Loop invariant of linear search

Question

As seen on Introduction to Algorithms (http://mitpress.mit.edu/algorithms), the exercise states the following:

Input: Array A[1..n] and a v

Answer 1

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.

.

Answer 2

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.

Answer 3

The invariant for linear search is that every element before i is not equal to the search key. A reasonable invariant for binary search might be for a range [low, high), every element before low is less than the key and every element after high is greater or equal. Note that there are a few variations of binary search with slightly different invariants and properties - this is the invariant for a "lower bound" binary search which returns the lowest index of any element equal to or greater than the key.

Source:https://www.reddit.com/r/compsci/comments/wvyvs/what_is_a_loop_invariant_for_linear_search/

Seems correct to me.

Answer 4

Loop invariant would be

forevery 0 <= i < k, where k is the current value of the loop iteration variable, A[i] != v

On loop termination:

if A[k] == v, then the loop terminates and outputs k

if A[k] != v, and k + 1 == n (size of list) then loop terminates with value nil

Proof of Correctness: left as an exercise

Answer 5

Assume that you have an array of length i, indexed from [0...i-1], and the algorithm is checking if v is present in this array. I'm not totally sure, but I think, the loop invariant is as follows: If j is the index of v, then [0..j-1] will be an array that does not have v.

Initialization : Before iterating from 0..i-1, the current array checked (none), does not consist of v.

Maintenance : On finding v at j, array from [0..j-1] will be an array without v.

Termination : As the loop terminates on finding v at j, [0..j-1] will be an array without j.

If the array itself does not have v, then j = i-1, and the above conditions will still hold true.

Answer 6

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?