Concept of algorithms Understand and use three tools to represent algorithms: Flowchart Pseudocode Programs

Transcription

1 Morteza Noferesti

2 Concept of algorithms Understand and use three tools to represent algorithms: Flowchart Pseudocode Programs

3 We want to solve a real problem by computers Take average, Sort, Painting, Web, Multimedia,.. We need a solution that Specifies how the real (complex) problem should be solved stepby-step using the basic operations The solution is the Algorithm of the problem Characteristics Specific Unambiguous Language independent

4 Algorithms are the problem solving steps/strategy in our mind!!! How can we document it (don t forget it)? How can explain/teach it to others peoples? How can explain it to computers? We need some methods to describe algorithms! Flow chart Pseudo-codes Codes/Programs

5 A typical 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

6 First produce a general algorithm (one can use pseudo code) Refine the algorithm successively to get step by step detailed algorithm that is very close to a computer language.

7 Pseudo code is an artificial and informal language that helps programmers develop algorithms. Pseudo code is very similar to everyday English. Employs 'programming-like' statements to depict the algorithm No standard format (language independent)

8 Write an algorithm to determine a student s final grade and indicate whether it is passing or failing. The final grade is calculated as the average of four marks.

9 Input a set of 4 marks Calculate their average by summing and dividing by 4 if average is below 50 Print FAIL else Print PASS

10 Step 1: Input M1,M2,M3,M4 Step 2: GRADE M1+M2+M3+M4)/4 Step 3: if (GRADE < 50) then Print FAIL else Print PASS endif

11 A graphical representation of the sequence of operations in an information system or program. A Flowchart shows logic of an algorithm emphasizes individual steps and their interconnections e.g. control flow from one action to the next

12 Name Symbol Use in Flowchart Oval Denotes the beginning or end of the program Parallelogram Denotes an input operation Rectangle Denotes a process to be carried out e.g. addition, subtraction, division etc. Diamond Denotes a decision (or branch) to be made. The program should continue along one of two routes. (e.g. IF/THEN/ELSE) Hybrid Denotes an output operation Flow line Denotes the direction of logic flow in the program

13 START Input M1,M2,M3,M4 GRADE (M1+M2+M3+M4)/4 Step 1: Input M1,M2,M3,M4 Step 2: GRADE (M1+M2+M3+M4)/4 Step 3: if (GRADE <50) then Print FAIL else Print PASS endif N IS GRADE<50 Y PRINT PASS PRINT FAIL STOP

14 Write an algorithm and draw a flowchart to convert the length in feet to centimeter. Pseudo code: Input the length in feet (Lft) Calculate the length in cm (Lcm) by multiplying LFT with 30 Print length in cm (LCM)

15 Flowchart Algorithm Step 1: Input Lft Step 2: Lcm Lft x 30 Step 3: Print Lcm START Input Lft Lcm Lft x 30 Print Lcm STOP

16 START Display message How many hours did you work? A flowchart for the pay-calculating program Read Hours Display message How much do you get paid per hour? Read Pay Rate Multiply Hours by Pay Rate. Store result in Gross Pay. Display Gross Pay END

17 Write an algorithm and draw a flowchart that will read the two sides of a rectangle and calculate its area. Pseudo code Input the width (W) and Length (L) of a rectangle Calculate the area (A) by multiplying L with W Print A

18 START Algorithm Step 1: Step 2: Step 3: Input W,L A L * W Print A Input W, L A L * W Print A STOP

19 Write an algorithm and draw a flowchart that will calculate the roots of a quadratic equation 2 ax bx c 2 Hint: d = sqrt ( b 4ac ), and the roots are: x1 = ( b + d)/2a and x2 = ( b d)/2a 0

20 Pseudo code: Input the coefficients (a, b, c) of the quadratic equation Calculate d Calculate x1 Calculate x2 Print x1 and x2

21 START 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 b b 4 a c Input a, b, c d sqrt(b x b 4 x a x c) x 1 ( b + d) / (2 x a) X 2 ( b d) / (2 x a) Print x 1,x 2 STOP

22 The structure is as follows If condition then true alternative//s1 else false alternative//s2 endif Y condition N S1 S2

23 The expression A>B is a logical expression 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 Y is A>B N Print A Print B

24 Conditions by comparisons; e.g., If a is greater then b If c equals to d Comparing numbers: Relational Operators

25 Relations produce a boolean value int a, b; bool bl; // #include <stdbool.h> bl = a == b; bl = a <= b; C Boolean operators and && or not!

