advanced data structures lab manual

5/26/2018 Advanced Data Structures Lab Manual

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UNITED INSTITUTE OF TECHNOLOGYPeriyanaickenpalayam, Coimbatore- 641 020.Department of Computer Science and Engineering

Advanced Data Structures Laboratory

Antonita Shilpa . J UIT ( FMCS02 ) 1

INDEX

S. No. PROGRAM Page No.

1 Graph Search 2

2 Breadth First Search 5

3 Depth First Search 8

4 Dijikstras Shortest Path Algorithm 12

5 Topological Sorting 18

6 Hill Climbing 22

7 Merge Sort 28

8 Quick Sort 33

9 Producer Consumer 36

10 Concurrency List 39

11 Concurrency Stack 44

12 Concurrency Queue 49

13 Dynamic Programming 54

14 Randomized Algorithm 58

15 Dining Philosophers Problem 61


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GRAPH SEARCH

ALGORITHM:

Step 1: Start

Step 2: Get an input graph, source node S, node to search N

Step 3: Add S to notHandled queue

Step 4: Take first node, U from notHandled

Step 4.1: If U is equal to N

Step 4.1.1: Return

Step 4.2: Else

Step 4.2.1: Add the child of U to notHandled

Step 4.2.2: Move U from notHandled to Handled queue

Step 5: Repeat step 4 till notHandled is not empty

Step 6: Return

Step 7: End the Algorithm

JAVA CODE

importjava.util.*;

publicclassGraphSearch

{

publicstaticvoidmain(String[] arg)

{ArrayList start = new ArrayList();

ArrayList end = newArrayList();

ArrayList nodes = newArrayList();

ArrayList notHandled = newArrayList();

ArrayList handled = newArrayList();

intedgeCount, i, hSize, nSize;

String node;

Scanner s= newScanner(System.in);

System.out.println( "NOTE: The first node will be taken as source");

System.out.println( "Enter the number of edges:");

edgeCount=s.nextInt();

for(i=0; i


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// get FROM node and TO node for each edge

System.out.println( "Enter the FROM node:");

start.add(s.next());

System.out.println( "Enter the TO node:");end.add(s.next());

while(start.get(i).equalsIgnoreCase(end.get(i)))

{

end.remove(i);

System.out.println( "Error: SAME AS START NODE");

System.out.println( "Enter the TO node:");

end.add(s.next());

}

if(!(nodes.contains(start.get(i))))

nodes.add(start.get(i));

if(!(nodes.contains(end.get(i))))

nodes.add(end.get(i));

}

nSize=nodes.size();

System.out.println( "Enter the node to be found:");

node=s.next();

notHandled.add(start.get(0)); // root node

handled.add(notHandled.get(0));

if(node.equalsIgnoreCase(notHandled.get(0)))

hSize=nSize;

else

hSize=handled.size();

nodes.remove(notHandled.get(0));

notHandled.remove(0);

// till all nodes are handled

while(hSize


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a

b c

ed

System.out.println("Nodes Not Handled = "+nodes);

}

}

GRAPH

OUTPUT

NOTE: The first node will be taken as source

Enter the number of edges:

4

Enter the FROM node:

a

Enter the TO node:

b


a

Enter the TO node:

c


b

Enter the TO node:

d

Enter the FROM node:b

Enter the TO node:

e

Enter the node to be found:

c

Nodes Handled = [a, b]

Node c Found !!!

Nodes Not Handled = [d, e]

RESULT Thus the code in written in Java and executed for finding a node in a graph.


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BREADTH FIRST SEARCH

ALGORITHM:

Step 1: Start

Step 2: Get an input graph G, source node S

Step 3: Add S to notHandled queue

Step 4: Take first node, U from notHandled

Step 4.1: Add the child of U to notHandled

Step 4.2: Move U from notHandled to Handled queue


Step 6: Return


JAVA CODE

importjava.util.*;publicclassBreadthFirstSearch

{


{

ArrayList start = new ArrayList();

ArrayList end = newArrayList();

ArrayList nodes = newArrayList();

ArrayList notHandled = newArrayList();

ArrayList handled = newArrayList();

intedgeCount, i, hSize;

Scanner s= newScanner(System.in);System.out.println( "NOTE: The first node will be taken as source");



for(i=0; i


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System.out.println( "Enter the TO node:");

end.add(s.next());

}

if(!(nodes.contains(start.get(i))))nodes.add(start.get(i));



}

notHandled.add(start.get(0)); // root node

System.out.println("Handled Not Handled");

System.out.println("[] "+ notHandled);



// till all nodes are handled

while(notHandled.size()>0)

{

notHandled.remove(0);

// for the rear node in handled queue, add its children in notHandled queue

for(i=0; i0)

if(!(handled.contains(notHandled.get(0))))



}

System.out.println("All Nodes are traversed in BFS");

}

}

GRAPH

OUTPUT



4


a

Enter the TO node:b

a

b d

ec


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b

Enter the TO node:c


a

Enter the TO node:

d


d

Enter the TO node:

e

Handled Not Handled

[] [a]

[a] [b, d]

[a, b] [d, c]

[a, b, d] [c, e]

[a, b, d, c] [e]

[a, b, d, c, e] []

All Nodes are traversed in BFS

RESULT Thus the code is written in Java and executed for implementing Breadth First Search.


