与多个不同尺寸的容器组合

时间:2011-02-22 07:44:16

标签: algorithm permutation combinations

如果你知道这个问题叫什么,请告诉我(除非你真的知道这个问题的答案)。

如果我有一组Z对象,是否有算法在一堆容器(每个容器中容纳一定数量的对象)之间进行潜水?

为了使问题稍微复杂一点,我们假设我们开始的对象集合有一个子集X.有X个容器,除了其他对象(如果它有空间),每个容器必须包含X的单个元素

我现在想到的最好的方法就是看Z和X的分离,我们称之为Y.然后我们可以生成z选择x组合,然后将其扩展为x的所有可能组合。

示例: 实际问题基本上是在空间中生成所有事件。假设我们有两个事件触发器(X)和两个事件参数(Y),其中Z = XU Y.每个事件必须有一个触发器,它可以有0 ... N个参数(取决于事件的类型,但是现在并不重要。触发器也可以是一个参数。显然,在这种情况下,我们可以有一个带有一个触发器和3个参数的事件(其中一个是第二个触发器)

我们的事件空间如下(Trigger [Arguments],+表示新事件):

X1[] + X2[]
X1[Y1] + X2[]
X1[Y2] + X2[]
X1[] + X2[Y1]
X1[] + X2[Y2]
X1[Y1] + X2[Y2]
X1[Y2] + X2[Y1]
X1[X2]
X1[X2,Y1]
X1[X2,Y2]
X1[X2,Y1,Y2]
X2[X1]
X2[X1,Y1]
X2[X1,Y2]
X2[X1,Y1,Y2]

我很确定这是所有的组合。

更新 在仔细考虑了这个问题之后,我对约束和内容有了一些想法:创建“事件”的规则: 1)每个触发器都有一个事件,每个事件都必须有一个触发器 2)活动必须具有> 0个参数 3)事件不能共享参数 4)触发器可以用作参数

对于暴力解决方案,也许可以生成触发器+事件的所有排列,然后消除与上述4条规则不匹配的结果,并将排序视为事件分组?

感谢您提出任何问题或想法!

2 个答案:

答案 0 :(得分:1)

算法:

For all nonempty subsets Triggers of X:
    For all maps from (X \ Triggers) to X:
        For all maps from Y to (X union {None}):
            print the combination, where an assignment of y in Y to None means y is omitted

在Python中:

def assignments(xs, ys):
    asgns = [[]]
    for x in xs:
        asgns1 = []
        for y in ys:
            for asgn in asgns:
                asgn1 = asgn[:]
                asgn1.append((x, y))
                asgns1.append(asgn1)
        asgns = asgns1
    return asgns

def combinations(xs, ys):
    xroleasgns = assignments(xs, ('argument', 'trigger'))
    for xroleasgn in xroleasgns:
        triggers = [x for (x, role) in xroleasgn if role == 'trigger']
        if (xs or ys) and not triggers:
            continue
        xargs = [x for (x, role) in xroleasgn if role == 'argument']
        for xargasgn in assignments(xargs, triggers):
            for yargasgn in assignments(ys, [None] + triggers):
                d = dict((x, []) for x in triggers)
                for xarg, t in xargasgn:
                    d[t].append(xarg)
                for yarg, t in yargasgn:
                    if t is not None:
                        d[t].append(yarg)
                print ' + '.join('%s[%s]' % (t, ','.join(args)) for (t, args) in d.iteritems())


"""
>>> assign.combinations(['X1','X2'],['Y1','Y2'])
X1[X2]
X1[X2,Y1]
X1[X2,Y2]
X1[X2,Y1,Y2]
X2[X1]
X2[X1,Y1]
X2[X1,Y2]
X2[X1,Y1,Y2]
X2[] + X1[]
X2[] + X1[Y1]
X2[Y1] + X1[]
X2[] + X1[Y2]
X2[] + X1[Y1,Y2]
X2[Y1] + X1[Y2]
X2[Y2] + X1[]
X2[Y2] + X1[Y1]
X2[Y1,Y2] + X1[]


"""

答案 1 :(得分:0)

这是我的java实现over9000解决原始问题的方法:

public static void main(String[] args) throws Exception {
    ArrayList xs = new ArrayList();
    ArrayList ys = new ArrayList();
    xs.add("X1");
    xs.add("X2");
    ys.add("Y1");
    ys.add("Y2");

    combinations(xs,ys);
}

private static void combinations(ArrayList xs, ArrayList ys) {
    ArrayList def = new ArrayList();
    def.add("argument");
    def.add("trigger");

    ArrayList<ArrayList> xroleasgns = assignments(xs, def);
    for(ArrayList xroleasgn:xroleasgns){
        // create triggers list
        ArrayList triggers = new ArrayList();
        for(Object o:xroleasgn){
            Pair p = (Pair)o;
            if("trigger".equals(p.b.toString()))
                triggers.add(p.a);              
        }

        if((xs.size()>0 || ys.size()>0) && triggers.size()==0)
            continue;

        // create xargs list
        ArrayList xargs = new ArrayList();
        for(Object o:xroleasgn){
            Pair p = (Pair)o;
            if("argument".equals(p.b.toString()))
                xargs.add(p.a);             
        }

        // Get combinations!
        for(ArrayList xargasgn:assignments(xargs,triggers)){
            ArrayList yTriggers = new ArrayList(triggers);
            yTriggers.add(null);

            for(ArrayList yargasgn:assignments(ys,yTriggers)){

                // d = dict((x, []) for x in triggers)
                HashMap<Object,ArrayList> d = new HashMap<Object,ArrayList>();
                for(Object x:triggers)
                    d.put(x, new ArrayList());

                for(Object o:xargasgn){
                    Pair p = (Pair)o;
                    d.get(p.b).add(p.a);    
                }

                for(Object o:yargasgn){
                    Pair p = (Pair)o;
                    if(p.b!=null){
                        d.get(p.b).add(p.a);
                    }
                }


                for(Entry<Object, ArrayList> e:d.entrySet()){
                    Object t = e.getKey();
                    ArrayList args = e.getValue();
                    System.out.print(t+"["+args.toString()+"]"+"+");
                }
                System.out.println();                   

            }
        }
    }
}

private static ArrayList<ArrayList> assignments(ArrayList xs, ArrayList def) {
    ArrayList<ArrayList> asgns = new ArrayList<ArrayList>();
    asgns.add(new ArrayList()); //put an initial empty arraylist
    for(Object x:xs){
        ArrayList asgns1 = new ArrayList();
        for(Object y:def){
            for(ArrayList<Object> asgn:asgns){
                ArrayList asgn1 = new ArrayList();
                asgn1.addAll(asgn);
                asgn1.add(new Pair(x,y));
                asgns1.add(asgn1);
            }
        }
        asgns = asgns1;
    }
    return asgns;
}