# Bubblesort loop invariant

```
def bubblesort(L):
'''
Pre: L is a list of numbers
Post: L is sorted
'''
k = 0
while k < len(L):
i = 0
while i < len(L) - k:
if L[i] > L[i+1]:
swap L[i] and L[i+1]
i += 1
k +=1
```

Let \(\mathbb{N}\) be the set of nonnegative integers. For \(a,b \in \mathbb{N}\), define

\[\mathtt{range}(a,b) = \{x \in \mathbb{N} : a \leq x < b\}\]If \(a \geq b\), then \(\mathtt{range}(a,b)=\emptyset\).

If \(a < b\), then \(\mathtt{range}(a,b)=\{a,\ldots,b-1\}\), e.g. \(\mathtt{range}(0,4)=\{0,1,2,3\}\).

Define the predicate \(P(i,j)\) by

- \(-1 \leq j < i \leq \mathtt{len}(L)\).
- If \(x \in \mathtt{range}(0,j+1)\), then \(L[x]=0\).
- If \(x \in \mathtt{range}(j+1,i)\), then \(L[x]=1\).

# Theorem: \(P(i,j)\) is a correct loop invariant

# Proof

# If list \(L\) is empty

## Entering loop

\(P(0,-1)\) is true because \(\mathtt{range}(0,0)=\emptyset\).

## Executing loop

The loop does not execute.

## Exiting loop

Suppose that \(P(i,j)\) is true when the loop exits. Since \(L\) is empty, this means \(-1 \leq j < i \leq 0\), i.e. \(j=-1\) and \(i=0\). We wish to show that the postcondition is true, namely, that \(L\) is sorted and that True is returned if 0 is an element of \(L\) and False otherwise.

Because \(L\) is empty, it is sorted.

Because \(L\) is empty, 0 is not an element of \(L\), so it is vacuously true that True is returned if 0 is an element of \(L\).

Because the loop does not execute, \(j=-1\). Therefore False is returned. So it is indeed the case that the postcondition is true.

# If list \(L\) is nonempty

## Entering loop

\(P(0,-1)\) is true because \(\mathtt{range}(0,0)=\emptyset\).

## Executing loop

Suppose that \(P(i_0,j_0)\) is true for some \(i_0\) and \(j_0\) at the start of a loop execution. The loop executing means that \(i_0 < \mathtt{len}(L)\), and combining this with \(P(i_0,j_0)\) being true gives

- \(-1 \leq j_0 < i_0 < \mathtt{len}(L)\).
- If \(x \in \mathtt{range}(0,j_0+1)\), then \(L[x]=0\).
- If \(x \in \mathtt{range}(j_0+1,i_0)\), then \(L[x]=1\).

Either \(L[i_0]=0\) or \(L[i_0]=1\).

### Case \(L[i_0]=0\)

Either \(j_0+1<i_0\) or \(j_0+1=i_0\).

#### Subcase \(j_0+1 < i_0\)

Here, \(L[j_0+1]=1\).

- \(j_1=j_0+1\).
- \(L[i_0]=0\) and \(L[j_1]=1\) are swapped: now \(L[i_0]=1\) and \(L[j_1]=0\).
- \(i_1=i_0+1\).

We want to show that \(P(i_1,j_1)\) is true, namely

- \(-1 \leq j_1 < i_1 \leq \mathtt{len}(L)\).
- If \(x \in \mathtt{range}(0,j_1+1)\), then \(L[x]=0\).
- If \(x \in \mathtt{range}(j_1+1,i_1)\), then \(L[x]=1\).

Using \(-1 \leq j_0\) and \(j_1=j_0+1\), we get \(-1 \leq j_1\).

Using \(j_0<i_0\) we get \(j_0+1<i_0+1\) and so \(j_1<i_1\).

Using \(i_0< \mathtt{len}(L)\) and \(i_1=i_0+1\) we get \(i_1 \leq \mathtt{len}(L)\). Hence

\[-1 \leq j_1 < i_1 \leq \mathtt{len}(L)\]If \(x \in \mathtt{range}(0,j_0+1)\), then \(L[x]=0\). And \(L[j_1]=0\). Hence if \(x \in \mathtt{range}(0,j_1+1)\), then \(L[x]=0\).

