**Proof.** Let us reduce the vertex cover (VC) problem to any approximation algorithm for CDW (reps., CMV). In VC, we are given an undirected graph $G=(V,E)$ and an integer k; the decision question is: Is there a set of nodes ${V}^{\prime}\subseteq V$ ($|{V}^{\prime}|\le k$) so that for each edge $(u,v)\in E$, at least one of its vertices are in ${V}^{\prime}$? Assume $\mathcal{I}(G,B)$ is a given instance for VC problem, where $G=(V,E)$ is the given graph, and B is an integer value. We create an instance ${\mathcal{I}}^{\prime}({G}^{\prime},B)$ for CDW (reps., CMV) so that ${G}^{\prime}=(V\cup {V}^{\prime}\cup {V}^{\u2033},{E}^{\prime})$ 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., $t=k=2,C={C}_{1}\cup {C}_{2},$ ${C}_{1}=\{{c}_{1}^{1},{c}_{2}^{1}\},{C}_{2}=\{{c}_{1}^{2},{c}_{2}^{2}\}$. We fix the order of candidates in the probability distribution of the voter v as ${\pi}_{v}=({\pi}_{v}\left({c}_{1}^{1}\right),{\pi}_{v}\left({c}_{2}^{1}\right),{\pi}_{v}\left({c}_{1}^{2}\right),{\pi}_{v}\left({c}_{2}^{2}\right))$, and build ${G}^{\prime}$ as follows.

For each undirected edge $(u,v)\in E$ add two directed edges $(u,v),(v,u)$ to ${E}^{\prime}$. Set the weight of each incoming edge to a node $v\in V$ as $\frac{1}{|{N}_{v}^{i}|}$. By this the sum over weight of all incoming edges is equal to one, i.e., $\forall v\in V:{\sum}_{u\in {N}_{v}^{i}}{b}_{u,v}=1$.

For each node $v\in V$, add two more nodes ${v}^{\prime},{v}^{\u2033}$ to ${V}^{\prime},{V}^{\u2033}$, respectively. Furthermore, add an edge $(v,{v}^{\prime})$ to ${E}^{\prime}$ with ${b}_{v,{v}^{\prime}}=1$. Formally, $\forall v\in V:{v}^{\prime}\in {V}^{\prime},{v}^{\u2033}\in {V}^{\u2033},(v,{v}^{\prime})\in {E}^{\prime}\phantom{\rule{4.pt}{0ex}}\mathrm{s}.\mathrm{t}.\phantom{\rule{4.pt}{0ex}}{b}_{v,{v}^{\prime}}=1$. Note that nodes in ${V}^{\u2033}$ are isolated.

Set the preferences list of the nodes as follows.

By this reduction, the score of candidates before any diffusion is $\mathcal{F}({c}_{1}^{1},\xd8)=\mathcal{F}({c}_{1}^{2},\xd8)=\left|V\right|,$ $\mathcal{F}({c}_{2}^{1},\xd8)=\mathcal{F}({c}_{2}^{2},\xd8)=\frac{1}{2}\left|V\right|$. Then $F({C}_{1},\xd8)=\mathcal{F}({C}_{2},\xd8)=1$.

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 $S\subseteq V$ 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 $S\subseteq V$ after the diffusion; otherwise, if there is a node ${v}^{\prime}\in {V}^{\prime}\cap S$ we can replace it with its incoming neighbor $v\in V$ such that $(v,{v}^{\prime})\in {E}^{\prime}$ and we get at least the same value for ${\mathrm{M}\mathrm{o}\mathrm{V}}_{c},{\mathrm{D}\mathrm{o}\mathrm{W}}_{c}$. Furthermore, if there exists a node ${v}^{\u2033}\in {V}^{\u2033}\cap S$ one of the following situations holds:

There exists an inactive node $v\in V\backslash {A}_{S}$ after the diffusion S. In this case, we can substitute v for ${v}^{\u2033}$ and then we get at least the same ${\mathrm{D}\mathrm{o}\mathrm{W}}_{c},{\mathrm{M}\mathrm{o}\mathrm{V}}_{c}$.

There is no inactive node $v\in V\backslash {A}_{S}$. In this case, according to the nodes’ probability distribution, when all nodes in V become active, the value of ${\mathrm{M}\mathrm{o}\mathrm{V}}_{c}$ and ${\mathrm{D}\mathrm{o}\mathrm{W}}_{c}$ is maximum. Then even if we remove ${v}^{\u2033}$ from S it does not change the value of ${\mathrm{M}\mathrm{o}\mathrm{V}}_{c}$ or ${\mathrm{D}\mathrm{o}\mathrm{W}}_{c}$. By the way, in this situation, if there exist any node $v\in V\backslash {A}_{S}$ we replace ${v}^{\u2033}$ with it, otherwise we replace it with a node $v\in V\backslash S$.

Then from now on, we assume $S\subseteq V$.

If all nodes in

V become active, since they have an outgoing edge to all nodes

${v}^{\prime}\in {V}^{\prime}$ with probability one, then all nodes in

$V\cup {V}^{\prime}$ will become active, and the score of the candidates will be as follows.

Then $F({C}_{1},S)=2,\mathcal{F}({C}_{2},S)=0$, ${\mathrm{D}\mathrm{o}\mathrm{W}}_{c}({C}_{1},S)>0$, ${\mathrm{M}\mathrm{o}\mathrm{V}}_{c}({C}_{1},S)>0$, and any approximation algorithm will return a positive value, then the answer of $\mathcal{I}$ will be YES.

On the other hand, if there is a node

$v\in V$, which is inactive after the diffusion, i.e.,

$\exists v\in V\backslash {A}_{S}$, the score of candidates will be as follows.

Then $F({C}_{1},S)=\mathcal{F}({C}_{2},S)=1,{\mathrm{D}\mathrm{o}\mathrm{W}}_{c}({C}_{1},S)={\mathrm{M}\mathrm{o}\mathrm{V}}_{c}({C}_{1},S)=0$, and any approximation algorithm will return zero, then the answer of $\mathcal{I}$ will be NO.

For the other direction, note that if we can find a set of nodes $S\subseteq V$, which is an answer for $\mathcal{I}$, using the same set of nodes, we can activate all nodes in $V\cup {V}^{\prime}$ and ${\mathrm{D}\mathrm{o}\mathrm{W}}_{c}({C}_{1},S)>0,{\mathrm{M}\mathrm{o}\mathrm{V}}_{c}({C}_{1},S)>0$.

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

$t,k>2$. The same as before, the order of the candidates in probability distribution of a voter

v is

${\pi}_{v}=({\pi}_{v}\left({c}_{1}^{1}\right),\cdots ,{\pi}_{v}\left({c}_{k}^{1}\right),{\pi}_{v}\left({c}_{1}^{2}\right),\cdots ,{\pi}_{v}\left({c}_{k}^{2}\right),\cdots ,{\pi}_{v}\left({c}_{1}^{t}\right),\cdots ,{\pi}_{v}\left({c}_{k}^{t}\right))$.

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The following theorem proves the same statement for the destructive case of the problem.