Implementation of Algorithms to Solve Problems: ISC Class 12 Computer Science

What Implementation of Algorithms to Solve Problems is about

Weightage: CISCE publishes no chapter or unit weightage for ISC Computer Science. Implementation of Algorithms to Solve Problems is part of Section B, 3 of the 2027 syllabus and can be asked anywhere in the 70-mark Paper 1 (Part I compulsory: ten 1-mark items and a 10-mark short-answer question; Part II: two of three 10-mark questions in each of Sections A and B, and two of three 5-mark questions in Section C).Source: CISCE ISC Computer Science (868) syllabus, examination year 2027; latest Paper 1 specimen (2025) and the 2025 board paper

Key concepts

From problem to algorithm

An algorithm is a finite sequence of clear, unambiguous steps that solves a problem for every valid input and stops.

Properties: finiteness (it ends), definiteness (each step is precise), input (zero or more), output (at least one), effectiveness (each step can be carried out).

Steps for designing one:
1. Understand the problem: identify inputs, outputs and the rule connecting them; work two or three examples by hand, including edge cases (0, 1, negative, single digit, empty string).
2. Break it into sub-tasks, each becoming a method (for example accept, isPrime, reverse, check).
3. Write the steps in structured English or pseudocode with sequence, selection and iteration (or recursion).
4. Check correctness on the hand-worked examples (a dry run).
5. Only then write the Java code.

Example (GCD by Euclid's algorithm):
Step 1: Read a, b.
Step 2: While b is not 0: r = a mod b; a = b; b = r.
Step 3: Output a (the GCD). LCM = a original × b original / GCD.
It is correct because GCD(a, b) = GCD(b, a mod b), and b strictly decreases, so the loop ends.

Worked example

[2] Write an algorithm to find the GCD of two positive integers by Euclid's method and state why it always stops. (Create)

Try it first, then show the solutionSolution

1. Steps: read a, b; while b ≠ 0 do r = a % b, a = b, b = r; print a - 1 mark.
2. It stops because the remainder is always smaller than b, so b decreases each time and must reach 0 - 1 mark.

Answer: Read a and b; while b is not zero, set r = a % b, a = b, b = r; when b becomes zero, a is the GCD. The loop must end because each new b (the remainder) is smaller than the previous b and cannot go below zero.

Dry runs and tracing

A dry run executes the algorithm by hand, recording the values of variables after each step in a trace table. It is the main tool for checking correctness and for answering 'find the output' questions.

Trace of the GCD loop for a = 84, b = 36 (program GcdTrace below):

a	b	a % b
84	36	12
36	12	0

Result: GCD = 12, LCM = 84 / 12 × 36 = 252.

// Euclid's algorithm with a printed trace, and LCM from GCD.
public class GcdTrace
{
    static int gcd(int a, int b)
    {
        while (b != 0)
        {
            System.out.println("a = " + a + ", b = " + b + ", a % b = " + (a % b));
            int r = a % b;
            a = b;
            b = r;
        }
        return a;
    }

    public static void main(String[] args)
    {
        int x = 84, y = 36;
        int g = gcd(x, y);
        System.out.println("GCD = " + g);
        System.out.println("LCM = " + (x / g * y));
    }
}

Output:

a = 84, b = 36, a % b = 12
a = 36, b = 12, a % b = 0
GCD = 12
LCM = 252

Tips for tracing questions:
1. Make one column per variable and one row per iteration or call.
2. Update variables in the order the statements execute.
3. For recursion, write each call on a new line and then unwind the returns in reverse order.
4. State the final answer and, if asked, what the method does in general (its purpose).

Worked example

[2] Trace the GCD loop for a = 45, b = 20 and give the GCD and LCM. (Apply)

Try it first, then show the solutionSolution

1. 45 % 20 = 5 → a = 20, b = 5; 20 % 5 = 0 → a = 5, b = 0 - 1 mark.
2. GCD = 5; LCM = 45 / 5 × 20 = 180 - 1 mark.

Answer: GCD = 5 and LCM = 180.

Correctness and test data

A program is correct if it produces the right output for every valid input. To check:
1. Normal cases: typical values (13, 4075).
2. Boundary cases: smallest and largest valid values (0, 1, single-digit numbers, an array of one element, an empty string).
3. Special cases suggested by the definition (for an Emirp number, a palindromic prime such as 11 must be rejected because its reverse is not a different prime).
4. Invalid input handling if the question asks for it (for example 'display an appropriate message if the input is out of range').

