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Comment by rahimnathwani | Hacker News Reader

  Does one stay locked in place? Unclear.

If you set C1=A1+B1 then, when you set a value for C1, A1 and B1 are each half of that value, even if they started off unbalanced.

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It would make more sense to preserve the ratio if possible.

Yeah, this concept is interesting but the fact that the simplest test case gives what's fundamentally a surprising result is very annoying.

It also doesn't help that in the example, the expected outcome of 53.3333/46.6667 isn't even considered.

You can do this with bidicalc already! You just have to model the problem correctly. If you expect the ratio to remain constant, what you actually want is a problem with a single free variable: the scale.

    A1 = 1.0       // the scale, your variable
    A2 = 6 * A1    // intermediate values
    A3 = 8 * A1 
    A4 = A2 + A3   // the sum

Now update A4 (or any other cell!) and the scale (A1, the only variable) will update as you expect.

To get that, you could pass the ratio explicitly. C = 5A + 7B

What is inputs a,b,c,d,and e are polynomial coefficients? I am hoping to get a fields medal plz respond.

You can actually solve fifth order polynomials with bidicalc! But it's a numerical solution, not an algebraic one, so no Fields medal.

I think it would be good if you could lock one of them.

You can.

With # (octothorpe) #Val

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