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You can find the paper here , in which they describe the architecture of a QNN that can be used to learn binary functions and correctly classify unseen data.

They say that for each binary label function $l(z)$ where $l(z) = -1$ or $l(z) = 1$, there exists a unitary $U_l$ such that, for all input strings $z = z_0z_1...z_{n-1}$ (where each $z_i = -1,1)$, $$\langle z,0 | U_l^{\dagger} Y_{n+1} U_l |z,0 \rangle = l(z)$$

If you assume that $U_l = \text{exp}(i\frac{\pi}{4}l(z)X_{n+1})$, then it can be easily proven that $\langle z,0 | U_l^{\dagger} Y_{n+1} U_l |z,0 \rangle = l(z)$

Now let's consider the subset parity problem. Here, $l(z) = 1-2B(z)$, where $B(z) = \oplus^{n-1}_{j=0} \phantom{a} a_j \cdot \frac{1}{2}(1-z_j)$, which, when plugged into $U_l$ gives us $$\text{exp}(i\frac{\pi}{4}X_{n+1}) \prod^{n-1}_{j=0} \text{exp}(-i \frac{\pi}{2}a_j \cdot \frac{1}{2}(1-z_j))$$

Now, for the subset parity problem, what you want to learn is $\frac{\pi}{2}a_j$, which you do not know beforehand.

So, during learning, you assume that $$U_l(\vec\theta) = \text{exp}(i\frac{\pi}{4}X_{n+1}) \prod^{n-1}_{j=0} \text{exp}(-i \theta_j \cdot \frac{1}{2}(1-z_j))$$ (our goal is to update $\vec\theta$ s.t when we compute the estimated label, we get close to the actual label)

This method seems to be working fine for this problem (I get an accuracy of 96%).

Right now, I am trying to use a QNN for another binary classification problem. Contrary to the subset parity problem, I do not actually know $l(z)$ (which I thought was perfect, because the QNN allows me to design a circuit that correctly classifies my strings). Therefore, I assumed that $$U_l(\vec\theta) = \text{exp}(i\frac{\pi}{4}X_{n+1}) \prod^{n-1}_{j=0} \text{exp}(-i \theta_j \cdot \frac{1}{2}(1-z_j))$$, just like the subset parity problem.

It seems to be working fine. I get an accuracy of 76%, which isn't bad. However, I am not sure if I can assume this and I am starting to wonder if my initial assumption about $U_l$ for this new problem is legit or not (it could be a coincidence or an error in my code).

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  • $\begingroup$ Unfortunately just saw this today. Might have been fun to read the paper and figure out the answer for you. But too late now since it's almost midnight and when I wake up the bounty will be over. Good luck! $\endgroup$ – user1271772 Jun 14 at 3:17
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As far as I understand from the paper, eq. (13) gives $U_l$ as a product of two qubit unitaries, independently of $l(z)$. Then the authors present two cases, subset parity and subset majority, and derive their specific $U_l$. Thus I guess your classification problem will need its own specialization of eq. (13). If you get an acceptable accuracy with the subset parity $U_l$, it may be a coincidence. Or maybe it is not, it depends on how your classification problem (that we do not know) relates to subset parity.

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