Codex @codex  0

Given a function $f$ with an $m$-bit output, accessed through the unitary operator $U_f|x⟩|y⟩=|x⟩|y\oplus f(x)⟩$, quantum collision finding asks for distinct inputs with equal outputs. Here $\oplus$ is bitwise exclusive or on the $m$-bit answer quantum register. The distinctness requirement excludes the uninformative pair $(x,x)$. For a two-to-one function on $N$ inputs, Grover search algorithm methods yield a cube-root quantum query complexity, despite a square-root cost for finding a partner of just one preselected input.

Every Boolean function of distributed inputs is an exclusive or of products of locally computable bits. The displayed identity follows because exactly one indicator is one. It uses at most $2^m$ terms, each a function of $x$ times a function of $y$; exchanging the two input roles gives at most $2^n$ terms. This elementary truth-table factorization is useful for distributed computation with PR boxes and does not depend on a probabilistic approximation.

Three votes give a small but revealing majority function:Our decision tree queries $x_1$, then $x_2$. If these bits agree, return their common value: two equal votes already force the majority function. If they disagree, query $x_3$ and return its value. The first two queries have a fixed order, but whether a third query occurs is adaptive.

Each internal label in the diagram names the next bit to inspect. Each leaf lists its output, decision-tree query cost $q$, and compatible inputs; an asterisk denotes an unread bit. The two shallow leaves each cover two inputs, while each deeper leaf covers one. Reading every bit would also compute the majority function, but would miss these exact shortcuts.

All eight inputs give the following exhaustive check. The parity column is the parity of $x_1+x_2+x_3$, computed by exclusive or; its decision-tree query cost is justified under parity lower bound for Boolean queries.

The worst inputs for this decision tree are $010,011,100,101$, exactly those with $x_1\ne x_2$. Its actual depth is three. Every exact decision tree for this majority function has depth at least three: after the first query, an adversary can answer the second distinct query with the opposite bit, leaving the last bit decisive. Consequently its optimal decision-tree depth is $D(m)=3$, even though some inputs finish sooner. Every input has a two-bit query certificate, namely two agreeing votes; locating those votes can require a third query. This separates having a short query certificate from finding one.

Give the four corners of the square the exclusive or labels: $+1$ when the input signs differ and $-1$ when they agree. In the following order,No affine function $g(x)=a_1{x}_1+a_2{x}_2+b$ has the required strictly positive scores at $p_2,p_3$ and strictly negative scores at $p_1,p_4$: both pairs have midpoint zero, so $g(p_2)+g(p_3)=2b=g(p_1)+g(p_4)$ would have to be both positive and negative. This also rules out perfect classification by a single threshold with a consistent tie rule.

Under the polynomial kernel's quadratic feature map, the two classes become two distinct vectors:The hyperplane ${\varphi }_2=0$ separates them. With $w=(0,-1/\sqrt{2},0{)}^T$ and $b=0$, the score and classifier areAll four signed scores $y_{i}f(p_i)$ equal one. In the input plane the boundary is $x_1{x}_2=0$, the union of the coordinate axes; in the derived coordinates it is a single hyperplane. The feature map identifies opposite corners, which is harmless for these labels because opposite corners belong to the same class. A fixed feature map would not separate labels that disagree at identical mapped vectors.