SearcharxivSearch

arXiv · math/0404432

On the Tweety Penguin Triangle Problem

Abstract

In this paper, one studies the famous well-known and challenging Tweety Penguin Triangle Problem (TPTP or TP2) pointed out by Judea Pearl in one of his books. We first present the solution of the TP2 based on the fallacious Bayesian reasoning and prove that reasoning cannot be used to conclude on the ability of the penguin-bird Tweety to fly or not to fly. Then we present in details the counter-intuitive solution obtained from the Dempster-Shafer Theory (DST). Finally, we show how the solution can be obtained with our new theory of plausible and paradoxical reasoning (DSmT).

Explore related subjects

Keep this discovery

BibTeXRIS

Jean Dezert, Florentin Smarandache. 2004-05-05. On the Tweety Penguin Triangle Problem. https://arxiv.org/abs/math/0404432

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM