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1. Polynomial Bounds for Chromatic Number. IV: A Near-polynomial Bound for Excluding the Five-vertex Path

   https://doi.org/10.1007/s00493-023-00015-w  

   Scott, Alex; Seymour, Paul; Spirkl, Sophie (October 2023, Combinatorica)

   Abstract A graphGisH-freeif it has no induced subgraph isomorphic toH. We prove that a$$P_5$$ $P_5$-free graph with clique number$$\omega \ge 3$$ $\omega \ge 3$has chromatic number at most$$\omega ^{\log _2(\omega )}$$ ${\omega }^{{log}_2\left(\omega \right)}$. The best previous result was an exponential upper bound$$(5/27)3^{\omega }$$ $(5/27)3^{\omega }$, due to Esperet, Lemoine, Maffray, and Morel. A polynomial bound would imply that the celebrated Erdős-Hajnal conjecture holds for$$P_5$$ $P_5$, which is the smallest open case. Thus, there is great interest in whether there is a polynomial bound for$$P_5$$ $P_5$-free graphs, and our result is an attempt to approach that. 

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2. Algorithms for covering multiple submodular constraints and applications

   https://doi.org/10.1007/s10878-022-00874-x  

   Chekuri, Chandra; Inamdar, Tanmay; Quanrud, Kent; Varadarajan, Kasturi; Zhang, Zhao (June 2022, Journal of Combinatorial Optimization)

   Abstract We consider the problem of covering multiple submodular constraints. Given a finite ground setN, a weight function$$w: N \rightarrow \mathbb {R}_+$$ $w:N\to R_{+}$,rmonotone submodular functions$$f_1,f_2,\ldots ,f_r$$ $f_1,f_2,\dots ,f_r$overNand requirements$$k_1,k_2,\ldots ,k_r$$ $k_1,k_2,\dots ,k_r$the goal is to find a minimum weight subset$$S \subseteq N$$ $S\subseteq N$such that$$f_i(S) \ge k_i$$ $f_i\left(S\right)\ge k_i$for$$1 \le i \le r$$ $1\le i\le r$. We refer to this problem asMulti-Submod-Coverand it was recently considered by Har-Peled and Jones (Few cuts meet many point sets. CoRR.arxiv:abs1808.03260Har-Peled and Jones 2018) who were motivated by an application in geometry. Even with$$r=1$$ $r=1$Multi-Submod-Covergeneralizes the well-known Submodular Set Cover problem (Submod-SC), and it can also be easily reduced toSubmod-SC. A simple greedy algorithm gives an$$O(\log (kr))$$ $O(log(kr\left)\right)$approximation where$$k = \sum _i k_i$$ $k={\sum }_i{k}_i$and this ratio cannot be improved in the general case. In this paper, motivated by several concrete applications, we consider two ways to improve upon the approximation given by the greedy algorithm. First, we give a bicriteria approximation algorithm forMulti-Submod-Coverthat covers each constraint to within a factor of$$(1-1/e-\varepsilon )$$ $(1-1/e-\epsilon )$while incurring an approximation of$$O(\frac{1}{\epsilon }\log r)$$ $O(\frac{1}{ϵ}logr)$in the cost. Second, we consider the special case when each$$f_i$$ $f_i$is a obtained from a truncated coverage function and obtain an algorithm that generalizes previous work on partial set cover (Partial-SC), covering integer programs (CIPs) and multiple vertex cover constraints Bera et al. (Theoret Comput Sci 555:2–8 Bera et al. 2014). Both these algorithms are based on mathematical programming relaxations that avoid the limitations of the greedy algorithm. We demonstrate the implications of our algorithms and related ideas to several applications ranging from geometric covering problems to clustering with outliers. Our work highlights the utility of the high-level model and the lens of submodularity in addressing this class of covering problems. 

