapsack has a weight

limitation

We study the 0-1 problem today.

Formal definition (0-1 problem)

• Knapsack has weight limit W• Items labeled 1, 2, …, n (arbitrarily)

• Items have weights w1, w2, …, wn

– Assume all weights are integers

– For practical reason, only consider wi < W

• Items have values v1, v2, …, vn

• Objective: find a subset of items, S, such that iS wi W and iS vi is maximal among all such (feasible) subsets

A DP algorithm

• Suppose you’ve find the optimal solution S

• Case 1: item n is included

• Case 2: item n is not included

Total weight limit:W

wn

Total weight limit:W

Find an optimal solution using items 1, 2, …, n-1 with weight limit W - wn

wn

Find an optimal solution using items 1, 2, …, n-1 with weight limit W

Recursive formulation

• Let V[i, w] be the optimal total value when items 1, 2, …, i are considered for a knapsack with weight limit w

=> V[n, W] is the optimal solution

V[n, W] = maxV[n-1, W-wn] + vn

V[n-1, W]

Generalize

V[i, w] = maxV[i-1, w-wi] + vi item i is taken

V[i-1, w] item i not taken

V[i-1, w] if wi > w item i not taken

Boundary condition: V[i, 0] = 0, V[0, w] = 0. Number of sub-problems = ?

Example

• n = 6 (# of items)• W = 10 (weight limit)• Items (weight, value):

2 24 33 35 62 46 9

0

0

0

0

0

0

00000000000

w 0 1 2 3 4 5 6 7 8 9 10

425

6

4

3

2

1

i

96

65

33

34

22

viwi

maxV[i-1, w-wi] + vi item i is taken



V[i, w] =

V[i, w]

V[i-1, w]V[i-1, w-wi]

6

wi

5

107400

1310764400

9633200

8653200

555532200

2222222200

00000000000

w 0 1 2 3 4 5 6 7 8 9 10

i wi vi

1 2 2

2 4 3

3 3 3

4 5 6

5 2 4

6 6 9

maxV[i-1, w-wi] + vi item i is taken



V[i, w] =

2

4

3

6

5

6

7

5

9

6

8

10

9 11

8

3

12 13

13 15

107400

1310764400

9633200

8653200

555532200

2222222200

00000000000

w 0 1 2 3 4 5 6 7 8 9 10

i wi vi

1 2 2

2 4 3

3 3 3

4 5 6

5 2 4

6 6 9

2

4

3

6

5

6

7

5

9

6

8

10

9 11

8

3

12 13

13 15

Item: 6, 5, 1

Weight: 6 + 2 + 2 = 10

Value: 9 + 4 + 2 = 15

Optimal value: 15

Time complexity

• Θ (nW)• Polynomial?

– Pseudo-polynomial– Works well if W is small

• Consider following items (weight, value):(10, 5), (15, 6), (20, 5), (18, 6)

• Weight limit 35– Optimal solution: item 2, 4 (value = 12). Iterate: 2^4 = 16 subsets– Dynamic programming: fill up a 4 x 35 = 140 table entries

• What’s the problem?– Many entries are unused: no such weight combination– Top-down may be better

Use DP algorithm to solve new problems

• Directly map a new problem to a known problem• Modify an algorithm for a similar task• Design your own

– Think about the problem recursively– Optimal solution to a larger problem can be computed

from the optimal solution of one or more subproblems– These sub-problems can be solved in certain

manageable order– Works nicely for naturally ordered data such as

strings, trees, some special graphs– Trickier for general graphs

• The text book has some very good exercises.

Greedy algorithm

• Proofs that show locally optimal selection leads to globally optimal solution

• Example problems that can be solved by greedy algorithm:– Uniform-profit restaurant location problem– Fractional knap-sack problem– Other similar problems (e.g. event scheduling

without weights/values for different event)

Back

Greedy algorithm for restaurant location problem

select t1

d = 0;

for (i = 2 to n)

d = d + dist(ti, ti-1);

if (d >= min_dist)

select ti

d = 0;

end

end

5 2 2 6 6 63 10 7

d 0 5 7 9 150

6 9 150

100

7

Complexity

• Time: Θ(n)

• Memory: – Θ(n) to store the input– Θ(1) for greedy selection

Knapsack problem

Three versions:

0-1 knapsack problem: take each item or leave it

Fractional knapsack problem: items are divisible

Unbounded knapsack problem: unlimited supplies of each item.

