在MatLab中迭代固定和中的值

假设我有n个弹珠，每个弹珠可以是8种可能颜色中的一种。我想迭代大理石颜色组合的所有可能值。如何在MatLab中实现这一点

例如，如果n=2，则我希望迭代：

200 0 0 0 0 0 0

1100 00 00

10010000

1001000

100100100

1 0 0 0 1 0 0

10 0 0 0 1 0

100 0 0 0 1

0 2 0 0 0 0 0 0 0

01101000

0110100

等等

编辑：

这是一些伪代码的外观，但正如您所看到的，它非常松散。我感兴趣的是一种更简洁的方法，一种不需要if语句的方法

for i8 = 0:1:n
    for i7 = 0:1:n - i8
        for i6 = 0:1:n - i8 - i7
            for i5 = 0:1:n - i8 - i7 - i6
                for i4 = 0:1:n - i8 - i7 - i6 - i5
                    for i3 = 0:1:n - i8 - i7 - i6 - i5 - i4
                        for i2 = 0:1:n - i8 - i7 - i6 - i5 - i4 - i3
                            for i1 = 0:1:n - i8 - i7 - i6 - i5 - i4 - i3 - i2
                                if i1 + i2 + i3 + i4 + i5 + i6 + i7 + i8 == n
                                    i = [i1, i2, i3, i4, i5, i6, i7, i8]
                                end
                            end
                        end
                    end
                end
            end
        end
    end
end

---

我认为这更容易，因为你想到的是大理石的状态，而不是颜色的状态。然后根据需要将大理石状态转换为颜色状态。例如，如果考虑所有可能的大理石状态的有序列表，您可以创建如下所示的函数（使用全零作为“全部完成”标志）：

然后，您可以编写一个快速循环来遍历所有大理石状态，如下所示：

nMarbles = 2;
nColors = 8;

marblestate = ones(1,nMarbles);
while sum(marblestate)>0
    disp(['Marblestate = [' num2str(marblestate) ']'])
    marblestate = nextmarblestate(marblestate, nMarbles, nColors);
end

或者，为了得到您想要的结果，编写一个从大理石状态到颜色状态的转换，如下所示（很抱歉这里过于简洁，可以扩展到更漂亮的版本）：

然后使用该转换函数，展开上面的while循环，如下所示：

marblestate = ones(1,nMarbles);
while sum(marblestate)>0
    disp(['Marblestate = [' num2str(marblestate) '],  Colorstate = [' num2str(marbleState2colorState(marblestate)) ']'])
    marblestate = nextmarblestate(marblestate, nMarbles, nColors);
end

