Coding–Decoding – Study Notes
Andaman & Nicobar Islands PSC – Reasoning Ability
Objective: Master the art of deciphering hidden relationships between symbols, letters, numbers, or objects.
Exam‑style: Focus on pattern recognition, logical deduction, and speed.
1. What is Coding‑Decoding?
| Coding | Decoding |
|---|---|
| A → 1 | 1 → A |
| B → 2 | 2 → B |
| … | … |
| X → 24 | 24 → X |
- Coding: Assigning a code (number, symbol, or word) to a given item.
- Decoding: Finding the original item from a given code.
Typical PSC format:
- 10 symbols in a row → find 6th symbol
- 4 numbers → find 5th number
- 3 objects → find 2nd object
2. Classification of Coding‑Decoding Questions
| Type | Example | Key Insight |
|---|---|---|
| Linear/Sequential | A B C D → 5 | Simple arithmetic progression |
| Arithmetic/Geometric | 2, 4, 8, 16 → 5 | Identify common ratio/difference |
| Pattern in Letters | A, C, E, G → ? | Skip‑letter pattern |
| Mixed Symbols | X, Y, Z, 1, 2, 3 → ? | Interleaved patterns |
| Cyclic | 1, 4, 7, 10 → 13 | Modulo arithmetic |
| Transposition | 5, 4, 3, 2, 1 → ? | Reverse order |
| Conditional | 1, 3, 5, 7 → 9 | Condition on parity, etc. |
| Coded Words | “APPLE” → “QTPOL” | Each letter shifted by a fixed amount |
3. General Strategy for Solving
| Step | What to Do | Why |
|---|---|---|
| 1. Identify the Domain | Letters, numbers, symbols, objects | Different methods apply |
| 2. Look for Simple Relations | +1, -1, ×2, ÷3, +3, –3, etc. | Most coding questions use basic arithmetic |
| 3. Check for Alternating Patterns | Even–odd positions, 1–3–5 pattern | Often hidden |
| 4. Test for Cycles | Modulo operations, 1–4–7–10 → 13 → 1 | Use mod to detect wrap‑around |
| 5. Verify Uniqueness | Does one rule fit all given items? | Avoid multiple answers |
| 6. Cross‑Check | Use the found rule to check all positions | Ensure consistency |
| 7. Record the Rule | Write the rule clearly | Useful for exam‑style speed |
4. Common Coding Patterns & Their “Formulas”
| Pattern | Formula | Example |
|---|---|---|
| Increment by k | a(n) = a(1) + (n-1)*k |
2, 5, 8, 11 → k=3 |
| Multiply by k | a(n) = a(1) * k^(n-1) |
3, 9, 27, 81 → k=3 |
| Alternating +, – | a(n) = a(1) + (-1)^(n-1) * k |
5, 8, 5, 8 → k=3 |
| Skip Letters | a(n) = Char(ASCII('A') + (n-1)*k) |
A, D, G → k=3 |
| Cyclic Modulo | a(n) = ((a(1)-1 + (n-1)*k) mod N) + 1 |
1,4,7,10 (mod 13) |
| Reverse Sequence | a(n) = a(1) - (n-1) |
9,8,7,6 → k=1 |
| Interleaved Pattern | Two sequences merged | 1, A, 2, B, 3, C |
5. Tricky Points & Common Pitfalls
| Pitfall | How to Avoid |
|---|---|
| Assuming Linear When It’s Not | Test alternate patterns before settling |
| Missing the Start Point | Verify if the first element follows the rule |
| Wrong Modulus | Remember to add 1 after modulo for 1‑based sequences |
| Ignoring Mixed Domains | Separate symbols into their own sub‑patterns |
| Skipping the “Zero” Index | Some patterns start with 0 (e.g., 0,3,6,9) |
| Forgetting About Repetition | Check if the sequence repeats after a few terms |
| Over‑Complicating | Start with simplest rule; only go deeper if necessary |
6. Step‑by‑Step Worked Example
Question:
2, 5, 11, 23, 47 → ?
Solution:
- Domain: Numbers.
- Check for simple arithmetic:
- Differences: 3, 6, 12, 24 → each difference doubles.
- Rule:
a(n+1) = a(n) + 3*2^(n-1)- Starting with
a(1)=2.
- Starting with
- Compute next term:
a(5)=47a(6)=47 + 3*2^4 = 47 + 48 = 95
- Answer: 95
7. Hard Practice Question
Question
X, 3, 7, Y, 19, Z, 43, 71, ?, 127