说我们有一个零和一个矩阵P，我们希望找到p的非重叠子膜的最佳(*(组合，以便每个subpatrix：
- 至少包含l zeros和l一个(最小面积= 2*l(;
- 最多包含h元素(最大区域= h(。
(*(最好的组合是最大化所有子元中元素总数的组合。
请注意，最佳组合可能不是唯一的，并且并非总是可以用最佳组合的一量覆盖P的所有元素，即使用
L = 1
H = 6
P =
0 0 0 0
1 1 0 0
0 0 0 0
最好的组合之一是两个子膜
给出的 1 2 2 4 % top left corner 1 2 - bottom right corner 2 4
1 3 1 1 % top left corner 1 1 - bottom right corner 3 1
仅涵盖p。
的12个元素中的9个我编写的解决问题的代码分为两个部分：
- 首先，它找到了使用内置函数conv2(以下代码中的更多信息(，找到满足以前两个属性(至少在最大h元素上(的P的所有可能的子膜；这部分只需要几秒钟，这是简单的；
- 然后，它使用递归技术分析了所有子膜的集合；这是问题的核心，可能需要几个小时才能找到解决方案(除非计算机首先冻结(。
递归是通过这种方式完成的：
- 将所有子膜和副本b的集合a的集合a用于
- 修复A的第一个子元素并将其添加到当前组合
- 找到所有非重叠子膜的集合C
- 使用A和C 来调用递归函数
- 修复C的第一个子序列，并将其添加到当前组合
- 找到所有非重叠子膜的集合D
- 使用作为输入A和D 调用递归函数
- 等等，直到非重叠的子膜的集合为空；最终组合是一个载体，其中包含所有子膜，计数器(即组合中的一量子量的数量(和矩阵中保留在矩阵中的元素数量，在删除了组合中包含的元素
- 然后y的第二个子质固定是固定的，然后将上一个组合的最后一个组替换为
- 如果一组非重叠的子媒体是空的，则将新的最终组合与上一组进行比较，如果一组非重叠的子媒体集合不空li>
- 当找到完美的组合或在A的所有子膜上的第一个周期结束时，该脚本将终止。
此方法仅适用于小矩阵(小于10x10(，并且可以成为较大矩阵的噩梦。我测试了一个200x200矩阵，几分钟后我的计算机冻结了。问题是，如果所有子膜的集合包含数千个元素，那么递归将产生数百个嵌套以进行循环，消耗大量的RAM和CPU。
。我想知道实现目标的最有效方法是什么，因为我的方法非常糟糕。
这是我的代码：
%% PROBLEM
%
% let P be a matrix whose elements are zeros and ones
% find the best(*) combination of non-overlapping submatrices of P
% so that each submatrix respect these properties:
% - contains at least L zeros and L ones (min area=2*L)
% - contains at most H elements (max area=H)
%
% (*) the best is the one which maximize the total number of elements in all the submatrices
%
% notices: the best combination could be not unique
% is not always possibile to cover all the elements of P with the submatrices of the best combination
%
%% INPUT
P=round(rand(8,8)); L=1; H=5;
%P=dlmread('small.txt'); L=1; H=5; % small can be found here https://pastebin.com/RTc5L8We
%P=dlmread('medium.txt'); L=2; H=8; % medium can be found here https://pastebin.com/qXJEiZTX
%P=dlmread('big.txt'); L=4; H=12; % big can be found here https://pastebin.com/kBFFYg3K
%P=[0 0 0 0 0 1;0 0 0 0 0 1;0 1 0 1 0 1;0 0 0 0 0 0;0 0 0 0 0 0]; L=1; H=6;
P=[0 0 0 0 0;0 1 1 1 0;0 0 0 0 0]; L=1; H=6;
%P=[1,0,0,0,0;1,1,1,1,1;1,0,0,0,0]; L=1; H=5;
%% FIND ALL THE SUBMATRICES OF AREA >= 2*L & <= H
%
% conv2(input_matrix,shape_matrix,'valid')
% creates a matrix, where each element is the sum of all the elements contained in
% the submatrix (contained in input_matrix and with the shape given by shape_matrix)
% having its top left corner at said element
%
% ex. conv2([0,1,2;3,4,5;6,7,8],ones(2,2),'valid')
% ans =
% 8 12
% 20 24
% where 8=0+1+3+4 12=1+2+4+5 20=3+4+6+7 24=4+5+7+8
%
s=[]; % will contain the indexes and the area of each submatrix
% i.e. 1 3 2 5 9 is the submatrix with area 9 and corners in 1 2 and in 3 5
for sH = H:-1:2*L
div_sH = divisors(sH);
fprintf('_________AREA %d_________n',sH)
for k = 1:length(div_sH)
a = div_sH(k);
b = div_sH(end-k+1);
convP = conv2(P,ones(a,b),'valid');
[i,j] = find((convP >= L) & (convP <= sH-L));
if ~isempty([i,j])
if size([i,j],1) ~= 1
% rows columns area
s = [s;[i,i-1+a , j,j-1+b , a*b*ones(numel(i),1)]];
else
s = [s;[i',i'-1+a,j',j'-1+b,a*b*ones(numel(i),1)]];
end
fprintf('[%dx%d] submatrices: %dn',a,b,size(s,1))
end
end
end
fprintf('n')
s(:,6)=1:size(s,1);
%% FIND THE BEST COMBINATION