Loop correctness: check the loop condition, the initial value and the update. Most errors are off-by-one errors (one iteration too many or too few) or a missing update that makes a loop infinite.

Recursion correctness: there must be a base case that stops the recursion, and each recursive call must move towards it.

Worked example

[2] Suggest four test values for a method that checks whether a number is an Emirp, and say why each is useful. (Evaluate)

Try it first, then show the solutionSolution

1. 13 and 97: normal Emirps (reverse 31 and 79 are prime); 23: prime whose reverse 32 is not prime - 1 mark.
2. 11: palindromic prime (reverse is the same number, so not an Emirp); 1: not prime (boundary) - 1 mark.

Answer: 13 (Emirp), 23 (prime but reverse not prime), 11 (palindromic prime, rejected) and 1 (not prime, a boundary case) together test every branch of the condition.

Debugging: syntax, run-time and logical errors

Types of error:
1. Syntax (compile-time) errors: break the rules of Java, so the compiler rejects the program: missing semicolon, unmatched braces, misspelt keyword, using an undeclared variable, type mismatch (int x = 2.5;).
2. Run-time errors (exceptions): the program compiles but fails while running: division by zero (ArithmeticException), array index out of bounds (ArrayIndexOutOfBoundsException), calling a method on null (NullPointerException), wrong input type (InputMismatchException).
3. Logical errors: the program runs but gives wrong results because the algorithm or code is wrong, such as a wrong loop condition. These are found only by testing and dry runs.

Example of a logical error: summing digits with while (n > 9) stops before the last digit.

// Intended: sum of the digits of n. The corrected loop is in sumDigits; the buggy one is shown for comparison.
public class BuggySum
{
    static int buggy(int n)
    {
        int s = 0;
        while (n > 9)          // bug: stops before the last digit
        {
            s = s + n % 10;
            n = n / 10;
        }
        return s;
    }

    static int sumDigits(int n)
    {
        int s = 0;
        while (n > 0)          // correct: runs until every digit is used
        {
            s = s + n % 10;
            n = n / 10;
        }
        return s;
    }

    public static void main(String[] args)
    {
        System.out.println("buggy(4075) = " + buggy(4075));
        System.out.println("sumDigits(4075) = " + sumDigits(4075));
        System.out.println("sumDigits(7) = " + sumDigits(7));
        System.out.println("sumDigits(0) = " + sumDigits(0));
    }
}

Output:

buggy(4075) = 12
sumDigits(4075) = 16
sumDigits(7) = 7
sumDigits(0) = 0

The buggy version returns 12 for 4075 because it never adds the leading digit 4; the correct condition n > 0 gives 16.

Debugging techniques: dry runs; printing intermediate values; testing small cases; checking boundary values; using the BlueJ debugger to step through code and inspect variables.

Worked example

[2] A method meant to count the digits of n uses the loop while (n > 9) { c++; n = n / 10; } with c starting at 0. Find the error and correct it. (Analyse)

Try it first, then show the solutionSolution

1. For n = 4075 the loop runs for 4075, 407, 40 and stops at 4, giving 3 instead of 4: a logical (off-by-one) error - 1 mark.
2. Correct by using while (n > 0) (or by starting c at 1); test with 7 → 1 and 4075 → 4 - 1 mark.

Answer: The condition n > 9 skips the last digit; use while (n > 0) so every digit is counted (4075 → 4).

Writing a Section B class: the standard pattern

Every Section B question follows the same shape. Marks are given for each method, for main( ) and for documentation.

1. Class header and data members exactly as named in the question (types as stated).
2. Constructor: initialise every data member to a legal initial value (0, "" or as specified). Constructors have no return type.
3. Input method: use Scanner to read values; follow any condition stated (for example lower case, positive integer).
4. Processing methods: implement the algorithm; respect the specified signature (parameters and return type); if the question says 'using recursive technique', the method must call itself with no loops.
5. Display or check method: call the processing methods as instructed (for example 'by invoking the function convert( )') and print results with clear messages.
6. main( ): create an object and call the methods in the right order.
7. Comments: describe each method's purpose and key variables (mnemonic names help).

Example: class Emirp. An Emirp is a prime number whose reverse is a different prime (13 → 31).

