Bump to xDSL supporting inplace bufferization

[PapyChacal]

Just bump to xDSL supporting inplace bufferization and support inplace

Stacked on:

• Unhack stencil.store result through bufferization #118

[PapyChacal]

We probably want to check correct computations here then 🤔

[PapyChacal]

Self-note: There's an issue with an unlowered memef.alloc
Which might mean the bufferization is not acting as expected, as I wouldn't expect an allocation in the in place case

[PapyChacal]

Okay so, Devito here is doing that (turning on OpenMP for clarity):

for (int x = x_M; x >= x_m; x -= 1)
    {
      #pragma omp parallel num_threads(nthreads)
      {
        #pragma omp for schedule(static,1)
        for (int y = y_m; y <= y_M; y += 1)
        {
          u[t0][x + 1][y + 1] = r0*(-u[t0][x + 1][y + 1]) + r0*u[t0][x + 2][y + 1];
        }
      }
    }

So it sees there's a dependency only on y and expresses the computation to be sequential over x so that it can indeed do this one in place.
This is just not something the stencil dialect can express at this point 🥲
@georgebisbas There are multiple ways from here onwards, let's discuss it when you have time. But indeed I don't see how to manage this without an extra allocation in xDSL in the time I have on my side

[georgebisbas]

Okay so, Devito here is doing that (turning on OpenMP for clarity):

for (int x = x_M; x >= x_m; x -= 1)
    {
      #pragma omp parallel num_threads(nthreads)
      {
        #pragma omp for schedule(static,1)
        for (int y = y_m; y <= y_M; y += 1)
        {
          u[t0][x + 1][y + 1] = r0*(-u[t0][x + 1][y + 1]) + r0*u[t0][x + 2][y + 1];
        }
      }
    }

So it sees there's a dependency only on y and expresses the computation to be sequential over x so that it can indeed do this one in place. This is just not something the stencil dialect can express at this point 🥲 @georgebisbas There are multiple ways from here onwards, let's discuss it when you have time. But indeed I don't see how to manage this without an extra allocation in xDSL in the time I have on my side

Thanks for that @PapyChacal , does xdsl produce one level of parallelism + correct results?
Let's discuss this yes.

[georgebisbas]

add tests remider

[georgebisbas]

probably no meaning for this

[PapyChacal]

Okay so, Devito here is doing that (turning on OpenMP for clarity):

for (int x = x_M; x >= x_m; x -= 1)
    {
      #pragma omp parallel num_threads(nthreads)
      {
        #pragma omp for schedule(static,1)
        for (int y = y_m; y <= y_M; y += 1)
        {
          u[t0][x + 1][y + 1] = r0*(-u[t0][x + 1][y + 1]) + r0*u[t0][x + 2][y + 1];
        }
      }
    }

So it sees there's a dependency only on y and expresses the computation to be sequential over x so that it can indeed do this one in place. This is just not something the stencil dialect can express at this point 🥲 @georgebisbas There are multiple ways from here onwards, let's discuss it when you have time. But indeed I don't see how to manage this without an extra allocation in xDSL in the time I have on my side

Thanks for that @PapyChacal , does xdsl produce one level of parallelism + correct results? Let's discuss this yes.

For the record, this does not seem to happen in TTI