一套完整的结合3一套组合

本文介绍了一套完整的结合3一套组合的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我需要生成组合的成套获得的结合三个不同的子集：

I need to generate the complete set of combinations obtained combining three different subset:

• 设置1 ：13元素的矢量选择任何4个号码
• 设置2 ：从3个元素的矢量选择任何2号。
• 设置3 ：从9个元素的矢量选择任何2号。

• Set 1: choosing any 4 numbers from a vector of 13 elements.
• Set 2: choosing any 2 numbers from a vector of 3 elements.
• Set 3: choosing any 2 numbers from a vector of 9 elements.

4 4 4 3 4 4 3 3 4 3 3 3 2 4 4 2 3 4 2 3 3 2 2 4 Set A = 2 2 3 2 2 2 1 4 4 1 3 4 1 3 3 1 2 4 1 2 3 1 2 2 1 1 4 1 1 3 1 1 2 1 1 1 3 3 2 3 Set B = 2 2 1 3 1 2 1 1 4 4 3 4 3 3 2 4 Set C = 2 3 2 2 1 4 1 3 1 2 1 1

[设置A] = [20×3] ， [B组] = [6×2] ， [设置C] = [10×2] 。然后，我需要从这些三套获得所有可能的组合： AllComb = [设置A]×[B组]×[设置C] = [1200×8] 。在 AllComb 矩阵将是这样的：

[Set A] = [20 x 3], [Set B] = [6 x 2], [Set C] = [10 x 2]. Then I need to obtain all possible combinations from these three sets: AllComb = [Set A] x [Set B] x [Set C] = [1200 x 8]. The AllComb matrix will be like this:

4 4 4 | 3 3 | 4 4 4 4 4 | 3 3 | 3 4 4 4 4 | 3 3 | 3 3 4 4 4 | 3 3 | 2 4 4 4 4 | 3 3 | 2 3 4 4 4 | 3 3 | 2 2 4 4 4 | 3 3 | 1 4 4 4 4 | 3 3 | 1 3 4 4 4 | 3 3 | 1 2 4 4 4 | 3 3 | 1 1 4 4 4 | 2 3 | 4 4 4 4 4 | 2 3 | 3 4 4 4 4 | 2 3 | 3 3 4 4 4 | 2 3 | 2 4 4 4 4 | 2 3 | 2 3 . . . 1 1 1 | 1 1 | 1 1

可惜我不能使用的集数相同，因为我需要替换的数字那样：

Unfortunately I can not use the same number for the sets since I need to substitute the numbers like that:

• 对于设定：1 = 10，2 = 25，3 = 30和4 = 45
• 对于 B组：1 = 5,2 = 20，3 = 35
• 对于设置C ：1 = 10，2 = 20，3 = 30和4 = 50

• For Set A: 1 = 10, 2 = 25, 3 = 30 and 4 = 45
• For Set B: 1 = 5, 2 = 20 and 3 = 35
• For Set C: 1 = 10, 2 = 20, 3 = 30 and 4 = 50

任何想法？实际案例设置往往会导致一个 AllComb 矩阵〜[491 400×8]因此量化的解决方案将被欣然接受。

Any ideas? Real case sets will often lead to an AllComb matrix ~[491 400 x 8] so vectorized solutions will be gladly accepted.

的注：每套用以下code获得：的

a = combnk(1:H+L-1, H); b = cumsum([a(:,1) diff(a,[],2) - 1],2);

什么是 ^ h 和→？

^ h 是MultiheadWeigher（MHW）机的料斗。我有一个 A MHW H = 8 ，我需要在每个漏斗一些材料来交付。如果我需要提供一种类型材料的所有不可能性的组合是（L + H-1）！/（H！（L-1）！）和我计算它们与code上面写（ A 和 b ）。现在，假设有3种不同的产品，那么我们有4个料斗产品A，在第4料斗2 B和2 C.产品A可以假设值 10：10：130 ，产品B 10点十分30秒和C 10:10:90 。随后的步数是A L = 13 ，B L = 3 和C L = 9

What is H and L?

H are the hoppers of a MultiheadWeigher (MHW) machines. I have a MHW with H=8 and I need to deliver in each of these hoppers some materials. If I need to deliver just one type of material all possibile combinations are (L+H-1)!/(H!(L-1)!) and i compute them with the code write above (a and b). Now, suppose to have 3 different product then we have 4 hoppers for product A, 2 for B and 2 for C. Product A in the first 4 hoppers can assume values 10:10:130, Product B 10:10:30 and c 10:10:90. Then the number of steps are for A L=13, B L=3 and C L=9

推荐答案

您基本上需要找到

为每组重复组合;

多版本（我不知道这一阶段1的结果的正确名称）。

Combinations with repetition for each set;

"Multi-variations" (I don't know the correct name for this) of the results of stage 1.

这两个阶段可以或多或少相同的逻辑来解决，从这里服用。

Both stages can be solved with more or less the same logic, taken from here.

%// Stage 1, set A LA = 4; HA = 3; SetA = cell(1,HA); [SetA{:}] = ndgrid(1:LA); SetA = cat(HA+1, SetA{:}); SetA = reshape(SetA,[],HA); SetA = unique(sort(SetA(:,1:HA),2),'rows'); %// Stage 1, set B LB = 3; HB = 2; SetB = cell(1,HB); [SetB{:}] = ndgrid(1:LB); SetB = cat(HB+1, SetB{:}); SetB = reshape(SetB,[],HB); SetB = unique(sort(SetB(:,1:HB),2),'rows'); %// Stage 1, set C LC = 4; HC = 2; SetC = cell(1,HC); [SetC{:}] = ndgrid(1:LC); SetC = cat(HC+1, SetC{:}); SetC = reshape(SetC,[],HC); SetC = unique(sort(SetC(:,1:HC),2),'rows'); %// Stage 2 L = 3; %// number of sets result = cell(1,L); [result{:}] = ndgrid(1:size(SetA,1),1:size(SetB,1),1:size(SetC,1)); result = cat(L+1, result{:}); result = reshape(result,[],L); result = [ SetA(result(:,1),:) SetB(result(:,2),:) SetC(result(:,3),:) ]; result = flipud(sortrows(result)); %// put into desired order

这给了

result = 4 4 4 3 3 4 4 4 4 4 3 3 3 4 4 4 4 3 3 3 3 4 4 4 3 3 2 4 4 4 4 3 3 2 3 4 4 4 3 3 2 2 4 4 4 3 3 1 4 4 4 4 3 3 1 3 4 4 4 3 3 1 2 4 4 4 3 3 1 1 4 4 4 2 3 4 4 4 4 4 2 3 3 4 4 4 4 2 3 3 3 4 4 4 2 3 2 4 4 4 4 2 3 2 3 4 4 4 2 3 2 2 4 4 4 2 3 1 4 4 4 4 2 3 1 3 4 4 4 2 3 1 2 ...