Tuesday, 28 April 2015
Wednesday, 22 April 2015
- Concept Of hashing• Need of Hashing• Hash Collision• Dealing with Hash Collision• Resolving Hash Collisions by Open Addressing• Primary clustering• Double Rehash
- 2. Concept Of hashing• Hashing: hashing is a technique for performing almost constant time in case of insertion deletion and find operation.• taking a very simple example, as array with its index as key is the example of table.• So each index (key) can be used for accessing values in the constant search time.• Mapping key must be simple to compute and must help in identifying the associated records.• Function that help us in generating such type of keys is termed as Hash Function.
- 3. Hashing• let h(key) is hashing function that returns the hash code. h(key) = key%1000, which can produce any value between 0 and 999. as shown in figure:
- 4. Need of Hashing• Hashing maps large data sets of variable length to smaller data sets of a fixed length. For example, an inventory file of a company having more than 100 items and the key to each record is a seven digit part number. To use direct indexing using entire seven digit key, an array of 10 million elements would be required. Which clearly is wastage of space, since company is unlikely to stock more than few thousand parts.• Hence hashing provides an alternative to convert seven digit key into an integer within limited range. The values returned by a hash function are called hash values, hash codes.
- 5. • Suppose two keys k1 and k2 hashes such that h(k1) = h(k2). Here two keys hashes into the same value and are supposed to occupy same slot in hash table ,which is unacceptable.• Such a situation is termed as hash collision.
- 6. Dealing with Hash Collision• Two methods to deal with hash collision are:• Rehashing and Chaining Rehashing: invokes a secondary hash function (say Rh(key)), which is applied successively until an empty slot is found, where a record can be placed. Chaining: builds a Linked list of items whose key hashes to same value. During search this short linked list is traversed sequentially for the desired key. This technique requires extra link field to each table position.
- 7. hashingAnalysis:• The worst case running time for insertion is O(1).• Deletion of an element x can be accomplished in O(1) time if the lists are doubly linked.• In the worst case behaviour of chain-hashing, all n keys hash to the same slot, creating a list of length n. The worst-case time for search is thus θ(n) plus the time to compute the hash function.
- 8. A good hash function is one that minimizes collision and spreads the records uniformly throughout the table. that is why it is desirable to have larger array size than actual number of records. More formally, suppose we want to store a set of size n in a table of size m. The ratio α = n/m is called a load factor, that is, the average number of elements stored in a Table.
- 1. Prims Algorithm on minimum spanning tree
- 2. What is Minimum Spanning Tree?• Given a connected, undirected graph, a spanning tree of that graph is a subgraph which is a tree and connects all the vertices together.• A single graph can have many different spanning trees.• A minimum spanning tree is then a spanning tree with weight less than or equal to the weight of every other spanning tree.
- 3. graph GSpanning Tree from Graph G2 2 4 3 4 51 1 1
- 4. Algorithm for finding Minimum Spanning Tree• The Prims Algorithm• Kruskals Algorithm• Baruvkas Algorithm
- 5. About Prim’s AlgorithmThe algorithm was discovered in 1930 bymathematician Vojtech Jarnik and later independentlyby computer scientist Robert C. Prim in 1957.The algorithm continuously increases the size of atree starting with a single vertex until it spans all thevertices. Prims algorithm is faster on densegraphs.Prims algorithm runs in O(n*n)But the running time can be reduceusing a simple binary heap data structureand an adjacency list representation
- 6. • Prims algorithm for finding a minimal spanning tree parallels closely the depth- and breadth-first traversal algorithms. Just as these algorithms maintained a closed list of nodes and the paths leading to them, Prims algorithm maintains a closed list of nodes and the edges that link them into the minimal spanning tree.• Whereas the depth-first algorithm used a stack as its data structure to maintain the list of open nodes and the breadth-first traversal used a queue, Prims uses a priority queue.
- 7. Let’s see an example to understand Prim’s Algorithm.
- 8. Lets…. At first we declare an array named: closed list. And consider the open list as a priority queue with min-heap. Adding a node and its edge to the closed list indicates that we have found an edge that links the node into the minimal spanning tree. As a node is added to the closed list, its successors (immediately adjacent nodes) are examined and added to a priority queue of open nodes.