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DEPTH FIRST SEARCH

ALGORITHM:

Step 1: Start

Step 2: Get an input graph G, source node S

Step 3: Push S to notHandled stack

Step 4: Pop top node, U from notHandled

Step 4.1: Push the child of U to notHandled

Step 4.2: Pop U from notHandled stack and add to Handled queue


Step 6: Return


JAVA CODE

importjava.util.*;

publicclassDepthFirstSearch

{

staticArrayList start= new ArrayList();

staticArrayList end= newArrayList();

staticString[] stack= newString[10];

staticArrayList handled= newArrayList();

staticArrayList notHandled= newArrayList();

staticArrayList child= newArrayList();

staticintedgeCount,top=-1;

publicstaticvoidmain(String[] arg){


System.out.println( "NOTE: The first node is assumed as ROOT");



for(inti=0; i


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{

end.remove(i);

System.out.println( "Error: SAME AS START NODE");

System.out.println( "Enter the TO node:");end.add(s.next());

}

}

System.out.println("Handled STACK");

check(start.get(0));

printStack();

System.out.println("All Nodes are traversed in DFS");

}

staticvoidcSort()

{

//since the child nodes are inserted in reverse order, reverse it

for(intq=child.size();q>0;q--)

push(child.get(q-1));

}

staticvoidpush(String node)

{

// add node to stack and notHandled

if(!(notHandled.contains(node)) && !(handled.contains(node)))

{

top++;

stack[top]=node;

notHandled.add(node);

}

}

staticvoidpop()

{

//move the top node from notHandled to handled

if(top>-1)

{

handled.add(stack[top]);

notHandled.remove(stack[top]);

top--;

}

}

staticvoidcheck(String node)

{

// insert the parent node in notHandled node

push(node);

printStack();

intflag=1;

for(intq=0; q


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a

b c

ed

pop();

flag=0;

}

//since they are not arranged, use child array ad temp arraychild.add(end.get(q));

}

//sort the child of current top of stack

cSort();

if(flag==1) // if no child nodes, then pop the top of stack

pop();

if(top>-1) //recurse the check function with new top of stack

check(stack[top]);

}

staticvoidprintStack()

{

//print the notHandled node in reverse order, since its stack

ArrayList sort = newArrayList();

for(intq=notHandled.size();q>0;q--)

sort.add(notHandled.get(q-1));

System.out.println(handled+" "+sort);

}

}

GRAPH

OUTPUT

NOTE: The first node is assumed as ROOT


4


a

Enter the TO node:b


a

Enter the TO node:

c


b

Enter the TO node:

e


c

Enter the TO node:

d

Handled STACK


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[] [a]

[a] [b, c]

[a, b] [e, c]

[a, b, e] [c][a, b, e, c] [d]

[a, b, e, c, d] []

All Nodes are traversed in DFS

RESULT Thus the code is written in Java and executed for implementing Depth First Search.


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DIJIKSTRAS SHORTEST PATH

ALGORITHM:

Step 1: Start

Step 2: Get an input graph G, source node S with cost of each edge c

Step 3: Assign path of each node as NIL

Step 4: Assign distance of each node except S as INFINITY

Step 5: Assign distance of S as 0

Step 6: Add S to notHandled

Step 7: Sort the nodes based on the distance

Step 8: Choose the node U, with minimum distance

Step 8.1: Add all children of U to notHandled

Step 8.2: For all the child nodes of U

Step 8.2.1: Assign path as node U

Step 8.2.2: If distance of current node greater than d(U) + c

Step 8.2.2.1: Assign distance as d(U) + c

Step 8.3: Move U from notHandled to Handled

Step 9: Go to step 7 if notHandled is not empty

Step 10: End Algorithm

JAVA CODE

importjava.util.*;

publicclassShortestPath_Dijikstra

{



staticArrayList cost= newArrayList();

staticArrayList nodes= newArrayList();staticString[]path= newString[20];

staticint[] length= newint[20];


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staticintedgeCount;


{

intnode, i;


// Get start node, end node, cost of each edge from user




for(i=0; i


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System.out.println(" ");

notHandled.clear();

handled.add(nodes.get(node));

}System.out.println("NODE KNOWN LENGTH PATH");

print();

System.out.println("");

}

staticintfindMinimum()

{

intcost=1000;

intnode=0;

for(intk=0; klength[k] && !(handled.contains(nodes.get(k))))

{

cost=length[k];

node=k;

}

returnnode;

}

staticvoidfindChild(intnode)

{

for(intj=0;j


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3

5

72

6

4

p

q r

ts

return0;

}

}

GRAPH

OUTPUT



6Enter the FROM node:

p

Enter the TO node:

q

Enter the cost of edge:

5


p

Enter the TO node:

r



q

Enter the TO node:

s


5


q

Enter the TO node:

t



r


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Enter the TO node:

s



r

Enter the TO node:

t


6

No. of nodes=5

NODE KNOWN LENGTH PATH

p 0 0 NIL

q 0 1000 null

r 0 1000 null

s 0 1000 null

t 0 1000 null


p 0 0 NIL

q 0 5 p

r 0 4 p

s 0 1000 null

t 0 1000 null


p 1 0 NIL

q 0 5 p

r 0 4 p

s 0 6 r

t 0 10 r


p 1 0 NIL

q 0 5 p

r 1 4 p

s 0 6 r

t 0 8 q


p 1 0 NIL

q 1 5 p

r 1 4 p

s 0 6 r

t 0 8 q


p 1 0 NIL

q 1 5 p

r 1 4 p

s 1 6 rt 0 8 q


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p 1 0 NIL

q 1 5 pr 1 4 p

s 1 6 r

t 1 8 q

RESULT Thus the code is written in Java and executed for implementing Dijikstras Shortest Path

Algorithm.


