use decision tree to solve 0-1 knapsack problem // optimal substructure

#quote from 'introduction to computation and programming using python,# revised edition, MIT'class Item(object):      def __init__(self, n, v, w):          self.name = n          self.value = float(v)          self.weight = float(w)      def getName(self):          return self.name      def getValue(self):          return self.value      def getWeight(self):          return self.weight      def __str__(self):          result = '<' + self.name + ', ' + str(self.value)\                 + ', ' + str(self.weight) + '>'          return result                     def maxVal(toConsider, avail):    """Assumes toConsider a list of items, avail a weigth       Returns a tuple of total weight of a solution to the       0/1 knapsack problem and the items of that solution"""    #simple case: no item left or no space left    if toConsider == [] or avail == 0:        result = (0, ())            #recursive decomposition    #the item is too big    elif toConsider[0].getWeight() > avail:        #Explore right branch only        result = maxVal(toConsider[1:], avail)    else:        nextItem = toConsider[0]        #Explore left branch: take        withVal, withToTake = maxVal(toConsider[1:],                                     avail - nextItem.getWeight())        withVal += nextItem.getValue()        #Explore right branch: not take        withoutVal, withoutToTake = maxVal(toConsider[1:], avail)        #Choose better branch        if withVal > withoutVal:            result = (withVal, withToTake + (nextItem,))        else:            result = (withoutVal, withoutToTake)    return result    def smallTest():    names = ['a', 'b', 'c', 'd']    vals = [6, 7, 8, 9]    weights = [3, 3, 2, 5]    Items = []    for i in range(len(vals)):        Items.append(Item(names[i], vals[i], weights[i]))    val, taken = maxVal(Items, 5)    for item in taken:        print item    print 'Total value of items taken =', val