Given the grammar below, provide a leftmost derivation for the given .pdf

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Given the grammar below, provide a leftmost derivation for the given input string: TOKENS: INT for integers, VAR for single letter, +. *, (, ) GRAMMAR: E::= T {+ E} T::= P{* T} P::= INT | VAR|(E) INPUT STRING: a*(a+b) Solution Left most derivation means that you start from the start and start making projection. At every point you resolve the left most nonterminal symbol. E => T => P*T => a*T => a*P => a*(E) => a*(T+E) => a*(P+E) => a*(a+E) => a*(a+T) => a*(a+P) => a*(a+b).

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What is a view? What is it used for? Give examples when a view can be used for your project. Solution View is way of portraying information in a database by arranging data items in certain order by highlighting certain items. Often thought of as Virtual table, doesn\'t store data itself but pulls out from one or more tables. Views are used according to purpose so that we query on selective data not whole of it. Using views saves us lot of querying resources. For example, people query telephone directory to search for Street Address associated with Phone Number. We can create View which stores 2 columns namely Phone number and Street Address ignoring all other fields..

What is a view What is it used for Give examples when a view can b.pdf

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^^^ Q3. I am trying to implement double linked list but I was failed to write the code so anyone gives the main Code in the main function THANK YOU FOR ADVANCE #include #include #include struct node { int info; struct node *lptr,*rptr; }; typedef struct node DL; DL *delete( ) , *insert ( ); void display(); DL *delete(DL *start,int x) { DL *left,*right,*curr; curr = start; if( start == NULL) { printf(\"\ Doubly Linked List is Empty\"); return(start); } while( (curr != NULL) && (curr->info != x) ) curr = curr->rptr; if(curr == NULL) { printf(\"%d does not Exist ... Invalid Deletion! \ \",x); return(start); } left = curr->lptr; right = curr->rptr; if(left == NULL) { start = right; right->lptr = NULL; } else { left->rptr = right; if(right != NULL) right->lptr = left; } printf(\"\ The item %d is Deleted\ \",curr->info); free(curr); return(start); } void display(DL *start) { DL *last,*temp = start; if( (temp != NULL) ) { printf(\"\ The Elements of Doubly Linked List from Left to Right \"); printf(\"\ ROOT-> \"); while(temp != NULL ) { printf(\"%d -> \",temp->info); last = temp; temp = temp->rptr; } printf(\"NULL\"); printf(\"\ The Elements of Doubly Linked List from Left to Right\"); printf(\"\ ROOT-> \"); while(last != NULL ) { printf(\"%d -> \",last->info); last = last->lptr; } printf(\"ROOT\ \ \"); } else printf(\"\ Empty List\ \"); } DL *insert(DL *start,int x) { DL *new,*left,*temp; new=(DL *) malloc(sizeof(DL)); new->info = x; new->lptr = NULL; new->rptr = NULL; if( start == NULL) start = new; else { temp = start; while(temp->rptr != NULL) temp = temp->rptr; temp->rptr = new; new->lptr = temp; } return(start); } Main() { ------------------Write the code here------------------------ } Solution main function for the given program. main() { DL *root = NULL; int item; char c; clrscr(); do { printf(\"Insert/Display/Remove/Quit <> I/D/R/Q \"); printf(\"\ Enter Choice : \"); fflush(stdin); c = getchar(); switch(c) { case \'i\': case \'I\': printf(\"Enter The Element you want to be Inserted : \"); scanf(\"%d\",&item); root=insert(root,item); break; case \'d\': case \'D\': display(root); break; case \'r\': case \'R\': printf(\"Enter the Item which you want to Remove :\"); scanf(\"%d\",&item); root = delete(root,item); break; case \'Q\': case \'q\': return; } } while(1); }.

^^^Q3. I am trying to implement double linked list but I was faile.pdf

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What is the evolutionary basis for sexual reproduction? Solution There are both asexual and sexual reproduction in nearly all groups of animals and plants, in neither kingdom sex developed as a more rapid means. When sexual reproduction first originated is still a question. But it is believed that sex originated in green algae. There are certain species like Ulothrix have different size of spores are produced. These spores comes in pair and become one and gives valuable new forms. In some groups of unicellular plants ,the Peridinea, the Conjugatae and the Diatomeae- there is no asexual reproduction. These are good example of the actual gametic origin. The whole process of evolution is understand by these lines that first there was a visible morphological differentiation of gametes, the one becoming a large inactive cell stored with food , the other becoming small and mobile. Then came the evolution of various sex organs, and finally the alternation of generation. There are 3 main categories of sexual life cycles: diploid-dominant, demonstrated by most animals; haploid - dominant, demonstrated by all fungi and some algae; and the alternation of generations , demonstrated by plants and some algae. The Red Queen Hyothesis- First proposed by Leigh Van Valen in 1973. According to this hypothesis are species co-evolved with other organisms like predators evolve with their prey and parasites evolve with their hosts. Each tiny advantages gained by favorable variations gives a secies an edge over close competitors, predators, parasites, or even prey. Species that cannot keep up become extinct. August Weismann proposed an explanation for the evolution of sex, where the advantages of sex is the creation of variation among siblings. A similar hypothesis proposed by Charles Darwin in his book Origin of species..

