## Friday, 15 November 2013

### Simon's problem modulo $m$

Throughout, the following notation and results will be used. Let $\omega_N=e^{i2\pi/N}$. The following sum can be evaluated as so:
$\SUM{x=0}{N-1}\omega_N^{xy} \left\{ \begin{array}{ll} N \ \ \text{if} \ \ y\equiv0 \ (\text{modulo} \ N) \\ 0 \ \ \text{if} \ \ y\equiv0 \ (\text{modulo} \ N). \end{array} \right.$

Let $m$ be some $n$-bit number $(2^{n-1}<m<2^m)$ and assume that we are given a black box computing $f : \mathbb{Z}_m\times\mathbb{Z}_m$ that is promised to have the property that $f(a_1, a_2)=f(b_1,b_2)$ if and only if $(a_1,a_2)-(b_1,b_2)\in S$, where $S=\{k(r,1) ; k\in \mathbb{Z}_m\}$ for some unknown $r\in\mathbb{Z}_m$. Let the goal be to compute $r$.

For each $a\in\mathbb{Z}_m$, define $S_a=S+(a,0)=\{(kr+a,k) : k\in\mathbb{Z}_m\}$. It will now be proved that $S_0, S_1, \dots, S_{m-1}$ for a partition of $\mathbb{Z}_m\times\mathbb{Z}_m$.

Here, it will be shown that $a\neq b$ implies that $S_a\cap S_b=\emptyset$, where without loss of generality $a,b\in\mathbb{Z}_m$ are such that $a,b\in\{0,1, \dots, m-1\}$. Moreover, for $x,y\in\mathbb{Z}$, write $x\equiv y$ if $x=y(\text{mod}\ m)$. Thus, assume that $a\neq b$. Consider any $s\in S_a$ so that $s=(kr+a,k)$ for some $k\in\mathbb{Z}_m$. Now suppose for the sake of contradiction that $s\in S_b$ so that $s=(k'r+b,k')$ for some $k'\in\mathbb{Z}_m$. Then $(kr+a, k)=(k'r+b, k')$, which implies that $k\equiv k' \ \text{and} \ kr+a\equiv k'r+b$. Furthermore, since $k\equiv k'$ implies $k-k'\equiv0$, then $kr+a\equiv k'r+a$ implies $a-b\equiv (k-k')r\equiv 0$. This then implies that $a\equiv b$ or that $a=b$ contradicting the original assumption that $a=b$. Thus, for every $s\in S_a$, $s\notin S_b$. By the same reasoning it can similarly be shown that for every $t\in S_b$, $t\notin S_a$. Therefore, if $a\neq b$, then it must be the case that $S_a\cap S_b=\emptyset$

Now it will be shown that $S_0\cup S_1\cup\dots\cup S_{m-1}=\mathbb{Z}_m\times\mathbb{Z}_m$. Since $S_a\subseteq\mathbb{Z}_m\times\mathbb{Z}_m$ for all $a\in\{0,1,\dots, m-1\}$, then if $s\in S_a$, for some $a$, it follows that $s\in\mathbb{Z}_m\times\mathbb{Z}_m$.  Instead, consider an arbitrary $s=(j,k)\in\mathbb{Z}_m\times\mathbb{Z}_m$. Let $c\in\{0,1,\dots, m-1\}$ be such that $c\equiv j-kr$. Observe that  $kr+c\equiv kr+j-kr\equiv j$, and therefore $s=(j,k)=(kr+c,k)\in S_c$. Hence, for any $s\in\mathbb{Z}_m\times\mathbb{Z}_m$ there exists some $S_c$ such that $s\in S_c$. Hence, $S_0\cup S_1\cup\dots\cup S_{m-1}=\mathbb{Z}_m\times\mathbb{Z}_m$.

The results of (i) and (ii) together imply that the sets $S_0, S_1, \dots, S_{m-1}$ form a partition of $\mathbb{Z}_m\times\mathbb{Z_m}$.

Here, we prove that $f(x_1,x_2)=f(y_1,y_2)$ if and only if $(x_1,x_2)$ and $(y_1,y_2)$ are in the same element of the above partition (in other words, $(x_1,x_2),(y_1,y_2)\in S_a$ for some $s$).

