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Design and Analysis of Algorithms
Problem 1. True or False.
Circle T or F for each of the following statements to indicate whether the statement is true
or false and briefly explain why.
(a) T F Using similar techniques used in Strassen’s matrix multiplication algorithm, the
Floyd–Warshall algorithm’s running time can be improved to O(V log2 7).
Solution: False. There is no way to define negation.
(b) T F For graphs G = (V, E) where E = O(V 1.5), Johnson’s algorithm is asymptotically faster
than Floyd–Warshall.
Solution: True. O(V E + V 2 log V ) = o(V 3) when E = o(V 2).
(c) T F Consider the directed graph where each vertex represents a subproblem in a
dynamic program, and there is an edge from p to q if and only if subproblem p depends on
(recursively calls) subproblem q. Then this graph is a directed rooted tree.
Solution: False. It is a Directed Acyclic Graphic (DAG).
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(d) T F In a connected, weighted graph, every lowest weight edge is always in some
minimum spanning tree.
Solution: True. It can be the first edge added by Kruskal’s algorithm.
(e) T F For a connected, weighted graph with n vertices and exactly n edges, it is possible
to find a minimum spanning tree in O(n) time.
Solution: True. This graph only contains one cycle, which can be found by a DFS. Just
remove the heaviest edge in that cycle.
(f) T F For a flow network with an integer capacity on every edge, the Ford–Fulkerson
algorithm runs in time O((V + E)|f|) where |f| is the maximum flow.
Solution: True. There can be O(|f|) iterations because each iteration increases the flow by
at least 1.
(g) T F Let C = (S, V S) be a minimum cut in a flow network. If we strictly increase the
capacity of every edge across C, then the maximum flow of the network must increase.
Solution: False. There could be another min cut whose capacity does not change. Then
the max flow remains the same.

(h) T F Every linear program has a unique optimal solution.
Solution: False. There can be many optimal solutions if the objective function is parallel to
one of the constrains.
Alternative solution: False. There could be no solutions at all.
Alternative solution: False. There could be no solutions at all.
(i) T F 3SAT cannot be solved in polynomial time, even if P = NP.
Solution: False. If P = NP, then all problems in P are also NP-hard, and these problems have
polynomial-time algorithms.
(j) T F Repeatedly selecting a vertex of maximum degree, and deleting the incident edges,
is a 2-approximation algorithm for Vertex Cover.
Solution: False: it can be as bad as a log-log approximation, see L17 notes.

Problem 2. Who Charged the Electric Car?
Prof. Musk is driving his Nikola electric car from Boston to New York. He wants to
take the shortest path, but his car can only drive m miles before needing to charge.
Fortunately, there are Furiouscharger charging stations on the way from Boston to
New York, which instantaneously charge the battery to full.
The road network is given to you as a weighted undirected graph G = (V, E,w) along
with the subset C ⊆ V of vertices that have charging stations. Each weight w(e)
denotes the (positive) length of road e. The goal is to find a shortest path from
node s ∈ V to node t ∈ V that does not travel more than m miles between charging
stations. Assume that s,t ∈ C.
(a) Draw the shortest path from Boston to New York in the following graph if m =
∞. Charging stations are marked as circles.

Solution:
(b) Draw the shortest path from Boston to New York in the following (identical) graph if m
= 100.
Solution:

(c) Give an algorithm to solve the problem. For full credit, your algorithm should run in O(V
E + V 2 log V ) time.
Solution: Our algorithm consists of two steps – the first step involves running Johnson’s
algorithm on the original graph G to obtain shortest path lengths for every pair of vertices.
Let δ(u, v) represent the length of the shortest path between vertices u and v in G.
For the second step, we build a graph G" with vertex set C. For every pair of vertices u and
v in the new graph G" , draw an edge between u and v with weight δ(u, v) if δ(u, v) ≤ m and
∞ otherwise.
Now, run Dijkstra’s algorithm on G" between Boston and New York to get the shortest path.
(Note that New York and Boston have charging stations and so are vertices in the graph G"
).
Running Johnson’s algorithm on the original graph G takes O(V E + V 2 log V ). Creating the
graph G" takes O(E) time, and running Dijkstra’s algorithm on G" takes O(V 2 + V log V )
time; this gives a total runtime complexity of O(V E + V 2 log V ).

