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Problem 1. Decision Problems
For each of the following questions, circle either T (True) or F (False), and briefly justify your
answer in the box provided (a single sentence or picture should be sufficient). Each problem is
worth 4 points: 2 points for your answer and 2 points for your justification. If you leave both
answer and justification blank, you will receive 1 point.
(a) T F22n ∈ Θ(2n).
(b) T F If T (n) = T n + n and T (1) = Θ(1)
, then T (n) = O(n ).
programmingexamhelp.com
Solution: False. This statement is equivalent to saying k2 ∈ O(k) for k = 2n.
Constants in exponents matter asymptotically!
Solution: True. This is the defintion of amortization.
Solution: False. This is an example of Case II of Master Theorem, since a = 9 ,
b = 3 , f (n) = n2 and n2 = Θ(nlog3/2 9/4
log0
n). Thus, the recurrence
evaluates to T (n) = Θ(n2 log n), which is not O(n2).
(c) T F Performing an O(1) amortized operation n times on an initially empty data structure
takes worst-case O(n) time.

Solution: False. As a counter example, during the final merge step between two sorted
halves of the array, each of size Θ(n), a single item from one array may get compared
to all the items from the other list.
(e) T F Given a binary min-heap storing n items with comparable keys, one can build a Set
AVL Tree containing the same items using O(n) comparisons.
Solution: False. If such an algorithm A existed, we would be able to sort an array of
comparable items in O(n) time, which would contradict the Ω(n log n) comparison
sort lower bound. Specifically, we could build a binary min-heap from the array using
O(n) comparisons, use A to construct a Set AVL Tree in O(n) comparisons, and then
return its traversal order.
(f) T F Given a directed graph G = (V, E), run breadth-first search from a vertex s ∈ V .
While processing a vertex u, if some v ∈ Adj+
(u) has already been processed, then
G contains a directed cycle.
(d) T F Given an array A containing n comparable items, sort A using merge sort.
While sorting, each item in A is compared with O(log n) other items of A.
Solution: False. BFS can’t be used to find directed cycles. A counterexample is V =
{s, a, b, t} and E = {(s, t), (s, a), (a, b), (b, t)}. Running BFS from s will first
process vertices in levels {s}, then {a, t}, then {b}. When processing vertex b,
vertex t ∈Adj+
(b) has already been processed, yet G is a DAG.

(g) T F Run Bellman-Ford on a weighted graph G = (V, E, w) from a vertex s ∈ V . If there is a
witness v ∈V , i.e., δ|V |(s, v) < δ|V |−1(s, v), then v is on a negative- weight cycle of G.
Solution: False. A witness is only guaranteed to be reachable from a
negative-weight cycle; it may not actually be on a negative-weight cycle.
(h) T F Floyd–Warshall and Johnson’s Algorithm solve all-pairs shortest paths in the same
asymptotic running time when applied to weighted complete graphs, i.e., graphs where every
vertex has an edge to every other vertex.
Solution: True. A complete graph is dense, i.e., |E| = Θ(|V |2), so
Johnson’s algorithm runs in O(|V |2 log |V | + |V ||E|) = O(|V |3) time,
which is the same as Floyd–Warshall.
(i) T F If there is an algorithm to solve 0-1 Knapsack in polynomial time, then there is also an
algorithm to solve Subset Sum in polynomial time.
(j) T F Suppose a decision problem A has a pseudopolynomial-time algorithm to solve A. If P
6= NP, then A is not solvable in polynomial time.
Solution: False. A problem could have a pseudopolynomial-time algorithm
and a polynomial-time algorithm. In fact, any polynomial-time algorithm is
also a pseudopolynomial-time algorithm!
Solution: True. Subset Sum is the special case of 0-1 Knapsack. Speciﬁcally, one
can (in linear time) convert an instance (A, T ) of Subset Sum into an equivalent
instance of 0-1 Knapsack, with an item i for each integer ai ∈A having size si = ai
and value vi = ai, needing to ﬁll a knapsack of size T ; and then solve the instance
via the polynomial-time algorithm for 0-1 Knapsack.

