Downloaded 36 times
![EXAMPLE 6
Step 1: Input N1, N2, N3
Step 2: if (N1>N2) then
if (N1>N3) then
MAX ← N1 [N1>N2, N1>N3]
else
MAX ← N3 [N3>N1>N2]
endif
else
if (N2>N3) then
MAX ← N2 [N2>N1, N2>N3]
else
MAX ← N3 [N3>N2>N1]
endif
endif
Step 3: Print “The largest number is”, MAX](https://image.slidesharecdn.com/programdesigntechniques-160913194130/75/Program-design-techniques-28-2048.jpg)
![EXAMPLE 6
Step 1: Input N1, N2, N3
Step 2: if (N1>N2) then
if (N1>N3) then
MAX ← N1 [N1>N2, N1>N3]
else
MAX ← N3 [N3>N1>N2]
endif
else
if (N2>N3) then
MAX ← N2 [N2>N1, N2>N3]
else
MAX ← N3 [N3>N2>N1]
endif
endif
Step 3: Print “The largest number is”, MAX](https://crownmelresort.com/image.slidesharecdn.com/programdesigntechniques-160913194130/75/Program-design-techniques-28-2048.jpg)
More Related Content
![Flow Chart & Input Output Statement [3] M](https://cdn.slidesharecdn.com/ss_thumbnails/flow-chart-inputoutput-statement-3-m-1233232763031429-3-thumbnail.jpg?width=640&height=640&fit=bounds)
Similar to Program design techniques
Program design techniques
- 1. Program Designing techniques Introductionto Computing Lecture# 24,25
- 2. PROGRAM DESIGNING TECHNIQUES Pseudocode Algorithm Flowchart
- 3. DESIGNING TECHNIQUES • Atypical programming task can be divided into two phases: • Problem solving phase • produce an ordered sequence of steps that describe solution of problem • this sequence of steps is called an algorithm • Implementation phase • implement the program in some programming language
- 4. STEPS IN PROBLEMSOLVING • First produce a general algorithm (one can use pseudocode) • Refine the algorithm successively to get step by step detailed algorithm that is very close to a computer language. • Pseudocode is an artificial and informal language that helps programmers develop algorithms. • Pseudocode is very similar to everyday English.
- 5. PSEUDOCODE & ALGORITHM •Example 1: Write a pseudocode and an algorithm to convert the length in feet to inches.
- 6. EXAMPLE 2 Pseudocode: • Inputthe length in feet • Calculate the length in inches by multiplying length in feet with 12 • Print length in inches.
- 7. EXAMPLE 2 Algorithm • Step1: Input L_ft • Step 2: L_inches ← L_ft x 12 • Step 3: Print L_inches
- 8. THE FLOWCHART • Aschematic representation of a sequence of operations, as in a manufacturing process or computer program. • It is a graphic representation of how a process works, showing, at a minimum, the sequence of steps. • A flowchart consists of a sequence of instructions linked together by arrows to show the order in which the instructions must be carried out.
- 9. CONT… Each instruction isput into a box. The boxes are different shapes depending upon what the instruction is. Different symbols are used to draw each type of flowchart.
- 10. CONT… A Flowchart • showslogic of an algorithm • emphasizes individual steps and their interconnections • e.g. control flow from one action to the next
- 11.
- 12. EXAMPLE 2 • Writean algorithm and draw a flowchart to convert the length in feet to centimeter. Pseudocode: • Input the length in feet (Lft) • Calculate the length in cm (Lcm) by multiplying LFT with 30 • Print length in cm (LCM)
- 13. EXAMPLE 2 Algorithm • Step1: Input L_ft • Step 2: Lcm ← L_ft x 30 • Step 3: Print L_cm START Input L_ft Lcm ← L_ft x 30 Print L_cm STOP Flowchart
- 14. EXAMPLE 3 Write analgorithm and draw a flowchart that will read the two sides of a rectangle and calculate its area. Pseudocode • Input the width (W) and Length (L) of a rectangle • Calculate the area (A) by multiplying L with W • Print A
- 15. EXAMPLE 3 Algorithm • Step1: Input W,L • Step 2: A ← L x W • Step 3: Print A START Input W, L A ← L x W Print A STOP
- 16. EXAMPLE 4 • Writean algorithm and draw a flowchart that will calculate the roots of a quadratic equation • Hint: d = sqrt ( ), and the roots are: x1 = (–b + d)/2a and x2 = (–b – d)/2a 2 0ax bx c+ + = 2 4b ac−
- 17. EXAMPLE 4 Pseudocode: • Inputthe coefficients (a, b, c) of the quadratic equation • Calculate d • Calculate x1 • Calculate x2 • Print x1 and x2
- 18. EXAMPLE 4 • Algorithm: •Step 1: Input a, b, c • Step 2: d ← sqrt ( ) • Step 3: x1 ← (–b + d) / (2 x a) • Step 4: x2 ← (–b – d) / (2 x a) • Step 5: Print x1, x2 4b b a c× − × × START Input a, b, c d ← sqrt(b x b – 4 x a x c) Print x1 ,x2 STOP x1 ←(–b + d) / (2 x a) X2 ← (–b – d) / (2 x a)
- 19. DECISION STRUCTURES • Theexpression A>B is a logical expression • it describes a condition we want to test • if A>B is true (if A is greater than B) we take the action on left • print the value of A • if A>B is false (if A is not greater than B) we take the action on right • print the value of B
- 20.
