IMPORTANT: Please do not post solutions, hints, or other spoilers until at least 60 hours after the date of this message. Thanks. IMPORTANTE: Por favor, no enviéis soluciones, pistas, o cualquier otra cosa que pueda echar a perder la resolución del problema hasta que hayan pasado por lo menos 60 horas desde el envío de este mensaje. Gracias. IMPORTANT: S'il vous plaît, attendez au minimum 60 heures après la date de ce message avant de poster solutions, indices ou autres révélations. Merci. Qing3 Zhu4Yi4: Qing3 Ning2 Deng3Dao4 Jie1Dao4 Ben3 Xin4Xi2 Zhi1Hou4 60 Xiao3Shi2, Zai4 Fa1Biao3 Jie3Da2, Ti2Shi4, Huo4 Qi2Ta1 Hui4 Xie4Lou4 Da2An4 De5 Jian4Yi4. Xie4Xie4. UWAGA: Prosimy nie publikowac rozwiazan, dodatkowych badz pomocniczych informacjii przez co najmniej 60 godzin od daty tej wiadomosci. Dziekuje. ---------------------------------------------------------------- In the 1960's, the grad students at the University of Chicago math department worked on a series of astoundingly useless and time-consuming puzzles. One of these follows. Consider the set of all possible strings of the alphabet ('a' .. 'z'). Let us agree to consider two strings "equivalent" if the following conditions hold: 1. They contain precisely the same letters, and 2. They both appear in Webster's Third International Dictionary. In such a case, the two strings are considered interchangeable in all contexts. For example, "am" and "ma" are equivalent, and this also implies that "amount" and "maount" are equivalent, as are "grammar" and "grmamar" and "gramamr" and "grmaamr". Moreover, equal letters can be cancelled from the front and back end of any string. For example, "abby" and "baby" are equivalent, and, cancelling the trailing "by", this implies that "ab" and "ba" are also equivalent, and can be exchanged anywhere. When two letters can be exchanged in this way, we say that they "commute". The third floor of the math building at UC had a huge 26x26 chart; the square in column i and row j contained a proof that letters i and j would commute. Sometimes these proofs can be rather elaborate. For example, the dictionary has "dire" and "ride", so, by cancelling the trailing "e"s, one has "dir" = "rid". The dictionary also has dirten = rident (No, I don't know what those mean.) Since "dir" = "rid" we have: rident = dirent and since "rident" = "dirten", dirten = dirent even though "dirent" is not a dictionary word. Cancelling the leading "dir" leaves: ten = ent but ten = net because "ten" and "net" are dictionary words, so ent = net and, cancelling the "t", en = ne and we've just proved that "en" and "ne" commute. This fact might be useful in later proofs. What's the point of all this? Well, the goal is to find out which letters commute with *every* other letter; such letters are said to be in the "center" of the system. As for the *point*, I'm not sure there is one. Apparently the math grads at UC didn't have enough to occupy their time. The chart in the UC math building has since been lost, so your task is to write a program whose input is a word list, with one word per line, and which makes appropriate deductions and eventually computes the center of the system. I don't have the headword list from Webster's Third, but I do have the list from Webster's Second, so let's use that. You can get a copy from http://perl.plover.com/qotw/words/Web2.bz2 http://perl.plover.com/qotw/words/Web2.gz