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October 19th, 2018, 02:23 PM  #1 
Member Joined: Jan 2012 From: Montreal, Canada Posts: 30 Thanks: 0  Formula from table with operation restrictions
I have a mostly automated profile/sheet/information that I'd like to automate even more as it allow custom formulas using limited data and operations. In the following table the column (col) 1 is entered (by the profile/sheet/information) then col 2 is automated and I'd like to do so for col 3; currently col 3 is entered and the user is supposed to verify its correct when col 1 is changed, but it's more than negligibly forgotten so the user doesn't benefit from having a col 3 as high as possible (using the table constraints). If possible I'd end up with a custom formula to generate col 3, the only information it could have is col 1 & 2 and the only available operations are arithmetic (+, , x & /) and rounding (round(), floor() & ceil()). There is no need to figure col 2, it's only there if it helps finding the formula for col 3 (with the restrictions of the previous sentence). I haven't figure out how to enter a table in this forum (BB code for table doesn't seem to work). The following is a TSV, which I imported into a spreadsheet then printed to the attached PDF. Code: 1 2 4 2 2 4 3 2 4 4 2 4 5 3 6 6 3 6 7 3 6 8 3 6 9 4 6 10 4 6 11 4 8 12 4 8 13 5 8 14 5 8 15 5 8 16 5 8 17 6 10 18 6 10 19 6 10 20 6 10 I'm not sure if this is the right forum for this request, moderators please change it if appropriate, or those that know one to send them a suggestion to do so. Thank you kindly for your help 
October 22nd, 2018, 09:54 PM  #2 
Member Joined: Jan 2012 From: Montreal, Canada Posts: 30 Thanks: 0 
The solution was shared with me on another math forum.

October 23rd, 2018, 12:18 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,757 Thanks: 2138 
What did you mean by "function incorrectly" and "the solution"? In the nth row of your table, the value in the second column is given by floor((n + 7)/4), and the value in the third column is given by 2floor((n + 13)/6). 
October 23rd, 2018, 01:10 AM  #4 
Member Joined: Jan 2012 From: Montreal, Canada Posts: 30 Thanks: 0 
Function incorrectly is about the system that await the # from col 3. The solution is Code: col3 = 2*floor( (col1 +1) /6)+4 
October 23rd, 2018, 02:48 AM  #5 
Member Joined: Jan 2012 From: Montreal, Canada Posts: 30 Thanks: 0 
Curiosity got the best of me, so I tried your formula and it does work. I have no idea how this work; do you know whether your formula or the one I posted is better?
Last edited by skipjack; October 23rd, 2018 at 07:40 AM. 
October 23rd, 2018, 07:36 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,757 Thanks: 2138 
As 2*floor( (col1 +1) /6)+4 = 2(floor((col1 +1) /6) + 2*6/6) = 2*floor((col1 + 1 + 2*6) /6)) = 2floor((col1 + 13)/6), the two are mathematically equivalent. I chose the shorter one.


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formula, operation, restrictions, table 
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