An Exact Algorithm for the Unanimous Vote Problem

TL;DR

This work resolves the Unanimous Vote problem by giving an exact $O(n \log n)$ algorithm that exploits structural monotonicity: the optimal non-adaptive ordering is nearly greedy, differing at most at one critical position. It places the problem among the rare polynomial-time cases of unit-cost SBFE by showing the optimal ordering can be obtained by a fast greedy procedure plus testing $O(n)$ tweaks, and it tightly bounds the adaptivity gap at $1.2\pm o(1)$. The results include a detailed structural analysis, a polynomial-time exact algorithm, a faster $O(n^2)$ variant, and a careful comparison to adaptive strategies, with implications for non-adaptive SBFE and broader symmetric function evaluation. Overall, the paper advances both the algorithmic and theoretical understanding of SBFE problems, yielding exact solutions where previously only approximations were known and clarifying the value of adaptivity in this setting.

Abstract

Consider $n$ independent, biased coins, each with a known probability of heads. Presented with an ordering of these coins, flip (i.e., toss) each coin once, in that order, until we have observed both a *head* and a *tail*, or flipped all coins. The Unanimous Vote problem asks us to find the ordering that minimizes the expected number of flips. Gkenosis et al. [arXiv:1806.10660] gave a polynomial-time $φ$-approximation algorithm for this problem, where $φ\approx 1.618$ is the golden ratio. They left open whether the problem was NP-hard. We answer this question by giving an exact algorithm that runs in time $O(n \log n)$. The Unanimous Vote problem is an instance of the more general Stochastic Boolean Function Evaluation problem: it thus becomes one of the only such problems known to be solvable in polynomial time. Our proof uses simple interchange arguments to show that the optimal ordering must be close to the ordering produced by a natural greedy algorithm. Beyond our main result, we compare the optimal ordering with the best adaptive strategy, proving a tight adaptivity gap of $1.2\pm o(1)$ for the Unanimous Vote problem.

Figures (5)

• Figure 1: Two example orderings for an instance of the Unanimous Vote problem. The second ordering has a smaller number of expected flips (cost). One intuitive strategy is to alternate between the highest and lowest bias coins, as in the first sequence. As we can see, this is not always optimal. The second sequence is the one returned by a natural greedy algorithm described below. For this instance, it happens to be optimal.
• Figure 2: Plot of the probability of heads of a coin versus its position in an ordering. Unbiased blocks are in gray, $0$-biased blocks are in blue, and $1$-biased blocks are in red. For example, the fourth coin in the ordering has probability of heads equal to 0.45 and is in a 0-biased block.
• Figure 3: Plot of the probability of heads of a coin versus its position in an example greedy ordering. This is the same instance as in \ref{['fig:biasedblocksunsorted']}. Unbiased blocks are in gray, $0$-biased blocks are in blue, and $1$-biased blocks are in red.
• Figure 4: Heuristic for the adaptivity gap. The maximum value over $0 \le x \le \frac{1}{2}$ is at least $1.2$ for all $k$, and at most $1.2$ when $k \ge 5$. At $k \to \infty$, the heuristic recovers the adaptivity gap of the instance $\{0, 1-x,1-x,1-x,\dots\}$.
• Figure 5: Schematic for the proof that the adaptivity gap is at most $1.2 + o(1)$ (\ref{['thm:adaptivitygap_body']}). We consider all choices of $(p_1, p_n, p_{k^{\star}})$ and crossover points $t$ of various orderings. The sets $A_2,B_2,A_3,B_3,A_4,B_4$, and the ranges of $p_{k^{\star}}$ are defined in the proof outline (\ref{['sub:proofoutline']}). Each level handles a set of instances the previous level could not handle. Note that if $t_{\ell} \ge \ell + 1$, then $t_{\ell+1} \ge \ell + 1$.

Theorems & Definitions (75)

• theorem 1.1
• claim 1.2: Greedy is near-optimal
• theorem 1.3: Bound on adaptivity gap
• claim 3.1: Cost of swapping adjacent positions
• proof
• corollary 3.2: Optimal ordering is sorted within each block
• proof
• claim 3.3: Optimal ordering uses coins on one side of $\frac{1}{2}$ in non-final blocks
• proof
• claim 3.4