Problem 3. Planning Ahead
You have N psets due right now, but you haven’t started any of them, so they are all going
to be late. Each pset requires di days to complete, and has a cost penalty of ci per day. So if
pset i ends up being finished t days late, then it incurs a penalty of t · ci. Assume that once
you start working on a pset, you must work on it until you finish it, and that you cannot
work on multiple psets at the same time.
For example, suppose you have three problem sets: 6.003 takes 3 days and has a penalty
of 12 points/day, 6.046 takes 4 days and has a penalty of 20 points/day, and 6.006 takes 2
days and has a peanlty of 4 points/day. The best order is then 6.046, 6.003, 6.006 which
results in a penalty of 20 · 4 + 12 · (4 + 3) + 4 · (3 + 4 + 2) = 200 points.
Give a greedy algorithm that outputs an ordering of the psets that minimizes the total
penalty for all the psets. Analyze the running time and prove correctness.
Solution: Sort by increasing di/ci and do the problem sets in that order. This takes O(N log
N) time.
Proof – If unsorted, we can improve by swapping.

Problem 4. Maze Marathoner
A group of m teens need to escape a maze, represented by a directed graph G = (V,
E). The teens all start at a common vertex s ∈ V , and all need to get to the single
exit at t ∈ V . Every night, each teen can choose to remain where they are, or
traverse an edge to a neighboring vertex (which takes exactly one night to
traverse). However, each edge e ∈ E has an associated capacity c(e), meaning that
at most c(e) teens can traverse the edge during the same night. The goal is to
minimize the number of nights required for all teens to escape by reaching the goal
t.
(a) First look at the special case where the maze is just a single path of length |E|
from s to t, and all the edges have capacity 1 (see below). Exactly how many nights
are required for the teens to escape?

Solution: |E| + m − 1 or |V | + m − 2. Em or V m will get partial credits.
(b) The general case is more complex. Assume for now that we have a “magic” algorithm
that calculates whether the teens can all escape using ≤ k nights. The magic algorithm
runs in polynomial time: kα T(V, E, m) where α = O(1). Give an algorithm to calculate
the minimum number of nights to escape, by making calls to the magic algorithm.
Analyze your time complexity in terms of V , E, m, α, and T(V, E, m).
Solution: Do a binary search. A sequential scan will get partial credits. The maximum
number of nights can be bounded by O(E+m) (or O(V +m), O(Em)) according to part(a).
Therefore, we need to run the “magic” algorithm O(log(E +m)) times. Each run takes no
more than O((E + m)αT(V, E, m)) time. So in total, the runtime is O((E + m)α log(E + m)T(V,
E, m)).
Common mistake 1: runtime O(kα log(E + m)T(V, E, m)). k should not appear in the
runtime. You need to find the bound on k.
Common mistake 2: using sequential scan runtime O((E + m)α n T(V, E, m)). The n α α+1
summation is calculated incorrectly, should be
(c) Now give the “magic” algorithm, and analyze its time complexity. Hint: Transform the
problem into a max-flow problem by constructing a graph G" = (V " , E" ) where V " = {(v, i)
| v ∈ V, 0 ≤ i ≤ k}. What should E" be?

Solution: Model this as a max flow problem. Construct a graph G" = (V " , E" ) where V "
= {(v,i) | v ∈ V, 0 ≤ i ≤ k}. For all 0 ≤ i ≤ k−1, connect (v, i) to (v, i+1) with capacity ∞ (or m);
this represents teens can stay at a vertex for the night. For every edge (u, v) in the
original graph, connect (u, i) to (v,i + 1) with capacity c((u, v)); this represents c((u, v))
teens can travel from u to v in a night. The new source s" is the vertex (s, 0) and the new
sink t " is the vertex (t, k − 1). If the max flow from s" to t " is no less than m, then people
can escape within k nights.
Runtime: Both of the following are accepted.
There are V = O(kV " ) vertices and E" = O(k(V + E)) edges in G" . Applying Edmonds-Karp
algorithm, the total time complexity is O(V E"2) = O(k3V (V + E)2). If using Ford-Fulkerson
runtime, notice that we can actually stop if the max flow reaches m. So at most m
iterations are needed. Runtime can be O(m(V " + E" )) = O(mk(V + E)).
Common mistake 1: connect (v, i) to (u, i) instead of (v, i) to (u, i + 1).
Common mistake 2: no edge from (v,i) to (v, i + 1).
Both solutions above are equivalent of running max flow on the original graph, because
there could be very long path (> k) with large capacity (> m) in the original graph. In that
case, max flow is larger than m, but teens cannot escape in k nights.

