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.
PBP code ...Code:public boolean isNumOdd(int num) { if (1 & num) return true; else return false; }
Has anyone got another way of doing this? I'm incredibly keen to hear your thoughts.Code:main: if 1 & num then oddNum = true else oddNum = false end if
Best Regards,
Trent Jackson
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Code:If num.0=1 then oddNum = true else oddNum = false endif
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GrandPa's way!
Code:OddNum = True IF Num // 2 = 0 THEN OddNum = False
Does Num = 1 work?
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"If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte
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Do me a favour and refresh my memory with what "//" does?Originally Posted by sayzer
Trent Jackson
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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...
Last edited by dhouston; - 11th June 2008 at 13:51.
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I knew it did something for sure; (well, at least at where I am it does).
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"If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte
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in VB
Code:If ((Text1.Text Mod 2) = 0) Then MsgBox "Even" Else MsgBox "Odd" End If
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That's pretty uncanny bud, because I just wrote this ...
I reckon that there's a recursive way of doing it too!Code:Dim num As Integer Dim oddNum As Boolean num = 5 If num Mod 2 = 0 Then oddNum = False Else oddNum = True End If
Trent Jackson
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I think only mod formula is easy way to find oddeven.
oddeven = number mod 2
if oddeven = 0 then page is even, otherwise odd
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How about the most rediculous, long winded, round-about way to do it?
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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
Last edited by T.Jackson; - 11th June 2008 at 14:17.
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Any solution that can be verified workable!
Trent Jackson
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Double subtraction
Should work, haven't tried it...but it's surely a waste of resources!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
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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.Should work, haven't tried it...but it's surely a waste of resources!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
Trent Jackson
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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:</i></font>TestNumber <font color="#000080"><b>VAR WORD </b></font>Index <font color="#000080"><b>VAR BYTE </b></font>Result1 <font color="#000080"><b>VAR BIT </b></font>Result2 <font color="#000080"><b>VAR BIT </b></font>FinalResult <font color="#000080"><b>VAR BIT </b></font>Even <font color="#000080"><b>CON </b></font><font color="#FF0000"><b>0 </b></font>Odd <font color="#000080"><b>CON </b></font><font color="#FF0000"><b>1 </b></font>Err <font color="#000080"><b>CON </b></font><font color="#FF0000"><b>2 </b></font>Begin: Result1 = Even <font color="#000080"><b>FOR </b></font>Index = <font color="#FF0000"><b>1 </b></font><font color="#000080"><b>TO </b></font><font color="#FF0000"><b>9 </b></font><font color="#000080"><b>STEP </b></font><font color="#FF0000"><b>2 </b></font><font color="#000080"><b>IF </b></font>TestNumber <font color="#000080"><b>DIG </b></font><font color="#FF0000"><b>0 </b></font>= Index <font color="#000080"><b>THEN </b></font>Result1 = Odd Index = <font color="#FF0000"><b>11 </b></font><font color="#000080"><i>' Exit loop. </i><b>ENDIF NEXT </b></font>Index <font color="#000080"><i>' To make sure, do another calculation !! </i></font>Result2 = Odd <font color="#000080"><b>IF </b></font>TestNumber <font color="#000080"><b>DIG </b></font><font color="#FF0000"><b>0 </b></font>= <font color="#FF0000"><b>0 </b></font><font color="#000080"><b>OR </b></font>TestNumber <font color="#000080"><b>DIG </b></font><font color="#FF0000"><b>0 </b></font>= <font color="#FF0000"><b>2 </b></font><font color="#000080"><b>OR </b></font>TestNumber <font color="#000080"><b>DIG </b></font><font color="#FF0000"><b>0 </b></font>= <font color="#FF0000"><b>4 </b></font><font color="#000080"><b>OR </b></font>TestNumber <font color="#000080"><b>DIG </b></font><font color="#FF0000"><b>0 </b></font>= <font color="#FF0000"><b>6 </b></font><font color="#000080"><b>OR </b></font>TestNumber <font color="#000080"><b>DIG </b></font><font color="#FF0000"><b>0 </b></font>= <font color="#FF0000"><b>8 </b></font><font color="#000080"><b>THEN </b></font>Result2 = Even <font color="#000080"><i>' And finally, compare the results; </i><b>IF </b></font>Result1 = Result2 <font color="#000080"><b>AND </b></font>Result1 = Odd <font color="#000080"><b>THEN </b></font>FinalResult = Odd <font color="#000080"><b>LCDOUT </b></font><font color="#FF0000"><b>$fe</b></font>,<font color="#FF0000"><b>1</b></font>, <font color="#008000"><b>"Result1 :"</b></font>,#Result1 <font color="#000080"><b>LCDOUT </b></font><font color="#FF0000"><b>$fe</b></font>,<font color="#FF0000"><b>$c0</b></font>,<font color="#008000"><b>"Result2 :"</b></font>,#Result2 <font color="#000080"><b>LCDOUT </b></font><font color="#FF0000"><b>$fe</b></font>,<font color="#FF0000"><b>$94</b></font>,<font color="#008000"><b>"FinalResult:"</b></font>,#FinalResult <font color="#000080"><b>ELSE </b></font>FinalResult = Err <font color="#000080"><b>LCDOUT </b></font><font color="#FF0000"><b>$fe</b></font>,<font color="#FF0000"><b>1</b></font>, <font color="#008000"><b>"Unknown Error Occured!" </b></font><font color="#000080"><b>LCDOUT </b></font><font color="#FF0000"><b>$fe</b></font>,<font color="#FF0000"><b>$c0</b></font>,<font color="#008000"><b>"Reporting to support page..." </b></font><font color="#000080"><b>HSEROUT </b></font>[<font color="#008000"><b>"weblink:www.picbasic.co.uk/forum"</b></font>] <font color="#000080"><b>PAUSE </b></font><font color="#FF0000"><b>100 </b></font><font color="#000080"><b>HSEROUT </b></font>[<font color="#008000"><b>"@Msg:alert@bug found"</b></font>] <font color="#000080"><b>ENDIF GOTO </b></font>Begin
Last edited by sayzer; - 11th June 2008 at 16:05.
"If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte
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Reasoning looks good. No need for the second test and comparisons though.
Trent Jackson
Last edited by T.Jackson; - 11th June 2008 at 16:16.
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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!
Last edited by skimask; - 11th June 2008 at 16:34.
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Here they call it "Stupiditer II".
"If the Earth were a single state, Istanbul would be its capital." Napoleon Bonaparte
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Rube Goldberg machines...Here they call it "Stupiditer II".
http://en.wikipedia.org/wiki/Rube_Goldberg_machine
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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
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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
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Excellent! That's called a recursive method.
Trent Jackson
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Aye lad...but if you'll notice...the direct bit checking is not used to determine the final result!A bit like using a pencil to design a machine to manufacture more pencils.
This could get fun...
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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.Aye lad...but if you'll notice...the direct bit checking is not used to determine the final result!
This could get fun...
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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
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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.
EDIT: This may be premature, I think I left a hole at the value of "65535". StandbyCode: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
Last edited by SteveB; - 11th June 2008 at 17:26.
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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
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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
Last edited by SteveB; - 11th June 2008 at 17:35.
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Recursive example ...
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.Code:public int divideBy(int n) { num = num / n if(num > 0) divideBy(n); }
Trent Jackson
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Hi Trent, I think...
There'r Out of 10's methods. You want's onlyup to 10th.
So should Listing now.
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