Prefix Matrix Products and Autograd

You are given a sequence of square integer matrices matrices with shape [N, D, D]. Define the inclusive prefix products
P[i] = matrices[0] @ matrices[1] @... @ matrices[i],
where @ is ordinary matrix multiplication.
Reverse-mode contract
You are also given upstream with shape [N, D, D]. Treat upstream[i] as the upstream gradient dL / dP[i] of a scalar loss L. Equivalently, the differential of the loss is
dL = sum(i = 0..N-1) <upstream[i], dP[i]>_F,
where <A, B>_F = sum(r, c) A[r][c] * B[r][c]. Compute every prefix product P[i] and every input gradient dL / dmatrices[i].
For the sequential backward step, keep the original input matrices available and use only a constant number of D x D scratch matrices beyond the inputs and returned prefixes and gradients.
Hillis-Steele follow-up
The intended final approach performs the forward pass as an out-of-place Hillis-Steele scan. Start with one snapshot containing the input matrices. For offsets 1, 2, 4,..., build a new snapshot from the previous one:
If i < offset, copy matrix i unchanged.
Otherwise, set matrix i to previous[i - offset] @ previous[i].
Keep the snapshots needed to reverse these multiplication rounds without mutating an earlier snapshot.
Return format
Return a long[][] with 2 * N rows and D * D columns. Flatten each matrix in row-major order:
Rows 0 through N - 1 contain P[0] through P[N - 1].
Rows N through 2 * N - 1 contain dL / dmatrices[0] through dL / dmatrices[N - 1].

Function
prefixProductAutograd(matrices: int[][][], upstream: int[][][]) → long[][]

Examples
Example 1
matrices = [[[1,2],[3,4]]]
upstream = [[[2,0],[1,-1]]]
return = [[1,2,3,4],[2,0,1,-1]]
There is one prefix, so P[0] = matrices[0]. Because that prefix is the input itself, its gradient is exactly upstream[0].
Example 2
matrices = [[[1,2],[0,1]],[[2,0],[1,3]],[[1,1],[2,0]]]
upstream = [[[1,0],[0,1]],[[0,1],[-1,0]],[[2,-1],[1,1]]]
return = [[1,2,0,1],[4,6,1,3],[16,4,7,1],[3,16,2,8],[1,5,3,12],[9,-3,15,-3]]
The first three rows are the row-major forms of P[0], P[1], and P[2]. The final three rows are the corresponding input gradients after contributions from every dependent prefix have been accumulated.
Example 3
matrices = [[[2]],[[-1]],[[3]],[[2]]]
upstream = [[[1]],[[2]],[[-1]],[[3]]]
return = [[2],[-2],[-6],[-12],[-16],[34],[-10],[-18]]
For D = 1, matrix multiplication becomes scalar multiplication. The four prefixes are 2, -2, -6, and -12; the remaining rows are their accumulated reverse-mode gradients.

Constraints
1 <= N <= 32
1 <= D <= 4
matrices.length = upstream.length = N.
Every matrix in both inputs has exactly D rows and D columns.
Every input entry is an integer from -10 through 10.
Every individual product, partial sum, prefix-product entry, scan-stage entry, reverse-mode adjoint, and final gradient fits in a signed 64-bit integer.
All multiplication and addition are exact; no modulus or floating-point tolerance is used.

The trick is the recurrence. P[i] = P[i-1] @ M[i]. Walk backward with a running adjoint G, where G starts as upstream[N-1]. For each i from N-1 down to 0, the gradient for M[i] is P[i-1]^T @ G (use identity-free handling when i = 0, where the gradient is just G). Then update G = upstream[i-1] + G @ M[i]^T. The pitfall is forgetting that upstream[i-1] adds in at every step, and mixing up transpose sides. You only need a constant number of DxD scratch matrices, so don't store every adjoint. Use long everywhere. Check Example 3 with D = 1 first, since it reduces to scalars and exposes sign mistakes fast. The Hillis-Steele follow-up needs saved snapshots per round, and you reverse each round in the opposite order. If the transpose order slips mid-assessment, StealthCoder is the hedge that reads the problem and hands you the working structure.

If you see this problem in your OA tomorrow, the play is to recognize the pattern in 30 seconds. StealthCoder buys you that recognition.