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Mini-Challenge #6: Code-breaking

Can you crack the codes to reveal the hidden answer?

1. Uijt jt b Dbftbs djqifs. Opx vtf b gsfrvfodz bobmztjt up efdpef uif xfmm-lopxo gjstu qbsbhsbqi, boe mppl gps b tfdsfu nfttbhf bu uif foe.

2. Ts tg d swjsu jvtkhwgdmme dfovbpmhnlhn, suds d gtvlmh cdv tv qbgghggtbv ba d lbbn abwsjvh cjgs ih tv pdvs ba d ptah. Ubphkhw mtssmh ovbpv suh ahhmtvlg bw kthpg ba gjfu d cdv cde ih bv utg atwgs hvshwtvl d vhtluibjwubbn, sutg swjsu tg gb phmm atrhn tv suh ctvng ba suh gjwwbjvntvl adctmthg, suds uh tg fbvgtnhwhn dg suh wtlusajm qwbqhwse ba gbch bvh bw bsuhw ba suhtw ndjlushwg. Jgh suh ntltsg ba qt sb gutas suh mhsshwg tv suh vhrs chggdlh abwpdwng.

3. qgamznyhntzjzlvauxjzqgfhfxebjzrm (Updated: 8th May 2013).

(The answer for the first and second versions of part (3), "qgazijuymfnsctejyqzemmhproake" and "qgamznyhntzjzldxtfkxuryu", can be found on Twitter, so I've updated it for new challengers).

(Hint: If you don't know how to get started, first read the Wikipedia article on the Caesar cipher.)


Email the answer to pi3challenge@sheffield.ac.uk or tweet to @pi3challenge

In Python, the ord() and chr() functions can be used for turning a character (e.g. 'a') into an integer, and back again.