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Let's say I have a formula like $ax + by + cz = N$. $a, b, c$, and $N$ are known and cannot be changed. $x, y$, and $z$ are known and can be changed.

The problem is that the equation is not true! My problem (for a program I'm writing) is: how can $x, y$, and $z$ be changed enough that they equal $N$ while differing from their previous values as little as possible?

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You could find all the possible values for $x,y,z$ that satisfy the (otherwise unchangeable) equation, and then choose the values that are closest to what you originally wanted. This is an optimization problem, and not terribly difficult to solve. However you need to specify an objective function, which means how do you measure the distance between $(x,y,z)$ that works and your original $(x_0,y_0,z_0)$. – hardmath Jan 5 at 20:05
Wouldn't it be easier to express $z$ as a function of $x$ and $y$? Alternatively, allow a couple to change while the other is fixed for a general idea here. – JB King Jan 5 at 20:07
Do you want $x,y,z$ real or integers????????????? – Will Jagy Jan 5 at 20:21
"while differing from their previous values as little as possible". There are more than one way to minimize three values.. You need to decide which norm you want L1 minimizes the sum of the three differences, L2 minimizes the sum of their squares, and Linfinity minimizes the maximum difference. Suppose that their values could be changed like any of the follow to meet N: (0,0,+2),(+0.5,+0.5,+1.1),(+1,+1,+1). Out of those, the first would be the best according to L1 (sum=2), but the second is the best L2 (sum of squares = 1.71), and the third is the best Linfinity (maximum difference is 1). – Paulpro Jan 5 at 23:31


This is called the "closest vector problem". Also the "nearest lattice point." Given one solution $X_0 = (x_0, y_0,z_0)$ to $ax+by+cz = N,$ you get a "lattice" by taking $X-X_0$ for every solution $X.$

You have a nonsolution, call it $Y = (x_1, y_1,z_1).$ You want the closest lattice point $X-X_0$ to $Y-X_0,$ so that $X-Y$ is shortest.

There is plenty of research on this, and probably some textbooks have good methods in dimensions 2 and 3 for the shortest vector problem and the related closest vector problem.

EDIT: the lattice given is two dimensional. This means that finding a reduced basis is just Gauss reduction for (positive) a binary quadratic form. Once you have a reduced basis, it becomes straightforward to produce a finite list of lattice vectors that must contain the best solution, then check them.

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Is this the discrete equivalent of "distance between a point and a plane"? Or am I missing a subtlety? – Eli Rose Jan 5 at 20:21
@EliRose I thought he wanted integer coordinates, otherwise the task is trivial. I looked at the question again, he makes no such specification; I put in a first line saying "In case..." – Will Jagy Jan 5 at 20:23
Sure, yeah no way of telling. It's funny that a multivariable calc. homework problem in the continuous case becomes the basis of some cryptosystems (as Wikipedia says) in the discrete case! – Eli Rose Jan 5 at 20:26
@EliRose now that I think of it, the lattice involved is two dimensional and so reduction (finding a lattice basis of very short vectors) is easy enough, it amounts to Gauss reduction for a positive binary quadratic form. – Will Jagy Jan 5 at 20:32

The solution space of the equation $ax+by+cz=N$ is a plane in three dimensions, thus for any given point, say, $(x_0,y_0,z_0)$, you may easily calculate the distance between this point and the plane of actual solutions (this is equivalent to the possibly-familiar "least-squares method", which I think is the answer you're after?)

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The least-squares criterion (objective function = sum of squared errors) leads to a solution of a system of linear equations (if $x,y,z$ are allowed to be any real values), which is easy to implement. – hardmath Jan 5 at 21:24

Let $x_0, y_0, z_0$ be the current values of $x, y, z$.

An equation of the form $ax + by + cz = N$ specifies a plane in space. Let your plane be $P$. So your problem is equivalent to the following one: in space, given a point $(x_0, y_0, z_0)$ and a plane $P$, find the point on $P$ that is closest to $(x_0, y_0, z_0)$.

The closest point will be the point you get by dropping a perpendicular line from $(x_0, y_0, z_0)$ to $P$ and seeing where it lands. (This must be the closest because a straight line is the shortest distance between any two points!)

What I just described is also called the orthogonal projection of a point onto a plane. It's solved e.g. here.

As charlestoncrabb mentions, this is equivalent to the method of least squares because we measure distance in space by $\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2}$. So that's the "default" meaning of "close to" in this context, oddly enough. If want to measure "close to" some other way, you can use more general optimization methods (like Lagrange multipliers).

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Define $A = [a,b,c]$ and $X=[x, y, z]^T$, so your equation is $A*X = N$. Assume your known point to be $ \bar X = [x_0, y_0,z_0]^T$. Define the residual to be $R = N - A*\bar X$, since the equation is not true. The question becomes $$\min|| X- \bar X ||,s.t. A*X = N$$

The solution $X^*$ that minimize the distance is $X^* = \bar X + A^+R$ where $A^+$ is the Moore_Penrose pseudo inverse of $A$ and is this case since $A$ has linearly independent rows, $A^+ = A^T(AA^T)^{-1}$. The distance is $||X^* - \bar X || = || A^+R||$, you can pick whatever norm you like.

This is also true if $A$ is a matrix instead of a row vector suppose you have a linear equation system.

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