376951072-3-Greedy-Method-new-ppt.ppt

2
• A greedy algorithm always makes the choice
that looks best at the moment.
• It makes a locally optimal choice in the hope
that this choice will lead to a globally optimal
solution.
• Greedy algorithms do not always yield
optimal solutions, but for many problems
they do.

3
Optimization problems
• An optimization problem is one in which you want
to find, not just a solution, but the best solution
• A “greedy algorithm” sometimes works well for
optimization problems
• A greedy algorithm works in phases. At each
phase:
– You take the best you can get right now, without regard
for future consequences
– You hope that by choosing a local optimum at each
step, you will end up at a global optimum

4
A scheduling problem
• You have to run nine jobs, with running times of 3, 5, 6,
10, 11, 14, 15, 18, and 20 minutes
• You have three processors on which you can run these jobs
• You decide to do the longest-running jobs first, on
whatever processor is available
20
18
15 14
11
10
6
5
3
P1
P2
P3
• Time to completion: 18 + 11 + 6 = 35 minutes
• This solution isn’t bad, but we might be able to do better

5
Another approach
• What would be the result if you ran the shortest job first?
• Again, the running times are 3, 5, 6, 10, 11, 14, 15, 18,
and 20 minutes
• That wasn’t such a good idea; time to completion is now
6 + 14 + 20 = 40 minutes
• Note, however, that the greedy algorithm itself is fast
– All we had to do at each stage was pick the minimum or maximum
20
18
15
14
11
10
6
5
3
P1
P2
P3

6
An optimum solution
• This solution is clearly optimal (why?)
• Clearly, there are other optimal solutions (why?)
• How do we find such a solution?
– One way: Try all possible assignments of jobs to processors
– Unfortunately, this approach can take exponential time
• Better solutions do exist:
20
18
15
14
11
10 6
5
3
P1
P2
P3

7
Fractional Knapsack Problem
We are given n objects and a bag or knapsack.
Object i has a weight w i
the knapsack has a capacity m.
If a fraction x I ,0  x i  1 of
Object i is placed into the bag
then a profit of pi x i is earned.
The objective is to obtain a filling of the knapsack
that maximizes the total profit earned.

8
Maximize
1
1
1, 1
i i
i n
i i
i n
i
p x
subject to
w x m
and
o x i n
 
 

   


A feasible solution is any set ( x 1 ,…. x n ) satisfying the above
inequalities. An optimal solution is a feasible solution for which
objective function is maximized.
1
1
1, 1
i i
i n
i i
i n
i
p x
subject to
w x m
and
o x i n
 
 

   


1
1
1, 1
i i
i n
i i
i n
i
p x
subject to
w x m
and
o x i n
 
 

   



9
0-1 knapsack problem
1
1
max
0 1, 1
.
i i
i n
i i
i n
i
imize p x
subject to w x m
and x i n
The prifits and weights are positivenumbers
 
 

   


Xi are o or 1

10
Greedy Algorithms for Knapsack
Problem
Several simple greedy strategies are there.
• We can fill the bag by including next object
with largest profit.

11
• Example: n= 3, m = 20,
• (p1 ,p 2 , p 3 ) = (25 , 24, 15)
• (w 1 , w 2 , w 3 ) = (18 , 15 , 10)
By above strategy solution is
x 1 = 1 , x 2 = 2/15, x 3 = 0
Total weight = 20, profit = 28.2

12
But this solution is not the best one.
(0,1,1/2) is a solution
having profit 31.5

13
Another greedy approach
To become greedy w.r.t. capacity:
Start with lightest object and so on:
(0 ,2/3 , 1) with profit = 31
Again not an optimal solution.

14
New greedy approach
With a balance between weight and profit
At each step we include that object which has
the maximum profit per unit of capacity used.
Using this strategy solution obtained is:
(0, 1 , ½) with profit = 31.5
Which is an optimal solution. For this example.

15
So this strategy gave optimal solution for this
data:
In fact we can prove that this strategy will give
optimal solution for any data.
.

16
There seem to be 3 obvious greedy strategies:
(Max value) Sort the objects from the highest value to the
lowest, then pick them in that order.
(Min weight) Sort the objects from the lowest weight to the
highest, then pick them in that order.
(Max value/weight ratio) Sort the objects based on the value
to weight ratios, from the highest to the lowest, then select.

