Search Results

The question is: "Suppose we perform a sequence of stack operations on a stack whose size never exceeds k. After every k operation, we make a copy of the entire stack for backup purposes. Show that the cost of n stack operations, including copying the stack, is O(n) by assigning suitable amortized costs to the various stack operations." We will assign the following amortized costs: Push uses one credit to pay for itself and saves one credit for future pops and one for copying the stack. Pop and Multipop pay for their operations using saved Push credits and save a credit for stack copying. After k operations, we have saved k credits exclusively for stack copying and can copy the stack for free. Since each operation costs at most O(1) amortized and the credits are nonnegative, the cost for n operations is O(n).

Let's take the left child of the node indexed by ⌊𝑛/2⌋+1⌊n/2⌋+1. LEFT(⌊𝑛/2⌋+1) =2(⌊𝑛/2⌋+1) >2(𝑛/2−1)+2 =𝑛−2+2 =n.​ Since the index of the left child is larger than the number of elements in the heap, the node doesn't have childrens and thus is a leaf. The same goes for all nodes with larger indices. Note that if we take element indexed by ⌊n/2⌋, it will not be a leaf. In case of even number of nodes, it will have a left child with index n and in the case of odd number of nodes, it will have a left child with index n−1 and a right child with index n. This makes the number of leaves in a heap of size n equal to ⌈n/2⌉.

The question is: "What is the minimum number of leaves in the decision tree for a comparison sort? Use this observation to derive a lower bound on the number of comparisons performed by a comparison sort in the worst case." In a comparison sort, the decision tree represents all the possible comparisons made between elements during the sorting process. Each leaf node in the decision tree represents a possible permutation of the input elements after sorting. For 𝑛n distinct elements, there are 𝑛! possible permutations. Each permutation corresponds to a unique leaf in the decision tree. The minimum number of leaves in the decision tree for a comparison sort is equal to the number of permutations of 𝑛n distinct elements, which is 𝑛!. Now, let's derive a lower bound on the number of comparisons performed by a comparison sort in the worst case based on this observation. Suppose a comparison sort algorithm constructs a decision tree to sort 𝑛n distinct elements. In the worst-case scenario, the algorithm must traverse the entire decision tree to find the correct permutation (i.e., the sorted order). Since each leaf in the decision tree represents a unique permutation of the input elements, the worst-case number of comparisons performed by the algorithm is at least equal to the depth of the decision tree. In a complete binary tree, the number of leaves is 2^d, where d is the depth of the tree. So, for n! leaves, we have: Taking the logarithm base 2 of both sides: Using Stirling's approximation, we have,

The smallest possible depth of a leaf in a decision tree for a comparison sort is θ(log n), where n is the number of elements being sorted. This depth corresponds to the best-case scenario in terms of the number of comparisons needed to sort the elements. A sorting technique that can achieve this best-case depth is Merge Sort. Merge Sort is a comparison-based sorting algorithm that divides the array into halves, recursively sorts each half, and then merges the sorted halves. The depth of the decision tree in Merge Sort is proportional to log⁡𝑛logn because the algorithm repeatedly divides the array in half, leading to a balanced binary recursion tree where the height (or depth) of the tree is log⁡𝑛logn.

The given question is: "Let G = (V,E) be a directed unweighted graph. Given two vertices s and t in V, what is the time required to determine if there exists at least one s-t path in G? Can we use the DFS algorithm to find the shortest-path distance from the s to t? If yes, justify, otherwise give a counter-example." Either Breadth First Search (BFS) or Depth First Search (DFS) can be used to find path between two vertices. Take the first vertex as a source in BFS (or DFS), follow the standard BFS (or DFS). If the second vertex is found in our traversal, then return true else return false. The time complexity of this algorithm is O(V+E), where V is the number of vertices and e is the number of edges. Breadth-First search can be useful to find the shortest path between nodes, and depth-first search may traverse one adjacent node very deeply before ever going into immediate neighbours. DFS may or may not find the shortest path from s to t. An example of this is shown below: Here, we have to find a path from A to D. Now, we know that the shortest path is A→E→D. However, Depth First Search might start exploring A→B→C→D path first and it will return this path, which is not the shortest path between A and D.

