C960 Discrete Mathematics II
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Free C960 Discrete Mathematics II Questions
Which of the following problems is NP-complete?
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Sorting a list of numbers
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Finding the shortest path in a graph
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3-SAT (Boolean satisfiability with clauses of size 3)
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Checking if a number is prime
Explanation
A problem is NP-complete if it is in NP (its solutions can be verified in polynomial time) and every problem in NP can be reduced to it in polynomial time. Among the options:
Sorting can be done in polynomial time, so it is in P, not NP-complete.
Shortest path can also be solved in polynomial time using Dijkstra’s or Bellman-Ford algorithms.
3-SAT is the canonical NP-complete problem: verifying a satisfying assignment is polynomial-time, and it is known that any NP problem can be reduced to 3-SAT.
Primality checking is in P (AKS algorithm), so it is not NP-complete.
How many onto functions are there from a set of 7 elements to a set of 4 elements?
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8400
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4096
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1800
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1200
Explanation
An onto function (also called a surjective function) from a set of 7 elements to a set of 4 elements means every element in the codomain (the set with 4 elements) must be mapped to by at least one element in the domain (the set with 7 elements). The number of onto functions from a set of size nnn to one of size mmm is given by the formula:
m!×S(n,m)
where S(n,m)S(n, m)S(n,m) is the Stirling number of the second kind, representing the number of ways to partition items into non-empty subsets. For =7 and =4:
What is the worst-case time complexity of Bubble Sort on an array of n elements?
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O(n)
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O(n log n)
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O(n²)
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O(n³)
Explanation
Explanation:
Bubble Sort is a comparison-based sorting algorithm that repeatedly steps through the list, compares adjacent elements, and swaps them if they are in the wrong order. In the worst case, the array is in reverse order.
For the first pass, Bubble Sort makes n−1n-1n−1 comparisons.
For the second pass, it makes n−2n-2n−2 comparisons.
This continues until the last pass, which makes 1 comparison.
The total number of comparisons is:
Which of the following graphs has an Eulerian circuit?
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A graph with exactly 2 vertices of odd degree
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A graph with all vertices of even degree
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A graph with exactly 4 vertices of odd degree
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A directed graph with unequal in/out degrees
Explanation
An Eulerian circuit is a closed trail that uses every edge exactly once. A classical theorem (Euler’s criterion) states that a finite graph has an Eulerian circuit iff it is connected (when restricted to edges that exist) and every vertex has even degree. If exactly two vertices have odd degree the graph has an Eulerian trail (open path) but not a circuit; if the graph is disconnected it cannot have a single circuit that traverses every edge. Therefore the only option that guarantees an Eulerian circuit is the connected graph where every vertex has even degree.
Which of the following is NOT a decidable language?
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The set of all strings accepted by a DFA
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The set of all Turing machines that halt on all inputs
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The set of all regular languages
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The set of all context-free languages
Explanation
A language is decidable if there exists a Turing machine that halts on every input and correctly decides membership in that language.
The language describing “the set of all Turing machines that halt on all inputs” corresponds to the totality problem, which asks whether a Turing machine halts for every possible input. This problem is undecidable, as it is more general than the Halting Problem itself.
In contrast, checking whether a DFA accepts a given string, or whether a grammar is regular or context-free, is decidable using algorithmic procedures.
What is the space complexity of depth-first search (DFS) on a graph with vertices and edges?
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O(V + E)
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O(V²)
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O(V)
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O(E)
Explanation
Depth-first search (DFS) uses a stack to keep track of vertices during traversal, which can be implemented either recursively (call stack) or iteratively (explicit stack). In addition to the stack, DFS may store the adjacency list or adjacency matrix of the graph. For a graph represented as an adjacency list, storing all edges takes space, and the stack can grow up to in the worst case. Combining these, the overall space complexity of DFS on a graph with vertices and edges is
Which of the following languages requires a linear bounded automaton (LBA)?
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The set of palindromes over {0,1}
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The set of strings of the form
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The set of all binary strings
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The set of all strings with an even number of 0s
Explanation
A linear bounded automaton (LBA) is a type of Turing machine whose tape usage is limited to a linear function of the input length. LBAs recognize context-sensitive languages, which are strictly more powerful than context-free languages but less powerful than general Turing machines.
Palindromes over {0,1} are context-free, so a pushdown automaton can handle them.
Strings of the form are context-sensitive but not context-free, so they require an LBA.
What is the minimal number of colors needed to color the vertices of the Mycielski graph with chromatic number 5?
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3
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4
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5
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6
Explanation
By definition, the chromatic number of a graph is the smallest number of colors needed to color its vertices so that no two adjacent vertices share the same color. If a particular Mycielski graph is given as having chromatic number 5, that directly means its minimal (i.e., least) number of colors required is 5 — the Mycielski construction produces graphs with a specified chromatic number while controlling other properties, but the chromatic number itself tells you the minimum coloring size.
How many 5-digit numbers have digits that sum to 20?
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106
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121
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126
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210
Explanation
A pushdown automaton that accepts must use the stack to:
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Count the number of a’s and ensure there are twice as many b’s.
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Match each a with exactly one b.
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Store the input symbols for reversal
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Recognize even-length strings only
Explanation
The language requires that for every aaa in the input, there are exactly two ’s following it. A pushdown automaton (PDA) can handle this by pushing one symbol for each read onto the stack. Then, as it reads ’s, it pops one stack symbol for every two ’s, ensuring that the total number of ’s is exactly double the number of ’s. This counting dependency between ’s and ’s is what makes the language context-free and not regular, and it is achieved by using the stack to count and verify the 2-to-1 ratio.
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