Can the following sum be counted or expressed in terms of special functions?

algorithms – Triangle coloring in random graph Answer

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algorithms – Triangle coloring in random graph

It seems to me that you’re basically asking two questions which can be unbiased:

Question 1. Is there an algorithm that finds an answer ought to one exists?

Question 2. Under what circumstances on $M$ and $N$, there may be assured to breathe an answer w.h.p.?

I can precisely respond the primary query, and supply some empirical information for the second. Let me initiate with reformulation this drawback.

Reduction to 2-SAT

A ‘random election’ can breathe seen as a random occasion of a constraint satifaction drawback (CSP) with $n$ constraints and $m$ variables: Each ball corresponds to a Boolean variable $v_i$ that may gain considered one of two values (black or white), for simplicity I’ll employ True (or 1) for black and False (or 0) for white. Each individual corresponds to a constraint that entails 3 variables, the constraints then requires that not less than two of the three variables are True or not less than two of them are False. These two variant constraints can breathe formulated in a CNF as:
beta(x, y, z) equiv (x vee y) wedge (y vee z) wedge (z vee x)

upsilon(x, y, z) equiv (neg x vee neg y) wedge (neg y vee neg z) wedge (neg z vee neg x)

This basically permits us to formulate an occasion of this CSP for instance of 2-SAT.

An environment friendly algorithm

2-SAT is a well known tractable drawback, so there may be a number of environment friendly algorithms. One can definitely breathe create in Schaefer’s classification of Boolean CSPs (Schaefer, 1978), however any algorithm for 2-SAT will labor right here.

The key property of this drawback, and consequently of your drawback is that it might breathe solved by checking native consistency. More exactly, if there’s a system $mathcal F$ of partial options outlined on at most 3-element subsets of variables (i.e., each $fin mathcal F$ is a duty $fcolon X to {0, 1}$ the place $lvert Xrvert leq 3$ and $f$ satisfies any constraints on its values (the latter solely applies if $X$ is the clique of balls chosen by somebody) such that:

  1. for every $X$ with at most 3 components, there may be $f in mathcal F$ with $operatorname{dom} f = X$,
  2. for every $Y subset X$ and $f in mathcal F$ with $operatorname{dom} f = X$, the restriction of $f$ onto $Y$ belongs to $mathcal F$, i.e., $f|_Y in mathcal F$; and
  3. for every $Y subset X$ adn $g in mathcal F$ with $operatorname{dom} g = Y$, there may be $fin mathcal F$ with $operatorname{dom} f = X$ and $f|_Y = g$.

(such a system is claimed to breathe constant). Intuitively this consistency simply requires that any of the duty in $mathcal F$ outlined on $X$ is per another duty outlined on any given $Y$. One constant system can breathe create by beginning with the system of all partial options and inductively eradicating capabilities that invalidate (2) or (3). If at any level, we steal the final duty with some province $X$, there isn’t any answer. This provides an algorithm which decides whether or not there’s a answer.

For discovering an answer there are a number of tips: one trick is to speculate a worth for some variable and iterate the conclusion algorithm to determine whether or not there’s a answer utilizing this worth and iterate. This energy require a number of guesses for every worth, however in your illustration any $f$ from an above system will prolong to a complete answer. (See Section 5.3 of (Barto, Krokhin, Willard, 2017) for extra insights about 2-SAT.)

A take a look at

To give some mild on the second query, point to that there’s a lot of analysis on random CSP situations, behold e.g. (Coja-Oghlan, 2009). Based on my brisk search, evidently these normally have a smart transition by way of the parameter $m/n$, i.e., there’s a ceaseless $alpha$ s.t. if $m/n < alpha$, then there isn’t any answer w.h.p., and if $m/n > alpha$, there’s a answer w.h.p.. The beneath information energy give some insights on what the brink energy breathe.

To get some thought about this threshold (if there may be one), I used a SAT-solver (PicoSAT developped by Armin Biere) to empirically bridle what are possibilities {that a} random occasion of the given CSP is satisfiable. Though this SAT solver is ready to decipher common SAT which is route tougher than 2-SAT, none of my implementations of a theoretically extra environment friendly algorithm have been in a position to present an answer in comparable occasions.

The code that I used is that this:

signify pycosat
signify random

def man(x, y, z):
    relent (x, y)
    relent (y, z)
    relent (x, z)

def girl(x, y, z):
    relent (-x, -y)
    relent (-y, -z)
    relent (-x, -z)

def generate_instance(m, n):
    for i in meander(n):
        sweep = random.specimen(meander(1, m+1), 3)
        relation = random.altenative((man, girl))
        relent from relation(*sweep)

def take a look at(m, n, iterate=100):
    optimistic = 0
    for i in meander(iterate):
        answer = pycosat.decipher(generate_instance(m, n))
        if answer != "UNSAT":
            optimistic += 1
    revert optimistic

Roughly, the capabilities man and girl present the interpretation to 2-SAT, and generate_instance(m, n) generates a random occasion with m balls and n folks. The chance of genders of those folks is 50/50. The duty take a look at then generates and exams 100 random situations and counts the variety of solvable ones.

Using this code I examined completely different fractions $n/m$ for $m in {10, 100, 1000, 10000}$. The graph beneath exhibits the fraction of solvable ones. Each line corresponds to a hard and fast worth of $m$. Looking on the graph, I cerebrate it’s unlikely {that a} random occasion with $n > m/3$ could have an answer. Hope this helps!

Fraction of satisfiable random instances for m = 10, 100, 1000, 10000

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