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What do you understand by stepwise refinement of the program?
The method of "Stepwise refinement" means to take an object and move it from a general perspective to a exact level of detail and this cannot be completed in one jump but in steps. The number of steps required to decompose an object into sufficient detail is in the end based on the inherent nature of the object. "Stepwise refinement" of program shows a "divide and conquer" approach to design. In other words, break a complex program into smaller, more manageable modules that can be reviewed and inspected before moving to the next level of feature. Step-wise refinement is a powerful paradigm for making a complex program from a simple program by adding features incrementally.
What is a universal gate? Give examples. Realize the basic gates with any one universal gate. Ans: Universal Gates: NAND and NOR are termed as Universal gates. The OR, AN
Distinguish between uniform scaling differential scaling
Convert binary number in two's compliment form 0100 1000. Converting the binary number into 2's compliment from 0100 1000 is given below: 01001000 => 10111000
Explain advantage of static storage class The second and more subtle use of 'static' is in connection with external declarations. With external constructs it provides a privacy
different types of Operating System
Assessing Heuristic Searches: Given a particular problem you want to build an agent to solve, so there may be more than one way of justifying it as a search problem, more than
Data buffering is quite helpful for purpose of smoothing out gaps in speed of processor and I/O devices. Data buffers are registers that hold I/O information temporarily. I/O is pe
describe briefly about the c token with suitable example program
#question.Smugglers are becoming very smart day by day. Now they have developed a new technique of sending their messages from one smuggler to another. In their new technology, the
Q. Prove using Boolean Algebra 1. AB + AC + BC' = AC + BC' 2. (A+B+C) (A+B'+C') (A+B+C') (A+B'+C)=A 3. (A+B) (A'+B'+C) + AB = A+B 4. A'C + A'B + AB'C + BC = C + A'B
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