- 9. Total Cost: 0Open List: dClose List:
- 10. Total Cost: 0Open List: a, f, e, bClose List: d
- 11. Total Cost: 5Open List: f, e, bClose List: d, a
- 12. Total Cost: 11Open List: b, e, gClose List: d, a, f
- 13. Total Cost: 18Open List: e, g, cClose List: d, a, f, b
- 14. Total Cost: 25Open List: c, gClose List: d, a, f, b, e
- 15. Total Cost: 30Open List: gClose List: d, a, f, b, e, c
- 16. Total Cost: 39Open List:Close List: d, a, f, b, e, c
- 17. PSEUDO-CODE FOR PRIMS ALGORITHM Designate one node as the start node Add the start node to the priority queue of open nodes. WHILE (there are still nodes to be added to the closed list) { Remove a node from priority queue of open nodes, designate it as current node. IF (the current node is not already in the closed list) { IF the node is not the first node removed from the priority queue, add the minimal edge connecting it with a closed node to the minimal spanning tree. Add the current node to the closed list. FOR each successor of current node IF (the successor is not already in the closed list OR the successor is now connected to a closed node by an edge of lesser weight than before) Add that successor to the priority queue of open nodes; } }
- 18. Sample C++ Implementation• void prim(graph &g, vert s) { • int minvertex(graph &g, int *d) { • int v;• int dist[g.num_nodes];• int vert[g.num_nodes]; • for (i = 0; i < g.num_nodes; i++) • if (g.is_marked(i, UNVISITED)) {• for (int i = 0; i < g.num_nodes; i++) { • v = i; break;• dist[i] = INFINITY; • }• dist[s.number()] = 0; • for (i = 0; i < g.num_nodes; i++) • if ((g.is_marked(i, UNVISITED)) && (dist[i] < dist[v])) v = i;• for (i = 0; i < g.num_nodes; i++) {• vert v = minvertex(g, dist); • return (v); • }• g.mark(v, VISITED);• if (v != s) add_edge_to_MST(vert[v], v);• if (dist[v] == INFINITY) return;• for (edge w = g.first_edge; g.is_edge(w), w = g.next_edge(w)) {• if (dist[g.first_vert(w)] = g.weight(w)) {• dist[g.second_vert(w)] = g.weight(w);• vert[g.second_vert(w)] = v;• }• }• }• }
- 19. Complexity Analysis Minimum edge weight data Time complexity (total) structureadjacency matrix, searching O(V*V)binary heap and adjacency O((V + E) log(V)) = O(Elist log(V))Fibonacci heap and O(E + V log(V))adjacency list
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20.
Application One practical application of a MST
would be in the design of a network. For instance, a group of
individuals, who are separated by varying distances, wish to be
connected together in a telephone network. Because the cost between two
terminal is different, if we want to reduce our expenses, Prims
Algorithm is a way to solve it Connect all computers in a computer
science building using least amount of cable. A less obvious
application is that the minimum spanning tree can be used to
approximately solve the traveling salesman problem. A convenient formal
way of defining this problem is to find the shortest path that visits
each point at least once. Another useful application of MST would be
finding airline routes. The vertices of the graph would represent
cities, and the edges would represent routes between the cities.
Obviously, the further one has to travel, the more it will cost, so MST
can be applied to optimize airline routes by finding the least costly
paths with no cycles.
Recovering unallocated space of a USB flash drive.
Today I played with the latest SUSE Linux Live. I had not have a DVD drive and used USB flash drive instead. I wanted to reformat my flash drive, but suddenly found it that it had not been possible. The most of the disk space had been unallocated, and my Windows 8 did not allow me to use it.Microsoft DiskPart version 6.2.9200
Copyright (C) 1999-2012 Microsoft Corporation.
On computer: COMPUTER
DISKPART> list disk
Disk ### Status Size Free Dyn Gpt
-------- ------------- ------- ------- --- ---
Disk 0 Online 298 GB 0 B
Disk 1 Online 7509 MB 6619 MB
DISKPART> select disk 1
Disk 1 is now the selected disk.
DISKPART> clean
DiskPart succeeded in cleaning the disk.
DISKPART> create partition primary
DiskPart succeeded in creating the specified partition.
DISKPART> exit
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