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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. Undecidable Translational Tilings with Only Two Tiles, or One Nonabelian Tile

   https://doi.org/10.1007/s00454-022-00426-4  

   Greenfeld, Rachel; Tao, Terence (January 2023, Discrete & Computational Geometry)

   Abstract We construct an example of a group$$G = \mathbb {Z}^2 \times G_0$$ $G=Z^2\times G_0$for a finite abelian group $$G_0$$ $G_0$, a subsetEof $$G_0$$ $G_0$, and two finite subsets$$F_1,F_2$$ $F_1,F_2$of G, such that it is undecidable in ZFC whether$$\mathbb {Z}^2\times E$$ $Z^2\times E$can be tiled by translations of$$F_1,F_2$$ $F_1,F_2$. In particular, this implies that this tiling problem isaperiodic, in the sense that (in the standard universe of ZFC) there exist translational tilings ofEby the tiles$$F_1,F_2$$ $F_1,F_2$, but no periodic tilings. Previously, such aperiodic or undecidable translational tilings were only constructed for sets of eleven or more tiles (mostly in $$\mathbb {Z}^2$$ $Z^2$). A similar construction also applies for$$G=\mathbb {Z}^d$$ $G=Z^d$for sufficiently large d. If one allows the group$$G_0$$ $G_0$to be non-abelian, a variant of the construction produces an undecidable translational tiling with only one tile F. The argument proceeds by first observing that a single tiling equation is able to encode an arbitrary system of tiling equations, which in turn can encode an arbitrary system of certain functional equations once one has two or more tiles. In particular, one can use two tiles to encode tiling problems for an arbitrary number of tiles. 

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5. Finding solutions with distinct variables to systems of linear equations over $$\mathbb {F}_p$$

   https://doi.org/10.1007/s00208-022-02391-y  

   Sauermann, Lisa (April 2022, Mathematische Annalen)

   Abstract Let us fix a primepand a homogeneous system ofmlinear equations$$a_{j,1}x_1+\dots +a_{j,k}x_k=0$$ $a_{j,1}{x}_1+\cdots +a_{j,k}{x}_k=0$for$$j=1,\dots ,m$$ $j=1,\cdots ,m$with coefficients$$a_{j,i}\in \mathbb {F}_p$$ $a_{j,i}\in F_p$. Suppose that$$k\ge 3m$$ $k\ge 3m$, that$$a_{j,1}+\dots +a_{j,k}=0$$ $a_{j,1}+\cdots +a_{j,k}=0$for$$j=1,\dots ,m$$ $j=1,\cdots ,m$and that every$$m\times m$$ $m\times m$minor of the$$m\times k$$ $m\times k$matrix$$(a_{j,i})_{j,i}$$ ${\left(a_{j,i}\right)}_{j,i}$is non-singular. Then we prove that for any (large)n, any subset$$A\subseteq \mathbb {F}_p^n$$ $A\subseteq F_p^n$of size$$|A|> C\cdot \Gamma ^n$$ $\left|A\right|>C·{\Gamma }^n$contains a solution$$(x_1,\dots ,x_k)\in A^k$$ $(x_1,\cdots ,x_k)\in A^k$to the given system of equations such that the vectors$$x_1,\dots ,x_k\in A$$ $x_1,\cdots ,x_k\in A$are all distinct. Here,Cand$$\Gamma $$ $\Gamma$are constants only depending onp,mandksuch that$$\Gamma $\Gamma <p$. The crucial point here is the condition for the vectors$$x_1,\dots ,x_k$$ $x_1,\cdots ,x_k$in the solution$$(x_1,\dots ,x_k)\in A^k$$ $(x_1,\cdots ,x_k)\in A^k$to be distinct. If we relax this condition and only demand that$$x_1,\dots ,x_k$$ $x_1,\cdots ,x_k$are not all equal, then the statement would follow easily from Tao’s slice rank polynomial method. However, handling the distinctness condition is much harder, and requires a new approach. While all previous combinatorial applications of the slice rank polynomial method have relied on the slice rank of diagonal tensors, we use a slice rank argument for a non-diagonal tensor in combination with combinatorial and probabilistic arguments. 

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