Proof-Number (PN) search is a best-first adversarial search algorithm especially suited for finding the game-theoretical value in game trees. The strategy of the algorithm may be described as developing the tree into the direction where the opposition characterised by value and branching factor is to expect to be the weakest. In this chapter we start by providing a short description of the orig...

abstract competition is a phenomenon that influences the size, structure and stability of insect communities, particularly on solitary endoparasitoids species. although, competition normally incurs costs, it might be adaptive for the surviving individual. to determine which factors affect the patch use strategy of foragers under competition, the effect of previous experience with hosts already-...

Denote by {Fn} and {Ln} the Fibonacci numbers and Lucas numbers, respectively. Let Fn = Fn × Ln and Ln = Fn + Ln. Denote by {Pn} and {Qn} the Pell numbers and Pell-Lucas numbers, respectively. Let Pn = Pn × Qn and Qn = Pn + Qn. In this paper, we give some determinants and permanent representations of Pn, Qn, Fn and Ln. Also, complex factorization formulas for those numbers are presented. Key–Wo...

This article introduces Evaluation-Function based Proof– Number Search (EF-PN) and its second-level variant EF-PN. It is a framework for setting the proof and disproof number of a leaf node with a heuristic evaluation function. Experiments in LOA and Surakarta show that compared to PN and PN, which use mobility to initialize the proof and disproof numbers, EF-PN and EF-PN take between 45% to 85...

The prime number theorem tells us that π(x), the number of primes below x, is ∼ x/ logx. Equivalently, if pn denotes the n-th smallest prime number then pn ∼ n logn. What is the distribution of the gaps between consecutive primes, pn+1 − pn? We have just seen that pn+1 − pn is approximately logn “on average”. How often do we get a gap of size 2 logn, say; or of size 1 2 logn? One way to make th...

Proof. First of all, we prove that if pn is the nth prime number then we have that pn ≤ 2 n−1 . Since there must be some pn+1 dividing the number p1p2 · · · pn− 1 and not exceeding it, it follows from the induction step that pn+1 ≤ 2 0 2 1 · · · 22n−1 = 220+21+···+2n−1 ≤ 22n . If x ≥ 2 is some real number, then we select the largest natural number n satisfying 22n−1 ≤ x, so that we have that 2 ...

In this note the split domination number of the Cartesian product of two paths is considered. Our results are related to [2] where the domination number of Pm¤Pn was studied. The split domination number of P2¤Pn is calculated, and we give good estimates for the split domination number of Pm¤Pn expressed in terms of its domination number.

For paths Pn, G. Chartrand, L. Nebeský and P. Zhang showed that ac′(Pn) ≤ ( n−2 2 ) +2 for every positive integer n, where ac′(Pn) denotes the nearly antipodal chromatic number of Pn. In this paper we show that ac′(Pn) ≤ ( n−2 2 ) − n2 − b 10 n c + 7 if n is even positive integer and n ≥ 10, and ac′(Pn) ≤ ( n−2 2 ) − n−1 2 − b 13 n c + 8 if n is odd positive integer and n ≥ 13. For all even pos...