Answer:
z = a.c' + a.b.d' + b.c'.d'
Explanation:
The truth table for this question is provided in the attachment to this question.
N.B - a' = not a!
The rows with output of 1 come from the following relations: 01 > 00, 10 > 00, 10 > 01, 11 > 00, 11 > 01, 11 > 10
This means that the Boolean expression is a sum of all the rows with output of 1.
z = a'bc'd' + ab'c'd' + ab'c'd + abc'd' + abc'd + abcd'
On simplification,
z = bc'd' + ab'c' + ac'd' + ac'd + abc' + abd'
z = ac' + abd' + bc'd'
Hope this helps!
Hi,
I changed your program using some of the concepts you were trying to use. Hopefully you can see how it works:
#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
short T;
cin >> T;
cin.ignore();
string str[100];
for(int i=0; i<T; i++)
{
getline(cin, str[i]);
}
for (int i = 0; i < T; i++)
{
stringstream ss(str[i]);
string tmp;
vector<string> v;
while (ss >> tmp)
{
// Let's capitalize it before storing in the vector
if (!tmp.empty())
{
transform(begin(tmp), end(tmp), std::begin(tmp), ::tolower);
tmp[0] = toupper(tmp[0]);
}
v.push_back(tmp);
}
if (v.size() == 1)
{
cout << v[0] << endl;
}
else if (v.size() == 2)
{
cout << v[0][0] << ". " << v[1] << endl;
}
else
{
cout << v[0][0] << ". " << v[1][0] << ". " << v[2] << endl;
}
}
return 0;
}
People who use Windows OS.
You can just put “/“ to represent dividing
