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  1. Free, publicly-accessible full text available November 1, 2025
  2. We conducted a comprehensive, systematic review and meta-analysis on the effects of the BSCS 5E Instructional Model—and its related variants—on science, math, and motivation outcomes. The 5E Instructional Model is a framework for delivering STEM instruction that is based on constructivist learning theory; it has been used throughout the U.S. and other countries, particularly in Turkey. Despite its wide usage, no comprehensive systematic review and meta-analysis on the effects of 5E and related models has yet been conducted. Our search and screening procedures yielded 61 randomized controlled trial studies, estimating 156 effect sizes; 70% of studies met WWC standards with or without reservations. We found that the 5E instructional model resulted in improved science outcomes ( g = 0.82, 95% CI [0.67, 0.97]), but a large amount of heterogeneity requires some caution ( t = 0.56). We explored numerous explanations for the effect heterogeneity and provided practical recommendations.

     
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  3. Free, publicly-accessible full text available August 1, 2025
  4. Free, publicly-accessible full text available May 13, 2025
  5. Abstract

    In a Merlin–Arthur proof system, the proof verifier (Arthur) accepts valid proofs (from Merlin) with probability 1, and rejects invalid proofs with probability arbitrarily close to 1. The running time of such a system is defined to be the length of Merlin’s proof plus the running time of Arthur. We provide new Merlin–Arthur proof systems for some key problems in fine-grained complexity. In several cases our proof systems have optimal running time. Our main results include:

    Certifying that a list ofnintegers has no 3-SUM solution can be done in Merlin–Arthur time$$\tilde{O}(n)$$O~(n). Previously, Carmosino et al. [ITCS 2016] showed that the problem has a nondeterministic algorithm running in$$\tilde{O}(n^{1.5})$$O~(n1.5)time (that is, there is a proof system with proofs of length$$\tilde{O}(n^{1.5})$$O~(n1.5)and a deterministic verifier running in$$\tilde{O}(n^{1.5})$$O~(n1.5)time).

    Counting the number ofk-cliques with total edge weight equal to zero in ann-node graph can be done in Merlin–Arthur time$${\tilde{O}}(n^{\lceil k/2\rceil })$$O~(nk/2)(where$$k\ge 3$$k3). For oddk, this bound can be further improved for sparse graphs: for example, counting the number of zero-weight triangles in anm-edge graph can be done in Merlin–Arthur time$${\tilde{O}}(m)$$O~(m). Previous Merlin–Arthur protocols by Williams [CCC’16] and Björklund and Kaski [PODC’16] could only countk-cliques in unweighted graphs, and had worse running times for smallk.

    Computing the All-Pairs Shortest Distances matrix for ann-node graph can be done in Merlin–Arthur time$$\tilde{O}(n^2)$$O~(n2). Note this is optimal, as the matrix can have$$\Omega (n^2)$$Ω(n2)nonzero entries in general. Previously, Carmosino et al. [ITCS 2016] showed that this problem has an$$\tilde{O}(n^{2.94})$$O~(n2.94)nondeterministic time algorithm.

    Certifying that ann-variablek-CNF is unsatisfiable can be done in Merlin–Arthur time$$2^{n/2 - n/O(k)}$$2n/2-n/O(k). We also observe an algebrization barrier for the previous$$2^{n/2}\cdot \textrm{poly}(n)$$2n/2·poly(n)-time Merlin–Arthur protocol of R. Williams [CCC’16] for$$\#$$#SAT: in particular, his protocol algebrizes, and we observe there is no algebrizing protocol fork-UNSAT running in$$2^{n/2}/n^{\omega (1)}$$2n/2/nω(1)time. Therefore we have to exploit non-algebrizing properties to obtain our new protocol.

    Certifying a Quantified Boolean Formula is true can be done in Merlin–Arthur time$$2^{4n/5}\cdot \textrm{poly}(n)$$24n/5·poly(n). Previously, the only nontrivial result known along these lines was an Arthur–Merlin–Arthur protocol (where Merlin’s proof depends on some of Arthur’s coins) running in$$2^{2n/3}\cdot \textrm{poly}(n)$$22n/3·poly(n)time.

    Due to the centrality of these problems in fine-grained complexity, our results have consequences for many other problems of interest. For example, our work implies that certifying there is no Subset Sum solution tonintegers can be done in Merlin–Arthur time$$2^{n/3}\cdot \textrm{poly}(n)$$2n/3·poly(n), improving on the previous best protocol by Nederlof [IPL 2017] which took$$2^{0.49991n}\cdot \textrm{poly}(n)$$20.49991n·poly(n)time.

     
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