26 p q p && q p q!p False False False False True False True False True True True False False True False True True True True False

27 Examples bool a = true, b=false, c; c =!a; //c=false c = a && b; //c=false c = a b; //c=true c =!a b; //c=false

28 Write an algorithm that reads two values, determines the largest value and prints the largest value with an identifying message. ALGORITHM Step 1: Step 2: Step 3: Input VALUE1, VALUE2 if (VALUE1 > VALUE2) then MAX VALUE1 else MAX VALUE2 endif Print The largest value is, MAX

29 START Input VALUE1,VALUE2 Y is VALUE1>VALUE2 N MAX VALUE1 MAX VALUE2 Print The largest value is, MAX STOP

30 NO x > min? YES Display x is outside the limits. NO x < max? YES Display x is outside the limits. Display x is within limits.

31 if (x > min) { if (x < max) printf("x is within the limits"); else printf("x is outside the limits"); } else printf("x is outside the limits");

32 Write and algorithm to a) read an employee name (NAME), overtime hours worked (OVERTIME), hours absent (ABSENT) and b) determine the bonus payment (PAYMENT). Bonus Schedule OVERTIME (2/3)*ABSENT >40 hours >30 but 40 hours >20 but 30 hours >10 but 20 hours 10 hours Bonus Paid $50 $40 $30 $20 $10

33 Step 1: Input NAME,OVERTIME,ABSENT Step 2: if (OVERTIME (2/3)*ABSENT > 40) then PAYMENT 50 else if (OVERTIME (2/3)*ABSENT > 30) then PAYMENT 40 else if (OVERTIME (2/3)*ABSENT > 20) then PAYMENT 30 else if (OVERTIME (2/3)*ABSENT > 10) then PAYMENT 20 else PAYMENT 10 endif Step 3: Print Bonus for, NAME is $, PAYMENT

34 Algorithm: calculate n 2 Input: n Output: sum sum 0 i 1 Repeat the following three steps while i n: sq i * i sum sum + sq i i + 1

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37 The flowchart reads three numbers and prints the value of the largest number.

38 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

39 #include <stdio.h> int main(void){ int number_to_test, remainder; printf("enter your number to be tested: "); scanf("%d", &number_to_test); remainder = number_to_test % 2; if(remainder == 0) else printf ("The number is even.\n ) } printf ("The number is odd.\n ) return 0;

40 Assign value according to conditions A ternary operator int i, j, k; bool b;... i = b? j : k; /*if(b) * i=j * else * i=k; */

41 y = abs(x) y = (x > 0)? x : -x; signum = (x < 0)? -1 : (x > 0? 1 : 0)

42 int d = numg / 25; charg = (d == 0)? D : ((d == 1)? C : (d == 2)? B : A );

43 Danger of assignment (=) and equality (==) int a = 10; int b = 20; if(a=b) // Logical Error Danger of similarity between C and mathematic if(a < b < c) // Logical Error if(a && b > 0) // Logical Error

44 ; is a null statement! #include <stdio.h> int main() { int num; printf("enter a number>>"); scanf("%d",&num); if (num>100); printf("this statement always is displayed!"); return 0; }

45 Multiple conditions If-else if-else if-. Select from alternative values of a variable switch-case Values should be constant not expression: i, Values & Variables should be int or char

46 Each switch-case can be rewritten by If-else if-else version of switch-case in the previous slide if(variable == value1)} <statements 1> <statements 2> } else if(variable == value2){ <statements 2> }

47 switch(variable){ case value1: <statements 1> break; case value2: <statements 2> break; default: <statements 3> } If(variable == value1) { <statements 1> } else if(variable == value2) { <statements 2> } else{ <statements 3> }

48 Get the operation (contains +,-,*,/,\) as a character (variable name: op)! Get two input numbers (variable names: num1, num2 ). Compute and display the expression of: num1 (op) num2 + - * /or\

