Gien.fr Cin; i>=l; i1=2){for Cjsl; j L= lugli); jt+){Some thing Ocn)33whart is the time complexity?

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Gien. fr Cin; i>=l; 1=2){ for Cjsl; j L= lugli); j++){ ocn) ; Some thing Ocn) 3 3 what is te time complexity?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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a) L= {anb²n, n>0} b) L= {a²nb, n>0} c) L= {ww], w={a,b}* d) L= {anb²n, n20} U {ww}, w={a,b}* Home work 1- Draw an NPDA for the following languages e) L= {a^b²n, n20} f) L= {a²nb", n>0} g) L= {ww), w={a,b}* h) L= {anb²n, n20} U {ww}, w={a,b}* 2- Write CFG for the following languages a) L= {anb²n, n>0} L= {a²nb", n>0} L= {ww}, w={a,b}* L= {anb²n, n>0} U {ww}, w={a,b}* b) c) d)
code in ada with Ada.Numerics.Generic_Elementary_Functions; package body Number_Theory is -- Instantiate the library for floating point math using Floating_Type. package Floating_Functions is new Ada.Numerics.Generic_Elementary_Functions(Floating_Type); use Floating_Functions; function Factorial(N : in Factorial_Argument_Type) return Positive is begin -- TODO: Finish me! -- -- 0! is 1 -- N! is N * (N-1) * (N-2) * ... * 1 return 1; end Factorial; function Is_Prime(N : in Prime_Argument_Type) return Boolean is Upper_Bound : Prime_Argument_Type; Current_Divisor : Prime_Argument_Type; begin -- Handle 2 as a special case. if N = 2 then return True; end if; Upper_Bound := N - 1; Current_Divisor := 2; while Current_Divisor < Upper_Bound loop if N rem Current_Divisor = 0 then return False; end if; Upper_Bound := N / Current_Divisor; end loop;…
Principles of programming languages: The C selection statement using EBNF is: <selection> → if ( <expression> ) <statement> [else <statement>] ; Without the brackets, how many BNF rules would describe this C selection statement? a. 3 b. 2 c. 4 d. 1
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convert this C++ codes into C language #include <bits/stdc++.h>using namespace std; // tree node is definedclass tree{ public: int data; tree *left; tree *right;}; void printSibling(tree* root){ //Declare queue using STL queue<tree*> q; //enqueue the root q.push(root); vector<int> store; tree* temp; //do the level order traversal & check for siblings while(!q.empty()){ //dequeue temp=q.front(); q.pop(); //if the current node has only one child //definitely the child has no sibling //store the child node value if(temp->left==NULL && temp->right!=NULL){ store.push_back(temp->right->data); } if(temp->left!=NULL && temp->right==NULL){ store.push_back(temp->left->data); } // do level order traversing if(temp->right) q.push(temp->right); if(temp->left) q.push(temp->left); } //if no node found without having sibling //vector size is zero //print -1 if(store.size()==0){ printf("-1, no…
Make this C language code into modular code (functions) #include #include main() { char choice; int num1, num2, result = 0; while(1) { printf("\nEnter Operator(+, -, *, /,f, s):");//f for floor division and s for square root //choice=getch(); scanf(" %c",&choice); if(choice=='s'){ printf("\nEnter Value to calculate square root:"); scanf("%ld",&num1); } else{ printf("\nEnter First Value:"); scanf("%d",&num1); printf("\nEnter Second Value:"); scanf("%d",&num2); } switch(choice) { case '+': result = num1 + num2; printf("\nSum is = %d",result); break; case '-': result = num1 - num2; printf("\nDifference is = %d",result); printf("\n\nPress Enter Again for New Input\n"); break; case '*': result = num1 * num2; printf("\nProduct is = %d",result); printf("\n\nPress Enter Again for New Input\n"); break; case '/': result = num1 / num2; printf("\nQuotient is = %d",result); printf("\n\nPress Enter Again…
check if the EBNF rules suffer from the Left Recursion Problem. You need to justify your answer and resolve the problem if any. <if_statment> => if <condition > <stats> <else_clouse>opt <condition> → <exp> | <exp> , <condition > <exp> → <ident> > <const> | <ident> > < <const> |<ident> >= <const> | <ident> >= <const> <else-clause> → else <stats> | else <if-statement> <stats> → <stat>opt
the code is in ada with Ada.Numerics.Generic_Elementary_Functions; package body Number_Theory is -- Instantiate the library for floating point math using Floating_Type. package Floating_Functions is new Ada.Numerics.Generic_Elementary_Functions(Floating_Type); use Floating_Functions; function Factorial(N : in Factorial_Argument_Type) return Positive is begin -- TODO: Finish me! -- -- 0! is 1 -- N! is N * (N-1) * (N-2) * ... * 1 return 1; end Factorial; function Is_Prime(N : in Prime_Argument_Type) return Boolean is Upper_Bound : Prime_Argument_Type; Current_Divisor : Prime_Argument_Type; begin -- Handle 2 as a special case. if N = 2 then return True; end if; Upper_Bound := N - 1; Current_Divisor := 2; while Current_Divisor < Upper_Bound loop if N rem Current_Divisor = 0 then return False; end if; Upper_Bound := N / Current_Divisor; end loop;…
Make this C language code into modular code (functions) #include #include main() { char choice; int num1, num2, result = 0; while(1) { printf("\nEnter Operator(+,-, *, /,f, s):");//f for floor division and s for square root //choice-getch(); scanf("%c",&choice); if(choice=='s') { printf("\nEnter Value to calculate square root:"); scanf("%ld",&num1); } else{ printf("\nEnter First Value:"); scanf("%d",&num1); printf("\nEnter Second Value:"); scanf("%d",&num2); } switch (choice) { case '+': result = num1 + num2; printf("\nSum is = %d", result); break; case '-': result = num1 - num2; printf("\nDifference is = %d", result); printf("\n\nPress Enter Again for New Input\n"); break; case '*': result = num1 * num2; printf("\nProduct is = %d", result); printf("\n\nPress Enter Again for New Input\n"); break; case '/': result = num1 / num2; printf("\nQuotient is = %d", result); printf("\n\nPress Enter Again for New Input\n"); break; case 'f'://floor division result = floor(num1/num2);…
Subject: Data Strucure and Algorithm in C++
Course : Data Structures & Algorithms in C++
Make this C language code into modular code (functions) #include #include main() { char choice; int num1, num2, result = 0; while(1) { printf("\nEnter Operator(+,-, *, /,f, s):");//f for floor division and s for square root //choice-getch(); scanf("%c",&choice); if(choice=='s') { printf("\nEnter Value to calculate square root:"); scanf("%ld",&num1); } else{ printf("\nEnter First Value:"); scanf("%d",&num1); printf("\nEnter Second Value:"); scanf("%d",&num2); } switch (choice) { case '+': result = num1 + num2; printf("\nSum is = %d", result); break; case '-': result = num1 - num2; printf("\nDifference is = %d", result); printf("\n\nPress Enter Again for New Input\n"); break; case '*': result = num1 * num2; printf("\nProduct is = %d", result); printf("\n\nPress Enter Again for New Input\n"); break; case '/': result = num1 / num2; printf("\nQuotient is = %d", result); printf("\n\nPress Enter Again for New Input\n"); break; case 'f'://floor division result = floor(num1/num2);…