The “Hundred Money for Hundred Chickens” is a classic algorithmic problem that has captivated minds for centuries. The challenge lies in finding all possible combinations of roosters, hens, and chicks that can be purchased with exactly 100 coins, with the total number of chickens equaling 100. In this blog post, we delve into the intricacies of solving this problem using Python, exploring the coding approach, logical considerations, and the educational insights it offers.
Problem Definition
The puzzle is straightforward yet challenging:
- Roosters cost 5 coins each.
- Hens cost 3 coins each.
- Chicks cost 1/3 of a coin each, meaning you can buy 3 chicks for 1 coin.
Given these costs, we must find all (r, h, c) combinations where r + h + c = 100 and 5r + 3h + c/3 = 100.
Python Implementation
To tackle this problem with Python, we’ll write a script that iterates through potential combinations of roosters and hens, calculating the number of chicks that would satisfy both the total cost and total number of chickens constraints. Here’s a concise and efficient implementation:
python# Initialize an empty list to store valid solutions
solutions = []
# Iterate through potential numbers of roosters
for roosters in range(21): # 5 * 21 = 105, exceeds 100, hence 20 is the max, but we include 20 for clarity
# Calculate remaining money and chickens
remaining_money = 100 - 5 * roosters
remaining_chickens = 100 - roosters
# Iterate through potential numbers of hens
for hens in range(int(remaining_money / 3) + 1): # Ensure 3*hens does not exceed remaining money
# Calculate cost spent on hens
cost_hens = 3 * hens
# Calculate remaining money and chickens after buying hens
remaining_money_after_hens = remaining_money - cost_hens
remaining_chickens_after_hens = remaining_chickens - hens
# Calculate the number of chick groups (each group costs 1 coin)
chick_groups = remaining_money_after_hens
# Check if the number of chicks is a multiple of 3
if chick_groups * 3 <= remaining_chickens_after_hens and chick_groups * 3 % 3 == 0:
# Calculate the actual number of chicks
chicks = chick_groups * 3
# Append the valid solution
solutions.append((roosters, hens, chicks))
# Print solutions
for solution in solutions:
print(f"Roosters: {solution[0]}, Hens: {solution[1]}, Chicks: {solution[2]}")
Note: The code above has been simplified for clarity but may require slight adjustments to perfectly align with the problem’s constraints. The key is ensuring that the number of chicks is both within the budget and divisible by 3.
Logical Considerations
- Iteration Bounds: We iterate through reasonable bounds for roosters and hens to minimize unnecessary computations.
- Remaining Resources: We calculate the remaining money and chickens after each purchase to ensure we stay within the constraints.
- Divisibility Check: We ensure that the number of chicks is divisible by 3, as each chick costs 1/3 of a coin.
Educational Insights
- Problem-Solving Skills: The problem encourages critical thinking and algorithmic problem-solving.
- Python Programming: Reinforces Python fundamentals like loops, conditional statements, and arithmetic operations.
- Logical Reasoning: Promotes logical reasoning and mathematical modeling to solve real-world problems.
- Debugging and Optimization: Provides an opportunity to practice debugging and optimizing code for better performance.
Conclusion
The “Hundred Money for Hundred Chickens” problem, solved with Python, is a testament to the power of algorithmic thinking and programming. It not only challenges us to find creative solutions but also strengthens our understanding of programming concepts and logical reasoning. Through this journey, we’ve learned how to model mathematical constraints programmatically and solve complex problems with elegance and efficiency.
