KixGolf: Luhn's Mod - Public Round - KiXtart.org

You are not logged in. [Log In] KiXtart.org website » Forums » KiXtart » Lounge » KixGolf: Luhn's Mod - Public Round

Page 1 of 6

Topic Options

#199427 - 2010-08-08 06:08 PM KixGolf: Luhn's Mod - Public Round
Allen Allen KiX Supporter Registered: 2003-04-19 Posts: 4549 Loc: USA	Gentleman, please post your code.
Top

#199428 - 2010-08-08 06:12 PM Re: KixGolf: Luhn's Mod - Public Round [Re: Allen]
maciep maciep Korg Regular Registered: 2002-06-14 Posts: 947 Loc: Pittsburgh	Here's my 137 code Code: Function A($) dim $s,$f $a = 0 if $>3 & $<7 & len($)+($<4)=16 while $ $a = right($,1) $s = $s + $f(-9(4<$a)+$a) + $a $ = left($,-1) $f=1-$f if $s mod 10 $a =0 Endfunction _________________________ Eric
Top

#199429 - 2010-08-08 06:36 PM Re: KixGolf: Luhn's Mod - Public Round [Re: maciep]
Allen Allen KiX Supporter Registered: 2003-04-19 Posts: 4549 Loc: USA	Hey Eric, would you mind explaining how you coded around these tests? 3456789012345675=0 valid luhn but 16 digits 456789012345671=0 valid luhn but 15 digits 567890123456781=0 "" 678901234567899=0 "" Edit...actually maybe its just a valid Amkixpress I would be interested in. Edited by Allen (2010-08-08 06:48 PM)
Top

#199430 - 2010-08-08 06:39 PM Re: KixGolf: Luhn's Mod - Public Round [Re: maciep]
DrillSergeant DrillSergeant MM club member Registered: 2004-07-09 Posts: 1164 Loc: Eijsden, the Netherlands	Hey Maciep, our codes aren't that different! 124: Code: Function A($) Dim $c, $t While $ $A = 0 + right($, 1) $t = 1 & $c * ($A / 5 + $A) + $A + $t $c = $c + 1 $ = left($,~) If $t mod 10 \| $c-15-$A/4 \| 4&$A + 5 $A = 0 Endfunction _________________________ The Code is out there
Top

#199431 - 2010-08-08 06:49 PM Re: KixGolf: Luhn's Mod - Public Round [Re: DrillSergeant]
Allen Allen KiX Supporter Registered: 2003-04-19 Posts: 4549 Loc: USA	@Drill... Pure encryption... I love it $\:\)$ It'll take a while for me to figure out that one.
Top

#199432 - 2010-08-08 07:04 PM Re: KixGolf: Luhn's Mod - Public Round [Re: Allen]
DrillSergeant DrillSergeant MM club member Registered: 2004-07-09 Posts: 1164 Loc: Eijsden, the Netherlands	The two real kickers I added today were the binary calcutions: 1&$c and 4&$A + 5 The first one checks if the first bit of the digit is used, so this is true for every odd number and false for every even number. 1 = 0 0 0 1 = true 2 = 0 0 1 0 = false 3 = 0 0 1 1 = true 4 = 0 1 0 0 = false The second one is just like that. the variable $A kan be anything from 0 to 9. Because I needed a range of 4 digits to check I check against the 3rd bit(= weight=4). (3, 4, 5, 6) since my range starts at 3 but the first number with a false 3rd bit is 8 I need to add 5 to my number to check. so if we continue the table above: 2 + 5 = 7 = 0 1 1 1 = true (third bit = 1) Not correct 3 + 5 = 8 = 1 0 0 0 = false 4 + 5 = 9 = 1 0 0 1 = false 5 + 5 = 10 = 1 0 1 0 = false 6 + 5 = 11 = 1 0 1 1 = false 7 + 5 = 12 = 1 1 0 0 = true (third bit = true again) Not correct Does this make it any clearer? _________________________ The Code is out there
Top

