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I have a 3x3 Convolution Function defined like this 

conv(x, y) = 0;
conv(x, y) += kernel(r.x, r.y) * in(x + r.x - 1, y + r.y - 1);

Size of the input buffer is 16 x 16
If I want to execute it with padding I can directly do
in = Halide::BoundaryConditions::constant_exterior(in_buffer, 0, 0, 16, 0, 16)
But I have to execute without padding and so I am trying to manually set the bounds on the function like this 

conv.bound(x, 1, 14);  
conv.bound(y, 1, 14);

This returns an error message

Error:
Bounds given for convolution in y (from 1 to 14) do not cover required region (from 0 to 15)

What should I do to set bounds on a Var in Func?

Arnab Mitra
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1 Answers1

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I think you need not to manually set the bounds using the *.bound function. Try this one:

Halide::Func conv("conv"), kernelF("kernel"), in("in");
Halide::Var x("x"), y("y");
Halide::RDom r(0, 3, 0, 3,"r");

in = Halide::BoundaryConditions::constant_exterior(in_buffer, 0, 
    0, 16, 0, 16);
kernelF = Halide::BoundaryConditions::constant_exterior(kernel_buffer, 0, 
    0, 3, 0, 3);

conv(x, y) = 0.0f;
conv(x, y) += kernelF(r.x, r.y) * in(x + r.x, y + r.y);

//conv.print_loop_nest();
Halide::Buffer<float_t> outputBuf = conv.realize(14, 14);

Look, we can set the bounds directly in *.realize() arguments, i.e. 14=16-3+1; Also, note that the convolution anchors are at the top-left of kernels.

Frank
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  • Yes, this works perfectly. Well I had solved the problem sometime back following this technique only. Thanks for the answer, hope it helps anyone else out there. I guess another approach can be using `set_min()` on a Halide Buffer. – Arnab Mitra Mar 14 '18 at 10:58