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Zorluk: Çok zorCoding and Decoding

In a certain code system, words are transformed according to the following rules:
1. Every vowel is replaced by its reverse alphabetical counterpart (where A↔ZA \leftrightarrow Z, E↔VE \leftrightarrow V, I↔RI \leftrightarrow R, O↔LO \leftrightarrow L, U↔FU \leftrightarrow F, such that the sum of the positional values of the original and transformed letter is 2727).
2. Every consonant located at an odd-numbered position in the word (1st, 3rd, 5th...) is replaced by the letter 33 steps ahead of it in the alphabetical series (+3+3).
3. Every consonant located at an even-numbered position in the word (2nd, 4th, 6th...) is replaced by the letter 22 steps behind it in the alphabetical series (−2-2).

If the word 'TRIANGLE' is coded as 'WPRZQEOV', which of the following represents the correct code for the word 'SOLUTION' in this language?

  1. VLOFWRLLCevap
  2. B
    VLOFWRLQ
  3. C
    QLJFRRLL
  4. D
    VKOEWQKL

Cevap

The correct code for the word 'SOLUTION' is 'VLOFWRLL'.
The correct code 'VLOFWRLL' is obtained by applying the specified rules systematically: 'S' (1st letter, consonant at odd position) becomes 'V' (+3+3); 'O' (2nd letter, vowel) becomes its reverse letter 'L' (27−15=1227-15=12); 'L' (3rd letter, consonant at odd position) becomes 'O' (+3+3); 'U' (4th letter, vowel) becomes its reverse letter 'F' (27−21=627-21=6); 'T' (5th letter, consonant at odd position) becomes 'W' (+3+3); 'I' (6th letter, vowel) becomes its reverse letter 'R' (27−9=1827-9=18); 'O' (7th letter, vowel) becomes its reverse letter 'L' (27−15=1227-15=12); and 'N' (8th letter, consonant at even position) becomes 'L' (−2-2).

Adım Adım Çözüm

1
Analyze the rule verification on the example word 'TRIANGLE'
Position 1 (T, Consonant, Odd) →T+3=W\rightarrow T+3 = W; Position 2 (R, Consonant, Even) →R−2=P\rightarrow R-2 = P; Position 3 (I, Vowel) →27−9=18(R)\rightarrow 27-9 = 18 (R); Position 4 (A, Vowel) →27−1=26(Z)\rightarrow 27-1 = 26 (Z); Position 5 (N, Consonant, Odd) →N+3=Q\rightarrow N+3 = Q; Position 6 (G, Consonant, Even) →G−2=E\rightarrow G-2 = E; Position 7 (L, Consonant, Odd) →L+3=O\rightarrow L+3 = O; Position 8 (E, Vowel) →27−5=22(V)\rightarrow 27-5 = 22 (V). Output: 'WPRZQEOV'.
Confirms that the three-part rule set applies strictly by letter type and 1-indexed letter position.
2
Apply rules to the target word 'SOLUTION' letter by letter
1st letter 'S' (Odd pos, Consonant): S(19)+3=22(V)S (19) + 3 = 22 (V).
2nd letter 'O' (Vowel): Reverse of O(15)O (15) is 27−15=12(L)27 - 15 = 12 (L).
3rd letter 'L' (Odd pos, Consonant): L(12)+3=15(O)L (12) + 3 = 15 (O).
4th letter 'U' (Vowel): Reverse of U(21)U (21) is 27−21=6(F)27 - 21 = 6 (F).
5th letter 'T' (Odd pos, Consonant): T(20)+3=23(W)T (20) + 3 = 23 (W).
6th letter 'I' (Vowel): Reverse of I(9)I (9) is 27−9=18(R)27 - 9 = 18 (R).
7th letter 'O' (Vowel): Reverse of O(15)O (15) is 27−15=12(L)27 - 15 = 12 (L).
8th letter 'N' (Even pos, Consonant): N(14)−2=12(L)N (14) - 2 = 12 (L).
Evaluates each character in 'SOLUTION' according to its specific classification and position.
3
Combine transformed letters to get final coded string
'VLOFWRLL'
Concatenates individual output characters into the complete coded string.

Anahtar Kavram

Multi-rule positional letter shifting combined with reverse alphabetical indexing (27−n27 - n rule).
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Konular

Coding and Decoding

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