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Variable-Based / Condition-Dependent Puzzle

Introduction

Variable-Based / Condition-Dependent puzzles में आपको ऐसे items को values (numbers, labels, या choices) असाइन करने होते हैं जिनके कुछ assignments clues में दिए गए variable conditions या thresholds पर depend करते हैं - जैसे “if X > Y then A”, “odd positions चुनें”, या “केवल जब condition Z पूरा हो तभी attribute Q असाइन करें”।

ये समस्याएँ advanced reasoning वाले सेक्शन्स में आती हैं क्योंकि ये conditional logic को standard elimination के साथ मिलाती हैं: अक्सर आपको केस में branching करना पड़ता है और अतिरिक्त constraints से invalid cases prune करने होते हैं।

Pattern: Variable-Based / Condition-Dependent Puzzle

Pattern

Conditional clauses को पहचानें, उन्हें explicit case splits में बदलें, और cross-constraints से eliminate करें।

  • Variables या thresholds पहचानें: उदाहरण के लिए numeric values, parity (odd/even), greater/less-than conditions।
  • Cases में विभाजित करें: हर branch को अलग से evaluate करें ( Case A: condition true; Case B: condition false )।
  • Deterministic clues लागू करें: हर केस के अंदर fixed items पहले भरें ताकि branching घटे।
  • Invalid cases prune करें: contradictions या impossibilities से branches हटाएँ और valid case(s) रखें।

Step-by-Step Example

Question

पाँच candidates - A, B, C, D, E - पाँच slots में टेस्ट देते हैं जिनके numbers 1 से 5 हैं (slot 1 earliest)। हर किसी का distinct score है {50, 60, 70, 80, 90}। Clues:

  1. अगर जो candidate slot 1 में है उसका score < 70 है तो A = 90; वरना A = 50।
  2. जिसने 80 score किया वह उसी के ठीक बाद 60-score करने वाले के तुरंत बाद बैठता है। (80 immediately after 60)
  3. B का score C से अधिक पर D से कम है।
  4. E = 70 और E एक odd-numbered slot में है।
Who scored 90?

Options:
A) A    B) B    C) C    D) D

Solution

  1. Step 1: Record deterministic facts

    E = 70 और E एक odd slot में → E ∈ {slot 1, 3, 5}.

  2. Step 2: Identify the conditional clause (branching)

    Clue (1) slot 1 के score पर depend करती है: दो branches बनते हैं -
    • Case A: slot 1 का score < 70 → slot 1 must be 50 → तब A = 90।
    • Case B: slot 1 का score ≥ 70 → slot 1 ∈ {70,80,90} → तब A = 50।

  3. Step 3: Apply fixed-chain clue

    Clue (2): 60 के ठीक बाद 80 आता है. यानी (…60,80…) किसी adjacent slot pair में होना चाहिए (1→2, 2→3, 3→4 या 4→5)। यह E=70 रखते हुए लगानी होगी।

  4. Step 4: Use ordering relation

    Clue (3): B > C और B < D, अतः इन तीनों का ordering C < B < D है।

  5. Step 5: Evaluate Case A (slot 1 = 50 → A = 90)

    1. E = 70 विषम slot में है; पर slot 1 = 50 होने से E ≠ slot 1. अतः E ∈ {slot3, slot5}.
    2. 60→80 pair कहाँ फिट होगा? बचे हुए scores (50 और 70 और A=90 के बाद) देखते हुए 60 और 80 को adjacent रखा जा सकता है; उपलब्ध स्लॉट्स में यह संभवता जाँची जा सकती है।
    3. उदाहरण: E = slot3 (70) लें - तो बाकी स्लॉट्स 2,4,5 में 60 और 80 adjacent के रूप में और बाकी एक score रखा जा सकता है; जल्दी से कोशिश करने पर एक valid filling मिलती है जिसमें A = 90 है। इसलिए Case A feasible है।

  6. Step 6: Evaluate Case B (slot 1 ∈ {70,80,90} → A = 50)

    1. अगर slot 1 = 70 तो E को वही slot लेना होगा (क्योंकि E=70 और E का slot विषम है) - पर इससे adjacency और ordering constraints पर आगे contradictions निकल सकती हैं।
    2. अगर slot 1 = 80 तो 80 के तुरंत पहले 60 होना चाहिए, पर 60 को slot0 में रखकर यह संभव नहीं - अतः 80 slot1 नहीं हो सकता।
    3. अगर slot 1 = 90 तो E = 70 फिर अन्य placements में contradictions आती हैं।
    4. कुल मिलाकर Case B में कोई भी full consistent assignment नहीं मिलती - इसलिए यह invalid है।

