#include<stdio.h>
void crazy(int,int);
void crazy(int m, int n)
{
int *ptr;
m=0;
ptr=&m;
n=*ptr;
*ptr=1;
printf("%d
%d",m,n);
}
void main()
{
crazy(2,2);
}
O/p:
struct add //abstract data type
{
int A,B;
}x,sum;
int main()
{
printf("Enter first no\n");
scanf("%d", &x.A);
printf("Enter second no \n");
scanf("%d", &x.B);
// adding nos
sum.A=x.A+x.B;
printf("Sum is = %d", sum.A);
return 0;
}
1. Write a program (using nltk toolkit in python
environment) to tokenize
a) Sentence
b) Multiple sentences
c) A paragraph
d) Information of a complete web page
Python code
import nltk
from nltk.tokenize import word_tokenize
from nltk.text import Text
mystring = "I am currently studying in XYZ UNIVERSITY atDELHI. Cuurently in 4th year studying placements and tyring for the masters too as i was tooearly to start the programming. Currently tyring to ops as datascientist at any product based copany as a data scientist i need to learn all the python programmming at the professionally and we have to be good at statistics and the machine learnng models which model will suit for which problem"
tokens = word_tokenize(mystring)
tokens
O/p:
['I', 'am', 'currently',
'studying', 'in', 'XYZ', 'UNIVERSITY', 'at', 'DELHI', '.', 'Cuurently', 'in',
'4th', 'year', 'studying', 'placements', 'and', 'tyring', 'for', 'the',
'masters', 'too', 'as', 'i', 'was', 'too', 'early', 'to', 'start', 'the',
'programming', '.', 'Currently', 'tyring', 'to', 'ops', 'as', 'datascientist',
'at', 'any', 'product', 'based', 'copany', 'as', 'a', 'data', 'scientist', 'i',
'need', 'to', 'learn', 'all', 'the', 'python', 'programmming', 'at', 'the',
'professionally', 'and', 'we', 'have', 'to', 'be', 'good', 'at', 'statistics',
'and', 'the', 'machine', 'learnng', 'models', 'which', 'model', 'will', 'suit',
'for', 'which', 'problem']
Python code
nltk.download('punkt')
O/p:
[nltk_data] Downloading package punkt to
[nltk_data]
C:\Users\Viru\AppData\AMK\nltk_data...
[nltk_data]
Package punkt is already up-to-date!
True
Tokenizing multiple sentences
Python code
print("Enter multiple sentences: ")
lines = []
tok = []
while True:
line = input()
if line:
lines.append(line)
else:
break
for t in lines:
t = word_tokenize(t)
tok.append(t)
print("Tokens for multiple sentences are as follows: ",tok)
O/p:
Enter multiple sentences:
Modern humans arrived on the Indian subcontinent
from Africa no later than 55,000 years ago
Their long occupation, initially in varying forms
of isolation as hunter-gatherers, has made the region highly diverse, second
only to Africa in human genetic diversity.
In the early medieval era, Christianity, Islam,
Judaism, and Zoroastrianism became established on India's southern and western
coasts
Tokens for multiple sentences are
as follows: [['Modern', 'humans',
'arrived', 'on', 'the', 'Indian',…
Python code
##Information of a complete web page
#bs4 module to crawl webpage
from bs4 import BeautifulSoup
import urllib.request
#requesting php.net for information
response = urllib.request.urlopen('https://en.wikipedia.org/wiki/India')
html = response.read()
#cleaning grabbed text
soup = BeautifulSoup(html,"html5lib")
text = soup.get_text(strip=True)
tokens_web = word_tokenize(text)
print("Tokens for this web page are: ",tokens_web[:])
#declare a dictionary
word_freq = {}
for tok in tokens_web:
tok = tok.split()
for t in tok:
if t in word_freq:
word_freq[t] += 1
else:
word_freq[t] = 1
O/p:
Tokens for this web
page are: ['India', '-',
'Wikipediadocument.documentElement.className=', "''", 'client-js',
"''", ';', 'RLCONF=', '{', '``', 'wgBreakFrames', "''",
':', 'false', ',', "''", 'wgSeparatorTransformTable', "''",
':', '[', '``', "''", ',', "''", "''", ']', ',',
"''", 'wgDigitTransformTable', "'
1. Create an array of 6 zeros
import numpy
as np np.zeros(6)
array([0., 0., 0., 0., 0., 0.])
