selection_sort..c#include stdio.hheader file input output func.pdf
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selection_sort..c #include //header file input output function #define N 10//default value N=10 void selection_sort(int a[], int n);//function declaration int *find_largest(int a[], int n);//function declaration void swap(int *p, int *q);//function declaration int main(void) {//main function int i; int a[N];//variable declaration int x,y; printf(\"Enter %d numbers to be sorted: \", N); for (i = 0; i < N; i++) scanf(\"%d\", &a[i]);//keyboard inputting selection_sort(a, N);//calling method printf(\"In sorted order:\"); for (i = 0; i < N; i++) printf(\" %d\", a[i]); printf(\"\ \"); printf(\"largest element in the array :\"); *find_largest(a,N);//calling method printf(\"\ Enter two variables for swapping :\"); scanf(\"%d\",&x); scanf(\"%d\",&y); swap(&x,&y); return 0; } void selection_sort(int a[], int n) {//function definition int i, largest = 0, temp; if (n == 1) return; for (i = 1; i < n; i++) if (a[i] > a[largest]) largest = i;//selection sort logic if (largest < n - 1) { temp = a[n-1]; a[n-1] = a[largest]; a[largest] = temp; } selection_sort(a, n - 1); } int *find_largest(int *a, int n) {//function definition for(int i=1;i if you want to find smallest element*/ *a=*(a+i); } printf(\"%d\",*a); } void swap(int *p, int *q) {//function definition *p = *p + *q; // x now becomes 15 *q = *p - *q; // y becomes 10 *p = *p - *q; // x becomes 5 printf(\"After swapping of values :\"); printf(\"%d %d\ \",*p,*q); } output Enter 10 numbers to be sorted: 562 1 344 6 4 3 2 4 67 8 In sorted order: 1 2 3 4 4 6 8 67 344 562 largest element in the array :562 Enter two variables for swapping :10 20 After swapping of values :20 10 Solution selection_sort..c #include //header file input output function #define N 10//default value N=10 void selection_sort(int a[], int n);//function declaration int *find_largest(int a[], int n);//function declaration void swap(int *p, int *q);//function declaration int main(void) {//main function int i; int a[N];//variable declaration int x,y; printf(\"Enter %d numbers to be sorted: \", N); for (i = 0; i < N; i++) scanf(\"%d\", &a[i]);//keyboard inputting selection_sort(a, N);//calling method printf(\"In sorted order:\"); for (i = 0; i < N; i++) printf(\" %d\", a[i]); printf(\"\ \"); printf(\"largest element in the array :\"); *find_largest(a,N);//calling method printf(\"\ Enter two variables for swapping :\"); scanf(\"%d\",&x); scanf(\"%d\",&y); swap(&x,&y); return 0; } void selection_sort(int a[], int n) {//function definition int i, largest = 0, temp; if (n == 1) return; for (i = 1; i < n; i++) if (a[i] > a[largest]) largest = i;//selection sort logic if (largest < n - 1) { temp = a[n-1]; a[n-1] = a[largest]; a[largest] = temp; } selection_sort(a, n - 1); } int *find_largest(int *a, int n) {//function definition for(int i=1;i if you want to find smallest element*/ *a=*(a+i); } printf(\"%d\",*a); } void swap(int *p, int *q) {//function definition *p = *p + *q; // x now becomes 15 *q = *p - *q; // y becomes 10 *p = *p - *q; // x bec.
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![selection_sort..c
#include //header file input output function
#define N 10//default value N=10
void selection_sort(int a[], int n);//function declaration
int *find_largest(int a[], int n);//function declaration
void swap(int *p, int *q);//function declaration
int main(void)
{//main function
int i;
int a[N];//variable declaration
int x,y;
printf("Enter %d numbers to be sorted: ", N);
for (i = 0; i < N; i++)
scanf("%d", &a[i]);//keyboard inputting
selection_sort(a, N);//calling method
printf("In sorted order:");
for (i = 0; i < N; i++)
printf(" %d", a[i]);
printf(" ");
printf("largest element in the array :");
*find_largest(a,N);//calling method
printf(" Enter two variables for swapping :");
scanf("%d",&x);
scanf("%d",&y);
swap(&x,&y);
return 0;
}
void selection_sort(int a[], int n)
{//function definition
int i, largest = 0, temp;
if (n == 1)
return;
for (i = 1; i < n; i++)
if (a[i] > a[largest])
largest = i;//selection sort logic](https://image.slidesharecdn.com/selectionsort-230410145707-aa9258b3/85/selection_sort-c-include-stdio-hheader-file-input-output-func-pdf-1-320.jpg)
![if (largest < n - 1) {
temp = a[n-1];
a[n-1] = a[largest];
a[largest] = temp;
}
selection_sort(a, n - 1);
}
int *find_largest(int *a, int n)
{//function definition
for(int i=1;i if you want to find smallest element*/
*a=*(a+i);
}
printf("%d",*a);
}
void swap(int *p, int *q)
{//function definition
*p = *p + *q; // x now becomes 15
*q = *p - *q; // y becomes 10
*p = *p - *q; // x becomes 5
printf("After swapping of values :");
printf("%d %d ",*p,*q);
}
output
Enter 10 numbers to be sorted: 562 1 344 6 4 3 2 4 67 8
In sorted order: 1 2 3 4 4 6 8 67 344 562
largest element in the array :562
Enter two variables for swapping :10 20
After swapping of
values :20 10
Solution
selection_sort..c](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
Write a program that reads in integer as many as the user enters from.docx
Write a program that reads in integer as many as the user enters from standard input and print out the Maximum and minimum values.
Solution
Answer
#include<stdio.h>
#include<conio.h>
//------main function-----//
main()
{
//------Variable Declaration--------//
int a[10],i,n,max,min;
//-----Enter the input value using printf() and scanf() function------//
printf(\"Enter the Value of N:\");
scanf(\"%d\",&n);
printf(\"\ \");
//----Input N Element using For Loop----//
for(i=0;i<n;i++)
{
printf(\"Enter the %d Element\",i+1);
scanf(\"%d\",&a[i]);
}
min=a[0]; //----Minimum Number Initialization-----//
max=a[0]; //----Maximum Number Initialization-----//
//----Minimum and Maximum Number Calculation using For Loop-----//
for(i=0;i<n;i++)
{
if(a[i]>max)
max=a[i];
if(a[i]<min)
min=a[i];
}
//---------- Dispaly Result-------//
printf(\"\ The Maximum Number is %d\ \",max);
printf(\"the Minimum Number is %d\ \",min);
getch(); //---Function for Hold Output Screen--//
return 0;
}
.
Recommended
Write a program that reads in integer as many as the user enters from.docx
Write a program that reads in integer as many as the user enters from standard input and print out the Maximum and minimum values.
Solution
Answer
#include<stdio.h>
#include<conio.h>
//------main function-----//
main()
{
//------Variable Declaration--------//
int a[10],i,n,max,min;
//-----Enter the input value using printf() and scanf() function------//
printf(\"Enter the Value of N:\");
scanf(\"%d\",&n);
printf(\"\ \");
//----Input N Element using For Loop----//
for(i=0;i<n;i++)
{
printf(\"Enter the %d Element\",i+1);
scanf(\"%d\",&a[i]);
}
min=a[0]; //----Minimum Number Initialization-----//
max=a[0]; //----Maximum Number Initialization-----//
//----Minimum and Maximum Number Calculation using For Loop-----//
for(i=0;i<n;i++)
{
if(a[i]>max)
max=a[i];
if(a[i]<min)
min=a[i];
}
//---------- Dispaly Result-------//
printf(\"\ The Maximum Number is %d\ \",max);
printf(\"the Minimum Number is %d\ \",min);
getch(); //---Function for Hold Output Screen--//
return 0;
}
.
Can you give an example of a binary heap programCan you give an .pdf
Can you give an example of a binary heap program
Can you give an example of a binary heap program
Solution
/*
* C Program to Implement a Heap & provide Insertion & Deletion Operation
*/
#include
int array[100], n;
main()
{
int choice, num;
n = 0;/*Represents number of nodes in the heap*/
while(1)
{
printf(\"1.Insert the element \ \");
printf(\"2.Delete the element \ \");
printf(\"3.Display all elements \ \");
printf(\"4.Quit \ \");
printf(\"Enter your choice : \");
scanf(\"%d\", &choice);
switch(choice)
{
case 1:
printf(\"Enter the element to be inserted to the list : \");
scanf(\"%d\", &num);
insert(num, n);
n = n + 1;
break;
case 2:
printf(\"Enter the elements to be deleted from the list: \");
scanf(\"%d\", &num);
delete(num);
break;
case 3:
display();
break;
case 4:
exit(0);
default:
printf(\"Invalid choice \ \");
}/*End of switch */
}/*End of while */
}/*End of main()*/
display()
{
int i;
if (n == 0)
{
printf(\"Heap is empty \ \");
return;
}
for (i = 0; i < n; i++)
printf(\"%d \", array[i]);
printf(\"\ \");
}/*End of display()*/
insert(int num, int location)
{
int parentnode;
while (location > 0)
{
parentnode =(location - 1)/2;
if (num <= array[parentnode])
{
array[location] = num;
return;
}
array[location] = array[parentnode];
location = parentnode;
}/*End of while*/
array[0] = num; /*assign number to the root node */
}/*End of insert()*/
delete(int num)
{
int left, right, i, temp, parentnode;
for (i = 0; i < num; i++) {
if (num == array[i])
break;
}
if (num != array[i])
{
printf(\"%d not found in heap list\ \", num);
return;
}
array[i] = array[n - 1];
n = n - 1;
parentnode =(i - 1) / 2; /*find parentnode of node i */
if (array[i] > array[parentnode])
{
insert(array[i], i);
return;
}
left = 2 * i + 1; /*left child of i*/
right = 2 * i + 2; /* right child of i*/
while (right < n)
{
if (array[i] >= array[left] && array[i] >= array[right])
return;
if (array[right] <= array[left])
{
temp = array[i];
array[i] = array[left];
array[left] = temp;
i = left;
}
else
{
temp = array[i];
array[i] = array[right];
array[right] = temp;
i = right;
}
left = 2 * i + 1;
right = 2 * i + 2;
}/*End of while*/
if (left == n - 1 && array[i]) {
temp = array[i];
array[i] = array[left];
array[left] = temp;
}
}
Output:.
a) Complete both insert and delete methods. If it works correctly 10.pdf
a) Complete both insert and delete methods. If it works correctly 100 points each.
b) Do not call the search method in delete method.
c) Make sure to write your name in main method.
d) DO NOT MODIFY PROCESS METHOD.
3- Create a proper input file with respect to process method to test all cases in both
insert and delete methods.
4- Compile and execute your program at the command prompt. Check your output, if it
is correct
//xxxxxH5.java is linked list implementation of lstinterface.
