Decision-tree depth ($D(f)$)

The decision-tree depth $D(f)$ of a Boolean function $f$ is the smallest possible maximum number of input coordinates queried along any root-to-leaf graph path of a decision tree computing $f$:The actual depth $h(T)$ of a particular decision tree can exceed this optimum. A constant function admits a depth-zero tree even though an inefficient algorithm could inspect its entire input before returning that constant. Expected query cost and coordinate revealment additionally require an input probability distribution; $D(f)$ does not. The three-bit majority decision tree and three-bit parity both have optimal decision-tree depth three, although only the former has the two-query shortcuts described above.

A query certificate fixes some input coordinates to their values in a chosen input and must force the output for every completion. The smallest such size is $C(f,x)$; maximizing over accepted or rejected inputs gives $C_1(f)$ or $C_0(f)$, and maximizing both gives $C(f)$. A one-sided maximum over an empty output class is defined as zero. This is query certificate complexity, distinct from polynomial verification of a witness in NP.

For a Boolean function $f$ and a chosen input $x$, a set of coordinates $S$ is a query certificate if every $y$ agreeing with $x$ on $S$ has $f(y)=f(x)$. It fixes actual input bits rather than supplying an independently guessed certificate to an NP verifier. Minimizing its size at each input and then maximizing gives query certificate complexity.

A Boolean function is evasive when every deterministic decision tree computing it has a worst-case path that queries all input bits. The alternating-sum criterion for decision-tree evasiveness is one sufficient method for proving this lower bound.

A leaf reached before all $n$ input bits are queried leaves at least one coordinate free. Flipping that coordinate pairs its inputs with opposite alternating signs and equal output. Thus every such leaf contributes zero. If the total alternating sum is nonzero, a depth below $n$ is impossible.

To prove that a threshold function requires every input bit in the worst case, an adversary answers queries while keeping completions on both sides of the threshold possible. If this remains true until the final unqueried bit, every decision tree has depth equal to the number of variables.