Problem 3. Sorting Sorts
(a)An integer array A is k-even-mixed if there are exactly k even integers in A, and the odd
integers in A appear in sorted order. Given a k-even-mixed array A containing n distinct
integers for k = dn/ lg ne, describe an O(n)-time algorithm to sort A.
Solution: Scan through A and put all even integers in order into an array AE and all odd
integers in order into an array AO (where |AE | = k and |AO| = n − k). AO is sorted by
deﬁnition, and we can sort AE in O(k log k) = O((n/ lg n) log(n/ lg n)) = O(n) time,
e.g., via merge sort. Then we can merge sorted AE and AO back into A in O(n) time using
the merge step of merge sort, using O(n) time in total.
Common Mistakes:
•Using insertion sort
•Splitting into a non-constant number of subarrays and trying to merge
•Using binary search to insert evens into odds (but with linear shifts)
•Using radix or counting sort
(b)let A be an array of n pairs of positive integers (xi, yi) with xi, yi < n2 for all i ∈ {0, . . . , n −
1}. The power of pair (x, y) is the integer x + ny. Describe an O(n)-time algorithm to sort the
pairs in A increasing by power.
Solution: First note that x < ny for any integer y > 1 and for any x ∈ {0, . . . , n2−1}. Scan
through A and put all pairs having y = 1 into array A1, and all other pairs into array A2. Sort
A1 directly by computing and comparing their respective powers x + n. Since these values are
bounded above by O(n2), sort A1 in O(n) time using Radix sort. To sort A2, use tuple sort,
sorting ﬁrst by x values and then by y values (since power is more sensitive to changes in y).
Since the x and y values are both bounded above by O(n2), we can use Radix

sort for tuple sort’s stable sorting algorithm to sort A2 in O(n) time. Then merge A1 and A2 back
into A in O(n) time using the merge step of merge sort, using O(n) time in total.
Common Mistakes:
•Not handling y = 1 (x may be larger or smaller than n1)
•Explicitly computing powers (which may have exponential size)
•Concatenating instead of merging two overlapping lists
Problem 4. Pythagorean Quad
A Pythagorean Quad consists of four integers (a, b, c, d) such that d = √
a2 + b2 + c2. Given an
array A containing n distinct positive integers, describe an O(n2)-time algorithm to determine
whether four integers from A form a Pythagorean Quad, where integers from A may appear more
than once in the Quad. State whether your running time is worst-case, expected, and/or amortized.
Solution: First, we observe that it sufﬁces to ﬁnd (a, b, c, d) such that a2 + b2 = d2 − c2. Let P be
the set of n2 ordered pairs of integers from A, where integers in A may be repeated in a pair.
Construct an empty hash table H, and for each pair (a, b) ∈ P , compute and insert value a2 + b2
into H. Then for each pair (c, d) ∈ P , compute and lookup value d2 − c2 in H. If the value is in H,
then some a2 + b2 equals some d2 − c2, so return that a Pythagorean Quad exists. Otherwise, if no
d2 − c2 exists in H, then return that a Pythagorean Quad does not exist. Each a2 + b2 or d2 − c2
value takes constant time to compute, so computing them all takes worst-case O(n2) time, while
inserting them into or looking them up in the hash table takes expected O(n2) time, so this
algorithm runs expected in O(n2) time in total.
Common Mistakes:
•Computing non-integers involving logarithms or square roots
•Saying a worst-case running time is amortized

