Hexa/tria lattice generation

How do we generate a regular hexagon / triangle combination of repeating cell geometry shown : (Plastic chair seat weave, Star of David).

Unlike a pure triangle lattice where a node has three intersecting lines this one has alternate sides disappearing when two lines intersect.

Can we modify this recent program? Recent Tria lattice generation

enter image description here

Above base pattern is made in a mechanical software but is cumbersome.

Thanks for all help.

Enumerating points on the integer lattice, within a sphere, sorted by angle, in O(1) space

Inspired by this StackOverflow question: https://stackoverflow.com/questions/63346135

(it was not clearly presented, and got closed)

Let’s say I wanted to enumerate all the 3D points on the integer lattice, within a sphere, in order of the angle between the vector to the point and the up vector (say z).

Could I do this in O(1) space efficiently?

All I can find is:

Remembering last point (init at (0,0,0)). (O(1) memory) while true     init best dot product to 0     going through all 3D points (three nested for's with radius range)         if this point has better dot product but least than last point             keep this point as best     if best dot product is still 0, exit      pick best point as current point (this is where the listing occurs)     update last point to best point 

Not only is this absolutely slow, it also needs the use of integer math for dot product and length so that numerical precision doesn’t mess with symmetrical points and it would also need a tweak to guarantee symmetrical points are listed in a known order.

Is there any good algorithms that would apply here?

Boolean Lattice Implementation

Let $ X = \{1,2,\dots n\}$ , and $ Y= \{ S\subset X: |S|\le 2\}$ . I am interested in "avoidance sets" $ A \subset Y$ which logically are supposed to correspond to $ f(A):=\bigwedge_{a\in A} \lnot a$ in the Boolean lattice defined over $ \mathcal{P}(X)$ . (if $ A = \{\}$ it shall correspond to $ X$ , the maximal element of our lattice)

Given a list of avoidance sets $ A_1,A_2,\dots A_k$ , I want to return an avoidance set $ A’$ such that $ f(A’)$ is the least upper bound of $ \bigvee_{1\le i \le k} f(A_i)$ . (by least upper bound I mean the least upper bound which can be represented by an avoidance set)

Can this be done in time linear to $ \sum_{1\le i \le k} |A_i|$ ? (you may assume that all the sets $ A_i$ are simplified, i.e. if $ \{a\}\in A_i$ and $ \{b\}\in A_i$ then $ \{a,b\} \not \in A_i$ )

Fully extended TQFT and lattice models

I often read that fully extended TQFTs are supposed to classify topological phases of matter. So I would like to understand the formal nature of fully extended TQFTs on a more direct physical level (without having to read up on a huge amount of category theory language):

1) Many algebraic structures are given by a finite set of tensors (linear maps if you wish) which fulfil a finite set of tensor-network equations (equations between different compositions of the linear maps if you wish). For example, fusion categories are given by a 10-index $ F$ -tensor satisfying the pentagon equation, or (ordinary axiomatic) 2-dimensional TQFTs are given by a bunch of tensors associated to the pair of pants and a few other cobordisms, satisfying the axioms of (something like) a Frobenius algebra.

Can $ n$ -dimensional fully extended TQFTs be formulated as a finite set of tensors obeying a finite set of tensor-network axioms? Is it known what these tensors and axioms are?

2) Is there any idea for a construction of a local partition function (similar to the Turaev-Viro construction) that takes data related to a fully extended TQFT as input? I guess for Turaev-Viro models (and any other topological state-sum construction) it’s possible to find a corresponding extended TQFT. Are there any examples of extended TQFTs that are conjectured to correspond to phases without known state-sum constructions (such as chiral topological phases in 2+1D)?

Which complete orthocomplemented lattices arise as the lattice of ‘regular opens’ in a closure space?

Every complete Boolean algebra arises as the lattice of regular open sets in some topological space, namely given a complete Boolean algebra $ B$ , the corresponding Stone space $ S(B)$ will be extremally disconnected, so in particular its regular open sets are precisely its clopen sets.

A semi-common generalization of topological spaces is the class of spaces $ (X,\tau)$ such that $ \varnothing,X\in\tau$ and such that $ \tau$ is closed under arbitrary unions (no assumption about intersections). This is equivalent to a weakening of the Kuratowski closure axioms to the axioms of a general closure operator. (Wikipedia says that these are sometimes called ‘closure spaces,’ but I’ve also seen them called ‘generalized topological spaces’ in some papers. If anyone knows another name for these things I would really like to know it because searching for relevant literature has been difficult.)

