@@ -19,7 +19,9 @@ The broadcast mode serves to calculate the rank of the corresponding output and
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@@ -19,7 +19,9 @@ The broadcast mode serves to calculate the rank of the corresponding output and
* {{{(accumulate, Accumulator)}}}
* {{{(accumulate, Accumulator)}}}
* output.rank = min(input.rank)
* output.rank = min(input.rank)
* for the inputs of greater rank, we use Accumulator (sum, product, etc.) to accumulate over the first dimensions
* for the inputs of greater rank, we use Accumulator (sum, product, etc.) to accumulate over the first dimensions
* e.g. {{{if Accumulator == sum, order == c, x.rank == 2, y.rank == 1 and z = f(x, y) then z[i] = f(sum_j(x[i, j]), y[i])}}}
* e.g. {{{if Accumulator == sum, order == c, x.rank == 2, y.rank == 1 and z = f(x, y) then z[i] = f(sum_j(x[i, j]), y[i])}}}
* if {{{order == f}}} ([3, 5], [5]) => [5] or ([7, 8, 9], [8, 9]) => [8, 9]
* if {{{order == f}}} ([3, 5], [5]) => [5] or ([7, 8, 9], [8, 9]) => [8, 9]
* if {{{order == c}}} ([3, 5], [3]) => [3] or ([7, 8, 9], [7, 8]) => [7, 8]
* if {{{order == c}}} ([3, 5], [3]) => [3] or ([7, 8, 9], [7, 8]) => [7, 8]
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@@ -27,6 +29,7 @@ The broadcast mode serves to calculate the rank of the corresponding output and
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@@ -27,6 +29,7 @@ The broadcast mode serves to calculate the rank of the corresponding output and
This does not cover all cases of broadcasting, but I believe they cover enough. Other cases of broadcasting can be emulated with proper transposition and/or slicing.
This does not cover all cases of broadcasting, but I believe they cover enough. Other cases of broadcasting can be emulated with proper transposition and/or slicing.
* Could you give some examples of what kinds of broadcasting are and are not covered by your proposed implementation?
* Could you give some examples of what kinds of broadcasting are and are not covered by your proposed implementation?
* For rank <= 2, I think only operations of the form {{{add(ones(3,1), ones(1,3)))}}} are missing. I actually didn't think of that one before now.
* For rank <= 2, I think only operations of the form {{{add(ones(3,1), ones(1,3)))}}} are missing. I actually didn't think of that one before now.
* In general, it only handles f(shape(head, ...), shape(head, ...), ...) and f(shape(..., tail), shape(..., tail), ...)
* In general, it only handles f(shape(head, ...), shape(head, ...), ...) and f(shape(..., tail), shape(..., tail), ...)
* Maybe I could add a general case later... the thing is that I think the ones I am considering here are easier to streamline.
* Maybe I could add a general case later... the thing is that I think the ones I am considering here are easier to streamline.
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@@ -71,8 +74,11 @@ An Optimizer should look at the operations in the graph and figure out whether t
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@@ -71,8 +74,11 @@ An Optimizer should look at the operations in the graph and figure out whether t
The input ranks become the output ranks and gradients of the same rank as the outputs are added to the input list. If an output was given mode {{{broadcast}}}, then all inputs used to calculate it had to be broadcasted to that shape, so we must sum over the broadcasted dimensions on the gradient. The mode that we give to those inputs is therefore {{{(accumulate, sum)}}}. Inversely, if an output was given mode {{{(accumulate, sum)}}}, then all inputs used to calculate it had to be summed over those dimensions. Therefore, we give them mode {{{broadcast}}} in grad. Other accumulators than sum might prove more difficult. For example, the ith gradient for product is grad*product/x_i. Not sure how to handle that automatically.
The input ranks become the output ranks and gradients of the same rank as the outputs are added to the input list. If an output was given mode {{{broadcast}}}, then all inputs used to calculate it had to be broadcasted to that shape, so we must sum over the broadcasted dimensions on the gradient. The mode that we give to those inputs is therefore {{{(accumulate, sum)}}}. Inversely, if an output was given mode {{{(accumulate, sum)}}}, then all inputs used to calculate it had to be summed over those dimensions. Therefore, we give them mode {{{broadcast}}} in grad. Other accumulators than sum might prove more difficult. For example, the ith gradient for product is grad*product/x_i. Not sure how to handle that automatically.
* I don't exactly follow this paragraph, but I think I catch the general idea and it seems to me like it will work very well.
* I don't exactly follow this paragraph, but I think I catch the general idea and it seems to me like it will work very well.
* In a nutshell for {{{broadcast}}} I calculate the gradient as normal assuming the shape is broadcasted and then I sum over what I had to broadcast.
* In a nutshell for {{{broadcast}}} I calculate the gradient as normal assuming the shape is broadcasted and then I sum over what I had to broadcast.
* Could you explain why the accumulator gradient (e.g. product) can be trickier?
* Could you explain why the accumulator gradient (e.g. product) can be trickier?
* I thought about it and I figured that the general case is {{{g_accum[N-i+1], g_m[i] = grad_fn(accum[i-1], m[i], g_accum[N-i])}}} where {{{g_accum}}} is the accumulated gradient wrt the accumulator {{{accum}}}. It can be short-circuited in sum and product's case: for sum, grad_fn is the identity on its last argument so {{{g_m[i] == g_accum[i] == g_accum[0] == g_z for all i}}}. In product's case, {{{accum[i-1] == product(m[1:i-1]) and g_accum[N-i] == g_z * product(m[i+1:N])}}}, multiply them together and you obtain {{{g_z * product(m)/m[i]}}} where obviously we only need to compute {{{product(m)}}} once. It's worth handling those two special cases, for the general case I don't know.
* I thought about it and I figured that the general case is {{{g_accum[N-i+1], g_m[i] = grad_fn(accum[i-1], m[i], g_accum[N-i])}}} where {{{g_accum}}} is the accumulated gradient wrt the accumulator {{{accum}}}. It can be short-circuited in sum and product's case: for sum, grad_fn is the identity on its last argument so {{{g_m[i] == g_accum[i] == g_accum[0] == g_z for all i}}}. In product's case, {{{accum[i-1] == product(m[1:i-1]) and g_accum[N-i] == g_z * product(m[i+1:N])}}}, multiply them together and you obtain {{{g_z * product(m)/m[i]}}} where obviously we only need to compute {{{product(m)}}} once. It's worth handling those two special cases, for the general case I don't know.