Is Number Odd?

Printable View

- 11th June 2008, 12:16
T.Jackson

Is Number Odd?

Is there more than one way to test to see if an integer is even or odd? So far I've only been abe to devise one solution. The solution makes use of the AND bitwise operator.

Code:

public boolean isNumOdd(int num) { if (1 & num) return true; else return false; }

PBP code ...

Code:

main: if 1 & num then oddNum = true else oddNum = false end if

Has anyone got another way of doing this? I'm incredibly keen to hear your thoughts.

Best Regards,

Trent Jackson
- 11th June 2008, 13:01
Melanie

Code:

If num.0=1 then oddNum = true else oddNum = false endif
- 11th June 2008, 13:18
sayzer

GrandPa's way!

Code:

OddNum = True IF Num // 2 = 0 THEN OddNum = False

Does Num = 1 work?
----------------------------
- 11th June 2008, 13:25
T.Jackson

Quote:

Originally Posted by sayzer

Code:

OddNum = True IF Num // 2 = 0 THEN OddNum = False

Do me a favour and refresh my memory with what "//" does?

Trent Jackson
- 11th June 2008, 13:44
dhouston
Page 32 of the PBP manual defines it as Remainder (Modulus). I believe Modulus or Modulo is the more generally used terminology. It's also more useful, generally, than the other methods suggested as you can use it to test whether any number is evenly divisible by any other number. See...
- http://en.wikipedia.org/wiki/Modulo_operation
- 11th June 2008, 13:49
sayzer

Quote:

Originally Posted by dhouston

Page 32 of the PBP manual defines it as Remainder (Modulus). I believe Modulus is the more generally used terminology. It's also more useful, generally, as you can use it to test whether any number is evenly divisible by any other number.

Quote:

Originally Posted by T.Jackson

Do me a favour and refresh my memory with what "//" does?

Trent Jackson

I knew it did something for sure; (well, at least at where I am it does).

--------------
- 11th June 2008, 13:50
precision

in VB

Code:

If ((Text1.Text Mod 2) = 0) Then MsgBox "Even" Else MsgBox "Odd" End If

.
- 11th June 2008, 13:54
T.Jackson

Quote:

Originally Posted by precision

in VB

Code:

If ((Text1.Text Mod 2) = 0) Then MsgBox "Even" Else MsgBox "Odd" End If

That's pretty uncanny bud, because I just wrote this ...

Code:

Dim num As Integer Dim oddNum As Boolean num = 5 If num Mod 2 = 0 Then oddNum = False Else oddNum = True End If

I reckon that there's a recursive way of doing it too!

Trent Jackson
- 11th June 2008, 14:06
precision

I think only mod formula is easy way to find oddeven.

oddeven = number mod 2
if oddeven = 0 then page is even, otherwise odd

.
- 11th June 2008, 14:06
skimask

How about the most rediculous, long winded, round-about way to do it?
- 11th June 2008, 14:07
T.Jackson

So to sum it up so far we've got the following, verified to be working solutions.

1. AND Bitwise operator (My initial approach)
2. Direct bit checking (Melanie's approach)
3. Modulus remainder operator (Precision & Sayzer's approach)

The goal is 10 solutions -- can we do it?

Trent Jackson
- 11th June 2008, 14:19
T.Jackson

Quote:

Originally Posted by skimask

How about the most rediculous, long winded, round-about way to do it?

Any solution that can be verified workable!

Trent Jackson
- 11th June 2008, 15:38
skimask

Double subtraction

Code:

testnumber var word main: if testnumber = 0 then lcdout "Undefined" stop endif testnumber = testnumber - 1 testnumber = testnumber - 1 if testnumber = 0 then lcdout "EVEN" stop endif if testnumber = $ffff then lcdout "ODD" stop endif if testnumber > 0 and testnumber < $ffff then lcdout "Not done yet" goto main end

Should work, haven't tried it...but it's surely a waste of resources! :)
- 11th June 2008, 15:44
T.Jackson

Quote:

Originally Posted by skimask

Code:

testnumber var word main: if testnumber = 0 then lcdout "Undefined" stop endif testnumber = testnumber - 1 testnumber = testnumber - 1 if testnumber = 0 then lcdout "EVEN" stop endif if testnumber = $ffff then lcdout "ODD" stop endif if testnumber > 0 and testnumber < $ffff then lcdout "Not done yet" goto main end

Should work, haven't tried it...but it's surely a waste of resources! :)

I agree that, that's workable. Continuously subtracting 2 from the target number until you either reach zero, which denotes an even num or -1 equating to an odd num.

Trent Jackson
- 11th June 2008, 16:03
sayzer

I also have another one.

Mine does two calculations and compares the results in case the processor is going weird.
And even more, has a bug report feature.

Ski, is this ridiculous enough? I can go deeper.