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TOPOLOGICAL SORT

ALGORITHM:

Step 1: Start

Step 2: Get an input graph G

Step 3: Find the in-degree of each node in graph G

Step 4: Sort the nodes with respect to their in-degrees

Step 5: Remove the node with in-degree 0 from graph

Step 6: Add the node to Handled queue

Step 7: Go to step 3 if graph is not empty

Step 8: Return Handled


JAVA CODE

packageGraph;importjava.util.*;

publicclassTopologicalSort

{



staticArrayList nodes= newArrayList();

staticint[] indegree= newint[20];


staticintedgeCount, i, nodeSize, k,j, count,pNodeSize;

staticbooleanflag=true;


{

intmin;





for(i=0; i


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while(start.get(i).equalsIgnoreCase(end.get(i)))

{

end.remove(i);//to remove node added before checking error

System.out.println( "Error: SAME AS START NODE");System.out.println( "Enter the TO node:");

end.add(s.next());

}

if(!(nodes.contains(start.get(i))))

nodes.add(start.get(i));



}

nodeSize=nodes.size();

for(k=0;k


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a

eb

cd

}

}

staticintminDegree(){

intmin=indegree[0],index=0;

for(i=0;iindegree[i])

{

min=indegree[i];

index=i;

}

}

returnindex;

}

staticbooleancheck()

{

for(i=0;i


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c

Enter the TO node:

d

Enter the FROM node:c

Enter the TO node:

e


a

Enter the TO node:

d

indegree of a is 0

indegree of b is 1

indegree of c is 1

indegree of d is 2

indegree of e is 1

Sorted queue [a]

-----------------------------------

indegree of b is 0

indegree of c is 1

indegree of d is 1

indegree of e is 1

Sorted queue [a, b]

-----------------------------------

indegree of c is 0

indegree of d is 1

indegree of e is 1

Sorted queue [a, b, c]

-----------------------------------

indegree of d is 0

indegree of e is 0

Sorted queue [a, b, c, d]

-----------------------------------

indegree of e is 0

Sorted queue [a, b, c, d, e]

-----------------------------------

RESULT Thus the code is written in Java and executed for implementing Topological sorting.


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HILL CIMBING

ALGORITHM:

Step 1: Start

Step 2: Get an input graph G, source node S, destination node D with flow_capacity of each edge,

rate of flow

Step 3: Find the various paths from source and destination

Step 4: Select a path from set of path

Step 4.1: Find min_cut of selected path

Step 4.2: Netflow is the sum of min_cut of all paths

Step 5: If the netFlow > rate of flow then

Step 5.1: Select a path from set of path

Step 5.2: Find min_cut of selected path

Step 5.3: For each edge of the path

Step 5.3.1: Calculate flow_capacitymin_cut

Step 5.3.2: Assign the result to flow_capacity of the edge

Step 5.3.3: Calculate and assign netFlow = netFlowmin_cut

Step 5.4: Goto Step 5.1 if netFlow > 0

Step 5.5: Else Return the flow through each edge of the graph

Step 6: Else Exit


JAVA CODE

importjava.util.*;

publicclassHillClimbing

{




staticArrayList capacity= newArrayList();

staticArrayList stack= newArrayList();

staticint[]flow= newint[20];


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staticString dest,source;

staticintedgeCount,i,j,netFlow,flag=0,top,netCapacity=0,capa;


{Scanner s= newScanner(System.in);



for(i=0; i


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else

queue.add(end.get(i));

}

while(!queue.isEmpty()){

findPath(queue.get(0));

stack.remove(stack.size()-1);

queue.remove(0);

}

}

staticvoidpathCapacity(ArrayList path)

{

intmin=1000;

ArrayList edges = newArrayList();

for(j=0;jmin)

{

netFlow=netFlow-min;

capa=capa+min;

System.out.println("Flow along the PATH "+stack+" with FLOW "+min);

for(intq=0;q


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8

16

7

12 42

17

15

10b

13

a

e

c

d

ts

}

}

}

staticvoidcheck()

{

if(netFlow>0)

{

System.out.println("The Maximum flow of the network is "+capa);

System.out.println("The requested flow cant be accomodated.."+netFlow);

}

}

staticvoidprint()

{


for(j=0;j


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s

Enter the TO node:

cEnter the capacity of the edge:

12


d

Enter the TO node:

e

Enter the capacity of the edge:

