# A funny way to prove that the set of primes is not regular

It is well known that $\text{Primes}= \{ a^p \mid p \text{ is prime}\}$ is not regular. The standard proof uses the pumping lemma for regular languages, but you can also use Parikh’s theorem or Myhill-Nerode theorem: see this question on cs.stackexchange.com. I tried to figure out another way to prove it, and came out with an alternate proof that uses a bit of number theory and Busy Beavers.

First of all: Theorem 1 (Prime gap theorem). There are gaps between primes that are arbitrarily large.

# The Power of One-State Turing Machines

A one-state Turing machine is a very weak device: it has no internal memory and it cannot even recognize the trivial language $L = \{1\}$. Its transition function is a simple map $\delta : \Sigma \to \Sigma \times \{L,R\}$, i.e. given the symbol under its head it can rewrite it with another one and move left or right, but the state remains the same. Nevertheless it can use its ability to write on the tape to “gain” some memory; in particular in each cell $C_i$ of the tape it can store:

• the number $n$ of times the head visited $C_i$ modulo a fixed constant $k$;
• if it has entered $C_i$ from the left or from the right.

This information is enough to build $k$-ary counters and also a sort of comparator between numbers written in different bases. So, some one-state Turing machines can “recognize” languages that are not Context-Free (the quotes are due to a different  accept/reject conditions that are necessary because we cannot distinguish between accepting and non accepting states).

# Busy Beavers are (allmost) incompressible

After a long time, here we are again … let’s resume with a small (and rather trivial) post …

Busy Beavers are allmost incompressible! … easy fact, but I didn’t find it anywhere.

# Using Kolmogorov complexity to solve the Halting problem

We assume that reader is familiar with the notions of undecidability, Turing reductions, Kolmogorov complexity, Halting problem, and related subjects.

The (easy) proof that the uncomputability of Kolmogorov complexity implies the undecidability of the Halting problem can be found in many lectures notes and books; usually the proof assumes that the Halting problem is decidable and derive the  computability of Kolmogorov complexity which is a contradiction. In other words given an oracle for the Halting problem, we can compute the Kolmogorov complexity of a string $x$.

But we can also derive the uncomputability of Kolmogorov complexity from the undecidability of the Halting problem; the proof is “less popular” but nevertheless can be found after a few searches on Google. For example the technical report: Gregory J. Chaitin, Asat Arslanov, Cristian Calude: Program-size Complexity Computes the Halting Problem. Bulletin of the EATCS 57 (1995) contains two different proofs, and the great book Li, Ming, Vitányi, Paul M.B.; An Introduction to Kolmogorov Complexity and Its Applications presents it as an exercise (with a hint on how to solve it that is credited to P. Gács by W. Gasarch in a personal communication Feb 13, 1992). Here we give an extended proof, with more details, that the Halting problem can be decided using an oracle that computes the Kolmogorov complexity of a string, i.e. that the Halting problem is Turing reducible to the Kolmogorov complexity. Continue reading

# Subway Shuffle is PSPACE-complete

Abstract: Subway shuffle is an addicting puzzle game created by Bob Hearn. It is played on a graph with colored edges that represent subway lines; colored tokens that represent subway cars are placed on the nodes of the graph. A token can be moved from its current node to an empty one, but only if the two nodes are connected with an edge of the same color of the token. The aim of the game is to move a special token to its final target position. We prove that deciding if the game has a solution is PSPACE-complete even when the game graph is planar.

# Weapon-target assignment problem

The decision version of the Weapon-Target Assignment (WTA) problem is as follows:

There are a number of weapons and a number of targets. The weapons are of type $i = 1, \ldots, m$ . There are $W_{i}$ available weapons of type $i$. Similarly, there are $j = 1, \ldots, n$ targets, each with a value of $V_{j}$ . Any of the weapons can be assigned to any target. Each weapon type has a certain probability of destroying each target, given by $p_{i,j}$ . Does there exist a weapon-target assignment such that:

$$D = \sum_{j=1}^n V_j \prod_{i=1}^m q_{i,j}^{x_{i,j}} \leq k$$

where $k$ is a given integer; $x_{i,j} = 1$ if weapon $i$ is assigned to target $j$, $x_{i,j} =0$ otherwise; $q_{i,j} = 1 – p_{i,j}$ is the survival probability of target $j$ when hit by weapon $i$. No more than $W_i$ weapons of type $i$ can be used, so we must add the following constraint:

$$\sum_{j=1}^n x_{i,j} \leq W_i \quad \mbox{ for } i = 1,\ldots,m$$

WTA problem is NP complete (see L. Chen, C. J. Ren, and S. Deng, “An efficient approximation for weapon-target assignment,” vol. 1, Computing, Communication, Control, and Management. ISES International Colloquium, 2008, pp. 764-767), we give an alternate hardness proof using a reduction from the NP-complete SUBSET PRODUCT problem. Continue reading

# The Crazy Frog Puzzle and Permutation Reconstruction from Differences

The Crazy Frog Puzzle (CFP) is the following: we have a $n \times n$ partially filled board, a crazy frog placed on an empty cell and a sequence of horizontal, vertical, and diagonal jumps; each jump has a fixed length and two possible opposite directions. The crazy frog must follow the sequence of jumps, and at each step it can only choose among the two available directions. Can the frog visit every empty cell of the board exactly once (it cannot jump on a blocked cell and on the same cell two times)?

