...being un-supervised. Other people in the community speed over the limit posted and children play there and may be hit or injured. Even though the children play in the street it cause a safety risk to drivers that don’t speed or drive wreck less do to the children running in and out of the streets without looking for cars. The children are our future and they need to be protected. When a person is looking to buy a future home they want to see a nice neighborhood that looks respectable and not as if it’s a bad neighborhood. The appearance of a home in my belief can say a lot about the people who live there. Someone who takes care of their home for example pressure washing, not trash in yard, paint nicely applied. Also there are some resident that don’t have the money but the residents should do as much as they can. To give an example of this a resident in my neighborhood has fixed the garage door which is tilted half way closed and it’s not very appealing. The appearance of a residents lawn as well as home its self make a community. A yard or lawn care should at the minimum have the grass mowed. The plants should be not over grown or deceased. Living in Florida the grass isn’t always...
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...The poor people usually depends of this places for basic things like food or sleep,without that help some of them don`t survive winter. Most of these institutions exist thanks to donations made it by average people and not government assistance. Every day there is more people living on the streets and the help is simply not enough, even when these kind of institutions exist. The people which suffer of poverty even when we see them like a kind of dog in the street, they are human and they want a decent life like normal people. Is normal to say that those poor people are living on the streets by their own choice, but like I said before, that is ignorance. That people suffer a lot of things, even things that we never going to feel, for example the hungry . The sad of these situations is that most of these people is...
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...Search Quiz ICS171 Name___________ ID____________ No text, no notes, no questions. Do the best that you can on each question. No questions will be answered about the quiz questions. If you think a question is ambiguous, write your interpretation and answer your modified question. Be reasonable. The following abbreviations are used: BF = branching factor, DFS = depth first search, BFS = breadth first search, IDS = iterative deepening search, A* = A* search, LI = local improvement search, HC = hill-climbing search. [pic] 1. For the 8-tile puzzle, what is the average branching factor, assuming the blank is equally likely to occur in any position. Show your work. (4*2 + 4*3 +1*4)/9 = 24/9 = 2 & 2/3. 2. Suppose that you are solving the 8-tile puzzle where it has solution. Which of the methods (DFS, BFS, IDS) is guaranteed to find a solution, assuming no computational limits are reached. List all that are correct. DFS, BFS, IDS 3. For the same puzzle, which of the methods (DFS,BFS,IDS) is guaranteed to find the shortest solution? List all that are correct. BFS, IDS 4. For the same puzzle, which methods are guaranteed to use no more than O(BF * length of solution) amount of memory. DFS, IDS 5. Suppose you apply the A* algorithm to the same problem. You decide to let f = current cost of the path. Would it be appropriate to let h = 0 for all states? Yes or no and why. ...
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...Data Structures & Algorithms Coursework Assignment 1 Q1. (a) Algorithm swap(x, y): Node n head While (n.getNext () != x ) do n n.getNext() Node v y.getNext () n.setNext(y) y.setNext(x) x.setNext(v) (b) Algorithm swap Doubly(x, y): DNode n x.getPrev() DNode v y.getPrev() n.setNext(y) y.setPrev(n) y.setNext(x) x.setPrev(y) x.setNext(v) v.setPrev(x) (c) The run time complexity for the singled linked algorithm is O (n) and for the doubly linked algorithm is O (1). Doubly linked list has the best time complexity. Time complexity in singly linked list take more time because we have to move from head to the node before x Q2. (b) RedBlueStack implements Stack{ protected Object A[]; Int capacity; int top = -1; RedBlueStack(int cap) { A = new Object [capacity]; capacity = cap; } int size() { return (top + 1); } void push(Object obj) throws FullStackException { if (size() == capacity) throws new FullStackException("Stack is full."); A[++top] = obj; } Object top() throws EmptyStackException { if (isEmpty()) throws new EmptyStackException("Stack is empty."); return A[top]; } Boolean isEmpty() { return (top < 0); } Object top() throws EmptyStackException { if (isEmpty()) throws new EmptyStackException("Stack is empty."); return A[top]; } Object pop() throws EmptyStackException { Object elem; if (isEmpty()) throws new EmptyStackException("Stack...
