CSE 101 Homework 1 Solutions
Winter 2021
Question 1 (Clone Maze, 35 points). Steven’s body has been split into two clones and he needs to reach the
recombiner to fix it. Unfortunately, the clones still have only one mind between them and will always move
in the same direction as each other. What is worse is that the laboratory that the clones are in is filled with
traps!
The laboratory is given as an n× n grid of squares. Some of these squares are marked as walls and are
impassible. Some are marked as traps and will kill Steven if a clone enters one. Two squares are marked
as the two pads for the recombiner, and two squares are marked as the current locations of Steven’s clones.
When moving, Steven selects one of the four directions (up, down, left, right) and both clones move one
square in that direction if possible (a clone does not move if there is a wall of the edge of the laboratory
one square in that direction). If a clone moves into a trap square, Steven will die. Give an algorithm to
determine if it is possible for Steven to get both of his clones to both of the recombiner pads. For full credit
your algorithm should run in time polynomial in n.
Solution 1.
The key idea in this problem is that we can treat the pair of positions of the two clones as a node of
a graph, G. Then every action that Steven takes(move up/down/left/right) corresponds to an edge in this
graph. If any clone enters in a trap point, then Steven can’t move in that direction. If one of the clones
hits a wall or the end of the maze, then that clone can’t move in that direction, whereas the other clone can
move, if possible. Therefore, first we define the CheckV alid function to check if the pair of points A,B are
legal locations for the clones. Also, we use another auxillary function, MoveDirection(A,B,direction) which
returns the next pair of points(node in G) that can be reached from (A,B). MoveDirection(A,B,direction)
returns NULL if one of the clone moves into a trap point. If either of the clone moves into a wall or the end
of the maze, then the MoveDirection(A,B,direction) function simply moves the other clone while keeping
the position of this clone fixed. If no clone can move, then MoveDirection(A,B,direction) simply returns the
current position.
CheckValid(A,B):
return 0 ≤ A.x < n and 0 ≤ B.x < n and 0 ≤ A.y < n and 0 ≤ B.y < n and !A.trap() and
!B.trap() and !A.wall() and !B.wall()
Now we can create the graph, G where a pair of points in the maze is a node in G and each node has at
most 4 neighbors(depending) on how many neighbours are valid nodes in the graph.
CreateGraph():
For point P1 in Maze:
For point P2 in Maze:
CurrNode=(P1,P2)
if CheckValid(CurrNode):
G.addNode(CurrNode)
LeftNode=MoveDirection(P1,P2,left)
RightNode=MoveDirection(P1,P2,right)
UpNode=MoveDirection(P1,P2,up)
DownNode=MoveDirection(P1,P2,down)
if LeftNode!=NULL and LeftNode!=CurrNode:
G.addEdge(CurrNode −→ LeftNode)
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if RightNode!=NULL and RightNode!=CurrNode:
G.addEdge(CurrNode −→ RightNode)
if UpNode!=NULL and UpNode!=CurrNode:
G.addEdge(CurrNode −→ UpNode)
if DownNode!=NULL and DownNode!=CurrNode:
G.addEdge(CurrNode −→ DownNode)
Now we can run Explore on this graph G starting from the initial positions of the two clones and with the
positions of the recombiner nodes as the target. Since the total number of possible pairs of points in an n×n
maze is n4 and we visit each pair at most once, the running time complexity is clearly polynomial in n.
Question 2 (Proportional Connectivity, 35 points). The nation of Graphania exists on a small chain of
islands. Recently, they have begun building (two-way) bridges connecting some of the islands. The government
of Graphania wants to measure the progress of this new bridge system and to do so, they want to measure the
probability that given two random Graphanian citizens, A and B that A will be able to reach B by travelling
along the newly built bridges.
You are given a list of Graphania’s islands along with the fraction of the population living on each island,
and which other islands each island has bridges to. Your goal is to compute the probability that a randomly
selected pair of citizens can reach each other using these bridges. You can assume that once on an island, a
citizen can reach anywhere else on that island.
(a) Give a linear time algorithm to solve this problem. [30 points]
(b) Does this algorithm work if some of the bridges are one-way instead of two way? If not, what goes wrong?
[5 points]
Solution 2.