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Why is it that the Gram\'s stain is not appropriate for acid fast bacterium? Solution During staining in microbiology, microbes are coloured or dyed so that their structure can be viewed against a lighter background. There are different staining techniques such as direct(Dye binds to the cell surface and all the microbes are stained alike),differential(based on their specific cell surface different microbes get stained differently) special (usually a particular organelle is targeted)and negative staining(back ground is stained instead of microbes) Gram stain is a differential stain where the bacteria with a thick peptidoglycan layer around the cells are stained with violet colour and those with an additional lipopolysaccharide layer will be stained pink. To understand this lets take a look at Gram staining procedure. First of all bacterial culture is smeared onto a glass slide and is heat fixed so that during the washes in the procedure the bacteria stay fixed onto the slide. Primary stain crystal violet is added onto the slide and the dye seperates into ions which penetrate or enter the cell membranes of both Gram positive and negative cells. Next iodine is added onto the slide which forms huge complexes with crystal violet. Then a dye remover usually called as decolouriser such as alchohol or acetone is used to remove the stain. In Gram negative bacteria the outer most membrane is lost during the treatment with alchohol whereas in the case of gram positive bacteria the dye is trapped inside the cell membrane. A secondary stain saffronin is used and which gives pink colour to Gram negative bacteria. Now coming to acid-fast bacteria, they have a compound called mycolic acid in their cell walls and are resistant to decolourisation after the acid treatment and alchohol and hence would not lose colour and appear violet. As we can see these organism can be easily confused with Gram positive bacteria which also do not lose stain not because they are resistant to acids but they have the crystal violet dye trapped in their cell membrane..

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Why does this version of the swap function fail to work? Is there a fix? (Choose 2) oid swap (int & lhs, int & rhs) Select one or more: It fails because the programmer forgot to make the parameters call-by-reference. Of course it works ! Just look at it It clearly swaps the two parameters! It fails OK, and we can fix it we can just reverse the order of the lines. To fix this, we must save the the value in a local variable before making the first assignment indicated, then instead of the second Iine, as int local = lhs; Ihs = rhs; rhs = local; It fails because the first line destroys the old value of ihs without saving it. Then both variables have the old value of rhs in them. Solution Choice d, and e are valid options. To swap the elements, we need to first copy the value of first variable to temporary variable, then copy the second variable to first variable, finally copy the value in temporary variable to second variable. So, int local = lhs; lhs = rhs; rhs = local; is the perfect way to swap two values. The last option is the actual problem with the given code. When we do lhs = rhs, the old value of lhs is destroyed, and now, both lhs and rhs are holding the same value, that is the value in rhs. And then, the second operation rhs = lhs, is of no use, as both of them are holding the same value. Therefore, options d, and e are correct..

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Which of the following is NOT a reason why heat is a bad method to overcome activation energy? A. Heat is not specific enough compared to enzymes B. Too much heat would denature enzymes C. Heat would allow for many enzymes to be activated, not just the ones that are needed D. Heat does not hold enough energy to overcome most activation energies Solution temperature is one the influence factor for enzyme acitvity. In the body of the biochemical reaction takes place for energy production. enzymes plays vital role by mediating the reactions. temperature influnece enzyme acticity. most the enzymes in general active between optimum temperature( 40-45c). in conditions where body temperature above 50 degree celcius enzymes gets denaturation which leads to derangment in the tertiary structure. in the become inactive at higher temperature (above 70c) answer B too much heat would denature the enzymes.

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What are the trends in browsers? Solution 1.chrome :Google chrome is a freeware web browser evolved by Google.a Webkit layout is used upto 27th version with the pecularity of IOS is declared. it is initially liberated in the form of beta version for the windows in the year 2008 september 2nd. 2.Firefox:Mozilla firefox is shortly called as firefox .it is an open source and also lavish in nature.moreover it was advanced for windows,linux and osx.firefox is periperal in nature.in firefox a Gecko layout engine is seen to provide web pages that applies present and predicted web attainments. 3.Opera:It is evolved by a opera software.the trendy version is obtainable for windows,linux and osx.Blink layout engine is seen in this opera web browser.versions of opera is obtained for widgets operating the IOS,Bada and blackberry. 4.Safari:Safari browser is evolved by Apple inc.containing the os x and IOS operarting systems.initially in the form of public beta in the year 2003 january 7 .it flatters as a Apple,s default browser starting with MAC OS X v 10.3 \"panther\".the congential browser of IOS is additionally called as safari.However it had a discrete Graphical user interface and utilizes a discrete webkit edition..