//Eliminate count form both insert and delete methods
// If works correctly, 200 points
import java.util.*;
import java.io.*;
//lstinterface is an interface
public class xxxxxH5 implements lstinterface {
// xxxxxH5 class variables
//head is a pointer to beginning of linked list
private node head = null;
//use prt for System.out to save typing
PrintStream prt = System.out;
// class node
private class node{
// class node variables
int data;
node rlink; //right link
// class node constructor
node(int x){
data = x;
rlink = null;
} // end class node constructor
}// class node
// insert x at position p,
// if successful return 1 otherwise return 0
public int insert(int x, int p){
prt.printf("\n\t\tInsert %5d at position %2d:", x, p);
// complete this method
return 0; // successful insertion
} // end insert
// delete x at position p
// if successful return 1 otherwise return 0
public int delete(int p){
prt.printf("\n\t\tDelete element at position %2d:", p);
// complete this method
return 0; // successful insertion
} // end delete
// sequential serach for x in the list
// if successful return its position otherwise return 0;
public int search(int x){
// Local variables
node cur;
prt.printf("\n\t\tSearch for %5d:", x);
// Complete the rest of the method
int pos = 0;
for (cur = head; cur != null ; cur = cur.rlink){
pos ++;
if (cur.data == x){
prt.printf(" Found at position %2d, ", pos);
return pos;
} // end if
} // end for
prt.printf(" Not Found.");
return 0; // x is not found.
} // end search
// print list elements formatted
public void printlst() {
// Local variables
node cur;
prt.printf("\n\tList contents: ");
for (cur = head; cur != null ; cur = cur.rlink)
prt.printf("%5d,", cur.data);
// end for
} // end printlst
// insert delete and search in the list
private void process(String fn){
// Local variables
int j, ins, del, srch, k, p;
int x;
prt.printf("\tLinked List implementation of int list without"+
//\n\tusing count, gets input file name from method argument, then
reads:"+
"\n\tinteger No. of elements to insert(ins)"+
" followed by elements to insert and their positions,"+
"\n\tinteger No. of elements to search(srch) followed by element to
search"+
"\n\tinteger No. of elements to delete(del) followed by position of"+
"elements to delete" +
"\n\t\tTo compile: javac xxxxxH5.java" +
"\n\t\tTo execute: java xxxxxH5 inputfilename");
try{
// open input file
Scanner inf = new Scanner(new File(fn));
//read no. of elements to insert
ins = inf.nextInt();
prt.printf("\n\tInsert %.
BINARY SEARCH TREE OPERATIONS #includestdio.h#includestdlib.pdf
BINARY SEARCH TREE OPERATIONS */
#include
#include
#define NULL 0
struct tree
{
int info;
struct tree *left;
struct tree *right;
};
struct tree *p;
struct tree *search(struct tree *,int);
struct tree *insert(struct tree *i,int);
void inorder(struct tree *);
void delete(struct tree *);
main()
{
int l,ch,key,x,y;
struct tree *z;
clrscr();
do
{
printf(\"\ BINARY SEARCH TREE OPERATIONS\ \");
printf(\"\ 1.INSERTING ELEMENT\");
printf(\"\ 2.DELETING ELEMENT\");
printf(\"\ 3.SEARCH FOR ELEMENT\");
printf(\"\ 4.DISPLAY\");
printf(\"\ 5.EXIT\");
printf(\"\ \ ENTER CHOICE:\");
scanf(\"%d\",&ch);
switch(ch)
{
case 1:printf(\"\ enter element to insertat end 0\ \");
scanf(\"%d\",&x);
while(x!=0)
{
p=insert(p,x);
printf(\"\ enter element to insert:\");
scanf(\"%d\",&x);
}
break;
case 2:printf(\"\ enter node to delete\");
scanf(\"%d\",&key);
z=search(p,key);
if(z!=NULL)
{
delete(z);
printf(\"\ given key node is deleted \");
}
else
printf(\"\ the given key node not found\");
break;
case 3:printf(\"\ enter node to be search\");
scanf(\"%d\",&y);
p=search(p,y);
if(p!=NULL)
printf(\"\ the key element is :%d\",p->info);
else
printf(\"\ the key is not found\");
break;
case 4:printf(\"\ tree elements in inorder\ :\");
inorder(p);
case 5:exit(0);
break;
}
}
while(ch>=1||ch<=5);
}
struct tree *insert(struct tree *s,int x)
{
struct tree *trail,*p,*q;
q=(struct tree *)malloc(sizeof(struct tree));
q->info=x;
q->left=NULL;
q->right=NULL;
p=s;
trail=NULL;
while(p!=NULL)
{
trail=p;
if(p->info>x)
p=p->left;
else
p=p->right;
}
if(trail==NULL)
{
s=q;
return(s);
}
if(trail->info>x)
trail->left=q;
else
trail->right=q;
return(s);
}
struct tree *search(struct tree *p,int x)
{
while(p!=NULL && p->info!=x)
{
if(p->info>x)
p=p->left;
else
p=p->right;
}
return(p);
}
void delete(struct tree *p)
{
struct tree *temp;
if(p==NULL)
printf(\"\ trying to delete a noexistent node \ \");
else if(p->left==NULL)
{
temp=p;
p=p->right;
free(temp);
}
else if(p->right==NULL)
{
temp=p;
p=p->left;
free(temp);
}
else if((p->left!=NULL) && (p->right!=NULL))
{
temp=p->right;
while(temp->left!=NULL)
temp=temp->left;
temp->left=p->left;
temp=p;
p=p->right;
free(temp);
}
}
void inorder(struct tree *p)
{
if(p!=NULL)
{
inorder(p->left);
printf(\"%d\\t\",p->info);
inorder(p->right);
}
}
Solution
BINARY SEARCH TREE OPERATIONS */
#include
#include
#define NULL 0
struct tree
{
int info;
struct tree *left;
struct tree *right;
};
struct tree *p;
struct tree *search(struct tree *,int);
struct tree *insert(struct tree *i,int);
void inorder(struct tree *);
void delete(struct tree *);
main()
{
int l,ch,key,x,y;
struct tree *z;
clrscr();
do
{
printf(\"\ BINARY SEARCH TREE OPERATIONS\ \");
printf(\"\ 1.INSERTING ELEMENT\");
printf(\"\ 2.DELETING ELEMENT\");
printf(\"\ 3.SEARCH FOR ELEMENT\");
printf(\"\ 4.DISPLAY\");
printf(\"\ 5.EXIT\");
printf(\"\ \ ENTER CHOICE:\");
scanf(\"%d\",&ch);
switch(ch)
{
case 1:printf(\"\ enter element to insertat end 0\ \");
scanf(\"%d\",&x);
while(x!=0)
{
p=insert(p,x);
printf(\"\ enter e.
Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
Data Structure using C
This document contains some programs of C using Data structures, like Stack, LinkedList, queue, Fibonacci series, addition and multiplication of two matrices,etc.
My C proggram is having trouble in the switch in main. Also the a co.pdf
My C proggram is having trouble in the switch in main. Also the a couple of funtion like delete
and display tree are failing
#include
#include
#include
#include
#include
#include
struct tree_node {
int data;
struct tree_node *left, *right;
};
typedef struct nodeT {
int key;
struct nodeT *left, *right;
int data;
} nodeT, *treeT;
//struct nodeT *t1, *t2;
void DisplayTree(nodeT*t)
{
if (t != NULL) {
DisplayTree(t->left);
printf(\"%c \", t->key);
DisplayTree(t->right);
}
}
nodeT *FindNode(nodeT *t, int key){
if(t==NULL) return 0;
if (key == t->key)
return t;
if(key < t->key){
return FindNode(t->left, key);;
} else {
return FindNode(t->right, key);;
}
}
int max(nodeT *p){
nodeT *t, *tmp;
while(tmp->right != NULL)
tmp=tmp->right;
return tmp->data;
}
int min(nodeT *t){
nodeT *tmp;
while(tmp->left != NULL)
tmp = tmp->left;
return tmp->data;
}
void ModifyPreOrderWalk(nodeT *t, int h)
{
int i;
if (t == NULL) return;
for(i=0; i<-1;i++){
printf(\"| \");
}
if(h>1)printf(\"+---\");
printf(\"%d\ “\", t->key);
ModifyPreOrderWalk(t->left, h+1);
ModifyPreOrderWalk(t->right, h+1);
}
void PrintTree(nodeT *t)
{
ModifyPreOrderWalk(t, 1);
}
void delete(nodeT **p){
nodeT *target, *imd_r, *plmd_r;
target=*p;
if (target->left==NULL && target->right==NULL) {
*p=NULL;
} else if (target->left == NULL) {
*p=target->right;
} else
if (target->right == NULL) {
*p=target->left;
} else {
plmd_r = target;
imd_r = target->right;
while( imd_r->left != NULL){
plmd_r = imd_r;
imd_r = imd_r->left;
}
if(plmd_r == target)
plmd_r->right = imd_r->right;
else
plmd_r->left = imd_r->right;
imd_r->left = target->left;
imd_r->right = target->right;
*p = imd_r;
}
free(target);
}
void listInorder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
printf(\"%d \", t->key);
DisplayTree(t->right);
}
}
void listPreorder(nodeT *t) {
if (t != NULL) {
printf(\"%d \", t->key);
DisplayTree(t->left);
DisplayTree(t->right);
}
}
void listPostOrder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
DisplayTree(t->right);
printf(\"%d\", t->key);
}
}
void InsertNode(nodeT **tptr, int key){
nodeT *t;
nodeT *tmp;
t=*tptr;
if (t == NULL) {
tmp= malloc(sizeof(nodeT));
tmp->key = key;
tmp->left=tmp->right=NULL;
*tptr=tmp;
return;
}
if (key < t->key) {
InsertNode
(&t->left, key);
} else {
InsertNode(&t->right, key);
}
}
int height(nodeT *t){
int lDepth = height(t->left);
int rDepth = height(t->right);
if(lDepth > rDepth)
return(lDepth+1);
else
return(rDepth+1);
}
int sum(nodeT *p){
if (p == NULL)
return 0;
else
return (p->data + sum(p->left) +sum(p->right) );
}
int count(nodeT *t){
if(t == NULL)
return 0;
int i = 1+count(t->left)+count(t->right);
return i;
}
/*nodeT *New(){
nodeT *tmp;
tmp = (nodeT *)malloc(sizeof(nodeT));
if (tmp==NULL) return NULL;
return tmp;
}*/
int main(){
nodeT *node;
nodeT *t=NULL, *findA;
int f;
int d, a, b, c, j, k, e, s, y;
while (1) {
printf(\"\ I - insert\ F - find node\ D - delete\ L - list in order\ R - list preorder\ P - list
PorstOrder\ M - max\ N - min\ H - height\ C - count\ S - sum.