17
w 10 20 30 40 50
v 20 30 66 40 60
v/w 2.0 1.5 2.2 1.0 1.2
select xi value
Max vi 0 0 1 0.5 1 146
Min wi 1 1 1 1 0 156
Max vi/wi 1 1 1 0 0.8 164
Example: Given n = 5 objects and a knapsack capacity
W = 100 as in Table I. The three solutions are given in
Table II.

18
The Optimal Knapsack Algorithm:
Algorithm (of time complexity O(n lgn))
(1) Sort the n objects from large to small based on the
ratios vi/wi . We assume the arrays w[1..n] and
v[1..n]
(2) store the respective weights and values after sorting.
(3) initialize array x[1..n] to zeros.
weight = 0; i = 1
(4) while (i  n and weight < W) do
(4.1) if weight + w[i]  W then x[i] = 1
(4.2) else x[i] = (W – weight) / w[i]
(4.3) weight = weight + x[i] * w[i]
(4.4) i++

19
Theorem: If p1 /w1  p2/ w2  …pn/wn
.then selecting items w.r.t. this ordering
gives optimal solution.
Proof: Let x = (x1 , …x n) be the
solution so generated.
By obvious reasons wi xi = m
If all the xi’s are one then clearly the solution
is optimal.

20
let j be the least index such that xj  1.
So x i = 1 for all 1  i < j and
xi = 0 for j < i  n and 0  xj <1.
Let y= (y1 , y2 ,…yn) be an optimal solution
1 1 1 ……1 Xj 0 0 0 0 0 0

21
1 1 1 ……1 Xj 0 0 0 0 0 0
y1 y2 yk y k+1……………..
Let k be the least index such that yk  xk.
It is easy to reason that yk < xk.
Now suppose we increase yk to xk and
decrease as many of (yk+1, …. Yn) as
necessary so that total capacity of bag is
still m.

22
This results in a new solution
Z = (z1 , z2, …zn)
zi = xi 1 i  k
And
∑ wi (yi - zi ) = wk( zk – yk)
k <i≤n

23
1 1
1
1
( ) /
( ) /
[( ) ( ) ] /
i n i n
k i n
i n k i n
i n
pizi piyi zk yk wk pk wk
yi zi wi pi wi
piyi zk yk wk yi zi wi pk wk
pi yi
   
 
   
 
  
 
    

 

 

If  pi zi >  pi yi then y could not have been optimal
solution.

24
• If these sums are equal then either z = x and
x is optimal or z  x.
• In the latter case repeated use of above
argument y can be transformed into x
without changing the utility value,and thus
x too is optimal

25
Optimal 2-way Merge patterns and Huffman Codes:
Example. Suppose there are 3 sorted lists L1, L2, and L3, of
sizes 30, 20, and 10, respectively, which need to be merged
into a combined sorted list, but we can merge only two at a
time.
We intend to find an optimal merge pattern which minimizes
the total number of comparisons.
For example, we can merge L1 and L2, which uses 30 + 20 =
50 comparisons resulting in a list of size 50.
We can then merge this list with list L3, using another 50 + 10
= 60 comparisons,
so the total number of comparisons 50 + 60 = 110.

26
Alternatively, we can first merge lists L2 and L3,
using 20 + 10 = 30 comparisons,
the resulting list (size 30) can then be merged with list L1,
for another 30 + 30 = 60 comparisons.
So the total number of comparisons is 30 + 60 = 90.
It doesn’t take long to see that this latter merge pattern is the
optimal one.

27
Binary Merge Trees:
We can depict the merge patterns using a binary tree, built from
the leaf nodes (the initial lists) towards the root in which each
merge of two nodes creates a parent node whose size is the sum of
the sizes of the two children. For example, the two previous merge
patterns are depicted in the following two figures:

28
Merge L1 and L2, then with L3
30 20
50 10
60
10
20
30 30
60
Merge L2 and L3, then with L1
merge cost = sum of all weighted external path lengths
Cost = 30*2 +
20*2 + 10*1
= 110
Cost = 30*1 +
20*2 + 10*2
= 90

29
Optimal Binary Merge Tree Algorithm:
Input: n leaf nodes each have an integer size, n  2.
Output: a binary tree with the given leaf nodes which
has a minimum total weighted external path lengths
Algorithm:
(1) create a min-heap T[1..n ] based on the n initial sizes.
(2) while (the heap size  2) do
(2.1) delete from the heap two smallest values, call
them a and b, create a parent node of size a + b
for the nodes corresponding to these two values
(2.2) insert the value (a + b) into the heap which
corresponds to the node created in Step (2.1)
When the algorithm terminates, there is a single value left in
the heap whose corresponding node is the root of the
optimal binary merge tree. The algorithm’s time complexity
is O(n lgn) because Step (1) takes O(n) time; Step (2) runs
O(n) iterations, in which each iteration takes O(lgn) time.

