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Other ES states are maximally entangled, such as the GHZ and W states [29]. 

Given any ES state, we would like to refer to the binary representation of each k ∈ S as k = j$_{1}$ . . . j$_{q}$ , where each j$_{m}$ is either 0 or 1 , and m = 1 , . . . , q is the index of the qubit. This notation in hand, along with the notation of Eq. (3), we can write down the overlap of the ES state with the product state | ϕ 〉 as a function of the x$_{m}$ ’s and θ$_{m}$ ’s. It is this overlap function we need to maximize in order to find P$_{max}$ : 

where 

To be more precise, we need to maximize P = | f q S | $^{2}$, which is equivalent to maximizing the magnitude of the complex function f q $_{S}$, ignoring its phase. In fact, we can fix the relative phases θ$_{m}$ to zero, since Eq. (11) then becomes: 

which is clearly not smaller in magnitude. This reduces the dimension of the search space from 2 q to q (this is actually a special case of an observation already made in Ref. [6]). Another simplification arises in an ES state for which there is one qubit, 1 ≤ m ≤ q , that all the basis states in the superposition “agree” on (that is, j$_{m}$ is the same for all k ∈ S ). This qubit is not entangled with the rest of the qubits and can be factored out. We shall call such an ES state reducible , since we can fix x$_{m}$ = j$_{m}$ and reduce the dimension of the search space by one. 

## 2. Examples: Special high symmetry states 

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We have so far reduced our problem to finding the values of x$_{1}$, . . . , x$_{q}$ for the nearest product state | ϕ 〉 , to a q -qubit nonreducible ES state | ψ q S 〉 , by maximizing the overlap function f q S in Eq. (13). Let us consider two special cases of this problem, which were already analyzed in Ref. [6]. The first one is the q -qubit GHZ state: 