A closure operator is equivalently specified by an interior operator, a function $ \mathrm{int}:\mathcal{P}(X)\rightarrow \mathcal{P}(X)$ satisfying:

  • $ \mathrm{int}(A)\subseteq A$
  • $ A\subseteq B \Rightarrow \mathrm{int}(A)\subseteq \mathrm{int}(B)$
  • $ \mathrm{int}(\mathrm{int}(A))=\mathrm{int}(A)$

(This definition makes sense in an arbitrary poset.) Under such an operator, the collection of open sets (sets such that $ \mathrm{int}(U)=U$ ) forms a complete lattice with arbitrary meets given by set theoretic union. Unlike in a topological space finite joins are given by $ U \wedge V = \mathrm{int}(U \wedge V)$ . (This lattice is dual to the lattice of closed sets, but I’m phrasing this question in terms of open sets since I find it a little more familiar and because there’s no standard name for the closure of complement operation.)

An interior operator gives rise to a natural exterior operator: $ \mathrm{ext}(A)=\mathrm{int}(X \setminus A)$ . Just like in a topological space one can show that for an open set $ U$ , $ \mathrm{ext}(\mathrm{ext}(\mathrm{ext}(U)))=\mathrm{ext}(U)$ , so call an open set $ U$ ‘regular’ if it satisfies $ \mathrm{ext}(\mathrm{ext}(U))=U$ . (Note that unlike in topology it seems that the exterior operator is not definable from the structure of the lattice of opens alone.)

The map $ U\mapsto \mathrm{ext}(\mathrm{ext}(U))$ gives a closure operator on the lattice of open sets, so we get that the collection of regular opens also forms a complete lattice whose join is the join of open sets (although again unlike in topology this is not set theoretic intersection) but whose meet is different, so in particular neither operation is set theoretic.

One can show that the collection of regular opens together with their meet, join, and the exterior operator form an orthocomplemented lattice, with the exterior playing the role of orthocomplementation, i.e. $ ext$ is an order-reversing involution such that $ U$ and $ \mathrm{ext}(U)$ are always complements. This is a purely algebraic fact about Boolean algebras with interior operators satisfying the above axioms. With a finite counterexample search I found that in general the lattice does not need to be orthomodular.

I convinced myself that every complete lattice arises as the lattice of opens of a closure space, although it’s easier to phrase in terms of closed sets and the dual lattice. Given a lattice $ (L,0,1,\wedge,\vee)$ we can consider the collection of subsets of $ L\setminus \{0\}$ and let a set be closed if it is of the form $ (0,a]$ for some $ a\in L\setminus \{0\}$ (so in particular the closure operator is $ \mathrm{cl}(A)=(0,\bigvee A]$ ). By completeness of the lattice any intersection of sets of this form is of this form or empty and in particular the intersection’s top is the greatest lower bound of the tops of the original sets, so the lattices agree (if we identify $ \varnothing$ with $ 0$ , this is why we need to remove $ 0$ , otherwise we get an extra bottom element). To get a lattice of opens we can just start with the dual lattice.

That construction always produces trivial lattices of regular opens (no non-trivial open sets are disjoint, so the exterior of any non-empty set is empty), which shows that the exterior operator really does depend on the underlying set and not just the lattice of opens (since there are closure spaces with non-trivial lattices of regular opens). So finally we can come to the question:

Which complete orthocomplemented lattices arise as the lattice of regular opens in a closure space?

Counting the Number of Lattice Points in an $n$-Dimensional Sphere

Let $ S_n(R)$ denote the number of lattice points in an $ n$ -dimensional “sphere” with radius $ R$ . For clarification, I am interested in lattice points found both strictly inside the sphere, and on its surface. I want to count exactly how many such points there are. This count corresponds to the amount of sets of integers that are solutions to the equation $ $ a_1^2+a_2^2+a_3^2+…+a_{n-1}^2+a_n^2= R^2$ $ where the $ a_i$ ‘s are required to be integers, not necessarily positive, and sets that differ just by the order of the summands are also considered distinct, i.e for the case $ n=3, R=5$ , the following are both counted: $ 3^2+(-4)^2+0^2$ and $ (-4)^2+0^2+3^2$ . This can also be interpreted as the sum of squares function$ r_d(k)$ denotes in how many such ways I can write $ k$ as the sum of $ d$ squares. This means that $ $ S_n(R) = \sum_{k=1}^{R^2}r_n(k)$ $

I am asking this question as a general one, but I am mostly concerned about the cases of $ n=2, n=3,n=4$ .

In order not to be too lengthy and remain focused, I won’t explain why but summing this way requires to factor each $ k$ first. This is very time consuming, so I searched for better ways.

For the case of $ n=2$ , which is essentially just the count of lattice points in a circle of radius $ R$ , there is a lot of information – this is known as the Gauss Circle Problem and I have managed to find that $ $ S_2(R)= 1+\sum_{i=1}^\infty\biggl(\biggl \lfloor\frac{R^2}{4i+1}\biggl \rfloor-\biggl \lfloor\frac{R^2}{4i+3}\biggl \rfloor\biggl )$ $ For the case of $ n=4$ , I found: $ $ S_4(R)= 1+8\sum_{k=1}^{R^2}\sum_{d|k \atop {4\nmid d}}d$ $ Unfortunately, for the case of a sphere ($ n=3$ ), I have found no formula, neither for the count of lattice points inside nor on the surface of the sphere.

