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Exponential sign bracketing finds an integer interval with opposite endpoint signs before sign-bracketing binary search. Suppose $f$ is defined on nonnegative integers, its signs can be evaluated exactly, $f(0)<0$, and $f(k)>0$ for all sufficiently large integers $k$. Test $1,2,4,\dots$. An exact zero can be returned immediately; otherwise stop at the first positive endpoint $2^t$ and retain the preceding negative endpoint, or zero when $t=0$.

Excluding the already known evaluation at zero, this takes $t+1$ endpoint tests. For $t\ge 1$, the final integer interval is $[2^{t-1},2^t]$, of width $2^{t-1}$. Repeated midpoint tests preserving opposite endpoint signs take $t-1$ further tests to reach adjacent integers, unless an exact zero is found earlier. For $t=0$, $[0,1]$ is already adjacent.

No monotonicity is needed to preserve opposite endpoint signs, though other crossings may be skipped. Adjacent opposite signs do not imply an integer root; for $f(k)=2k-1$, neither endpoint of $[0,1]$ is a root. A real-valued extension requires continuity to infer a real root by the intermediate value theorem. A machine implementation must handle integer overflow and unreliable floating-point arithmetic; mathematical termination does not ensure representable endpoints.