On Fri, Sep 6, 2013 at 6:50 PM, James Bergstra <bergstrj@iro.umontreal.ca> wrote:

>

> Thanks, this is exactly what I was looking for. I'll look into what this Diophantine equation is.

Let's say we have two arrays with shape tuples `shape0` and `shape1`, stride tuples `stride0` and `stride1` and memory offsets `offset0` and `offset1`. Without loss of generality, let's assume that itemsize=1 since it is trivial to convert the general case to itemsize=1. Now, if you will permit Einstein summation notation, you can generate the memory address of every item in the first array like so:

index0[i]*stride0[i] + offset0

0 <= i < len(shape0)

0 <= index0[i] < shape0[i]

There exists an overlap between the two arrays iff there exists two tuples `index0` and `index1` such that

>

> Thanks, this is exactly what I was looking for. I'll look into what this Diophantine equation is.

Let's say we have two arrays with shape tuples `shape0` and `shape1`, stride tuples `stride0` and `stride1` and memory offsets `offset0` and `offset1`. Without loss of generality, let's assume that itemsize=1 since it is trivial to convert the general case to itemsize=1. Now, if you will permit Einstein summation notation, you can generate the memory address of every item in the first array like so:

index0[i]*stride0[i] + offset0

0 <= i < len(shape0)

0 <= index0[i] < shape0[i]

There exists an overlap between the two arrays iff there exists two tuples `index0` and `index1` such that

index0[i]*stride0[i] + offset0 = index1[j]*stride0[j] + offset1

0 <= i < len(shape0)

0 <= index0[i] < shape0[i]

0 <= j < len(shape1)

0 <= index1[j] < shape1[j]

0 <= index0[i] < shape0[i]

0 <= j < len(shape1)

0 <= index1[j] < shape1[j]

This is a bounded linear Diophantine equation. Diophantine because the variables `index0[i]` and `index1[j]` are integer-valued, linear because the variables only appear multiplied by a constant, and bounded because each variable must stay within the size of its corresponding axis.

Unbounded linear Diophantine equations can be solved using an extended version of the Euclidean GCD algorithm. Bounded linear Diophantine equations are NP-<mumble> to solve. With the right heuristics, a branch-and-bound approach could probably work well.

> Also, relatedly, a few months ago Julian Taylor at least wrote what was there in C, which made it faster, if not better.

The main benefit of that was to make it available at the C level for other C-implemented routines, IIRC, not speed.

--

Robert Kern

The main benefit of that was to make it available at the C level for other C-implemented routines, IIRC, not speed.

--

Robert Kern