# Equation solving in c++

Discussion in 'Programming & Webmastering' started by caleb, Mar 16, 2011.

1. ### caleb

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Hello!

The business problem is calculating certain Task time's of workers. I have only their total time per day and the amount task's they did in that time those task are split into types (Its that 18 variable's I already mentioned). An example:
8x+4y+0z+...+5n=4300[seconds]
8x+0y+0z+....+0n=1000[seconds]

Yes some workers have only 1 task done and other don't but that's not a problem as I can zero-fill those spot's or easily ignore unsolvable equations.

I'm trying to solve 2000 equations (2000workers/day) with 15 variables (15 diffrent tasks) . I found an algorithm that solves an equation (Gauss elimination) so that part I have behind now I'm puzzled how to calculate all combination's of those equations. I need to match each equation with all solvable combination's of others. I want have a all the possible variable values as output then I can run some statistical program to find my sweet spot for the task time but that's another story.
Any nice c++ trick you could suggest ?

2. ### FourstaffModeratorStaff Member

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If you have 15 unknowns, you will need at most 15 equations to solve the "system".

With that in mind, it means that you will have to readjust the order of your 2000 equations so that you run your Gaussian elimination for every combination of 15 equations, and its 2000C15 so you will need to test for 23773578694372656339690207516593086800 number of different combinations, but that will get all of your answers, with plenty of repeats.

You might possibly eliminate the 2000 equations by repeats (ie 1 of them is x times another), leaving you with a much more manageable number of equations to solve. Out of ideas for now, will come back if I have any more.

Edit:
If you have learned "basis" sets in linear algebra, then you might be able to use the concept, find all the basis sets in the 2000 equations, and then run the Gaussian elimination on each of the basis set to get your answers. Problem is that there are just about as many iterations you will need to do

3. ### Drone

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