一套完整的结合3一套组合 [英] Complete set of combinations combining 3 set
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
- 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
...
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