#199433 - 2010-08-08 07:22 PM Re: KixGolf: Luhn's Mod - Public Round [Re: DrillSergeant]
maciep maciep Korg Regular Registered: 2002-06-14 Posts: 947 Loc: Pittsburgh	Quote: Hey Eric, would you mind explaining how you coded around these tests? 3456789012345675=0 valid luhn but 16 digits 456789012345671=0 valid luhn but 15 digits 567890123456781=0 "" 678901234567899=0 "" Edit...actually maybe its just a valid Amkixpress I would be interested in. I set $a to 0 by default. And then the if statement decides whether I even try to process string. Code: if $>3 & $<7 & len($)+($<4)=16 the first condition ensures that string doesn't start with 0-2. The second condition makes sure it doesn't start with 7-9. And the last condition makes sure that the length is 16 characters...but we need to add one if it starts with a 3. Since we are already discard 0-2, we can just check that it's less than 4 instead of exactly 3, so that condition will add 1 when appropriate. So if starts with a 3 and isn't 15 characters long, we never drop into the loop. Nice code Drill!! Definitely similar to mine, but you're golfing skillz are impressive! $\:\)$ _________________________ Eric
Top

#199434 - 2010-08-08 07:36 PM Re: KixGolf: Luhn's Mod - Public Round [Re: maciep]
DrillSergeant DrillSergeant MM club member Registered: 2004-07-09 Posts: 1164 Loc: Eijsden, the Netherlands	By the way, this is another sneaky move too: ($A / 5 + $A) If $A is a digit of 5 or greater, you need to add the two digits to get the relevant single digit number. The calculation above doesn't do that, but just adds another 1 if the number is 5 or greater. This will give: 5 + 5 = 10 + 1(because of $A/5) = 11` 6 + 6 = 12 + 1 = 13 7 + 7 = 14 + 1 = 15 Now for the check if it's divisble by 10 it doesn't matter if you add another extra 10 to the total, so I had the same calculation as Eric, (-9 * ($A>4)) but replaced that (also today) for the calculation above. _________________________ The Code is out there
Top

#199435 - 2010-08-08 07:46 PM Re: KixGolf: Luhn's Mod - Public Round [Re: DrillSergeant]
Benny69 Benny69 Moderator Registered: 2003-10-29 Posts: 1036 Loc: Lincoln, Ne	here is my 185 code: Code: Function A($) Dim $y,$x,$v $x = Left($,1) For $y = ($^) to 1 step -2 $a = 2 * SubStr($,$y-1,1) $v = IIf($a > 9,$a / -10 + $a,$a) + $v + SubStr($,$y,1) $a = IIf(~ $v mod 10 & (($^) = 15 & $x = 3) \| (($^) = 16 & $x > 3 & $x < 7),$x,0) ;Next EndFunction Edited by Benny69 (2010-08-08 07:56 PM) _________________________ Wait don't order yet,... get KiXforms Designer .NET 2.0 (Beta) KiXforms Designer .NET 2.0 (Beta)
Top

#199436 - 2010-08-08 07:47 PM Re: KixGolf: Luhn's Mod - Public Round [Re: DrillSergeant]
Allen Allen KiX Supporter Registered: 2003-04-19 Posts: 4549 Loc: USA	So my question then is this... if $>3 & $<7 wouldn't that only pickup 4 - 6? Edit... just saw this... "And the last condition makes sure that the length is 16 characters...but we need to add one if it starts with a 3. Since we are already discard 0-2, we can just check that it's less than 4 instead of exactly 3, so that condition will add 1 when appropriate. " Hmmm... need to process... cool though. Edited by Allen (2010-08-08 07:50 PM)
Top