Code:

TestNumber VAR WORD Index VAR BYTE Result1 VAR BIT Result2 VAR BIT FinalResult VAR BIT Even CON 0 Odd CON 1 Err CON 2 Begin: Result1 = Even FOR Index = 1 TO 9 STEP 2 IF TestNumber DIG 0 = Index THEN Result1 = Odd Index = 11 ' Exit loop. ENDIF NEXT Index ' To make sure, do another calculation !! Result2 = Odd IF TestNumber DIG 0 = 0 OR TestNumber DIG 0 = 2 OR TestNumber DIG 0 = 4 OR TestNumber DIG 0 = 6 OR TestNumber DIG 0 = 8 THEN Result2 = Even ' And finally, compare the results; IF Result1 = Result2 AND Result1 = Odd THEN FinalResult = Odd LCDOUT $fe,1, "Result1 :",#Result1 LCDOUT $fe,$c0,"Result2 :",#Result2 LCDOUT $fe,$94,"FinalResult:",#FinalResult ELSE FinalResult = Err LCDOUT $fe,1, "Unknown Error Occured!" LCDOUT $fe,$c0,"Reporting to support page..." HSEROUT ["weblink:www.picbasic.co.uk/forum"] PAUSE 100 HSEROUT ["@Msg:alert@bug found"] ENDIF GOTO Begin
- 11th June 2008, 16:13
T.Jackson

Quote:

Originally Posted by sayzer

I also have another one.

Mine does two calculations and compares the results in case the processor is going weird.
And even more, has a bug report feature.

Ski, is this ridiculous enough? I can go deeper.

Code:

TestNumber VAR WORD Index VAR BYTE Result1 VAR BIT Result2 VAR BIT FinalResult VAR BIT Even CON 0 Odd CON 1 Err CON 2 Begin: Result1 = Even FOR Index = 1 TO 9 STEP 2 IF TestNumber DIG 0 = Index THEN Result1 = Odd Index = 11 ' Exit loop. ENDIF NEXT Index ' To make sure, do another calculation !! Result2 = Odd IF TestNumber DIG 0 = 0 OR TestNumber DIG 0 = 2 OR TestNumber DIG 0 = 4 OR TestNumber DIG 0 = 6 OR TestNumber DIG 0 = 8 THEN Result2 = Even ' And finally, compare the results; IF Result1 = Result2 AND Result1 = Odd THEN FinalResult = Odd LCDOUT $fe,1, "Result1 :",#Result1 LCDOUT $fe,$c0,"Result2 :",#Result2 LCDOUT $fe,$94,"FinalResult:",#FinalResult ELSE FinalResult = Err LCDOUT $fe,1, "Unknown Error Occured!" LCDOUT $fe,$c0,"Reporting to support page..." HSEROUT ["weblink:www.picbasic.co.uk/forum"] PAUSE 100 HSEROUT ["@Msg:alert@bug found"] ENDIF GOTO Begin

Reasoning looks good. No need for the second test and comparisons though.

Trent Jackson
- 11th June 2008, 16:25
skimask

Quote:

Originally Posted by sayzer

Ski, is this ridiculous enough? I can go deeper.

Keep going!!!

(What do they call those 'machines' that take the most indirect route to accomplish something? Remember that kids game 'Mousetrap'?)

My next submission:

Code:

testnumber var word onescount var byte onescount1 var byte loopvar var byte testnumber = 12345 main: for loopvar = 0 to 15 'counts bits set in the testnumber if testnumber.0[loopvar] = 1 then onescount = onescount + 1 endif next loopvar testnumber = testnumber >> 1 'shifts the LSB out testnumber = testnumber << 1 'shifts a zero bit back in for loopvar = 0 to 15 'recounts bits set in the testnumber if testnumber.0[loopvar] = 1 then onescount1 = onescount1 + 1 endif next loopvar if onescount = onescount1 then lcdout "EVEN" 'if the number of set bits is the same, its even if onescount <> onescount1 then lcdout "ODD" STOP 'please stop!
- 11th June 2008, 16:31
sayzer

Quote:

Originally Posted by skimask

Keep going!!!

(What do they call those 'machines' that take the most indirect route to accomplish something? Remember that kids game 'Mousetrap'?)

Here they call it "Stupiditer II".

:)
- 11th June 2008, 16:35
skimask

Quote:

Originally Posted by sayzer

Here they call it "Stupiditer II".
:)

Rube Goldberg machines...

http://en.wikipedia.org/wiki/Rube_Goldberg_machine
- 11th June 2008, 16:49
SteveB

Here's another approach:

Code:

TestNumber as word TestVal as word If TestNumber = 0 then Zero TestVal = 2 Test: If TestNumber = TestVal Then LCDOUT "Even" EndProgram Else If TestNumber = TestVal - 1 then LCDOUT "Odd" EndProgram Else TestVal = TestVal + 2 Test EndiF Endif Zero: LCDOUT "Zero" EndProgram EndProgram: end
- 11th June 2008, 16:50
T.Jackson

Quote:

Originally Posted by skimask

Keep going!!!

(What do they call those 'machines' that take the most indirect route to accomplish something? Remember that kids game 'Mousetrap'?)

