On the parenthesisations of matrix chains: All are useful, few are essential

Abstract The product of a matrix chain consisting of n matrices can be computed in $$C_{n-1}$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:msub> <mml:mi>C</mml:mi> <mml:mrow> <mml:mi>n</mml:mi> <mml:mo>-</mml:mo> <mml:mn>1</mml:mn> </mml:mrow> </mml:msub> </mml:math> (Catalan’s number) different ways, each identified by a distinct parenthesisation of the chain. The best algorithm to select a parenthesisation that minimises the cost runs in $$O(n \log n)$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>O</mml:mi> <mml:mo>(</mml:mo> <mml:mi>n</mml:mi> <mml:mo>log</mml:mo> <mml:mi>n</mml:mi> <mml:mo>)</mml:mo> </mml:mrow> </mml:math> time. Approximate algorithms run in O ( n ) time and find solutions that are guaranteed to be within a certain factor from optimal; the best factor is currently 1.155. In this article, we first prove two results that characterise different parenthesisations, and then use those results to improve on the best known approximation algorithms. Specifically, we show that (a) each parenthesisation is optimal somewhere in the problem domain, and (b) exactly $$n + 1$$ <mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML"> <mml:mrow> <mml:mi>n</mml:mi> <mml:mo>+</mml:mo> <mml:mn>1</mml:mn> </mml:mrow> </mml:math> parenthesisations are essential in the sense that the removal of any one of them causes an unbounded penalty for an infinite number of problem instances. By focusing on essential parenthesisations, we improve on the best known approximation algorithm and show that the approximation factor is at most 1.143.