•Trying to write an incorrect dynamic program
•Trying to write an incorrect four-ﬁnger algorithm
Problem 5. Animal Counting
PurpleRock Park is a wildlife reserve, divided into zones, where each zone has a park ranger
who records current sightings of animals of different species over time. Old animal sightings are
periodically removed from the database. A species s is common if current park records contain at
least 100 sightings of species s within any single zone of the park.
Describe a database to store animal sightings, supporting the following four operations, where n
is the number of sightings stored in the database at the time of the operation. State whether your
running times are worst-case, expected, and/or amortized.
initialize() Initialize an empty database in O(1) time
add sighting(s, i) Record a newest sighting of species sin zone iin O(log n) time
remove oldest() Remove the oldest sighting stored in the database in O(log n) time
is common(s) Return whether species sis common based on sightings
that have not yet been removed from the database in O(1) time
Solution: To implement the database, maintain the following data structures:
•A hash table H mapping each species s to a Set AVL tree Ts
•Each Set AVL Tree Ts stores pairs (i, ci) of zone numbers i and the count ci representing the
number of sightings of species s in zone i, keyed by zone number.
•Augment each node x in each Ts by the maximum number of sightings x.m of any zone in
the subtree of x. x.m can be maintained in O(1) time from the augmentations of x’s children,
speciﬁcally x.m = max{x.left.m, x.key, x.right.m}.
•A doubly-linked list L of all current sightings (s, i) in the order in which they were added to
the database (oldest at the front). programmingexamhelp.com

To implement initialize(), initialize an empty H and empty L in worst-case O(1) time.
To implement add sighting(s, i), lookup s in H to find Ts in expected O(1) time (if s does not
exist in H, insert s mapping to an empty Ts in expected amortized O(1) time). Then find zone i
in Ts. If zone i is not in Ts, insert (i, 1) into Ts. Otherwise, i is in Ts, so remove (i, ci) from Ts
and reinsert (i, ci + 1). It takes worst-case O(log n) time to remove or insert items from Ts
while maintaining augmentations (since at most n sightings could exist for species s). Lastly,
Insert (s, i) to the back of L in worst-case O(1) time. Thus this operation takes O(log n)
expected amortized time, and maintains the invariants of the database directly.
To implement remove oldest(), remove the oldest pair (s, i) from the front of L in worst-case
O(1) time. Lookup s in H to find Ts in expected O(1) time; then lookup i in Ts and decrease
ci by one. If ci is decreased to zero, remove i from Ts. If Ts becomes empty, remove s from
H in expected amortized O(1) time. This operation takes O(log n) expected amortized time,
and maintains the invariants of the database directly.
To implement is common(s), simply lookup s in H and return whether s is in H and the stored
max at the root of Ts is 100 or greater in expected O(1) time. This operation is correct based on
the invariants of the data structure.
Common Mistakes: Continued on S1
Problem 6. Limited Connections
For any weighted graph G = (V, E, w) and integer k, define Gk to be the graph that results
from removing every edge in G having weight k or larger.
Given a connected undirected weighted graph G = (V, E, w), where every edge has a
unique
integer weight, describe an O(|E| log |E|)-time algorithm to determine the largest value of k
such

that Gk is not connected.
Solution:Construct an array A containing the |E| distinct edge weights in G, and sort it in O(|E|
log |E|) time, e.g., using merge sort. We will binary search to find k. Specifically, consider an edge
weight k0 in A (initially the median edge weight), and run a reachability algorithm (e.g., Full-BFS
or Full-DFS) to compute the reachability of an arbitrary vertex x ∈V in O(|E|) time. If exactly |V
| vertices are reachable from x, then Gk0 is connected and k > k0; recurse on strictly larger values
for k0. Otherwise, Gk0 is not connected, so k ≤ k0; recurse on non-strictly smaller values for k0. By
dividing the search range by a constant fraction at each step (i.e., by always choosing the median
index weight of the unsearched space), binary search will terminate after O(log |E|) steps,
identifying the largest value of k such that Gk is not connected. This algorithm takes O(|E| log |E|)
time to sort, and computes reachability of a vertex in O(|E|) time, O(log |E|) times, so this
algorithm runs in O(|E| log |E|) time in total.
Common Mistakes:
•Filtering linearly on all edge weights, rather than using binary search
•Using Dijkstra/Bellman-Ford to inefficiently solve reachability
•Simply stating ‘Use Dijkstra’

