determine if an integer value is odd or is even

[Qube]

Code: [Select]

Function isEven(num%)
For i = -2147483648 To +2147483647 Step 2
If num = i Then Return True
Next
Return False
End Function

Function isOdd(num%)
For i = -2147483647 To +2147483647 Step 2
If num = i Then Return True
Next
Return False
End Function


That's my boy! true brute force at it's best ( even with a bug )

[Steve Elliott]

lol 

[TomToad]

Ohh no ... seems to contain a mistake.

That's my boy! true brute force at it's best ( even with a bug )

I don't see the bug.  Tried replacing the For loops with
For i = -648 to +647 step 2
For i = -647 to +647 step 2
So that I wouldn't have to wait a month for a result, and the values return correct results.

Edit: Here's another way to determine if a number is odd or even

Code: [Select]

Function isEven(num%)
If Instr("02468",Right(Str(num),1)) > 0 Then Return True
Return False
End Function

Function isOdd(num%)
If Instr("13579",Right(Str(num),1)) > 0 Then Return True
Return False
End Function

Rube Goldberg use to find complex ways of solving simple problems.  I guess this is the programmer's version of it?

[Derron]

Yepp, no bug - just compared the numbers ("- ...8 to ..." and "- ...7 to ..." and did not grasp the "roll over" at "2^31 -1" +1).

See - your algorithm was already over my head, so it must be the perfect one

bye
Ron

[iWasAdam]

or...

Code: [Select]

method IsEven:bool( input:int )
  local check:float = input / 2
  return int(check) = check
end method

@Qube - Brilliant! 

[Derron]

@IwasAdam
"int/2" = int (at least in BlitzMax)

so better write "int/2.0" to return the float value.

Yet your

Code: [Select]

method IsEven:bool( input:int )
  local check:float = input / 2.0
  return int(check) = check
end method
might even be faster than the modulo


alternative to yours:

Code: [Select]

method IsEven:bool( input:int )
  return int(input / 2.0)*2 = input
end method


Back to serious mode:
https://stackoverflow.com/questions/2229107/what-is-the-fastest-way-to-find-if-a-number-is-even-or-odd

wich exposes that "number & 1" might fail on "one complements" computers - and that "number % 2" could be optimized correctly by the final compiler then ...


And the thread there contains another funny option - recursion:

Code: [Select]

Function IsOdd:int(n:int)
   if n = 0
      return 0
   elseif n = 1
      return 1
   else
      return not IsOdd(n - 1)
   endif
End Function


bye
Ron

[RemiD]

it seems that you have the topic for the next competition :
each participant much create 2 procedures to do something,
one very optimized and minimalist (and fast)
one very bullshity and bulky (and slow)