Which one is easiest to solve?

•Each item has a value and a weight•Objective: maximize value•Constraint: knapsack has a weight

limitation

We can solve the fractional knapsack problem using greedy algorithm

Greedy algorithm for fractional knapsack problem

• Compute value/weight ratio for each item• Sort items by their value/weight ratio into

decreasing order– Call the remaining item with the highest ratio the most

valuable item (MVI)

• Iteratively: – If the weight limit can not be reached by adding MVI

• Select MVI

– Otherwise select MVI partially until weight limit

Example• Weight limit: 10

1.5

2

1.2

1

0.75

1

$ / LB

966

425

654

333

342

221

Value ($)

Weight (LB)

item

Example• Weight limit: 10

• Take item 5– 2 LB, $4

• Take item 6– 8 LB, $13

• Take 2 LB of item 4– 10 LB, 15.4

item Weight (LB)

Value ($)

$ / LB

5 2 4 2

6 6 9 1.5

4 5 6 1.2

1 2 2 1

3 3 3 1

2 4 3 0.75

Why is greedy algorithm for fractional knapsack problem valid?

• Claim: the optimal solution must contain the MVI as much as possible (either up to the weight limit or until MVI is exhausted)

• Proof by contradiction: suppose that the optimal solution does not use all available MVI (i.e., there is still w (w < W) units of MVI left while we choose other items)– We can replace w pounds of less valuable items by MVI– The total weight is the same, but with value higher than the

“optimal”– Contradiction

w w w w

Graphs

• Representations– Adjacency list vs adjacency matrix– Cons and pros

• MST– Kruskal’s– Prim’s

Back

Representing Graphs

• Assume V = {1, 2, …, n}• An adjacency matrix represents the graph as a n

x n matrix A:– A[i, j] = 1 if edge (i, j) E

= 0 if edge (i, j) E

• For weighted graph– A[i, j] = wij if edge (i, j) E

= 0 if edge (i, j) E

• For undirected graph– Matrix is symmetric: A[i, j] = A[j, i]

Graphs: Adjacency Matrix

• Example:

1

2 4

3

A 1 2 3 4

1

2

3 ??4


• Example:

1

2 4

3

A 1 2 3 4

1 0 1 1 0

2 0 0 1 0

3 0 0 0 0

4 0 0 1 0

How much storage does the adjacency matrix require?A: O(V2)


• Example:

1

2 4

3 4

3

2

0100

1011

0101

01101

4321A

Undirected graph


• Example:

1

2 4

3

5

6

9 4

4

3

2

0400

4096

0905

06501

4321A

Weighted graph


• Time to answer if there is an edge between vertex u and v: Θ(1)

• Memory required: Θ(n2) regardless of |E|– Usually too much storage for large graphs– But can be very efficient for small graphs

• Most large interesting graphs are sparse– E.g., road networks (due to limit on junctions)– For this reason the adjacency list is often a

more appropriate representation

Graphs: Adjacency List

• Adjacency list: for each vertex v V, store a list of vertices adjacent to v

• Example:– Adj[1] = {2,3}– Adj[2] = {3}– Adj[3] = {}– Adj[4] = {3}

• Variation: can also keep a list of edges coming into vertex

1

2 4

3

Graph representations

• Adjacency list

1

2 4

3

2 3

3

3

How much storage does the adjacency list require?A: O(V+E)


• Undirected graph

1

2 4

3 4

3

2

0100

1011

0101

01101

4321A

2 3

1

3

3

1 2 4


• Weighted graph

1

2 4

3

5

6

9 4 4

3

2

0400

4096

0905

06501

4321A

2,5 3,6

1,5 3,9

3,4

1,6 2,9 4,4

Tradeoffs between the two representations

Adj Matrix Adj List

test (u, v) E Θ(1) O(n)

Degree(u) Θ(n) O(n)

Memory Θ(n2) Θ(n+m)

Edge insertion Θ(1) Θ(1)

Edge deletion Θ(1) O(n)

Graph traversal Θ(n2) Θ(n+m)

|V| = n, |E| = m

Both representations are very useful and have different properties, although adjacency lists are probably better for more problems

Good luck with your exam!

cs 3343: analysis of algorithms review for exam 2

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