tic
[R,C]=size(P); % rows and columns of P
no_ones=sum(P(:)); % counts how many ones are in P
% a combination of submatrices cannot contain more than max_no_subm submatrices
if no_ones <= R*C-no_ones
max_no_subm=floor(no_ones/L);
else
max_no_subm=floor(R*C-no_ones/L);
end
comb(2,1)=R*C+1; % will contain the best combination
s_copy=s; % save a copy of s
[comb,perfect]=recursion(s,s_copy,comb,0,0,R,C,0,false,H,[],size(s,1),false,[0,0,0],0,0,0,0,0,0,max_no_subm);
fprintf('ntime: %2.2fsnn',toc)
if perfect
disp('***********************************')
disp('*** PERFECT COMBINATION FOUND ***')
disp('***********************************')
end
%% PRINT RESULTS
if (R < 12 && C < 12)
for i = 1:length(find(comb(2,3:end)))
optimal_comb_slices(i,:)=s(comb(2,i+2),:);
end
optimal_comb_slices(:,1:5)
P
end
用
给出的功能function [comb,perfect,counter,area,v,hold_on,ijk,printed,first_for_i,second_for_i,third_for_i] = recursion(s,s_copy,comb,counter,area,R,C,k,hold_on,H,v,size_s,perfect,ijk,size_s_ovrlppd,size_s_ovrlppd2,printed,third_for_i,second_for_i,first_for_i,max_no_subm)
%
% OUTPUT (that is actually going to be used in the main script)
% comb [matrix] a matrix of two rows, the first one contains the current combination
% the second row contains the best combination found
% perfect [boolean] says if the combination found is perfect (a combination is perfect if
% the submatrices cover all the elements in P and if it is made up with
% the minimum number of submatrices possible)
%
% OUTPUT (only needed in the function itself)
% counter [integer] int that keeps track of how many submatrices are in the current combination
% area [integer] area covered with all the submatrices of the current combination
% v [vector] keeps track of the for loops that are about to end
% hold_on [boolean] helps v to remove submatrices from the current combination
%
% OUTPUT (only needed to print results)
% ijk [vector] contains the indexes of the first three nested for loops (useful to see where the function is working)
% printed [boolean] used to print text on different lines
% first_for_i second_for_i third_for_i [integers] used by ijk
%
%
% INPUT
% s [matrix] the set of all the submatrices of P
% s_copy [matrix] the set of all the submatrices that don't overlap the ones in the current combination
% (is equal to s when the function is called for the first time)
% R,C [integers] rows and columns of P
% k [integer] area of the current submatrix
% H [integer] maximum number of cells that a submatrix can contains
% size_s [integer] number of rows of s i.e. number of submatrices in s
% size_s_ovrlppd [integer] used by ijk
% size_s_ovrlppd2 [integer] used by ijk
% max_no_subm [integer] maximum number of submatrices contained in a combination
%
%
% the function starts considering the first submatrix (call it sub1) in the set 's' of all the submatrices
% and adds it to the combination
% then it finds 's_ovrlppd' i.e. the set of all the submatrices that don't overlap sub1
% and the function calls itself considering the first submatrix (call it sub2) in the set 's_ovrlppd'
% and adds it to the combination
% then it finds the set of all the submatrices that don't overlap sub2 and
