Finding MAX of numbers without conditional IF statements c++ [duplicate] - c++

Question

This question already has answers here:
Mathematically Find Max Value without Conditional Comparison
(18 answers)
Closed 9 years ago.
So i have too get two numbers from user input, and find the max of the two numbers without using if statements.
The class is a beginner class, and we have too use what we already know. I kinda worked something out, but it only works if the numbers are inputted with the max number first.
#include <iostream>
using namespace std;
int main()
{
int x = 0, y = 0, max = 0;
int smallest, largest;
cout << "Please enter 2 integer numbers, and i will show you which one is larger: ";
cin >> x >> y;
smallest = (x < y == 1) + (x - 1);
smallest = (y < x == 1) + (y - 1);
largest = (x < y == 1) + (y - 1);
largest = (y > x == 1) + (x + 1 - 1);
cout << "Smallest: " << smallest << endl;
cout << "Largest: " << largest << endl;
return 0;
}
Thats what i have so far, but after putting different test data in, i found out it only works for numbers such as 4,5 or 6,7. But numbers with more then 2 spaces between eachother they dont such as, 4,8 or 5, 7. Any help would be appreciated.

Answer 1

I saw this question in Cracking the Coding interview book.
Let’s try to solve this by “re-wording” the problem We will re-word the problem until we get something that has removed all if statements
Rewording 1: If a > b, return a; else, return b
Rewording 2: If (a - b) is negative, return b; else, return a
Rewording 3: If (a - b) is negative, let k = 1; else, let k = 0 Return a - k * (a - b)
Rewording 4: Let c = a - b Let k = the most significant bit of c Return a - k * c
int getMax(int a, int b) {
int c = a - b;
int k = (c >> ((sizeof(int) * CHAR_BIT) - 1)) & 0x1;
int max = a - k * c;
return max;
}
Source: http://www.amazon.com/Cracking-Coding-Interview-Programming-Questions/dp/098478280X
Edit: This code works even when a-b overflows.
Let k equal the sign of a-b such that if a-b >=0, then k is 1, else k=0.Let q be the inverse of k. Above code overflows when a is positive or b is negative, or the other way around. If a and b have different signs, then we want the k to equal sign(a).
/* Flips 1 to 0 and vice-versa */
public static int flip(int bit){
return 1^bit;
}
/* returns 1 if a is positive, and 0 if a is negative */
public static int sign(int a){
return flip((a >> ((sizeof(int) * CHAR_BIT) - 1)) & 0x1);
}
public static int getMax(int a, int b){
int c = a - b;
int sa = sign(a-b); // if a>=0, then 1 else 0
int sb = sign(a-b); // if b>=1, then 1 else 0
int sc = sign(c); // depends on whether or not a-b overflows
/* If a and b have different signs, then k = sign(a) */
int use_sign_of_a = sa ^ sb;
/* If a and b have the same sign, then k = sign(a - b) */
int use_sign_of_c = flip(sa ^ sb);
int k = use_sign_of_a * sa + use_sign_of_c * sc;
int q = flip(k); //opposite of k
return a * k + b * q;
}

Answer 2

Here is a funny solution:
int max_num = (x>y)*x + (y>=x)*y;

Answer 3

Assuming that you have covered bitwise operators already you can do this:
max = a-((a-b)&((a-b)>>(sizeof(int)*8-1)));
This is based off of the solution from Mathematically Find Max Value without Conditional Comparison that #user93353 pointed out in the comments above.
This may be overkill if you really are just trying to avoid if statements, not comparisons in general.

Answer 4

You can try this code to find max and min for two input variables.
((a > b) && (max = a)) || (max=b);
((a < b) && (min = a)) || (min=b);
For three input variables you can use similar method like this:
int main()
{
int a = 10, b = 9 , c = 8;
cin >> a >> b >> c;
int max = a, min = a;
// For Max
((a > b) && (a > c) && (max=a)) ||
((b > c) && (b > a) && (max=b)) ||
(max=c) ;
// For min
((a < b) && (a < c) && (min=a)) ||
((b < c) && (b < a) && (min=b)) ||
(min=c) ;
cout << "max = " << max;
cout << "and min = " << min;
return 1;
}
One run is:
:~$ ./a.out
1
2
3
max = 3 and min = 1
Edit
Thanks to #Tony D: This code will fail for negative numbers.
One may try this for negative numbers for two inputs to find max(not sure for this):
((a > b) && ( a > 0 && (max = a))) || ((b > a) && (max = b)) || (max = a);