Mathematical Puzzles by Risse T.

By Risse T.

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The two agree on the following procedure: 1. Bob sends the box locked by his lock to Alice. 2. Alice additionally locks the box she received by her lock and sends it back to Bob. 3. Bob removes his lock from the box and sends the box locked by only Alice’s lock back to Alice. Solutions to Problems 50 Problem 10(a) They agree – say per e-mail – on the following procedure: 1. e. a one–to–one function f : IN ⊃ D → W ⊂ IN, so that f (x) is easily and f inv (y) is extremely hard to compute. 2. Now, say Alice starts and chooses an odd or even x ∈ D.

Hence there are a total of at most eight keys for three locks and for four persons: Solutions to Problems 53 There is no lock with only one key, because without the owner of that one key pairs of persons cannot open the treasure box. Hence there is either one person with keys to four locks or there are two persons with keys to three locks. In both cases a contradiction! Finally with four locks, it is not sufficient to have two keys per person because then any two persons together might not have keys to each of the four locks!

C) xn+1 = (a xn + c) mod m, mit xo = 1 is periodic – why? and with which maximal/minimal periodic length? Section 5: Probability & Intuition 32 • What is Randomness? Criteria for the quality of pseudo random number generators have to be established, especially of generators of evenly distributed, continuous pseudo random numbers in the unit interval. These criteria are to be assessed in tests. But, randomness has no definition, no specification. Therefore, there can be tests only for certain features of randomness.

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