CS 5800 Midterm Exam | Questions and
Solutions | 2026 Updated | 100% Correct -
Northeastern University.
1. Which of the following statements about the asymptotic relationship between the
functions f(n) = n^2 log n and g(n) = n^2.001 is true?
A. f(n) = O(g(n))
B. f(n) = (g(n))
C. f(n) = (g(n))
D. f(n) and g(n) are asymptotically incomparable
Answer: A. f(n) = O(g(n))
Rationale: For large n, n^2.001 grows faster than n^2 log n because any positive power of
n eventually dominates any polylogarithmic factor. Thus f(n) = O(g(n)), but the reverse is
not true, so and are incorrect. Incomparability is false because both are positive functions
and one dominates.
2. Consider the recurrence T(n) = 2T(n/4) + n. What is the tight asymptotic bound for
T(n)?
A. (n)
B. (n log n)
C. (n)
D. (n log n)
Answer: B. (n log n)
Rationale: Using the Master Theorem: a=2, b=4, f(n)=n. log_b a = log_4 2 = 0.5, so
n^{log_b a} = n. Since f(n) = (n), we are in Case 2, giving T(n) = (n log n). Other options
misapply the theorem or ignore the logarithmic factor.
Page 1 of 76
,3. In the 0/1 knapsack problem, a dynamic programming solution uses a table dp[i][w]
representing the maximum value achievable using the first i items with capacity w. If
item i has weight w_i and value v_i, which recurrence correctly computes dp[i][w]?
A. dp[i][w] = max(dp[i-1][w], v_i + dp[i-1][w - w_i])
B. dp[i][w] = max(dp[i-1][w], v_i + dp[i][w - w_i])
C. dp[i][w] = max(dp[i-1][w], v_i + dp[i-1][w])
D. dp[i][w] = dp[i-1][w] + v_i
Answer: A. dp[i][w] = max(dp[i-1][w], v_i + dp[i-1][w - w_i])
Rationale: For 0/1 knapsack, each item can be used at most once, so we must refer to the
previous row (i-1) when including the item. Option B allows reusing the same item
(unbounded knapsack), Option C ignores the weight constraint, and Option D always adds
the value without considering capacity.
4. Which of the following problems can be solved optimally by a greedy algorithm?
A. 0/1 knapsack problem
B. Activity selection problem
C. Traveling salesman problem
D. Longest common subsequence
Answer: B. Activity selection problem
Rationale: The activity selection problem exhibits the greedy-choice property and optimal
substructure, allowing a greedy algorithm that selects the activity with the earliest finish
time. The 0/1 knapsack and TSP are NP-hard, and longest common subsequence requires
dynamic programming.
Page 2 of 76
,5. In a directed graph with non-negative edge weights, which algorithm correctly
computes single-source shortest paths and has a time complexity of O(E + V log V)
when implemented with a Fibonacci heap?
A. Bellman-Ford algorithm
B. Dijkstra's algorithm
C. Floyd-Warshall algorithm
D. Prim's algorithm
Answer: B. Dijkstra's algorithm
Rationale: Dijkstra's algorithm with a Fibonacci heap achieves O(E + V log V) for
non-negative weights. Bellman-Ford handles negative weights but is O(VE);
Floyd-Warshall solves all-pairs; Prim's algorithm finds minimum spanning trees, not
shortest paths.
6. Which of the following statements about the NP-completeness of the Boolean
satisfiability problem (SAT) is correct?
A. SAT is NP-complete, and 3-SAT is also NP-complete.
B. SAT is NP-complete, but 3-SAT is in P.
C. SAT is in P, but 3-SAT is NP-complete.
D. Both SAT and 3-SAT are in P.
Answer: A. SAT is NP-complete, and 3-SAT is also NP-complete.
Rationale: The Cook-Levin theorem establishes SAT as NP-complete. 3-SAT is also
NP-complete because any SAT instance can be reduced to 3-SAT in polynomial time. The
other options incorrectly claim 3-SAT or SAT is in P.
Page 3 of 76
, 7. Consider a randomized algorithm that, on any input, has a probability of at most
1/2 of giving an incorrect answer. If we run the algorithm independently 10 times and
take the majority answer, what is the probability that the majority answer is
incorrect?
A. At most 1/2
B. At most 1/4
C. At most (1/2)^10
D. At most 1/1024
Answer: C. At most (1/2)^10
Rationale: This question is under revision. The correct answer is C based on the
assumption that the majority vote reduces error exponentially, but the exact bound depends
on the error probability. In standard randomized algorithms, repeating independent runs
and taking majority reduces error probability exponentially if the original error probability
is strictly less than 1/2. Here, with error 1/2, the bound (1/2)^10 is a loose upper bound.
However, the question is flawed; a better version would specify error 1/3. For now, C is the
intended answer.
8. In the context of NP-completeness, which of the following is a polynomial-time
reduction from problem A to problem B?
A. A function f that maps instances of A to instances of B such that x A if and only if
f(x) B, and f can be computed in polynomial time.
B. A function f that maps instances of A to instances of B such that x A if and only if
f(x) B, and f can be computed in exponential time.
C. A function f that maps instances of B to instances of A such that x B if and only if
f(x) A, and f can be computed in polynomial time.
D. A function f that maps instances of A to instances of B such that x A implies f(x) B,
and f can be computed in polynomial time.
Answer: A. A function f that maps instances of A to instances of B such that x A if and
only if f(x) B, and f can be computed in polynomial time.
Rationale: A polynomial-time reduction from A to B requires a polynomial-time
computable function f such that x A iff f(x) B. Option B uses exponential time, which is not
a polynomial-time reduction. Option C reverses the direction, and Option D only requires
one direction, missing the 'if and only if' condition.
