What do this functions do??

[AlbaLed]

Can you find out what do this functions do? Can you prove it? Don't worry about parameter passing methods, just what happens inside the function

1.
f(x,y)
{
    x:=x+y
    y:=x-y
    x:=x-y
}

2. what's returned?
f(y,z)
{
    x:=1
    while z>0 do
        x:=x*y
        z:=z-1
   return (x)
}

3. what's returned?
f(y,z)
{
    x:=1
    while z>0 do
       if x is odd then x:=x*y
        z:= floor(z/2)
        y:=y^2
   return (x)
}

4. what's returned?
f(y,z)
{
    x:=0
    while z>0 do
        x:=x+y*(z mod 2)
        y:=2y
        z:=floor(z/2)
   return (x)
}

Enjoyy!!!!!!!!!

AlbaLed

[jaideep]

1) Swapping
2) y power z
3) This is my guess y (even) then x = y^(2^(|_ lg n _| + 1) - 1)
y = odd then x = y.
4) Convert the z value to its binary equivalent. Start from right side as y then move left as 2.y further left as y.2^2 ans so on. Each time you encounter a one in the binary multiply by its equivalent y factor and add it to the sum.

e.g 1 1 1 1
2.2.2.y 2.2.y 2.y y

This gives me another idea to state the same thing.
Write down the expression for converting a binary number represented by z to its equivalent decimal number.
Then instead of multiplying by 2^i at the places multiply it by y.2^i.

Nice one.
Jaideep

[AlbaLed]

Anyone else want to give 'em a try before I post the answers, Jaideep got some right, not necessarily all, I will not point which though. I'll post the answers in about 2-3 hours

AlbaLed

[crackit]

answer to 4

is 2*(number of 1 in Z)

[crackit]

Originally posted by crackit

answer to 4

y*z

[AlbaLed]

Answers are

1. swaping
2. x power y

typo on 3, sorry

f(y,z)
{
    x:=1
    while z>0 do
       if z is odd then x:=x*y
        z:= floor(z/2)
        y:=y^2
   return (x)
}

3. y power z
4. y*z

Good job jaideep and crackit

AlbaLed

[bb]

For swapping 1. has to have

x:= x + y
y:= x - y
x:= x - x

[bb]

#2.

Let's say z:=5

z:= 5 ( >0) x := 1*y z:=5-1 = 4
z:=4 ( >0) x ..> 1*y*y z:=4-1 = 3
z:=3 ( >0) x ..> 1*y*y*y z:=3-1 = 2
z:=2 ( >0) x ..> 1*y*y*y*y z:=2-1 = 1
z:=1 ( >0) x ..> 1*y*y*y*y*y z:=1-1 = 0
z:=0 not >0 OUT

x = y^z (not x^y)

[AlbaLed]

wouldn't

x=x-x be 0 ???

it is not x-x because now the previous value of x is in y.
AlbaLed

[bb]

Sorry for #1. :o)