Stack using array in C
#include<stdio.h>
#include<stdlib.h>
#define
Size 4
int
Top=-1, inp_array[Size];
void
Push();
void
Pop();
void
show();
int
main()
{
int choice;
while(1)
{
printf("\nOperations performed by Stack");
printf("\n1.Push the element\n2.Pop the element\n3.Show\n4.End");
printf("\n\nEnter the choice:");
scanf("%d",&choice);
switch(choice)
{
case 1: Push();
break;
case 2: Pop();
break;
case 3: show();
break;
case 4: exit(0);
default: printf("\nInvalid choice!!");
}
}
}
void
Push()
{
int x;
if(Top==Size-1)
{
printf("\nOverflow!!");
}
else
{
printf("\nEnter element to be inserted to the stack:");
scanf("%d",&x);
Top=Top+1;
inp_array[Top]=x;
}
}
void Pop()
{
if(Top==-1)
{
printf("\nUnderflow!!");
}
else
{
printf("\nPopped element: %d",inp_array[Top]);
Top=Top-1;
}
}
void
show()
{
if(Top==-1)
{
printf("\nUnderflow!!");
}
else
{
printf("\nElements present in the stack: \n");
for(int i=Top;i>=0;--i)
printf("%d\n",inp_array[i]);
}
}
#include<stdio.h>
#include<string.h>
int i,j,m,n,c[20][20];
char x[20],y[20],b[20][20];
void printFunction(int i,int j)
{
if(i==0 || j==0)
return;
if(b[i][j]=='c')
{
printFunction(i-1,j-1);
printFunctionf("%c",x[i-1]);
}
else if(b[i][j]=='u')
printFunction(i-1,j);
else
printFunction(i,j-1);
}
void LongComSub()
{
m=strlen(x);
n=strlen(y);
for(i=0;i<=m;i++)
c[i][0]=0;
for(i=0;i<=n;i++)
c[0][i]=0;
//c, u and l denotes cross, upward and downward directions respectively
for(i=1;i<=m;i++)
for(j=1;j<=n;j++)
{
if(x[i-1]==y[j-1])
{
c[i][j]=c[i-1][j-1]+1;
b[i][j]='c';
}
else if(c[i-1][j]>=c[i][j-1])
{
c[i][j]=c[i-1][j];
b[i][j]='u';
}
else
{
c[i][j]=c[i][j-1];
b[i][j]='l';
}
}
}
int main()
{
printf("Provide first sequence:");
scanf("%s",x);
printf("Provide second sequence:");
scanf("%s",y);
printf("\nThe LongComSub is ");
LongComSub();
printFunction(m,n);
return 0;
}
Find the output of the following recursive algorithm.
#include<stdio.h>
int counter = 0;
int calc(int a, int b) {
int c;
counter++;
if (b==3) {
return (a*a*a);
}
else {
c = calc(a, b/3);
return (c*c*c);
}
}
int main (){
calc(4, 81);
printf ("%d", counter);
}
Output: 4
Job Sequencing using Greedy method in C
#include<stdio.h>
#define MAX 100
typedef struct Job {
char id[6];
int deadline;
int profit;
} Job;
void JobSeq(Job jobs[], int n);
// minimum of two jobs.
int Min(int x, int y) {
if(x < y) return x;
return y;
}
int main(void) {
/*variables*/
int i, j;
/*jobs with deadline and profit*/
Job jobs[6] = {
{"j1", 60, 200},
{"j2", 40, 180},
{"j3", 30, 190},
{"j4", 20, 300},
{"j5", 40, 120},
{"j6", 70, 120},
};
/*need one temp variable*/
Job temp;
//total jobs
int n = 6;
/*sort profit in descending order*/
for(i = 1; i < n; i++) {
for(j = 0; j < n - i; j++) {
if(jobs[j+1].profit > jobs[j].profit) {
temp = jobs[j+1];
jobs[j+1] = jobs[j];
jobs[j] = temp; //
}
}
}
printf("%10s %10s %10s\n", "Job", "Deadline", "Profit");
for(i = 0; i < n; i++) {
printf("%10s %10i %10i\n", jobs[i].id, jobs[i].deadline, jobs[i].profit);
}
JobSeq(jobs, n);
return 0;
}
void JobSeq(Job jobs[], int n) {
/*variables*/
int i, j, k, maxprofit;
//free time slots
int timeslot[MAX];
//filled time slots
int filledTimeSlot = 0;
//find max deadline value
int dmax = 0;
for(i = 0; i < n; i++) {
if(jobs[i].deadline > dmax) {
dmax = jobs[i].deadline;
}
}
//Initially let us keep all free empty slot as -1
for(i = 1; i <= dmax; i++) {
timeslot[i] = -1;
}
printf("dmax: %d\n", dmax);
for(i = 1; i <= n; i++) {
k = Min(dmax, jobs[i - 1].deadline);
while(k >= 1) {
if(timeslot[k] == -1) {
timeslot[k] = i-1;
filledTimeSlot++;
break;
}
k--;
}
//if all time slots are filled then stop
if(filledTimeSlot == dmax) {
break;
}
}
/*Sequenced jobs*/
printf("\nSequenced Jobs: ");
for(i = 1; i <= dmax; i++) {
printf("%s", jobs[timeslot[i]].id);
if(i < dmax) {
printf(" --> ");
}
}
// profit
maxprofit = 0;
for(i = 1; i <= dmax; i++) {
maxprofit += jobs[timeslot[i]].profit;
}
printf("\nMax Profit: %d\n", maxprofit);
}
Modify the following Floyd-Warshall algorithm by introducing a π( pi ) table as per the class’s discussion. Here you can write one function...