30
Example of the optimal merge tree algorithm:
2 3 5 7 9
2 3
5
5 7 9
2 3
5 5
10
7 9
Initially, 5 leaf nodes with sizes
Iteration 1: merge 2 and 3 into 5
Iteration 2:
merge 5 and
5 into 10
16 Iteration 3: merge 7 and
9 (chosen among 7, 9,
and 10) into 16
2 3
5
10
5 7 9
16
26
Iteration 4: merge
10 and 16 into 26
Cost = 2*3 + 3*3 + 5*2 + 7*2
+ 9*2 = 57.

31
Proof of optimality of the binary merge tree algorithm:
We use induction on n  2 to show that the binary merge tree is optimal in that it gives
the minimum total weighted external path lengths (among all possible ways to merge
the given leaf nodes into a binary tree).
(Basis) When n = 2. There is only one way to merge two nodes.
(Induction Hypothesis) Suppose the merge tree is optimal when there are k leaf
nodes, for some k  2.
(Induction) Consider (k + 1) leaf nodes. Call them a1, a2, …, and ak+1. We may
assume nodes a1, a2 are of the smallest values, which are merged in the first step of the
merge algorithm into node b. We call the merge tree T, the part excluding a1, a2 T’ (see
figure). Suppose an optimal binary merge tree is S. We make two observations.
(1) If node x of S is a deepest internal node, we may swap its two children with
nodes a1, a2 in S without increasing the total weighted external path lengths. Thus, we
may assume tree S has a subtree S’ with leaf nodes x, a2, …, and ak+1.
(2) The tree S’ must be an optimal merge
tree for k nodes x, a2, …, and ak+1. By
induction hypothesis, tree S’ has a
total weighted external path lengths equal
to that of tree T’. Therefore, the total weighted
external path lengths of T equals to that of tree S,
proving the optimality of T.
T
T’
S
S’
x
a1 a2 a1 a2
b

32
Minimum Spanning Tree problem
• Kruskal’s Algorithm
• Prim’s algorithm
Both are based on greedy algorithm
Algorithm and proof of getting optimal
solution

33
Minimum Spanning Tree
Spanning subgraph
– Subgraph of a graph G containing
all the vertices of G
Spanning tree
– Spanning subgraph that is itself a
(free) tree
Minimum spanning tree (MST)
– Spanning tree of a weighted graph
with minimum total edge weight
• Applications
– Communications networks
– Transportation networks
ORD
PIT
ATL
STL
DEN
DFW
DCA
10
1
9
8
6
3
2
5
7
4

34
Greedy Algorithms
• We are trying to solve a problem in an optimal
way.
• We start with a set of candidates
• As the algorithm proceeds, we keep two sets,
one for candidates already chosen, and one for
candidates rejected.
• Functions exist for choosing the next element
and testing for solutions and feasibility.
• Let’s identify these points for the Making
Change problem.

35
Minimum Spanning Trees
• Given G={N,A }, G is undirected and connected
• Find T, a subset of A, such that the nodes are connected
and the sum of the edge weights is minimized. T is a
minimum spanning tree for G.
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

36
Minimum Spanning Trees
• How many edges in T?
• Let’s come up with some algorithms to
compute MSTs.
1 2 3
4 5 6
7
1
3
4
4
2
3

37
Algorithm #1
• Start with an empty set.
• From all the unchosen and unrejected edges,
select the shortest edge.
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

38
Algorithm #1
• Note we actually have multiple “correct”
answers.
• Sometimes, not just one MST.
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

39
MST
• We keep two sets, one for candidates already chosen,
and one for candidates rejected.
• Functions exist for choosing the next element and
testing for solutions and feasibility.
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

40
Kruskal’s Algorithm
• Each node is in its own set
• Sort edges in increasing order
• Add shortest edge that connects two sets
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

41
S = {1} {2} {3} {4} {5} {6} {7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (2-5) (4-7) (3-5) (2-4) (3-
6) (6-7)
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

42
S = {1, 2} {3} {4} {5} {6} {7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (2-5) (4-7) (3-5) (2-4) (3-
6) (6-7)
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

43
S = {1, 2, 3} {4} {5} {6} {7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (2-5) (4-7) (3-5) (2-4) (3-
6) (6-7)
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

