⚡️ Speed up function fibonacci by 6,659%

[codeflash-ai]

📄 6,659% (66.59x) speedup for fibonacci in code_to_optimize/js/code_to_optimize_js_esm/fibonacci.js

⏱️ Runtime : 3.26 milliseconds → 48.2 microseconds (best of 250 runs)

📝 Explanation and details

Primary benefit — dramatic runtime improvement: the optimized version reduces total runtime from ~3.26 ms to ~48.2 μs (a ~6658% speedup) by eliminating the huge amount of duplicated work in the naive recursive algorithm.

What changed (specific optimizations)

• Added per-call memoization using a Map. The optimized fibonacci creates a Map cache and an inner fib(k) that:
  • returns k for the base case k <= 1 (same as before),
  • checks cache.get(k) and returns a cached result when present,
  • computes result = fib(k-1) + fib(k-2), stores it with cache.set(k, result), and returns it.
• The recursion structure is preserved (same call shape/stack depth), but repeated subcalls are avoided by caching each distinct k once.

Why this speeds things up

• The original naive recursion recomputes the same Fibonacci numbers exponentially many times: time complexity ~O(phi^n). The memoized version computes each distinct Fibonacci value up to n only once, giving an overall time complexity ~O(n) for distinct inputs (with O(n) extra space for the cache).
• Map.get / Map.set are O(1) on average, so the overhead per computed Fibonacci number is tiny compared to re-running exponential subtrees.
• Practical result in your tests: large inputs (e.g. n=20, n=25) go from milliseconds of work to microseconds. This is where the optimization yields its huge win.

Key behavioral/dep changes and trade-offs

• Behavior: Function semantics are preserved for the tested inputs (including floats, negative numbers, and JS coercions), and the cache is local to each fibonacci(n) call (no global state). Idempotence across calls is preserved.
• Memory: uses O(n) extra memory for the cache per call.
• Overhead on tiny inputs: microbenchmarks for very small n (0, 1, 2, etc.) show small slowdowns (roughly ~40–50% slower in some of your microtests). This is expected and reasonable because each call now pays the cost of creating a Map, an inner function closure and doing Map.get/set checks — overhead that is negligible for medium/large n but visible for trivial calls.
• Trade-off framed positively: the small per-call overhead is an acceptable trade for massive runtime gains on larger inputs and in hot codepaths that call fibonacci with non-trivial n (as evidenced by the 38705% faster result on n=20 in the annotated tests).

Impact on workloads and hot paths

• If fibonacci(n) is called with moderate-to-large n or inside loops, the optimization will substantially increase throughput and reduce latency because repeated subcomputations are removed.
• If the hot path is lots of very small calls and you need minimal per-call overhead, consider:
  • reusing a shared cache across calls (if safe for your application), or
  • switching to an iterative implementation to avoid closure/Map allocation overhead.
• Recursion depth is unchanged — this keeps existing stack characteristics intact (no tail-call elimination).

Tests this optimization is good for

• Large/medium n: huge wins (annotated tests show ms→μs).
• Multiple calls with overlapping inputs in the same call: big wins because cached values within a single fibonacci(n) invocation are reused.
• Cases that rely on purity / no shared state: preserved because cache is per-call.

Summary

• The optimized version turns exponential wasted work into linear work by memoizing subresults with a Map. That yields the large runtime improvement measured in your benchmarks. The only notable trade-off is a small fixed cost per call (Map + closure) that may slow down trivial inputs slightly; this is generally acceptable given the orders-of-magnitude improvement for real workloads and hot paths that exercise non-trivial n.

✅ Correctness verification report:

Test	Status
⚙️ Existing Unit Tests	✅ 13 Passed
🌀 Generated Regression Tests	✅ 58 Passed
⏪ Replay Tests	🔘 None Found
🔎 Concolic Coverage Tests	🔘 None Found
📊 Tests Coverage	100.0%

⚙️ Click to see Existing Unit Tests

Test File::Test Function	Original ⏱️	Optimized ⏱️	Speedup
fibonacci.test.js::fibonacci returns 0 for n=0	500ns	1.00μs	-50.0%⚠️
fibonacci.test.js::fibonacci returns 1 for n=1	333ns	625ns	-46.7%⚠️
fibonacci.test.js::fibonacci returns 1 for n=2	458ns	750ns	-38.9%⚠️
fibonacci.test.js::fibonacci returns 233 for n=13	2.25μs	1.96μs	14.9%✅
fibonacci.test.js::fibonacci returns 5 for n=5	667ns	1.04μs	-36.0%⚠️
fibonacci.test.js::fibonacci returns 55 for n=10	750ns	1.83μs	-59.1%⚠️

