This problem is inspired by the single-player game Minesweeper, generalized to an arbitrary graph. Let $\mathit{G}$be an undirected graph, where each node either contains a single, hidden mine or is empty. The player chooses nodes, one by one. If the player chooses a node containing a mine, the player loses. If the player chooses an empty node, the player learns the number of neighboring nodes containing mines. (A neighboring node is one connected to the chosen node by an edge.) The player wins if and when all empty nodes have been so chosen.

In the mine consistency problem, you are given a graph$\mathit{G}$ along with numbers labeling some of $\mathit{G}$’s nodes. You must determine whether a placement of mines on the remaining nodes is possible, so that any node v that is labeled m has exactly m neighboring nodes containing mines. Formulate this problem as a language and show that it is$\mathit{N}\mathit{P}\mathbf{}\mathit{c}\mathit{o}\mathit{m}\mathit{p}\mathit{l}\mathit{e}\mathit{t}\mathit{e}$.

Consideris a Boolean formula, now user want to convert it to $G$, which is an instance of the$MCP$ problem.

Here, suppose $\$c\_\left\{1\right\}, c\_\{2, \backslash idots\}\$$are used to denote the clause appearing in $\$\backslash varphi\$$. Now, suppose $\$x\_\left\{1\right\}, x\backslash \mathrm{user}1\_\{2, \backslash \mathrm{user}1\backslash idots\}\$$denotes the variable used, Here, a variable $\$x\_\left\{i\right\}\$$'appears' in $\$\backslash varphi\$$if one of $\$x\_\left\{i\right\}\$$or $\$\backslash bar\left\{x\right\}\_\left\{i\right\}\$$appear in minimum single clause in.

For all the variables $\$x\_\left\{i\right\}\$$which appears in $\$\backslash varphi\$$:

1. Three nodes are created as$\$x\_\left\{i\right\}$,$\$x\_\left\{i\right\}^\left\{f\right\}\$$and$\$x\_\left\{i\right\}^\left\{t\right\}\$$.

2. Edges are added role="math" localid="1663224975793" $\$\backslash left(x\_\{i\},x\_\{i\}^\{t\}\backslash right)\$$and $\$\backslash left(x\_\{i\},x\_\{i\}^\{f\}\backslash right)\$$and

3. Now, set$\$N\backslash left(x\_\{i\}\backslash right)=1,N\backslash left(x\_\{i\}^\{t\}\backslash right)=X\$$and $N\backslash left(x\_\{i\}^\{f\}\backslash right)=X\$$

For all the clause$\$c\_\left\{i\right\}\$$which appears in $\$\backslash varphi\$$:

1. Three nodes are created as$\$c\_\left\{i\right\}$ ,$c\_\left\{i\right\}^\left\{1\right\}\$$and $\$c\_\left\{i\right\}^\left\{2\right\}\$$.

2. Edges are added from$\$c\_\left\{i\right\}\$$to the nodes corresponding to the three literals in$\$c\_\left\{i\right\}\$$

3. Now, set $\$N\backslash left(c\_\{i\}\backslash right)=3, N\backslash left(c\_\{i\}^\{1\}\backslash right)=X\$$and $\$N\backslash left(c\_\{i\}^\{2\}\backslash right)=X\$$.

All instances of the 3$SAT$ problem can be reduced to an instance of the Circuit-$SAT$ problem in polynomial time in a trivial manner by changing the Boolean operators to a circuit of logic gates. The validity of this reduction needs to be proven.

If there is an instance of the $3SAT$problem it can be converted to an instance of the Circuit-$SAT$ problem by simply mapping Boolean operators to logic gates and connections between the gates.

Setting the corresponding logic inputs to the Boolean values that satisfy the $3SAT$problem will satisfy this instance of the Circuit-$SAT$ problem.

Using an instance of the Circuit-$SAT$ problem, an equivalent instance of the$SAT$ problem is constructed by interchanging logic gates/wires with Boolean variables and operators. If the logical values that satisfy the instance of Circuit-$SAT$ are changed into Boolean values, then the$SAT$ instance will be satisfied.

The Circuit-$SAT$ problem is reducible in polynomial time to an$NP-complete$ problem, so it is also in$NP-complete$ . Hence, the MCP problem is $NP-complete$.