16


d

Enter the TO node:

b


7


a

Enter the TO node:

b


15


e

Enter the TO node:

c


8


b

Enter the TO node:

c


17


c

Enter the TO node:

t


42

Enter the Source Node

s

Enter the Destination Node

t

Enter the Flow/Rate of the network

35

Flow along the PATH [s, a, b, c, t] with FLOW 10

Flow along the PATH [s, d, e, c, t] with FLOW 8

Flow along the PATH [s, d, b, c, t] with FLOW 5

Flow along the PATH [s, c, t] with FLOW 12

Flow across the edge s - a is : 10/10


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Flow across the edge s - d is : 13/13

Flow across the edge s - c is : 12/12

Flow across the edge d - e is : 8/16

Flow across the edge d - b is : 5/7Flow across the edge a - b is : 10/15

Flow across the edge e - c is : 8/8

Flow across the edge b - c is : 15/17

Flow across the edge c - t is : 35/42

RESULT Thus the code is written in Java and executed for implementing Hill Climbing.


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MERGE SORT

ALGORITHM:

Step 1: Start

Step 2: Get n inputs elements

Step 3: If the number of elements >2 then

Step 3.1: Split the elements into 2 halves, right and left using the index of the element

Step 3.2: Recurse the Step 3 with right array

Step 3.3: Recurse the Step 3 with left array

Step 3.4: Call merge function

Step 4: Else if the number of elements ==2 then

Step 4.1: Add first element to right array n second element to left array

Step 4.2: Call the merge function

Step 5: In the merge function

Step 5.1: Compare each element of the right and left array

Step 5.2: Add the sorted list to a global array


JAVA CODE

importjava.util.*;

publicclassMergeSort{

staticintn,i;

staticArrayList element= newArrayList();

staticArrayList sorted= newArrayList();


{


System.out.println( "NOTE: Duplicate terms will be removed");

System.out.println( "Enter the number of elements to be sorted:");

n=s.nextInt();

for(i=0;i


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element.add(s.nextInt());

}

splitting(element);

System.out.println(" ");System.out.println("Sorted List of given List is : "+sorted);

}

staticvoidsplitting(ArrayList entire)

{

intsize=entire.size()/2;

ArrayList right = newArrayList();

ArrayList left = newArrayList();

// if n>2 then recurse

if(entire.size()>2)

{

//divide into 2

for(i=0;i


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if(!merged.contains(left.get(y)))

merged.add(left.get(y));

y++;

i++;}

elseif(ls==y && rs==x && ss>z)//only sort

{

if(!merged.contains(sorted.get(z)))

merged.add(sorted.get(z));

z++;

i++;

}

elseif(ls>y && rs==x && ss>z)//left n sort

{

if(left.get(y)x && ss>z)//right n sort

{

if(right.get(x)y && rs>x && ss==z)//right n left

{

if(right.get(x)


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}

else

{

if(!merged.contains(left.get(y)))merged.add(left.get(y));

y++;

i++;

}

}

elseif(ls>y && rs>x && ss>z)//right, sort n left

{

if(left.get(y)


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sorted=merged;

}

}

OUTPUT

NOTE: Duplicate terms will be removed

Enter the number of elements to be sorted:

5

Enter the element 1 :

1


3Enter the element 3 :

5


2


4

Left : [3]

Previously Sorted : []

Right : [1]

-----------------------------

Locally Merged Sorted List : [1, 3]

Left : [4]

Previously Sorted : [1, 3]

Right : [2]

-----------------------------

Locally Merged Sorted List : [1, 2, 3]

Left : [2, 4]

Previously Sorted : [1, 2, 3]

Right : [5]

-----------------------------

Locally Merged Sorted List : [1, 2, 3, 4, 5]

Left : [5, 2, 4]

Previously Sorted : [1, 2, 3, 4, 5]

Right : [1, 3]

-----------------------------

Locally Merged Sorted List : [1, 2, 3, 4, 5]

Sorted List of given List is : [1, 2, 3, 4, 5]

RESULT Thus the code is written in Java and executed for implementing Merge Sort.


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QUICK SORT

ALGORITHM:

Step 1: Start

Step 2: Get n elements to be sorted

Step 3: If element size>2

Step 3.1: Find the average of n elements, pivot

Step 3.2: if element lesser than pivot

Step 3.2.1: Add element to lesser queue

Step 3.2.2: Recurse step 3 with lesser queue

Step 3.3: else

Step 3.3.1: Add element to bigger queue

Step 3.3.2: Recurse step 3 with bigger queue

Step 4: else

Step 4.1: if element[0] < element[1]

Step 4.1.1: Swap elements

Step 4.1.2: Join lesser and bigger list continuously


JAVA CODE

importjava.util.*;

publicclassQuickSort

{

staticintn,i;

staticArrayList element= newArrayList();

staticArrayList sorted= newArrayList();


{


//get elements

System.out.println( "Enter the number of elements to be sorted:");

n=s.nextInt();

for(i=0;i


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System.out.println( "Enter the element "+(i+1)+" :");

element.add(s.nextInt());

}

if(n>1){

System.out.println("Pivot -- Smaller -- Larger");

quickSort(element);

System.out.println("Sorted list : "+sorted);

}

else

{

System.out.println("List is : "+element);

}

}

staticvoidquickSort(ArrayList entire)

{

intsum=0,pivot=0;

ArrayList smaller = newArrayList();

ArrayList higher = newArrayList();

// if n>2 then recurse

if(entire.size()>2)

{

//find pivot element

for(i=0;i


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System.out.println(" "+pivot+" ["+entire.get(0)+"] ["+entire.get(1)+"]");

}

elseif(entire.size()==1)

{sorted.add(entire.get(0));

}

}

}

}

OUTPUT

Enter the number of elements to be sorted:

5

Enter the element 1 :46


25


47


36


12

Pivot -- Smaller -- Larger

33 [25, 12] [46, 47, 36]

18 [25] [12]

43 [36] [46, 47]

46 [46] [47]

Sorted list : [12, 25, 36, 46, 47]

RESULT Thus the code is written in Java and executed for implementing Merge Sort.