#include iostream #include deque #include stdio.h scan.pdf
#include
#include
#include //scanf,printf
#include
#include //istringstream
using namespace std;
//maps
#include \"MapGeneric.h\"
#include \"MapTriple.h\"
#include \"MapAbsoluteValue.h\"
#include \"MapSquare.h\"
//filters
#include \"FilterGeneric.h\"
#include \"filterEven.h\"
#include \"filterPositive.h\"
#include \"filterPositiveTwoDigit.h\"
//reduce
#include \"ReduceGeneric.h\"
#include \"ReduceMinimum.h\"
#include \"ReduceGCD.h\"
int main()
{
deque L,L1,L2;
int i;
string line;
getline(cin,line);
if(line.size() == 0)
return 0;
line.erase(remove(line.begin(), line.end(), \',\'), line.end());
istringstream s(line);
while(s>>i)
{
L.push_back(i);
}
if(L.size() != 20)
return 0;
//converting L -> 3*|L| aka L1
MapTriple triple; MapAbsoluteValue abs;
L1 = triple.map(abs.map(L));
//converting L1 -> positive two digit integers -> even intergers aka L2
filterEven even; filterPositiveTwoDigit p2;
L2 = p2.filter(L1);
if(L2.size() <2)
return 0;
L2 = even.filter(L2);
if(L2.size() <2)
return 0;
//finding the minimum and GCD from L2 respectively
ReduceMinimum min; ReduceGCD gcd;
printf(\"%d %d\ \",min.reduce(L2),gcd.reduce(L2));
}
FilterGeneric.cpp
#include
#include
#include \"FilterGeneric.h\"
using namespace std;
deque FilterGeneric::filter(deque deq)
{
//calling recursive function with deq and its length
select(deq,deq.size()-1);
return filterDeq;
}
//int i is the index that will start from the length of deque till 0
void FilterGeneric::select(deque deq, int i)
{
if(i == 0)
{
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return;
}
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return select(deq,i-1);
}
FilterGeneric.h
#ifndef FILTERGENERIC_H
#define FILTERGENERIC_H
#include
class FilterGeneric
{
public:
std::deque filter(std::deque);
void select(std::deque,int);
protected:
//this deque will store the integers after filtration.
std::deque filterDeq;
private:
virtual bool f(int)=0;
};
#endif
MapAbsoluteValue.cpp
#include
#include
#include
#include \"MapGeneric.h\" //base
#include \"MapAbsoluteValue.h\"
using namespace std;
int MapAbsoluteValue::f(int num)
{
return abs(num);
}
MapAbsoluteValue.h
#ifndef MAPABSOLUTEVALUE_H
#define MAPABSOLUTEVALUE_H
#include
//this class will map a given input N, using the function map(input), to |N|
class MapAbsoluteValue: public MapGeneric
{
public:
//overwriting pure virtual function making N -> |N|
int f(int);
private:
};
#endif
MapGeneric.cpp
#include
#include
#include \"MapGeneric.h\"
using namespace std;
deque MapGeneric::map(deque deq)
{
//calling recursive function with deq and its length
calculate(deq,deq.size()-1);
return mapDeq;
}
//int i is the index that will start from the length of deque till 0
void MapGeneric::calculate(deque deq, int i)
{
if(i == 0)
{
int val = f(deq[i]);
mapDeq.push_front(val);
return;
}
int val = f(deq[i]);
mapDeq.push_front(val);
return calculate(deq,i-1);
}
MapGeneric.h
#ifndef MAPGENERIC_H
#define MAPGENERIC_H
#include
class MapGeneric
{
public:
std::deque ma.
IN C LANGUAGE- I've been trying to finish this program for the last fe.docx
IN C LANGUAGE: I've been trying to finish this program for the last few days, and I'm stuck on one part of it, and I can't figure out what to do.
The goal is to read in a file (separated by commas), put the information into a linked list, and then sort that linked list by either first or last name using function pointers.
Everything I have so far works, aside from the fact that my output is never sorted. I'm doing something wrong with the listSort function, and I don't know what it is.
Can anyone please just take a look at what I have and offer a suggestion or solution? It might not even be an issue with the function itself, but something somewhere else.
I would greatly appreciate it.
Please do not make any changes to the variables in the structs. Month has to be a string, it cannot be an integer.
driver.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
int main(int argc, char* argv[]) {
//creating input and output files
FILE* in = NULL;
FILE* out = NULL;
node_t* head = NULL;
node_t* temp = NULL;
//checking for 4 command line arguments
assert(argc == 4);
//opening input file
in = fopen(argv[1], "r");
assert(in != NULL);
//opening output file
out = fopen(argv[2], "w");
assert(out != NULL);
int sortBy = atoi(argv[4]);
temp = createList(in, &head);
if(sortBy == 1) {
sortList(&head, compare_by_firstname);
}
else if(sortBy == 2) {
sortList(&head, compare_by_lastname);
}
printList(out, temp);
deleteList(&head);
//closing files
fclose(in);
fclose(out);
return 0;
}
functions.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
#include <stdbool.h>
bool isLeapYear(int year) {
if(year % 4 != 0) {
return false;
}
else if(year % 100 != 0) {
return true;
}
else if(year % 400 != 0) {
return false;
}
else {
return true;
}
}
void printBorder(FILE* out)
{
int i = 0;
for(i = 0; i < 80; i++)
fprintf(out,"*");
fprintf(out, "\n");
}
bool checkInvalidDate(node_t *node) {
int year = (int)node->birthday.year;
if(node->birthday.day < 1 || node->birthday.year < 1) {
return true;
}
if(strcmp(node->birthday.month, "February") == 0) {
if(isLeapYear(year)) {
if(node->birthday.day > 29) {
return true;
}
}
else if(node->birthday.day > 28) {
return true;
}
}
else if((strcmp(node->birthday.month, "April") == 0) || (strcmp(node->birthday.month, "June") == 0) ||
(strcmp(node->birthday.month, "September") == 0) || (strcmp(node->birthday.month, "November") == 0)) {
if(node->birthday.day > 30) {
return true;
}
}
else {
if(node->birthday.day > 31) {
return true;
}
}
return false;
}
void add(node_t **node, node_t **head) {
node_t* current = *head;
while(current != NULL) {
//comparing first and last names to check for duplicates; if duplicate the info isn't added to the list
if((strcmp(current->firstName, (*node)->firstName) == 0) && (strcmp(current->lastName, (*node)->lastName) == 0)) {
return;
}
current = current->next;
}
//adding information as a node to the list
(*node.
888678123317368915236 #include stdio.h #include stdlib.h.pdf
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}
Solution
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}.
Hi,Please fidn the Answer.Sorting,h is Header .pdf
Hi,
Please fidn the Answer.
//////////////////Sorting,h is Header file which contians all Function decalartion
#include
#include // std::cout
#include // std::shuffle
#include // std::array
#include // std::default_random_engine
#include
using namespace std;
class Sorting
{
private:
int num;
public:
Sorting::Sorting() {
};
///Buble Sort Input array and n Number of Elemnets
void BubbleSorting(int array[], int n);
void SelectionSorting(int array[], int n);
void ShuffleArray(int array[], int n);
void DisplayArray(int array[], int n);
void displayMenu();
int choice;
};
//////////////////////////Sorting.cpp is class fuction Which Implemnets the Function
#include \"Sorting.h\"
#include
using namespace std;
///Sorting the Array Using Bubble Sort...
void Sorting::BubbleSorting(int Array[],int n)
{
int temp = 0;
cout << \"\ ------------ BUBBLE SORT ------------ \ \ \";
for (int i = 1; iArray[j + 1])
{
temp = Array[j];
Array[j] = Array[j + 1];
Array[j + 1] = temp;
}
}
cout << \"\ ------------ RESULTS BUBBLE SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
void Sorting::SelectionSorting(int Array[], int n)
{
int i, n, p, k, min, loc, temp;
for (p = 1; p <= n - 1; p++) // Loop for Pass
{
min = Array[p]; // Element Selection
loc = p;
for (k = p + 1; k <= n; k++) // Finding Min Value
{
if (min > Array[k])
{
min = Array[k];
loc = k;
}
}
temp = Array[p]; // Swap Selected Element and Min Value
Array[p] = Array[loc];
Array[loc] = temp;
}
cout << \"\ ------------ RESULTS SELECTION SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
///exchange each element with a randomly chosen element. It\'s possible that an element will be
exchanged with itself, but there is no problem with that
void Sorting::ShuffleArray(int Array[], int n)
{
cout << \"\ ------------ ShuffleArray SORT ------------ \ \ \";
for (int i = 0; i
#include \"Sorting.h\"
using namespace std;
int main()
{
bool SelectionOn = true;
Sorting Sorting;
int choice;
int n = 10;
int *Arry;
while (SelectionOn != false) {
cout << \"*******************************\ \";
cout << \" 1 - Enter User Input Array of Elements.\ \";
cout << \" 2 - Display User Input.\ \";
cout << \" 3 - Shuffle User Input the Elements i.\ \";
cout << \" 4 - Bubble Sorting.\ \";
cout << \" 5 - Selection Sorting.\ \";
cout << \" 6 - Exit.\ \";
cout << \" 7 - Display Menu Again. \ \";
cout << \" Enter your choice and press return: \";
cin >> choice;
switch (choice)
{
case 1:
cout << \"Enter User Input Array of Elements\ \";
cout << \"Enter No. of Elements=\";
cin >> n;
Arry = new int[n];
cout << \"\ Enter Elements=\ \";
for (int i = 1; i <= n; i++)
{
cin >> Arry[i];
}
break;
case 2:
cout << \"DISPLAY OF ARRAY\ \";
if (Arry != NULL)
{
Sorting.DisplayArray(Arry, n);
}
else
{
cout << \"ENTER ELEMENTS TO ARRAY BY SELCTING CHOICE 1\ \";
}
break;
case 3:
cout << \"SHUFFLE OF A.
There are many ways to do the Lewis symbol for yo.pdf
There are many ways to do the Lewis symbol for you to draw a Lewis dot diagram.
As long as you make sure to put a dot in for each valence electron it should be just fine. Some
teachers like to make things more complicated by drawing Lewis dots in regards to electron pairs
or bonding electrons. I\'ve seen some even go so far as to draw in charges on each of the dots of
a Lewis structure. Either way, these are unnecessary and, in my opinion, a waste of time. Just
put dots around the symbol and you should be fine. This also happens to be the best way to draw
it when you need to put together larger, more complicated Lewis structures and figure out a
\"formal charge\" (later on.) You did it exactly as I would do it and I\'m taking upper-division
chemistry. Just put one dot on each side, if there\'s more than four, put double dots on each side
until either all the sides are filled or all valence electrons are noted. You can also put a small line
(connects the two dots with a line) to designate an electron pair, either way is valid, just pick one
you think looks nicer.
Solution
There are many ways to do the Lewis symbol for you to draw a Lewis dot diagram.
As long as you make sure to put a dot in for each valence electron it should be just fine. Some
teachers like to make things more complicated by drawing Lewis dots in regards to electron pairs
or bonding electrons. I\'ve seen some even go so far as to draw in charges on each of the dots of
a Lewis structure. Either way, these are unnecessary and, in my opinion, a waste of time. Just
put dots around the symbol and you should be fine. This also happens to be the best way to draw
it when you need to put together larger, more complicated Lewis structures and figure out a
\"formal charge\" (later on.) You did it exactly as I would do it and I\'m taking upper-division
chemistry. Just put one dot on each side, if there\'s more than four, put double dots on each side
until either all the sides are filled or all valence electrons are noted. You can also put a small line
(connects the two dots with a line) to designate an electron pair, either way is valid, just pick one
you think looks nicer..
the compound is heptane note the molecular ion.pdf
the compound is: heptane note: the molecular ion peak is at m/z = 100 --->
indicates the molecular weight of the compound = 100 only heptane has a molecular weight of
100
Solution
the compound is: heptane note: the molecular ion peak is at m/z = 100 --->
indicates the molecular weight of the compound = 100 only heptane has a molecular weight of
100.