Member	Purpose
int n, int rev	the number and its reverse
Emirp( )	constructor: n = 0, rev = 0
void accept( )	read n
boolean isPrime(int x)	true if x is prime
void reverse( )	store the reverse of n in rev
void check( )	use reverse( ) and isPrime( ) and print the result

import java.util.Scanner;

// An Emirp number is a prime whose reverse is a different prime (13 -> 31).
public class Emirp
{
    int n;      // the number to test
    int rev;    // its reverse

    Emirp()
    {
        n = 0;
        rev = 0;
    }

    void accept()
    {
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter a positive integer:");
        n = sc.nextInt();
    }

    // returns true if x is prime
    boolean isPrime(int x)
    {
        if (x < 2)
            return false;
        for (int d = 2; d * d <= x; d++)
            if (x % d == 0)
                return false;
        return true;
    }

    // reverses n and stores the result in rev
    void reverse()
    {
        int t = n;
        rev = 0;
        while (t > 0)
        {
            rev = rev * 10 + t % 10;
            t = t / 10;
        }
    }

    void check()
    {
        reverse();
        if (isPrime(n) && isPrime(rev) && rev != n)
            System.out.println(n + " is an Emirp number (reverse " + rev + " is also prime)");
        else
            System.out.println(n + " is not an Emirp number");
    }

    public static void main(String[] args)
    {
        Emirp ob = new Emirp();
        ob.accept();
        ob.check();
    }
}

Sample input 13 gives:

Enter a positive integer:
13 is an Emirp number (reverse 31 is also prime)

Other runs: 23 → not an Emirp (32 is not prime); 17 → Emirp (71); 11 → not an Emirp (reverse is the same number); 97 → Emirp (79).

Worked example

[2] In the Emirp program, why does check( ) test rev != n as well as both numbers being prime? (Analyse)

Try it first, then show the solutionSolution

1. By definition the reverse must be a different prime - 1 mark.
2. Palindromic primes such as 11 or 101 are prime and equal to their reverse, so without rev != n they would wrongly be accepted - 1 mark.

Answer: Because an Emirp's reverse must be a different prime; the extra test rejects palindromic primes such as 11.

Implementation of Algorithms to Solve Problems in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Debugging Errors and the Section B Class Pattern
• Designing Algorithms, Dry Runs and Choosing Test Data

Implementation of Algorithms to Solve Problems: syntax and key facts

Euclid's GCD

gcd(a, b) = gcd(b, a % b); gcd(a, 0) = a

LCM = a × b / GCD.

Digit extraction

last digit = n % 10; remove it: n = n / 10

Loop while n > 0.

Reverse a number

rev = rev × 10 + n % 10; n = n / 10

rev starts at 0.

Prime test

x ≥ 2 and no divisor d with 2 ≤ d and d × d ≤ x

Checking up to the square root is enough.

Error types

Syntax (compile time) / run time (exception) / logical (wrong output)

Logical errors need testing.

How the board asks Implementation of Algorithms to Solve Problems

The question patterns to expect, and the method that works for each.

Tricks to remember Implementation of Algorithms to Solve Problems

Short memory aids. Say each one aloud once, then use it on a question the same day.

• C-A-P-D-M

  Constructor, Accept, Process, Display, Main: the order of every Section B answer.

  Use it when: Structure of a 10-mark program.

• Three bugs

  Compiler catches syntax; JVM catches run-time; only you catch logic.

  Use it when: Error types.

• Edge first

  Test 0, 1, one-digit, largest and just-too-large values first.

  Use it when: Correctness checks.

Implementation of Algorithms to Solve Problems: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

int countDigit(int n, int k)
{
    int c = ?1?;
    while (n > 0)
    {
        if (?2?)
            c++;
        ?3?;
    }
    return c;
}

int n = 1234, s = 0, k = 1;
while (n > 0)
{
    s = s + (n % 10) * k;
    k = -k;
    n = n / 10;
}
System.out.println(s);

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 7-mark question split into 2 + 2 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Implementation of Algorithms to Solve Problems

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using a loop where recursion is demanded

  CISCE examiners' comments, 2024 and 2025: candidates 'Used a loop instead of recursion, or a mix of both' and had a 'Missing or wrong base case and badly structured recursive case; used loops instead of recursion'.

  ✓Do this: When the question says 'using recursive technique', the method must call itself, with a base case and no loop.

  Usually costs: Most of that method's marks.

• ✗Mistake: Wrong or missing main( ) and object calls

  CISCE examiners' comments: 2024 'object creation or method calls in main wrong'; 2025 'Object creation and method invocation in main( ) wrong, or main( ) omitted; no documentation'.

  ✓Do this: Always end with public static void main(String[] args): create the object with new, call the methods in order; add comments.

  Usually costs: 1-2 marks.

• ✗Mistake: Not invoking a method when told to

  CISCE examiners' comments, 2024: in display( ) candidates 'did not call convert( ), printed hexa before convert, or did not print both values'; 'used return in a void method'.

