Penrose patterns are almost entirely determined by two points (Q1199587)

A Penrose pattern \(\pi\) is a tiling of the real plane by rhombs of two types (with an acute angle of \(36^ \circ\) and \(72^ \circ\), respectively) with particularly directed and marked edges. There are infinitely many Penrose tilings. The author calls two Penrose tilings \(\pi_ 1\), \(\pi_ 2\) equal up to \(\varepsilon\) (\(\varepsilon > 0\)) if \(S_ 1\cap S_ 2\) is \(\varepsilon\)-complete with respect to \(S_ 1\cup S_ 2\) (where \(S_ i\) is the set of vertices of \(\pi_ i\)). Here, a subset \(T'\) of a set \(T\) is called \(\varepsilon\)-complete if \(\text{card}(T'\cap D)>(1- \varepsilon)\cdot\text{card}(T\cap D)\) for any disc \(D\) large enough. The author shows that to any Penrose tiling \(\pi\) and to any \(\varepsilon > 0\) there exist two vertices \(P\), \(Q\) of \(\pi\) such that any Penrose tiling \(\pi'\) containing \(P\), \(Q\) and generated from the same non-rotated rhombs as \(\pi\) is equal to \(\pi\) up to \(\varepsilon\).