More Related Content
Similar to selection_sort..c#include stdio.hheader file input output func.pdf
Can you give an example of a binary heap programCan you give an .pdf
Can you give an example of a binary heap program
Can you give an example of a binary heap program
Solution
/*
* C Program to Implement a Heap & provide Insertion & Deletion Operation
*/
#include
int array[100], n;
main()
{
int choice, num;
n = 0;/*Represents number of nodes in the heap*/
while(1)
{
printf(\"1.Insert the element \ \");
printf(\"2.Delete the element \ \");
printf(\"3.Display all elements \ \");
printf(\"4.Quit \ \");
printf(\"Enter your choice : \");
scanf(\"%d\", &choice);
switch(choice)
{
case 1:
printf(\"Enter the element to be inserted to the list : \");
scanf(\"%d\", &num);
insert(num, n);
n = n + 1;
break;
case 2:
printf(\"Enter the elements to be deleted from the list: \");
scanf(\"%d\", &num);
delete(num);
break;
case 3:
display();
break;
case 4:
exit(0);
default:
printf(\"Invalid choice \ \");
}/*End of switch */
}/*End of while */
}/*End of main()*/
display()
{
int i;
if (n == 0)
{
printf(\"Heap is empty \ \");
return;
}
for (i = 0; i < n; i++)
printf(\"%d \", array[i]);
printf(\"\ \");
}/*End of display()*/
insert(int num, int location)
{
int parentnode;
while (location > 0)
{
parentnode =(location - 1)/2;
if (num <= array[parentnode])
{
array[location] = num;
return;
}
array[location] = array[parentnode];
location = parentnode;
}/*End of while*/
array[0] = num; /*assign number to the root node */
}/*End of insert()*/
delete(int num)
{
int left, right, i, temp, parentnode;
for (i = 0; i < num; i++) {
if (num == array[i])
break;
}
if (num != array[i])
{
printf(\"%d not found in heap list\ \", num);
return;
}
array[i] = array[n - 1];
n = n - 1;
parentnode =(i - 1) / 2; /*find parentnode of node i */
if (array[i] > array[parentnode])
{
insert(array[i], i);
return;
}
left = 2 * i + 1; /*left child of i*/
right = 2 * i + 2; /* right child of i*/
while (right < n)
{
if (array[i] >= array[left] && array[i] >= array[right])
return;
if (array[right] <= array[left])
{
temp = array[i];
array[i] = array[left];
array[left] = temp;
i = left;
}
else
{
temp = array[i];
array[i] = array[right];
array[right] = temp;
i = right;
}
left = 2 * i + 1;
right = 2 * i + 2;
}/*End of while*/
if (left == n - 1 && array[i]) {
temp = array[i];
array[i] = array[left];
array[left] = temp;
}
}
Output:.
a) Complete both insert and delete methods. If it works correctly 10.pdf
a) Complete both insert and delete methods. If it works correctly 100 points each.
b) Do not call the search method in delete method.
c) Make sure to write your name in main method.
d) DO NOT MODIFY PROCESS METHOD.
3- Create a proper input file with respect to process method to test all cases in both
insert and delete methods.
4- Compile and execute your program at the command prompt. Check your output, if it
is correct
//xxxxxH5.java is linked list implementation of lstinterface.
//Eliminate count form both insert and delete methods
// If works correctly, 200 points
import java.util.*;
import java.io.*;
//lstinterface is an interface
public class xxxxxH5 implements lstinterface {
// xxxxxH5 class variables
//head is a pointer to beginning of linked list
private node head = null;
//use prt for System.out to save typing
PrintStream prt = System.out;
// class node
private class node{
// class node variables
int data;
node rlink; //right link
// class node constructor
node(int x){
data = x;
rlink = null;
} // end class node constructor
}// class node
// insert x at position p,
// if successful return 1 otherwise return 0
public int insert(int x, int p){
prt.printf("\n\t\tInsert %5d at position %2d:", x, p);
// complete this method
return 0; // successful insertion
} // end insert
// delete x at position p
// if successful return 1 otherwise return 0
public int delete(int p){
prt.printf("\n\t\tDelete element at position %2d:", p);
// complete this method
return 0; // successful insertion
} // end delete
// sequential serach for x in the list
// if successful return its position otherwise return 0;
public int search(int x){
// Local variables
node cur;
prt.printf("\n\t\tSearch for %5d:", x);
// Complete the rest of the method
int pos = 0;
for (cur = head; cur != null ; cur = cur.rlink){
pos ++;
if (cur.data == x){
prt.printf(" Found at position %2d, ", pos);
return pos;
} // end if
} // end for
prt.printf(" Not Found.");
return 0; // x is not found.
} // end search
// print list elements formatted
public void printlst() {
// Local variables
node cur;
prt.printf("\n\tList contents: ");
for (cur = head; cur != null ; cur = cur.rlink)
prt.printf("%5d,", cur.data);
// end for
} // end printlst
// insert delete and search in the list
private void process(String fn){
// Local variables
int j, ins, del, srch, k, p;
int x;
prt.printf("\tLinked List implementation of int list without"+
//\n\tusing count, gets input file name from method argument, then
reads:"+
"\n\tinteger No. of elements to insert(ins)"+
" followed by elements to insert and their positions,"+
"\n\tinteger No. of elements to search(srch) followed by element to
search"+
"\n\tinteger No. of elements to delete(del) followed by position of"+
"elements to delete" +
"\n\t\tTo compile: javac xxxxxH5.java" +
"\n\t\tTo execute: java xxxxxH5 inputfilename");
try{
// open input file
Scanner inf = new Scanner(new File(fn));
//read no. of elements to insert
ins = inf.nextInt();
prt.printf("\n\tInsert %.
BINARY SEARCH TREE OPERATIONS #includestdio.h#includestdlib.pdf
BINARY SEARCH TREE OPERATIONS */
#include
#include
#define NULL 0
struct tree
{
int info;
struct tree *left;
struct tree *right;
};
struct tree *p;
struct tree *search(struct tree *,int);
struct tree *insert(struct tree *i,int);
void inorder(struct tree *);
void delete(struct tree *);
main()
{
int l,ch,key,x,y;
struct tree *z;
clrscr();
do
{
printf(\"\ BINARY SEARCH TREE OPERATIONS\ \");
printf(\"\ 1.INSERTING ELEMENT\");
printf(\"\ 2.DELETING ELEMENT\");
printf(\"\ 3.SEARCH FOR ELEMENT\");
printf(\"\ 4.DISPLAY\");
printf(\"\ 5.EXIT\");
printf(\"\ \ ENTER CHOICE:\");
scanf(\"%d\",&ch);
switch(ch)
{
case 1:printf(\"\ enter element to insertat end 0\ \");
scanf(\"%d\",&x);
while(x!=0)
{
p=insert(p,x);
printf(\"\ enter element to insert:\");
scanf(\"%d\",&x);
}
break;
case 2:printf(\"\ enter node to delete\");
scanf(\"%d\",&key);
z=search(p,key);
if(z!=NULL)
{
delete(z);
printf(\"\ given key node is deleted \");
}
else
printf(\"\ the given key node not found\");
break;
case 3:printf(\"\ enter node to be search\");
scanf(\"%d\",&y);
p=search(p,y);
if(p!=NULL)
printf(\"\ the key element is :%d\",p->info);
else
printf(\"\ the key is not found\");
break;
case 4:printf(\"\ tree elements in inorder\ :\");
inorder(p);
case 5:exit(0);
break;
}
}
while(ch>=1||ch<=5);
}
struct tree *insert(struct tree *s,int x)
{
struct tree *trail,*p,*q;
q=(struct tree *)malloc(sizeof(struct tree));
q->info=x;
q->left=NULL;
q->right=NULL;
p=s;
trail=NULL;
while(p!=NULL)
{
trail=p;
if(p->info>x)
p=p->left;
else
p=p->right;
}
if(trail==NULL)
{
s=q;
return(s);
}
if(trail->info>x)
trail->left=q;
else
trail->right=q;
return(s);
}
struct tree *search(struct tree *p,int x)
{
while(p!=NULL && p->info!=x)
{
if(p->info>x)
p=p->left;
else
p=p->right;
}
return(p);
}
void delete(struct tree *p)
{
struct tree *temp;
if(p==NULL)
printf(\"\ trying to delete a noexistent node \ \");
else if(p->left==NULL)
{
temp=p;
p=p->right;
free(temp);
}
else if(p->right==NULL)
{
temp=p;
p=p->left;
free(temp);
}
else if((p->left!=NULL) && (p->right!=NULL))
{
temp=p->right;
while(temp->left!=NULL)
temp=temp->left;
temp->left=p->left;
temp=p;
p=p->right;
free(temp);
}
}
void inorder(struct tree *p)
{
if(p!=NULL)
{
inorder(p->left);
printf(\"%d\\t\",p->info);
inorder(p->right);
}
}
Solution
BINARY SEARCH TREE OPERATIONS */
#include
#include
#define NULL 0
struct tree
{
int info;
struct tree *left;
struct tree *right;
};
struct tree *p;
struct tree *search(struct tree *,int);
struct tree *insert(struct tree *i,int);
void inorder(struct tree *);
void delete(struct tree *);
main()
{
int l,ch,key,x,y;
struct tree *z;
clrscr();
do
{
printf(\"\ BINARY SEARCH TREE OPERATIONS\ \");
printf(\"\ 1.INSERTING ELEMENT\");
printf(\"\ 2.DELETING ELEMENT\");
printf(\"\ 3.SEARCH FOR ELEMENT\");
printf(\"\ 4.DISPLAY\");
printf(\"\ 5.EXIT\");
printf(\"\ \ ENTER CHOICE:\");
scanf(\"%d\",&ch);
switch(ch)
{
case 1:printf(\"\ enter element to insertat end 0\ \");
scanf(\"%d\",&x);
while(x!=0)
{
p=insert(p,x);
printf(\"\ enter e.
