Let us reduce the vertex cover (VC) problem to any approximation algorithm for CDW (reps., CMV). In VC, we are given an undirected graph and an integer k; the decision question is: Is there a set of nodes () so that for each edge , at least one of its vertices are in ? Assume is a given instance for VC problem, where is the given graph, and B is an integer value. We create an instance for CDW (reps., CMV) so that is the graph build from G, and B is also the budget for our problem. Let us consider a case where there are two parties and four candidates, i.e., . We fix the order of candidates in the probability distribution of the voter v as , and build as follows.
For each undirected edge add two directed edges to . Set the weight of each incoming edge to a node as . By this the sum over weight of all incoming edges is equal to one, i.e., .
For each node , add two more nodes to , respectively. Furthermore, add an edge to with . Formally, . Note that nodes in are isolated.
Set the preferences list of the nodes as follows.
By this reduction, the score of candidates before any diffusion is . Then .
Note that in this reduction a node v will become active deterministically, if either it is selected as a seed node, or all of its incoming neighbors are selected as the seed nodes. Then if we can find a set of seed nodes so that it activates all nodes in V deterministically, the seed set S is also an answer for the corresponding VC problem.
In any approximation algorithm, we know that after the diffusion; otherwise, if there is a node we can replace it with its incoming neighbor such that and we get at least the same value for . Furthermore, if there exists a node one of the following situations holds:
There exists an inactive node after the diffusion S. In this case, we can substitute v for and then we get at least the same .
There is no inactive node . In this case, according to the nodes’ probability distribution, when all nodes in V become active, the value of and is maximum. Then even if we remove from S it does not change the value of or . By the way, in this situation, if there exist any node we replace with it, otherwise we replace it with a node .
Then from now on, we assume .
If all nodes in V
become active, since they have an outgoing edge to all nodes
with probability one, then all nodes in
will become active, and the score of the candidates will be as follows.
Then , , , and any approximation algorithm will return a positive value, then the answer of will be YES.
On the other hand, if there is a node
, which is inactive after the diffusion, i.e.,
, the score of candidates will be as follows.
Then , and any approximation algorithm will return zero, then the answer of will be NO.
For the other direction, note that if we can find a set of nodes , which is an answer for , using the same set of nodes, we can activate all nodes in and .
To extend the proof for any number of parties (t
) and candidates (k
), we need to assign the probability distribution as follows, and the same approach concludes the proof for any
. The same as before, the order of the candidates in probability distribution of a voter v
The following theorem proves the same statement for the destructive case of the problem.