Using (3) and \(j_1 \geq j_0\), if \(x \in \mathtt{range}(j_1+1,i_0)\), then \(L[x]=1\). \(L[i_0]=1\), thus if \(x \in \mathtt{range}(j_1+1,i_1)\), then \(L[x]=1\).

We have established in this case that \(P(i_1,j_1)\) is true.

#### Subcase \(j_0+1 = i_0\)

Here, \(L[j_0+1]=0\).

- \(j_1=j_0+1\); so \(j_1=i_0\).
- \(L[i_0]=0\) and \(L[j_1]=0\) are swapped: now \(L[i_0]=0\) and \(L[j_1]=0\).
- \(i_1=i_0+1\).

Using \(-1 \leq j_0\) and \(j_0 \leq j_1\) we get \(-1 \leq j_1\).

Using \(j_1=i_0\) and \(i_1=i_0+1\), we get \(j_1 < i_1\).

Using \(i_0< \mathtt{len}(L)\) and \(i_1=i_0+1\) we get \(i_1 \leq \mathtt{len}(L)\). Hence

\[-1 \leq j_1 < i_1 \leq \mathtt{len}(L)\]If \(x \in \mathtt{range}(0,j_0+1)\), then \(L[x]=0\). And \(L[j_1]=0\). Hence if \(x \in \mathtt{range}(0,j_1+1)\), then \(L[x]=0\).

Since in this case \(j_1+1=i_1\), here \(\mathtt{range}(j_1+1,i_1)=\emptyset\). Thus it is vacuously true that if \(x \in \mathtt{range}(j_1+1,i_1)\), then \(L[x]=1\).

We have established in this case that \(P(i_1,j_1)\) is true.

### Case \(L[i_0]=1\)

- \(j_1=j_0\).
- \(L[i_0]=1\) and \(L[j_1]=0\)
- \(i_1=i_0+1\).

Using \(-1 \leq j_0\) and \(j_1=j_0\), we get \(-1 \leq j_1\). Using \(j_0<i_0\), we get \(j_1 = j_0 < i_0 < i_0+1 = i_1\) so \(j_1<i_1\). Using \(i_0< \mathtt{len}(L)\) and \(i_1=i_0+1\) we get \(i_1 \leq \mathtt{len}(L)\). Hence

\[-1 \leq j_1 < i_1 \leq \mathtt{len}(L)\]If \(x \in \mathtt{range}(0,j_0+1)\), then \(L[x]=0\). But \(j_1=j_0\). Hence if \(x \in \mathtt{range}(0,j_1+1)\), then \(L[x]=0\).

If \(x \in \mathtt{range}(j_0+1,i_0)\), then \(L[x]=1\). Using \(j_1=j_0\), if \(x \in \mathtt{range}(j_1+1,i_0)\) then \(L[x]=1\). And \(L[i_0]=1\), so if \(x \in \mathtt{range}(j_1+1,i_1)\) then \(L[x]=1\).

We have established in this case that \(P(i_1,j_1)\) is true.

Therefore we have shown that if \(P(i_0,j_0)\) is true for \(i_0,j_0\) at the beginning of the loop execution, then \(P(i_1,j_1)\) is true for \(i_1,j_1\) at the end of the loop execution.

## Exiting loop

Suppose that \(P(i,j)\) is true when the loop exits. Because \(i \leq \mathtt{len}(L)\), when the loop exits it is the case that \(i = \mathtt{len}(L)\). Thus \(P(i,j)\) being true means

- \(-1 \leq j < \mathtt{len}(L)\).
- If \(x \in \mathtt{range}(0,j+1)\), then \(L[x]=0\).
- If \(x \in \mathtt{range}(j+1,\mathtt{len}(L))\), then \(L[x]=1\).

It follows from this that \(L\) is sorted in non-descending order.

\(j \geq 0\) if and only if then there is at least one \(x\) such that \(L[x]=0\).

Thus when the loop exits, the postcondition is true.

This completes the proof that \(P(i,j)\) is a correct loop invariant for the given code.