The question is stated as follows: A student was asked to sort a list of n numbers in decreasing order. The student writes an algorithm that works iteratively as follows. In every iteration, the following two steps are done: Linear search is used to find the maximum element in the portion of the array which is not yet sorted. The maximum element found in step 1 is placed at the beginning of the not-yet-sorted portion of the array. This algorithm is given as input a list already sorted in descending order. What would be the time complexity of the algorithm on this input? Explain. The algorithm described by the student is a variant of the selection sort algorithm, adapted to sort in decreasing order. To understand its time complexity when given a list that is already sorted in descending order, let's analyze the steps involved. In the first iteration, the algorithm searches through the entire list of 𝑛n elements to find the maximum, which is the first element (since the list is already sorted in descending order). In the second iteration, the algorithm searches through the remaining n−1 elements. This continues until the last iteration, where only one element is left. Since the list is already sorted, the maximum element found in each iteration is already in its correct position. Thus, the placement step involves no movement or change in the list. Even though the input list is already sorted in descending order, the algorithm still performs a linear search for the maximum element in each iteration. Therefore, the time complexity remains O(n²) This inefficiency arises because the algorithm does not take advantage of the already sorted order and continues to perform redundant searches.

Let us first discuss about merge sort. Merge sort is a stable sorting algorithm. Stability in sorting algorithms means that two equal elements retain their relative positions after sorting. Merge sort achieves this by ensuring that when two elements are compared and found to be equal, the algorithm does not change their order in the merged array. This is particularly useful when the data has multiple fields and the sorting is performed based on one of these fields while preserving the relative order of equal elements. Merge sort is not an in-place sorting algorithm. An in-place algorithm sorts the elements within the original data structure using only a constant amount of extra space. Merge sort, however, requires additional space proportional to the size of the array being sorted because it creates temporary arrays for merging. The space complexity of merge sort is 𝑂(𝑛)O(n), where 𝑛n is the number of elements in the array. Now, let us discuss about insertion sort. insertion sort is a stable sorting algorithm. As elements are inserted into their correct positions, the algorithm ensures that equal elements retain their relative order. This stability is achieved because when inserting an element into its proper position, the algorithm shifts all larger elements one position to the right, preserving the relative order of equal elements. Insertion sort is an in-place sorting algorithm. It sorts the array within the original data structure using a small, constant amount of extra space. The algorithm only requires a few additional memory locations for temporary variables used during the insertion process. Thus, its space complexity is O(1), which qualifies it as an in-place algorithm.

Let us look at the entire question: "Consider the scheduling problem wherein you are given a single resource and a set of requests having deadlines. A request is said to be late be late if it misses the deadline, Your goal is to minimize the maximum lateness. With respect to a schedule S, idle time is defined as the time during which the resource is idle, in between two requests. S is said to have an inversion when request i has been scheduled before j and d(i) > d(j), where d(i) and d(j) are the deadlines of the requests i and j respectively. Argue that all schedules with no idle time and no inversions have the same maximum lateness." To minimize maximum lateness, we use the EDD Rule. The Earliest Due Date (EDD) rule is a scheduling heuristic used to minimize the maximum lateness in a set of jobs. The rule is straightforward: it schedules jobs in order of their due dates, starting with the job that has the earliest due date and proceeding in ascending order of due dates. Consider a scenario with three jobs: Job A: deadline = 5, Processing Time=2 Job B: deadline = 3, Processing Time=3 Job C: deadline = 6, Processing Time=1 According to EDD rule, the correct order should be B, A, C. Here, scheduling A before B would be an inversion since it leads to higher maximum lateness compared to the optimal order of B, A, C. An inversion occurs when a job that is scheduled earlier should ideally be scheduled later to minimize maximum lateness. It often means a job with a later due date is scheduled before a job with an earlier due date, contrary to the optimal ordering. Time during which the resource is idle between two requests. A schedule with no idle time means the resource is continuously processing requests without any gaps. Now, let's consider two schedules S1​ and S2​ that have no idle time and no inversions. In both schedules, requests are processed continuously, and they are ordered by non-decreasing deadlines. Since there are no inversions, both schedules will have the same ordering of requests. Since both schedules are continuously working and following the same order of deadlines, they effectively do the same thing. They start with the task that has the earliest deadline and move on to the next, always picking the next task that is due the soonest. Because they follow these same rules, the order in which tasks are completed will be the same in both schedules. As a result, the latest any task finishes relative to its deadline (the maximum lateness) will be the same in both schedules.

