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No if() Only abs()

Given two variable a and b, construct a function f(a,b) which return the maximum one in {a,b}.

You can only use +- · / and abs() (to take the absolute value). NO if() is allowed in your construction.

Pure arithmetic ; )


Also, {a+b-abs(a-b)}/2 is the smaller one in {a,b}

Here's an intuitive explanation by @Grant Goodman (in comments): "(a+b)/2 gives you the average, which is directly in between a and b, and it is 1/2 the difference from either a or b. So to get to the larger one, just add the difference of the two divided by two, giving you (a+b)/2 + abs(a-b)/2, or (a+b+abs(a-b))/2"



sure it would b wrong...but i tried something...

int res=(a-b)>0?a:b;

it too doesn't have if()



That's a good point, but I think the idea is to avoid conditionals altogether. Your compiler turns the ternary ( ? : ) operator into the same code that it turns an if statement into (i.e. a conditional branch with two blocks).


You didn't say you could use a 2 in there.

So you'll need to make one with (a/a + b/b) or something like that.



That's a good point. More verbosely, {a+b+abs(a-b)} / [{a+b+abs(a-b)}+{a+b+abs(a-b)}]



Good point. Nod. But we have to exclude 0.

Whatever, the question should just emphasize on "pure arithmetic" or ask for an algebra expression.


Or: {a + b + abs(a-b)}/(a/a + a/a). Slightly shorter version.



This, however, of course wouldn't work if a was 0. So perhaps the parent's reply is actually more general.



Jeez, I'm going in circles here. ravi's reply is incorrect. That whole expression simply equals 1/2. As long as a and b are not both equal to 0, the following would suffice: {a + b + abs(a-b)}/{(a + a + b + b)/(a+b)}



Agreed, unless both a and b are zero...

Aashka Kotecha

this won't hold true if a and b both are odd

Jay Elliott

Isn't that just 1/2?



Eduard Nicodei

Wohooo touring-completeness achieved once again! Time to play with this in Haskell.

Anurag Poddar

max(a,b) = ((a+b)+|a-b|)/2

Grant Goodman

Here's an intuitive derivation of the solution:

(a+b)/2 gives you the average, which is directly in between a and b, and it is 1/2 the difference from either a or b. So to get to the larger one, just add the difference of the two divided by two, giving you (a+b)/2 + abs(a-b)/2, or (a+b+abs(a-b))/2



That's an awesome explanation. MoralMachine, I suggest adding it to the puzzle solution.

Martyn Shutt

This is the solution I came up with in C++;

int largerNum(int x, int y){

int pt1 = (x - y);
int pt2 = (pt1 - abs(pt1)) / (x/x + y/y); // 2
int pt3 = abs(pt2);
return pt3 + x;


I decided it would be easier for me to get to the solution if I broke it up into individual parts. But working from that, as a single expression;

return (abs((x - y - abs(pt1)) / (x/x + y/y))) + x;


Martyn Shutt

But as highlighted in other posts, it doesn't work if either number is zero. It actually crashes my compiled application. Probably due to dividing by zero.


Martyn Shutt

Correction; as a single return statement it should be

return (abs((x - y - abs(x - y)) / (x/x + y/y))) + x;

I accidentally left variable pt1 in there.

Kaushik Nayak

x = a - b

y = b - a

absx = abs( a - b ) absy = abs( b - a )

s = absx + absy -x -y

max = s/2


Kaushik Nayak

sorry..the correct one is

x = a - b absx = abs(x)

s = absx + a + b

max = s/2


Great question! Thoroughly enjoyed thinking about this one.

Oliver Matthews

Thinking about this, I'm fairly sure it is impossible unless you do one of the following: 1. allow numbers in the algorithm (as per solution) 2. assume at least one of the values is not 0. (as per below) 3. allow ^ 4. use a system that defines 0/0 as 1 (i.e. is wrong).

the last two would allow you to implicitly get a 1 by doing (a^a)/(a^a) or a/a respectively (and thus a 2 through addition).


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Submitted by
about 5 years ago
Difficulty 5.8 ?