# Counterexample of Abusing Simultaneous Assignments in Python

When learning python, I was amazed by a, b = b, a and thought, “I don’t need to keep track of or caching any variables now when I want to do simultaneous assignment thanks to python!”

However, things are not so simple, and I was stuck when I try to simultaneously assign value to linked list.

# Reverse Nodes in k-Group

## reverse the pointing of adjacent nodes

The critical step to solve this problem is to reverse the pointing of adjacent nodes. Which is:

change

$\overset{l}{1}\rightarrow2\rightarrow\overset{r}{3}$

to

$\overset{l}{2}\rightarrow1 \rightarrow \overset{r}{3}$

here the numbers denote nodes and the alphabets on top of the node denote pointers that keep track of these nodes (l for left and r for right. In particular, I need the r because the problem is actually not of size 3, but size k)

## Counterexample of simultaneous assignment

I attempted to achieve it with the following code:

l, l.next, l.next.next = l.next, r, l

Here is my idea.

1. l: a pointer that is pointing at 1 should now points to 2, which is l.next
2. l.next: the next of l (1) should be set to 3
3. l.next.next: the next of l.next (2) should be set to 1

But it does not work. Actually the program is stuck in a dead loop.

## Limitation of simultaneous assignment

It is not valid to perform the line because python only do the cache for you when you are doing simultaneous assignment (the things on the right-hand side of the =) but it will never remember your variable settings (the things on the left-hand side of the =).

To be specific, what python really did was:

1. l: a pointer that is pointing at 1 is now pointing to 2. No problem.
2. l.next: the next of l (which is now 3!) is set to 3(nothing happened in effect)
3. l.next.next: the next of l.next (which is still 3!!) is set to l, which is 1 (notice that python helped me here. It cached the original l = 1 for me)

The result turns out to be a dead circle in the linked list!

$\rightarrow1\rightarrow2\rightarrow3\rightarrow1\rightarrow\dots$

## The correct solution

Just don’t believe python can do anything for you. Cache the nodes if necessary, and all set.


m = l.next
l.next = r
m.next = l



However, actually there is a tidier solution.

# ‘Order’ of ‘Simultaneous’ Assignment

If you thought “it is actually not simultaneous!” then you are on the right track. The order does matter.

## An example from the above Counterexample

Actually I just used another simultaneous assignment in the same piece of code, as following.

l.next, l = self.kth_later, r


Why is this one correct? The code tells l to point to a node called $k^{th} later$ and change the pointer that points to l now to r.

Therefore, it seems that it is possible to assign value to the property of the object before assigns value to the pointer to the object itself.

## A simpler analogous example

Let assume we have the class Anime with property name, and we do the following.


a = Anime()
b = Anime()
saekano.name = "saekano"


Now

a.name, a = "eromanga-sensei", saekano


will define a’s name to “eromanga-sensei”, and let a points to saekano. So a.name will be “saekano”

While

a, a.name = saekano, "eromanga-sensei"


change both a.name and b.name to “eromanga-sensei” in effect.

## back to the original problem

Noticing the order help us to come up with a tidier solution for the original Reverse Nodes problem.

If we try

l.next.next, l.next, l = l, r, l.next

the result will be 2->1->3

What happens is, Python first cache the l, r, l.next, which are 1, 3, 2. Then we let 2 points to 1, 1 points to 3, and set l to point to 2.

## the Takeaway

• Be careful in simultaneous assignment when the left-hand side (the variables) is under possible changes during the assignment.

• Changing the order of simultaneous assignment might help, but more importantly, think of the assignment process as “ caching the right, and assign to left, one by one”. If the logic is clear, then go fo the simpler code.