Poor Pigs - Problem

🐷 The Poor Pigs Problem

You work in a laboratory where there are buckets buckets of liquid, and exactly one of them is poisonous. Your mission is to identify which bucket contains the poison using the minimum number of pigs possible.

The Testing Process:

  • You have minutesToTest minutes total to find the poisonous bucket
  • Each pig takes minutesToDie minutes to die after consuming poison
  • You can feed multiple buckets to the same pig simultaneously
  • You can feed the same bucket to multiple pigs
  • After feeding, you must wait minutesToDie minutes to see results
  • You can repeat this process until time runs out

Goal: Return the minimum number of pigs needed to guarantee finding the poisonous bucket within the time limit.

Example: With 1000 buckets, 15 minutes to die, and 60 minutes to test, you can conduct 4 tests (60/15 = 4). Each pig can have 5 states (alive after 0, 1, 2, 3, or 4 tests). Since 5³ = 125 < 1000 < 5⁴ = 625, you need at least 4 pigs.

Input & Output

example_1.py — Basic Case
$ Input: buckets = 1000, minutesToDie = 15, minutesToTest = 60
› Output: 5
💡 Note: We can conduct 60/15 = 4 tests. Each pig can have 5 states (die after test 1, 2, 3, 4, or never die). With 4 pigs: 5^4 = 625 < 1000, so we need 5 pigs where 5^5 = 3125 ≥ 1000.
example_2.py — Single Bucket
$ Input: buckets = 1, minutesToDie = 1, minutesToTest = 1
› Output: 0
💡 Note: With only 1 bucket, there's no poison to find, so 0 pigs are needed.
example_3.py — Limited Time
$ Input: buckets = 4, minutesToDie = 15, minutesToTest = 15
› Output: 2
💡 Note: Only 1 test possible (15/15 = 1). Each pig has 2 states (die or survive). With 2 pigs: 2^2 = 4 buckets can be distinguished.

Constraints

  • 1 ≤ buckets ≤ 1000
  • 1 ≤ minutesToDie ≤ minutesToTest ≤ 100
  • minutesToTest is always divisible by minutesToDie

Visualization

Tap to expand
🐷 Pig States Encoding SystemExample: 4 buckets, 2 test rounds → Each pig has 3 states🐷 Pig A: State 0=Dies Round 1, State 1=Dies Round 2, State 2=Never Dies🐷 Pig B: State 0=Dies Round 1, State 1=Dies Round 2, State 2=Never DiesBucket 1Pig A: State 0Pig B: State 0Bucket 2Pig A: State 0Pig B: State 1Bucket 3Pig A: State 1Pig B: State 0Bucket 4Pig A: State 1Pig B: State 1🎯 Key Insight: Multi-dimensional EncodingEach pig represents a 'digit' in base-(tests+1)2 pigs with 3 states each = 3² = 9 unique combinationsFormula: ceiling(log(buckets) / log(states)) pigs needed
Understanding the Visualization
1
Calculate Test Rounds
Determine how many complete test cycles fit in the time limit
2
Determine States
Each pig can die after round 1, 2, 3... or never die (poison-free)
3
Encode Buckets
Map each bucket to a unique combination of pig death timings
4
Apply Formula
Use logarithms to find minimum pigs needed
Key Takeaway
🎯 Key Insight: This problem brilliantly demonstrates how information theory and multi-dimensional encoding can dramatically reduce resource usage - from potentially needing 1000 pigs to just 5!

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