Algorithms presentation on Path Matrix, Bell Number and Sorting

Group 5

Algorithms Presentation

Agenda Items and Presenters

Bell NumbersAll Pairs Shortest PathShell Sort and Radix Sort Psuedocode

Bell Numbers

In how many ways, counting ties, can 8 horses cross a finish line?

A Use Case

Bell NumbersLet’s consider just four horses for now and develop a recurrence relation, a systematic way of counting the possible outcomes.

In mathematics, a recurrence relation is an equation that recursively defines a sequence, one or more initial terms are given: each further term of the sequence is defined as a function of the preceding terms.

Four horses may finish in one, two, three or four “blocks”. Define “blocks”…

A block in this case is a pattern of possible ways the horses can finish.

If we have fours horses as follows…

Horse Racing Example

Bell NumbersAlfred

Bell NumbersBoxtrot

Bell NumbersYisele

Bell NumbersXray

Bell NumbersThree block example:

(Alpha) (Boxtrot) (Yisele, Xray) have finished separately.Yisele and Xray have tied.

More Background Info

Bell Numbers

Three blocks – (Alpha), (Beta), (Yisele, Xray) may be arranged in 3! = 6 ways. 

The arrangements for a block of three ways in which the horses can finish.

Note: This does not take into account places (first, second, third, fourth).

(Alpha) (Boxtrot) (Yisele, Xray), (Alpha) (Yisele) (Boxtrot, Xray),(Alpha) (Xray) (Boxtro, Yisele), (Boxtrot) (Yisele) (Alpha, Xray), (Boxtrot) (Xray) (Alpha, Yisele), (Yisele) (Xray) (Alpha, Beta)

Blocks

Bell Numbers Blocks, PartitionsArrangements and Outcomes

The number of ways a set of n elements can be partitioned into m nonempty subsets S(n,m) is the Bell number. In this case 4 horses (elements) can be partitioned in 15 different ways. In this case we added it up manually but we can determine a formula and also a graphical way to calculate as follows….

Bell Numbers Calculate Graphically

The numbers can be constructed by using the Bell Triangle. Start with a row with the number one. Afterward each row begins with the last number of the previous row and continues to the right adding each number to the number above it to get the next number in the row..

Bell Numbers The Formula

S(n+1) = S(n,m-1) + m * S(n,m)

n = elementsm = blocks

B(5) = S(5,1) + S(5,2) + S(5,3) + S(5,4) + S(5,5)

Answer to the ProblemH8 =

Where S(n,k) denotes the # ways n horses can cross in k blocks. H8 = 1×1! + 127×2! + 966×3! + 1701×4! +

1050×5! + 266×6! + 28×7! + 1×8!  

= 545835.

H8 denotes the number of ways 8 horses can cross the finish line.

Bell Numbers Psuedocode

Technical Information:Technical Information:

Language:Language:Objects:Objects:

How do you represent a partitioning of a set of n elements?

(This example is somewhat more challenging that the previous ones. Feel free to skim it, and go on.) The n’th Bell number Bn is the number of ways of partitioning n (distinct) objects. One way of computing the Bell numbers is by using the following double recursive algorithm. This algorithm computes numbers with two arguments: B(i,j). The n’th Bell number, Bn is computed as

B(n,n). For example to find B3 compute B(3,3).1

B(1,1) = 1.

B(n,1) = B(n-1,n-1) for n > 1.

B(i,j) = B(i-1,j-1) + B(i,j-1) for n > 1 and 1 < j # i.

All Pairs Shortest PathGiven a weighted graph G(V,E,w), the all-pairs shortest paths problem is to find the shortest paths between all pairs of vertices vi, vj V. ∈

A number of algorithms are known for solving this problem.

All Pairs Shortest PathConsider the multiplication of the weighted adjacency matrix with itself - except, in this case, we replace the multiplication operation in matrix multiplication by addition, and the addition operation by minimization. Notice that the product of weighted adjacency matrix with itself returns a matrix that contains shortest paths of length 2 between any pair of nodes. It follows from this argument that An contains all shortest paths.

Transitive Closure : of R is the smallest transitive relation containing R.

All Pairs Shortest Path0 0 1 0

0 1 0 1

0 0 0 1

0 1 0 0

0 0 1 0

0 1 0 1

0 0 0 1

0 1 0 0

R P1

R[i,j] = {1

0Pk [i,j] = {

1

0

If there is a path of exactly K from i-> j

All Pairs Shortest Path0 0 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 0

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 0 1

0 1 1 1

0 1 0 1

0 1 0 1

0 1 0 1

P2 P3

P4 P = P1 + P2+P3+P4

All Pairs Shortest PathAn is computed by doubling powers - i.e., as A, A2, A4, A8, and so on. We need log n matrix multiplications, each taking time O(n3). The serial complexity of this procedure is O(n3log n). This algorithm is not optimal, since the best known algorithms have complexity O(n3).

All Pairs Shortest PathParallel Formulation

Each of the log n matrix multiplications can be performed in parallel. We can use n3/log n processors to compute each matrix-matrix product in time log n. The entire process takes O(log2n) time.

All Pairs Shortest PathParallel Formulation

def adj2(i,j): if adj(i,j) == 1: return 1 else: for k in range(0,n): # where n is the number of vertices in G if adj(i,k) == 1 and adj(k,j) == 1: return 1 return 0

80 93 60 6812 8542 30 10

Initial Segmenting Gap = 4

10 30 60 6812 8542 93 80

Shell Sort – Example 1

10 30 60 6812 8542 93 80

Resegmenting Gap = 2

10 12 42 6830 9360 85 80

Shell Sort – Example 2

10 12 30 8042 8560 68 93

10 12 42 6830 9360 85 80

Resegmenting Gap = 1

Shell Sort – Example 3

Shell Sort Psuedocode

Technical Information:Technical Information:

Language:Language:Programming Constructs:Programming Constructs:Efficiency:Efficiency:

# Sort an array a[0...n-1].gaps = [701, 301, 132, 57, 23, 10, 4, 1]

foreach (gap in gaps){ # Do an insertion sort for each gap size. for (i = gap; i < n; i += 1) { temp = a[i] for (j = i; j >= gap and a[j - gap] > temp; j -= gap) { a[j] = a[j - gap] } a[j] = temp }

}

Radix Sort

For simplicity, say you want to use the decimal radix (=10) for sorting. Then you would start by separating the numbers by units and then putting them together again; next you would separate the numbers by tens and then put them together again; then by hundreds and so on until all the numbers are sorted. Each time you loop, just read the list from left to right. You can also imagine you are separating the numbers into buckets. Here is an illustration using

Radix Sort

5, 213, 55, 21, 2334, 31, 20, 430

Separate by units:

zeros: 20, 430

ones: 21, 31

twos:

threes: 213

fours: 2334

fives: 5, 55

Back together: 20, 430, 21, 31, 213, 2334, 5, 55

To put them back together, first read the zeroes bucket, then the ones bucket, then so on, until you read the nines bucket.

Radix Sort

Separate by tens:

zeros: 05

ones: 213

twos: 20, 21

threes: 430, 2334,

fours:

fives: 55

Back together: 5, 213, 20, 21, 430, 2334, 55

Radix Sort

Separate by hundreds:

zeros: 005, 020, 021, 055

ones:

twos: 213

threes: 2334

fours: 430

fives:

Back together: 5, 20, 21, 55, 213, 2334, 430

Radix Sort

Separate by thousands:

zeros: 0005, 0020, 0021, 0055,0213, 0430

ones:

twos: 2334

threes:

fours:

fives:

Back together: 5, 20, 21, 55, 213, 430, 2334

Thank You