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...Task 1 LO 1.1 All high-level programming languages support the concept of data types. A data type defines a set of values that a variable can store along with a set of operations. Data types are used to store various types of data which is managed by program. Data type attaches with variable to determine the number of bytes to be allocate to variable and valid operations which can be performed on that variable. Although C has several built-in data types, it is not a strongly typed language, as are Pascal and Ada. C supports various data types and here some common data type as character, integer and floating-point types. C defines five foundational data types as defined below: ▪ character ▪ integer ▪ floating-point ▪ double floating-point ▪ valueless These are declared as by char, int, float, double, and void, respectively. These types form the basis for some other types also. The extent and choice of these and data types may contrast amongst processor natures and compilers. However, in all belongings an object of type char is 1 byte. C stores character type inside as an integer. Each character has 8 bits so, we can have 256 different characters values (0-255). Character set is used to map between an integer value and a character. The size of an int is ordinarily the same as the word length of the implementation setting of the program. C has 3 classes of integer storage namely short int, int and long int. All of these data types...
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...Write the following queries in SQL: Display a list of all instructors, showing their ID, name, and the number of sections that they have taught. Make sure to show the number of sections as 0 for instructors who have not taught any section. Your query should use an outer join, and should not use scalar subqueries. By using the university schema provided by db-book.com the following queries were done on the university database. The first query uses an outer join which works similar to the join operation but it keeps the rows that don’t match between the two tables that would be lost in a join operation. There are three forms of outer join: a) Left outer join displays the results from the left table even if the condition does not find any matching record in the right table. b) Right outer join will displays the results from the right table regardless if there is matching data in the left table. c) Full outer join will retain all rows from both tables, regardless if the data matches or not. The group by clause when used in a select statement collects data from multiple records and groups the results into one or more columns. The below is the query using left outer join operation: select ID, name, count(sec_id) as Number_of_sections from instructor natural left outer join teaches group by ID, name; The next query was written by using a scalar subquery without using an outer join operation. A scalar subquery is where the output of a subquery returns only...
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...i SASTRA UNIVERSITY Shanmugha Arts, Science, Technology and Research Academy Thirumalaisamudram Thanjavur – 613 402 BCSCCS705 PARALLEL COMPUTING AND ALGORITHMS LAB B.TECH (COMPUTER SCIENCE AND ENGINEERING) 7TH SEMESTER ii List of Exercises: 1. Basic arithmetic operations in parallel 2. Find out factorial of a number 3. Generation of Fibonacci series, finding prime numbers in an interval. 4. Evaluate the integral of a function 5. Merging of two sorted lists 6. Parallel tree traversals 7. Matrix multiplication 8. Enumeration sort 9. Odd-Even transposition sort 10. Bitonic merge 11. Quick sort 12. Single source shortest path iii Exercise 1 Aim: Basic Arithmetic Operations in parallel To perform a set of arithmetic operations in parallel using a cluster of computers Procedure: Step 1: Identify a set of numbers over which arithmetic operations are to be done Step 2: Identify the rank in the cluster Step 3: Split the set of numbers into domains and assign each domain to their corresponding processor Step 4: Get the result from each computer in the cluster and assimilate them at the master Step 5: Verify the time taken for the completion of each task. Algorithm:function sum(+,identity,a) = if #a == 1 then [identity] else let e = even_elts(a); o = odd_elts(a); s = scan_op(op,identity,{op(e,o): e in e; o in o}) in interleave(s,{op(s,e): s in s; e in e}); iv Input:- An array of integers Output: - The sum 3,2,7,6 P1 18 0,5,4,8 P2 17 62 2,0,1,5 P3 8...
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...*****Library for OleDb command and Connection Imports System.Data.OleDb ** *****Declaration for command, connection and Reader Public dcom As OleDbCommand Public dcon As OleDbConnection Public dbrd As OleDbDataReader Public sql As String ** ******Connection String *******2000 – 2003 Version (.mdb format) **A. dcon = New OleDbConnection("provider=microsoft.jet.oledb.4.0;data source = " & Application.StartupPath & "\maindata.mdb") dcon.Open() **B. dcon = New OleDbConnection("provider=microsoft.jet.oledb.4.0;data source = C:\Folder\DSource.mdb") dcon.Open() *******2007-2013 Version (.accdb format) dcon = New OleDbConnection ("Provider=Microsoft.ACE.Oledb.12.0;Data Source= " & Application.StartupPath & "\maindata.accdb") dcon.Open() ******2007-2003 Version (.xlsx format or excel) dcon = New OleDbConnection ("Provider=Microsoft.ACE.Oledb.12.0;Data Source= C:\Folder\DSource.xlsx"; Extended Properties=""Excel 12.0;HDR=Yes;IMEX=1"";") dcon.Open() SQL Query ******Add sql = "insert into TableName (Field1, Field2, Field3) values ('" & ValueIsString & "', #" & ValueIsDate & "#," & ValueIsNumber & ")" dcom = New OleDbCommand(sql, dcon) dcom.ExecuteNonQuery() *****Update sql = "update TableName set Field1= '" & ValueIsString & "', Field2= #" & ValueIsDate & "#, Field3= "#," & ValueIsNumber & " where ID = '" & ID & "'" dcom = New OleDbCommand(sql, dcon) dcom.ExecuteNonQuery() *****Delete sql = "DELETE FROM TableName WHERE...
