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Each run of Simon's algorithm uses one oracle call and returns one uniform sample from $s^{\perp }$. Collect samples as rows of a matrix $Z$ and solve $Zs=0$ over the two-element field using Gaussian elimination. Once the rows have matrix rank $n-1$, the kernel is $\{0,s\}$, so the promised nonzero vector is uniquely determined. Dependent samples, including $z=0$, must be allowed for; exactly $n-1$ runs need not suffice.

For a rough quantitative estimate, put $d=n-1$. When the existing samples span an $r$-dimensional vector subspace of $s^{\perp }$, the next sample increases the matrix rank with probability $1-2^{r-d}$. The random spanning of a binary vector space estimate givesAlternatively, $m=d+t$ samples fail to span with probability less than $2^{-t}$: failure means some nonzero linear functional on the $d$-dimensional space vanishes on every sample, and a union bound gives $(2^d-1)2^{-m}$. ConsequentlyThe Gaussian elimination requires only a polynomial amount of classical work, for example $O(n^3)$ bit operations for $O(n)$ samples. The degenerate case $n=1$ needs no calls because the promise already specifies $s=1$.

Classically, distinct inputs with equal function values reveal $s=x\oplus x^{\prime }$. Uniformly sampling distinct inputs finds such a function collision after roughly $\sqrt{2^n}$ calls, by the birthday square-root scale. For $q$ sampled inputs, the expected number of colliding pairs is $\left(\genfrac{}{}{0ex}{}{q}{2}\right)/(2^n-1)$. The usual randomized classical query complexity of Simon's problem is therefore exponential in $n$, on the scaleA straightforward exhaustive classical search instead uses $O(2^n)$ calls. The square-root collision estimate is the sharper classical comparison; the quantum advantage concerns access to the promised quantum oracle, rather than an explicit table of all function values.

The first Hadamard transform prepares a uniform superposition in the first register while the second stays zero. The Boolean quantum oracle, now with an $n$-bit output register and bitwise addition, therefore givesThis state is normalized because distinct $x$ labels are orthogonal, even when their oracle outputs coincide. Addition in the oracle is bitwise exclusive-or, making the oracle reversible; replacing the first register by $f(x)$ would not define a unitary gate for a two-to-one function.

For the later recovery steps we use the intended nonzero hidden-period promise $s\ne 0$. Then two-to-one behavior and $f(x+s)=f(x)$ force each fibre to be exactly $\{x,x+s\}$. The printed allowance $s\in \{0,1{\}}^n$ does not explicitly exclude zero. If zero were allowed without a nonzero-period promise, the displayed identity would hold trivially for every function and would not specify Simon's problem. That distinction is addressed again in the query count.

The promise is Simon's problem. If $z=0^n$, then $f(x)=f(y)$ implies $x=y$, so $f$ is an injective function. If $z\ne 0^n$, then each value has exactly the two preimages $x$ and $x\oplus z$; thus $f$ is two-to-one and is constant on the two-element cosets of $\{0^n,z\}\le ({\mathbb{Z}}_2{)}^n$.

Regard bit strings as the elementary abelian group $G=({\mathbb{Z}}_2{)}^n$ under bitwise exclusive or. The promise saysThus $f$ is constant exactly on the cosets of $H$ and distinct between them. Determining the hidden subgroup determines its nonzero element $p$, so this is Simon's problem as an instance of the hidden subgroup problem.

Simon's algorithm prepares hidden-subgroup coset states, applies the quantum Fourier transform over $({\mathbb{Z}}_2{)}^n$, and samples vectors $y$ satisfying $y\cdot p=0$ over ${\mathbb{F}}_2$. Repeated samples and Gaussian elimination recover $p$ with $O(n)$ oracle queries and polynomial computation.

To solve Simon's problem, prepare a uniform quantum superposition, evaluate the function coherently into a second quantum register, and perform Simon Fourier sampling on the first. Each run produces a uniform binary string $z$ with $z\cdot s=0$ over the two-element field. Once the observed rows have matrix rank $n-1$, Gaussian elimination recovers $s$ as the unique nonzero element of their kernel. The random spanning of a binary vector space bounds the required number of runs.

Simon's problem asks for the hidden string $p\in \{0,1{\}}^n$ promised by $f(x)=f(y)$ exactly when $x\oplus y\in \{0^n,p\}$. It is the hidden subgroup problem on $({\mathbb{Z}}_2{)}^n$ with hidden subgroup $\{0^n,p\}$.