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数独
四独游戏
解的存在和唯一性
算法
常微分方程 数独 采用蛮力试凑法来解决数独问题。#xff08;采用单选数#xff0c;以及计算机科学技术中的递推回溯法#xff09; 以上的数独是图14-2的两个矩阵的和#xff0c;左侧的矩阵可以由kron和magic函…目录
数独
四独游戏
解的存在和唯一性
算法
常微分方程 数独 采用蛮力试凑法来解决数独问题。采用单选数以及计算机科学技术中的递推回溯法 以上的数独是图14-2的两个矩阵的和左侧的矩阵可以由kron和magic函数建立起来前一个函数用来求Kronecker乘积后者生成幻方矩阵。
X kron(eye(3), magic(3)) 四独游戏 使用4×4网格可能的备选项用小数标注出来。将单选数填在网格内如果没有单选数则采用递归回溯法来求解。左有单选数 右无单选数 Xdiag(1:4) Yshidoku(diag(1:4))
Zshidoku(diag(1:4)) %可能的另一个解 解的存在和唯一性 可以用sudoku_all程序来寻找数独全部的解。 %生成数独谜题
Xkron(eye(3), magic(3))
Xsudoku_puzzle(13) 数独网格的一些运算可能改变其在图形用户界面中的显示但不能改变其基本特性。所有变化基本上都是相同的数字谜。这些等效的运算可以用matlab的数组运算表示。 %重新排列表示数字的文字
p randperm(9), zfind(X0). X(z)p(X(z))
%其它运算
X, rot90(X, k)
flipud(X), fliplr(X)X([4:9 1:3], :)
X(:, randperm(3) 4:9) function L sudoku_all(X,L)
% SUDOKU_ALL Enumerate all solutions to a Sudoku puzzle.
% L sudoku_all(X), for a 9-by-9 array X, is a list of all solutions.
% L{k} is the k-th solution.
% L{:} will print all the solutions.
% length(L) is the number of solutions. A valid puzzle must have only one.
% See also sudoku, sudoku_basic, sudoku_puzzle, sudoku_assist.if nargin 2% Initialize the list on first entry.L {};end% Fill in all singletons, the cells with only one candidate.% C is the array of candidates for each cell.% N is the vector of the number of candidates for each cell.% s is the index of the first cell with the fewest candidates.[C,N] candidates(X);while all(N0) any(N1)s find(N1,1);X(s) C{s};[C,N] candidates(X);end% Add a solution to the list.if all(X(:)0)L{end1} X;end% Enumerate all possible solutions.if all(N0)Y X;s find(Nmin(N),1);for t [C{s}] % Iterate over the candidates.X Y;X(s) t; % Insert a value.L sudoku_all(X,L); % Recursive call.endend% ------------------------------function [C,N] candidates(X)% C candidates(X) is a 9-by-9 cell array of vectors.% C{i,j} is the vector of allowable values for X(i,j).% N is a row vector of the number of candidates for each cell.% N(k) Inf for cells that already have values.tri (k) 3*ceil(k/3-1) (1:3);C cell(9,9);for j 1:9for i 1:9if X(i,j)0z 1:9;z(nonzeros(X(i,:))) 0;z(nonzeros(X(:,j))) 0;z(nonzeros(X(tri(i),tri(j)))) 0;C{i,j} nonzeros(z);endendendN cellfun(length,C);N(X0) Inf;N N(:);end % candidates
end % sudoku_all算法 基本流程 ① 填入所有的单选数 ②如果某个单元没有备选项停止程序 ③在某个空白的网格中填入一个试探的值 ④ 递归式调用此程序 function [X,steps] sudoku(X,steps)
% SUDOKU Solve a Sudoku puzzle using recursive backtracking.
% sudoku(X), for a 9-by-9 array X, solves the Sudoku puzzle for X.
% [X,steps] sudoku(X) also returns the number of steps.