Page 4 of 76
Solutions | 2026 Updated | 100% Correct -
Northeastern University.
1. Which of the following statements about the asymptotic relationship between the
functions f(n) = n^2 log n and g(n) = n^2.001 is true?
A. f(n) = O(g(n))
B. f(n) = (g(n))
C. f(n) = (g(n))
D. f(n) and g(n) are asymptotically incomparable
Answer: A. f(n) = O(g(n))
Rationale: For large n, n^2.001 grows faster than n^2 log n because any positive power of
n eventually dominates any polylogarithmic factor. Thus f(n) = O(g(n)), but the reverse is
not true, so and are incorrect. Incomparability is false because both are positive functions
and one dominates.
2. Consider the recurrence T(n) = 2T(n/4) + n. What is the tight asymptotic bound for
T(n)?
A. (n)
B. (n log n)
C. (n)
D. (n log n)
Answer: B. (n log n)
Rationale: Using the Master Theorem: a=2, b=4, f(n)=n. log_b a = log_4 2 = 0.5, so
n^{log_b a} = n. Since f(n) = (n), we are in Case 2, giving T(n) = (n log n). Other options
misapply the theorem or ignore the logarithmic factor.
Page 1 of 76
,3. In the 0/1 knapsack problem, a dynamic programming solution uses a table dp[i][w]
representing the maximum value achievable using the first i items with capacity w. If
item i has weight w_i and value v_i, which recurrence correctly computes dp[i][w]?
A. dp[i][w] = max(dp[i-1][w], v_i + dp[i-1][w - w_i])
B. dp[i][w] = max(dp[i-1][w], v_i + dp[i][w - w_i])
C. dp[i][w] = max(dp[i-1][w], v_i + dp[i-1][w])
D. dp[i][w] = dp[i-1][w] + v_i
Answer: A. dp[i][w] = max(dp[i-1][w], v_i + dp[i-1][w - w_i])
Rationale: For 0/1 knapsack, each item can be used at most once, so we must refer to the
previous row (i-1) when including the item. Option B allows reusing the same item
(unbounded knapsack), Option C ignores the weight constraint, and Option D always adds
the value without considering capacity.
4. Which of the following problems can be solved optimally by a greedy algorithm?
A. 0/1 knapsack problem
B. Activity selection problem
C. Traveling salesman problem
D. Longest common subsequence
Answer: B. Activity selection problem
Rationale: The activity selection problem exhibits the greedy-choice property and optimal
substructure, allowing a greedy algorithm that selects the activity with the earliest finish
time. The 0/1 knapsack and TSP are NP-hard, and longest common subsequence requires
dynamic programming.
Page 2 of 76
,5. In a directed graph with non-negative edge weights, which algorithm correctly
computes single-source shortest paths and has a time complexity of O(E + V log V)
when implemented with a Fibonacci heap?
A. Bellman-Ford algorithm
B. Dijkstra's algorithm
C. Floyd-Warshall algorithm
D. Prim's algorithm
Answer: B. Dijkstra's algorithm
Rationale: Dijkstra's algorithm with a Fibonacci heap achieves O(E + V log V) for
non-negative weights. Bellman-Ford handles negative weights but is O(VE);
Floyd-Warshall solves all-pairs; Prim's algorithm finds minimum spanning trees, not
shortest paths.
6. Which of the following statements about the NP-completeness of the Boolean
satisfiability problem (SAT) is correct?
A. SAT is NP-complete, and 3-SAT is also NP-complete.
B. SAT is NP-complete, but 3-SAT is in P.
C. SAT is in P, but 3-SAT is NP-complete.
D. Both SAT and 3-SAT are in P.
Answer: A. SAT is NP-complete, and 3-SAT is also NP-complete.
Rationale: The Cook-Levin theorem establishes SAT as NP-complete. 3-SAT is also
NP-complete because any SAT instance can be reduced to 3-SAT in polynomial time. The
other options incorrectly claim 3-SAT or SAT is in P.
Page 3 of 76
, 7. Consider a randomized algorithm that, on any input, has a probability of at most
1/2 of giving an incorrect answer. If we run the algorithm independently 10 times and
take the majority answer, what is the probability that the majority answer is
incorrect?
A. At most 1/2
B. At most 1/4
C. At most (1/2)^10
D. At most 1/1024
Answer: C. At most (1/2)^10
Rationale: This question is under revision. The correct answer is C based on the
assumption that the majority vote reduces error exponentially, but the exact bound depends
on the error probability. In standard randomized algorithms, repeating independent runs
and taking majority reduces error probability exponentially if the original error probability
is strictly less than 1/2. Here, with error 1/2, the bound (1/2)^10 is a loose upper bound.
However, the question is flawed; a better version would specify error 1/3. For now, C is the
intended answer.
8. In the context of NP-completeness, which of the following is a polynomial-time
reduction from problem A to problem B?
A. A function f that maps instances of A to instances of B such that x A if and only if
f(x) B, and f can be computed in polynomial time.
B. A function f that maps instances of A to instances of B such that x A if and only if
f(x) B, and f can be computed in exponential time.
C. A function f that maps instances of B to instances of A such that x B if and only if
f(x) A, and f can be computed in polynomial time.
D. A function f that maps instances of A to instances of B such that x A implies f(x) B,
and f can be computed in polynomial time.
Answer: A. A function f that maps instances of A to instances of B such that x A if and
only if f(x) B, and f can be computed in polynomial time.
Rationale: A polynomial-time reduction from A to B requires a polynomial-time
computable function f such that x A iff f(x) B. Option B uses exponential time, which is not
a polynomial-time reduction. Option C reverses the direction, and Option D only requires
one direction, missing the 'if and only if' condition.
Page 4 of 76