49 Header file: <ctype.h> Function isalnum isalpha iscntrl isdigit isgraph islower isprint ispunct isspace isupper isxdigit tolower toupper Work Of Function Tests whether a character is alphanumeric or not Tests whether a character is aplhabetic or not Tests whether a character is control or not Tests whether a character is digit or not Tests whether a character is grahic or not Tests whether a character is lowercase or not Tests whether a character is printable or not Tests whether a character is punctuation or not Tests whether a character is white space or not Tests whether a character is uppercase or not Tests whether a character is hexadecimal or not Converts to lowercase if the character is in uppercase Converts to uppercase if the character is in lowercase

50 #include <stdio.h> #include <stdlib.h> int main(void){ int res, opd1, opd2; char opr; printf("operand1 : "); scanf("%d", &opd1); printf("operand2 : "); scanf("%d", &opd2); printf("operator : "); scanf(" %c", &opr); switch(opr){ case '+': res = opd1 + opd2; break; case '-': res = opd1 - opd2; break; case '/': res = opd1 / opd2; break;

51 } case '*': res = opd1 * opd2; break; default: printf("invalid operator \n"); return -1; } printf("%d %c %d = %d\n", opd1, opr, opd2, res); return 0;

52 switch(variable){ case value1: case value2: <statements 1> break; case value3: <statements 2> } If( (variable == value1) (variable == value2) ){ <statements 1> } else if (variable == value3) { <statements 2> }

53 if-else is more powerful than switch-case switch-case is only for checking the values of a variable and the values must be constant Some if-else cannot be rewritten by switch-case double var1, var2; if(var1 <= 1.1) <statements 1> if(var1 == var2) <statements 2>

54 bool b; switch (x){ case 0: b = 0; break; case 1: switch(y){ case } } break; //b = x && y case 0: b = 0; break; 1: b = 1; break;

55 All values used in case should be different switch(i){ //Error case 1: case 2: case 1:

56 All values must be value, not expression of variables switch(i){ //Error case j: case 2: case k+10:

57 While Do while For

58 Example: Write a program that read 3 integer and compute average It is easy. 3 scanf, an addition, a division and, a printf Example: Write a program that read 3000 integer and compute average?? 3000 scanf!!! Example: Write a program that read n integer and compute average N??? scanf Repetition in algorithms

59 A loop is a group of instructions the computer executes repeatedly while some loop-continuation condition remains true. We have discussed two means of repetition: Counter-controlled repetition Sentinel-controlled repetition Counter-controlled repetition is sometimes called definite repetition because we know in advance exactly how many times the loop will be executed. Sentinel-controlled repetition is sometimes called indefinite repetition because it s not known in advance how many times the loop will be executed.

60 In counter-controlled repetition, a control variable is used to count the number of repetitions. Counter-controlled repetition requires: The initial value of the control variable. The increment (or decrement) by which the control variable is modified each time through the loop. The condition that tests for the final value of the control variable (i.e., whether looping should continue).

61 When we know the number of iteration Average of 10 number Initialize counter = 0 Initialize other variables While (counter < number of loop repetition) do something (e.g. read input, take sum) counter = counter + 1

62 Consider the following problem statement: A class of ten students took a quiz. The grades (integers in the range 0 to 100) for this quiz are available to you. Determine the class average on the quiz.

63 Sentinel values are used to control repetition when: The precise number of repetitions is not known in advance, and The loop includes statements that obtain data each time the loop is performed. The sentinel is entered after all regular data items have been supplied to the program. Sentinels must be distinct from regular data items.

64 When we do NOT know the number of iteration But we know, when loop terminates E.g. Average of arbitrary positive numbers ending with <0 Get first input as n While (n is not sentinel) do something (sum, ) get the next input as n if (there is not any valid input) else S2 then S1

65 Consider the following problem: Develop a class averaging program that will process an arbitrary number of grades each time the program is run. One way to solve this problem is to use a special value called a sentinel value (also called a signal value, a dummy value, or a flag value) to indicate end of data entry.