- 21. IF–THEN–ELSE STRUCTURE • Thestructure is as follows If condition then true alternative else false alternative endif
- 22. IF–THEN–ELSE STRUCTURE If A>Bthen print A else print B endif is A>B Print B Print A Y N
- 23. RELATIONAL OPERATORS Relational Operators OperatorDescription > Greater than < Less than = Equal to ≥ Greater than or equal to ≤ Less than or equal to ≠ Not equal to
- 24. EXAMPLE 5 • Writean algorithm that reads two values, determines the largest value and prints the largest value with an identifying message. ALGORITHM Step 1: Input VALUE1, VALUE2 Step 2: if (VALUE1 > VALUE2) then MAX ← VALUE1 else MAX ← VALUE2 endif Step 3: Print “The largest value is”, MAX
- 25. EXAMPLE 5 MAX ←VALUE1 Print “The largest value is”, MAX STOP Y N START Input VALUE1,VALUE2 MAX ← VALUE2 is VALUE1>VALUE2
- 26. NESTED IFS • Oneof the alternatives within an IF–THEN–ELSE statement • may involve further IF–THEN–ELSE statement
- 27. EXAMPLE 6 • Writean algorithm that reads three numbers and prints the value of the largest number.
- 28. EXAMPLE 6 Step 1:Input N1, N2, N3 Step 2: if (N1>N2) then if (N1>N3) then MAX ← N1 [N1>N2, N1>N3] else MAX ← N3 [N3>N1>N2] endif else if (N2>N3) then MAX ← N2 [N2>N1, N2>N3] else MAX ← N3 [N3>N2>N1] endif endif Step 3: Print “The largest number is”, MAX
- 29. EXAMPLE 6 • Flowchart:Draw the flowchart of the above Algorithm.
- 30.
- 31. EXAMPLE 7 • Writeand algorithm and draw a flowchart to read an employee name (NAME), overtime hours worked (OVERTIME), hours absent (ABSENT) and determine the bonus payment (PAYMENT).
- 32. EXAMPLE 7 Bonus Schedule OVERTIME– (2/3)*ABSENT Bonus Paid >40 hours >30 but ≤ 40 hours >20 but ≤ 30 hours >10 but ≤ 20 hours ≤ 10 hours $50 $40 $30 $20 $10
- 33. Step 1: InputNAME,OVERTIME,ABSENT Step 2: if (OVERTIME–(2/3)*ABSENT > 40 ) then PAYMENT ← 50 else if (OVERTIME–(2/3)*ABSENT > 30 && OVERTIME–(2/3)*ABSENT<= 40 ) then PAYMENT ← 40 else if (OVERTIME–(2/3)*ABSENT > 20 && OVERTIME–(2/3)*ABSENT<= 30 ) then PAYMENT ← 30 Algorithm
- 34. else if (OVERTIME–(2/3)*ABSENT> 10 && OVERTIME–(2/3)*ABSENT<= 20) then PAYMENT ←20 else PAYMENT ← 10 endif Step 3: Print “Bonus for”, NAME “is $”, PAYMENT
- 35. EXAMPLE 7 • Flowchart:Draw the flowchart of the above algorithm?
- 36. LOOPS • Computers areparticularly well suited to applications in which operations are repeated many times. • If the same task is repeated over and over again a loop can be used to reduce program size and complexity
- 37. FLOWCHART FOR FINDINGTHE SUM OF FIRST FIVE NATURAL NUMBERS ( I.E. 1,2,3,4,5): Example 8:
- 38. EXAMPLE 9 Flowchart tofind the sum of first 50 natural numbers.
- 39. EXAMPLE 10 • Writedown an algorithm and draw a flowchart to find and print the largest of N (N can be any number) numbers. (Assume N to be 5 and the following set to be the numbers {1 4 2 6 8 })
- 40. ALGORITHM: • Step 1:Input N • Step 2: Input Current • Step 3: Max Current • Step 4: Counter 1 • Step 5: While (Counter < N) Repeat steps 5 through 8 • Step 6: Counter Counter + 1 • Step 7: Input Next • Step 8: If (Next > Max) then Max Next endif • Step 9: Print Max
- 41. START Input N, Current Max Current Print Max STOP Y Counter < N N Counter 1 Counter Counter +1 Input Next Next >Max Y N Max Next Flowchart
- 42.