44
S = {1, 2, 3} {4} {5} {6, 7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (2-5) (4-7) (3-5) (2-4) (3-
6) (6-7)
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

45
S = {1, 2, 3} {4, 5} {6, 7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (2-5) (4-7) (3-5) (2-4) (3-
6) (6-7)
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

46
S = {1, 2, 3, 4, 5} {6, 7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (2-5) (4-7) (3-5) (2-4) (3-
6) (6-7)
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

47
S = {1, 2, 3, 4, 5, 6, 7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (2-5) (4-7) (3-5) (2-4) (3-
6) (6-7)
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

48
S = {1, 2, 3, 4, 5, 6, 7}
E = (1-2) (2-3) (6-7) (4-5) (1-4) (4-7)
Total = 1 + 2 + 3 + 3 + 4 + 4 = 17
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

49
E = (5-6), (5-7), (6-3), (4-2), (1-4), (2-5)
Total = 2 + 3 + 4 + 4 + 5 + 6 = 24
1 2 3
4 5 6
7
9
4
7
6
6 6 4
8
5
2
7
3

50
Another possible Minimum Spanning Tree:
E = (5-6), (5-7), (6-3), (4-2), (1-4), (4-7)
Total = 2 + 3 + 4 + 4 + 5 + 6 = 24
1 2 3
4 5 6
7
9
4
7
6
6 6 4
8
5
2
7
3

51
Timing Kruskal’s Algorithm
• Each node is in its own set
• Sort edges in increasing order
• Add shortest edge that connects two sets
1 2 3
4 5 6
7
1
6
3
4
4 4 6
2
5
8
3
7

52
Prim-Jarnik’s Algorithm
• We pick an arbitrary vertex s and we grow the MST as a cloud
of vertices, starting from s
• We store with each vertex v a label d(v) = the smallest weight
of an edge connecting v to a vertex in the cloud
At each step:
 We add to the cloud the
vertex u outside the cloud
with the smallest distance
label
 We update the labels of the
vertices adjacent to u

53
Example
B
D
C
A
F
E
7
4
2
8
5
7
3
9
8
0
7
2
8 

B
D
C
A
F
E
7
4
2
8
5
7
3
9
8
0
7
2
5 
7
B
D
C
A
F
E
7
4
2
8
5
7
3
9
8
0
7
2
5 
7
B
D
C
A
F
E
7
4
2
8
5
7
3
9
8
0
7
2
5 4
7

54
Example (contd.)
B
D
C
A
F
E
7
4
2
8
5
7
3
9
8
0
3
2
5 4
7
B
D
C
A
F
E
7
4
2
8
5
7
3
9
8
0
3
2
5 4
7

55
Kruskal’s Algorithm generates
optimal spanning tree

56
Let
T: Sp.tree obtained by Kruskal’s algorithm
T’: Optimal(min) spanning tree
| E(T) | = | E(T’) | = n-1
where E(T) = edge set in T
If C(T) = C(T’) Nothing to prove T is also
optimal.

57
So assume E(T)  E(T’)
Let e be a min cost edge such that e is in E(T)
but not in E(T’)
It implies that all the edges of Sp tree T which are of
lesser weight than e are also in T’.
Add e to T’
A cycle containing e
is formed in T’

58
Let e, e1, e2…. ek be this cycle.
At least one of these edges say ej will not be in
T ?
If C(ej) < C(e) we can argue that it is not possible
If so Kruskal’s algorithm must have included it as
ej must have been tried before . But ej is not in T
so it means that ej was not selected there reason
being it must be forming a cycle with edges
already selected.Which is not true since all the
preceding edges together with ej
are also present in T’ which is an acyclic
graph..…..

59
So C(ej) >= C(e)
consider the sp tree
E(T’’)= E( T’) +{e)-{ej}
C(T”)<= C(T’) which implies C(T’)=C(T’’)
so T’’ is of same cost as T’
T’’ is more similar to T than T’
By repeated use of above argument T’’ can be
modified to T’’’and so on and finally same
as T
Thus C(T) =C(T’)

60
Huffman Codes
• Very effective technique for
compressing data; saving of 20% to
90%.
• It uses a table of frequencies of
occurrence of characters.

61
Fixed length code
• If only six different characters are used in a
text then we need 3 bits to represent six
characters.
• a = 000; b = 001; …f = 101.
• This method thus requires 300,000 bits to
code the entire file having 100,000
characters.

