
How about just moving the '11000000' strings into an 8element array and xFoot the array? If the result if 2, you've got a 2bit number of it is 3 you've got your 3bit numbers. Original Message From: mi400bounces@xxxxxxxxxxxx [mailto:mi400bounces@xxxxxxxxxxxx] On Behalf Of Dan Bale Sent: Monday, September 20, 2004 9:42 AM To: MI400@xxxxxxxxxxxx Subject: [MI400] odd sort of bitcounter... Esteemed listers: I have a scenario where I would like to test various combinations of numbers in an array and sum them to find a total amount to solve a reconciliation problem. So, for example, I have the following array of numbers: 185.69 11,134.60 500.03 4,841.65 1,500.02 419.85 etc. In this example, I am looking for any combination that adds up to 13,134.65. A good tool would be able to find that the 2nd, 3rd, and 5th amounts add up to the total amount I'm looking for. In a reallife application, this problem would involve an array of hundreds, if not thousands, of amounts, usually to find a small number of amounts to sum up to the difference needed to reconcile. Joe Pluta, in a rpg400l thread, had suggested a solution that is somewhat workable, but I haven't figured out how to sum all of the 2amount combinations before I start with the 3amount combinations. In essence, Joe suggests creating an array (ARR1) loaded with all of the amounts in the list, a second array (ARR2) whose elements will contain either 1 or 0, and a third array (ARR3) that contains the multiplication result of the first two. So, using the example above: ARR1 ARR2 ARR3 185.69 0 0.00 11,134.60 1 11,134.60 500.03 1 500.03 4,841.65 0 0.00 1,500.02 1 1,500.02 419.85 0 0.00 ========= 13,134.65 ARR2 would be a parsed out binary value with element 1 having the loworder digit, i.e. '010110' in this case. ARR2 seems like it should be some sort of a binarycalculated counter, but it isn't as easy if I'm incrementing by one and I want to exhaust all 2amount combinations before I test any 3amount combinations. So, I would like to try utilize something like: 00000011 = 3 00000101 = 5 00000110 = 6 00001001 = 9 00001010 = 10 00001100 = 12 ... 11000000 = 192 <= Highest of the 2digit binary for 8 bits 00000111 = 7 <= The first 3digit binary 00001011 = 11 00001101 = 13 00001110 = 14 00010011 = 19 ... 11100000 = 224 <= Highest of the 3digit binary for 8 bits (Of course, I'd need as many bits as the number of amounts I'm dealing with.) Is there an easy way to "imcrement" like this? I haven't been able to think my way out of this box, so if there are any "outside" thinkers, please weigh in. It seems that this type of problem can't be easily solved in RPG. I was hoping that MI has something that facilitate the type of solution I'm thinking of. tia, db _______________________________________________ This is the MI Programming on the AS400 / iSeries (MI400) mailing list To post a message email: MI400@xxxxxxxxxxxx To subscribe, unsubscribe, or change list options, visit: http://lists.midrange.com/mailman/listinfo/mi400 or email: MI400request@xxxxxxxxxxxx Before posting, please take a moment to review the archives at http://archive.midrange.com/mi400.
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