Python - 如何将二叉树转换为保持相同信息的 N 叉树

Question

I have a binary tree that represents a parsed logical formula. For example, f = a & b & -c | d is represented by a list of lists in prefix notation, where the first element is the operator (unary or binary) and the next elements their arguments:

f = [ |, [&, a, [&, b, [-, c]]], d]

But if you translate (by recursion) to the classical infix notation the result is the same.

f = (((-c & b) & a) | d) = a & b & -c | d

What I'm trying to do is to convert it into an N-ary tree that retains the same information, that is, if you translate it into formula again, the result must be the same. Something like this:

f = {l: [{&: [a,b,{-:[c]}]}, d]}

Which in infixed notation is the following.

f = ((a & b & -c) | d) = a & b & -c | d

I haven't found any library so I tried to do it by myself recursively. However, I have only achieved this code that fails in some cases and it's not very elegant...

    def explore_tree(self,tree, last_symbol, new_tree):
        if type(tree) != list: # This true means that root is an atom
            new_tree[last_symbol].append(tree)
            return
        root = tree[0]
        if is_operator(root):
            if root != last_symbol:
                branch = {root: []}
                new_tree[last_symbol].append(branch)
                #This line is to search the index of branch object and expand by them
                self.explore_branches(tree, root, new_tree[last_symbol]
                                     [new_tree[last_symbol].index(branch)]) 
            else:
                self.explore_branches(tree,root,new_tree)

The functions explore_branches() call recursively to explore tree from left and right (if exist), and is_operator() return true if the given string is one logic operator, for example & or |.

Any other idea of how can I do this?

Thanks you in advance.

Answer 1

The only touchy case is the negation. Apart from it, one can simply write your algo or similar such as

from functools import reduce
def op(tree):
    return 0 if type(tree)!=list else tree[0]

def bin_to_n(tree):
    if op(tree)==0:
        return tree
    op_tree = tree[0]
    out = [op_tree]
    for node in tree[1:]:
        flat_node = bin_to_n(node)
        if op(node) != op_tree:
            out.append(flat_node)
        else:
            out += flat_node[1:]
    return out

Now regarding the negation. The failing case of above algorithm is when flattening -(-(1)) which gives -1 instead of 1

• A very basic fix is thus

< if op(node) != op_tree
---
> if op(node) != op_tree or op(node)=="-"

meaning that if a "minus" is found, you never "concatenate" it. Thus this lets -(-(1)) as is.

Now we can simplify more but those simplifications could have been done beforehand on the input list. So it "semantically" changes the tree (even though evaluation stays identical).

• Just handling the double negation:

op_tree = tree[0]
> if op_tree == '-' and op(tree[1]) == '-':
>    return bin_to_n2(tree[1][1])
out = [op_tree]

• Or going monkas and actually applying DeMorgan's law whenever a negation is found

#really invert according to demorgan's law
def bin_to_n3(tree, negate=False):
    if op(tree)==0:
        return tree

    op_tree = tree[0]

    if negate:
        if op_tree == '-':
            #double neg, skip the node
            return bin_to_n3(tree[1])

        #demorgan
        out = [ '+' if op_tree == '*' else '*' ]
        for node in tree[1:]:
            flat_node = bin_to_n3(node, True)
            #notice that since we modify the operators we have 
            #to take the operator of the resulting tree
            if op(flat_node) != op_tree:
                out.append(flat_node)
            else:
                out += flat_node[1:]
        return out

    if op_tree == '-' and op(op_tree)==0:
        #do not touch the leaf
        return tree

    #same code as above, not pun to factorize it
    out = [op_tree]
    for node in tree[1:]:
        flat_node = bin_to_n3(node)
        if op(flat_node) != op_tree:
            out.append(flat_node)
        else:
            out += flat_node[1:]
    return out

---

Below some random checks to ensure transformation keeps the tree's value intact

from functools import reduce
def op(tree):
    return 0 if type(tree)!=list else tree[0]

def bin_to_n(tree):
    if op(tree)==0:
        return tree
    op_tree = tree[0]
    out = [op_tree]
    for node in tree[1:]:
        flat_node = bin_to_n(node)
        if op(node) != op_tree or op(node)=='-':
            out.append(flat_node)
        else:
            out += flat_node[1:]
    return out

def bin_to_n2(tree):
    if op(tree)==0:
        return tree

    op_tree = tree[0]
    if op_tree == '-' and op(tree[1]) == '-':
        return bin_to_n2(tree[1][1])
    out = [op_tree]
    for node in tree[1:]:
        flat_node = bin_to_n2(node)
        if op(node) != op_tree:
            out.append(flat_node)
        else:
            out += flat_node[1:]
    return out

#really invert according to demorgan's law
def bin_to_n3(tree, negate=False):
    if op(tree)==0:
        return tree

    op_tree = tree[0]

    if negate:
        if op_tree == '-':
            #double neg, skip the node
            return bin_to_n3(tree[1])

        #demorgan
        out = [ '+' if op_tree == '*' else '*' ]
        for node in tree[1:]:
            flat_node = bin_to_n3(node, True)
            #notice that since we modify the operators we have 
            #to take the operator of the resulting tree
            if op(flat_node) != op_tree:
                out.append(flat_node)
            else:
                out += flat_node[1:]
        return out

    if op_tree == '-' and op(op_tree)==0:
        #do not touch the leaf
        return tree

    #same code as above, not pun to factorize it
    out = [op_tree]
    for node in tree[1:]:
        flat_node = bin_to_n3(node)
        if op(flat_node) != op_tree:
            out.append(flat_node)
        else:
            out += flat_node[1:]
    return out

def calc(tree):
    if op(tree) == 0:
        return tree
    s = 0
    subtree = tree[1:]
    if op(tree)=='+':
        s = reduce(lambda x,y: x or calc(y), subtree, False)
    elif op(tree) == '-':
        s = not calc(subtree[0])
    else:
        s = reduce(lambda x,y: x and calc(y), subtree, True)
    return s

#adaptated from https://stackoverflow.com/questions/6881170/is-there-a-way-to-autogenerate-valid-arithmetic-expressions
def brute_check():
    import random
    random.seed(3)
    def make_L(n=3):
        def expr(depth):
            if depth==1 or random.random()<1.0/(2**depth-1): 
                return random.choice([0,1])
            if random.random()<0.25:
                return ['-', expr(depth-1)]
            return [random.choice(['+','*']), expr(depth-1), expr(depth-1)]
        return expr(n)

    for i in range(100):
        L = make_L(n=10)
        a = calc(L)
        b = calc(bin_to_n(L))
        c = calc(bin_to_n2(L))
        d = calc(bin_to_n3(L))
        if a != b:
            print('discrepancy', L,bin_to_n(L),  a, b)

        if a != c:
            print('discrepancy', L,bin_to_n2(L),  a, c)

        if a != d:
            print('discrepancy', L,bin_to_n3(L),  a, d)
brute_check()