  7. Step 7: Conclude from case elimination

    केस elimination से केवल Case A ही consistent बचता है ( slot 1 = 50 → A = 90 )। अतः A ने 90 score किया।

  8. Final Answer:

    A → Option A

  9. Quick Check:

    Case split सही से handle किया गया; slot 1 low होने पर A=90, E=70 odd slot में, 60/80 adjacent फिट होते हैं, और ordering C < B < D working arrangement में satisfy होता है → consistent ✅

Quick Variations

1. Parity-based conditional assignments (odd/even slot)।

2. Threshold rules: जैसे “यदि दो slots का sum > X हो तो special label असाइन करो”।

3. Multi-condition branching (nested if-then chains) - इनमें deeper case trees बनते हैं।

4. Hybrid puzzles: condition-dependent numeric values को seating/position constraints के साथ जोड़ना।

Trick to Always Use

  • Step 1: Conditional clauses को पहचान कर explicit case splits पहले लिख लें।
  • Step 2: हर केस के अंदर deterministic facts पहले भरें - opposites, fixed labels, parity आदि।
  • Step 3: जल्दी से contradictions के लिए हर case टेस्ट करें (contradictions से branch को जल्दी prune करें)।

Summary

Summary

  • हर conditional clue को explicit case(s) में बदलें - उन्हें नाम दें (Case A / Case B)।
  • प्रत्येक case में fixed facts पहले लागू करें, फिर adjacency और ordering rules लगाएँ।
  • Impossible cases को जल्दी eliminate करें ताकि बेकार की खोज न हो।
  • जब केवल एक case बचे तो वही unique, exam-ready solution देता है।

याद रखने योग्य उदाहरण:
पहले condition पर branch करें - फिर हर branch को एक सामान्य puzzle की तरह सुलझाएँ और aggressive pruning करें।

Practice

(1/5)
1. Five students - A, B, C, D, E - occupy exam slots 1 to 5 (1 = earliest). Scores available: 50, 60, 70, 80, 90. Clues: 1) If the score in slot 1 is less than 60 then A scored 90; otherwise A scored 50. 2) The student who scored 60 sits immediately before the student who scored 80. 3) E scored 70 and sits in an odd-numbered slot. 4) B scored higher than C. Who scored 90?
easy
A. A
B. B
C. C
D. D

Solution

  1. Step 1: Identify fixed facts

    Record fixed facts: E = 70 and E is in slot 1, 3 or 5 (odd slot).

  2. Step 2: Branch on the conditional

    Case I - slot 1 score < 60 ⇒ slot1 = 50 ⇒ then A = 90. Case II - slot 1 score ≥ 60 ⇒ slot1 ∈ {60,70,80,90} ⇒ then A = 50.

  3. Step 3: Eliminate infeasible case

    Case II becomes infeasible because adjacency 60→80 cannot be maintained with E in odd slots and A = 50.

  4. Step 4: Accept feasible case

    Case I works: A = 90, E = 70 (odd slot), B > C satisfied.

  5. Final Answer:

    A → Option A

  6. Quick Check:

    A = 90 satisfies all constraints ✅
Hint: Convert 'if-then' statements into two cases and test adjacency first - it usually invalidates one branch immediately.
Common Mistakes: Not testing the alternate branch - adjacency and order constraints often break it.
2. Seven tasks (T1-T7) are scheduled across days Monday-Sunday (one per day). Durations available (hours): 2, 3, 4, 5, 6, 7, 8. Clues: 1) If the duration on Monday is ≤3 hours then task X took 8 hours; otherwise X took 2 hours. 2) The 3-hour task is scheduled immediately before the 5-hour task. 3) Task X is scheduled later in the week than the 4-hour task. Who took 8 hours?
easy
A. Task X
B. The 7-hour task
C. The 6-hour task
D. Task scheduled on Wednesday

Solution

  1. Step 1: Translate the conditional

    Case A - Monday ≤ 3 ⇒ X = 8. Case B - Monday ≥ 4 ⇒ X = 2.

  2. Step 2: Apply adjacency

    The 3-hour task immediately before the 5-hour task limits placement (e.g., Mon→Tue, Tue→Wed).