66 The user types in grades until all legitimate grades have been entered. The user then types the sentinel value to indicate that the last grade has been entered. Clearly, the sentinel value must be chosen so that it cannot be confused with an acceptable input value.

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68 Repetition is performed by loops Put all statements to repeat in a loop Don t loop to infinity Stop the repetition Based on some conditions (counter, sentinel) C has 3 statements for loops while statement do-while statement for statement

69 A loop tests a condition, and if the condition exists, it performs an action. Then it tests the condition again. If the condition still exists, the action is repeated. This continues until the condition no longer exists x < y? YES Process A

70 while ( <expression> ) <statements>

71 <stdio.h> برنامه ای بنویسید که عدد n را از کاربر بگیرد و اعداد 0 تا n را چاپ کند. int main(void){ int n, number; number = 0; printf("enter n: "); scanf("%d", &n); while(number <= n){ printf("%d \n", number); number++; } return 0; }

72 #include <stdio.h> int main(void){ int negative_num, positive_num; int number; negative_num = positive_num = 0; printf("enter Zero to stop \n"); printf("enter next number: "); scanf("%d", &number); while(number!= 0){ if(number > 0) positive_num++; else negative_num++; } printf("enter next number: "); scanf("%d", &number); } printf("the number of positive numbers printf("the number of negative numbers return 0; = %d\n", positive_num); = %d\n", negative_num);

73 Consider a program segment designed to find the first power of 3 larger than 100. When the following while repetition statement finishes executing, product will contain the desired answer: product = 3; while ( product <= 100 ) { product = 3 * product; } /* end while */

74 do <statements> while (<expression>);

75 #include <stdio.h> int main(void){ int n; double number, sum; printf("enter n > 0: "); scanf("%d", &n); if(n < 1){printf("wrong input"); return -1;} sum = 0; number = 0.0; do{ number++; sum += number / (number + 1.0); }while(number < n); printf("sum = %f\n", sum); return 0; }

76 #include <stdio.h> int main(void){ int negative_num=0, positive_num=0; int number; printf("enter Zero to stop \n"); do{ printf("enter next number: "); scanf("%d", &number); if(number > 0) positive_num++; else if(number < 0) negative_num++; }while(number!= 0); printf("the number of positive printf("the number of negative return 0; numbers numbers = %d\n", positive_num); = %d\n", negative_num); }

77 for(<expression1>;<expression2>; <expression3>) <statements>

78 counter 1, sum 0 false counter < 6 true input n sum sum + n counter++ output sum int x, sum, i; sum = 0; for (i = 1; i < 6; i++) { scanf( %d,&x); sum = sum + x; } printf( %d,sum);

79 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); _

80 Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); num 1 _

81 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); 1 _

82 Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); num 1 1 _

83 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); 2 1 _

84 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); 2 1 _

85 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); _

86 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); _

87 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); _

88 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); _

89 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); _

90 num Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); _

91 4 Example: for (num = 1; num <= 3; num++ ) printf( %d\t, num); printf( have come to exit\n ); have come to exit_

92 int grade, count, i; double average, sum; sum = 0; printf("enter the number of students: "); scanf("%d", &count); for(i = 0; i < count; i++){ printf("enter the grade of %d-th student: scanf("%d", &grade); sum += grade; ", (i + 1)); } average = sum / count; printf("the average of your class is %0.3f\n", average); return 0; }

93 int n, number; printf("enter n: "); scanf("%d", &n); for(number = 1; number <= n; number++) if((number % 2) == 0) printf("%d \n", number); } return 0;

94 int n, number; printf("enter n: "); scanf("%d", &n); for(number = 2; number <= n; number += 2) printf("%d \n", number); } return 0;

95 The following examples show methods of varying the control variable in a for statement. Vary the control variable from 1 to 100 in increments of 1. for ( i = 1; i <= 100; i++ ) Vary the control variable from 100 to 1 in increments of -1 (decrements of 1). for ( i = 100; i >= 1; i-- ) Vary the control variable from 7 to 77 in steps of 7. for ( i = 7; i <= 77; i += 7 ) Vary the control variable from 20 to 2 in steps of -2. for ( i = 20; i >= 2; i -= 2 ) Vary the control variable over the following sequence of values: 2, 5, 8, 11, 14, 17. for ( j = 2; j <= 17; j += 3 ) Vary the control variable over the following sequence of values: 44, 33, 22, 11, 0. for ( j = 44; j >= 0; j -= 11 )