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PRODUCER CONSUMER

ALGORITHM:

Step 1: Start

Step 2: Get choice

Step 3: If choice is to produce

Step 3.1: Get the amount to be produced

Step 3.2: Add the amount to total amount

Step 4: Else if choice is to consume

Step 4.1: Get the amount to be consumed

Step 4.2: Subtract the amount from total amount

Step 5: Else

Step 5.1: Exit

Step 6: If total amount > threshold

Step 7: Exit


JAVA CODE

importjava.util.*;

publicclassProducerConsumer

{

staticintamount=0, currAmt;

staticbooleanflag=true;staticString choice;


{

Scanner s=newScanner(System.in);

System.out.println( "If u want to produce enter p/P");

System.out.println( "If u want to consume enter c/C");

System.out.println( "Others if u want to exit");

while(flag)

{

System.out.println( "Enter your Choice..");

System.out.println( "Amount present in warehouse = "+amount);

choice=s.next();if(choice.equalsIgnoreCase("p"))


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{

System.out.println( "What quantity do u want to produce?");

currAmt=s.nextInt();

amount=amount+currAmt;if(amount>500)flag=false;

}

elseif(choice.equalsIgnoreCase("c"))

{

if(amount==0)

System.out.println( "No production to be consumed..");

else

{

System.out.println( "What quantity do u want to consume?");

currAmt=s.nextInt();

if(currAmt>amount)

System.out.println( "Can't consume more than production !");

else

amount=amount-currAmt;

}

}

else

{

System.out.println( "Wrong choice ! ! ");

flag=false;

}

System.out.println( " ");

}

System.out.println( "The quantity in warehouse is "+amount);

if(amount>500)

System.out.println( "More production and few/no consumption");

}

}

OUTPUT

If u want to produce enter p/P

If u want to consume enter c/C

Others if u want to exit

Enter your Choice..

Amount present in warehouse = 0

p

What quantity do u want to produce?

34

Enter your Choice..


p

What quantity do u want to produce?

56

Enter your Choice..


cWhat quantity do u want to consume?


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100

Can't consume more than production !

Enter your Choice..Amount present in warehouse = 90

c

What quantity do u want to consume?

90

Enter your Choice..


c

No production to be consumed..

Enter your Choice..


s

Wrong choice ! !

The quantity in warehouse is 0

RESULT Thus the code is written in Java and executed for implementing Producer and consumer

problem.


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CONCURRENT LIST

ALGORITHM:

Step 1: Start

Step 2: Run all the threads simultaneously

Step 3: Lock the list to access it with the thread one and make other threads if present to wait

Step 3.1: Get choice as input

Step 3.2: To add an element list

Step 3.2.1: Get the element and add to the end of the list

Step 3.3: To delete the element

Step 3.3.1: Get the element and remove from the list

Step 3.4: To display the list, print the list

Step 3.5: when choice is wrong, terminate the current thread and unlock the list

Step 4: If a thread is waiting then continue till list is locked

Step 4.1: Check if list is free for each second then

Step 4.2: If list is free then go to Step3

Step 4.3: Else go to Step 4

Step 5: End Algorithm when no thread is waiting

JAVA CODE

List_Thread.java

packageConList;

classRunnableList implementsRunnable

{

Thread runner;

booleanflag=true;

publicRunnableList() {

}

publicvoidrun()

{

System.out.println("Now executing thread----"+ Thread.currentThread());List l = newList();


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l.myList();

}

}

publicclassList_Thread

{

publicstaticvoidmain(String[] args)

{

Thread thread1 = newThread(newRunnableList(), "thread1");

Thread thread2 = newThread(newRunnableList(), "thread2");

//Start the threads

thread1.start();

if(List.choice>3)

{

thread2.start();

}

else

{


System.out.println("Some thread is accessing the List...");

System.out.println("Now "+Thread.currentThread() +" is waiting......");

create(thread2);

}

}

publicstaticvoidcreate(Thread thread)

{

try{

//delays for one second

Thread.sleep(1000);

} catch(InterruptedException e) {

}

finally

{

if(List.choice>3)

{

thread.start();

}

else

{

create(thread);

}

}

}

}

List.java

packageConList;

importjava.util.*;

publicclassList

{

staticArrayList list1=newArrayList();publicstaticintchoice;


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staticScanner s= newScanner(System.in);

publicvoidmyList()

{

intitem;System.out.println("1. add element");

System.out.println("2. delete element");

System.out.println("3. display list");

System.out.println("Enter choice...");

choice=s.nextInt();

while(choice0)

{

if(choice==1)

