Building a high-precision π calculator in Scratch
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MathWizard_Alex
Posted on March 14, 2024 • Advanced
🥧 Need help creating the fastest π calculator possible
Hey math enthusiasts! I’m working on a high-precision π calculator project and need help optimizing it for speed and accuracy. My current approach is too slow for multiple iterations.
My goals:
- Calculate π to at least 10-15 decimal places
- Use iterative methods for better precision
- Optimize for speed - need thousands of iterations
- Compare different mathematical approaches
- Display results in real-time
I know the basic circumference/diameter approach, but that’s limited by Scratch’s decimal precision. I need algorithms that can converge quickly to π with high accuracy. Any advanced mathematical techniques would be amazing! 🔢
Current project: My π Calculator Attempt
NumericalMethods_Sarah
Replied 2 hours later • ⭐ Best Answer
Excellent question @MathWizard_Alex! Creating high-precision π calculators is a fantastic way to explore numerical methods. Here are several powerful algorithms you can implement:
🧮 π Calculation Algorithm Comparison
Here’s how different methods compare for speed and accuracy:
📊 Method 1: Leibniz Series (Educational)
Great for understanding convergence, but slow:
when flag clicked // Leibniz series: π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ... set [pi estimate v] to [0] set [iterations v] to [0] set [sign v] to [1] set [denominator v] to [1] repeat [10000] change [pi estimate v] by ((sign) / (denominator)) set [sign v] to ((sign) * [-1]) change [denominator v] by [2] change [iterations v] by [1] // Display progress every 100 iterations if <((iterations) mod [100]) = [0]> then set [current pi v] to ((pi estimate) * [4]) say (join [π ≈ ] (current pi)) for [0.1] seconds end end set [final pi v] to ((pi estimate) * [4]) say (join [Final π = ] (final pi))
🚀 Method 2: Machin Formula (Fast Convergence)
Much faster convergence using arctangent series:
// Machin's formula: π/4 = 4*arctan(1/5) - arctan(1/239) when flag clicked set [pi result v] to [0] // Calculate 4*arctan(1/5) arctan series [1] [5] [4] set [term1 v] to (arctan result) // Calculate arctan(1/239) arctan series [1] [239] [1] set [term2 v] to (arctan result) // Combine using Machin's formula set [pi result v] to (((term1) - (term2)) * [4]) say (join [Machin π = ] (pi result)) // Custom block for arctan series define arctan series (numerator) (denominator) (multiplier) set [arctan result v] to [0] set [x v] to ((numerator) / (denominator)) set [x power v] to (x) set [factorial v] to [1] set [sign v] to [1] repeat [50] // 50 terms gives excellent precision set [term v] to (((sign) * (x power)) / (factorial)) change [arctan result v] by (term) // Update for next iteration set [x power v] to ((x power) * (x) * (x)) change [factorial v] by [2] set [sign v] to ((sign) * [-1]) end set [arctan result v] to ((arctan result) * (multiplier))
🎲 Method 3: Monte Carlo Estimation
Probabilistic approach - great for visualization:
when flag clicked // Monte Carlo method using random points in a circle set [total points v] to [0] set [points in circle v] to [0] set [pi estimate v] to [0] repeat [100000] // Generate random point in unit square set [x v] to (pick random [-100] to [100]) set [y v] to (pick random [-100] to [100]) // Convert to unit circle coordinates set [x coord v] to ((x) / [100]) set [y coord v] to ((y) / [100]) // Check if point is inside unit circle set [distance squared v] to (((x coord) * (x coord)) + ((y coord) * (y coord))) change [total points v] by [1] if <(distance squared) < [1]> then change [points in circle v] by [1] // Optional: draw point in green pen up go to x: (x) y: (y) set pen color to [#00ff00] pen down move [1] steps pen up else // Optional: draw point in red pen up go to x: (x) y: (y) set pen color to [#ff0000] pen down move [1] steps pen up end // Calculate π estimate: π ≈ 4 * (points in circle / total points) if <(total points) > [0]> then set [pi estimate v] to ([4] * ((points in circle) / (total points))) end // Update display every 1000 points if <((total points) mod [1000]) = [0]> then say (join [Monte Carlo π ≈ ] (pi estimate)) for [0.1] seconds end end
⚡ Method 4: Optimized Chudnovsky-Style Algorithm
Extremely fast convergence (each term adds ~14 digits):