62
Variable Length code
We can do considerable better by giving
frequent characters short code words and
infrequent characters log code words.
frequency 45 13 12 16 9 5
Fixed len
codeword
Var. length
code
000 001 010 011 100 101
0 101 100 111 1101 1100

63
Thus using variable length code it requires
(45.1 +13.3 12.3 16.3 9.4 5.4)1000
= 224,000 Bits
Thus saving approx 25%.

64
Prefix code: no code word is also a prefix of
some other codeword.
These prefix codes are desirable because they
simplify decoding.
Like in our example
001011101 can be parsed uniquely as
0-0-101-1101 which decodes as aabe

65
Huffman Encoding
• Compression
– Typically, in files and messages,
• Each character requires 1 byte or 8 bits
• Already wasting 1 bit for most purposes!
• Question
– What’s the smallest number of bits that can be used to
store an arbitrary piece of text?
• Idea
– Find the frequency of occurrence of each character
– Encode Frequent characters short bit strings
– Rarer characters longer bit strings

66
Huffman Encoding
• Encoding
– Use a tree
– Encode by following
tree from root to leaf
– eg
• E is 00
• S is 011
– Frequent characters
E, T 2 bit encodings
– Others
A, S, N, O 3 bit encodings

67
Huffman Encoding
• Encoding
– Use a tree
• Inefficient in practice
– Use a direct-addressed lookup
table
? Finding the optimal encoding
– Smallest number of bits to
represent arbitrary text
A 010
E 00
B
:
:
N
:
S
T
110
001
10

68
Huffman Encoding - Operation
Initial sequence
Sorted by frequency
Combine lowest two
into sub-tree
Move it to correct
place

69
After shifting sub-tree
to its correct place ...
Combine next lowest
pair
Move sub-tree to
correct place

70
Move the new tree
to the correct place ...
Now the lowest two are the
“14” sub-tree and D
Combine and move to
correct place

71
Move the new tree
to the correct place ...
Now the lowest two are the
the “25” and “30” trees
Combine and move to
correct place

74
Huffman Encoding - Time
Complexity
• Sort keys O(n log n)
• Repeat n times
– Form new sub-tree O(1)
– Move sub-tree O(logn)
(binary search)
– Total O(n log n)
• Overall O(n log n)

75
Theorem
• Let C be an alphabet in which each
character c of C has frequency f(c ) .
• Let x and y be two characters in C having
lowest frequencies.
• Then there exists an optimal prefix code for
C in which the code words for x and y have
the same length and differ only in the last
bit.

76
y
a b
x
T:
y
x b
a
T’:
Let a and b be two characters that are sibling leaves of
maximum depth in T.
We can assume f(a) ≤ f(b) and f(x) ≤ f(y)
Since x and y are of lowest frequencies
f(x) ≤ f(a) and f(y) ≤ f(b)

78
• The number of bits required to encode a file is
• B(T) = ∑f (c ) d T(c)
• we can call it the cost of the tree T.
• B(T) – B(T’) = ∑f (c ) d T(c) - ∑f (c ) d T’(c)
= f(x) dT(x) + f(a) dT(a) - f(x) dT’(x) – f(a) dT’(a)
= f(x) dT(x) + f(a) dT(a) - f(x) dT(a) – f(a) dT(x)
=(f(a) – f(x))(dT(a) – dT(x))
>= 0 because both
f(a)-f(x) and dT(a) – dT(x) are nonnegative.

79
• Similarly exchanging y and b does not increase
the cost so
B(T’) - B(T’’) is non negative.
There fore
B(T’’) ≤ B(T)
• And since T was taken to be optimal
• B(T) ≤ B(T’’)
• Which implies
• B(T’’) =B(T)
• Thus T’’ is an optimal tree in which x and y
appear as sibling leaves of maximum depth.

80
Theorem - Let C be a given alphabet with frequency
f(c). Let x and y be two characters in C with
minimum frequencies. Let C’ be the alphabet
obtained from C as
C’ = C-{x,y}{z}.
Frequencies for new set is same as for C except that
f(z) = f(x) + f(y).
Let T’ be any optimal tree representing optimal code
for C’, then the tree T obtained from T’ by
replacing the leaf node for z with an internal node
having x and y as children, represents an optimal
prefix code for the alphabet C

81
Proof: For any c ε C – {x ,y}
dT(c ) = dT’(c ) but for x and y
dT(x) = dT(y) = dT’(z) + 1
We have
f(x)dT(x) + f(y) dT(y) = (f(x) + f(y) )(dT(z) + 1)
= f(z) dT(z) + (f(x) + f(y))
B(T) = B(T’) + f(x) + f(y)