% so on until there are no more non-overlapping submatrices
% then it replaces the last submatrix in the combination with the second one of the last set of non-overlapping
% submatrices and saves the combination which covers more elements in P
% and so on for all the submatrices of the set 's'
%
% DOWNSIDE OF THIS METHOD
% if 's' contains thousands of submatrices, the function will create hundreds of nested for loops
% so both time and space complexities can be really high and the computer might freeze
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%% SAVE AND RESET COMBINATIONS %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% s_copy is empty when no more submatrices can be added to the current
% combination, in this case we have to check if this combination is
% better than the best combination previosly found, if so then we overwrite it
%
% then we have to remove one or more submatrices from the combination (depending on
% how many nested for loops are about to be closed)
% and compute another combination
% to 'remove one or more submatrices from the combination' it is necessary to do these things:
% - reduce the area
% - reduce the combination
% - reduce the counter
%
if isempty(s_copy)
comb(1,2)=counter; % final no of submatrices in the combination
comb(1,1)=R*C-area; % no. of cells remained in P after removing the cells contained in the submatrices of the combination
% if the combination just found is better than the previous overwrite it
if comb(1,1)<comb(2,1) || (comb(1,1)==comb(2,1) && comb(1,2)<comb(2,2))
comb(2,:)=comb(1,:);
disp(['[area_left] ', num2str(comb(2,1)), ' [slices] ', num2str(comb(2,2))])
printed=true;
end
area=area-k; % tot area of the combination excluding the last sumatrix
if ~isempty(v) && ~hold_on % more than one submatrix must be removed
i=size(v,2);
if i>length(find(v)) % if v contains at least one 0
while v(i)==1 % find the index of the last 0
i=i-1;
end
last_i_counter=size(v(i+1:end),2); % no. of consecutive for loop that are about to end
v=v(1:i-1);
else
last_i_counter=i;
v=[];
end
for i=1:last_i_counter
area=area-s(comb(1,2+counter-i),5); % reduce the area
end
comb(1,2+counter-last_i_counter:2+counter)=0; % remove submatrices from the combination
counter=counter-(last_i_counter+1); % reduce the counter
hold_on=true;
else % exactly one submatrix must be removed
comb(1,2+counter)=0;
counter=counter-1;
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% FIND COMBINATIONS %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
else
for i = 1:size(s_copy,1) % fix the i-th submatrix
% if the combination cover all elements of P & the no. of submatrices is the minimum possibile
if comb(2,1)==0 && comb(2,2)==ceil(R*C/H)
perfect=true;
return
end
pos=s_copy(i,6); % position of current submatrix in s
comb(1,3+counter)=pos; % add the position to the combination
k=s(pos,5); % save the area of the current submatrix
area=area+k; % area covered with all the submatrices of the combination up now
counter=counter+1;
% if the area in P covered by the current combination could not
% be bigger than the best combination found up now, then discard