Although these formulas do indeed provide the count of lattice points, they are very slow in computational terms, as their running time complexity is $ O(R^2)$ . I was wondering if a more efficient way exists to count the number of such lattice points, perhaps $ O(R)$ or $ O(R^{1/2+\epsilon})$ , so that I can work with radii as big as $ 10^9$ or even more. I suspect there is a way, but I can’t find it. Not necessarily a different approach to the problem, but rather just a more clever way to reduce the order of the sum. Also, any insight about $ S_3(R)$ will be appreciated.

Reference request: lower sets of a preorder form a lattice

Consider a set $ S$ with a preorder $ \preceq$ (a preorder is a reflexive and transitive relation). A lower set $ A$ of $ S$ is defined as a subset of $ S$ such that for all $ x \in S$ and $ y \in A$ , if $ x \preceq y$ then $ x \in A$ .

I believe the set of all lower sets of $ S$ form a complete, distributive lattice. Is there a reference which states this?

Minimize number of lattice paths below a given path

Every north-east lattice path (NE-path) $ v$ from $ (0,0)$ to $ (k, \lambda _k)$ can be identified with a sequence $ 0 \le \lambda_1 \le \lambda_2 \le . . . \le \lambda_k$ , that represent the hight of each step. There is a well known formula for the number of NE-paths from $ (0,0)$ to $ (k, \lambda _k)$ below the $ v$ , namely $ $ \ell(v)= \det_{1 \le i,j \le k} \left( \binom{\lambda_i+1}{j-i+1} \right).$ $ Now my question is, for a given number $ n$ , is it known which shape $ 0 \le \lambda_1 \le \lambda_2 \le . . . \le \lambda_k$ with $ \sum_{i=1}^k \lambda_i=n$ that gives the smallest value for $ \ell(v)$ ? I. e. which shape sould I give my path, so that there is as few paths below it as possible?

Feynman-Kac formula for lattice heat equation with non-diagonal potential

Suppose that $ X$ is the continuous-time simple symmetric random walk on the lattice $ \mathbb Z^d$ (i.e., a simple symmetric random walk with i.i.d. exponential jump times), and let $ $ u(t,x):=\mathbf E\left[\exp\left(\int_0^tV(X_s)~ds\right)f(X_t)\bigg|X_0=x\right] \tag{1}$ $ for $ (t,x)\in[0,\infty)\times\mathbb Z^d$ , where $ V,f:\mathbb Z^d\to\mathbb R$ are functions.

According to the Feynman-Kac formula, we know that $ u(t,x)$ solves the lattice/matrix heat equation $ $ \partial_tu=\tfrac12\Delta u+Vu,\qquad u(0,x)=f(x),\tag{2}$ $ where $ $ \Delta:=\left[ \begin{array}{ccccc} &\ddots&\ddots&\ &\ddots&-2&1&\ &&1&-2&1&\ &&&1&-2&\ddots\ &&&&\ddots&\ddots& \end{array}\right]$ $ is the discrete Laplacian, and we think of $ V$ as the diagonal matrix $ $ V=\left[ \begin{array}{ccccc} &\ddots&&\ &&V(-1)&&\ &&&V(0)&&\ &&&&V(1)&&\ &&&&&\ddots&& \end{array}\right].$ $


As an alternative to $ (2)$ , a common model for a lattice heat equation is to consider $ $ \partial_tu=\tfrac12\Delta u+\tilde Vu,\qquad u(0,x)=f(x),\tag{3}$ $ where the potential $ \tilde V$ is instead of the form $ $ \tilde V=\left[ \begin{array}{ccccc} &\ddots&\ddots&\ &\ddots&0&V(-1)&\ &&V(-1)&0&V(0)&\ &&&V(0)&0&V(1)&\ &&&&V(1)&0&\ddots\ &&&&&\ddots&\ddots& \end{array}\right],$ $ or a more general tridiagonal matrix $ $ \tilde V=\left[ \begin{array}{ccccc} &\ddots&\ddots&\ &\ddots&U(-1)&V(-1)&\ &&V(-1)&U(0)&V(0)&\ &&&V(0)&U(1)&V(1)&\ &&&&V(1)&U(2)&\ddots\ &&&&&\ddots&\ddots& \end{array}\right].$ $


Question. Does there exist a Feynman-Kac formula similar to $ (1)$ for lattice operators with non-diagonal potential such as $ (3)$ ?

To clarify a bit what I mean by similar to $ (1)$ : It’s easy enough to come up with some probabilistic representation of the solution of $ (3)$ (for example by using the Trotter-Kato theorem: $ e^{\Delta/2+V}\approx(e^{\Delta/2n}e^{V/n})^n$ for large $ n$ ), but I can’t get anything nice like $ (1)$ , and It’s not clear to me if we should/shouldn’t expect such a nice representation in those cases.