## Another puzzle, another prize!

It has been some time since I have given something away.  So I think it is time for another contest.  This time you are competing for a PSoC 5 DEV KIT  (-050). First to answer correctly wins!

The function y = f(x) where both x and y are integers and x is a non negative value.

The function is implemented with the following code;

Given a non negative integer x:

for(y = 0; x > 2*y; x -= 2*y++);

1) What is this function?

2) Prove it works for all non negative values of x.

3) What is the answer for a negative value of x?

Hi Dave,

Not sure if I'm right, but let's try!

1) What is this function?
A: It is an inverter.
Ex: input = 50
output = -50

2) Prove it works for all non negative values of x.

A: Using matlab.
clc
clear all
close all

%input
y=0;
temp = 0;

hold on
for x=1:50
if (x > (2*y))
x = (2*y)-x;
temp = temp+1;
end
stem(temp,x);
end

result graph demonstrates all input values of x inverted.

3) What is the answer for a negative value of x?

A: Assuming that test condition is x>2*y; y=0, the statement will be skiped, so input value will be unchanged.

Best regards,
Andre Fernandes
Brazil

You always ask such questions on a friday evening when I'm away for the whole weekend. But I cannot leave them unanswered :)

1) it calculates y=round(sqrt(x))

2)
My math is a little bit rusty, so I don't have a real, mathematical proof. But
I can explain how it works:

* the definition of y=floor(sqrt(x)) is the number y, where y**2<=x<(y+1)**2 (but this is _not_ what is calculated here)

* n**2 can be calculated as the sum of the first n _odd_ integers

* the code above calculates the sum of the first y _even_ integers (this is the 2*y++ part)

* the sum of the first n even integers lies always just 0.5 below the
middle of the first n odd integers and the sum of the first n+1 odd integers

* the code stops, when this sum is close enough to x (this is the x>2*y part)

* this criteria is true when the next addition to the sum of even integers
would go above x

* so for calculating floor(sqrt(x)), we would just add the first y odd integers,
(which gives us the y-th square number) and stop when we reach
(or overflow) x, with y as the result

* but since we add even integers, the sum is always halfways between two square
numbers, so we increase y sooner - which effects in rounding up the square root

* we can never hit, with our calculation a number 'n.5', because 'n.5'**2 ends up
in the form 'm.25'

3) for negative numbers, it just returns 0 (since x is always smaller than 2*y),
and this is OK since the square root of negative numbers is not defined for
rational / real numbers

1) This function finds the integer square root of x with rounding.
2)the simplification of above function will like as below
x(n)=x(n-1) - 2*n for x(n)>2n
so Convergence of this function will be Y;
Z transform of above series will be -2/(1-z^(-1)) So for non negative number ROC will be always |z|<1 so it will converges at some finite value.

3) For negative value of x the answer y will hold 0 according to "for" loop.

rftgrf

Hi,
integer = 16bit? 32 bit?
1) Y = 50 * SQRT(1 - [(X +(2550 - Xi))/2550]) where Xi is the first value of X (value given)
2) Not possible to work with negative values because you have a strict condition. For Xi=-32768, Y=0 and next value for X will be 2 (thinking in 16 bit, with sign) but with you strict condition, result is false so, not possible to continue. Other thing is the response of the curve
3) Curve is displaced to left.
Meaning of this function? I don't know (by now ;-) ).

Hi Dave,

I have another try:

1) Its an inverter, the statement is executed just once cause for x>0 the output always will be -x;

2) As the Y will be always zero, we have the follow condition:
X>2Y -> X>0 (condition);
Now, let's call the output value as A, so if Y=0, we have:
A=(2Y)-X -> A=-X;
Finally for X>0 -> A=-X;

3) If the input value (X) is negative it does not meet the condition (X>0), in other words, will just skip the statement and X will be unchanged.

Thank you,

Andre P. Fernandes

1) C-Code is implemented for x=-y(y+1) + Xi, where Xi is the initial value of x (or Xi is the value of x, when y=0)

2) 2y is subtracted from x, if and only if x>2y, therefore x is always positive. Otherwise for-loop will terminate.

3) for a negative value of x, for-loop will not execute, because x>2y condition will not satisfied for positive values of y, and after for-loop x=Xi, and y=0.

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