% the current combination and consider the next one
if R*C-area-(max_no_subm-counter)*H > comb(2,1) && i < size(s_copy,1)
% R*C-area-(max_no_subm-counter)*H can be negative if ceil(R*C/H) < max_no_subm
counter=counter-1;
comb(1,3+counter)=0;
area=area-k;
else
s_ovrlppd=s_copy; % initializing the set of non-overlapping submatrices
s_ovrlppd(s_copy(:,1)<=s_copy(i,2) & s_copy(:,2)>=s_copy(i,1) & s_copy(:,3)<=s_copy(i,4) & s_copy(:,4)>=s_copy(i,3),:)=[]; % delete submatrices that overlap the i-th one
s_ovrlppd(s_ovrlppd(:,6)<pos,:)=[]; % remove submatrices that will generate combinations already studied
% KEEP TRACK OF THE NESTED 'FOR' LOOPS ENDS REACHED
if i==size(s_copy,1) % if i is the last cycle of the current for loop
v(size(v,2)+1)=1; % a 1 means that the code entered the last i of a 'for' loop
if size(s_ovrlppd,1)~=0 % hold on until an empty s_ovrlppd is found
hold_on=true;
else
hold_on=false;
end
elseif ~isempty(v) && size(s_ovrlppd,1)~=0
v(size(v,2)+1)=0; % a 0 means that s_ovrlppd in the last i of a 'for' loop is not empty => a new 'for' loop is created
end
%%%%%%%%%%%%%%%%%%%%%%%%
%%% PRINT STATUS %%%
%%%%%%%%%%%%%%%%%%%%%%%%
if size(s_copy,1)==size_s
ijk(1)=i;
ijk(2:3)=0;
fprintf('[%d,%d,%d]n',ijk)
size_s_ovrlppd=size(s_ovrlppd,1);
first_for_i=i;
second_for_i=0;
elseif size(s_copy,1)==size_s_ovrlppd
ijk(2)=i;
ijk(3)=0;
if ~printed
fprintf(repmat('b',1,numel(num2str(first_for_i))+numel(num2str(second_for_i))+numel(num2str(third_for_i))+2+2+1)) % [] ,, return
else
printed=false;
end
fprintf('[%d,%d,%d]n',ijk)
size_s_ovrlppd2=size(s_ovrlppd,1);
second_for_i=i;
third_for_i=0;
elseif size(s_copy,1)==size_s_ovrlppd2
ijk(3)=i;
if ~printed
fprintf(repmat('b',1,numel(num2str(first_for_i))+numel(num2str(second_for_i))+numel(num2str(third_for_i))+2+2+1))
else
printed=false;
end
fprintf('[%d,%d,%d]n',ijk)
third_for_i=i;
end
[comb,perfect,counter,area,v,hold_on,ijk,printed,first_for_i,second_for_i,third_for_i]=recursion(s,s_ovrlppd,comb,counter,area,R,C,k,hold_on,H,v,size_s,perfect,ijk,size_s_ovrlppd,size_s_ovrlppd2,printed,third_for_i,second_for_i,first_for_i,max_no_subm);
end
end
end
end
您基本上是在尝试解决非线性整数编程问题，如果可能的话，通常很难(非常！(解决。在这种情况下，200*200并不是一个小问题，它很大。
我最好的建议是使用一些方法，以减少搜索空间或应用一些启发式方法，如果您可以接受近似解决方案。我尚未测试它，但是我相信某些树搜索算法可以表现出色，因为许多子膜将重叠，因此可以从搜索树中删除。
我尝试使用MATLAB构建遗传算法，ga，该算法也可以执行，但可能存在更好的解决方案。
在ga算法中，您定义了目标函数：
function [left] = toMin(rect,use)
left = numel(rect(1).mat);
for i = 1:length(rect)
left = left - use(i)*sum(rect(i).mat(:)==1);
end
您可以最小化。约束功能
function [val,tmp] = constraint(rect,use)
tot = zeros(size(rect(1).mat));
totSum=0;
for i = 1:length(rect)
if use(i)==1
tot = tot|rect(i).mat;
totSum = totSum + sum(rect(i).mat(:)==1);
end
end
tmp = [];
val = abs(sum(tot(:)==1)-totSum);
我用字段.mat制作了 rect的矩形结构，这是矩阵，表示其所在的位置。
我使用了它(s与您的算法相同(
rect(size(s,1)).size = s(i,end);
rect(size(s,1)).mat = [];
for i = 1:size(s,1)
rect(i).size = s(i,end);
cmp = zeros(size(P));
[x,y] = meshgrid(s(i,1):s(i,2),s(i,3):s(i,4));
cmp(x,y) = 1;
rect(i).mat=cmp;
end
现在您可以应用ga
sol = ga(@(use)toMin(rect,use),length(rect),[],[],[],[],zeros(length(rect),1),ones(length(rect),1),@(x) constraint(rect,x),1:length(rect));