Program of sorting using shell sort #include stdio.h #de.pdf
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\"Increment=%d \ \",incr);
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
incr=incr-2; /*Decrease the increment*/
}/*End of while*/
printf(\"Sorted list is :\ \");
for (i = 0; i < n; i++)
printf(\"%d \", arr[i]);
printf(\"\ \");
}/*End of main()*/
redix sort
/*Program of sorting using radix sort*/
# include
# include
struct node
{
int info ;
struct node *link;
}*start=NULL;
main()
{
struct node *tmp,*q;
int i,n,item;
printf(\"Enter the number of elements in the list : \");
scanf(\"%d\", &n);
for(i=0;iinfo=item;
tmp->link=NULL;
if(start==NULL) /* Inserting first element */
start=tmp;
else
{
q=start;
while(q->link!=NULL)
q=q->link;
q->link=tmp;
}
}/*End of for*/
printf(\"Unsorted list is :\ \");
display();
radix_sort();
printf(\"Sorted list is :\ \");
display ();
}/*End of main()*/
display()
{
struct node *p=start;
while( p !=NULL)
{
printf(\"%d \", p->info);
p= p->link;
}
printf(\"\ \");
}/*End of display()*/
radix_sort()
{
int i,k,dig,maxdig,mindig,least_sig,most_sig;
struct node *p, *rear[10], *front[10];
least_sig=1;
most_sig=large_dig(start);
for(k = least_sig; k <= most_sig ; k++)
{
printf(\"PASS %d : Examining %dth digit from right \",k,k);
for(i = 0 ; i <= 9 ; i++)
{
rear[i] = NULL;
front[i] = NULL ;
}
maxdig=0;
mindig=9;
p = start ;
while( p != NULL)
{
/*Find kth digit in the number*/
dig = digit(p->info, k);
if(dig>maxdig)
maxdig=dig;
if(diglink = p ;
rear[dig] = p ;
p=p->link;/*Go to next number in the list*/
}/*End while */
/* maxdig and mindig are the maximum amd minimum
digits of the kth digits of all the numbers*/
printf(\"mindig=%d maxdig=%d\ \",mindig,maxdig);
/*Join all the queues to form the new linked list*/
start=front[mindig];
for(i=mindig;ilink=front[i+1];
else
rear[i+1]=rear[i];
}
rear[maxdig]->link=NULL;
printf(\"New list : \");
display();
}/* End for */
}/*End of radix_sort*/
/* This function finds number of digits in the largest element of the list */
int large_dig()
{
struct node *p=start ;
int large = 0,ndig = 0 ;
while(p != NULL)
{
if(p ->info > large)
large = p->info;
p = p->link ;
}
printf(\"Largest Element is %d , \",large);
while(large != 0)
{
ndig++;
large = large/10 ;
}
printf(\"Number of digits in it are %d\ \",ndig);
return(ndig);
} /*End of large_dig()*/
/*This function returns kth digit of a number*/
int digit(int number, int k)
{
int digit, i ;
for(i = 1 ; i <=k ; i++)
{
digit = number % 10 ;
number = number /10 ;
}
return(digit);
}/*End of digit()*/
Solution
/* Program of sorting using shell sort */
#include
#define MAX 20
main()
{
int arr[MAX], i,j,k,n,incr;
printf(\"Enter the number of elements : \");
scanf(\"%d\",&n);
for(i=0;i=1)
{
for(j=incr;j= 0 && k < arr[i]; i = i-incr)
arr[i+incr]=arr[i];
arr[i+incr]=k;
}
printf(\.
Data Structure using C
This document contains some programs of C using Data structures, like Stack, LinkedList, queue, Fibonacci series, addition and multiplication of two matrices,etc.
My C proggram is having trouble in the switch in main. Also the a co.pdf
My C proggram is having trouble in the switch in main. Also the a couple of funtion like delete
and display tree are failing
#include
#include
#include
#include
#include
#include
struct tree_node {
int data;
struct tree_node *left, *right;
};
typedef struct nodeT {
int key;
struct nodeT *left, *right;
int data;
} nodeT, *treeT;
//struct nodeT *t1, *t2;
void DisplayTree(nodeT*t)
{
if (t != NULL) {
DisplayTree(t->left);
printf(\"%c \", t->key);
DisplayTree(t->right);
}
}
nodeT *FindNode(nodeT *t, int key){
if(t==NULL) return 0;
if (key == t->key)
return t;
if(key < t->key){
return FindNode(t->left, key);;
} else {
return FindNode(t->right, key);;
}
}
int max(nodeT *p){
nodeT *t, *tmp;
while(tmp->right != NULL)
tmp=tmp->right;
return tmp->data;
}
int min(nodeT *t){
nodeT *tmp;
while(tmp->left != NULL)
tmp = tmp->left;
return tmp->data;
}
void ModifyPreOrderWalk(nodeT *t, int h)
{
int i;
if (t == NULL) return;
for(i=0; i<-1;i++){
printf(\"| \");
}
if(h>1)printf(\"+---\");
printf(\"%d\ “\", t->key);
ModifyPreOrderWalk(t->left, h+1);
ModifyPreOrderWalk(t->right, h+1);
}
void PrintTree(nodeT *t)
{
ModifyPreOrderWalk(t, 1);
}
void delete(nodeT **p){
nodeT *target, *imd_r, *plmd_r;
target=*p;
if (target->left==NULL && target->right==NULL) {
*p=NULL;
} else if (target->left == NULL) {
*p=target->right;
} else
if (target->right == NULL) {
*p=target->left;
} else {
plmd_r = target;
imd_r = target->right;
while( imd_r->left != NULL){
plmd_r = imd_r;
imd_r = imd_r->left;
}
if(plmd_r == target)
plmd_r->right = imd_r->right;
else
plmd_r->left = imd_r->right;
imd_r->left = target->left;
imd_r->right = target->right;
*p = imd_r;
}
free(target);
}
void listInorder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
printf(\"%d \", t->key);
DisplayTree(t->right);
}
}
void listPreorder(nodeT *t) {
if (t != NULL) {
printf(\"%d \", t->key);
DisplayTree(t->left);
DisplayTree(t->right);
}
}
void listPostOrder(nodeT *t){
if (t != NULL) {
DisplayTree(t->left);
DisplayTree(t->right);
printf(\"%d\", t->key);
}
}
void InsertNode(nodeT **tptr, int key){
nodeT *t;
nodeT *tmp;
t=*tptr;
if (t == NULL) {
tmp= malloc(sizeof(nodeT));
tmp->key = key;
tmp->left=tmp->right=NULL;
*tptr=tmp;
return;
}
if (key < t->key) {
InsertNode
(&t->left, key);
} else {
InsertNode(&t->right, key);
}
}
int height(nodeT *t){
int lDepth = height(t->left);
int rDepth = height(t->right);
if(lDepth > rDepth)
return(lDepth+1);
else
return(rDepth+1);
}
int sum(nodeT *p){
if (p == NULL)
return 0;
else
return (p->data + sum(p->left) +sum(p->right) );
}
int count(nodeT *t){
if(t == NULL)
return 0;
int i = 1+count(t->left)+count(t->right);
return i;
}
/*nodeT *New(){
nodeT *tmp;
tmp = (nodeT *)malloc(sizeof(nodeT));
if (tmp==NULL) return NULL;
return tmp;
}*/
int main(){
nodeT *node;
nodeT *t=NULL, *findA;
int f;
int d, a, b, c, j, k, e, s, y;
while (1) {
printf(\"\ I - insert\ F - find node\ D - delete\ L - list in order\ R - list preorder\ P - list
PorstOrder\ M - max\ N - min\ H - height\ C - count\ S - sum.
#include iostream #include deque #include stdio.h scan.pdf
#include
#include
#include //scanf,printf
#include
#include //istringstream
using namespace std;
//maps
#include \"MapGeneric.h\"
#include \"MapTriple.h\"
#include \"MapAbsoluteValue.h\"
#include \"MapSquare.h\"
//filters
#include \"FilterGeneric.h\"
#include \"filterEven.h\"
#include \"filterPositive.h\"
#include \"filterPositiveTwoDigit.h\"
//reduce
#include \"ReduceGeneric.h\"
#include \"ReduceMinimum.h\"
#include \"ReduceGCD.h\"
int main()
{
deque L,L1,L2;
int i;
string line;
getline(cin,line);
if(line.size() == 0)
return 0;
line.erase(remove(line.begin(), line.end(), \',\'), line.end());
istringstream s(line);
while(s>>i)
{
L.push_back(i);
}
if(L.size() != 20)
return 0;
//converting L -> 3*|L| aka L1
MapTriple triple; MapAbsoluteValue abs;
L1 = triple.map(abs.map(L));
//converting L1 -> positive two digit integers -> even intergers aka L2
filterEven even; filterPositiveTwoDigit p2;
L2 = p2.filter(L1);
if(L2.size() <2)
return 0;
L2 = even.filter(L2);
if(L2.size() <2)
return 0;
//finding the minimum and GCD from L2 respectively
ReduceMinimum min; ReduceGCD gcd;
printf(\"%d %d\ \",min.reduce(L2),gcd.reduce(L2));
}
FilterGeneric.cpp
#include
#include
#include \"FilterGeneric.h\"
using namespace std;
deque FilterGeneric::filter(deque deq)
{
//calling recursive function with deq and its length
select(deq,deq.size()-1);
return filterDeq;
}
//int i is the index that will start from the length of deque till 0
void FilterGeneric::select(deque deq, int i)
{
if(i == 0)
{
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return;
}
bool val = f(deq[i]);
if(val)
filterDeq.push_front(deq[i]);
return select(deq,i-1);
}
FilterGeneric.h
#ifndef FILTERGENERIC_H
#define FILTERGENERIC_H
#include
class FilterGeneric
{
public:
std::deque filter(std::deque);
void select(std::deque,int);
protected:
//this deque will store the integers after filtration.
std::deque filterDeq;
private:
virtual bool f(int)=0;
};
#endif
MapAbsoluteValue.cpp
#include
#include
#include
#include \"MapGeneric.h\" //base
#include \"MapAbsoluteValue.h\"
using namespace std;
int MapAbsoluteValue::f(int num)
{
return abs(num);
}
MapAbsoluteValue.h
#ifndef MAPABSOLUTEVALUE_H
#define MAPABSOLUTEVALUE_H
#include
//this class will map a given input N, using the function map(input), to |N|
class MapAbsoluteValue: public MapGeneric
{
public:
//overwriting pure virtual function making N -> |N|
int f(int);
private:
};
#endif
MapGeneric.cpp
#include
#include
#include \"MapGeneric.h\"
using namespace std;
deque MapGeneric::map(deque deq)
{
//calling recursive function with deq and its length
calculate(deq,deq.size()-1);
return mapDeq;
}
//int i is the index that will start from the length of deque till 0
void MapGeneric::calculate(deque deq, int i)
{
if(i == 0)
{
int val = f(deq[i]);
mapDeq.push_front(val);
return;
}
int val = f(deq[i]);
mapDeq.push_front(val);
return calculate(deq,i-1);
}
MapGeneric.h
#ifndef MAPGENERIC_H
#define MAPGENERIC_H
#include
class MapGeneric
{
public:
std::deque ma.
IN C LANGUAGE- I've been trying to finish this program for the last fe.docx
IN C LANGUAGE: I've been trying to finish this program for the last few days, and I'm stuck on one part of it, and I can't figure out what to do.
The goal is to read in a file (separated by commas), put the information into a linked list, and then sort that linked list by either first or last name using function pointers.
Everything I have so far works, aside from the fact that my output is never sorted. I'm doing something wrong with the listSort function, and I don't know what it is.
