Introduction:
In this lab, I had done some practices about basic structures and operations of MATLAB what I had learnt in last semester. By reviewing these efficient tips on using MATLAB, my ability to program in MATLAB had been improved.
Solutions and Analysis:
Lab1.1:
In this problem, I just use a variable called varargin to read all uncertain variables. Then we can do a for loop to get the sum of these input variables. The function that can sum all input is shown as following:
| function [Sum]=SumALL(varargin) Sum=0; for i=1:nargin Sum=Sum+varargin{i}; end end |
Then we can use an example to verify this function. By obverting the result of the function. We can find that the function is right.
Lab1.2:
This problem is verifying the random search algorithm. The principle of this algorithm is just add a random increment, and judge or compare whether the increment is suitable.
The result of this algorithm is:
Lab1.3:
Exercise1:
This excise is using MATLAB to construct a simple database, and achieve four basic operations in database like insert, delete, list and clear. For this problem, I use structure and some basic operation in files.
The code of the function called people was shown like following:
| function people(db_file,operation,varargin) % beacause the number of the input is unceartain switch operation %use switch to jundge the input opeartion case 'reset' %clear opeartion delete('db_file.mat'); case 'list' %just load all terms in the file and display all terms with some special format load('db_file.mat'); [~,b]=size(person); for k=1:b disp(['Name: ',person(k).Name,' Age:', num2str(person(k).Age)]); end case 'insert' % insert some new terms
j=1; i=1; while(i<=nargin-3) %record the people who need to be inserted in the list person(j) = struct('Name', varargin{i}, 'Age',varargin{i+1}); i=i+2; j=j+1; end save('db_file.mat', 'person') case 'remove' % delete some terms in the list j=1; i=1; while(i<=nargin-3) %record the terms which need to be deleted in the database DeletedPeople(j) = struct('Name', varargin{i}, 'Age',varargin{i+1}); i=i+2; j=j+1; end load('db_file'); [a,b]=size(person); count =0; check=zeros(1,20);% this array is to record if this person had been deleted. for k=1:b for p=1:((nargin-2)/2+1) if(p>=(nargin-2)/2+1) p1=0; p2=0; else p1=(strcmp(person(k).Name, DeletedPeople(p).Name )); p2=(person(k).Age== DeletedPeople(p).Age); end if p1&&p2 % Only when name and age are both equal to the people we want to delete, check will change. check(k)=1; end end end for k=1:b if check(k)~=1 new(k-count)=person(k); %if the check not equal to one, this indicate that this peeson had not been deleted else count=count+1; end end person=new; save('db_file.mat', 'person') end end
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The result of this function is:
We can find that the result is right.
Exercise2:
The purpose of this exercise is rotating the polygon, we can use some basic math knowledge to get the transform matrix. The brief process of finding matric was shown in this figure:
By this matrix, we can rotate any polygon conveniently.
The code of this exercise was shown as this:
| function RotatePlygon(inputAngle) X = [-5 -3 -4 2 1 8 6 3]; Y = [-6 0 10 8 2 3 -5 -2]; figure plot(X,Y); title('b'); hSquare = fill(X,Y,'k'); thetad = 1; R = [cosd(thetad) -sind(thetad); sind(thetad) cosd(thetad)]; C = repmat([0 0], 8, 1)'; axis([-15 15 -15 15]) for k=1:inputAngle V = get(hSquare,'Vertices')'; % get the current set of vertices V = R*(V - C) + C; % do the rotation relative to the centre of the square set(hSquare,'Vertices',V'); % update the vertices pause(0.01); end end |
Then we can test this function’
Figure 1 The original figure
Figure 2 The figure rotated 20 degrees
Exercise3:
For this problem, we can find this maze is a kind of graph. The problem description asked us to find the shortest path to through the maze. Apparently this problem is an easy shortest path problem. We can use some classical algorithm to get the solution. But this problem is not as easy as we imaged. And I haven’t found a very good algorithm to solve this problem.
The reasons why I think this problem is difficult are:
- The maze may construct a loop; we need to judge whether the graph has a loop. Otherwise our algorithm will go wrong.
- The positions of entrances and exits are arbitrary position that is not an obstacle. These position will increase our computation.
At last, I use a naive way which is using DFS to every pair of entrance and exit. And by this method, we can get all doable ways, we can find the shortest way.
We can also use BFS and SPFA.
In DFS, we can use stacks to store data, but I don’t have enough time to do that.
This function is to span the maze to big matrices with different entrances and exits.
| function [AllMaze]=GenerateAllMaze(maze) [n,n]= size(maze); NumOfEntrance=0; NumOfExit=0; for i =1:n if (maze(1,i)~=1) NumOfEntrance=NumOfEntrance+1; end if (maze(n,i)~=1) NumOfExit=NumOfExit+1; end end k=0; for i=1:n maze = [0 1 0 0 1 0 1 1 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 0]; if(maze(1,i)~=1) maze(1,i)=2; for j=1:n if(maze(n,j)~=1) maze(n,j)=3; k=k+1; AllMaze(:,:,k)=maze; maze(n,j)=0; end end end end end |
Then we will write the DFS.
| function [Solution, PathLength]=DFS_Maze(maze,InputSolution,InputShortestPathLength) % This function can get all path f the mazze by using DFS % maze: 0 represents that this point can go % 1 represents that this point is obstacles % 2 represents that this point is the entrance of this maze % 3 represents that this point is the exit of this maze % 5 represents that this point is on the path which was found in this % time % 0 2 0 0 1 % 0 1 1 0 1 % 0 1 0 0 1 % 0 1 0 0 1
% Defined the four directions directions = kron(eye(2),[-1,1]); % The number of the path which were found in this time NumberOfPath = 0; % Find the entrance for this time [I,J] = find(maze == 2); [Solution,PathLength]=search(I,J); %disp(NumberOfPath); % Jundge whether the point can pass function z = WhetherGo(x,y) % I use a constructure called try to jundge z = true; try % jundge whether this point is a obstacal or the point which we had passed if ismember(maze(x,y),[1,2,5]) z = false; end catch z = false; end end
function [Solution,PathLength]=search(x,y) if maze(x,y) == 3 % Find the exit for this maze disp(maze); % print the path NumberOfPath = NumberOfPath + 1; PathLength=length(find(maze==5))+1; disp(PathLength); Solution=maze; return else PathLength=10; Solution=maze; end
% Search four different directions for i = 1 : 4 if WhetherGo(x + directions(1,i),y + directions(2,i)) if maze(x + directions(1,i),y + directions(2,i)) ~= 3 maze(x + directions(1,i),y + directions(2,i)) = 5;
end search(x + directions(1,i),y + directions(2,i)); % countine to find if maze(x + directions(1,i),y + directions(2,i)) ~= 3 maze(x + directions(1,i),y + directions(2,i)) = 0; % come back to this point and find another direction end end end end end
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Then we can test our program.
We can find that the program is right.
转载：https://blog.csdn.net/zzx2015/article/details/102488138