Loop invariant of linear search

In the case of linear search, the loop variant will be the backing store used for saving the index(output) .

Lets name the backing store as index which is initially set to NIL.The loop variant should be in accordance with three conditions :

• Initialization : This condition holds true for index variable.since, it contains NIL which could be a result outcome and true before the first iteration.
• Maintenance : index will hold NIL until the item v is located. It is also true before the iteration and after the next iteration.As, it will be set inside the loop after comparison condition succeeds.
• Termination : index will contain NIL or the array index of the item v.

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After you have looked at index i, and not found v yet, what can you say about v with regard to the part of the array before i and with regard to the part of the array after i?

LINEAR-SEARCH(A, ν)
1  for i = 1 to A.length
2      if A[i] == ν 
3          return i
4  return NIL 

Loop invariant: at the start of the ith iteration of the for loop (lines 1–4),

∀ k ∈ [1, i) A[k] ≠ ν.  

Initialization:

i == 1 ⟹ [1, i) == Ø ⟹ ∀ k ∈ Ø A[k] ≠ ν,

which is true, as any statement regarding the empty set is true (vacuous truth).

Maintenance: let's suppose the loop invariant is true at the start of the ith iteration of the for loop. If A[i] == ν, the current iteration is the final one (see the termination section), as line 3 is executed; otherwise, if A[i] ≠ ν, we have

∀ k ∈ [1, i) A[k] ≠ ν and A[i] ≠ ν ⟺ ∀ k ∈ [1, i+1) A[k] ≠ ν,

which means that the invariant loop will still be true at the start of the next iteration (the i+1th).

Termination: the for loop may end for two reasons:

1. return i (line 3), if A[i] == ν;

2. i == A.length + 1 (last test of the for loop), in which case we are at the beginning of the A.length + 1th iteration, therefore the loop invariant is

   ∀ k ∈ [1, A.length + 1) A[k] ≠ ν ⟺ ∀ k ∈ [1, A.length] A[k] ≠ ν
   

   and the NIL value is returned (line 4).

In both cases, LINEAR-SEARCH ends as expected.