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1. Photo-response of the $$N=Z$$ nucleus $$^{24}$$Mg

   https://doi.org/10.1140/epja/s10050-023-01111-7  

   Deary, J; Scheck, M; Schwengner, R; O’Donnell, D; Bemmerer, D; Beyer, R; Hensel, Th; Junghans, A R; Kögler, T; Müller, S E; et al (September 2023, The European Physical Journal A)

   Abstract The electricE1 and magneticM1 dipole responses of the$$N=Z$$ $N=Z$nucleus$$^{24}$$ ${}^{24}$Mg were investigated in an inelastic photon scattering experiment. The 13.0 MeV electrons, which were used to produce the unpolarised bremsstrahlung in the entrance channel of the$$^{24}$$ ${}^{24}$Mg($$\gamma ,\gamma ^{\prime }$$ $\gamma ,{\gamma }^{\prime }$) reaction, were delivered by the ELBE accelerator of the Helmholtz-Zentrum Dresden-Rossendorf. The collimated bremsstrahlung photons excited one$$J^{\pi }=1^-$$ $J^{\pi }=1^{-}$, four$$J^{\pi }=1^+$$ $J^{\pi }=1^{+}$, and six$$J^{\pi }=2^+$$ $J^{\pi }=2^{+}$states in$$^{24}$$ ${}^{24}$Mg. De-excitation$$\gamma $$ $\gamma$rays were detected using the four high-purity germanium detectors of the$$\gamma $$ $\gamma$ELBE setup, which is dedicated to nuclear resonance fluorescence experiments. In the energy region up to 13.0 MeV a total$$B(M1)\uparrow = 2.7(3)~\mu _N^2$$ $B\left(M1\right)\uparrow =2.7\left(3\right){\mu }_N^2$is observed, but this$$N=Z$$ $N=Z$nucleus exhibits only marginalE1 strength of less than$$\sum B(E1)\uparrow \le 0.61 \times 10^{-3}$$ $\sum B\left(E1\right)\uparrow \le 0.61\times {10}^{-3}$ e$$^2 \, $$ ${}^2$fm$$^2$$ ${}^2$. The$$B(\varPi 1, 1^{\pi }_i \rightarrow 2^+_1)/B(\varPi 1, 1^{\pi }_i \rightarrow 0^+_{gs})$$ $B(\Pi 1,1_i^{\pi }\to 2_1^{+})/B(\Pi 1,1_i^{\pi }\to 0_{\mathrm{gs}}^{+})$branching ratios in combination with the expected results from the Alaga rules demonstrate thatKis a good approximative quantum number for$$^{24}$$ ${}^{24}$Mg. The use of the known$$\rho ^2(E0, 0^+_2 \rightarrow 0^+_{gs})$$ ${\rho }^2(E0,0_2^{+}\to 0_{\mathrm{gs}}^{+})$strength and the measured$$B(M1, 1^+ \rightarrow 0^+_2)/B(M1, 1^+ \rightarrow 0^+_{gs})$$ $B(M1,1^{+}\to 0_2^{+})/B(M1,1^{+}\to 0_{\mathrm{gs}}^{+})$branching ratio of the 10.712 MeV$$1^+$$ $1^{+}$level allows, in a two-state mixing model, an extraction of the difference$$\varDelta \beta _2^2$$ $\Delta {\beta }_2^2$between the prolate ground-state structure and shape-coexisting superdeformed structure built upon the 6432-keV$$0^+_2$$ $0_2^{+}$level. 

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2. Improved Merlin–Arthur Protocols for Central Problems in Fine-Grained Complexity

   https://doi.org/10.1007/s00453-023-01102-6  

   Akmal, Shyan; Chen, Lijie; Jin, Ce; Raj, Malvika; Williams, Ryan (February 2023, Algorithmica)

