Using decision oracle to solve optimization problem of maximum polyomino tiling

So, this problem is a kind of variant of polyomino packing which has been discussed frequently elsewhere, but I haven’t been able to find anything on my particular problem. Suppose we have a list of polyominos $ p_1, p_2, …, p_n$ (not necessarily distinct), and we want to find a tiling of a rectangle of dimension $ a \times b$ with $ a, b \leq n$ that maximizes the number of squares covered, where we can use each $ p_i$ at most once, and polyominos must be fully contained within the rectangle. Now, we have the decision problem which tells us, for a given $ t$ , if there is some tiling covering at least $ t$ squares, and the optimization problem which is finding a tiling that covers the maximum number of squares. There are two parts: first, if you can solve the optimization problem in polynomial time, can you solve the decision problem in polynomial time? And secondly, if you can solve the decision problem in polynomial problem, can you solve the optimization problem in polynomial time?

If we have an oracle that solves the optimization in polynomial time, solving the decision problem in polynomial time is easy. However, given an oracle for the decision problem, I was unable to find a way to solve the optimization problem in polynomial time. The main issue I’m facing is that the decision oracle only works for rectangular boards, which means we can’t just place pieces and then use the oracle to see if the placement works, since we won’t have a rectangular board if we want to exclude the piece we just placed. It isn’t hard to determine the actual maximum number of tiles you can cover, and you can even find the actual pieces you need to use, but I haven’t been able to figure out a way to find an arrangement of the pieces in polynomial time using the oracle. I assume there is some trick here, but I don’t see it.

Number of ways of tiling a 3*N board with 2*1 dominoes problem

I came across this problem, Tiling with Dominoes and initially I faced difficulty in understanding the logic behind recurrence relation, but after reading it from here , I understood it. But I had a doubt, why aren’t we considering the below cases for the above problem?

In the solution only these cases are mentioned,

********   AA*******   AA******   A******* ******** = BB******* + B******* + A******* ********   CC*******   B*******   BB******   f(n)   =  f(n-2)   +  g(n-1)  +  g(n-1) 

But my question is there can be other case as well, like below, why aren’t we adding them, because the problem asks for the number of possible cases-

********   AA*******   AA******   A*******   AB******   CC****** ******** = BB******* + B******* + A******* + AB****** + AB****** ********   CC*******   B*******   BB******   CC******   AB******   f(n)   =  f(n-2)   +  g(n-1)  +  g(n-1)  +   f(n-2) +  f(n-2) 

Why aren’t we adding them? Thanks for your help.

Tiling $\mathbb{R}^n$ with a single shape of positive measure

Let $ n>1$ be an integer. For $ A\subseteq \mathbb{R}^n$ and $ x\in\mathbb{R}^n$ we set $ x+A = \{x+a: a\in A\}$ .

Is there a measurable subset $ A\subseteq \mathbb{R}^n$ with positive $ n$ -dimensional Lebesgue measure and the following properties?

  1. $ \mathbb{R}^n = \bigcup\{z+A: z\in \mathbb{Z}^n\}$ , and
  2. if $ z_1\neq z_2\in \mathbb{Z}^n$ then $ (z_1+A)\cap (z_2+A) = \emptyset$ .

Tiling the surface of a hypersphere with regular simplices

Let $ S^{n-1} = \{x \in \mathbb{R}^n : x_1^2 + \cdots + x_n^2 = 1\}$ . Consider a regular spherical simplex, obtained e.g. by taking a hyperspherical cap, picking $ n$ equally-spaced points $ P = \{p_1, \dots, p_n\}$ on the boundary (which is isomorphic to $ S^{n-2}$ ), then taking the boundaries of the simplex to be the great (hyper)circles through each $ P \setminus \{p_i\}$ .

For $ n \geq 4$ , we know that $ \mathbb{R}^{n-1}$ cannot be tiled by congruent regular simplices. Despite this, I see two ways to tile $ S^{n-1}$ with congruent regular spherical simplices: (1) the orthants are regular spherical simplices of side length $ \pi/2$ (there are $ 2^n$ of them); (2) inscribing a regular simplex in $ S^{n-1}$ , then taking the natural partition of the surface tiles $ S^{n-1}$ with $ n+1$ spherical simplices.

Are there other ways to tile $ S^{n-1}$ with congruent regular spherical simplices? Feel free to assume $ n \geq 4$ .

Computational complexity of Wang tiling of a finite plane

For a given set of Wang tiles on an $ L \times L$ plane (where a tile is $ 1\times 1$ ) we first need to determine whether a tiling of the plane is possible. If so, we can then consider the problem of how to construct such a tiling.

Is it known what the computational complexity of these problems are (i.e., how they scale with $ L$ )?

I’ve searched and found a lot of interesting info about the problem. But I’ve seen no statement of a complexity.

Computational complexity of Wang tiling of a finite plane

For a given set of Wang tiles on an $ L \times L$ plane (where a tile is $ 1\times 1$ ) we first need to determine whether a tiling of the plane is possible. If so, we can then consider the problem of how to construct such a tiling.

Is it known what the computational complexity of these problems are (i.e., how they scale with $ L$ )?

I’ve searched and found a lot of interesting info about the problem. But I’ve seen no statement of a complexity.

Is there a tiling window manager, like ChunkWM, that doesn’t require disabling SIP?

After spending a long time trying to find some software that worked for me on OSX and enabled me to be productive, I found ChunkWM. Now my company’s new jamf policy requires SIP be enabled which basically means that the scripting additions get disabled and all the cross-“Spaces” functionality provided by the Scripting Additions is lost.

Is there any thing else for OSX, like ChunkWM, that can work with SIP?

Can the (infinite) graph of a tiling of the plane have no two vertices of the same degree?

I mean for each region to have three or more edges in its boundary. If there’s a term for that I’d appreciate a comment as to the term. Another way to state this requirement is that the graph not have loops or parallel edges, sort of an infinite version of a simple graph.

I tried starting with a triangular region and making its vertices have degrees $ 3,4,5$ by its three adjoining regions, but quickly got a rather ungainly mess as I tried to make more regions while avoiding two same-degree vertices.

Any insight (or reference) appreciated. Thanks