将生成以下输出：

Marblestate = [1  1],  Colorstate = [2  0  0  0  0  0  0  0]
Marblestate = [1  2],  Colorstate = [1  1  0  0  0  0  0  0]
Marblestate = [1  3],  Colorstate = [1  0  1  0  0  0  0  0]
Marblestate = [1  4],  Colorstate = [1  0  0  1  0  0  0  0]
Marblestate = [1  5],  Colorstate = [1  0  0  0  1  0  0  0]
Marblestate = [1  6],  Colorstate = [1  0  0  0  0  1  0  0]
Marblestate = [1  7],  Colorstate = [1  0  0  0  0  0  1  0]
Marblestate = [1  8],  Colorstate = [1  0  0  0  0  0  0  1]
Marblestate = [2  1],  Colorstate = [1  1  0  0  0  0  0  0]
Marblestate = [2  2],  Colorstate = [0  2  0  0  0  0  0  0]
Marblestate = [2  3],  Colorstate = [0  1  1  0  0  0  0  0]
Marblestate = [2  4],  Colorstate = [0  1  0  1  0  0  0  0]
Marblestate = [2  5],  Colorstate = [0  1  0  0  1  0  0  0]
Marblestate = [2  6],  Colorstate = [0  1  0  0  0  1  0  0]
Marblestate = [2  7],  Colorstate = [0  1  0  0  0  0  1  0]
Marblestate = [2  8],  Colorstate = [0  1  0  0  0  0  0  1]
Marblestate = [3  1],  Colorstate = [1  0  1  0  0  0  0  0]
Marblestate = [3  2],  Colorstate = [0  1  1  0  0  0  0  0]
Marblestate = [3  3],  Colorstate = [0  0  2  0  0  0  0  0]
Marblestate = [3  4],  Colorstate = [0  0  1  1  0  0  0  0]
Marblestate = [3  5],  Colorstate = [0  0  1  0  1  0  0  0]
Marblestate = [3  6],  Colorstate = [0  0  1  0  0  1  0  0]
Marblestate = [3  7],  Colorstate = [0  0  1  0  0  0  1  0]
Marblestate = [3  8],  Colorstate = [0  0  1  0  0  0  0  1]
Marblestate = [4  1],  Colorstate = [1  0  0  1  0  0  0  0]
Marblestate = [4  2],  Colorstate = [0  1  0  1  0  0  0  0]
Marblestate = [4  3],  Colorstate = [0  0  1  1  0  0  0  0]
Marblestate = [4  4],  Colorstate = [0  0  0  2  0  0  0  0]
Marblestate = [4  5],  Colorstate = [0  0  0  1  1  0  0  0]
Marblestate = [4  6],  Colorstate = [0  0  0  1  0  1  0  0]
Marblestate = [4  7],  Colorstate = [0  0  0  1  0  0  1  0]
Marblestate = [4  8],  Colorstate = [0  0  0  1  0  0  0  1]
Marblestate = [5  1],  Colorstate = [1  0  0  0  1  0  0  0]
Marblestate = [5  2],  Colorstate = [0  1  0  0  1  0  0  0]
Marblestate = [5  3],  Colorstate = [0  0  1  0  1  0  0  0]
Marblestate = [5  4],  Colorstate = [0  0  0  1  1  0  0  0]
Marblestate = [5  5],  Colorstate = [0  0  0  0  2  0  0  0]
Marblestate = [5  6],  Colorstate = [0  0  0  0  1  1  0  0]
Marblestate = [5  7],  Colorstate = [0  0  0  0  1  0  1  0]
Marblestate = [5  8],  Colorstate = [0  0  0  0  1  0  0  1]
Marblestate = [6  1],  Colorstate = [1  0  0  0  0  1  0  0]
Marblestate = [6  2],  Colorstate = [0  1  0  0  0  1  0  0]
Marblestate = [6  3],  Colorstate = [0  0  1  0  0  1  0  0]
Marblestate = [6  4],  Colorstate = [0  0  0  1  0  1  0  0]
Marblestate = [6  5],  Colorstate = [0  0  0  0  1  1  0  0]
Marblestate = [6  6],  Colorstate = [0  0  0  0  0  2  0  0]
Marblestate = [6  7],  Colorstate = [0  0  0  0  0  1  1  0]
Marblestate = [6  8],  Colorstate = [0  0  0  0  0  1  0  1]
Marblestate = [7  1],  Colorstate = [1  0  0  0  0  0  1  0]
Marblestate = [7  2],  Colorstate = [0  1  0  0  0  0  1  0]
Marblestate = [7  3],  Colorstate = [0  0  1  0  0  0  1  0]
Marblestate = [7  4],  Colorstate = [0  0  0  1  0  0  1  0]
Marblestate = [7  5],  Colorstate = [0  0  0  0  1  0  1  0]
Marblestate = [7  6],  Colorstate = [0  0  0  0  0  1  1  0]
Marblestate = [7  7],  Colorstate = [0  0  0  0  0  0  2  0]
Marblestate = [7  8],  Colorstate = [0  0  0  0  0  0  1  1]
Marblestate = [8  1],  Colorstate = [1  0  0  0  0  0  0  1]
Marblestate = [8  2],  Colorstate = [0  1  0  0  0  0  0  1]
Marblestate = [8  3],  Colorstate = [0  0  1  0  0  0  0  1]
Marblestate = [8  4],  Colorstate = [0  0  0  1  0  0  0  1]
Marblestate = [8  5],  Colorstate = [0  0  0  0  1  0  0  1]
Marblestate = [8  6],  Colorstate = [0  0  0  0  0  1  0  1]
Marblestate = [8  7],  Colorstate = [0  0  0  0  0  0  1  1]
Marblestate = [8  8],  Colorstate = [0  0  0  0  0  0  0  2]

---

如果您有2个大理石，并且每个大理石可以是8种可能颜色中的一种，那么这就足够了：

>> dec2base(0:63,8)
ans =
00
01
02
03
04
05
06
07
10
11
12
13
14
15
16
17
20
21
22
23
24
25
26
27
30
31
32
33
34
35
36
37
40
41
42
43
44
45
46
47
50
51
52
53
54
55
56
57
60
61
62
63
64
65
66
67
70
71
72
73
74
75
76
77