Problem 7. On the Street
Friends Dal and Sean want to take a car trip across the country from Yew Nork to Fan
Sancrisco by driving between cities during the day, and staying at a hotel in some city each
night. There are n cities across the country. For each city ci, Dal and Sean have compiled:
•the positive integer expense h(ci) of staying at a hotel in city ci for one night; and
•a list Li of the at most 10 other cities they could drive to in a single day starting from city
ci, along with the positive integer expense g(ci, cj ) required to drive directly from ci to cj
for each cj ∈Li.
Describe an O(nd)-time algorithm to determine whether it is possible for Dal and Sean to drive
from Yew Nork to Fan Sancrisco in at most d days, spending at most b on expenses along the
way.
Solution: Let C = {c0, . . . , cn−1}, and let cs denote Yew Nork and let ct denote Fan
Sancrisco. Construct a graph G with:
•a vertex (ci, d0) for each city ci ∈C and day d0∈{0, . . . , d}, representing
staying the night in city ci on the night before day d0; and
•a directed weighted edge ((ci, d0), (cj, d0+ 1)) with weight g(ci, cj ) + h(cj )
for each city ci ∈C , cj ∈L i and d0 ∈{0, . . . , d − 1}.
Then the weight of any path in G from vertex (cs, 0) to any vertex (ct, d0) for d0∈ {0, . . . , d}
corresponds to the expenses incurred along a driving route from Yew Nork to Fan Sancrisco
in at most d days (assuming they stay the night upon reaching ct; other assumptions are also
okay). G is acyclic, since each edge always connects a vertex from a smaller day to a larger
day, so run DAG Relaxation to compute single-source shortest paths from (cs, 0) in G. Then
return whether δ((cs, 0), (ct, d0)) ≤ b for any d0∈ {0, . . . d}. G has O(nd) vertices and O(nd)
edges (since)

|Li| ≤ 10 for all i ∈{0, . . . , n − 1}), so DAG relaxation runs in O(nd) time and checking all
destination values takes O(d) time, leading to O(nd) time in total.
Common Mistakes:
•Not considering paths on fewer than d days (or considering paths on more than d days)
•Not enforcing staying at a hotel every night
•Not including hotel costs
•Problem 8. RGB Graph
•Let G = (V, E, w) be a weighted directed graph. Let c : V → {r, g, b} be an assignment of each
vertex v to a color c(v), representing red, green, or blue respectively. For x ∈{r, g, b},
•let Vx be the set of vertices with color x, i.e., Vx = {v ∈V | c(v) = x}; and
•let Ex be the set of edges outgoing from vertices in Vx, i.e., Ex = {(u, v) ∈E | u ∈Vx}.
Suppose graph G and coloring c have the following properties:
1.Every edge in E either connects two vertices of the same color,
goes from a red vertex to a green vertex, or goes from a green vertex to a blue vertex.
2.|Vr| = |Er| = O(|V |), and edges in Er have identical positive integer weight wr.
3.|Vg| = |Eg | = O(|V |0.99), and edges in Eg have nonnegative integer weights.
4.|Vb| = |Eb| = O(p
|V |), and edges in Eb can have positive or negative integer weights.

S
0
vertices appearing in those edges, specifically
V =
{u, v} (which contains vertices from
(u,v)∈E
r
Vr and Vg). Run breadth-first search from s in G0 to compute unweighted distances. Then the
minimum weight distance in G from s to any green vertex in V 0∩ Vg is wr times the
unweighted
distance computed. G0 has size O(|V |), so this step takes O(|V |) time.
Step 2: Now construct weighted graph G00 = (V 00,E00) composed of vertex s with a new
directed
Given G, c, a red vertex s ∈ Vr, and a blue vertex t ∈ Vb, describe an O(|V |)-time algorithm to
compute δ(s, t), the minimum weight of any path from s to t.
Solution: Any path from s to t is a path through edges in Er, followed by a path through edges
in Eg , followed by a (possibly empty) path through edges in Eg . So we compute minimum
weight distances in G incrementally, first using edges in Er, then using edges in Eg , then edges
in Eb.
Step 1: Construct unweighted graph G0 = (V 0,E0) composed of the edges E0 = Er and the