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The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 5.1% per day. Find the half-life of this substance (that is, the time it takes for one-half the original amount in a given sample of this substance to decay). Solution Half life = log2/ decay constant = 0.693/ decay constant Decay constant 5.1% = 0.051 So, Half life =0.693 /0.051 = 13.58 days.

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The groin is also known as the _____ region. crural inguinal perineum popliteal The upper midportion of the abdomen is called the _____ region. hypochondriac iliac hypogastric epigastric The thoracic cavity lies where In relationship to the abdominopelvic cavity? dorsal (posterior) ventral (anterior) superior inferior The urinary bladder is in the _____ region. inguinal epigastric umbilical hypogastric All of the following are found in the thoracic cavity EXCEPT the lungs. pleurae. meninges. heart. Which of the following organs is retroperitoneal in location? stomach liver heart kidney Solution 1.b) Groin is also calles as inguinal region which is between the abdomen and the thigh. 2.d)epigastric region-It is the upper central region of the abdomen.It is located between the costal margins and the subcostal plane. 3.c)superior 4.d)hypogastric region-It is a region of the abdomen located below the umbilical region. 5.c)meninges Thoracic cavity contains trachea,bronchi,lungs,oesophagus,heart and great blood vessels,thymus gland,lymph nodes,and nerves. It also contains two smaller cavities-pleural cavities surround each lung and pericardial cavity covers the heart. 6.d)kidney Retroperitoneal organs include- the kidneys,adrenal glands,ureters,urinary bladder,part of the oesophagus,rectum,ovaries,uterus,aorta and caudal vena cava..

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short answer List and describe two advances during the Industrial Revolution that changed how businesses delivered goods and services during the 18th and 19th centuries. What management tactics did John D. Rockefeller employ to gain control of the country\'s oil industry? Give an example of Theory X management style. Give an example of Theory Y management style. What prompted new challenges for management? What does a country develop during an industrial revolution? Solution 19. innovative technology, rapid growth 20. concentrative tactics 21. people are not interested to work- autocratic management style is fit 22. people are interested and self motivated. consultative model 23. high competiton 24. its production and output will increase, and it generates employment..

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Sheila/Salmonella agar is _ because _ breaks down amino acids that contain _, such as _ and_. This releases _ which reacts with _ to produce a _ color. Solution Shigella / salmonella agar is differential media because bacteria beaks down amino acids that contains sulfurgroup such as methionine and cystiene.This releases H2S which reacts with indicator ferric ammonium sulfate to produce a black colour.

SheilaSalmonella agar is _ because _ breaks down amino acids that co.pdf

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Security Challenges\" Please respond to the following: Distributed applications and cloud computing have become a viable option within the LAN-to- WAN Domain. Discuss the primary challenges related to maintaining the security of both applications and data in such an environment. Describe the controls that are needed to ensure the security of distributed applications and data. Identify the auditing approach that you believe is best suited to this arrangement. Provide a rationale for your response. No plagiarism please. Solution Challenges in Distributed applications : 1) Computational problems:- Those kind of problems which cannot be resolved within a specific amount of time on a single computer. 2) Scalability:- The system has to remain effective enough so that it can handle the increase in the number of resources and users. 3) Security:- In Distributed applications, the possible threats could be the information leakage, integrity violation, DoS(Denial Of Services) , illegitimate usage of the system. 4) Concurrency:- The shared access to the resources should be handled properly. Challenges in Cloud computing : 1) Security & Privacy:- The fact that the organization\'s data will reside outside the organization\'s firewall really raises a serious concern. Various types of attacks including hacking may affect multiple clients even if only one site is attacked.These risks has to be mitigated by using proper security applications, encrypted file-systems, using a proper storage devices to track unsual activity across the servers . 2) Loss of control over sensitive data:- The sensitive data that are stored in the servers has to be properly controlled. 3) Reliability and Availability:- The frequent outages has to be avoided here. It is very important to monitor the activity of servers using monitoring tool. The controls that are essential to prevent security threats in distributed applications and data include: 1) The control should protect data and integrity of system in case of invalid operation on secured data:- In case, a user tries to enter invalid data into the system, the control should protect the system from such operations. 2) Unauthorized invocations should be rejected by security control. 3) Efficient protection against unauthorized users:- Unauthorized access request by non-system users should be rejected by the control. The auditing approach should firstly start by ensuring data integrity that is associated with application. Control should specify the operation that can be invoked only by authorized person while rejecting access to unauthorized users when the data resources are accessed. Irrespective of operation, control should select only that user which can be allowed access while restricting others non-system users. Rules should be strictly followed by control and on the basis of defined rules, users access should be granted or denied..

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