1. Tirgul 10 Solving parts of the first two theoretical exercises: –T1 Q.1(a,b) –T1 Q.4(a,e) –T2 Q.2(a) –T2 Q.3(b) –T2 Q.4(a,b)

2. T1 Q.1(a) - pseudo-code for selection sort Selection-Sort(array A) for j = 0 to A.length - 1 k = smallest(A, j) swap(A, j, k) // swaps A[j] with A[k] return; // Returns the index of the smallest element of A[j..n] int smallest(array A, int j) smallest = j for i = j+1 to A.length - 1 if A[i] < A[smallest] smallest = i return(smallest)

3. T1 Q.1(b) - proof of correctness The loop invariant: induction plus “termination claim”. Induction Claim: After performing the j’th iteration, A[0..j-1] contains the j smallest elements in A, sorted in a non- decreasing order: For j = 1: The algorithm finds the smallest element and places it in A[0] The induction step: Suppose A[0..j-1] are the j smallest elements. We find the smallest element in A[j..N-1], and thus after the swap A[0..j] are the j+1 smallest elements. Since A[j] is larger than all elements in A[0..j-1] by assumption, then A[0..j] is still sorted. Termination: After n iterations, A[0..n-1] (the entire array) is sorted.

4. T1 Q.4(a) - solving recurrences We would like to prove a matching lower bound:

5. T1 Q.4(e) - solving recurrences

6. T1 Q.4(e) (another perspective) How did we guess that T(n) = O(n) ?? One way is to look at the recursion tree: for any node with work x, the work of all its children is (7/8)x. So the total work of some level is 7/8 times the total work of one level higher, and the first level has work n. So we get an upper bound to the total amount of work:

7. T2 Q.2(a) - implement queue using two stacks Denote the two stacks “L” and “R”. Enqueue: push the item to stack L. Dequeue: –If stack R is empty: while stack L is not empty, pop an element from stack L and push it to stack R. When stack L is emptied, pop an element from stack R and return it. –If stack R is not empty: pop an element from stack R and return it. Explanation: 1) If stack R is empty, then after moving the elements from L to R they are in a FIFO order. 2) As long as stack R is not empty, the top element is the first element entered the d.s. (additional enqueus does not change this), so it should be dequeued.

8. T2 Q.2(a) (continued) What is the running time? “Most” operations take O(1), occasionally we have O(n) when stack R is empty and we want a dequeue. If we do amortized analysis (the accounting method) it is enough to pay 3 for every enqueue and 1 for every dequeue, since, after being pushed to stack L, each element has to pay exactly 2 more before being dequeued - 1 for being popped from L and 1 for being pushed to R. Notice the run-time difference with respect to the more simple solution - always keep the elements in L, and for a dequeue, move the elements to R, return its top, and move the elements back to L. Here, all dequeues are O(n).

9. T2 Q.3(b) - the selection problem The selection problem: Given an array A, find the k’th smallest element in linear time (on average). Reminder from Quicksort: the procdure partition(A,p,r) first chooses a pivot element x from A[p..r], and then rearranges A[p..r] so that all elements before x are smaller than x and all elements after x are larger than x. It returns the new index of x (called q). RandomizedPartition(A,p,r) chooses the pivot randomly. This always takes O(n) time. What is the size of A[p..q-1] (on average)? If we chose the i’th smallest element then the size will be i-1. Since every element is chosen with probability 1/n then each size 0,..,n-1 has probability 1/n.

10. T2 Q.3(b) - the algorithm // Returns the k’th element in A[p..r] Select(A, p, r, k) q = RandomizedPartition(A, p, r) if (k == q-p+1) return A[k] else if (k < q-p+1) return Select(A, p, q-1, k) else return Select(A, q+1, r, k-(q-p+1))

11. T2 Q.3(b) - Average run time Suppose that the size of A[p..q-1] is i, and k causes us to enter the larger part of A. Since the probability of i is 1/n for i=0..n-1 and the partition takes about n operations: Assume by induction that T(n) < cn, so (using the arithmetic sum formula):

12. T2 Q.4 - check the heap property of a given (pointer based) complete binary tree A recursive code: boolean isHeap(Node t) if (t == null) then return TRUE boolean l = isHeap(t.leftChild()) boolean r = isHeap(t.rightChild()) return ( l && r && checkNode(t) ) boolean checkNode(Node t) if ( t.leftChild().val() > t.val() ) return FALSE if ( t.rightChild().val() > t.val() ) return FALSE return TRUE

13. T2 Q.4 (continued) An iterative code (using a stack): boolean isHeap(Node t) Stack s s.push(t) While ( ! s.empty() ) t = s.pop() if (t != null) if ( ! checkNode(t) ) return FALSE s.push(t.leftChild()) s.push(t.rightChild()) return TRUE