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eimsori [14]
3 years ago
11

In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set M

INIMUM to X.” Read this STATEMENT very carefully and Answer the below Questions.
A. Will this Algorithm RUN?? Yes or No with a justification

B. Will X Replace Any Value or Not?
Computers and Technology
1 answer:
Dvinal [7]3 years ago
4 0

Answer:

The answer to this question is given below in the explanation section.

Explanation:

The given problem scenario is:

In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set MINIMUM to X.

The answer to question A is: this algorithm will not run because:

If you a list of a positive number and you have already set it MINIMUM to 1, the comparison result with MINIMUM will not obtain. For example, lets suppose a list of numbers L. L={1,2,3,1,1,5,4,3,2}. You have already set the Minimum to 1. When you compare each number X in the list, if X >MINIMUM, then set MINIMUM to X.  

This will not work, because there are positive numbers in the list and the minimum positive number is 1. So the algorithm, will not execute the if-part.

In short, the pseudo-code of this problem is given below:

<em>L={list of positive number e.g. 1,2,3,.......Xn}</em>

<em>MINIMUM=1</em>

<em>foreach ( X in L)</em>

<em>{</em>

<em>(if X<MINIMUM)</em>

<em>    { </em>

<em>        MINIMUM=X</em>

<em>    }</em>

<em>}</em>

The answer to question B is X will not be replaced because X  replaced with MINIMUM only when if-part of this algorithm get executed. However, it is noted that X will not replace any value, if if-part will get executed, the Variable MINIMUM will get replaced with any value.

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Locker doors There are n lockers in a hallway, numbered sequentially from 1 to n. Initially, all the locker doors are closed. Yo
kow [346]

Answer:

// here is code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int n,no_open=0;

   cout<<"enter the number of lockers:";

   // read the number of lockers

   cin>>n;

   // initialize all lockers with 0, 0 for locked and 1 for open

   int lock[n]={};

   // toggle the locks

   // in each pass toggle every ith lock

   // if open close it and vice versa

   for(int i=1;i<=n;i++)

   {

       for(int a=0;a<n;a++)

       {

           if((a+1)%i==0)

           {

               if(lock[a]==0)

               lock[a]=1;

               else if(lock[a]==1)

               lock[a]=0;

           }

       }

   }

   cout<<"After last pass status of all locks:"<<endl;

   // print the status of all locks

   for(int x=0;x<n;x++)

   {

       if(lock[x]==0)

       {

           cout<<"lock "<<x+1<<" is close."<<endl;

       }

       else if(lock[x]==1)

       {

           cout<<"lock "<<x+1<<" is open."<<endl;

           // count the open locks

           no_open++;

       }

   }

   // print the open locks

   cout<<"total open locks are :"<<no_open<<endl;

return 0;

}

Explanation:

First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.

Output:

enter the number of lockers:9

After last pass status of all locks:

lock 1 is open.

lock 2 is close.

lock 3 is close.

lock 4 is open.

lock 5 is close.

lock 6 is close.

lock 7 is close.

lock 8 is close.

lock 9 is open.

total open locks are :3

5 0
3 years ago
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