We have three containers whose sizes are 10 pints, 7 pints, and 4 pints, respectively. The 7-pint and 4-pint containers start ou
t full of water, but the 10-pint container is initially empty. - We are allowed one type of operation: pouring the contents of one container into another, stopping only when the source container is empty or the destination container is full. - We want to know if there is a sequence of pourings that leaves exactly 2 pints in the 7- or 4-pint container.
2 answers:
Answer:
Here’s a way to solve this:
(7/7,4/4,0/10) Æ (7/7, 0/4, 4/10) Æ(1/7, 0/4, 10/10) Æ (1/7, 4/4, 6/10), (5/7, 0/4, 6/10) Æ
(5/7, 4/4, 2/10) Æ (7/7, 2/4, 2/10), hence we end up with 2 pints in the 4‐pint container .
To model this as a graph problem, we represent each potential state of the three containers with an ordered pair (A,B), where A represents the amount of water in the 7‐pint container and B represents the amount of water in the 4‐pint container.
Hence the initial state would be(7,4). Note that we do not need an ordered triple as we know the total amount of water in all three containers (11 pints). Therefore, the total state is completely determined by water in the 7 and 4 pint containers.
Hence, we have 5 x 8 = 40 possible states for the system. We model these states as nodes in a directed graph and draw an edge from one node to another node if we can go from the configuration represented by one node to the configuration represented by another node through one pour.
For example, there would be an edge from (7,4) to (0,4) but no edge from (1,3) to (5,1).
Thus, if we can find a path from one node to another, then we can go from one state to another
through a series of pourings. The nodes are represented in the attached file (the edges are left out).
Explanation:
Since we want to find a path from (7,4) to one of the nodes contained in one of the rectangles. Our solution above is depicted in the attached file
Looking at the graph in attachment, we will start at node (7,4) and run DFS to see if we can reach one of the nodes boxed by a rectangle (i.e. has 2 as one of the coordinates). If we can, then there is a solution.
Otherwise, there is no way to have two liters in one of the containers through a series of pouring from
the original state of (7/7,4/4,0/10).
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Answer:
Here is the C++ program:
#include <iostream> //to use input output functions
using namespace std; //to identify objects like cin cout
int main(){ //start of main function
int MinFloors = 1; //minimum number of floors
int MinRooms = 10; //minimum number of rooms
int NoOfFloors; //stores number of floors
int NoOfRooms; //stores number of rooms
int OccupiedRooms; //stores number of rooms occupied
double TotalRooms=0, TotalOccupied=0; //stores computed total number of rooms and total number of occupied rooms
cout<<"How many floors does the hotel have? "; //prompts user to enter number of floors
do{ //iterates until the user enters valid number of floors
cout<<"(cannot be less than "<<MinFloors<<"): "; //error message
cin >>NoOfFloors; //reads number of floors from user
}while(NoOfFloors<MinFloors); //repeats as long as number of floors is less than minimum number of floors
for(int floor=1; floor <= NoOfFloors; floor++){ //iterates through the floors to skip third iteration
if(floor == 3){ //if floor is third floor
continue; }
cout<<"How many rooms are on floor " <<floor; //prompts user to enter number of floors
do{ //start of do while loop
cout<<"(cannot be less than "<<MinRooms<<"): "; //error message
cin >>NoOfRooms; //reads number of rooms from user
}while(NoOfRooms<MinRooms); //iterates as long as number of rooms are less than valid minimum number of rooms
TotalRooms += NoOfRooms; //adds number of rooms to the count of total number of rooms
cout<<"How many of those rooms are occupied?"; //prompts user to enter number of rooms occupied
do{ //start of do while loop
cout<<"(cannot be less than 0 or greater than "<<NoOfRooms<<"): "; //generates error message
cin >>OccupiedRooms; //reads number of rooms occupied by user
}while(OccupiedRooms<0 || OccupiedRooms>NoOfRooms); //iterates as long as the number of occupied rooms are less than 0 or greater than number of rooms
TotalOccupied += OccupiedRooms; } //adds number of rooms occupied to the count of total number of occupied rooms
cout<<"\nThe hotel has a total of "<<TotalRooms<<" rooms"<<endl; //displays the total number of rooms in the hotel
cout<<TotalOccupied<<" are occupied."<<endl; //displays the total number of occupied rooms
cout<<(TotalRooms-TotalOccupied)<<" are unoccupied."<<endl; //displays the total number of unoccupied rooms
cout<<"The occupancy rate is: "<<100*(TotalOccupied/TotalRooms)<<"%"<<endl; } //computes and displays the occupancy rate
Explanation:
The program first prompts the user to enter number of floors in the hotel. Lets say user enters 6. This is stored in NoOfFloors So
NoOfFloors = 6
So the loop runs for 6 times
Next it asks user to enter the number of rooms in the floor 1. Lets say user enters 12 so this is stored in NoOfRooms so
NoOfRooms = 12
TotalRooms += NoOfRooms;
this statement keeps adding number of rooms to TotalRooms so
TotalRooms = 12
Next program asks user about the number of occupied rooms. Lets say user enters 10 so this is stored in OccupiedRooms so
OccupiedRooms = 10
this statement keeps adding number of rooms to TotalOccupied so
TotalOccupied += OccupiedRooms;
TotalOccupied = 10
At next iteration program asks user again to enter number of rooms in floor 2. Suppose user enters 14 so
NoOfRooms = 12
TotalRooms += NoOfRooms;
TotalRooms = 12+14
TotalRooms = 26
Program asks again to enter number of occupied rooms so it becomes:
OccupiedRooms = 8
this statement keeps adding number of rooms to TotalOccupied so
TotalOccupied += OccupiedRooms;
TotalOccupied = 10+8
TotalOccupied = 18
Next is skips floor 3 and iteration 3. and asks user to enter number of rooms in floor 4. Suppose user enters 14
Number of rooms become:
TotalRooms = 12+14+14
TotalRooms = 40
and suppose user enters 14 as occupied rooms so total occupied become:
TotalOccupied = 10+8 + 14
TotalOccupied = 32
For floor 5: Suppose user enters 13
TotalRooms = 12+14+14+13
TotalRooms = 53
For floor 5: Suppose user enters 10
TotalOccupied = 10+8 + 14+10
TotalOccupied = 42
For floor 6: Suppose user enters 12
TotalRooms = 12+14+14+13+12
TotalRooms = 65
For floor 6: Suppose user enters 11
TotalOccupied = 10+8 + 14+10+11
TotalOccupied = 53
Now the loop breaks
Hence
TotalRooms = 65
TotalOccupied = 53
total unoccupied = TotalRooms-TotalOccupied = 65-53 = 12
The occupancy rate is: 100*(TotalOccupied/TotalRooms) = 100*(53/65) = 81.5385
The output of the program is attached in a screenshot.
Answer:
Here the code is given as follows,
Explanation:
Code:-
#include <stdio.h>
int isSorted(int *array, int n) {
if (n <= 1) {
return 1;
} else {
return isSorted(array, n-1) && array[n-2] <= array[n-1];
}
}
int main() {
int arr1[] = {3, 6, 7, 7, 12}, size1 = 5;
int arr2[] = {3, 4, 9, 8}, size2 = 4;
printf("%d\n", isSorted(arr1, size1));
printf("%d\n", isSorted(arr2, size2));
return 0;
}
Output:-