Problem 5. 6.046 Carpool
The n people in your dorm want to carpool to 34-101 during the m days of 6.046.
On day i, some subset Si of people actually want to carpool (i.e., attend lecture),
and the driver di must be selected from Si. Each person j has a limited number of
days fj they are willing to drive.
Give an algorithm to find a driver assignment di ∈ Si for each day i such that no
person j has to drive more than their limit fj . (The algorithm should output “no” if
there is no such assignment.) Hint: Use network flow.
For example, for the following input with n = 3 and m = 3, the algorithm could
assign Penny to Day 1 and Day 2, and Leonard to Day 3.

Solution: First, we create a graph with following vertices:
1.a super source s and a super sink t
2.vertex pi for each person who wants to carpool
3.vertex dj for each day of the class.
Then create the following edges:
1.s to pi with capacity of fj
2.pi to dj with capacity of 1 if person i needs to carpool on day j
3.dj to t with weight 1 for all j.
Finally, run max flow from s to t, and find f. If |f| = m, return that person i will drive on
day j if the edge (pi, dj ) has non-zero flow. If |f| < m, then return no valid assignment.
At a high level, the graph represents a matching between the driver and the days
he/she will be driving. The capacity from s to pi will ensure that no pi drives more than li
days (flow conservation). The non-zero flows from pi to dj means that pi will drive on
day dj . The capacity of dj to t will ensure that no more than 1 driver will be assigned to
a particular day (flow conservation). If |f| = m, then all vertices dj has an incoming flow,
and therefore all dj has a valid driver assignment. If |f| < m, then there is at least one dj
that has no incoming flow, and therefore does not have a driver assigned to it. By
maximality of the flow, this means that there does not exist a valid driver assignment.

Ford-Fulkerson algorithm would run in O(nm2) since there are at most nm + 2 edges
(bipartite graph), and the max flow is bounded by m. Running Edmonds-Karp would run
in O((n + m)(nm)2) = O(n3m2 + n2m3), which is slower in this case.
Problem 6. Paths and/or Cycles
A Hamiltonian path on a directed graph G = (V, E) is a path that visits each
vertex in V exactly once. Consider the following variants on Hamiltonian path:
(a) Give a polynomial-time algorithm to determine whether a directed graph G
contains either a cycle or a Hamiltonian path (or both).
Solution: To solve the problem, we simply run DFS on G. If a cycle exists, DFS
will traverse a vertex twice and can report the cycle. If no cycle exists, then the
graph is a DAG.
If the graph is a DAG, then we can run a topological sort on the graph. If there is
a complete, or unique, ordering of every vertex in the graph, the graph has a
Hamiltonian Path, and we accept the graph.

(b) Show that it is NP-hard to decide whether a directed graph G" contains both a cycle and
a Hamiltonian Path, by giving a reduction from the HAMILTONIAN PATH problem: given a
graph G, decide whether it has a Hamiltonian path. (Recall from recitation that the
HAMILTONIAN PATH problem is NP-complete.)
Solution: We construct a graph G" = (V " , E" ) from G, where
G" always has a cycle of length 3 - (u1, u2, u3). For any Hamiltonian Path P in G, (u2, u3, u1,
P) is a Hamiltonian Path in G" . For any Hamiltonian Path P" in G" , P" must be of the form
(u2, u3, u1, P), where P is a Hamiltonian path for G. Thus in all cases, solving B(G" ) is
equivalent to solving Hamiltonian Path for G.
An alternate solution used the algorithm for cycle and Hamiltonian Path as an oracle. An
input graph to the Hamiltonian Path is used as an input to the cycle and Hamiltonian Path
algorithm. If the cycle and Hamiltonian Path algorithm accepts, then the original graph is
also accepted. However, if the cycle and Hamiltonian path algorithm rejects the input, we
use DFS to determine if the graph contains a cycle and the cycle or Hamiltonian Path
algorithm from part a. If DFS cannot find a cycle but the cycle or Hamiltonian path algorithm
accepts, then we accept the graph, and reject the graph otherwise.

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