tinygrad/docs-legacy/reshape_without_symbolic.md

"View.reshape without symbolic"

This section contains the sketch proof of "Complete, Fast and Correct View.reshapes without using Symbolic". The goal is to reduce multi-views which cost runtime.

1. old_shape = (s₁,s₂,...,s_i,s_(i+1),...,s_n)
2. old_stride = (st₁, st₂, ... ,st_i, st_(i+1), ..., st_n)
3. merge_old_shape = (p₁, p₂), where p₁ = s₁ * ... * s_i & p₂ = s_(i+1) * ... * s_n,
4. new_shape = (k₁, ..., k_p, k_(p+1), ..., k_l)
5. prod(new_shape) = p₁ * p₂ (trivial)
6. mask and new_mask represent valid indexes before & after reshape respectively.

Assumption

p₁ & p₂ individually are mergeable (we will discuss later on this) & we cannot merge p₁ & p₂.

Claim

If prod([k₁ ... k_p]) < p₁ and prod([k₁ ... k_(p+1)]) > p₁, reshape is not possible.

Proof

k_(p+1) will require some dimensions from p₁ & some from p₂, which means p₁ & p₂ should be mergeable, but they are not.

Conclusion

Hence, reshape is only possible if ∃ a p, where prod([k₁ .. k_p]) = p₁.

Conditions for mergeability

Case 1 - All non-zero strides

They will merge if st_x = st_(x+1) * s_(x+1), where x ∈ [1, ..., i-1, i+1, ..., n-1].

Proof

Lets consider merging of (s₁ ... s_i) -> p₁, here we have to get a single new stride corresponding to p₁. For which it has to be contiguous.

Case 2 - Some stride is zero

Let st_j = 0 & st_(j+1) != 0 & s_(j+1) > 1, where 1 < j < i.

If s_j = 1 , reshape is trivial.

If s_j > 1,

• If mask_j has range > 1, reshape is not possible, because s_(j+1) will need to be repeated at-least once and a single stride can't capture repetition.
• If mask_j has range = 1, reshape is possible, since it is virtually shape = 1, with some offset.

Conditions for reshaping mask

Case 1 - Splitting Dimension - Mask shouldn't be cut for successful reshape.

• Example - [1,2,3,4,5,6,7,8] -> 1,2,3,4], [5,6,7,8 ; mask = ((2,6)) ; new_mask[0] = (0,2) (trivial split).

• new_mask[1] = not possible. It is only possible if mask spans [1-8] or lies within a single dimension [1-4] or [5-8].

Case 2 - Combining Dimension - Mask should unfold continuously.

• Example - 1,2],[3,4],[5,6 -> [1,2,3,4,5,6]; mask = ((0,2),(0,2)).

• new_mask = (0,4); only possible because mask₁ span the whole dimension.

• If mask₁ did not span the whole dimension, the only way combining would be possible is if mask₀ had range 1 as shown below.

  • 1,2,3],[4,5,6 -> [1,2,3,4,5,6]; mask = ((1,2),(0,2)); new_mask = ((3,5))