🌀 Click to see Generated Regression Tests

import { jest, describe, it, expect, beforeEach, afterEach, beforeAll, test } from '@jest/globals'
// imports
import { fibonacci } from '../fibonacci.js';

// Increase default timeout for the potentially heavier performance test
jest.setTimeout(10000);

// unit tests
describe('fibonacci', () => {
    // Basic Test Cases
    describe('Basic functionality', () => {
        test('should handle small integers correctly (0 and 1)', () => {
            // 0 and 1 are base cases in this implementation and should return themselves
            expect(fibonacci(0)).toBe(0);  // 1.67μs -> 3.25μs (48.7% slower)
            expect(fibonacci(1)).toBe(1);
        });

        test('should compute small Fibonacci numbers correctly', () => {
            // Standard checks for first few Fibonacci numbers
            expect(fibonacci(2)).toBe(1); // 0,1,1
            expect(fibonacci(3)).toBe(2); // 0,1,1,2
            expect(fibonacci(4)).toBe(3);  // 6.67μs -> 9.25μs (27.9% slower)
            expect(fibonacci(5)).toBe(5);
            expect(fibonacci(6)).toBe(8);
        });

        test('should compute a moderately larger Fibonacci number correctly', () => {
            // Verify known value for n=10
            expect(fibonacci(10)).toBe(55);  // 1.83μs -> 3.50μs (47.7% slower)
        });
    });

    // Edge Test Cases
    describe('Edge cases', () => {
        test('should return the input for n <= 1 (including negative integers)', () => {
            // This implementation returns n when n <= 1. Verify negative integers follow that branch.
            expect(fibonacci(-1)).toBe(-1);  // 1.83μs -> 3.58μs (48.9% slower)
            expect(fibonacci(-5)).toBe(-5);
            // Also verify the boundary n=1 still returns 1 (already covered above, but explicit here for edge semantics)
            expect(fibonacci(1)).toBe(1);
        });

        test('should handle non-integer numeric inputs (floats) according to recurrence', () => {
            // The implementation uses arithmetic and compares n <= 1, so non-integers eventually reduce to <=1
            // fibonacci(1.5) = fibonacci(0.5) + fibonacci(-0.5) = 0.5 + (-0.5) = 0
            expect(fibonacci(1.5)).toBeCloseTo(0);  // 2.08μs -> 4.25μs (51.0% slower)

            // fibonacci(2.5) = fibonacci(1.5) + fibonacci(0.5) = 0 + 0.5 = 0.5
            expect(fibonacci(2.5)).toBeCloseTo(0.5);

            // fibonacci(0.5) should directly return 0.5 because 0.5 <= 1
            expect(fibonacci(0.5)).toBeCloseTo(0.5);
        });

        test('should coerce numeric strings and return correct Fibonacci number', () => {
            // A numeric string should be coerced in arithmetic and lead to correct numeric result
            expect(fibonacci('5')).toBe(5); // '5' -> behaves like 5
            expect(fibonacci('7')).toBe(13); // '7' -> behaves like 7
        });

        test('should reflect JS coercion rules for null input', () => {
            // null is coerced to 0 in numeric comparisons; since n <= 1 is true, function returns n (null)
            // The implementation will return null (not the numeric 0). We assert that behavior explicitly.
            expect(fibonacci(null)).toBeNull();  // 582ns -> 1.00μs (41.8% slower)
        });
    });

    // Large Scale Test Cases
    describe('Performance tests', () => {
        test('should compute larger Fibonacci numbers correctly (moderate scale)', () => {
            // This is a heavier test to ensure correctness at larger n.
            // We keep n small enough to avoid unreasonable runtime for the naive recursive implementation.
            // Known value: fibonacci(20) = 6765
            expect(fibonacci(20)).toBe(6765);  // 3.23ms -> 8.33μs (38705% faster)

            // Another moderately larger check. fibonacci(25) = 75025
            // This exercises more recursion but remains within reasonable runtime for most test environments.
            expect(fibonacci(25)).toBe(75025);
        });

        test('should produce consistent results across multiple calls (idempotence / purity)', () => {
            // Call several inputs in a small array (keeps loop iterations well under 1000)
            const inputs = [5, 8, 12, 15, 20];
            const expected = [5, 21, 144, 610, 6765];

            // Ensure repeated calls return the same results and do not rely on hidden state
            inputs.forEach((n, idx) => {
                const first = fibonacci(n);
                const second = fibonacci(n);
                expect(first).toBe(expected[idx]);  // 1.67μs -> 3.00μs (44.5% slower)
                expect(second).toBe(expected[idx]);
            });
        });
    });
});

import { jest, describe, it, expect, beforeEach, afterEach, beforeAll, test } from '@jest/globals'
// imports
import { fibonacci } from '../fibonacci.js';

