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 [color=Red]z[/color] 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)

[jaideep]

Alba,

 

Nice on. But I still fell that you answer to second is wrong. Can you recheck it again.

I feel answer should be y pow z.

Jaideep

[crackit]

I agree with jaideep

 

answer to 2 should be y^z.

[wood]

True. Answer is definitely y^z.

 

Wood

[AlbaLed]

True, sorry for the typo

 

 

AlbaLed