Can anyone please just take a look at what I have and offer a suggestion or solution? It might not even be an issue with the function itself, but something somewhere else.
I would greatly appreciate it.
Please do not make any changes to the variables in the structs. Month has to be a string, it cannot be an integer.
driver.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
int main(int argc, char* argv[]) {
//creating input and output files
FILE* in = NULL;
FILE* out = NULL;
node_t* head = NULL;
node_t* temp = NULL;
//checking for 4 command line arguments
assert(argc == 4);
//opening input file
in = fopen(argv[1], "r");
assert(in != NULL);
//opening output file
out = fopen(argv[2], "w");
assert(out != NULL);
int sortBy = atoi(argv[4]);
temp = createList(in, &head);
if(sortBy == 1) {
sortList(&head, compare_by_firstname);
}
else if(sortBy == 2) {
sortList(&head, compare_by_lastname);
}
printList(out, temp);
deleteList(&head);
//closing files
fclose(in);
fclose(out);
return 0;
}
functions.c:
#include "functions.h"
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
#include <stdbool.h>
bool isLeapYear(int year) {
if(year % 4 != 0) {
return false;
}
else if(year % 100 != 0) {
return true;
}
else if(year % 400 != 0) {
return false;
}
else {
return true;
}
}
void printBorder(FILE* out)
{
int i = 0;
for(i = 0; i < 80; i++)
fprintf(out,"*");
fprintf(out, "\n");
}
bool checkInvalidDate(node_t *node) {
int year = (int)node->birthday.year;
if(node->birthday.day < 1 || node->birthday.year < 1) {
return true;
}
if(strcmp(node->birthday.month, "February") == 0) {
if(isLeapYear(year)) {
if(node->birthday.day > 29) {
return true;
}
}
else if(node->birthday.day > 28) {
return true;
}
}
else if((strcmp(node->birthday.month, "April") == 0) || (strcmp(node->birthday.month, "June") == 0) ||
(strcmp(node->birthday.month, "September") == 0) || (strcmp(node->birthday.month, "November") == 0)) {
if(node->birthday.day > 30) {
return true;
}
}
else {
if(node->birthday.day > 31) {
return true;
}
}
return false;
}
void add(node_t **node, node_t **head) {
node_t* current = *head;
while(current != NULL) {
//comparing first and last names to check for duplicates; if duplicate the info isn't added to the list
if((strcmp(current->firstName, (*node)->firstName) == 0) && (strcmp(current->lastName, (*node)->lastName) == 0)) {
return;
}
current = current->next;
}
//adding information as a node to the list
(*node.
888678123317368915236 #include stdio.h #include stdlib.h.pdf
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}
Solution
8886781233
17368915236
#include
#include
int main() {
int i, n, *data , p;
/* get the number of inputs from the user */
printf(\"Enter the number of inputs:\");
scanf(\"%d\", &n);
/* dynamically allocate memory to store i/p values */
data = (int *) malloc(sizeof(int) * n);
/* get the input data from the user */
for (i = 0; i < n; i++) {
printf(\"data[%d]: \", i);
scanf(\"%d\", &data[i]);
}
/* sorts the given numbers */
bubbleSort(data, n ,p );
}
void bubbleSort(int *data, int n , int p) {
int i, temp, k = 1, l,j;
if (n > 0) {
for (i = n; i > 0; i--) {
if (data[i] > data[i - 1])
{
temp = data[i];
data[i] = data[i - 1];
data[i - 1] = temp;
}
if( p = = 0)
{
p++;
l = n;
for ( i = 0 ; i < n ; i++)
{
prinft ( %d, data[i])
}
bubbleSort(data, i - 2 , p );
}
else
{
for ( j = n - 1 ; j < l ; j++)
{
if (data[j] > data[j - 1])
{
temp = data[j];
data[j] = data[j - 1];
data[j - 1] = temp;
}
}
for ( j = 0 ; j < l ; j++)
{
prinft ( %d, data[j])
}
bubbleSort(data, i - 2,p);
}
}
}
return;
}.
Hi,Please fidn the Answer.Sorting,h is Header .pdf
Hi,
Please fidn the Answer.
//////////////////Sorting,h is Header file which contians all Function decalartion
#include
#include // std::cout
#include // std::shuffle
#include // std::array
#include // std::default_random_engine
#include
using namespace std;
class Sorting
{
private:
int num;
public:
Sorting::Sorting() {
};
///Buble Sort Input array and n Number of Elemnets
void BubbleSorting(int array[], int n);
void SelectionSorting(int array[], int n);
void ShuffleArray(int array[], int n);
void DisplayArray(int array[], int n);
void displayMenu();
int choice;
};
//////////////////////////Sorting.cpp is class fuction Which Implemnets the Function
#include \"Sorting.h\"
#include
using namespace std;
///Sorting the Array Using Bubble Sort...
void Sorting::BubbleSorting(int Array[],int n)
{
int temp = 0;
cout << \"\ ------------ BUBBLE SORT ------------ \ \ \";
for (int i = 1; iArray[j + 1])
{
temp = Array[j];
Array[j] = Array[j + 1];
Array[j + 1] = temp;
}
}
cout << \"\ ------------ RESULTS BUBBLE SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
void Sorting::SelectionSorting(int Array[], int n)
{
int i, n, p, k, min, loc, temp;
for (p = 1; p <= n - 1; p++) // Loop for Pass
{
min = Array[p]; // Element Selection
loc = p;
for (k = p + 1; k <= n; k++) // Finding Min Value
{
if (min > Array[k])
{
min = Array[k];
loc = k;
}
}
temp = Array[p]; // Swap Selected Element and Min Value
Array[p] = Array[loc];
Array[loc] = temp;
}
cout << \"\ ------------ RESULTS SELECTION SORT ------------ \ \ \";
DisplayArray(Array, n);
}
//////SelectionSorting input Array and n is number of Elemnets//////////////////////
///exchange each element with a randomly chosen element. It\'s possible that an element will be
exchanged with itself, but there is no problem with that
void Sorting::ShuffleArray(int Array[], int n)
{
cout << \"\ ------------ ShuffleArray SORT ------------ \ \ \";
for (int i = 0; i
#include \"Sorting.h\"
using namespace std;
int main()
{
bool SelectionOn = true;
Sorting Sorting;
int choice;
int n = 10;
int *Arry;
while (SelectionOn != false) {
cout << \"*******************************\ \";
cout << \" 1 - Enter User Input Array of Elements.\ \";
cout << \" 2 - Display User Input.\ \";
cout << \" 3 - Shuffle User Input the Elements i.\ \";
cout << \" 4 - Bubble Sorting.\ \";
cout << \" 5 - Selection Sorting.\ \";
cout << \" 6 - Exit.\ \";
cout << \" 7 - Display Menu Again. \ \";
cout << \" Enter your choice and press return: \";
cin >> choice;
switch (choice)
{
case 1:
cout << \"Enter User Input Array of Elements\ \";
cout << \"Enter No. of Elements=\";
cin >> n;
Arry = new int[n];
cout << \"\ Enter Elements=\ \";
for (int i = 1; i <= n; i++)
{
cin >> Arry[i];
}
break;
case 2:
cout << \"DISPLAY OF ARRAY\ \";
if (Arry != NULL)
{
Sorting.DisplayArray(Arry, n);
}
else
{
cout << \"ENTER ELEMENTS TO ARRAY BY SELCTING CHOICE 1\ \";
}
break;
case 3:
cout << \"SHUFFLE OF A.
Similar to selection_sort..c#include stdio.hheader file input output func.pdf (20)
Can you give an example of a binary heap programCan you give an .pdf
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a) Complete both insert and delete methods. If it works correctly 10.pdf
a) Complete both insert and delete methods. If it works correctly 10.pdf
BINARY SEARCH TREE OPERATIONS #includestdio.h#includestdlib.pdf
BINARY SEARCH TREE OPERATIONS #includestdio.h#includestdlib.pdf
Program of sorting using shell sort #include stdio.h #de.pdf
Program of sorting using shell sort #include stdio.h #de.pdf
My C proggram is having trouble in the switch in main. Also the a co.pdf
My C proggram is having trouble in the switch in main. Also the a co.pdf
#include iostream #include deque #include stdio.h scan.pdf
#include iostream #include deque #include stdio.h scan.pdf
IN C LANGUAGE- I've been trying to finish this program for the last fe.docx
IN C LANGUAGE- I've been trying to finish this program for the last fe.docx
888678123317368915236 #include stdio.h #include stdlib.h.pdf
888678123317368915236 #include stdio.h #include stdlib.h.pdf
Hi,Please fidn the Answer.Sorting,h is Header .pdf
Hi,Please fidn the Answer.Sorting,h is Header .pdf
More from anton291
There are many ways to do the Lewis symbol for yo.pdf
There are many ways to do the Lewis symbol for you to draw a Lewis dot diagram.
As long as you make sure to put a dot in for each valence electron it should be just fine. Some
teachers like to make things more complicated by drawing Lewis dots in regards to electron pairs
or bonding electrons. I\'ve seen some even go so far as to draw in charges on each of the dots of
a Lewis structure. Either way, these are unnecessary and, in my opinion, a waste of time. Just
put dots around the symbol and you should be fine. This also happens to be the best way to draw
it when you need to put together larger, more complicated Lewis structures and figure out a
\"formal charge\" (later on.) You did it exactly as I would do it and I\'m taking upper-division
chemistry. Just put one dot on each side, if there\'s more than four, put double dots on each side
until either all the sides are filled or all valence electrons are noted. You can also put a small line
(connects the two dots with a line) to designate an electron pair, either way is valid, just pick one
you think looks nicer.
Solution
There are many ways to do the Lewis symbol for you to draw a Lewis dot diagram.
As long as you make sure to put a dot in for each valence electron it should be just fine. Some
teachers like to make things more complicated by drawing Lewis dots in regards to electron pairs
or bonding electrons. I\'ve seen some even go so far as to draw in charges on each of the dots of
a Lewis structure. Either way, these are unnecessary and, in my opinion, a waste of time. Just
put dots around the symbol and you should be fine. This also happens to be the best way to draw
it when you need to put together larger, more complicated Lewis structures and figure out a
\"formal charge\" (later on.) You did it exactly as I would do it and I\'m taking upper-division
chemistry. Just put one dot on each side, if there\'s more than four, put double dots on each side
until either all the sides are filled or all valence electrons are noted. You can also put a small line
(connects the two dots with a line) to designate an electron pair, either way is valid, just pick one
you think looks nicer..
the compound is heptane note the molecular ion.pdf
the compound is: heptane note: the molecular ion peak is at m/z = 100 --->
indicates the molecular weight of the compound = 100 only heptane has a molecular weight of
100
Solution
the compound is: heptane note: the molecular ion peak is at m/z = 100 --->
indicates the molecular weight of the compound = 100 only heptane has a molecular weight of
100.