  3. Step 3: Test branches

    Case B fails after placement checks because it violates 'X after 4-hour' condition. Case A works smoothly.

  4. Step 4: Confirm consistent branch

    Case A → Monday ≤ 3, X = 8 satisfies all conditions.

  5. Final Answer:

    Task X → Option A

  6. Quick Check:

    All adjacency and timing constraints satisfied ✅
Hint: Start with the 'else' branch in scheduling problems - adjacency usually breaks it.
Common Mistakes: Forgetting to check task order when applying time-based conditions.
3. Four projects - P, Q, R, S - are assigned priority numbers 1-4 (1 = highest). Clues: 1) If project P’s priority number is odd then Q’s priority = 4; otherwise Q’s priority = 1. 2) R’s priority is numerically between Q and S. 3) P does not have priority 2. Which project has priority 4?
easy
A. P
B. Q
C. R
D. S

Solution

  1. Step 1: Branch by parity

    P odd ⇒ Q = 4; P even ⇒ Q = 1.

  2. Step 2: Eliminate invalid case

    P ≠ 2, so even case means P = 4 → Q = 1. That breaks the 'R between Q and S' condition because no valid number lies between 1 and 2 or 3 simultaneously.

  3. Step 3: Accept valid case

    P odd → Q = 4 fits all remaining rules.

  4. Final Answer:

    Q → Option B

  5. Quick Check:

    R between Q(4) and S valid only under odd P case ✅
Hint: Test numeric constraints after each branch - 'between' clues eliminate false cases fast.
Common Mistakes: Ignoring the numeric gap between ranks; it invalidates many branches quickly.
4. Six contestants - A-F - receive badge colours from {Red, Blue, Green, Yellow, Black, White} and sit in positions 1-6. Rules: 1) If position 2 is Red or Blue then A gets Black; otherwise A gets White. 2) The Green badge holder sits immediately after the Yellow badge holder. 3) B's colour is alphabetically earlier than C's colour. 4) D sits in a higher-numbered position than E. Who wears Black?
medium
A. A
B. B
C. C
D. D

Solution

  1. Step 1: Analyze conditional case

    Case X - pos2 ∈ {Red,Blue} ⇒ A = Black; Case Y - pos2 ∉ {Red,Blue} ⇒ A = White.

  2. Step 2: Apply adjacency of colours

    Green immediately after Yellow forms a consecutive pair Y→G; restricts placement of position 2.

  3. Step 3: Test Case X

    Case X leads to contradiction in Y→G placement and alphabetical rule B<C.

  4. Step 4: Accept Case Y

    Case Y allows a valid placement where C gets Black and all constraints hold.

  5. Final Answer:

    C → Option C

  6. Quick Check:

    All conditions hold only under Case Y ✅
Hint: For colour-based seating, always test adjacency before alphabetical constraints.
Common Mistakes: Overlooking that Y→G means consecutive seats, not random positions.
5. Five managers - P, Q, R, S, and T - receive different bonuses of ₹1,000, ₹2,000, ₹3,000, ₹4,000, and ₹5,000. Conditions: 1) If P’s bonus is less than ₹3,000 then Q gets the highest bonus; otherwise Q gets the lowest bonus. 2) R’s bonus is double S’s bonus. 3) T’s bonus is less than R’s bonus. 4) P < ₹3,000. Who receives ₹5,000?
medium
A. P
B. Q
C. R
D. S

Solution

  1. Step 1: List available values

    Bonuses = ₹1k, ₹2k, ₹3k, ₹4k, ₹5k.

  2. Step 2: Identify fixed relation

    R = 2×S; T < R. Valid (S,R) pairs = (₹1k,₹2k) or (₹2k,₹4k).

  3. Step 3: Apply given P < ₹3k

    This activates IF branch → Q = ₹5,000.

  4. Step 4: Assign remaining logically

    P = ₹1k, S = ₹2k, R = ₹4k, T = ₹3k. All clues satisfied.

  5. Final Answer:

    Q - ₹5,000 → Option B

  6. Quick Check:

    P<₹3k⇒Q=₹5k ✅ R=2×S ✅ T<R ✅
Hint: Use the numeric relation (×2) first; it quickly prunes inconsistent options.
Common Mistakes: Mixing up who ‘gets highest’ when IF condition changes based on inequality.

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