Code-language inference: “Sue Re Nik” = “She is brave”; “Pi Sor re nik” = “She is always smiling”; “Sor Re Zhi” = “is always cheerful”. What is the code for the word “smiling”?

Difficulty: Easy

Correct Answer: Pi

Explanation:


Introduction / Context:
This is a classic code-to-word mapping problem. We compare multiple encoded sentences that share some words and differ in others, then align overlaps to deduce which code token maps to which English word. The focus word to decode here is “smiling”.


Given Data / Assumptions:

  • “Sue Re Nik” → “She is brave”.
  • “Pi Sor re nik” → “She is always smiling”.
  • “Sor Re Zhi” → “is always cheerful”.
  • Capitalization in code tokens is irrelevant; code tokens are distinct symbols.


Concept / Approach:
Find common words across sentences and match them to common code tokens. “is” appears in all three English sentences, so its code must appear in all three code lines. “She” appears in the first and second; “always” appears in the second and third. “smiling” is present only in the second, so after identifying the shared tokens, we can isolate the remaining token as the code for “smiling”.


Step-by-Step Solution:

Common across all three English lines: “is”. Common across all three code lines: “Re”. Therefore, Re = is.Common across lines 2 and 3 in English: “always”. Common across lines 2 and 3 in code: “Sor”. Therefore, Sor = always.Common across lines 1 and 2 in English: “She”. Common across lines 1 and 2 in code: “Nik”. Therefore, Nik = She.In line 2, the unmatched English word is “smiling”. The unmatched code token in line 2 is “Pi”. Hence, Pi = smiling.


Verification / Alternative check:
Substitute all discovered mappings back into the three lines: they reconstruct the English sentences exactly (She/Nik, is/Re, always/Sor), leaving Pi solely for “smiling”.


Why Other Options Are Wrong:

  • Nik: Maps to “She”, not “smiling”.
  • Re: Maps to “is”.
  • Sor: Maps to “always”.
  • None of these: Incorrect because “Pi” fits uniquely.


Common Pitfalls:
Assuming one-to-many mappings or ignoring the strict overlap logic; in these problems, one-to-one mapping is typically enforced by the dataset.


Final Answer:
Pi

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