  ✓Do this: If the question says 'by invoking', call that method inside; match return types exactly.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong starting value in fill-in-the-blank code

  CISCE examiners' comments, 2024: on the smallest-digit method candidates 'Took initial min as 0 or -1' and were 'Unclear about the ? : operator'.

  ✓Do this: Ask what the variable must be before the loop so the first comparison works (minimum starts at 9 or the number itself), then test by tracing.

  Usually costs: 1 mark per blank.

Practice questions on Implementation of Algorithms to Solve Problems

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · Very short answerBoard level

   Name the error caused by a missing semicolon.

   Hint

   Compiler.

   Answer

   A syntax (compile-time) error.

2. Q2 · 1 mark · Very short answerBoard level

   What is a dry run?

   Hint

   By hand.

   Answer

   Executing a program or algorithm by hand, recording the values of variables at each step, to check its output or correctness.

3. Q3 · 1 mark · Very short answerBoard level

   Which exception occurs when an integer is divided by zero in Java?

   Hint

   Arithmetic.

   Answer

   ArithmeticException.

4. Q4 · 1 mark · Very short answerHarder board level

   What is the output of the GCD loop for a = 17, b = 5?

   Hint

   17 % 5 = 2.

   Answer

   1 (17 % 5 = 2, 5 % 2 = 1, 2 % 1 = 0).

5. Q5 · 2 marks · Short answerHarder board level

   [2] Write the loop to count how many digits of n are even.

   Hint

   n % 10 % 2.

   Answer

   int c = 0; while (n > 0) { if ((n % 10) % 2 == 0) c++; n = n / 10; } (for n = 4075 this gives 2: digits 0 and 4).

6. Q6 · 3 marks · Short answerHarder board level

   [3] A method should return true if n is a perfect number (sum of proper divisors equals n). Fill in the blanks: int s = ?1?; for (int d = 1; d < n; d++) if (?2?) s += d; return ?3?;

   Hint

   Divisor test.

   Answer

   ?1? = 0; ?2? = n % d == 0; ?3? = s == n. For n = 28: 1 + 2 + 4 + 7 + 14 = 28, so it returns true.

7. Q7 · 5 marks · Long answerHOTS / competency

   [5] Write an algorithm and a Java method boolean isArmstrong(int n) that checks whether a 3-digit number equals the sum of the cubes of its digits. Give two test values.

   Hint

   153.

   Answer

   Algorithm: copy n to t, s = 0; while t > 0: d = t % 10, s = s + d × d × d, t = t / 10; return s == n.

   boolean isArmstrong(int n)
   {
       int t = n, s = 0;
       while (t > 0)
       {
           int d = t % 10;
           s = s + d * d * d;
           t = t / 10;
       }
       return s == n;
   }

   Tests: 153 → true (1 + 125 + 27); 154 → false (1 + 125 + 64 = 190).

8. Q8 · 10 marks · Long answerHOTS / competency

   [10] Design a class Twin with data members a and b (two integers); a constructor; void accept( ); boolean prime(int x); and void show( ) that prints whether a and b are twin primes (both prime and differing by 2), using prime( ). Write main( ).

   Hint

   Math.abs(a - b) == 2.

   Answer

   Constructor sets a = b = 0; accept( ) reads a and b with Scanner; prime(int x) returns false for x < 2 and checks divisors d with d × d ≤ x; show( ) prints 'Twin primes' if prime(a) && prime(b) && Math.abs(a - b) == 2, else 'Not twin primes'; main( ) creates an object and calls accept( ) and show( ). Tests: (11, 13) → twin primes; (13, 17) → not (difference 4); (7, 9) → not (9 is not prime).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Five properties of an algorithm?Show answer

   Finiteness, definiteness, input, output, effectiveness.

2. 2.Three kinds of error?Show answer

   Syntax, run-time, logical.

3. 3.Order of methods in a Section B answer?Show answer

   Constructor, input, processing, display, main( ).

4. 4.Why test boundary values?Show answer

   Off-by-one and limit errors appear at the edges.

5. 5.GCD of 84 and 36?Show answer

   12.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Algorithm: finite, definite, input, output, effective.
• ☐Design: understand, examples, sub-tasks, steps, dry run, code.
• ☐Trace table: one column per variable, one row per step.
• ☐Test normal, boundary and special cases.
• ☐Errors: syntax (compiler), run-time (exception), logical (wrong answer).
• ☐Section B: exact names and signatures, recursion when asked, main( ), comments.

Implementation of Algorithms to Solve Problems: frequently asked questions