(a) The key insight here is that any pair of citizens can only reach each other if they are in the same
connected component of islands. If we first compute the connected components of the islands, then
the problem reduces to ”Given citizens A and B, what is the probability that they are in the same
connected component?” For each connected component Cj containing islands xj1 , xj2 , ..., xjnJ , let pxjk
be the fraction of the population living on island xjk (this is given), and let pCj =
∑n
k=1 pxjk be the
fraction of the population living in connected component Cj (this is just the sum of the fraction of the
population living on each island in the component). Then we have P (A,B ∈ Cj) = P (A ∈ Cj) · P (B ∈
Cj) = pCj ·pCj = pCj 2. Summing this up over all connected components yields P (A,B are connected) =∑n
j=1 P (A,B ∈ Cj) =
∑n
j=1 pCj
2, which is our answer.
Here is pseudocode for our algorithm:
ProportionalConnectivity(G, island_pops): // island_pops[v] = fraction of pop. on island v
component_pops <- list of zeros, length |V| // component_pops[c] stores fraction of
// pop. in component c
ConnectedComponents(G) // run algorithm from class
For v in G:
component_pops[v.CC] += island_pops[v]
total_prob <- 0
For c in {1..component_pops.size}: // iterate through components
total_prob += component_pops[c]^2
Return total_prob
We know from class that ConnectedComponents(G) runs in linear time. The rest of our algorithm
involves twice performing a constant-time operation on (at most) each vertex of G. Thus, our algorithm
is clearly linear time, as required.
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(b) Note that in a directed graph, it is possible for A to be able to reach B even if A and B are not in the
same strongly connected component (i.e. if B is unable to reach A due to some of the bridges on the
path from A to B being one-way). Thus, the probability that A is able to reach B and the probability
that A and B are in the same strongly connected component are no longer equal, and so our algorithm
fails to work.
Question 3 (DFS Tree from Pre- and Post- Orders, 30 points). Given the pre- and post- order numbers of
each vertex of a graph G in a particular execution of DFS, give a polynomial time algorithm that computes
the DFS tree for that execution.
Solution 3.
The key to solve this question is to understand the meaning of Pre and Post number of a vertex. The
Pre number indicates when we start exploring a vertex and its neighbours. The Post number indicates when
we finish exploring a vertex and its neighbours. Here is the pseudo-code for the algorithm:
This algorithm will try to find the all the children of vertex u in the DFS tree. if the u.post == u.pre+1,
then it means that when we applied the DFS algorithm on vertex u, all the neighbours of u have already
been visited, which makes u a leaf. Otherwise, the vertex v such that v.pre == u.pre + 1 will be the next
vertex to visit in the DFS algorithm. This makes v the first child of u. we add v to the children of u and then
explore v until time v.post. After that we return to u to check whether all the neighbours of u have been
explored, if not then the vertex w such that w.pre = v.post + 1 is the next children of u. We will continue
until all the children of u has been found. If we repeat this algorithm for all the vertices in the Graph G,
then we can construct the DFS tree.
// for each vertex u we want to find all the children of u
DFS_Tree(u, pre_list, post_list):
if u.pre+1 == u.post: // vertex u is a leaf
return u.post+1
next <- u.pre + 1 // pre_number of the first children of u
while next != u.post: // hasn’t finish exploring u
v <- vertex such v.pre == next // takes O(n) time
add v to u’s children
next <- DFS_Tree(v,pre_list,pos_list)
return u.post+1
The time complexity of of this algorithm is O(n2) where n is the total number of vertices. Since for each
vertex u in Graph G, we need to call DFS Tree recursively to explore all the children of u. Let’s denote |uc|
as the number of children of vertex u in the DFS tree. Then the time complexity for DFS Tree(u,...) will be
O(n) ∗ |uc| since it takes O(n) to find vertex v such that v.pre == next. Since in a DFS tree, the number
of edges is at most n− 1. So
∑
u |uc| = O(n). As a result, the time complexity to reconstruct the DFS tree
will be
∑
uO(n) ∗ |uc| = O(n) ∗
∑
u |uc| = O(n
2). If we use a hash-map to store the pre/post numbers, then
the time complexity can be optimized to O(n) since right now it takes O(1) to find a vertex with a specified
pre number.
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