从Smalltalk中的集合中生成所有组合 [英] Generating all combinations from collections in Smalltalk

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问题描述

我看到C#和其他语言可以解决此问题,而Smalltalk却无法解决。我有3个收藏集,例如:

I've seen this problem resolved for C# and other languages but not for Smalltalk. I have 3 collections, for example:

a := #(3 4 5).
b := #(4 1 2).
c := #(5 2 3).

我需要进行所有可能的组合,即e。:

and I need to make all possible combinations, i. e.:

#(3 4 5)
#(3 4 2)
#(3 4 3)

#(3 1 5)
#(3 1 2)
#(3 1 3)

#(3 2 5)
#(3 2 2)
#(3 2 3)

#(4 4 5)
...

我在Squeak和Pharo中看到过组合:atATimeDo:但是在这种情况下,我不知道如何使用它。这不是功课。有帮助吗?

I have seen in Squeak and Pharo there is combinations:atATimeDo: but I don't get how to use it for this case. This is not homework. Any help?

推荐答案

这是Smalltalk / X的类库(在SequentialCollection中)的代码。
请参阅最后的示例使用注释。

here is the code from Smalltalk/X's class library (in SequentialCollection). See the example-use comments at the end.

combinationsDo: aBlock
    "Repeatly evaluate aBlock with all combinations of elements from the receivers elements. 
     The receivers elements must be collections of the individuals to be taken for the combinations"

    self combinationsStartingAt:1 prefix:#() do:aBlock







combinationsStartingAt:anInteger prefix:prefix do:aBlock
    "a helper for combinationsDo:"

    |loopedElement|

    loopedElement := self at:anInteger.

    anInteger == self size ifTrue:[
        loopedElement do:[:el | aBlock value:(prefix copyWith:el)].
        ^ self.
    ].

    loopedElement do:[:el |
        |newPrefix|

        newPrefix := (prefix copyWith:el).
        self combinationsStartingAt:anInteger+1 prefix:newPrefix do:aBlock
    ].

    "
     (Array 
            with:($a to:$d)
            with:(1 to: 4)) 
        combinationsDo:[:eachCombination | Transcript showCR: eachCombination]
    "
    "
     (Array 
            with:#(1 2 3 4 5 6 7 8 9)
            with:#(A)) 
        combinationsDo:[:eachCombination | Transcript showCR: eachCombination]
    "
    "
     #( (3 4 5) 
        (4 1 2)
        (5 2 3) 
     ) combinationsDo:[:eachCombination | Transcript showCR: eachCombination]
    "

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