edge to each green vertex in V 0∩ Vg weighted by its distance found in Step 1 (i.e., the minimum
weight of any path from s to that vertex), along with weighted edges Eg and the vertices appearing
in those edges. All the weights in G00 are positive, so run Dijkstra from s in G00 to compute
minimum weight distances. Then the computed distance to any blue vertex v in V 0
0∩ Vb is the
minimum weight of any path from s to v in G that traverses only red or green edges. G00has size
O(1 + |Vg|+ |Eg |) = O(|V |0.99), so this step takes O(|V |0.99 log |V |0.99) = O(|V |) time.
Step 3: Now construct a weighted graph G000= (V 000,E000) composed of vertex s with a new directed
edge to each blue vertex in V 0
0∩ Vb weighted by its distance found in Step 2 (i.e., the minimum
weight of any path from s to that vertex), along with weighted edges Eb and the vertices appearing
in those edges. Weights in G000may be positive or negative, so run Bellman-Ford from s in G000to
compute weighted minimum weight distances. Then the computed distance to t is the minimum
weight of any path from s to t in G, as desired. G000has size O(1 + |Vb| + |Eb|) = O(p
|V |), so
p p
this step takes O( |V | |V |) = O(|V |) time, leading to O(|V |) time in total.
Common Mistakes: Continued on S1.
Problem 9. Separated Subsets
For any set S of integers and for any positive integers m and k, an (m,k)-separated subset of S
is any subset S0 ⊆ S such that S0 sums to m and every pair of distinct integers a, b ∈ S0 satisﬁes
|a − b| ≥ k. Given positive integers m and k, and a set S containing n distinct positive integers,
describe an O(n2m)-time algorithm to count the number of (m, k)-separated subsets of S.
(When solving this problem, you may assume that a single machine word is large enough to
hold any integer computed during your algorithm.)
Solution:

describe an O(n2m)-time algorithm to count the number of (m, k)-separated subsets of S.
(When solving this problem, you may assume that a single machine word is large enough to hold
any integer computed during your algorithm.)
Solution:
1.Subproblems
•First sort the integers in S increasing into array A in O(n log n) time, e.g., via merge sort
•where A = (a0, . . . , an−1)
•x(i, j): the number of (j, k)-separated subsets of sufﬁx A[i :]
•for i ∈{0, . . . , n} and j ∈{0, . . . , m}
2.Relate
•Sum the number of (j, k)-separated subsets using A[i] with the ones that do not use A[i]
•If A[i] ≤ j is used:
– Then no integer in A[i :] smaller than A[i] + k may be used
– Let f (i) be the smallest index greater than i such that A[f (i)] − A[i] ≥ k
– Then recrusively count x(f (i), j − A[i])
•Otherwise, A[i] is not used and we can recursively count
x(i + 1, j)
P
x(f (i), j − A[i]) if A[i] ≤ j,
•x(i, j) =
3. Topo
x(i + 1, j) always

•Subproblem x(i, j) only depends on strictly larger i, so acyclic
4.Base
•x(n, 0) = 1, the empty subset can always be acheived
•x(n, j) = 0 for j > 0, empty sets cannot sum to a positive number
5.Original
•x(0, m), the number of (m, k)-separated subsets of A
6.Time
•# subproblems: (n + 1)(m + 1) = O(nm)
•Work per subproblem: O(n) to ﬁnd f (i) by linear scan
•O(n m) time in total
2
•(Note that it is possible to compute f (i) in O(log n) time via binary search, or in amor-
tized O(1) time from f (i −1), but these optimizations are not necessariy for full points.)
Problem 10. A Feast for Crowns
Ted Snark is arranging a feast for the Queen of Southeros and her guests, and has been tasked
with seating them along one side of a long banquet table.
•Ted has a list of the 2n guests, where each guest i has a known distinct positive integer fi
denoting the guest’s favor with the Queen.
•Ted must seat the guests respectfully: the Queen must be seated in the center with n guests
on either side so that guests’ favor monotonically decreases away from the Queen, i.e., any
guest seated between a guest i and the Queen must have favor larger than fi.