A binary heap is a complete binary tree which satisfies the heap property. For a max-heap, this property is: For every node 𝑖, the value of 𝑖 is greater than or equal to the values of its children. Formally, if 𝐴A is an array representation of the heap and 𝑖i is a node, then: 𝐴[𝑖]≥𝐴[2𝑖+1] and 𝐴[𝑖]≥𝐴[2𝑖+2] (assuming a 0-based array indexing). To prove that the root of any subtree in a max-heap contains the largest value in that subtree, let's consider a node 𝑖 in the max-heap and the subtree rooted at 𝑖. According to the max-heap property, for the root node 𝑖, 𝐴[𝑖]≥𝐴[2𝑖+1] and 𝐴[𝑖]≥𝐴[2𝑖+2] This means 𝐴[𝑖] is greater than or equal to its immediate children. Since the subtree rooted at any child of 𝑖 is also a max-heap, the root of these subtrees will be greater than or equal to all their respective children. Thus, by recursive application of the heap property: 𝐴[2𝑖+1]≥all descendants of 2𝑖+1 𝐴[2𝑖+2]≥all descendants of 2𝑖+2 Thus, A[i]≥all descendants of i Thus, the root of any subtree in a max-heap contains the largest value occurring anywhere in that subtree.

The theorem is that if edge weights are distinct then a graph will have only one minimum spanning tree. However, if the edge weights are not necessarily distinct, then the graph may have more than one spanning tree. For example, let us consider the following graph: This graph has three minimum-spanning trees. Let us consider two of them With this example, we have proved that minimum spanning tree need not be unique.

The worst-case scenario for bucket sort occurs when the distribution of elements is highly skewed, causing most of the elements to be placed in a single bucket. If all elements fall into one bucket, the sorting algorithm applied within that bucket determines the overall time complexity. If that sorting algorithm is insertion sort, the time complexity for sorting n elements within the bucket would be θ(n²). Instead of using insertion sort (which has a worst-case time complexity of θ(n²)), we can use a more efficient sorting algorithm like merge sort or quicksort. Both merge sort and quicksort have an average-case time complexity of O(nlogn), and merge sort has a worst-case time complexity of O(nlogn). This change ensures that even if one bucket ends up with most of the elements, the sorting within that bucket is efficient.

The given question is: Consider the following graph: Specify whether the above graph is bipartite or not. If yes, give the partition, else justify. A bipartite graph is a graph in which the vertices can be divided into two disjoint sets, such that no two vertices within the same set are adjacent. In other words, it is a graph in which every edge connects a vertex of one set to a vertex of the other set. If all the vertices of the graph can be colored only using two colors, such that no two adjacent vertex have the same colour, the graph is said to be bipartite. Let's use the colors - RED and BLUE. As we can see, the graph can be colored using only two colors and no two adjacent vertex have the same color. Hence, we can say that the given graph is bipartite. The graph can be partitioned on basis of colour - all red vertices together and all blue vertices together. The partition is given as follows: SET A: 1,6,7,4 SET B: 2,5,3,8