在 MatLab 中从固定总和中迭代值答案

【问题标题】：Iterating over values from fixed sum, in MatLab在 MatLab 中从固定总和中迭代值
【发布时间】：2012-03-19 22:04:19
【问题描述】：

假设我有 n 个弹珠，每个弹珠可以是 8 种可能的颜色之一。我想遍历弹珠颜色组合的所有可能值。如何在 MatLab 中做到这一点？

例如，如果n = 2，那么我想遍历：

2 0 0 0 0 0 0 0

1 1 0 0 0 0 0 0

1 0 1 0 0 0 0 0

1 0 0 1 0 0 0 0

1 0 0 0 1 0 0 0

1 0 0 0 0 1 0 0

1 0 0 0 0 0 1 0

1 0 0 0 0 0 0 1

0 2 0 0 0 0 0 0

0 1 1 0 0 0 0 0

0 1 0 1 0 0 0 0

等等

编辑：

这是一些伪代码的样子，但正如您所见，它非常草率。我对一种更简洁的方法感兴趣，并且不需要那个 if 语句......

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

【问题讨论】：

标签： matlab combinatorics loops

【解决方案1】：

这里是解决方案（代码在 GNU/Octave 中测试过，它也应该在 Matlab 中工作）。此代码来源于Octave-Forge multinom_exp.m

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);

【讨论】：

你好牙买加虫，这段代码适用于任意数量的颜色和弹珠。它也是矢量化的，你不会有内存问题。输出的格式完全符合您的需要。什么问题，为什么不测试？
它使用 stars and bars 方法的属性来计算弹珠的位置（每个 bin 都有与之关联的颜色状态）。就是这样，剩下的就是产生你想要的输出格式。如果您查看 GNU/Octave 中的 multinom_exp 函数，您会发现 JordiGH 是提出这个想法的人。我在网上找不到太多解释，但您可以获取 Feller 的书以了解更多信息。

【解决方案2】：

我认为这更容易是您考虑弹珠的状态，而不是颜色的状态。然后根据需要从大理石状态转换为颜色状态。例如，如果考虑所有可能的弹珠状态的有序列表，您可以创建如下所示的函数（使用全零作为“全部完成”标志）：

function state = nextmarblestate(state, nMarbles, nColors)
%Increment the marble state by 1

%Add one to the last marble
state(end) = state(end)+1;

%Propogate change towards the first marble as colors overflow
for ixCurrent = nMarbles:-1:2
    if state(ixCurrent) > nColors;
        state(ixCurrent) = 1;
        state(ixCurrent-1) = state(ixCurrent-1)+1;
    else
        return;
    end
end

%Check to see if we are done (reset to 0)
if state(1) > nColors
    state = zeros(1,nMarbles);
end

然后您可以编写一个快速循环来遍历所有弹珠状态，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，为了得到你想要的结果，写一个从大理石状态到颜色状态的转换，像这样（对不起这里的简洁，这可以扩展为更漂亮的版本）：

marbleState2colorState = @(marblestate) arrayfun(@(color)sum(marblestate==color), 1:nColors);

然后使用那个转换函数，像这样展开上面的while循环：

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

产生以下输出：

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]

【讨论】：

【解决方案3】：

如果你有 2 个弹珠，每个可以是 8 种可能的颜色之一，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，但存在相同的信息。如果需要，它应该很容易转换为您的表单。

【讨论】：

对，这只是迭代了 8^n，不是吗？如何转换我的问题中的格式（即总和为 n 的 8 个向量）？

【解决方案4】：

我相信以下代码 sn-p 适用于 n = 2 的情况。

这是用 Octave 编写的，因为我目前无法访问 Matlab。

我已经针对手动生成的结果使用各种低颜色值（2、3、4、5、6 和 8）进行了测试。

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

我仍在为任意数量的颜色和任意数量的弹珠的一般情况编写代码 sn-p。

【讨论】：

谢谢！如果你能让它适用于任何 n，那就太好了。（到目前为止，它不起作用，因为例如对于 n=3，它不会让像 1 1 1 0 0 0 0 0 这样的东西的可能性。）

【解决方案5】：

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.

【讨论】：

这很好，但是 MatLab 的内存不足 n > 9。我正在寻找类似 n = 500 的迭代，所以为了不耗尽内存，我不'不想将每个可能的组合存储在一个数组中——我只需要遍历每个可能的组合（这样我就可以为每个组合执行一个简单的操作）。