My Math Forum Formula from table with operation restrictions

 Algebra Pre-Algebra and Basic Algebra Math Forum

October 19th, 2018, 03:23 PM   #1
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Joined: Jan 2012

Posts: 22
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
Step by step: In the profile/sheet/information, a user goes row-by-row, hopefully going all the way to the end but might stop along the way, stopping would have nothing to do with the data. The user currently have to enter col 1 & 3 (as they go along, thus row-by-row), in the profile/sheet/information what's related to col 2 is automated and will always reflect what's in the table (whatever is in col 1, it will become col 2 of the corresponding row). As mentioned currently the data in the profile/sheet/information related to col 3 has to be entered manually with the corresponding disadvantages; it will function incorrectly if a user forget to modify it after changing the data corresponding to col 1, or do so incorrectly, it will be less efficient if it's lower than the corresponding row col 3, it will be breaking the rules if it's higher, so it's better if it's lower than higher, but it would be nice to have the exact data. I would like an algorithm to automatically modify the data related to col 3 so there's no chance of a user forgetting to change it after changing col 1; the only operations allowed are in the 4th sentence, italicized.

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
Attached Files
 print-1-4_4,5-10_6,11-16_8,17-20_10-181019-1-prepress.pdf (9.7 KB, 2 views)

 October 22nd, 2018, 10:54 PM #2 Newbie   Joined: Jan 2012 From: Montreal, Canada Posts: 22 Thanks: 0 The solution was shared with me on another math forum.
 October 23rd, 2018, 01:18 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,868 Thanks: 1833 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). Thanks from DynV
 October 23rd, 2018, 02:10 AM #4 Newbie   Joined: Jan 2012 From: Montreal, Canada Posts: 22 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, 03:48 AM #5 Newbie   Joined: Jan 2012 From: Montreal, Canada Posts: 22 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 08:40 AM.
 October 23rd, 2018, 08:36 AM #6 Global Moderator   Joined: Dec 2006 Posts: 19,868 Thanks: 1833 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. Thanks from DynV

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