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...Public Class RBTree Public root As RBTNode Dim null As RBTNode Public Sub New(ByVal data As Integer) null = New RBTNode("-1") null.color = "black" root = New RBTNode(data) root.color = "black" root.Left = null root.Left.parent = root root.right = null root.right.parent = root End Sub Public Sub New() End Sub Public Sub Insert(ByVal data As Integer) null = New RBTNode("-1") null.color = "black" 'step one, insert a red node in the tree If root Is Nothing Then root = New RBTNode(data) root.color = "black" root.Left = null root.Left.parent = root root.right = null root.right.parent = root Else Dim newNode As New RBTNode(data) newNode.Left = null newNode.Left.parent = newNode newNode.right = null newNode.right.parent = newNode Dim current As RBTNode current = root Dim parent As RBTNode While (True) parent = current If (data < current.data) Then current = current.Left If (current.data < 0) Then parent.Left = newNode newNode.parent = parent Exit While End...
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...Table Name : Employee EMPLOYEE _ID FIRST_NA ME LAST_NA ME SALA RY JOINING_D ATE DEPARTME NT 1 John Abraham 1000000 01-JAN-13 12.00.00 AM Banking 2 Michael Clarke 800000 01-JAN-13 12.00.00 AM Insurance 3 Roy Thomas 700000 01-FEB-13 12.00.00 AM Banking 4 Tom Jose 600000 01-FEB-13 12.00.00 AM Insurance 5 Jerry Pinto 650000 01-FEB-13 12.00.00 AM Insurance 6 Philip Mathew 750000 01-JAN-13 12.00.00 AM Services 7 TestName1 123 650000 01-JAN-13 12.00.00 AM Services 8 TestName2 Lname% 600000 01-FEB-13 12.00.00 AM Insurance Table Name : Incentives EMPLOYEE_REF_ID INCENTIVE_DATE INCENTIVE_AMOUNT 1 01-FEB-13 5000 2 01-FEB-13 3000 3 01-FEB-13 4000 1 01-JAN-13 4500 2 01-JAN-13 3500 SQL Queries Interview Questions and Answers on "SQL Select" 1. Get all employee details from the employee table Select * from employee 2. Get First_Name,Last_Name from employee table Select first_name, Last_Name from employee 3. Get First_Name from employee table using alias name “Employee Name” Select first_name Employee Name from employee 4. Get First_Name from employee table in upper case Select upper(FIRST_NAME) from EMPLOYEE 5. Get First_Name from employee table in lower case Select lower(FIRST_NAME) from EMPLOYEE 6. Get unique DEPARTMENT from employee table select...
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...the rows that are in the first query but not the second by removing the rows that are only found in the second query. There are three primary SQL commands involved when implementing a Union, Intersection and difference relational operators. As you may know, SQL data manipulation commands are set-oriented which are involved in operating over entire sets of rows and columns in tables at once. The UNION, INTERSECT, and MINUS statements make sure these operations occur. Union, Intersect and Minus only work properly if relations are Union-Compatible, which is based on the names of the relation attributes that must be the same and their data types must be alike. Being compatible does not mean the data types have to be exactly the same. For example, both data types can be used to store numeric values such as NUMBER and SMALLINT as well as character (string) values such as VARCHAR and CHAR. An SQL IN statement can be used with where clause to list a set of matching records of a table. We can use SQL IN query with a NOT combination to filter out some records. We can link or collect data from two different tables using...
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...Cursors A cursor is a variable that runs through the tuples of some relation. This relation can be a stored table, or it can be the answer to some query. By fetching into the cursor each tuple of the relation, we can write a program to read and process the value of each such tuple. If the relation is stored, we can also update or delete the tuple at the current cursor position. The example below illustrates a cursor loop. It uses our example relation T1(e,f) whose tuples are pairs of integers. The program will delete every tuple whose first component is less than the second, and insert the reverse tuple into T1. 1) DECLARE /* Output variables to hold the result of the query: */ 2) a T1.e%TYPE; 3) b T1.f%TYPE; /* Cursor declaration: */ 4) CURSOR T1Cursor IS 5) SELECT e, f 6) FROM T1 7) WHERE e < f 8) FOR UPDATE; 9) BEGIN 10) OPEN T1Cursor; 11) LOOP /* Retrieve each row of the result of the above query into PL/SQL variables: */ 12) FETCH T1Cursor INTO a, b; /* If there are no more rows to fetch, exit the loop: */ 13) EXIT WHEN T1Cursor%NOTFOUND; /* Delete the current tuple: */ 14) DELETE FROM T1 WHERE CURRENT OF T1Cursor; /* Insert the reverse tuple: */ 15) INSERT INTO T1 VALUES(b, a); 16) END LOOP; /* Free cursor used by the...
