Download Das Java-Praktikum: Aufgaben und Lösungen zum by Reinhard Schiedermeier, Klaus Köhler PDF

By Reinhard Schiedermeier, Klaus Köhler

Hauptbeschreibung Dieses Buch richtet sich an Studenten und Autodidakten, die das Programmieren mit Java lernen möchten. Es bietet eine breit gefächerte Auswahl von Aufgaben mit vollständigen Lösungen, die stufenweise immer neue Sprachstrukturen einbeziehen. Das Buch ist dabei kein Lehrbuch und stellt die Elemente von Java nur sehr knapp vor. Das Hauptgewicht liegt auf der Pragmatik der Programmiersprache, die Read more...

summary: Hauptbeschreibung Dieses Buch richtet sich an Studenten und Autodidakten, die das Programmieren mit Java lernen möchten. Es bietet eine breit gefächerte Auswahl von Aufgaben mit vollständigen Lösungen, die stufenweise immer neue Sprachstrukturen einbeziehen. Das Buch ist dabei kein Lehrbuch und stellt die Elemente von Java nur sehr knapp vor. Das Hauptgewicht liegt auf der Pragmatik der Programmiersprache, die am eingängigsten durch die ausführlich erklärten Lösungsbeispiele illustriert wird. Alle Aufgabenstellungen sind kompakt und weitgehend unabhängig voneinander. Bi

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Extra info for Das Java-Praktikum: Aufgaben und Lösungen zum Programmierenlernen

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Therefore, Peggy can know 2 out of the 3 rk ’s. There is 1/3 that the rk that Peggy does not know is the one that Victor does not ask for. 01, five repetitions are enough. Another way: Victor asks for only one of the rk ’s. Then the probability is 2/3 that Peggy can cheat in a round. 01, twelve repetitions suffice. Chapter 15 - Exercises 1. There are four possible outcomes for the pair (X1 , X2 ), namely HH, TH, HT, TT. Each occur with probability 1/4. Hence the entropy H(X1 , X2 ) is −4 1 1 log2 4 4 =2 Observe that H(X1 ) = H(X2 ) = 1, and that H(X1 , X2 ) = H(X1 ) + H(X2 ).

These calculations are all in Z2 [X] (mod X 4 − 1). The fact that X 4 ≡ 1 and therefore X 5 ≡ X was used. By the theorem, we find that 1 + X + X 2 + X 3 is in C and the other two polynomials are not in C. 14. The proposition on page 334 says that the matrix H defined on that page is a parity check matrix for C. This means that cH T = 0 for all c ∈ C, so the rows of H generate a subspace of C ⊥ . The rank of H is clearly k, since it contains the upper triangular submatrix with bℓ = 1 down the diagonal.

For example, she can take k = 1, r = α, and s = m and have a valid signature (m, r, s). 27 Chapter 10 - Exercises 1. First calculate aA , aB , aC , bA , bB and bC to get aA = 10, aB = 17, aC = 14, bA = 14, bB = 6 and bC = 5. Using gA (x) = aA + bA x, and similarly for gB (x), we may calculate the keys by KAB = gA (rB ) = 21, KAC = gA (rC ) = 7 and KBC = gB (rC ) = 29. 2. (a) To show KAB = a + b(rA + rB ) + crA rB simply substitute the definition for aA and bA into KAB = gA (rB ). (b) Using part (a) it is clear that KAB = KBA .

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