{

System.out.println("Enter item to be added..");

item=s.nextInt();

add(item);

}

elseif(choice==2)

{

System.out.println("Enter item to be removed..");

item=s.nextInt();

remove(item);

}

elseif(choice==3)

{

System.out.println("The List is.."+list1);

}


choice=s.nextInt();

}

System.out.println("---------Thread Execution completed-----------");

}

staticvoidadd(intitem)

{

list1.add(item);

}

staticvoidremove(intitem)

{

if(!list1.isEmpty())

{

if(list1.contains(item))

list1.remove(find(item));

else

System.out.println("element not found");

}

else

System.out.println("List is empty....");

}

staticintfind(intitem)

{

for(inti=0;i


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returni;

return0;

}

}

OUTPUT

Now executing thread----Thread[thread1,5,main]

1. add element

Some thread is accessing the List...

2. delete elementNow Thread[main,5,main] is waiting......

3. display list

Enter choice...

1

Enter item to be added..

1

Enter choice...

1


2

Enter choice...

1


3

Enter choice...

3

The List is..[1, 2, 3]

Enter choice...

2

Enter item to be removed..

1

Enter choice...

4

---------Thread Execution completed-----------


1. add element

2. delete element

3. display list

Enter choice...

3

The List is..[2, 3]

Enter choice...

1


5

Enter choice...

3

The List is..[2, 3, 5]


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Enter choice...

2

Enter item to be removed..

1Enter choice...

5


RESULT Thus the code is written in Java and executed for accessing list concurrently.


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CONCURRENT STACK

ALGORITHM:

Step 1: Start


Step 3: Lock the stack to access it with the thread one and make other threads if present to wait


Step 3.2: To push an element list

Step 3.2.1: Get the element and add to the top of stack

Step 3.3: To pop an element

Step 3.3.1: Remove the element from top of the stack

Step 3.4: To display the stack, print the stack

Step 3.5: when choice is wrong, terminate the current thread and unlock the stack






JAVA CODE

Stack_Thread.java

packageConStack;

classRunnableStack implementsRunnable

{

Thread runner;

booleanflag=true;

publicRunnableStack() {

}

publicvoidrun()

{

System.out.println("Now executing thread----" +Thread.currentThread());

Stack s = newStack();s.myStack();


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}

}

publicclassStack_Thread{


{

Thread thread1 = newThread(newRunnableStack(), "thread1");

Thread thread2 = newThread(newRunnableStack(), "thread2");

//Start the threads

thread1.start();

if(Stack.choice>3)

thread2.start();

else

{

System.out.println(" Some thread is accessing the Stack...");

System.out.println(" Now "+

Thread.currentThread() +" is waiting......");

create(thread2);

}

}


{

try{


Thread.sleep(1000);


}

finally

{

if(Stack.choice>3)

thread.start();

else

create(thread);

}

}

}

Stack.java

packageConStack;

importjava.util.*;

publicclassStack

{

staticint[] stack1= newint[25];

publicstaticintchoice,top=-1;


publicvoidmyStack()

{

intitem;System.out.println("1. Push element");


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System.out.println("2. Pop element");

System.out.println("3. Display Stack");


choice=s.nextInt();while(choice0)

{

if(choice==1)

{

if(top>24)

{

System.out.println("Stack is full...");

}

else

{

System.out.println("Enter item to be pushed..");

item=s.nextInt();

push(item);

}

}

elseif(choice==2)

{

if(top==-1)

System.out.println("Stack is empty..");

else

pop();

}

elseif(choice==3)

{

System.out.print("The stack is..");

printStack();

}


choice=s.nextInt();

}

System.out.println("-----------Thread Execution completed----------");

}

staticvoidpush(intitem)

{

top++;

stack1[top]=item;

}

staticvoidpop()

{

System.out.println("The element popped is "+stack1[top]);

top--;

}

staticvoidprintStack()

{

if(top==-1)

System.out.print("empty");

else{


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System.out.print("[ ");

for(inti=top;i>-1;i--)

{

System.out.print(stack1[i]+" ");}

System.out.println("]");

}

}

}

OUTPUT


Some thread is accessing the Stack...

1. Push element

2. Pop element

3. Display Stack

Enter choice...

Now Thread[main,5,main] is waiting......

1

Enter item to be pushed..

1

Enter choice...

1


2

Enter choice...

1


3

Enter choice...

3

The stack is..[ 3 2 1 ]

Enter choice...

4

-----------Thread Execution completed----------


1. Push element

2. Pop element

3. Display Stack

Enter choice...

3


Enter choice...

2

The element popped is 3

Enter choice...

1


4


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Enter choice...

3


Enter choice...5

-----------Thread Execution completed----------

RESULT Thus the code is written in Java and executed for accessing stack concurrently.


