2. Loop invariants, Merge Sort, Inductive Proofs

Datastructures and Algorithms Spring-Summer 2011/12
Lab
Valeria Fionda, Mouna Kacimi,
Werner Nutt, Simon Razniewski
2. Loop invariants, Merge Sort, Inductive Proofs
1. Loop invariants: SelectKth
1. Inner Loop:
• mini points to the smallest element in the subarray [i, .., j − 1]
• Formally,
∀l : i ≤ l ≤ j − 1 =⇒ A[mini] ≤ A[l]
2. Outer Loop
• All elements to the left of i are ordered
• All elements to the left of i are smaller than elements to the right of i
(or, A[i − 1] is the smallest among A[i − 1]..A[n])
• Formally,
∀l : 1 < l ≤ i − 1 =⇒ A[l − 1] ≤ A[l] ∧
∀m : 1 ≤ i − 1 ≤ m ≤ n =⇒ A[i − 1] ≤ A[m]
3. Prove that at the end of the algorithm the k-smallest element of the input
array is the one at position k.
• Initialization: i = 1, j = 2. The invariants holds because there are no
elements to the left of i. i = 1 yields ∀i ∈ [1..0] : A[i] < A[j], ∀j > i.
• Maintenance Consider the iteration i, the invariants of the outer loop
say that ∀l : 1 < l ≤ i − 1 =⇒ A[l − 1] ≤ A[l] ∧
∀m : 1 ≤ i − 1 ≤ m ≤ n =⇒ A[i − 1] ≤ A[m]
At the end of the inner loop we have that j = n+1, and thus according
to the invariant of the inner loop (∀l : i ≤ l ≤ j − 1 =⇒ A[mini] ≤
A[l]) we have that ∀l : i ≤ l ≤ n =⇒ A[mini] ≤ A[l]. At the
end of the loop we exchange the element in position i with the one
in position mini thus, by exchanging A[i] with A[mini] it must hold
∀k : i ≤ k ≤ n =⇒ A[i] ≤ A[k]. Moreover from the first invariant
of the outer loop we have that ∀l : 1 < l ≤ i =⇒ A[l − 1] ≤ A[l].

• Termination At the end i = k + 1. Thus according to the invariants
of the outer loop we have ∀l : 1 < l ≤ k =⇒ A[l − 1] ≤ A[l] ∧
∀m : 1 ≤ k ≤ m ≤ n =⇒ A[k] ≤ A[m]
As for the invariant of the inner loop, considering i = k, we have:
∀m : k ≤ m ≤ n =⇒ A[k] ≤ A[m] (result of inner loop),
2. Algorithm Correctness: Merging of Arrays
1. Pseudocode:
MERGE (A, B)
1 i := 1
2 j := 1
3 k := 1
4 n1 := A.length
5 n2 := B.length
6 while i

1 i := 1
2 j := 1
3 for k := 1 to n1 + n2 do
4 if A[i] n1
or j > n2, and thus, when evaluating the condition “if A[i] n2 or (i

3. Inductive Proofs: Fibonacci numbers
• Base case: n = 1 → 3n = 3 → fib(3n) = fib(3) = 2 is even.
• Inductive step: Suppose fib(3k) be even. We have that fib(3(k + 1)) =
fib(3k + 3) = fib(3k + 2) + fib(3k + 1) = fib(3k + 1) + fib(3k) +
fib(3k + 1) = 2 · fib(3k + 1) + fib(3k). Since fib(3k) is even due to the
induction hypothesis, fib(3(k + 1)) = fib(3k + 3) is even.