C960 Discrete Mathematics II
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Free C960 Discrete Mathematics II Questions
A fair six-sided die is rolled until the running total (sum) of outcomes reaches at least 15. What is the expected number of rolls?
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4.5
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4.76
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5
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6
Explanation
Let be the expected additional rolls needed when the current sum is sss. For any s≥15 we have E(s)=0. For s<15 each roll adds 1 plus the average of the six possible next states, so . Solving these linear recurrences backwards from s=14 down to s=0 (with E(s)=0 for s≥15yields E(0)≈4.7600083766. Thus the expected number of rolls to reach a sum of at least 15 is about 4.76.
Which of the following is true about the Post Correspondence Problem (PCP)?
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It is decidable for all instances.
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It is undecidable in general.
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It can be solved in polynomial time.
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It is solvable only for unary alphabets.
Explanation
The Post Correspondence Problem (PCP) is a classical problem in computability theory. It asks whether, given two lists of strings over some alphabet, there exists a sequence of indices such that concatenating the corresponding strings from both lists produces the same string. Emil Post proved that PCP is undecidable in general, meaning there is no algorithm that can solve all instances of the problem. However, for some restricted cases (like unary alphabets), it can be decidable. It is not solvable in polynomial time for the general case.
Using extended Euclidean algorithm, find the inverse of 19 modulo 47
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5
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12
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23
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37
Explanation
To find the modular inverse of 19 mod 47, we need an integer such that 19≡1(mod47) . The extended Euclidean algorithm expresses 1 as a linear combination of 19 and 47:
Using Euler’s theorem, compute 2¹⁰⁰⁰ mod 101
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1
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100
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50
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10
Explanation
Solve the recurrence aₙ = 4aₙ₋₁ − 4aₙ₋₂ with a₀=1, a₁=4
Explanation
How many 3×3 reduced Latin squares are there?
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12
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24
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36
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48
Explanation
A Latin square of order 3 is a 3×3 grid filled with the numbers 1, 2, and 3 such that each number appears exactly once in each row and each column. A reduced Latin square is one where the first row and first column are fixed in natural order (1, 2, 3).
The total number of 3×3 Latin squares is 12, but since fixing the first row and column removes symmetries, only one unique reduced Latin square remains. All other Latin squares can be obtained from this one by permuting rows, columns, or symbols.
In a single-elimination tournament with 16 teams, how many games are played until a champion is determined?
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15
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16
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8
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31
Explanation
In a single-elimination tournament, each game eliminates exactly one team. Starting with 16 teams, we need to eliminate 15 teams to determine a champion. Since each game eliminates one team, the total number of games played is equal to the number of teams eliminated, which is 16−1=15. This formula holds for any single-elimination tournament: total games = total teams − 1.
Decrypt the ciphertext using the RSA private key .
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12
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3
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4
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5
Explanation
Using Miller-Rabin, is 91 a prime number?
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Yes
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No, it is 7×13
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No, it is 3×31
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No, it is 7×11
Explanation
The number 91 is not prime, as it can be factored into integers greater than 1. In fact, 91=7×13. Applying the Miller-Rabin test would also identify it as composite. Therefore, the correct statement specifies its factorization as 7 times 13.
What is the time complexity of Kruskal’s algorithm using Union-Find with path compression and union-by-rank?
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O(E log V)
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O(E log E)
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O(V + E log V)
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O(E + V log V)
Explanation
Kruskal’s algorithm finds a Minimum Spanning Tree (MST) by sorting all edges by weight and adding them one by one, avoiding cycles. The main steps are:
Sort all edges: Sortingedges takes time.
Union-Find operations: With path compression and union-by-rank, each union or find operation takes nearly (more precisely O(α(V)), where α is the inverse Ackermann function, which grows extremely slowly). Since there are at most operations, this part is effectively O(E).
Thus, the overall time complexity is dominated by the sorting step: O(E log E). For connected graphs, since , sometimes O(E log V)is also used as a bound, but the more precise standard bound is O(E log E).
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