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I'm trying to solve this. But I'm not figuring the right
solution. Can some one help what the answer is for the
question below?
You can use as many variables as you need, there are no
negative numbers, all numbers are integers. You do not know
the size of the integers, they could be infinitely large, so
you can't count on truncating at any point. There are NO
comparisons allowed, no if statements or anything like that.
There are only four operations you can do on a variable.
1) You can set a variable to 0.
2) You can set a variable = another variable.
3) You can increment a variable (only by 1), and it's a post
increment.
4) You can loop. So, if you were to say loop(v1) and v1 =
10, your loop would execute 10 times, but the value in v1
wouldn't change so the first line in the loop can change
value of v1 without changing the number of times you loop.
You need to do 3 things.
1) Write a function that decrements by 1.
2) Write a function that subtracts one variable from another.
3) Write a function that divides one variable by another.
4) See if you can implement all 3 using at most 4 variables.
Meaning, you're not making function calls now, you're making
macros. And at most you can have 4 variables. The
restriction really only applies to divide, the other 2 are
easy to do with 4 vars or less. Division on the other hand
is dependent on the other 2 functions, so, if subtract
requires 3 variables, then divide only has 1 variable left
unchanged after a call to subtract. Basically, just make
your function calls to decrement and subtract so you pass
your vars in by reference, and you can't declare any new
variables in a function, what you pass in is all it gets.



I'm trying to solve this. But I'm not figuring the right solution. Can some one help what..

Answer / 1337

1. Dec:

var DEC(var x)
{
var y = 0;
loop(x)
{
x = y;
y++;
}
return x;
}

var SUB(var x, var y)
{
loop(y)
{
x = DEC(x);
}
return x;
}

// Too lazy to do the rest - you get the idea...
}

Is This Answer Correct ?    0 Yes 0 No

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