void remove(int); void findSol(int, int); void displaySol(); class Maze{ public: Maze(int); int find_root(int); void union_cell(int, int); }; /* precondition: n must be a positive integer postcondition: attributes of each cell are assigned values */ Maze::Maze(int n){ - determine each cell’s location and initialize the data members } /* precondition: n must be >= 0 and < the number of all cells postcondition: return the root of n */ int Maze::find_root(int n){ - return the root of n } /* precondition: root1 and root2 both must be >= 0 and < the number of all cells postcondition: root1 and root2 belong to the same set */ void Maze::union_cell(int root1, int root2){ - make root1 and root2 belong to the same set } int main(int argc, char** argv){ - exit if something is missing in the command line - exit if the user provides unacceptable information - initialize the random number generator - store critical data in variables - push all elements into a vector except the last one because it has no walls to knock down - use a while loop to construct the maze - inside the while loop randomly choose a cell deal with the case when the cell has 2 neighbors deal with the case when the cell has only 1 neighbor - generate code for Matlab to display the maze - find and put solution code in the solution file } /* precondition: victim should, but not must, be an element in the vector that holds all cells postcondition: victim is erased from the vector */ void remove(int victim){ - remove victim the vector because victim is already processed } /* precondition: from and to must be >= 0 and < numOfCells postcondition: find the solution */ void findSol(int from, int to){ - push from to the solution vector - mark from visited - if from equals to, that means we’ve found the solution, so we call displaySol() to write code to output files - now go through each surrounding wall - inside an if block if the right wall is down and the cell on the other side is not visited yet, use that cell to call findSol recursively if control returns here, that means this cell has led to a deadlock, so pop the element in the back of the solution vector - inside an if block if the lower wall is down and the cell on the other side is not visited yet, use that cell to call findSol recursively if control returns here, that means this cell has led to a deadlock, so pop the element in the back of the solution vector - inside an if block if the upper wall is down and the cell on the other side is not visited yet, use that cell to call findSol recursively if control returns here, that means this cell has led to a deadlock, so pop the element in the back of the solution vector - inside an if block if the left wall is down and the cell on the other side is not visited yet, use that cell to call findSol recursively if control returns here, that means this cell has led to a deadlock, so pop the element in the back of the solution vector } /* precondition: none postcondition: put code for displaying the maze in file2 */ void displaySol(){ - generate code for Matlab to display the solution to the second output file }Wow, it looks like a pretty big program. Well, it is big compared to the programming exercises you have done earlier, but it is not that big.