My next submission:

Code:

testnumber var word onescount var byte onescount1 var byte loopvar var byte testnumber = 12345 main: for loopvar = 0 to 15 'counts bits set in the testnumber if testnumber.0[loopvar] = 1 then onescount = onescount + 1 endif next loopvar testnumber = testnumber >> 1 'shifts the LSB out testnumber = testnumber << 1 'shifts a zero bit back in for loopvar = 0 to 15 'recounts bits set in the testnumber if testnumber.0[loopvar] = 1 then onescount1 = onescount1 + 1 endif next loopvar if onescount = onescount1 then lcdout "EVEN" 'if the number of set bits is the same, its even if onescount <> onescount1 then lcdout "ODD" STOP 'please stop!

That submission is void. The method for direct bit testing has already recognized. You were actually in the running for first prize of ($100,000) for your first submission. But unfortunately you have been suspended from the competition for acts plagiarism.

Naughty, naughty

Trent Jackson
- 11th June 2008, 16:57
T.Jackson

Quote:

Originally Posted by SteveB

Here's another approach:

Code:

TestNumber as word TestVal as word If TestNumber = 0 then Zero TestVal = 2 Test: If TestNumber = TestVal Then LCDOUT "Even" EndProgram Else If TestNumber = TestVal - 1 then LCDOUT "Odd" EndProgram Else TestVal = TestVal + 2 Test EndiF Endif Zero: LCDOUT "Zero" EndProgram EndProgram: end

Excellent! That's called a recursive method.

Trent Jackson
- 11th June 2008, 17:00
skimask

Quote:

Originally Posted by T.Jackson

The method for direct bit testing has already recognized.

Aye lad...but if you'll notice...the direct bit checking is not used to determine the final result! :D A bit like using a pencil to design a machine to manufacture more pencils.
This could get fun...
- 11th June 2008, 17:06
SteveB

Quote:

Originally Posted by skimask

Aye lad...but if you'll notice...the direct bit checking is not used to determine the final result! :D A bit like using a pencil to design a machine to manufacture more pencils.
This could get fun...

My approach is also somewhat indirect, in that it does not directly check if the number is odd. Rather, it sequentially checks to see if the test number matches a value which is known to be either even or odd.
- 11th June 2008, 17:13
T.Jackson

Quote:

Originally Posted by SteveB

My approach is also somewhat indirect, in that it does not directly check if the number is odd. Rather, it sequentially checks to see if the test number matches a value which is known to be either even or odd.

You're pretty much doing the same as Sayzer (comparing with known to be odd / even numbers) -- but the way in which you have it configured to call itself repeatedly until the operation has completed is considered to be recursion. However, normally recursive methods perform the work on the subject and not references.

Trent Jackson
- 11th June 2008, 17:22
SteveB

Another recursive method. However this time it uses the unique feature of Variable Rollover. Also, if nothing is found at the end of checking all possible values for a WORD, the outcome must be Zero.

Code:

TestNumber as word TestVal as word For TestVal = 65534 to 2 step -2 If (TestNumber + TestVal)= 0 Then LCDOUT "Even" EndProgram Else If (TestNumber + TestVal) + 1 = 0 then LCDOUT "Odd" EndProgram Else Endif Next TestVal LCDOUT "Zero" EndProgram: end

EDIT: This may be premature, I think I left a hole at the value of "65535". Standby
- 11th June 2008, 17:28
T.Jackson

Quote:

Originally Posted by SteveB

Another recursive method. However this time it uses the unique feature of Variable Rollover. Also, if nothing is found at the end of checking all possible values for a WORD, the outcome must be Zero.

Code:

TestNumber as word TestVal as word For TestVal = 65534 to 2 step -2 If (TestNumber + TestVal)= 0 Then LCDOUT "Even" EndProgram Else If (TestNumber + TestVal) + 1 = 0 then LCDOUT "Odd" EndProgram Else Endif Next TestVal LCDOUT "Zero" EndProgram: end

I don't consider that to be recursive. To be recursive means to have a procedure (method in Java's case) -- that repeatedly calls itself until the operation is complete.

Kinda like getting a pie that you need to cut up into (n) pieces. Instead of cutting out each piece you cut the whole thing repeadedly until you have (n) pieces.

Trent Jackson
- 11th June 2008, 17:31
SteveB

OK, This is better. It has to do one final check after the FOR..NEXT loop.

Code:

TestNumber as word TestVal as word For TestVal = 65534 to 2 step -2 If (TestNumber + TestVal)= 0 Then LCDOUT "Even" EndProgram Else If (TestNumber + TestVal) + 1 = 0 then LCDOUT "Odd" EndProgram Else Endif Next TestVal If (TestNumber + TestVal) - 1 = 0 then LCDOUT "Odd" EndProgram Else LCDOUT "Zero" Endif EndProgram: end
- 11th June 2008, 17:40
T.Jackson

Recursive example ...

Code:

public int divideBy(int n) { num = num / n if(num > 0) divideBy(n); }

It keeps calling itself until we have arrived at something. In PBP you would of course need to exchange the "public int divideBy(int n)" with a label and swap "divideBy(n)" with a goto.

Trent Jackson
- 11th June 2008, 17:53
precision

Hi Trent, I think...

There'r Out of 10's methods. You want's onlyup to 10th.
So should Listing now.

.