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Zorluk: ZorCoding and Decoding

In a certain code system, words are transformed according to the following rules based on letter positions indexed 1,2,3,,n1, 2, 3, \dots, n from left to right:
1. Letters at odd position indices (1,3,5,1, 3, 5, \dots) are replaced by their reverse alphabetical counterparts (where AZA \leftrightarrow Z, BYB \leftrightarrow Y, CXC \leftrightarrow X, etc.).
2. Letters at even position indices (2,4,6,2, 4, 6, \dots) are replaced by the letter shifted forward in the English alphabet by a number of positions equal to their 1-based index (wrapping around from ZZ to AA if necessary).

Based on these rules, what is the correct code for the word 'JOURNEY'?

  1. QQFVMKBCevap
  2. B
    KQXVSKF
  3. C
    PQEVLKA
  4. D
    KLXISVF

Cevap

QQFVMKB
The correct answer 'QQFVMKB' is obtained by applying the two positional rules step-by-step: at odd positions (1, 3, 5, 7), the reverse alphabetical positions (27rank27 - \text{rank}) yield Q, F, M, and B respectively. At even positions (2, 4, 6), adding the letter's 1-based index to its rank gives Q (15 + 2 = 17), V (18 + 4 = 22), and K (5 + 6 = 11). Combining these yields 'QQFVMKB'.

Adım Adım Çözüm

1
Identify the 1-based position index and standard alphabetical rank of each letter in 'JOURNEY'
J (pos 1, rank 10), O (pos 2, rank 15), U (pos 3, rank 21), R (pos 4, rank 18), N (pos 5, rank 14), E (pos 6, rank 5), Y (pos 7, rank 25).
Establishing letter positions and numerical ranks is required before applying positional transformation rules.
2
Apply reverse alphabetical substitution (27rank27 - \text{rank}) to letters at odd positions (1, 3, 5, 7)
Position 1: 2710=17Q27 - 10 = 17 \rightarrow \text{Q}; Position 3: 2721=6F27 - 21 = 6 \rightarrow \text{F}; Position 5: 2714=13M27 - 14 = 13 \rightarrow \text{M}; Position 7: 2725=2B27 - 25 = 2 \rightarrow \text{B}.
Letters at odd position indices must be replaced by their reverse counterparts.
3
Apply forward shift by position index (rank+index)(\text{rank} + \text{index}) to letters at even positions (2, 4, 6)
Position 2: 15+2=17Q15 + 2 = 17 \rightarrow \text{Q}; Position 4: 18+4=22V18 + 4 = 22 \rightarrow \text{V}; Position 6: 5+6=11K5 + 6 = 11 \rightarrow \text{K}.
Letters at even position indices are shifted forward by their respective 1-based index.
4
Combine the transformed letters in order from position 1 to 7
The encoded string is QQFVMKB.
Assembling the results from all letter positions yields the complete final code.

Anahtar Kavram

Positional Hybrid Coding (Reverse Indexing combined with Index-Based Forward Shift)
Tahmini Süre:1m 30s
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