Changing local variable using recursion

Sisoviromol 0
Viewed 31

I am using a recursive function implemented in a python class. Am I right that the local variable (passed through the methods attributes) isn't changing through a recursive change?

I have the following tree: example tree

My current method is the following:

def readSequence(self, linkTable, partTable, seq, element):
    #clearFlags returns an array of objects with flags set to 0
    partTable = self.clearFlags(partTable)
    #readFlags returns an array of objects with updated flags. (flag > 0 if object is in sequence)
    partTable = self.readFlags(partTable, seq)
    #readChildren returns an array of objects with all children of the current element
    children = self.readChildren(linkTable, partTable, element)

    if len(children) == 0:
        self.subsequences.append(seq)
    else:
        for child in children:
            if child.flag == 1:
                self.subsequences.append(seq)
                return
            else:
                seq.append(child)
                self.readSequence(linkTable, partTable, seq, child)

In my understanding the sequence seq should grow as follows:

  1. [1]
  2. [1, 2]
  3. [1, 2, 4] --> appended to subsequences
  4. [1, 2, 5] --> appended to subsequences
  5. [1, 3] --> appended to subsequences

But instead it does this:

  1. [1]
  2. [1, 2]
  3. [1, 2, 4] --> appended to subsequences
  4. [1, 2, 4, 5] --> appended to subsequences
  5. [1, 2, 4, 5, 3] --> appended to subsequences

The problem is clearly that the sequence seq is changed like a global variable and doesn't stay the same to add the other child.

Hope you can help me! Thanks in advance!

local-variables recursion python
1 Answers

list is a mutable type in python.

Essentially, that means that lists are passed by reference rather than by value; if two variables refer to the same list, then modifying one will reflect on the other.

For instance:

seq = [1,2,3]
b = seq
b.append(4)
print(seq) # [1,2,3,4]

Likewise if you pass seq to a function, and the function modifies it, then seq is really modified, not just a copy:

def f(l):
    l.append(4)

seq = [1,2,3]
f(l)
print(seq) # [1,2,3,4]

If you don't want this behaviour, then there are three possible solutions:

  • Use a non-mutable type, such as a tuple, instead of a list; or
  • explicitly make a copy, using list(seq) or seq.copy(); or
  • undo the changes before you return from the recursive call.

Use a non-mutable type or explicitly make a copy

For instance:

# USING A NON-MUTABLE TYPE

def f(l):
    l = (*l, 4)

seq = (1,2,3) # tuple
f(seq)
print(seq) # (1,2,3)

# MAKING A COPY

def f(l):
    l.append(4)

seq = [1,2,3]
f(list(seq)) # explicitly pass a copy rather than the original
print(seq) # [1,2,3]

Undo the changes before returning from the recursive call

The two options above would add to the complexity of your function, because making a copy at every recursive call takes time linear in the length of the list at every recursive call.

Instead, you can undo your list.append by following it with a list.pop.

Replace:

                seq.append(child)
                self.readSequence(linkTable, partTable, seq, child)

with:

                seq.append(child)
                self.readSequence(linkTable, partTable, seq, child)
                seq.pop()
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