•Additionally, every guest hates every other guest: for every two guests i, j, Ted knows the
positive integer mutual hatred d(i, j) = d(j, i) between them.
Given Ted’s guest information, describe an O(n3)-time algorithm to determine a respectful
seating order that minimizes the sum of mutual hatred between pairs of guests seated next to
each other. Signiﬁcant partial credit will be awarded to correct O(n4)-time algorithms.
Solution:
1.Subproblems
•Sort the guests increasing by favor in O(n log n) time into F = (f0, . . . , f2n−1)
•Any partition of F into two length-n subsequences corresponds to a respectful seating
•x(i, jL, jR, nL): minimum total hatred of adjacent guests possible by respectfully
seating the n − i guests from sufﬁx F [i :] next to the Queen, with nL guests to
the left and nR = (n − i) − nL guests to the right, where guest jL < i has already
been seated nL + 1 places to the left, and guest jR < i has already been seated nR +
1 places to the right.
•for i ∈{0, . . . , 2n}, jL, jR ∈{−1, . . . , 2n − 1} and nL ∈{0, . . . , n}
where either jL = i − 1 or jR = i − 1
•Let d(−1, i) = d(i, −1) = 0 for all i ∈{0, . . . , 2n − 1} (no hatred at the end of table)
2.Relate
•Guess whether guest i is seated on the right or left

•Sitting next to jL costs hatred d(i, jL); sitting next to jR costs hatred d(i, jR)
L R L
d(i, j ) + x(i + 1, i, j , n − 1) if nL
> 0,
L R L
•x(i, j , j , n ) = min
d(i, jR ) + x(i + 1, jL , i, nL ) if (n − i) − nL > 0
3.Topo: Subproblem x(i, jL, jR, nL) only depends on strictly larger i, so acyclic
4.Base: x(2n, jL, jR, 0) = 0 for all jL, jR ∈{0, . . . , 2n} (no hatred if no guests)
5.Original: x(0, −1, −1, n), min hatred of adjacent guests by respectfully seating all guests
6.Time
•# subproblems: though there are four parameters, there are only O(n3) subproblems for
which either jL = i − 1 or jR = i − 1
•Work per subproblem: O(1), so O(n3) time in total
You can use this paper to write a longer solution if you run out of space, but be sure to write
“Continued on S1” on the problem statement’s page.
Common Mistakes: (for Problem 5)
•Not removing records for sightings or zones containing zero sightings
•Not accounting for existence or non-existence of a key in a data structure
•Algorithm is linear in number of zones or does account for zones at all
•Forgetting to say amortized for dynamic hash table operations
•Confusing Sequence/Set AVL Trees, or dynamic/direct-access arrays
•Not maintaining or updating number of sightings per zone per species correctly

Common Mistakes:
•Correctly finding shortest paths within each color, but not connecting them properly
•Assuming an optimal path from s to t goes through the green vertex closest to s
•Solving APSP on the red or green graph (inefficient)
•Trying to apply DAG relaxation to a possibly cyclic graph
•Assuming that if |V | = |E|, graph has at most one cycle (only true if connected)
Common Mistakes:
•Maintaining a subset of used items (yields exponential state)
•Unconrstrained subproblem for relation that relies on a constraint
•Solving the decision problem rather than the counting problem (or max instead of sum)
•Missing a base case
Common Mistakes:
•Not ensuring that n guests are seated to either side of Queen
•Not keeping track of guests on either end
•Defining Ω(n3) subproblems whose parameters are not independent
•Maintaining a subset of used items (yields exponential state)

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