96 Expression1 and Expression3 can be any number of expressions for(i = 0, j = 0; i < 10; i++, j--) Expression2 at most should be a single expression for(i = 0, j = 0; i < 10, j > -100; i++, j--) //ERROR Any expression can be empty expression for(;i<10;t++) for(;;)//infinite loop

97 The three expressions in the for statement are optional. for(;;) One may omit expression1 if the control variable is initialized elsewhere in the program. If expression2 is omitted, C assumes that the condition is true, thus creating an infinite loop. expression3 may be omitted if the increment is calculated by statements in the body of the for statement or if no increment is needed.

98 <statement> in loops can be empty while(<expression>) ; E.g., while(i++ <= n) ; for(<expression1>; <expression2>;<expression3>); E.g., for(i = 0; i < 10; printf("%d\n",i), i++) ;

99 <statement> in loops can be loop itself while(<expression0>) for(<expression1>; <expression2>;<expression3>) <statements> for(<expression1>; <expression2>;<expression3>) do <statements> while(<expression>);

100 A program that takes n and m and prints ***.* (m * in each line) ***.* ***.* (n lines)

101 int main(void){ int i, j, n, m; printf("enter n & scanf("%d%d", &n, m: "); &m); for(i = 0; i < n; i++){ for(j = 0; j < m; j++) printf("*"); } printf("\n"); } return 0;

102 A program that takes n and prints * ** *** (i * in i-th line) ***.* (n lines)

103 #include <stdio.h> int main(void){ int i, j, n; printf("enter n: "); scanf("%d", &n); i = 1; while(i <= n){ for(j = 0; j < i; j++) printf("*"); printf("\n"); i++; } return 0; }

104 A program that takes a number and generates the following pattern Input = 5 * * ** *** **** ***** **** *** ** for(i= 1; i <= n; i++){ for(j= 0; j < i-1;j++) printf(" "); for(j= 1; j <= i; j++) printf("*"); printf("\n"); } for(i=n-1; i >= 1; i--){ for(j= 1; j < i; j++) printf(" "); for(j = 1; j <= i; j++) printf("*"); printf("\n"); }

105 The break and continue statements are used to alter the flow of control. The break statement, when executed in a while, for, do while or switch statement, causes an immediate exit from that statement. Program execution continues with the next statement.

106 Exit from loop based on some conditions do{ scanf("%d", &a); scanf("%d", &b); if(b == 0) break; res = a / b; printf("a /= %d\n", res); }while(b > 0);

107

108 int i,j; for(i=1; i<6;i++){ for(j =1; j<6;j++) { printf("%d %d\n", i,j); if(j==3) break; } }

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110 int n, i, flag = 0; printf("enter a positive integer: "); scanf("%d",&n); for(i=2; i<=sqrt(n); ++i) { if(!n%i) { flag=1; break; } } if (!flag) printf("%d is a prime number.",n); else printf("%d is not a prime number.",n);

111 Jump to end of loop and continue repetition The continue statement, when executed in a while, for or do while statement, skips the remaining statements in the body of that control statement and performs the next iteration of the loop. do{ scanf("%f", &a); scanf("%f", &b); if(b == 0) continue; res = a / b; }while(a> 0); printf("a / b= %f\n", res);

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113 When you know the number of repetition Counter-controlled loops Usually, for statements When you don t know the number of repetitions (sentinel loop) Some condition should be check before starting loop Usually, while statement The loop should be executed at least one time Usually, do-while statement

114 Loop should terminate E.g., in for loops, after each iteration, we should approach to the stop condition for(i = 0; i < 10; i++) //OK for(i = 0; i < 10; i--) //Bug Initialize loop control variables int i; for( ; i < 10; i++) //Bug

115 Don t modify forloop controller in loop body for(i = 0; i < 10; i++){... i--; //Bug } Take care about wrong control conditions < vs. <= = vs. == int b = 10; while(a = b){//it means while(true) scanf("%d",&a) {