First, suppose that $(x_1,x_2),(y_1,y_2)\in S_a$ for some $a$. Then there exists $k,k'\in\mathbb{Z}_m$ such that $(x_1,x_2)=(kr+a, k)$ and $(y_1,y_2)=(k'r+a,k')$. This implies that
\begin{align*} (x_1,x_2)-(y_1,y_2)&=(kr+a, k)-(k'r+a,k')\\ &=(kr+a-k'r-a,k-k')\\ &=((k-k')r,k-k')\\ &=(k-k')(r,1)\in S. \end{align*}
Then provided with the promise on $f$, it follows that $f(x_1,x_2)=f(y_1,y_2)$.

Suppose instead that $f(x_1,x_2)=f(y_1,y_2)$. The promise on $f$ then implies that $(x_1,x_2)-(y_1,y_2)\in S$, which means that $(x_1,x_2)-(y_1,y_2)=k(r,1)$ for some $k\in \mathbb{Z}_m$. Since
$(x_1,x_2)-(y_1,y_2)=(x_1-y_1,x_2-y_2)=k(r,1)=(kr,k),$
this implies that $x_2-y_2=k$. Moreover, since $kr=(x_2-y_2)r=x_1-y_1$ this implies that $x_2r-x_1=y_2r-y_1$. Let $c=x_2r-x_1=y_2r-y_1$. Now note that for any $(z_1,z_2\in S_a$ for some $a$, $(z_1,z_2)=(nr+a,n)$ for some $n$. This is equivalent to the condition $z_1=z_2r+a$. Therefore, since $x_2r-x_1=c$ and $y_2r-y_1=c$, it follows that $x_1=x_2r+c$ and $y_1=y_2r+c$. Hence, $(x_1,x_2),(y_1,y_2)\in S_c$ showing that if $f(x_1,x_2)=f(y_1,y_2)$, $(x_1,x_2)$ and $(y_1,y_2)$ belong to the same set $S_c$.

It has now been shown that $f(x_1,x_2)=f(y_1,y_2)$ if and only if $(x_1,x_2)$ and $(y_1,y_2)$ are in the same element of the above partition.

Consider the state
$\ket{\psi}=\frac{1}{m}\displaystyle\sum_{x_1=0}^{m-1}\SUM{x_2=0}{m-1}\ket{x_1}\ket{x_2}\ket{f(x_1,x_2)}.$
Since $f(x_1,x_2)=f(y_1,y_2)$ if $(x_1,x_2),(y_1,y_2)\in S_a$ for some $a$. The state $\ket{\psi}$ is an entangled state which can be equivalently expressed as
\begin{align*} \ket{\psi}&=\frac{1}{m}\SUM{a=0}{m-1}\SUM{k=0}{m-1}\ket{kr+a}\ket{k}\ket{f(kr+a,k)}\\ &=\frac{1}{m}\SUM{a=0}{m-1}\SUM{k=0}{m-1}\ket{kr+a}\ket{k}\ket{f_a}. \end{align*}
Here the state of the third register has been labelled with $f_a$, which denotes the value of $f(kr+a,k)$, since this value does not depend on $k$ due to the promise that $f(kr+a,k)=f(k'r+a,k')$ for any $k$ and $k'$.

Then applying the Fourier transform $F_m^\dagger$ to the first two registers yields the state:
\begin{align*} \ket{\overline{\psi}}:=(F_m^\dagger\otimes F_m^\dagger\otimes I)\ket{\psi}&=\frac{1}{m}\SUM{a=0}{m-1}\SUM{k=0}{m-1}F_m^\dagger\ket{kr+a}F_m^\dagger\ket{k}\ket{f_a}\\ &=\frac{1}{m^2}\SUM{a=0}{m-1}\SUM{k=0}{m-1}\SUM{s_1=0}{m-1}\SUM{s_2=0}{m-1}\omega_m^{-(kr+a)s_1}\omega_m^{-ks_2}\ket{s_1}\ket{s_2}\ket{f_a} \\ &=\frac{1}{m^2}\SUM{a=0}{m-1}\SUM{k=0}{m-1}\SUM{s_1=0}{m-1}\SUM{s_2=0}{m-1}\omega_m^{-(kr+a,k)\cdot(s_1,s_2)}\ket{s_1}\ket{s_2}\ket{f_a} \\ &=\frac{1}{m^2}\SUM{a=0}{m-1}\SUM{k=0}{m-1}\SUM{s_1=0}{m-1}\SUM{s_2=0}{m-1}\omega_m^{-k(r,1)\cdot(s_1,s_2)-as_1}\ket{s_1}\ket{s_2}\ket{f_a} \end{align*}