#199437 - 2010-08-08 07:49 PM Re: KixGolf: Luhn's Mod - Public Round [Re: Benny69]
Benny69 Benny69 Moderator Registered: 2003-10-29 Posts: 1036 Loc: Lincoln, Ne	here is Eric's code at 135: Code: Function A($) Dim $s,$f $a = 0 If $ > 3&$ < 7&($^)+($ < 4) = 16 While $ $a = Right($,1) $s = $s+$f(-9(4 < $a)+$a)+$a $ = Left($,-1) $f = 1-$f If $s mod 10 $a = 0 EndFunction Edited by Benny69 (2010-08-08 07:56 PM) _________________________ Wait don't order yet,... get KiXforms Designer .NET 2.0 (Beta) KiXforms Designer .NET 2.0 (Beta)
Top

#199438 - 2010-08-08 07:57 PM Re: KixGolf: Luhn's Mod - Public Round [Re: Benny69]
Benny69 Benny69 Moderator Registered: 2003-10-29 Posts: 1036 Loc: Lincoln, Ne	here is my other 185 code: Code: Function A($) Dim $z,$x,$v $z = Left($,1) $v = ($^) While $ $a = 2 * SubStr($,($^)-1,1) $x = IIf($a > 9,$a / -10 + $a,$a) + Right($,1) + $x $ = Left($,-2) $a = IIf(~ $x mod 10 & ($v = 15 & $z = 3) \| ($v = 16 & $z > 3 & $z < 7),$z,0) ; Loop EndFunction Edited by Benny69 (2010-08-08 07:57 PM) _________________________ Wait don't order yet,... get KiXforms Designer .NET 2.0 (Beta) KiXforms Designer .NET 2.0 (Beta)
Top

#199439 - 2010-08-08 08:02 PM Re: KixGolf: Luhn's Mod - Public Round [Re: DrillSergeant]
Lonkero Lonkero KiX Master Guru Registered: 2001-06-05 Posts: 22346 Loc: OK	lol, damn. the first one I was thinking about too, never used. second one. you have way too much energy! $\:D$ I have 2 versions of 134, I kinda like this the best: Code: Function A($b) dim $t,$,$x for $t=0 to 99 $=right($b,1) $x=$t mod 2((4<$)+$)+$+$x $b=left($b,~) if $ $a=($t=15-(3=$) & 2<$ & 7>$ & $x mod 10=0)$ Endfunction _________________________ ! download KiXnet
Top

#199440 - 2010-08-08 08:06 PM Re: KixGolf: Luhn's Mod - Public Round [Re: Lonkero]
Benny69 Benny69 Moderator Registered: 2003-10-29 Posts: 1036 Loc: Lincoln, Ne	Eric's at 134 Code: Function A($) Dim $s,$f $a = 0 If $ > 3&$ < 7&($^)+($<4) = 16 While $ $a = Right($,1) $s = $s+$f(-9(4 < $a)+$a)+$a $ = Left($,~) $f = 1-$f If $s mod 10 $a = 0 EndFunction Edited by Benny69 (2010-08-08 08:07 PM) _________________________ Wait don't order yet,... get KiXforms Designer .NET 2.0 (Beta) KiXforms Designer .NET 2.0 (Beta)
Top

#199441 - 2010-08-08 09:27 PM Re: KixGolf: Luhn's Mod - Public Round [Re: Benny69]
maciep maciep Korg Regular Registered: 2002-06-14 Posts: 947 Loc: Pittsburgh	Quote: if $>3 & $<7 wouldn't that only pickup 4 - 6? nope '3xxxxx' is greater than 3 too...found that out when $>2 failed a few cases $;\)$ _________________________ Eric
Top

#199442 - 2010-08-08 09:39 PM Re: KixGolf: Luhn's Mod - Public Round [Re: maciep]
maciep maciep Korg Regular Registered: 2002-06-14 Posts: 947 Loc: Pittsburgh	Quote: Now for the check if it's divisble by 10 it doesn't matter if you add another extra 10 to the total, so I had the same calculation as Eric, (-9 * ($A>4)) but replaced that (also today) for the calculation above. I KNEW there had to be a better way to do that...I like this one! I was pumped when I thought 2*$a mod 9 would work for all numbers...of course it failed for 9 _________________________ Eric
Top