% See also sudoku_all, sudoku_assist, sudoku_basic, sudoku_puzzle. if nargin 1X sudoku_puzzle(1);endif nargin 2steps 0;gui_init(X);endsudoku_gui(X,steps);% Fill in all singletons, the cells with only one candidate.% C is the array of candidates for each cell.% N is the vector of the number of candidates for each cell.% s is the index of the first cell with the fewest candidates.[C,N] candidates(X);while all(N0) any(N1)sudoku_gui(X,steps,C);s find(N1,1);X(s) C{s};steps steps 1;sudoku_gui(X,steps,C);[C,N] candidates(X);endsudoku_gui(X,steps,C);% Recursive backtracking.if all(N0)Y X;s find(Nmin(N),1);for t [C{s}] % Iterate over the candidates.X Y;sudoku_gui(X,steps,C);X(s) t; % Insert a tentative value.steps steps 1;sudoku_gui(X,steps,C,s); % Color the tentative value.[X,steps] sudoku(X,steps); % Recursive call.if all(X(:) 0) % Found a solution.breakendsudoku_gui(X,steps,C,-s); % Revert color of tentative value.endendif nargin 2gui_finish(X,steps);end% ------------------------------function [C,N] candidates(X)% C candidates(X) is a 9-by-9 cell array of vectors% C{i,j} is the vector of allowable values for X(i,j).% N is a row vector of the number of candidates for each cell.% N(k) Inf for cells that already have values.tri (k) 3*ceil(k/3-1) (1:3);C cell(9,9);for j 1:9for i 1:9if X(i,j)0z 1:9;z(nonzeros(X(i,:))) 0;z(nonzeros(X(:,j))) 0;z(nonzeros(X(tri(i),tri(j)))) 0;C{i,j} nonzeros(z);endendendN cellfun(length,C);N(X0) Inf;N N(:);end % candidates% ------------------------------function gui_init(X)% Initialize gui% H is the structure of handles, saved in figure userdata.dkblue [0 0 2/3];dkgreen [0 1/2 0];dkmagenta [1/3 0 1/3];grey [1/2 1/2 1/2];fsize get(0,defaulttextfontsize);fname Lucida Sans Typewriter;clfshgset(gcf,color,white)axis squareaxis offfor m [2 3 5 6 8 9]line([m m]/11,[1 10]/11,color,grey)line([1 10]/11,[m m]/11,color,grey)endfor m [1 4 7 10]line([m m]/11,[1 10]/11,color,dkmagenta,linewidth,4)line([1 10]/11,[m m]/11,color,dkmagenta,linewidth,4)endH.a zeros(9,9);for j 1:9for i 1:9if X(i,j) 0string int2str(X(i,j));color dkblue;elsestring ;color dkgreen;endH.a(i,j) text((j1/2)/11,(10.5-i)/11,string, ...units,normal,fontsize,fsize6,fontweight,bold, ...fontname,fname,color,color,horizont,center);endendstrings {step,slow,fast,finish};H.b zeros(1,4);for k 1:4H.b(k) uicontrol(style,toggle,string,strings{k}, ...units,normal,position,[(k3)*0.125,0.05,0.10,0.05], ...background,white,value,0, ...callback, ...Hget(gcf,user); H.sfind(H.bgco); set(gcf,user,H));endset(H.b(1),style,pushbutton)H.s 1;H.t title(0,fontweight,bold);set(gcf,userdata,H)drawnowend % gui_init% ------------------------------function sudoku_gui(X,steps,C,z)H get(gcf,userdata);if H.s 4if mod(steps,50) 0set(H.t,string,int2str(steps))drawnowendreturnelseset(H.t,string,int2str(steps))endk [1:H.s-1 H.s1:4];set(H.b(k),value,0);dkblue [0 0 2/3];dkred [2/3 0 0];dkgreen [0 1/2 0];cyan [0 2/3 2/3];fsize get(0,defaulttextfontsize);% Update entire array, except for initial entries.for j 1:9for i 1:9if ~isequal(get(H.a(i,j),color),dkblue) ...