Comparing strings from the back?
Johannes Bauer
dfnsonfsduifb at gmx.de
Tue Sep 4 12:32:57 EDT 2012
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Tue Sep 4 12:32:57 EDT 2012
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On 04.09.2012 04:17, Steven D'Aprano wrote: > On average, string equality needs to check half the characters in the > string. How do you arrive at that conclusion? When comparing two random strings, I just derived n = (256 / 255) * (1 - 256 ^ (-c)) where n is the average number of character comparisons and c. The rationale as follows: The first character has to be compared in any case. The second with a probability of 1/256, the third with 1/(256^2) and so on. Best regards, Johannes -- >> Wo hattest Du das Beben nochmal GENAU vorhergesagt? > Zumindest nicht öffentlich! Ah, der neueste und bis heute genialste Streich unsere großen Kosmologen: Die Geheim-Vorhersage. - Karl Kaos über Rüdiger Thomas in dsa <hidbv3$om2$1 at speranza.aioe.org>
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