Questions: What will be the first from and to towers, given that there is an odd (not even) number of rings? Once the towers are initialized, what rings will be on towers 1, 2, and 3? Given the C++ code: // toh.cpp #include #include using namespace std; int main() { vector t[3]; int n; cout << "Enter the number of rings: "; cin >> n; int from = 0; int candidate = 1; int to, move = 0; if ( n % 2 == 0) to = 2; // (from + to)%mod3, counter clockwise else to = 1; // (from + to)%mod3, clockwise for ( int i = n + 1 ; i >= 1; i-- ) t[0].push_back(i); t[1].push_back(n+1); t[2].push_back(n+1); if ( n % 2 == 0 ) { while ( t[1].size() < (n+1) ) { cout << "move " << ++move << ":" << " Move candidate " << candidate; cout << " from tower " << (char) (from + 'A'); cout << " to tower " << (char)(to + 'A') << endl; t[to].push_back(t[from].back()); t[from].pop_back(); if (t[(to+2)%3].back() < t[(to+1)%3].back()) from = (to + 2)%3; else from = (to + 1)%3; if (t[(from)%3].back() < t[(from+2)%3].back()) to = (from + 2)%3; else to = (from + 1)%3; candidate = t[from].back(); } return 0; } else { while ( t[1].size() < ( n+1 ) ) { cout << "move " << ++move << ":" << " Move candidate " << candidate; cout << " from tower " << (char) (from + 'A'); cout << " to tower " << (char)(to + 'A') << endl; t[to].push_back(t[from].back()); t[from].pop_back(); if (t[(to+1)%3].back() < t[(to+2)%3].back()) from = ( to + 1 )%3; else from = ( to + 2 )%3; //cout << "from = " << from << endl; if (t[(from)%3].back() < t[(from+1)%3].back()) to = (from + 1)%3; else to = (from + 2)%3; //cout << "to = " << to <

Questions: What will be the first from and to towers, given that there is an odd (not even) number of rings? Once the towers are initialized, what rings will be on towers 1, 2, and 3? Given the C++ code: // toh.cpp #include #include using namespace std; int main() { vector t[3]; int n; cout << "Enter the number of rings: "; cin >> n; int from = 0; int candidate = 1; int to, move = 0; if ( n % 2 == 0) to = 2; // (from + to)%mod3, counter clockwise else to = 1; // (from + to)%mod3, clockwise for ( int i = n + 1 ; i >= 1; i-- ) t[0].push_back(i); t[1].push_back(n+1); t[2].push_back(n+1); if ( n % 2 == 0 ) { while ( t[1].size() < (n+1) ) { cout << "move " << ++move << ":" << " Move candidate " << candidate; cout << " from tower " << (char) (from + 'A'); cout << " to tower " << (char)(to + 'A') << endl; t[to].push_back(t[from].back()); t[from].pop_back(); if (t[(to+2)%3].back() < t[(to+1)%3].back()) from = (to + 2)%3; else from = (to + 1)%3; if (t[(from)%3].back() < t[(from+2)%3].back()) to = (from + 2)%3; else to = (from + 1)%3; candidate = t[from].back(); } return 0; } else { while ( t[1].size() < ( n+1 ) ) { cout << "move " << ++move << ":" << " Move candidate " << candidate; cout << " from tower " << (char) (from + 'A'); cout << " to tower " << (char)(to + 'A') << endl; t[to].push_back(t[from].back()); t[from].pop_back(); if (t[(to+1)%3].back() < t[(to+2)%3].back()) from = ( to + 1 )%3; else from = ( to + 2 )%3; //cout << "from = " << from << endl; if (t[(from)%3].back() < t[(from+1)%3].back()) to = (from + 1)%3; else to = (from + 2)%3; //cout << "to = " << to <

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Question

Questions:

What will be the first from and to towers, given that there is an odd (not even) number of rings?
Once the towers are initialized, what rings will be on towers 1, 2, and 3?

Given the C++ code:

// toh.cpp
#include <iostream>
#include <vector>
using namespace std;

int main() {
vector<int> t[3];
int n;
cout << "Enter the number of rings: ";
cin >> n;
int from = 0;
int candidate = 1;
int to, move = 0;

if ( n % 2 == 0)
to = 2; // (from + to)%mod3, counter clockwise
else
to = 1; // (from + to)%mod3, clockwise

for ( int i = n + 1 ; i >= 1; i-- )
t[0].push_back(i);
t[1].push_back(n+1);
t[2].push_back(n+1);

if ( n % 2 == 0 )
{

while ( t[1].size() < (n+1) ) {

cout << "move " << ++move << ":" << " Move candidate " << candidate;
cout << " from tower " << (char) (from + 'A');
cout << " to tower " << (char)(to + 'A') << endl;

t[to].push_back(t[from].back());
t[from].pop_back();
if (t[(to+2)%3].back() < t[(to+1)%3].back())
from = (to + 2)%3;
else
from = (to + 1)%3;

if (t[(from)%3].back() < t[(from+2)%3].back())
to = (from + 2)%3;
else
to = (from + 1)%3;
candidate = t[from].back();
}
return 0;

} else {

while ( t[1].size() < ( n+1 ) ) {
cout << "move " << ++move << ":" << " Move candidate " << candidate;
cout << " from tower " << (char) (from + 'A');
cout << " to tower " << (char)(to + 'A') << endl;
t[to].push_back(t[from].back());
t[from].pop_back();
if (t[(to+1)%3].back() < t[(to+2)%3].back())
from = ( to + 1 )%3;
else
from = ( to + 2 )%3;
//cout << "from = " << from << endl;
if (t[(from)%3].back() < t[(from+1)%3].back())
to = (from + 1)%3;
else
to = (from + 2)%3;
//cout << "to = " << to <<endl;

candidate = t[from].back();
}
return 0;
}

return 0;
}

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