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HP Forum Archive 19

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Re: Mini Challenge
Message #1 Posted by Thomas Klemm on 28 Jan 2009, 9:30 p.m.

First I was looking for LBL B in your program for the HP 15C but then I realized that this is not the only problem. I tried to fix it leaving as much as possible unchanged with the hope it's what you had in mind:

001 - 42,21,11   LBL A
002 -    44  0   STO 0
003 -        1   1
004 -       26   EEX
005 -        6   6
006 -       16   CHS
007 -    44  1   STO 1
008 - 42,21,12   LBL B
009 -    45  0   RCL 0
010 -   43  11   x2
011 -        1   1
012 -       40   +
013 -        2   2
014 -       10   ÷
015 -    44  2   STO 2
016 -       11   SQRT x
017 -    45  1   RCL 1
018 -       40   +
019 -   43  44   INT
020 -   43  11   x2
021 -    45  2   RCL 2
022 - 43,30, 6   TEST 6
023 -   22  13   GTO C
024 -        1   1
025 -       30   -
026 -        2   2
027 -       20   ×
028 -       31   R/S
029 - 42,21,13   LBL C
030 -        2   2
031 - 44,40, 0   STO + 0
032 -   22  12   GTO B

It has to be started with:

1 GSB A

Since internally a digit by digit method is used to calculate the square root I'm confident that taking the root of a square-number won't show any rounding errors. That's different from using an iterative algorithm and might be even worse when a binary representation for the numbers is used.
Therefore you can confidently omit lines 003-007 and 017-018 from the program above.

      
Re: Mini Challenge
Message #2 Posted by Chris Dean on 29 Jan 2009, 2:54 a.m.,
in response to message #1 by Thomas Klemm

Thanks for pointing out the typos.


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