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I have two pieces of code that produce a bunch of real numbers, say $A$ and $B$ respectively. (It is not relevant to the question, but $A$ consists of eigenvalues of the Hamiltonian of some physical system, so these are all energies, and $B$ are some energies computed based on an ansatz.) In theory, it should be the case that $B \subset A$.

Because these sets seem to grow quite fast, and because the code is prone to bugs (like $+1$ instead of $-1$, or need for assumptions that some parameter does not become $0$), I would very much like to have an automated way of checking if the inclusion $B \subset A$ really holds for the produced sets.

If $A$ and $B$ were just sets of, say, integers, it would suffice to check that: Complement[B,A]=={} evaluates to True. However, we are dealing with reals in finite precision arithmetic, so A and B consist merely of approximately equal numbers. I tried a workaround like first projecting them to lower precision, and then doing the check: Complement[N[B,4],N[A,4]]=={} (the number of digits that agree is always much greater than $4$ so far, and if there is a difference because of a bug, it has always been within the first couple of digits). Unfortunately, this does not work reliably (I suspect because $A$ and $B$ might be rounded differently if close to some boundary value). Also, this approach isn't too elegant. Could someone suggest a better way?

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Did you happen to notice the SameTest option of Complement[]? That should allow you to accommodate fuzzier comparisons. – J. M. May 29 '13 at 7:52
@J.M. I did not, thank you for pointing this out. I will try this and report if it worked. – Jakub Konieczny May 29 '13 at 8:02

1 Answer 1

up vote 2 down vote accepted

It seems like what you want to know is, for every element of B, is there an element of A that is very close. This can be answered straightforwardly using the Nearest function. First, create some data A and pick a small number of those to be B (we add some small noise to B so that they are not identical).

a = RandomReal[{-3, 100}, 100];
ind = RandomChoice[Range[100], 7];
b = a[[ind]] + RandomReal[{-0.1, 0.1}, Length[ind]];

Now you can build a nearest function and do the check:

nf = Nearest[a];
Flatten[nf /@ b] - b

The output will be something like this:

{-0.05925, 0.0654228, -0.00150679, -0.00979743, -0.016876, -0.0564278, 0.0121807}

which shows you exactly how far away from each of the B's the corresponding element of A is. You can then decide if any of these are too far.

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