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...| Database systems | Assignment # 1 | Areeba kamil (34421) | BSCS 4A | 10/4/2015 | 4.7. Consider the LIBRARY relational database schema shown in Figure 4.6. Choose the appropriate action (reject,cascade,set to NULL,set to default) for each referential integrity constraint, both for the deletion of a referenced tuple and for the update of a primary key attribute value in a referenced tuple.Justify your choices. CREATE DATABASE library; CREATE TABLE book( book_id int, title varchar(50), publisher_name varchar(50), primary key(book_id), CONSTRAINT "fk_book_publisher" FOREIGN KEY(publisher_name) REFRENCES publisher(name) ON DELETE SET NULL ON UPDATE CASCADE ); ON DELETE SET NULL is chosen in case the name of a publisher needs to be removed, it can be deleted and set to NULL in all the tables containing publisher’s name ON UPDATE CASCADE is chosen here so that all the tables containing that publisher’s name are updated in accordance to the new publisher’s name CREATE TABLE book_authors( book_id int, author_name varchar(50), CONSTRAINT "fk_book_authors_book" FOREIGN KEY(book_id) REFRENCES book(book_id) ON DELETE CASCADE ON UPDATE CASCADE ); ON DELETE CASCADE is chosen because in case a book is to be removed from the LIBRARY database, all the rows corresponding to that deleted book_id will be deleted from all tables containing book_id. Since this could be dangerous, we could have also used ON DELETE RESTRICT ON UPDATE CASCADE is chosen...
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...both tables. JoinKeyword is used in SQL queries for joining two or more tables. Minimum required condition for joining table, is(n-1) where n, is number of tables. A table can also join to itself known as, Self Join. Types of Join The following are the types of JOIN that we can use in SQL. * Inner * Outer * Left * Right Cross JOIN or Cartesian Product This type of JOIN returns the cartesian product of rows from the tables in Join. It will return a table which consists of records which combines each row from the first table with each row of the second table. Cross JOIN Syntax is, SELECT column-name-list from table-name1 CROSS JOIN table-name2; Example of Cross JOIN The class table, ID | NAME | 1 | abhi | 2 | adam | 4 | alex | The class_info table, ID | Address | 1 | DELHI | 2 | MUMBAI | 3 | CHENNAI | Cross JOIN query will be, SELECT * from class, cross JOIN class_info; The result table will look like, ID | NAME | ID | Address | 1 | abhi | 1 | DELHI | 2 | adam | 1 | DELHI | 4 | alex | 1 | DELHI | 1 | abhi | 2 | MUMBAI | 2 | adam | 2 | MUMBAI | 4 | alex | 2 | MUMBAI | 1 | abhi | 3 | CHENNAI | 2 | adam | 3 | CHENNAI | 4 | alex | 3 | CHENNAI | INNER Join or EQUI Join This is a simple JOIN in which the result is based on matched data as per the equality condition specified in the query. Inner...
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...approach to the database design. It would be different because the scope of the system would be larger and require different techniques to be tied in at different stages. One thing that is the same is that all of the teams tied in with implementation of a larger warehouse would use the findings of their research and different states to help build the development of the system for a larger warehouse. Answer the Following Questions 1. Cross Join is identical to the product relational operator also known as the Cartesian product of two tables. An example of the syntax would be having two tables, one called MANAGER with 5 rows and EMPLOYEES with 10 rows, the cross join would have 50 rows including all of the columns from both tables. SELECT * FROM MANAGER CROSS JOIN EMPLOYEES 2. Correlated Subquery is a subquery that executes once for each row in the outer query. The process is similar to the typical nested loop in a programming language. An example would be product line sale where listing all the average units sold value, is lesser than the units sold value SELECT INV_NUMBER, P_CODE, LINE_UNITS FROM LINE LS WHERE LS.LINE_AVG<(SELECT UNITS(LINE_AVG) FROM LINE LA WHERE LA.P_CODE= LS.P_CODE); 3. Information system is a system that provides the conditions for data collection, storage, and retrieval. The information system helps transform data into information and allows for the management of both data and information. The purpose of information system is making good decisions to decision...
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