Figure 1: an example of the Crazy Frog Puzzle on the left and its solution on the right.

We prove that the Crazy Frog Puzzle is $\sf{NP}$-complete even if restricted to 1 dimension and even without blocked cells. The 1-D CFP without blocked cells corresponds to the problem of reconstructing a permutation from its differences:

Permutation Reconstruction from Differences problem:
Input: a set of $n-1$ distances $a_1,a_2,…,a_{n-1}$
Question: does exist a permutation $\pi_1,…,\pi_n$ of the integers $[1..n]$ such that $| \pi_{i+1} – \pi_i| = a_i$, $i=1,…,n-1$ ?

For example, given the differences $(2,1,2,1,5,3,1,1)$ a valid permutation of $[1..9]$ is $(5,7,6,8,9,4,1,2,3)$

Update 2013-12-19: a new version of the paper is available.

# Complexity of the Hidden Polygon Puzzle

The Hidden Polygon Puzzle (for brevity HPP) decision problem  is:

Input: a set $P$ of $m$ integer points on a $n \times n$ square grid and an integer $k \leq m$;

Question: does exist a simple rectilinear polygon with $k$ or more vertices $(v_1,v_2, …, v_t), \; v_i \in P, t \geq k$?

The following figure shows an example of a HPP puzzle.

Figure 1: Given the 21 points on the right, can we find
a simple rectilinear polygon with at least 16 vertices?
A possible solution is shown on the right.

The problem is a slight variant of the $\sf{NP}$-complete puzzle game Hiroimono; we prove that the Hidden Polygon Puzzle is  $\sf{NP}$-complete, too using a completely different reduction.

# Hidato is NP-complete

Hidato (also known as Hidoku) is a logic puzzle game invented by Dr. Gyora Benedek, an Israeli mathematician. The rules are simple: given a grid with $n$ cells some of which are already filled with a number between $1$ and $n$ (the first and the last number are circled), the player must completely fill the board with consecutive numbers that connect horizontally, vertically, or diagonally.

Figure 1: An Hidato game (that fits on a $8 \times 8$ grid) and its solution on the right.

We prove that the corresponding decision problem $\sf{HIDATO}$ : “Given a Hidato game that fits in a $m \times n$ grid, does a valid solution exist?” is $\sf{NP}$-complete.

# Binary Puzzle is NP-complete

Binary Puzzle (also known as Binary Sudoku) is an addictive puzzle played on a $n \times n$ grid; intially some of the cells contain a zero or a one; the aim of the game is to fill the empty cells according to the following rules:

• Each cell should contain a zero or a one and no more than two similar numbers next to or below each other are allowed
• Each row and each column should contain an equal number of zeros and ones
• Each row is unique and each column is unique

We prove that the decision version of Binary Puzzle is NP-complete.

# Three easy deletion games on paths

The node deletion game NODE KAYLES is a two persons perfect information game played on a graph $G$. Players alternate picking a node from the graph $G$; the node and its adjacent nodes are deleted. The first player unable to move loses the game. Deciding the winner of a NODE KAYLES game is $\mathsf{PSPACE}$-complete [1]. Similarly in the EDGE KAYLES [1] game the two players must pick an edge; the edge is deleted along with the two endpoint nodes and the edges incident to its two endpoint nodes. The complexity of finding the winner of a EDGE KAYLES game is unknwon. On the Q&A site cstheory.stackexchange.com I submitted a mix of the two games; I call it the BRIDGES AND ISLANDS game: at each turn each player must pick an edge (a “bridge”) or an isolated node (an “island”); if a player picks a node the node is deleted along with its incident edges, if it picks an edge, the edge is deleted along with its two endpoint nodes and the edges incident to them. Like the other two games, the first player unable to move loses the game. I didn’t succeed in finding the complexity of BRIDGES AND ISLANDS.

However decideng the winner is polynomial time solvable when the three games are played on a path. Here I give a quick proof for EDGE KAYLES; the proof for the other two games is similar. Continue reading

# Two bits are enough for a “hard” sum

On Jan 17 I asked the following question on cs.stackexchange.com:

What is the complexity of the following variant of the SUBSET-SUM decision problem?

2-BIT SUBSET SUM: Given an integer $m \geq 0$, and a set of nonnegative integers $A = \{x_1, x_2, …, x_n\}$ such that every $x_i$ has at most $k=2$ bits set to $1$ ($x_i = 2^{b_{i_1}}+2^{b_{i_2}},\;\; b_{i_1},b_{i_2}\geq 0$); is there a subset $A’ \subseteq A$ such that the sum of its elements is equal to $m$ ?

# The complexity of the puzzle game Net: rotating wires can drive you crazy

An amateur proof that the puzzle game Net is NP-complete.

Abstract
The puzzle game Net, also known as FreeNet or NetWalk, is played on a grid filled with terminals and wires; each tile of the grid can be rotated and the aim of the game is to connect all the terminals to the central power unit avoiding closed loops and open-ended wires. We prove that Net is NP-complete.