请注意，它以与您不同的方式存储信息，但相同的信息仍然存在。如果您需要的话，转换为您的表单应该很容易。

下面是解决方案（代码在GNU/Octave中进行了测试，它也应该在Matlab中工作）。 此代码源于

---

我相信下面的代码片段适用于n=2的情况

这是用八度音阶写的，因为我目前无法访问Matlab

我用各种低颜色值（2、3、4、5、6和8）与手工生成的结果进行了对比测试

colours = 4;
marbles = 2;

for i=1:colours
   a = zeros(1,colours);
   a(i) = marbles
   for j=i:colours-1
      a(j) = a(j) -1;
      a(j+1) = a(j+1)+1  
   endfor
endfor 

---

我仍然在为任意数量的颜色和任意数量的大理石编写代码片段。

对，这只是在8^n上迭代，不是吗？我如何转换问题中的格式（即8个向量，其和为n）？谢谢！如果你能让它为任何n工作，那就太好了。（到目前为止，它不起作用，因为对于n=3，例如，它不允许有类似于1 1 0 0 0 0 0的可能性。）这很好，但MatLab在n>9时内存不足。我希望对n=500这样的东西进行迭代，所以为了不耗尽内存，我不想在数组中存储每个可能的组合——我只需要迭代每个可能的组合（这样我就可以对每个组合执行一个简单的操作）。Hello jamaicaworm，这段代码适用于任意数量的颜色和大理石。它也是矢量化的，您不会有内存问题。输出的格式完全符合您的需要。问题是什么，为什么不测试它？它使用方法的属性来计算大理石的位置（每个箱子都有一个与之相关的颜色状态）。就这样，剩下的就是生成您想要的输出格式。如果你看看GNU/Octave中的multinom_exp函数，你会发现JordiGH就是这个想法的始作俑者。我在网上找不到太多的解释，但你可以从费勒的书中了解更多。

nc = 8;    % number colors
nm = 2;    % number marbles

% For now let marbles be unique. There are then nc^nm assignments of
% marbles to colors
nmc = nc^nm;

% Enumerate the assignments of marbles to colors: i.e., sub{k} is a row
% array giving the color of marble k for each of the nmc assignments
[sub{1:nm,1}] = ind2sub(nc*ones(1,nm),1:nmc);

% We'll use accumarray to map the marble color assignments in sub to number
% of marbles of each color
% Bring sub into form required for application of accumarray
s = mat2cell(cell2mat(sub),nm,ones(nmc,1));

% apply accumarray to each assignement of colors to marbles
a = cellfun(@(c)accumarray(c,1,[nc,1]),s,'UniformOutput',false);

% Each element of a is a column vector giving the number of marbles of each
% color. We want row vectors, and we want only the unique rows
a = unique(cell2mat(a)','rows'); 

% a is as desired. 

function conf = marbin (nmar,nbin)
%% Returns the position of nmar in nbis, allowing the marbles to be in the same bin.

%% This is standard stars and bars.
 numsymbols = nbin+nmar-1;
 stars = nchoosek (1:numsymbols, nmar);

%% Star labels minus their consecutive position becomes their index
%% position!
 idx = bsxfun (@minus, stars, [0:nmar-1]);

%% Manipulate indices into the proper shape for accumarray.
 nr = size (idx, 1);
 a = repmat ([1:nr], nmar, 1);
 b = idx';
 conf = [a(:), b(:)];
 conf = accumarray (conf, 1);

colours = 4;
marbles = 2;

for i=1:colours
   a = zeros(1,colours);
   a(i) = marbles
   for j=i:colours-1
      a(j) = a(j) -1;
      a(j+1) = a(j+1)+1  
   endfor
endfor 

nc = 8;    % number colors
nm = 2;    % number marbles

% For now let marbles be unique. There are then nc^nm assignments of
% marbles to colors
nmc = nc^nm;

% Enumerate the assignments of marbles to colors: i.e., sub{k} is a row
% array giving the color of marble k for each of the nmc assignments
[sub{1:nm,1}] = ind2sub(nc*ones(1,nm),1:nmc);

% We'll use accumarray to map the marble color assignments in sub to number
% of marbles of each color
% Bring sub into form required for application of accumarray
s = mat2cell(cell2mat(sub),nm,ones(nmc,1));

% apply accumarray to each assignement of colors to marbles
a = cellfun(@(c)accumarray(c,1,[nc,1]),s,'UniformOutput',false);

% Each element of a is a column vector giving the number of marbles of each
% color. We want row vectors, and we want only the unique rows
a = unique(cell2mat(a)','rows'); 

% a is as desired.