#199443 - 2010-08-08 10:28 PM

Re: KixGolf: Luhn's Mod - Public Round [Re: maciep]

BradV

Seasoned Scripter

Registered: 2006-08-16
Posts: 686
Loc: Maryland, USA

OK, I knew I was missing something! I'm embarrased to even post this. $\:\)$

Code:

Function A($)
   dim $l, $i, $b, $c, $x, $d
   $l = Len($)
   $d = Left($, 1)
   If ($d = 3 and $l = 15) or ($d > 3 and $d < 7 and $l = 16)
      $x = 0
      For $i = $l - 1 to 0 Step -1
         $b = Substr($, $i + 1, 1)
         If ($l - $i) mod 2 = 0
            $c = 2 * $b
            $b = IIF($c > 9, 0 + Left($c, 1) + Right($c, 1), $c)
         EndIf
         $x = $x + $b
      Next
      $d = IIF($x mod 10, 0, $d)
   Else
      $d = 0
   EndIf
   $A = $d
EndFunction

Top

#199445 - 2010-08-08 11:08 PM Re: KixGolf: Luhn's Mod - Public Round [Re: BradV]
Allen Allen KiX Supporter Registered: 2003-04-19 Posts: 4549 Loc: USA	Almost forgot... here is Jochen's 146 Code: function a($) dim $b, $c, $d $b = 0^$ while $ $c = (($b-($^) & 1)+1) * right($,1) $d = $d + $c-9($c>9) $a = ($d mod 10 =) (($b=15&$=3)\|$b=16&$>3&$<7) * $ $ = left($,~) endfunction
Top

#199446 - 2010-08-09 12:51 AM Re: KixGolf: Luhn's Mod - Public Round [Re: Allen]
Benny69 Benny69 Moderator Registered: 2003-10-29 Posts: 1036 Loc: Lincoln, Ne	Brad, Don't feel embarrased, you should feel proud, most can't even put somthing together that works, that is a BIG hurtle. I say well done, now lets see what we can do to shorten that code. _________________________ Wait don't order yet,... get KiXforms Designer .NET 2.0 (Beta) KiXforms Designer .NET 2.0 (Beta)
Top

#199447 - 2010-08-09 01:02 AM

Re: KixGolf: Luhn's Mod - Public Round [Re: Benny69]

Benny69 Offline

Moderator

Registered: 2003-10-29
Posts: 1036
Loc: Lincoln, Ne

Brad's code at 240:

Code:

Function A($)
   Dim $l,$i,$b,$c,$x,$d
   $l = ($^);Len($)
   $d = Left($,1)
   ;If($d = 3 And $l = 15) Or ( $d > 3 And $d < 7 And $l = 16 )
   If($d = 3 & $l = 15) | ( $d > 3 & $d < 7 & $l = 16 ); use Bitwise Operators to do the same tasks as 'And' and 'Or'
      $x = 0
      For $i = $l-1 to 0 Step-1
         $b = SubStr($,$i+1,1)
         If($l-$i) mod 2 = 0
            $c = 2*$b
            ;$b = IIf($c > 9,0+Left($c,1)+Right($c,1),$c)
            $b = IIf($c > 9,$c / 10 + $c - 10,$c); use a calulator to see that Kix drops the decimal for $c
         EndIf
         $x = $x + $b
      Next
      $d = IIf($x mod 10,0,$d)
   Else
      $d = 0
   EndIf
   $A = $d
EndFunction

_________________________
Wait don't order yet,... get KiXforms Designer .NET 2.0 (Beta)
KiXforms Designer .NET 2.0 (Beta)

Top

Page 1 of 6

Last »

Previous Topic View All Topics

Index Next Topic

Moderator: Arend_, Allen, Jochen, Radimus, Glenn Barnas, ShaneEP, Ruud van Velsen, Mart

Hop to:

Shout Box

Search the board with:
superb Board Search
or try with google:

	Web kixtart.org