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CONCURRENT QUEUE

ALGORITHM:

Step 1: Start


Step 3: Lock the queue to access it with the thread one and make other threads if present to wait


Step 3.2: To enqueue an element to the queue

Step 3.2.1: Get the element and add to the end of the queue

Step 3.3: To dequeue an element

Step 3.3.1: remove the element which is in the beginning of the queue

Step 3.4: To display the queue, print the queue

Step 3.5: when choice is wrong, terminate the current thread and unlock the queue






JAVA CODE

Queue_Thread.java

packageConQueue;

classRunnableQueue implementsRunnable

{

Thread runner;

booleanflag=true;

publicRunnableQueue() {

}

publicvoidrun() {

System.out.println("Now executing thread----"+Thread.currentThread());

Queue l = newQueue();l.myQueue();


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}

}

publicclassQueue_Thread{


{

Thread thread1 = newThread(newRunnableQueue(), "thread1");

Thread thread2 = newThread(newRunnableQueue(), "thread2");

//Start the threads

thread1.start();

if(Queue.choice>3)

thread2.start();

else

{


System.out.println("Some thread is accessing the Queue...");

System.out.println("Now "+Thread.currentThread() +" is waiting......");

create(thread2);

}

}


{

try{


Thread.sleep(1000);


}

finally

{

if(Queue.choice>3)

thread.start();

else

create(thread);

}

}

}

Queue.java

packageConQueue;

importjava.util.*;

publicclassQueue

{

staticint[] queue1= newint[25];

publicstaticintchoice,rear=0,front=0;


publicvoidmyQueue()

{

intitem;

System.out.println("1. Push element");

System.out.println("2. Pop element");System.out.println("3. Display Stack");


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choice=s.nextInt();

while(choice0)

{if(choice==1)

{

if(rear>24)

System.out.println("Modifications on 25 elements processed..");

else

{

System.out.println("Enter item to be added..");

item=s.nextInt();

enqueue(item);

}

}

elseif(choice==2)

{

if(rear==front)

System.out.println("Queue is empty");

else

dequeue();

}

elseif(choice==3)

{

System.out.println("The queue is..");

printQueue();

}


choice=s.nextInt();

}

System.out.println("-----------Thread Execution completed----------");

}

staticvoidenqueue(intitem)

{

queue1[rear]=item;

rear++;

}

staticvoiddequeue()

{

System.out.println("The element dequeued is "+queue1[front]);

front++;

}

staticvoidprintQueue()

{

if(rear==front)

System.out.print("empty");

else

{

System.out.print("[ ");

for(inti=front;i


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}

}

}

OUTPUT


Some thread is accessing the Queue...

Now Thread[main,5,main] is waiting......1. Push element

2. Pop element

3. Display Stack

Enter choice...

1


1

Enter choice...

1


2

Enter choice...

1


3

Enter choice...

3

The queue is..

[ 1 2 3 ]

Enter choice...

2

The element dequeued is 1

Enter choice...

4



1. Push element

2. Pop element

3. Display Stack

Enter choice...

3

The queue is..

[ 2 3 ]

Enter choice...

1


4Enter choice...


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3

The queue is..

[ 2 3 4 ]

Enter choice...5


RESULT Thus the code is written in Java and executed for accessing queue concurrently.


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DYNAMIC PROGRAMMING

ALGORITHM:

Step 1: Start

Step 2: Get an input graph G, source node S with cost of each edge c

Step 3: Assign path of each node as NIL

Step 4: Assign distance of each node except S as INFINITY

Step 5: Assign distance of S as 0

Step 6: Add S to notHandled

Step 7: Sort the nodes based on the distance

Step 8: Choose the node U, with minimum distance

Step 8.1: Add all children of U to notHandled

Step 8.2: For all the child nodes of U

Step 8.2.1: Assign path as node U

Step 8.2.2: If distance of current node greater than d(U) + c

Step 8.2.2.1: Assign distance as d(U) + c

Step 8.3: Move U from notHandled to Handled

Step 9: Go to step 7 if notHandled is not empty

Step 10: Return all the nodes and their cost from the source node S


JAVA CODE

importjava.util.*;

publicclassDynamicPrgm

{



staticArrayList cost= newArrayList();


staticint[] length= newint[20];




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staticintedgeCount;

publicstaticvoidmain(String[] arg){

intnode;

inti;


// Get start node, end node, cost of each edge from user




for(i=0; i


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3

5

72

6

4

p

q r

ts

{

intcost=1000;

intnode=0;

for(intk=0; klength[k] && !(handled.contains(nodes.get(k))))

{

cost=length[k];

node=k;

}

returnnode;

}

staticvoidfindChild(intnode)

{

for(intj=0;j


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OUTPUT


Enter the number of edges:6


p

Enter the TO node:

q


5


p

Enter the TO node:

r

Enter the cost of edge:4


q

Enter the TO node:

s


7


q

Enter the TO node:

t


3


r

Enter the TO node:

s


2


r

Enter the TO node:

t


6

NODE LENGTH

p - 0

q - 5

r - 4

s - 6

t - 8

RESULT Thus the code is written in Java and executed for performing dynamic programming in

calculating the cost of each node from source node.


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RANDOMIZED ALGORITHM

ALGORITHM:

Step 1: Start

Step 2: Get the choice from user either Combination or Permutation

Step 3: If choice is Combination

Step 3.1: Get n and r

Step 3.2: Calculate n! / [(n-r)! r!]

Step 3.3: Return result

Step 4: If choice is Permutation

Step 4.1: Get n and r

Step 4.2: Calculate n! / (n-r)!

