 Binary search

Binary search finds a target $x$ in a sorted array $A$ by inspecting the middle of the remaining candidate range. If the middle entry is smaller than $x$, every entry to its left is also too small; if larger, every entry to its right is too large. One comparison therefore rules out many entries. Equal entries are allowed; this version returns any matching index, without promising the first occurrence.

Use indices starting at zero and inclusive endpoints $L,R$. Initially $L=0$, $R=n-1$, where $n$ is the array length. While $L\le R$, chooseThe brackets denote the floor function. Compare $A_m$ with $x$: return $m$ on equality; set $L=m+1$ if $A_m<x$; otherwise set $R=m-1$. If $L>R$, report absence. The updates exclude the already inspected middle entry. This also handles an empty array, whose initial endpoints are $0,-1$. Sedgewick and Wayne specify a sorted array as input and return a matching index or $-1$ for absence.^[ref]

For a fully worked example, take the sorted arraySearch for $x=23$:

For $x=24$, the first two steps agree. Next $24>23$ retains only index $6$; inspecting $A_6=38$ sets $R=5$. Now $L=6>5=R$, so the answer is "absent" after four comparisons.

The loop invariant is: the array stays sorted, $0\le L\le R+1\le n$, and every index containing $x$ lies between $L$ and $R$. It holds initially. The sorted array property justifies each exclusion. Returning on equality gives a match; an empty candidate range implies absence. Crucially, this loop invariant allows $x$ to be absent from the beginning.

For a nonempty range of size $s=R-L+1$, an unsuccessful comparison leaves at most $\lfloor s/2\rfloor$ candidates. Thus sizes satisfy $s_k\le \lfloor n/2^k\rfloor$ after $k$ unsuccessful comparisons. This proves termination and, for $n\ge 1$, at most $\lfloor {\log }_{2}n\rfloor +1$ middle-entry comparisons. The final singleton still needs inspection: $n=1$ requires one comparison, not zero. For $n=15$, the bound is four. This is the halving argument behind logarithmic time complexity.^[ref]

A decision tree with $q$ binary tests has at most $2^q$ leaves. Distinguishing $n$ possible matching positions therefore requires $q\ge \lceil {\log }_{2}n\rceil$. This lower bound counts binary tests, whereas the bound above counts three-way comparisons.