Invariant

• An invariant is a statement that is true during all steps of a procedure.
• In other words, your invariant should be true at Step 0, Step 1, and Step 1000 of your algorithm.
• If the invariant holds for an arbitrary step of your algorithm, this means it’ll always hold! (What does that sound like?)
• If the invariant always holds, this means that it’ll hold at the last step, which tells us something about our output!

Proving our Invariant

• We’ve discussed that if we can prove our invariant, we can prove something about our output! So how do we prove our invariant?
• We can prove our invariant by induction!
• We are going to induct on our loop variable \(i\) and show that for every trip through the loop our invariant holds!

Step 1

We want to prove: \(\textrm{For input list } A,\)
\(\forall i \in [0, \verb|length(A)|],\)
\(\exists ~ \verb" max_idx" \in [0,i-1],\)
\(\forall j \in [0,i-1], \verb|A[max_idx]| \geq \verb|A[j]|\)

Step 2

We prove this by induction on \(i\).

1. INPUT: list A
2. max_idx = 0
3. max_elem = A[0]
4. n = length(A)
5. FOR i in [1,n-1]:
6.    IF A[i] > max_elem:
7.        max_elem = A[i]
8.        max_idx = i
9. RETURN max_idx

Step 3 (Base case)

• Before we enter the loop. (\(i=1\))
• \[\exists \verb" max_idx" \in [0,0], \textrm{ and } \forall j \in [0,0], \verb|A[max_idx]| \geq \verb|A[j]|\]
• Possible values for \(\verb"max_idx"\): \(0\)
• Possible values for \(j\): \(0\)
• \[\verb|A[0]| \geq \verb|A[0]| \square\]

Step 4

For a given \(i > 1\),

Step 5

Induction Hypothesis: I can assume, for all \(k\), with \(1 \leq k < i\), that \(\exists \verb" max_idx" \in [0,k-1],\) and \(\forall j \in [0,k-1],\) \(\verb|A[max_idx]| \geq \verb|A[j]|\)

Step 6

I want to prove that \(\exists \verb" max_idx" \in [0,i-1],\) and \(\forall j \in [0,i-1], \verb|A[max_idx]| \geq \verb|A[j]|\)

1. INPUT: list A
2. max_idx = 0
3. max_elem = A[0]
4. n = length(A)
5. FOR i in [1,n-1]:
6.    IF A[i] > max_elem:
7.        max_elem = A[i]
8.        max_idx = i
9. RETURN max_idx

• Line 6, we have an IF statement, so let’s do a proof by cases!
• Case 1: \(\verb"A[i]" > \verb"max_elem"\)
• Case 2: \(\verb"A[i]" \leq \verb"max_elem"\)

1. INPUT: list A
2. max_idx = 0
3. max_elem = A[0]
4. n = length(A)
5. FOR i in [1,n-1]:
6.    IF A[i] > max_elem:
7.        max_elem = A[i]
8.        max_idx = i
9. RETURN max_idx

Case 1: \(\verb"A[i]" > \verb"max_elem"\)

\[\begin{equation} \begin{array}{l l l} 1. & \forall k, 1 \leq k < i, \exists \verb" max_idx" \in [0,k-1], \textrm{ and } \forall j \in [0,k-1], \verb|A[max_idx]| \geq \verb|A[j]| & \textrm{Ind. Hypothesis}\\ 2. & \exists \verb" max_idx" \in [0,i-2], \textrm{ and } \forall j \in [0,i-2], \verb|A[max_idx]| \geq \verb|A[j]| & \textrm{Ind. Hypothesis with } k = i-1 \\ 3. & \verb|A[i] > max_elem| & \textrm{Case 1}\\ 4. & \verb|max_elem = A[i]| \land \verb|max_idx = i| & \textrm{Defined in code lines 4-6}\\ 5. & \forall j \in [0,i-1], \verb|A[max_idx]| \geq \verb|A[j]| & \textrm{Combined 2, 3, and 4} \end{array} \end{equation}\]

Case 2: \(\verb"A[i]" \leq \verb"max_elem"\)

\[\begin{equation} \begin{array}{l l l} 1. & \forall k, 1 \leq k < i, \exists \verb" max_idx" \in [0,k-1], \textrm{ and } \forall j \in [0,k-1], \verb|A[max_idx]| \geq \verb|A[j]| & \textrm{Ind. Hypothesis}\\ 2. & \exists \verb" max_idx" \in [0,i-2], \textrm{ and } \forall j \in [0,i-2], \verb|A[max_idx]| \geq \verb|A[j]| & \textrm{Ind. Hypothesis with } k = i-1 \\ 3. & \verb"A[i]" \leq \verb"max_elem" & \textrm{Case 2}\\ 4. & \forall j \in [0,i-1], \verb|A[max_idx]| \geq \verb|A[j]| & \textrm{Combined 2 and 3} \square \end{array} \end{equation}\]

Step 8

Therefore, we have proved
\(\textrm{For input list } A,\)
\(\forall i \in [0, \verb|length(A)|],\)
\(\exists \verb" max_idx" \in [0,i-1],\) \(\forall j \in [0,i-1], \verb|A[max_idx]| \geq \verb|A[j]|\) by induction.