The file organization which allows us to read records that would satisfy the join condition by using one block read is a. Heap file organization b. Sequential file organization c. Clustering file organization d. Hash file organization
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| 167. | The file organization which allows us to read records that would satisfy the join condition by using one block read is |
| a. | Heap file organization |
| b. | Sequential file organization |
| c. | Clustering file organization |
| d. | Hash file organization |
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- Compute total no of blocks} Consider a file of 8192 records. Each record is 16 bytes long and its key field Is of size 6 bytes. The file is ordered on a key field, and the file organization is unspanned. The file is stored in a file system with block size 512 bytes, and the size of a block pointer is 10 bytes. If the primary index is built on the key field of the file, and a multilevel index scheme is used to store the primary index, number of first-level and second level blocks in the multilovel index are the respectively 16 and 1 b. 32 and 1 16 and 2 8 and 1 c. d.A Ring, refers to a record chain, the last of which refers to the first record, in the chain, is called a pointer wxplain?Program in Java: topic: insertion sort The state and variable display: Unsorted Partition Reference Index: 7, Traversing Index: 1, Current Traversing Index Value: avocado, Condition Value Index: 0 Current Condition Value: apple, Swapping Condition: FalseCurrent Array: ['apple', 'avocado', 'orange', 'banana', 'strawberry', 'pineapple', 'plum', 'mango'] The Input must be:["apple", "avocado", "orange", "banana", "strawberry", "pineapple", "plum", "mango"]The assumptions and conditions: Sorted partition is on the right side.The order is increasing alphabetically but the strings that has an odd length should be the first to be sorted then the strings with the even length goes next The result or Output must be:['apple', 'avocado', 'mango', 'pineapple', 'banana', 'orange', 'plum', 'strawberry']
- 14. The unit of storage that can store one are more records in a hash file organization are a. Buckets b. Disk pages c. Blocks d. NodesA sorted file has the following details: Number of records = 13,10,720; record length = 200 bytes; size of the primary key field = 13 bytes; size of block pointer = 7 bytes; block size = 2048 bytes. For this file, we need to build a multi-level index on the primary key. The number of blocks in the first level index file isGive the benefits of employing pointers in file organization over sequential file structure.
- 20. The file organization that provides very fast access to any arbitrary record of a file is a. Ordered file b. Unordered file c. Hashed file d. B-tree1:34 b a 2. Different operating systems use different file allocation techniques to store files in the secondary storage device such as HDD. One of the techniques is "Linked allocation". In this technique, each file is divided into several data blocks of same size. Each data block has a link to its next block like a singly linked list. So, the data blocks are scattered on the disk. The directory contains a pointer to the starting and the ending data block of a file. The diagram below depicts a sample linked allocation system of a file Abc.txt: Directory entry File Start End Abc.txt 100 900 Data 900 Data Data block 1 block 2 Data block 3 block N Write a Java program to implement the linked allocation of a given text file Assgn1_2.txt (uploaded with the assignment) who's each block will contain maximum 50 characters. Read the content of the given file and store it in data blocks (maximum 50 characters per block). Now show a menu that will ask the user to i) add new content at the end of…1. A hashing function converts a large ___ to a small ___. 2. A ___ occurs when two keys hash to the same address. 3. For an open hashing scheme, the records within each bucket are stored in key order ( true / false ). 4. For an open hashing scheme, the most recently inserted record will be on the front end of the bucket list ( true / false )
- MCQ: Which of the following statements is FALSE? a. The internal fragmentation can be used to grow and shrink the file. b. External fragmentation is a disadvantage of both Linked and FAT allocation algorithms. c. Internal and external fragmentation can occur in the contiguous Allocation method. d. Chaining through the FAT is faster than chaining through a linked list.Part Two: Implementing a rainbow table You are to write a program, in C/C++, Java or Python, that should run using the following instruction: $ ./Rainbow Passwords.txt where the file Passwords.txt contains a list of possible passwords. The password file contains a password per line, as in the provided words file and consists of strings of printable characters. Any password used must be taken from this file, so the only stored hash information needs to relate to those entries in the file. The program is used to find pre-images for given hash values. Rainbow tables can be used to solve pre-image problems for hash functions. At the simplest level they can simply be a list of hash values and the corresponding pre-images, often from some dictionary. This can be expensive in terms of storage space however, and a more efficient way of identifying pre-images involves the use of the hash function and reduction functions. First step The process is as follows: 1. Read in the list of possible…Please write in C/C++Write a program that spawn 2 thread: a producer and a consumer The producer thread MUST open an input file and repeatedly copy values to a CIRCULAR BUFFER The consumer should open an output file and repeatedly copy values from the same circular buffer to the file. If the producer is unable to write to the buffer (because it does not contain enough empty elements), or if the consumer is unable to read from the buffer (because it does not contain enough unread items), then it should proceed to the next iteration, choosing a new random number of bytes to copy.