   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$$\tilde{O}(n)$$ $\tilde{O}\left(n\right)$. Previously, Carmosino et al. [ITCS 2016] showed that the problem has a nondeterministic algorithm running in$$\tilde{O}(n^{1.5})$$ $\tilde{O}\left(n^{1.5}\right)$time (that is, there is a proof system with proofs of length$$\tilde{O}(n^{1.5})$$ $\tilde{O}\left(n^{1.5}\right)$and a deterministic verifier running in$$\tilde{O}(n^{1.5})$$ $\tilde{O}\left(n^{1.5}\right)$time).Counting the number ofk-cliques with total edge weight equal to zero in ann-node graph can be done in Merlin–Arthur time$${\tilde{O}}(n^{\lceil k/2\rceil })$$ $\tilde{O}\left(n^{\lceil k/2\rceil }\right)$(where$$k\ge 3$$ $k\ge 3$). 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$${\tilde{O}}(m)$$ $\tilde{O}\left(m\right)$. 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$$\tilde{O}(n^2)$$ $\tilde{O}\left(n^2\right)$. Note this is optimal, as the matrix can have$$\Omega (n^2)$$ $\Omega \left(n^2\right)$nonzero entries in general. Previously, Carmosino et al. [ITCS 2016] showed that this problem has an$$\tilde{O}(n^{2.94})$$ $\tilde{O}\left(n^{2.94}\right)$nondeterministic time algorithm.Certifying that ann-variablek-CNF is unsatisfiable can be done in Merlin–Arthur time$$2^{n/2 - n/O(k)}$$ $2^{n/2-n/O\left(k\right)}$. We also observe an algebrization barrier for the previous$$2^{n/2}\cdot \textrm{poly}(n)$$ $2^{n/2}·\text{poly}\left(n\right)$-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$$2^{n/2}/n^{\omega (1)}$$ $2^{n/2}/n^{\omega \left(1\right)}$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$$2^{4n/5}\cdot \textrm{poly}(n)$$ $2^{4n/5}·\text{poly}\left(n\right)$. 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$$2^{2n/3}\cdot \textrm{poly}(n)$$ $2^{2n/3}·\text{poly}\left(n\right)$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$$2^{n/3}\cdot \textrm{poly}(n)$$ $2^{n/3}·\text{poly}\left(n\right)$, improving on the previous best protocol by Nederlof [IPL 2017] which took$$2^{0.49991n}\cdot \textrm{poly}(n)$$ $2^{0.49991n}·\text{poly}\left(n\right)$time. 

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3. On a Traveling Salesman Problem for Points in the Unit Cube

   https://doi.org/10.1007/s00453-024-01257-w  

   Balogh, József; Clemen, Felix Christian; Dumitrescu, Adrian (September 2024, Algorithmica)

   Abstract LetXbe ann-element point set in thek-dimensional unit cube$$[0,1]^k$$ ${[0,1]}^k$where$$k \ge 2$$ $k\ge 2$. According to an old result of Bollobás and Meir (Oper Res Lett 11:19–21, 1992) , there exists a cycle (tour)$$x_1, x_2, \ldots , x_n$$ $x_1,x_2,\dots ,x_n$through thenpoints, such that$$\left( \sum _{i=1}^n |x_i - x_{i+1}|^k \right) ^{1/k} \le c_k$$ ${\left({\sum }_{i=1}^n,{|x_i-x_{i+1}|}^k\right)}^{1/k}\le c_k$, where$$|x-y|$$ $|x-y|$is the Euclidean distance betweenxandy, and$$c_k$$ $c_k$is an absolute constant that depends only onk, where$$x_{n+1} \equiv x_1$$ $x_{n+1}\equiv x_1$. From the other direction, for every$$k \ge 2$$ $k\ge 2$and$$n \ge 2$$ $n\ge 2$, there existnpoints in$$[0,1]^k$$ ${[0,1]}^k$, such that their shortest tour satisfies$$\left( \sum _{i=1}^n |x_i - x_{i+1}|^k \right) ^{1/k} = 2^{1/k} \cdot \sqrt{k}$$ ${\left({\sum }_{i=1}^n,{|x_i-x_{i+1}|}^k\right)}^{1/k}=2^{1/k}·\sqrt{k}$. For the plane, the best constant is$$c_2=2$$ $c_2=2$and this is the only exact value known. Bollobás and Meir showed that one can take$$c_k = 9 \left( \frac{2}{3} \right) ^{1/k} \cdot \sqrt{k}$$ $c_k=9{\left(\frac{2}{3}\right)}^{1/k}·\sqrt{k}$for every$$k \ge 3$$ $k\ge 3$and conjectured that the best constant is$$c_k = 2^{1/k} \cdot \sqrt{k}$$ $c_k=2^{1/k}·\sqrt{k}$, for every$$k \ge 2$$ $k\ge 2$. Here we significantly improve the upper bound and show that one can take$$c_k = 3 \sqrt{5} \left( \frac{2}{3} \right) ^{1/k} \cdot \sqrt{k}$$ $c_k=3\sqrt{5}{\left(\frac{2}{3}\right)}^{1/k}·\sqrt{k}$or$$c_k = 2.91 \sqrt{k} \ (1+o_k(1))$$ $c_k=2.91\sqrt{k}(1+o_k\left(1\right))$. Our bounds are constructive. We also show that$$c_3 \ge 2^{7/6}$$ $c_3\ge 2^{7/6}$, which disproves the conjecture for$$k=3$$ $k=3$. Connections to matching problems, power assignment problems, related problems, including algorithms, are discussed in this context. A slightly revised version of the Bollobás–Meir conjecture is proposed. 