// Simplified Chudnovsky-inspired algorithm when flag clicked set [pi inverse v] to [0] set [k v] to [0] repeat [10] // Just 10 iterations for high precision! // Calculate factorial terms (simplified) set [factorial 6k v] to [1] set [temp k v] to (k) repeat ((k) * [6]) set [factorial 6k v] to ((factorial 6k) * (temp k)) change [temp k v] by [-1] end // Calculate the series term set [numerator v] to ((factorial 6k) * ((13591409) + ((k) * [545140134]))) set [denominator v] to [1] // Calculate 3^k and other terms set [power 3k v] to [1] repeat (k) set [power 3k v] to ((power 3k) * [3]) end set [denominator v] to ((power 3k) * [640320]) // Add term to series if <((k) mod [2]) = [0]> then change [pi inverse v] by ((numerator) / (denominator)) else change [pi inverse v] by ([0] - ((numerator) / (denominator))) end change [k v] by [1] end // Convert to π set [pi result v] to ([426880] / ((pi inverse) * [10005])) say (join [Chudnovsky π = ] (pi result))
📈 Performance Optimization Tips
Speed Optimization:
// Use efficient variable updates define optimize calculation speed // Pre-calculate constants set [constant 1 v] to [3.14159265359] set [constant 2 v] to [1.41421356237] // Use integer arithmetic when possible set [scaled value v] to ((original value) * [1000000]) // Do calculations with integers set [result scaled v] to ((scaled value) * (other scaled value)) // Scale back down set [final result v] to ((result scaled) / [1000000000000]) // Minimize expensive operations if <(expensive calculation needed) = [true]> then expensive calculation set [expensive calculation needed v] to [false] end
Precision Display:
// Format π to desired decimal places define format pi to (decimal places) set [multiplier v] to [1] repeat (decimal places) set [multiplier v] to ((multiplier) * [10]) end set [rounded pi v] to ([round v] of ((pi result) * (multiplier))) set [formatted pi v] to ((rounded pi) / (multiplier)) // Create display string set [pi string v] to (formatted pi) if <(length of (pi string)) < ((decimal places) + [2])> then // Add trailing zeros if needed set [pi string v] to (join (pi string) [0]) end
🎯 Real-Time Comparison Display
// Compare multiple methods simultaneously when flag clicked clear set [comparison mode v] to [true] // Start all methods broadcast [start leibniz v] broadcast [start machin v] broadcast [start monte carlo v] // Display comparison forever go to x: [-200] y: [150] set pen color to [#000000] write (join [Leibniz: ] (leibniz pi)) go to x: [-200] y: [100] write (join [Machin: ] (machin pi)) go to x: [-200] y: [50] write (join [Monte Carlo: ] (monte carlo pi)) go to x: [-200] y: [0] set pen color to [#ff0000] write (join [True π: ] [3.14159265358979323846]) wait [0.5] seconds clear end
🏆 Advanced Features
- Convergence Analysis: Track how quickly each method approaches π
- Error Calculation: Show the difference from the true value
- Iteration Timing: Measure performance of different algorithms
- Visual Representation: Graph the convergence over time
The Machin formula is your best bet for speed and accuracy in Scratch! It converges much faster than simple series and gives excellent precision with just 50-100 terms. 🎯
MathWizard_Alex
Replied 5 hours later
@NumericalMethods_Sarah This is absolutely incredible! 🤯 Thank you so much!
I implemented the Machin formula and it’s giving me 12 decimal places of accuracy with just 30 iterations! The Monte Carlo visualization is also really cool - you can actually see π emerging from the random points.
Quick question: How can I optimize the factorial calculations for the Chudnovsky method? They seem to be the bottleneck.
OptimizationExpert_David
Replied 1 hour later
@MathWizard_Alex Great question! For factorial optimization, use incremental calculation:
// Optimized factorial calculation define calculate factorial incrementally (n) if <(n) = [0]> then set [factorial result v] to [1] else if <(previous n) = ((n) - [1])> then // Use previous result set [factorial result v] to ((previous factorial) * (n)) else // Calculate from scratch set [factorial result v] to [1] set [i v] to [1] repeat (n) set [factorial result v] to ((factorial result) * (i)) change [i v] by [1] end end end set [previous n v] to (n) set [previous factorial v] to (factorial result)
This avoids recalculating the entire factorial each time! Also consider using Stirling’s approximation for very large factorials. 📈
Vibelf_Community
Pinned Message • Moderator
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