Succinic acid is oxidized C goes from +34 to +1 .pdf
Succinic acid is oxidized C goes from +3/4 to +1 state b)FAD is reduced it goes 2
more -ve state
Solution
Succinic acid is oxidized C goes from +3/4 to +1 state b)FAD is reduced it goes 2
more -ve state.
pH = pK +log ([salt][acid]) = pH = -log(1.8 x .pdf
pH = pK +log ([salt]/[acid]) => pH = -log(1.8 x 10-04) + log(0.050 /0.150) => pH
= 7.52
Solution
pH = pK +log ([salt]/[acid]) => pH = -log(1.8 x 10-04) + log(0.050 /0.150) => pH
= 7.52.
Step1 25 ml of1M HCl w ill neutralise 25 ml 1 M N.pdf
Step1 25 ml of1M HCl w ill neutralise 25 ml 1 M NaOH Step2 Moles of H+=0 ;
Moles of OH-= 50x10^-3 x1=.05 Step3 Moles of H2O= (100/18)+ .025 =5.581
Solution
Step1 25 ml of1M HCl w ill neutralise 25 ml 1 M NaOH Step2 Moles of H+=0 ;
Moles of OH-= 50x10^-3 x1=.05 Step3 Moles of H2O= (100/18)+ .025 =5.581.
Moseley used X-ray diffraction to show that each .pdf
Moseley used X-ray diffraction to show that each element has an atomic number.
With this tool he was able to predict elements for the six remaining gaps in the periodic table and
to resolve identities of rare earths. neohumanism.org/h/he/henry_moseley.html
Solution
Moseley used X-ray diffraction to show that each element has an atomic number.
With this tool he was able to predict elements for the six remaining gaps in the periodic table and
to resolve identities of rare earths. neohumanism.org/h/he/henry_moseley.html.
Welding is a permanent joining of two similar metals by melting the .pdf
Welding is a permanent joining of two similar metals by melting the point where they join and
(often) adding additional \"filler\" material to facilitate. If done properly, the join becomes
homogeneous and stronger than the surrounding metal. The only way of removing a weld is
destructively.
Rivet jointing of material is a semi-permanent jointing method. two components are placed
alongside - one normally with a flange - clamped in place and then a hole drilled directly through
both metals. A rivet is then fitted and hammered in place - normally forming a dome, but other
rivet forms are available and can be used....even \"hidden rivets\" where the rivet is forged
(repeatedly hit by a hammer) into a countersunk hole and then machined afterwards to create an
invisible join.
There are advantages and disadvantages to both. Welding can affect the cellular structure of the
metal creating internal stresses and weak points, adds material and can only be undertaken in
similar metals.
Riveting on the other hand can be undertaken in dissimilar metals, and can be separated
afterwards (with some effort), however the joint is not homogenous and significant attention has
to be paid to the stresses within the rivets in all likely situations. Poorly chosen rivet material
(and plate material) combinations can result in unintended galvanic corrosion and early failure,
thermal expansion of the plate material overstressing the rivets and breaking down the join (this
is where in high temperatures the plates expand, causing a tensile loading to present within the
rivet and a compressive loading on the plate around the rivet head that may exceed the Young\'s
Modulus -of either the rivet or, compressively, the plate itself - causing a loosening of the joint
and premature failure)
Solution
Welding is a permanent joining of two similar metals by melting the point where they join and
(often) adding additional \"filler\" material to facilitate. If done properly, the join becomes
homogeneous and stronger than the surrounding metal. The only way of removing a weld is
destructively.
Rivet jointing of material is a semi-permanent jointing method. two components are placed
alongside - one normally with a flange - clamped in place and then a hole drilled directly through
both metals. A rivet is then fitted and hammered in place - normally forming a dome, but other
rivet forms are available and can be used....even \"hidden rivets\" where the rivet is forged
(repeatedly hit by a hammer) into a countersunk hole and then machined afterwards to create an
invisible join.
There are advantages and disadvantages to both. Welding can affect the cellular structure of the
metal creating internal stresses and weak points, adds material and can only be undertaken in
similar metals.
Riveting on the other hand can be undertaken in dissimilar metals, and can be separated
afterwards (with some effort), however the joint is not homogenous and significant attention has
to be paid to the stresses w.
Use of oxygen for ATP synthesis is called as Aerobic respiration. Th.pdf
Use of oxygen for ATP synthesis is called as Aerobic respiration. The bacteria which uses
molecules other than bacteria for the production of ATP are known as Anaerobic bacteria.
1. Facultative anaerobes and obligate anaerobes-use the Iron reduction pathway where, Fe III is
the electron acceptor, eg-Desulfuromonadales(Geobacter,Geothermobacter) and Shewanella
species.
2. Facultative anaerobes and obligate anaerobes-use manganese pathway-Mn IV is the electron
acceptor-eg- Desulfuromonadales and Shewanella species.
3. Facultative anaerobes and obligate anaerobes- Cobalt reduction pathway- Co III is the electron
acceptor,eg-Geobacter sulfurreducens
4. -Facultative anaerobes and obligate anaerobe-Uranium reduction U VI is the electron
acceptor- eg. Geobacter metallireducens
5. Nitrate reduction (denitrification)- Facultative anaerobes -NO3- is the electron acceptor- eg
Ecoli
6.Fumarate respiration- Facultative anaerobes-Fumarate is the electron acceptor- eg Ecoli
7. Sulfate respiration- Obligate anaerobes-Sulfate SO42- is the electron acceptor- Eg- Many
Gammaproteobacteria, Desulfovibrionales
8.Methanogenesis (carbonate reduction)-Methanogens-CO2 is the electron acceptor and CH4 is
the product-Eg Methanosarcina barkeri
9. Sulfur respiration- Facultative anaerobes and obligate anaerobes-S0 is the elctron acceptor-Eg-
Desulfuromonadales
10.Acetogenesis(carbonate reduction)- obligate anaerobes-CO2 is the electron acceptor-Eg
Acetobacterium woodii
11.Dehalirespiration-Facultative anaerobes and obligate anaerobes-Halogenated organic
compound is the electron acceptor- Eg-Dehalococcoides and Dehalobacter species.
Solution
Use of oxygen for ATP synthesis is called as Aerobic respiration. The bacteria which uses
molecules other than bacteria for the production of ATP are known as Anaerobic bacteria.
1. Facultative anaerobes and obligate anaerobes-use the Iron reduction pathway where, Fe III is
the electron acceptor, eg-Desulfuromonadales(Geobacter,Geothermobacter) and Shewanella
species.
2. Facultative anaerobes and obligate anaerobes-use manganese pathway-Mn IV is the electron
acceptor-eg- Desulfuromonadales and Shewanella species.
3. Facultative anaerobes and obligate anaerobes- Cobalt reduction pathway- Co III is the electron
acceptor,eg-Geobacter sulfurreducens
4. -Facultative anaerobes and obligate anaerobe-Uranium reduction U VI is the electron
acceptor- eg. Geobacter metallireducens
5. Nitrate reduction (denitrification)- Facultative anaerobes -NO3- is the electron acceptor- eg
Ecoli
6.Fumarate respiration- Facultative anaerobes-Fumarate is the electron acceptor- eg Ecoli
7. Sulfate respiration- Obligate anaerobes-Sulfate SO42- is the electron acceptor- Eg- Many
Gammaproteobacteria, Desulfovibrionales
8.Methanogenesis (carbonate reduction)-Methanogens-CO2 is the electron acceptor and CH4 is
the product-Eg Methanosarcina barkeri
9. Sulfur respiration- Facultative anaerobes and obligate anaerobes-S0 is the elctron acceptor-Eg-
Desulfuromonadales
10.A.
the solute potential of the plant cell is Greater than the pure wate.pdf
the solute potential of the plant cell is Greater than the pure water. there for greater water
potential is in the solution.
if we palce the plant cell in Pure water the plant cell have more solute concentration than pure
water. water generally moves towards higher solute concetration (or) water moves from higer
water potential to lower potential. then the plant cell become Turgid.
Solution
the solute potential of the plant cell is Greater than the pure water. there for greater water
potential is in the solution.
if we palce the plant cell in Pure water the plant cell have more solute concentration than pure
water. water generally moves towards higher solute concetration (or) water moves from higer
water potential to lower potential. then the plant cell become Turgid..
The Human Genome Project (HGP) determines the sequence of nucleotide.pdf
The Human Genome Project (HGP) determines the sequence of nucleotide base pairs of human
DNA, It is used to identifies and mapping all of the genes of the human genome.HGP was
proposed by Robert Sinshelmer in May1884. This project formally launched in 1990, and it was
completed in 2003. The goal of HGP is to identify the DNA sequence of entire human genome. 3
billion DNA base pairs are found in DNA that make up the human genome. Because of HGP, the
name and bases of nitrogenous bases are known.
US government through the National Institutes of Health (NIH) provide the fund for the project.
In 1990 DOE and NIH, shows interest to developed a memorandum of understanding and give
support and coordinate plans and set the clock to initiate the Project in 1990. David Galas was
Director of the renamed “Office of Biological and Environmental Research” in the U.S.
Department of Energy’s Office of Science and James Watson headed the NIH Genome Program.
In 1993, U.S. National Institutes of Health (NIH) and National Center for Human Genome
Research ( now known as National Human Genome Research Institute) worked on the project.
The United States, the United Kingdom, France, Germany, Japan and China are participated to
complete this project.
The genomic sequencing was conducted at numerous universities and research center given
below:-
The time period of this project is 13-years. It is initiated in 1990 and completed in 2003. Robert
Sinsheimer proposed this project in May 1985, President Regan\'s in 1988 submitted budget for
this project. In 1990,DOE and NIH, provided the fund for the project. In 2000 working draft of
the genome was announced and in the February 2001, papers were published . And more
sequence were identified and lateron, complete draft was published in 2003, and genome
\"finishing\" work continued for more than a decade. It was published on May 27, 2004 with
99.99% accurate sequence.
The research techniques that play major role in this project are Mapping the human genome,
sequencing the genome. Mapping the human genome includes two types of map:-A gene linkage
map and physical map.
Solution
The Human Genome Project (HGP) determines the sequence of nucleotide base pairs of human
DNA, It is used to identifies and mapping all of the genes of the human genome.HGP was
proposed by Robert Sinshelmer in May1884. This project formally launched in 1990, and it was
completed in 2003. The goal of HGP is to identify the DNA sequence of entire human genome. 3
billion DNA base pairs are found in DNA that make up the human genome. Because of HGP, the
name and bases of nitrogenous bases are known.
US government through the National Institutes of Health (NIH) provide the fund for the project.
In 1990 DOE and NIH, shows interest to developed a memorandum of understanding and give
support and coordinate plans and set the clock to initiate the Project in 1990. David Galas was
Director of the renamed “Office of Biological and Environmental Research” in the U.S.