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4. Enumeration of Rooted Binary Unlabeled Galled Trees

   https://doi.org/10.1007/s11538-024-01270-8  

   Agranat-Tamir, Lily; Mathur, Shaili; Rosenberg, Noah A. (March 2024, Bulletin of Mathematical Biology)

   Abstract Rooted binarygalledtrees generalize rooted binary trees to allow a restricted class of cycles, known asgalls. We build upon the Wedderburn-Etherington enumeration of rooted binary unlabeled trees withnleaves to enumerate rooted binary unlabeled galled trees withnleaves, also enumerating rooted binary unlabeled galled trees withnleaves andggalls,$$0 \leqslant g \leqslant \lfloor \frac{n-1}{2} \rfloor $$ $0⩽g⩽\lfloor \frac{n-1}{2}\rfloor$. The enumerations rely on a recursive decomposition that considers subtrees descended from the nodes of a gall, adopting a restriction on galls that amounts to considering only the rooted binarynormalunlabeled galled trees in our enumeration. We write an implicit expression for the generating function encoding the numbers of trees for alln. We show that the number of rooted binary unlabeled galled trees grows with$$0.0779(4.8230^n)n^{-\frac{3}{2}}$$ $0.0779(4.{8230}^n)n^{-\frac{3}{2}}$, exceeding the growth$$0.3188(2.4833^n)n^{-\frac{3}{2}}$$ $0.3188(2.{4833}^n)n^{-\frac{3}{2}}$of the number of rooted binary unlabeled trees without galls. However, the growth of the number of galled trees with only one gall has the same exponential order 2.4833 as the number with no galls, exceeding it only in the subexponential term,$$0.3910n^{\frac{1}{2}}$$ $0.3910n^{\frac{1}{2}}$compared to$$0.3188n^{-\frac{3}{2}}$$ $0.3188n^{-\frac{3}{2}}$. For a fixed number of leavesn, the number of gallsgthat produces the largest number of rooted binary unlabeled galled trees lies intermediate between the minimum of$$g=0$$ $g=0$and the maximum of$$g=\lfloor \frac{n-1}{2} \rfloor $$ $g=\lfloor \frac{n-1}{2}\rfloor$. We discuss implications in mathematical phylogenetics. 

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5. Transversality of holomorphic maps into hyperquadrics

   https://doi.org/10.1007/s00208-025-03134-5  

   Huang, Xiaojun; Zhu, Weixia (March 2025, Mathematische Annalen)

   Abstract We study holomorphic mapsFfrom a smooth Levi non-degenerate real hypersurface$$ M_{\ell }\subset {\mathbb {C}}^n $$ $M_{\ell }\subset C^n$into a hyperquadric$$ {\mathbb {H}}_{\ell '}^N $$ $H_{{\ell }^{\prime }}^N$with signatures$$ \ell \le (n-1)/2 $$ $\ell \le (n-1)/2$and$$ \ell '\le (N-1)/2,$$ ${\ell }^{\prime }\le (N-1)/2,$respectively. Assuming that$$ N - n < n - 1,$$ $N-n<n-1,$we prove that if$$ \ell = \ell ',$$ $\ell ={\ell }^{\prime },$thenFis either CR transversal to$$ {\mathbb {H}}_{\ell }^N $$ $H_{\ell }^N$at every point of$$ M_{\ell },$$ $M_{\ell },$or it maps a neighborhood of$$ M_{\ell } $$ $M_{\ell }$in$$ {\mathbb {C}}^n $$ $C^n$into$$ {\mathbb {H}}_{\ell }^N.$$ $H_{\ell }^N.$Furthermore, in the case where$$ \ell ' > \ell ,$$ ${\ell }^{\prime }>\ell ,$we show that ifFis not CR transversal at$$0\in M_\ell ,$$ $0\in M_{\ell },$then it must be transversally flat. The latter is best possible. 

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