AskSia

Plus

Let

G=(V, E, w)

be a directed and weighted graph with edge weights

w(e)

, whi...

Feb 11, 2024

Let

G=(V, E, w)

be a directed and weighted graph with edge weights

w(e)

, which can be negative or positive or zero, for each edge

e \in E

. You may assume that

G

has no negatively weighted cycles. Additionally, each vertex is coloured either red or blue. Recall that a walk is a sequence of edges joining a sequence of vertices; there are no restrictions on the edges or vertices (i.e. they may repeat or not repeat). Given an integer

k \geq 2

, describe an

O(k n(k m+n))

algorithm that returns a walk with the smallest weight, satisfying the following conditions: - The walk starts and ends on a vertex coloured red, and - The number of vertices coloured blue in the walk is divisible by

k

. Note. For a directed and weighted graph

G=(V, E, w)

where the edge weights can be negative, BellmanFord will compute the shortest distance from a vertex

u

to every vertex in time

O(|V| \cdot|E|)

. We will see how it works later but for now, you may use it as a black box. Hint. Try constructing your edges in layers modulo

k

. You should have

k m+2 n

edges and

k n+2

vertices in your new graph.

Solution by Steps

step 1

Construct a new graph

G'

with

k

layers to account for the number of blue vertices modulo

k

. Each layer represents a different remainder when dividing the count of blue vertices by

k

step 2

For each vertex

v

in the original graph

G

, create

k

copies in

G'

, one for each layer. This results in

k \cdot |V|

vertices. Additionally, add two special vertices: a source

s

and a sink

t

step 3

For each edge

e = (u, v)

G

, create

k

directed edges in

G'

connecting the corresponding vertices in consecutive layers, with the same weight

w(e)

. This results in

k \cdot |E|

edges

step 4

Add edges with weight 0 from the source

s

to every copy of the red vertices in the first layer and from every copy of the red vertices in the last layer to the sink

t

. This adds

2 \cdot |V|

edges

step 5

Use the Bellman-Ford algorithm to find the shortest path from

s

t

G'

. Since Bellman-Ford runs in

O(|V| \cdot |E|)

, and

G'

has

k \cdot |V| + 2

vertices and

k \cdot |E| + 2 \cdot |V|

edges, the total time complexity is

O(k \cdot |V| \cdot (k \cdot |E| + 2 \cdot |V|))

Answer

The algorithm described above will return a walk with the smallest weight that starts and ends on a vertex coloured red, with the number of vertices coloured blue in the walk being divisible by

k

, in

O(k \cdot |V| \cdot (k \cdot |E| + 2 \cdot |V|))

time.

Key Concept

Graph Transformation and Layering

Explanation

The key concept is transforming the original graph into a layered graph that encodes the condition of having a number of blue vertices divisible by

k

and then applying a shortest path algorithm to find the desired walk.