// Increase default timeout for the potentially heavier performance test
jest.setTimeout(10000);

// unit tests
describe('fibonacci', () => {
    // Basic Test Cases
    describe('Basic functionality', () => {
        test('should handle small integers correctly (0 and 1)', () => {
            // 0 and 1 are base cases in this implementation and should return themselves
            expect(fibonacci(0)).toBe(0);  // 1.67μs -> 3.25μs (48.7% slower)
            expect(fibonacci(1)).toBe(1);
        });

        test('should compute small Fibonacci numbers correctly', () => {
            // Standard checks for first few Fibonacci numbers
            expect(fibonacci(2)).toBe(1); // 0,1,1
            expect(fibonacci(3)).toBe(2); // 0,1,1,2
            expect(fibonacci(4)).toBe(3);  // 6.67μs -> 9.25μs (27.9% slower)
            expect(fibonacci(5)).toBe(5);
            expect(fibonacci(6)).toBe(8);
        });

        test('should compute a moderately larger Fibonacci number correctly', () => {
            // Verify known value for n=10
            expect(fibonacci(10)).toBe(55);  // 1.83μs -> 3.50μs (47.7% slower)
        });
    });

    // Edge Test Cases
    describe('Edge cases', () => {
        test('should return the input for n <= 1 (including negative integers)', () => {
            // This implementation returns n when n <= 1. Verify negative integers follow that branch.
            expect(fibonacci(-1)).toBe(-1);  // 1.83μs -> 3.58μs (48.9% slower)
            expect(fibonacci(-5)).toBe(-5);
            // Also verify the boundary n=1 still returns 1 (already covered above, but explicit here for edge semantics)
            expect(fibonacci(1)).toBe(1);
        });

        test('should handle non-integer numeric inputs (floats) according to recurrence', () => {
            // The implementation uses arithmetic and compares n <= 1, so non-integers eventually reduce to <=1
            // fibonacci(1.5) = fibonacci(0.5) + fibonacci(-0.5) = 0.5 + (-0.5) = 0
            expect(fibonacci(1.5)).toBeCloseTo(0);  // 2.08μs -> 4.25μs (51.0% slower)

            // fibonacci(2.5) = fibonacci(1.5) + fibonacci(0.5) = 0 + 0.5 = 0.5
            expect(fibonacci(2.5)).toBeCloseTo(0.5);

            // fibonacci(0.5) should directly return 0.5 because 0.5 <= 1
            expect(fibonacci(0.5)).toBeCloseTo(0.5);
        });

        test('should coerce numeric strings and return correct Fibonacci number', () => {
            // A numeric string should be coerced in arithmetic and lead to correct numeric result
            expect(fibonacci('5')).toBe(5); // '5' -> behaves like 5
            expect(fibonacci('7')).toBe(13); // '7' -> behaves like 7
        });

        test('should reflect JS coercion rules for null input', () => {
            // null is coerced to 0 in numeric comparisons; since n <= 1 is true, function returns n (null)
            // The implementation will return null (not the numeric 0). We assert that behavior explicitly.
            expect(fibonacci(null)).toBeNull();  // 582ns -> 1.00μs (41.8% slower)
        });
    });

    // Large Scale Test Cases
    describe('Performance tests', () => {
        test('should compute larger Fibonacci numbers correctly (moderate scale)', () => {
            // This is a heavier test to ensure correctness at larger n.
            // We keep n small enough to avoid unreasonable runtime for the naive recursive implementation.
            // Known value: fibonacci(20) = 6765
            expect(fibonacci(20)).toBe(6765);  // 3.23ms -> 8.33μs (38705% faster)

            // Another moderately larger check. fibonacci(25) = 75025
            // This exercises more recursion but remains within reasonable runtime for most test environments.
            expect(fibonacci(25)).toBe(75025);
        });

        test('should produce consistent results across multiple calls (idempotence / purity)', () => {
            // Call several inputs in a small array (keeps loop iterations well under 1000)
            const inputs = [5, 8, 12, 15, 20];
            const expected = [5, 21, 144, 610, 6765];

            // Ensure repeated calls return the same results and do not rely on hidden state
            inputs.forEach((n, idx) => {
                const first = fibonacci(n);
                const second = fibonacci(n);
                expect(first).toBe(expected[idx]);  // 1.67μs -> 3.00μs (44.5% slower)
                expect(second).toBe(expected[idx]);
            });
        });
    });
});

To edit these changes git checkout codeflash/optimize-fibonacci-mkyfql53 and push.