Form a $\mathbb{Z}^d$ lattice cycle from given lengths

Suppose you have a machine that solves the problem in $\mathbb{Z}^{d+1}$ quickly, and you want to use it to solve the problem in $\mathbb{Z}^d$. Suppose that your input is a list $(a_1,\ldots,a_k)$ of $k$ integers each of $n$ bits, so that your total input size is $kn$. Feed the list $(a_1,k2^n,a_2,k2^n,a_3,\ldots k2^n,a_k,k(k-1)2^n)$ to your machine. The new input size is $kn + (k-1)(\log k + n) + 2\log k + n$. All the entries $k2^n,k2^n,\ldots,k2^n,k(k-1)2^n$ must apear along the same dimension, so solving this new problem in $\mathbb{Z}^{d+1}$ amounts to solving the original problem in $\mathbb{Z}^d$. Does something like this work?

Suppose we are given a sequence $a_1,\dots, a_n$ (input to a partition problem). It seems that this can be encoded as a cycle problem in $\mathbb{Z}^3$ by finding two large numbers $M$ and $N$ and taking the sequence to be $$a_1, M, N, a_2, M, N, \dots, a_n, (n-1)M, (n-1)N.$$ The numbers $M$ and $N$ should be chosen so that the two last terms cannot be canceled in any other way than the obvious one. Then by the orthogonality rule, the only remaining possibility is to let the terms $a_1,\dots,a_n$ go in the third direction, and we are back to the partition problem.

This ought to work also for $d\geq 4$, although admittedly some details remain to be filled in.