Now define the sets $R^\perp, R\subseteq \mathbb{Z}_m\times\mathbb{Z}_m$ as
\begin{align*} R^\perp&:=\{(s_1,s_2)\in\mathbb{Z}_m\times\mathbb{Z}_m : (r,1)\cdot(s_1,s_2)=0\} \\ R&:=\{ (s_1,s_2)\in\mathbb{Z}_m\times\mathbb{Z}_m : (r,1)\cdot(s_1,s_2)\neq0\}. \end{align*}
These two sets $R$ and $R^\perp$ partition $\mathbb{Z}_m\times\mathbb{Z}_m$ since for every element $(s_1,s_2)\in\mathbb{Z}_m\times\mathbb{Z}_m$ either $(r,1)\cdot(s_1,s_2)=0$ or $(r,1)\cdot(s_1,s_2)\neq0$. This allows the sum over $s_1$ and $s_2$ in the state for $\ket{\overline{\psi}}$ to be split into two sums. That is,
\begin{align*} \ket{\overline{\psi}}&=\frac{1}{m^2}\SUM{k=0}{m-1}\SUM{a=0}{m-1}\SUM{(s_1,s_2)\in R^\perp}{}\omega_m^{-k(r,1)\cdot(s_1,s_2)-as_1}\ket{s_1}\ket{s_2}\ket{f_a}+\frac{1}{m^2}\SUM{k=0}{m-1}\SUM{a=0}{m-1}\SUM{(s_1,s_2)\in R}{}\omega_m^{-k(r,1)\cdot(s_1,s_2)-as_1}\ket{s_1}\ket{s_2}\ket{f_a}. \end{align*}

Consider just the first summation ranging over$(s_1,s_2)\in R^\perp$ contained in $\ket{\overline{\psi}}$. In this case, $(r,1)\cdot(s_1,s_2)=0$, so the summation becomes
\begin{align*} &\frac{1}{m^2}\SUM{k=0}{m-1}\SUM{a=0}{m-1}\SUM{(s_1,s_2)\in R^\perp}{}\omega_m^{-k(r,1)\cdot(s_1,s_2)-as_1}\ket{s_1}\ket{s_2}\ket{f_a} \\ =&\frac{1}{m^2}\SUM{k=0}{m-1}\SUM{a=0}{m-1}\SUM{(s_1,s_2)\in R^\perp}{}\omega_m^{-as_1}\ket{s_1}\ket{s_2}\ket{f_a} \\ =&\frac{1}{m}\SUM{a=0}{m-1}\SUM{(s_1,s_2)\in R^\perp}{}\omega_m^{-as_1}\ket{s_1}\ket{s_2}\ket{f_a} \end{align*}
Now consider the second summation contained in $\ket{\overline{\psi}}$, which ranges over $(s_1,s_2)\in R$. This implies that $(r,1)\cdot(s_1,s_2)\neq0$, so that
\begin{align*} &\frac{1}{m^2}\SUM{k=0}{m-1}\SUM{a=0}{m-1}\SUM{(s_1,s_2)\in R}{}\omega_m^{-k(r,1)\cdot(s_1,s_2)-\overline{a}s_1}\ket{s_1}\ket{s_2}\ket{f_a}=0, \end{align*}
which implies that no states of the form $\ket{s_1}\ket{s_2}$ where $(s_1,s_2)\in R$ appear in the superposition in $\ket{\overline{\psi}}$. Hence the final state is of the form
$\ket{\overline{\psi}}=\frac{1}{m^2}\SUM{a=0}{m-1}\SUM{(s_1,s_2)\in R^\perp}{}\omega_m^{-as_1}\ket{s_1}\ket{s_2}\ket{f_a},$
and measuring $\ket{\overline{\psi}}$ will yield some state $\ket{s_1}\ket{s_2}$ such that $(s_1,s_2)\in R^\perp$ with probability
$\left|\frac{1}{m}\SUM{a=0}{m-1}\omega_m^{-as_1}\right|^2=\frac{1}{m^2}\SUM{a=0}{m-1}|\omega_m^{-as_1}|^2=\frac{1}{m^2}\SUM{a=0}{m-1}|1=\frac{m}{m^2}=\frac{1}{m}.$