82
From which we conclude that
B(T) = B(T’) + f(x) + f(y)
Or
B(T’) = B(T) – f(x) – f(y)
We now prove by contradiction.
Suppose that T is not optimal for C then there is
another optimal tree T’’ such that
B(T’’) < B(T)
Without any loss of generality we can assume that x
and y are siblings here.
Let T’’’ be the tree obtained from T’’with the
common parent of x and y replaced by a leaf z
with freq f(z) = f(x) +f(y)
then

83
• B(T’’’) = B(T’’) – f(x) – f(y)
< B(T) – f(x) – f(y)
=B(T’)
This contradicts the assumption that T’
represents an optimal prefix code for C’
Thus T must represent an optimal prefix code
for the alphabet C.

84
Job Sequencing with dead lines
We are given n jobs,
associated with each job i, here is an (integer)
deadline di ≥ 0 and a profit pi > 0.
This profit pi will be earned only if job is completed
before its deadline.
Each job needs processing for one unit time on a
single machine available.
Feasible solution-is a subset of jobs each of which
can be completed without crossing any deadline.
Optimal solution is a feasible solution with maximum
profit

85
Example: n=4 , p=(100,10,15,27) and
d=(2,1,2,1)
Possible feasible solutions:
Subset sequence value
{1,2} 2,1 110
{1,3} 1,3 or 3,1 115
{1,4} 4,1 127
{2,3} 2,3 25
{3,4} 4,3 42
{1} 1 100

86
Greedy approach
objective is to maximize ∑pi
So next job to select in J is the one that
increases ∑pi the most,
subject to the restriction that set J remains
feasible.
Feasibility test becomes very easy through the
following result

87
Theorem: Let J be a set of k jobs and
S= i1,i2,….ik a permutation of jobs in J such that
di1 ≤ di2 ≤ ≤ dik. Then J is feasible iff the jobs
in J can be processed in the order S without
violating any deadline.
Proof:
We need to show that if J is feasible set then a
permutation S =i1,i2 …ik with di1 ≤ di2 ≤ dik is also
feasible.
If J is taken to be a feasible set then there exists S` =
r1, r2….rk such that drq ≥ q, 1 ≤ q ≤ k.
Assume S ≠ S`,

88
Let a be the least index such that ra ≠ ia.
Let ia = rb for some b > a
In S’ we can interchange ra and rb .since dra drb the resulting
permutation s’’
S: i1 i2 . . . ia … ik
S’: r1 r2 ra rb= ia rk
(b)
dia<=dra
Since S is a permutation of jobs with nondecreasing
deadlines.and ra is positioned later than ia
If we interchange ra and rb we get another feasible sequence.
And new sequence is closer to S.
Continuing in this way S` can be transformed into S without
violating any dead line.
b

89
Theorem: greedy approach always obtains an
optimal solution for job sequencing with dead
lines
Proof: Let I be the set of jobs obtained by
greedy method.
And let J be the set of jobs in an optimal
solution.
We will show that I and J both have same
profit values.
We assume that I ≠ J.
J cannot be subset of I as J is optimal
Also I can not be subset of J by the algo.

90
So there exists a job a in I such that a is
not in J
And a job b in J Which is not in I
Let us take a “a highest profit job” such
that a is in I and not in J.
Clearly pa ≥ pb for all jobs that are in J
but not in I.
Since if pb > pa then greedy approach
would consider job b before job a and
included into I.
Let Si and Sj be the feasible sequences
for feasible sets I and J.

91
We will assume that common jobs of I
and J can be processed in same time
intervals.
Let i be a job scheduled in t to t+1 in S i
and t’ to t’+1 in Sj
If t < t’ then interchange the job if any in
[t’ , t’+1] in Si with job i.
Similarly if t’ < t similar transformation can
be done in Sj

92
Now consider the interval [ ta, ta+1] in Si in
which job a is scheduled.
Let b the job(if any) scheduled in Sj in this
interval.
pa ≥ pb from the choice of a
So scheduling a from ta to ta+1 in Sj and
discarding job b gives a feasible schedule for
the set J’ = J- {b} + {a}.
Clearly J’ has profit no less than that of J and
differs from I in one less job than J does.
By repeatedly using this approach J can be
transformed into I without decreasing the
profit value.

376951072-3-Greedy-Method-new-ppt.ppt

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