# Is this generalized 2D cluster state still a universal resource?

Let $$G = (V,E)$$ be a graph that defines the graph state $$|G\rangle = \prod_{(i,j)\in E} CZ_{i,j}|+\rangle^{\otimes |V|}.$$ Alternatively, we can write $$| G \rangle = \sum_{x \in \{0,1\}^{|V|}} f_G(x)|x\rangle,$$ for some function $$f_G$$ that can be determined by expanding the definition of $$|G\rangle$$.

In the case where $$G$$ is a 2d lattice graph, then $$|G\rangle$$ is a cluster state, and $$|G\rangle$$ is also a universal resource for measurement-based quantum computing. Thus, with post-selection, one can simulate any BQP circuit.

My question is the following. Let $$G$$ be a 2d lattice graph and consider the state $$| G' \rangle = \sum_{x \in \{0,1\}^{|V|}} f_G(x)|x\rangle \otimes | x\rangle.$$ Is this universal for MBQC? What if I restrict to post-selecting on only the upper $$|V|$$ registers?

TLDR: Yes, it's universal. If you only measure on one of the two registers, No.

1. prepare all qubits in the $$|0\rangle$$ state
2. take one register, and apply Hadamards on every qubit. Now we have $$\sum_x|x\rangle|0\rangle$$.
3. Apply controlled nots controlled off each qubit of the first register targeting the equivalent qubit in the second register. This produces $$\sum_x|x\rangle|x\rangle$$