~isequal(get(H.a(i,j),color),cyan)if X(i,j) 0set(H.a(i,j),string,int2str(X(i,j)),fontsize,fsize6, ...color,dkgreen)elseif nargin 3set(H.a(i,j),string, )elseif length(C{i,j}) 1set(H.a(i,j),string,char3x3(C{i,j}),fontsize,fsize-4, ...color,dkred)elseset(H.a(i,j),string,char3x3(C{i,j}),fontsize,fsize-4, ...color,dkgreen)endendendendif nargin 4if z 0set(H.a(z),color,cyan)elseset(H.a(-z),color,dkgreen)returnendend% Gui action single step, brief pause, or no pauseswitch H.scase 1H.s 0;set(gcf,userdata,H);while H.s 0;drawnowH get(gcf,userdata);endcase 2pause(0.5)case 3drawnowendif nargin 4if z 0set(H.a(z),color,cyan)elseset(H.a(-z),color,dkgreen)returnendend% ------------------------------function s char3x3(c)% 3-by-3 character array of candidates.b blanks(5);s {b; b; b};for k 1:length(c)d c(k);p ceil(d/3);q 2*mod(d-1,3)1;s{p}(q) int2str(d);endendend % gui% ------------------------------function gui_finish(X,steps)H get(gcf,userdata);H.s 2;set(H.b(1:3),vis,off)set(gcf,userdata,H)set(H.b(4),string,close,value,0, ...callback,close(gcf))sudoku_gui(X,steps)end % gui_finishend % sudoku备选项计算数独题目生成
%% Candidates备选项计算% C candidates(X) 为向量数组构成的单元结构
% C{i,j} 为 X(i,j)构成的集合C cell(9,9);tri (k) 3*ceil(k/3-1) (1:3);for j 1:9for i 1:9if X(i,j)0z 1:9;z(nonzeros(X(i,:))) 0;z(nonzeros(X(:,j))) 0;z(nonzeros(X(tri(i),tri(j)))) 0;C{i,j} nonzeros(z);endendendC%% First singleton and first empty. 第一个单选数第一个空白网格% N number of candidates in each cell.
% s first cell with only one candidate.
% e first cell with no candidates.N cellfun(length,C)s find(X0 N1,1)e find(X0 N0,1)%% Sudoku puzzles 数独题目生成help sudoku_puzzlefor p 1:16sudoku_puzzle(p)end 常微分方程 matlab提供了很多求给定常微分方程数组近似解的函数这一常微分方程组数值解的函数包括ode23、ode45、ode113、ode23s、ode15s、ode23t、ode23tb。函数名中的数字表示所用算法的阶次阶次和算法的复杂程度和精度有关。所有这些函数都会自动选择近似步长来保证预先选择的精度要求。所选择的阶次越高每一步计算量越大但是所用的步长也越大。比如ode23算法比较二阶和三阶算法来估计计算步长而ode45比较的是四阶和五阶算法。 s表示为stiff 刚性微分方程的求解函数。 微分方程求解程序库提供的求解函数都至少需要下面三个输入变元
1F为定义微分方程组的函数
2tspan为描述积分区间的向量
3 y0为初始值的向量 ode1的算法-误差较大 function [t,y] ode1(F,tspan,y0)
% ODE1 Worlds simplest ODE solver.
% ODE1(F,[t0,tfinal],y0) uses Eulers method to solve
% dy/dt F(t,y)
% with y(t0) y0 on the interval t0 t tfinal.t0 tspan(1);
tfinal tspan(end);
h (tfinal - t0)/200;
y y0;
for t t0:h:tfinalydot F(t,y);y y h*ydot;
end 用匿名函数生成微分方程 acircle (t,y) [y(2); -y(1)]; ode23求解 %% ODE23 Automatic Plotting.figuretspan [0 2*pi];y0 [0; 1];ode23(acircle,tspan,y0)%% Phase Plot.figuretspan [0 2*pi];y0 [0; 1];[t,y] ode23(acircle,tspan,y0)plot(y(:,1),y(:,2),-o)axis squareaxis([-1.1 1.1 -1.1 1.1])%% ODE23 Automatic Phase Plot.opts odeset(outputfcn,odephas2)ode23(acircle,tspan,y0,opts)axis squareaxis([-1.1 1.1 -1.1 1.1])odeset