Step 4.3: Return result


JAVA CODE

packageDynamicProgram;

importjava.util.*;

publicclassRandomizedAlgm

{


staticString choice;

staticintn;

staticintr;publicstaticvoidmain(String[] arg)

{

Scanner s=newScanner(System.in);

System.out.println( "If u want to find Permutations p/P");

System.out.println( "If u want to find Combinations c/C");

System.out.println( "Others if u want to exit");

while(flag)

{


System.out.println( "Enter your choice..");

choice=s.next();

if(choice.equals("p")){

System.out.println( "TO CALCULATE nPr :");


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System.out.println( "Enter the total no. of items, n :");

n=s.nextInt();

System.out.println( "Enter the no. of items in one sample, r :");

r=s.nextInt();if(n>=r)

findP(n,r);

else

System.out.println( "r should be lesser than or equal to n");

}

elseif(choice.equals("c"))

{

System.out.println( "TO CALCULATE nCr :");

System.out.println( "Enter the total no. of items, n :");

n=s.nextInt();

System.out.println( "Enter the no. of items in one sample, r :");

r=s.nextInt();

if(n>=r)

findC(n,r);

else

System.out.println( "r should be lesser than or equal to n");

}

else

{

System.out.println( "Wrong Choice...");

flag=false;

}

}

}

publicstaticvoidfindP(intn, intr)

{

// npr = n!/(n-r)!

intresult=fact(n)/fact(n-r);

System.out.println( n+" P "+r+" = "+result);

}

publicstaticvoidfindC(intn, intr)

{

// ncr = n!/(n-r)!r!

intresult=fact(n)/(fact(n-r)*fact(r));

System.out.println( n+" C "+r+" = "+result);

}

publicstaticintfact(intn)

{

intfact=1;

for(inti=1;i


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OUTPUT

If u want to find Permutations p/P

If u want to find Combinations c/C

Others if u want to exit

Enter your choice..

p

TO CALCULATE nPr :

Enter the total no. of items, n :

10

Enter the no. of items in one sample, r :

5

10 P 5 = 30240

Enter your choice..

c

TO CALCULATE nCr :

Enter the total no. of items, n :

7

Enter the no. of items in one sample, r :

4

7 C 4 = 35

Enter your choice..

h

Wrong Choice...

RESULT Thus the code is written in Java and executed for a randomized algorithm to calculate

combination and permutation.


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DINING PHILOSOPHERS PROBLEM

Algorithm:

Step 1: Start

Step 2: Assume all philosophers are reading

Step 3: Input a philosopher and his action

Step 3.1: If action is to eat

Step 3.1.1: Check if both chopsticks are available for the philosopher then

Step 3.1.1.1: Lock the chopsticks

Step 3.1.2: Else keep the philosopher waiting

Step 3.2: if action is to read

Step 3.2.1: Check if the chopsticks are locked then

Step 3.2.1.1: Unlock the chopsticks

Step 3.3: Else Exit


JAVA CODE

packageDynamicProgram;

importjava.util.*;

publicclassPhilosopherPrblm

{

staticScanner s=newScanner(System.in);

staticint[] reader={0,0,0,0,0};staticint[] chopStick={0,0,0,0,0};

staticintphilo;

staticString choice;



{

System.out.println( "NOTE: Initally all philosophers are reading...");

while(flag)

{

System.out.println( "Select a philosopher..");

philo=s.nextInt();if(philo0)

{


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System.out.println( "Does he want to eat(e/E) / read(r/R)?");

choice=s.next();

if(choice.equalsIgnoreCase("e"))

{switch(philo)

{

case1:

lock(0,1);

break;

case2:

lock(1,2);

break;

case3:

lock(2,3);

break;

case4:

lock(3,4);

break;

case5:

lock(4,0);

break;

}

}

elseif(choice.equalsIgnoreCase("r"))

{

switch(philo)

{

case1:

unlock(0,1);

break;

case2:

unlock(1,2);

break;

case3:

unlock(2,3);

break;

case4:

unlock(3,4);

break;

case5:

unlock(4,0);

break;

}

}

else

{

System.out.println( "Wrong choice.. The philosopher can either

read or eat..");

flag=false;

}

job();

System.out.println( ".........................................");

}else


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{

System.out.println( "There are only 5 philosophers on the table..");

flag=false;

}}

}

staticvoidlock(intm, intn)

{

if(chopStick[m]==0 && chopStick[n]==0)

{

chopStick[m]=1;

chopStick[n]=1;

reader[m]=1;

}

else

{

System.out.println( "Wait till philosopher's neighbour finishes eating..");


}

}

staticvoidunlock(intm,intn)

{

chopStick[m]=0;

chopStick[n]=0;

reader[m]=0;

}

staticvoidjob()

{

for(inti=0;i


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1

2

3

5

4

1

2

3

4

5

PICTURIZATION

OUTPUT

NOTE: Initally all philosophers are reading...

Select a philosopher..

2

Does he want to eat(e/E) / read(r/R)?

e

Philosopher 1 is reading

Philosopher 2 is eating




.........................................


4


e






.........................................



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5


e

Wait till philosopher's neighbour finishes eating..






.........................................


6

There are only 5 philosophers on the table..

RESULT Thus the code is written in Java and executed for the implementation of diningphilosophers problem.

advanced data structures lab manual

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