Oracle Certification /Reasoning Ability

Coding-decoding

Verbal Reasoning · free study notes

Coding‑Decoding in Oracle Certification – Reasoning Ability

Advanced, MIT‑level treatment (first principles → theory → practice)


1. What is Coding‑Decoding?

Aspect Definition
Coding A deterministic mapping  (f:\mathcal{A}\rightarrow\mathcal{B})  between two alphabets (usually the same alphabet).
Decoding The inverse operation (f^{-1}) (when it exists). In exams we’re typically asked to infer the mapping from a few examples and then apply it.
Goal Recognise the hidden pattern, reconstruct the mapping, and use it to decode new items.

Typical exam format:

  • A set of letter/word pairs is given.
  • One or more of the pairs is missing.
  • You must supply the missing code/decoded word.

2. First Principles – How to Formalise the Problem

  1. Alphabet
    [ \mathcal{A} = {A,B,\dots ,Z} \quad (|\mathcal{A}|=26) ]

  2. Mapping
    [ f:\mathcal{A}\rightarrow\mathcal{A} ] Often a bijection (injective & surjective).
    If not bijective, at least injective (no two letters have the same code).

  3. Encoding of a word
    For word (w=w_1w_2\ldots w_n), [ \operatorname{enc}(w)=f(w_1)f(w_2)\ldots f(w_n). ]

  4. Decoding
    If (f) is bijective, decode by (f^{-1}).
    If only partial information is available, we work symbolically:
    [ f(A) = X,; f(B)=Y,\dots ]


3. Rigorous Reasoning – Step‑by‑Step Strategy

Step Action Rationale
1. Collect all given pairs Write them in a table. Avoid missing clues.
2. Identify pattern types Check for:
• Arithmetic (shift)
• Reverse alphabet
• Position‑based (1↔26)
• Mixed permutations
Narrow the search space.
3. Test hypotheses Apply each candidate mapping to all given pairs. Verify consistency (no contradictions).
4. Resolve ambiguities If multiple mappings fit, look for additional constraints:
• Repeated letters in words
• Unmapped letters in the alphabet
Force uniqueness.
5. Decode missing items Once mapping is fixed, apply (f) or (f^{-1}). Final answer.

3.1 Common Mapping Families

Family Formula Example
Caesar Shift (f(X)=X+k \ (\text{mod }26)) Shift 3: A→D, B→E
Reverse (f(X)=\text{Reverse}(X)=\text{Z}-(X-\text{A})) A→Z, B→Y
Symmetric (pairwise) (f(A)=Z, f(B)=Y, \dots) Same as reverse
Position‑Based (mirror) (f(\text{letter at pos }i)=\text{letter at pos }(27-i)) A↔Z, B↔Y
Custom Permutation (f) defined by a scrambled alphabet A→Q, B→V, …

4. Edge Cases & Pitfalls

Edge Case What Happens How to Spot Fix
Repeated source letter mapping to different codes Contradicts injectivity Check each pair for repeated letters Reject mapping
Incomplete alphabet Multiple possible bijections Use any unused letters to complete Enumerate possibilities
Non‑bijection (many‑to‑one) No unique decoding Accept that decoding may be ambiguous

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