.
solutionlist intersect(list L1, list L2){list result;positi.pdf
solution:
list intersect(list L1, list L2)
{
list result;
position L1= first(L1), L2 = first(L2), resultPos = first(result);
while( L1 != NULL && L2 != NULL )
{
if( L1->element < L2->element )
L1 = L1->next;
else if( L1->element > L2->element )
L2 = L2->next;
else
{
insert(L1->element, result, resultPos);
L1= L1->next;
L2 = L2->next;
resultPos = resultPos->next;
}
}
return result;
}
Solution
solution:
list intersect(list L1, list L2)
{
list result;
position L1= first(L1), L2 = first(L2), resultPos = first(result);
while( L1 != NULL && L2 != NULL )
{
if( L1->element < L2->element )
L1 = L1->next;
else if( L1->element > L2->element )
L2 = L2->next;
else
{
insert(L1->element, result, resultPos);
L1= L1->next;
L2 = L2->next;
resultPos = resultPos->next;
}
}
return result;
}.
Sorry for not seeing const temp, The answers are-The reaction abso.pdf
Sorry for not seeing const temp, The answers are
-The reaction absorbs energy.
-[H2O] increases.
-[CO] increases.
-[CO2] decreases.
-[H2] decreases.
Solution
Sorry for not seeing const temp, The answers are
-The reaction absorbs energy.
-[H2O] increases.
-[CO] increases.
-[CO2] decreases.
-[H2] decreases..
SO2 = 18IF4+ = 34I3- = 22The easiest way to think about this i.pdf
SO2 = 18
IF4+ = 34
I3- = 22
The easiest way to think about this is to use the periodic table. The elements all the way to the
right have 8 valence electrons, (noble gases) then each column you go to the left, has once less
valence electron
F, Cl,Br,I,Ar = 7
O,S,Se,Te,Po = 6
and so on.
Then when the molecule has a charge on it, you take that into account, if it says \"+\", subtract a
valence electron, if it says \"-\", add a valence electron. (Remember, electrons are negatively
charged, so if it has a positive charge, you are short an electron, thats why you subtract for \"+\")
So: SO2: [S=6 valence, 2x O = 6 x 2 = 12 total valence, 6+ 12 = 18 valence electrons]
IF4+ [I= 7 valence, 4x F = 7x4=28 valence, 7 + 28 = 35, but it has a + charge, so subtract one, 34
valenece electrons]
I3- [I=7 valenece, 3x7 = 21 valence, minus charge so add another = 22 valence electrons]
Solution
SO2 = 18
IF4+ = 34
I3- = 22
The easiest way to think about this is to use the periodic table. The elements all the way to the
right have 8 valence electrons, (noble gases) then each column you go to the left, has once less
valence electron
F, Cl,Br,I,Ar = 7
O,S,Se,Te,Po = 6
and so on.
Then when the molecule has a charge on it, you take that into account, if it says \"+\", subtract a
valence electron, if it says \"-\", add a valence electron. (Remember, electrons are negatively
charged, so if it has a positive charge, you are short an electron, thats why you subtract for \"+\")
So: SO2: [S=6 valence, 2x O = 6 x 2 = 12 total valence, 6+ 12 = 18 valence electrons]
IF4+ [I= 7 valence, 4x F = 7x4=28 valence, 7 + 28 = 35, but it has a + charge, so subtract one, 34
valenece electrons]
I3- [I=7 valenece, 3x7 = 21 valence, minus charge so add another = 22 valence electrons].
Q2.AnswerThe Toll -like receptor encodes a protein which plays .pdf
Q2.
Answer:
The Toll -like receptor encodes a protein which plays an important and fundamental role in the
pathogen recognition and activation of innate immunity. These TLRs are highly conserved
regions ranging from the well model organism of genetics Drosophila to humans and share
structural and functional similarities. They recognize pathogen-associated molecular patterns that
are expressed on infectious agents, and mediate the production of cytokines necessary for the
development of effective immunity. The various TLRs exhibit different patterns of expression
and these receptor has been implicated in signal transduction events induced by
lipopolysaccharide (LPS) found in most gram-negative bacteria. Mutations in this gene have
been associated with differences in LPS responsiveness.
The lysosome protects the cell by acting upon the infectious pathogen as soon as its entry in to
the cell and before reaching the cytoplasm it will be degraded. The pathogens often hijack
endocytotic pathways such as pinocytosis and gain entry into the cell. The acid phospatases
present in the lysosome hydrolyze the biomolecules of pathogens necessary for their replication
strategies. The reduced Lysosomal activity results in an increase in viral infectivity, including
HIV infection. The toxins such as AB5 toxins such as cholera hijack the endosomal pathway
while evading lysosomal degradation.
Solution
Q2.
Answer:
The Toll -like receptor encodes a protein which plays an important and fundamental role in the
pathogen recognition and activation of innate immunity. These TLRs are highly conserved
regions ranging from the well model organism of genetics Drosophila to humans and share
structural and functional similarities. They recognize pathogen-associated molecular patterns that
are expressed on infectious agents, and mediate the production of cytokines necessary for the
development of effective immunity. The various TLRs exhibit different patterns of expression
and these receptor has been implicated in signal transduction events induced by
lipopolysaccharide (LPS) found in most gram-negative bacteria. Mutations in this gene have
been associated with differences in LPS responsiveness.
The lysosome protects the cell by acting upon the infectious pathogen as soon as its entry in to
the cell and before reaching the cytoplasm it will be degraded. The pathogens often hijack
endocytotic pathways such as pinocytosis and gain entry into the cell. The acid phospatases
present in the lysosome hydrolyze the biomolecules of pathogens necessary for their replication
strategies. The reduced Lysosomal activity results in an increase in viral infectivity, including
HIV infection. The toxins such as AB5 toxins such as cholera hijack the endosomal pathway
while evading lysosomal degradation..
Nucleus is the dense central part of an atom. .pdf
Nucleus is the dense central part of an atom.
Solution
Nucleus is the dense central part of an atom..
Its demorgans theorem. Draw truth table to verify.pdf
Its demorgans theorem. Draw truth table to verify
Solution
Its demorgans theorem. Draw truth table to verify.
Ionic Bonds (between ionic molecules) - Medium strong bond - In an.pdf
Ionic Bonds (between ionic molecules) - Medium strong bond
- In an ionic solid, adjacent ions have electrostatic attractions and are arranged in a lattice
structure.
- Ionic solids have strong bonds resulting in high melting and boiling point temperatures.
Metallic Bonds (bonds between metal atoms) - Weakest strong bond
- Metallic solids are often described as \"a group of nuclei surrounded by a sea of mobile
electrons.\" In other words, the electrons in metallic solids are delocalized and are free to move
about, making metals good conductors of electricity.
- Metallic bonds are still relatively strong forces of attraction and most metals have high melting
and boiling point temperatures.
Mercury is the only metal that is not solid at room temperature.
Hydrogen Bonding - Strongest weak bond
- In hydrogen bonds, the positively charged hydrogen end of one molecule is attracted to the
negatively charged end of another molecule which must be an extremely electronegative element
(fluorine, oxygen, or nitrogen - FON)
Dipole-Dipole Forces - Medium weak bond
- Dipole-dipole forces exist between neutral, polar molecules where the positive end of one
molecule is attracted to the negative end of another molecule.
London Dispersion Forces - Weakest weak bond
- London dispersion forces (LDF) occur between neutral, nonpolar molecules. LDF occur due to
the \"random motion of electrons.\" At any moment, one atom may be surrounded by an extra
electron from a neighboring atom, resulting in an instantaneous polarity on the atom. During that
instant, the \"polarized\" atom will act as a very weak dipole.
Solution
Ionic Bonds (between ionic molecules) - Medium strong bond
- In an ionic solid, adjacent ions have electrostatic attractions and are arranged in a lattice
structure.
- Ionic solids have strong bonds resulting in high melting and boiling point temperatures.
Metallic Bonds (bonds between metal atoms) - Weakest strong bond
- Metallic solids are often described as \"a group of nuclei surrounded by a sea of mobile
electrons.\" In other words, the electrons in metallic solids are delocalized and are free to move
about, making metals good conductors of electricity.
- Metallic bonds are still relatively strong forces of attraction and most metals have high melting
and boiling point temperatures.
Mercury is the only metal that is not solid at room temperature.
Hydrogen Bonding - Strongest weak bond
- In hydrogen bonds, the positively charged hydrogen end of one molecule is attracted to the
negatively charged end of another molecule which must be an extremely electronegative element
(fluorine, oxygen, or nitrogen - FON)
Dipole-Dipole Forces - Medium weak bond
- Dipole-dipole forces exist between neutral, polar molecules where the positive end of one
molecule is attracted to the negative end of another molecule.
London Dispersion Forces - Weakest weak bond
- London dispersion forces (LDF) occur between neutral, nonpolar molecules. LDF occur due to
the \"random moti.
Dicer it is also referred as endoribonuclease Dicer. It has activit.pdf
Dicer: it is also referred as endoribonuclease Dicer. It has activity of cleaving double-stranded
RNA and pre –microRNA into specific fragments of RNA 20 to 15 kb (double stranded). This
enzyme is also known as helicase with RNase. This enzyme called dicer and large RNA
molecules produces small interfering RNA and micro RNA.
Gene expression & dicer: The process of generation of messenger RNA from genome of DNA in
the presence of RNA polymerase and transcription factors is known as “gene expression”. Dicer
influences gene expression by binding to “nascent mRNA” to cut into small fragements result in
negative feedback of gene expression
Example:
In fission yeast, defective RNA interference (RNAi) is leading to loss of H3K9 methylation at
cetromere result in loss of centromere function. Through a series of transcriptive and Dicer
related activities, siRNAs are formed that are required for recruiting RITS complex to the
centromere and heterochromatin formation. RNA-dependent RNA polymerase II is mainly
responsible for the generation of short- lived siRNA finally a double stranded dsRNA is
generated from this siRNA by dicer. These siRNA are complexed with argonaute result in Ago1
blockade finally RITS (RNA-induced initiation of transcriptional silencing) is recruited for co-
transcriptional and post-transcriptional silencing result in assembly of heterochromatin to the
centromere
Solution
Dicer: it is also referred as endoribonuclease Dicer. It has activity of cleaving double-stranded
RNA and pre –microRNA into specific fragments of RNA 20 to 15 kb (double stranded). This
enzyme is also known as helicase with RNase. This enzyme called dicer and large RNA
molecules produces small interfering RNA and micro RNA.
Gene expression & dicer: The process of generation of messenger RNA from genome of DNA in
the presence of RNA polymerase and transcription factors is known as “gene expression”. Dicer
influences gene expression by binding to “nascent mRNA” to cut into small fragements result in
negative feedback of gene expression
Example:
In fission yeast, defective RNA interference (RNAi) is leading to loss of H3K9 methylation at
cetromere result in loss of centromere function. Through a series of transcriptive and Dicer
related activities, siRNAs are formed that are required for recruiting RITS complex to the
centromere and heterochromatin formation. RNA-dependent RNA polymerase II is mainly
responsible for the generation of short- lived siRNA finally a double stranded dsRNA is
generated from this siRNA by dicer. These siRNA are complexed with argonaute result in Ago1
blockade finally RITS (RNA-induced initiation of transcriptional silencing) is recruited for co-
transcriptional and post-transcriptional silencing result in assembly of heterochromatin to the
centromere.
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