Number guessing game - c++

Question

I've been given a 'broken algorithm' to fix. It's for the game 'Guess a number between 1-100' Where the computer answers within 7 questions/iterations.
The brief i've been given suggests only minimal changes need to be made to the algorithm, sorry if that's vague, i'm thinking the same thing.
Anyway the algorithm was full of silly mistakes which i've cleaned up. For a test case of 33 the algorithm assigns the following medians
50,25,37,19 << 19 is obviously incorrect.
I'm aware the last_median = current median is not in the right place. It's been a long day and if anyone could shed any light onto this i'd be grateful.
const int MAX_VALUE = 100;
int current_median = MAX_VALUE /2;
int last_median = 0;
while (true)
{
last_median = current_median;
if(number >= current_median)
{
if(number == current_median)
{
//Check for equality
cout << endl << number << endl;
break;
}
current_median += last_median /2;
}
else if(number <= current_median)
{
if(number == current_median)
{
// Check for equality
cout<<endl<<number<<endl;
break;
}
current_median -= last_median /2;
}
}

Answer 1

Two hints:
You can move the equality comparison outside the two ifs.
You should keep track of the range that you know contains the number. Once you do that, the logic will be very straightforward.

Answer 2

The algorithm is basically doing a binary search. One problem you have is with your conditionals:
if(number >= current_median)
// ...
else if(number <= current_median)
// ...
If the number is equal, you want to exit the loop, not continue it:
if(number > current_median)
// ...
else if(number < current_median)
// ...
else // we are equal
{
break;
}
You should also adjust your conditionals to check for a (low, high) range as it should (quickly) get smaller as your median value approaches the real value.
A simple example:
#include <iostream>
int main()
{
std::cout << "Pick a number between [0, 100] (Don't tell me what it is!)";
unsigned int low = 0;
unsigned int high = 100;
do
{
unsigned int median = (low + high) / 2;
std::cout << "Is your number > " << median << "? ";
char answer;
std::cin >> answer;
if (answer == 'Y' || answer == 'y')
{
low = median;
continue;
}
else
{
std::cout << "Is your number < " << median << "? ";
std::cin >> answer;
if (answer == 'Y' || answer == 'y')
{
high = median;
continue;
}
else // we are equal
{
low = high = median;
}
}
} while (low != high);
std::cout << "Your number is " << low << std::endl;
return 0;
}

Answer 3

If you're looking for a number between 0 and 100 you're going to find it in 7 or less steps. If you want to be more precise, it's exactly 7 steps if you're looking for an odd number and it's going to be 6 or less in some cases if you're looking for an even number. I'll put below a simple code for this. I didn't put in all the checks (entering numbers not letters, or other stuff), it's just written in a couple of minutes:
#include <iostream>
int main(){
std::cout << "Think of a number between 1 and 100" << std::endl ;
std::cout << "Press 0 if my number is correct. 1 if YOUR number is smaller. 2 if bigger" << std::endl;
int response = 0;
int middle = 64;
int low = 0;
int high = 128;
int counter = 1;
while(response!=0)
{
std::cout << counter << " Guess " << counter << ": " << middle << std::endl;
std::cout <<"Press 0(bingo) 1(smaller) 2(greater): " ;
std::cin >> response ;
std::cout << std::endl ;
counter++;
switch (response){
case 0:
std::cout << "Your number is: " << middle << std::endl;
break;
case 1:
high = middle;
middle = (low+high)/2;
break;
case 2:
low = middle;
middle = (low+high)/2;
break;
} //end swhitch
}; //end while
return 0;
}

Answer 4

int main()
{
char uput = '?';
int maxx = 100;
int minn = 0;
while (minn != maxx)
{
cout << (minn+maxx)/2 << ": [h]igher, [l]ower or [e]qual?";
cin >> uput;
if (uput == 'l')
{
maxx = (minn+maxx)/2;
}
else if (uput == 'h')
{
minn = (minn+maxx)/2;
}
else if (uput == 'e')
{
cout << "Your number is " << (minn+maxx)/2 << ". Yay.";
return 0;
}
}
return 0;
}
If the number is lower than the median, the median is the new maximum.
If the number is higher than the median, the median is the new minimum.
Create a loop. For each time around the loop, find the median between the minn and the maxx. Since you are narrowing the gap between the numbers between the minimum and the maximum each time, you will eventually arrive with only 1 possibility remaining, and that's your number.
The code above isn't a great example but hopefully it helps illustrate.