# A personal view of APL by Iverson K.E.

By Iverson K.E.

Marie Curie and the science of radioactivity

Examines the lifetime of the Polish-born scientist who, along with her husband Pierre, was once offered a 1903 Nobel Prize for locating radium.

Erfolgreiche Organisationsentwicklung im Krankenhaus: Mehr Personal spart Kosten ! Gelebte Investition in Qualitat, Know-how und Skills am Beispiel der Radiologie

Qualität ist der Weg aus der Kostenfalle. Das weist der Autor in seinem Buch nach. Sein Fazit lautet: Investitionen in Qualität, information und talents lohnen sich. Es genügt aber nicht, die technische Ausstattung auf modernstem Stand zu halten, die enterprise an den aktuellsten Konzepten auszurichten, die Mitarbeiter fachlich ständig weiter zu qualifizieren und sie intestine zu bezahlen.

Interface Fantasy: A Lacanian Cyborg Ontology (Short Circuits)

At the back of our computing device monitors we're all cyborgs: via fable we will be able to comprehend our involvement in digital worlds.

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The remainder of the insertion algorithm is now almost routine. We insert the new point a, walk back to the root and merge the pIS as we go along. More precisely, we first compute P(a), then merge it with P(S(vk )), then with p(S(brother(vk ))), . . The time bound derived in lemma 2 applies again except that we forgot about rotations and doublerotations. Suppose that we have to rotate at node vi. ]. Rotation 24 and assume that v i + 1 is the root of subtree D1 . As we walk back to the root we have already computed P(S(v i + 1 )).

First, the sequence a(v) of actions executed to compute 0 applied to P(S(x)) and P(S(y)). This sequence has length O(C(n)). Second, the piece p* (S (v)) which is leftover from P (S (v)) when P (S (father (v))) is computed by applying 0 to P (S (v)) and P (S (brother(v)). We call tree D augmented by this additional information an augmented tree Lemma 1: An au~ented tree D for set S has space requirement T( lSI) and can be constructed in time Sort(ISI) + T(ISI) where T(n) = max [T(Sn) + T((1-S)n) + O(C(n)) J.

However it is trivial to improve upon this result by using O(dln) storage. ~ D = Do x ••. x Dd - 1 • For any of the d! possible orderings of the attributes build a search tree as follows: Let S Order S lexicographically and build a standard one-dimensional search tree for S. A partial match query with s specified components is then easily answered in time O(d log n + IAI). g. e. R with ~i = hi for 0 Search for key ( ~ o' ~ i < s and ···'~s-1' ~i -=, = -=, = [~o,ho] x ... x [~d-1,hd-1] h. = += for s ~ ~ i < d.