From 59505f1c76b93a7b58d778ded02e6171e84d67b9 Mon Sep 17 00:00:00 2001 From: "codeflash-ai[bot]" <148906541+codeflash-ai[bot]@users.noreply.github.com> Date: Wed, 28 Jan 2026 04:39:47 +0000 Subject: [PATCH] Optimize fibonacci MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Primary benefit — runtime improvement: the optimized version cuts execution time from ~7.95 ms to ~113 μs (≈6908% speedup) for the measured cases. That dramatic win is what drove acceptance. What changed (specific optimizations) - For integer inputs the implementation replaces the naive exponential recursion with the fast-doubling algorithm (fibPair) which computes F(n) in O(log n) arithmetic steps instead of O(φ^n) recursive calls. - For finite non-integer inputs it replaces deep recursive branching with a small iterative DP: it steps downward until reaching the base region (x ≤ 1) then computes upward with the 2-term recurrence in O(steps) time and O(1) extra memory. - The code keeps the original base-case behavior (return n for n ≤ 1) and intentionally preserves the original infinite-recursion behavior for non-finite values (NaN, Infinity) so error/looping behavior remains consistent. Why this speeds things up (how it affects Python/JS runtime cost) - Algorithmic complexity: moving integers from exponential time to logarithmic time is the dominant improvement. Eliminating exponential branching removes the huge number of function calls and duplicated subcomputation that made the original blow up for moderately large n. - Function-call overhead and recursion depth: fast-doubling uses a small number of deterministic calls and local arithmetic, avoiding the heavy call-stack pressure and interpreter overhead of deep recursion. - Iterative handling of fractional finite inputs replaces many recursive calls with tight loops and constant-space updates (fewer allocations and less interpreter overhead). - Overall this reduces both CPU-time (fewer operations) and interpreter overhead (fewer calls/frames), producing the measured microsecond-level runtimes in tests. Behavioral and dependency notes (key differences and trade-offs) - Correctness: base-case behavior for n ≤ 1 is preserved. Integer results match the canonical Fibonacci values; fractional finite inputs still follow the same numeric recurrence and converge via the iterative path. - Non-finite inputs (NaN/Infinity): left on the original recursive path to preserve the same failure/infinite-recursion semantics. - Coercion/strings: the optimized code uses Number.isFinite/Number.isInteger which do not coerce strings to numbers; arithmetic (e.g., n - 1) still coerces strings, so correctness for string inputs is preserved but the fast integer path won't kick in for numeric strings. The tests reflect this: some string-input cases are slightly slower under the optimized code (expected and acceptable trade-off given the large overall runtime win). - Stack safety / memory: eliminating deep recursion avoids potential stack overflows and reduces memory pressure for large n. Impact on workloads and hot paths - Any workload that computes F(n) for integer n, especially medium-to-large n, benefits hugely (logarithmic vs exponential). The performance tests and recurrence-loop tests show very large relative speedups (examples in tests: huge reductions from ms to μs). - Workloads that call fibonacci repeatedly over a range (0..N) benefit because each integer call is cheap (fast-doubling or O(1) per small n), and the function is now safe for larger inputs in tight loops. - Workloads relying heavily on numeric-string input may see smaller or negative micro-regressions because those inputs bypass the fast path; this was visible in annotated tests and is a reasonable trade-off given the large win for numeric inputs. What test cases this optimization is good for - Integer-heavy tests and performance/stress tests (large n or many repeated calls): huge wins, as seen in the annotated tests (sequence builds and recurrence checks). - Correctness tests involving fractional finite inputs: still handled and tested by the iterative fallback. - Edge tests for NaN/Infinity: preserved behavior. In short: the optimized code replaces exponential recursion with a logarithmic fast-doubling algorithm for integers and an iterative linear fallback for fractional finite inputs. That algorithmic change removes massive redundant recursion and call overhead, delivering the measured 6908% speedup while keeping test-correctness and preserving original failure semantics for non-finite inputs. The only notable trade-off is slightly different fast-path behavior for string inputs (minor perf regression), which is acceptable given the large runtime improvement for numeric inputs. --- code_to_optimize_js_esm/fibonacci.js | 49 +++++++++++++++++++++++++++- 1 file changed, 48 insertions(+), 1 deletion(-) diff --git a/code_to_optimize_js_esm/fibonacci.js b/code_to_optimize_js_esm/fibonacci.js index 0ee526315..a0e5a234b 100644 --- a/code_to_optimize_js_esm/fibonacci.js +++ b/code_to_optimize_js_esm/fibonacci.js @@ -13,7 +13,54 @@ export function fibonacci(n) { if (n <= 1) { return n; } - return fibonacci(n - 1) + fibonacci(n - 2); + + // Preserve original infinite-recursion behavior for non-finite inputs + // (NaN, Infinity) so they produce the same RangeError in the same circumstances. + if (!Number.isFinite(n)) { + return fibonacci(n - 1) + fibonacci(n - 2); + } + + // Fast path for integer n: use fast-doubling (O(log n)) + if (Number.isInteger(n)) { + // fibPair(k) returns [F_k, F_{k+1}] + const fibPair = (k) => { + if (k === 0) { + return [0, 1]; + } + const [a, b] = fibPair(k >> 1); + const c = a * (2 * b - a); + const d = a * a + b * b; + if ((k & 1) === 0) { + return [c, d]; + } + return [d, c + d]; + }; + return fibPair(n)[0]; + } + + // Non-integer finite n: iterative DP using O(steps) time and O(1) memory. + // Build downwards until we reach the first value <= 1 (base region), + // then compute upwards using the two-term recurrence. + let x = n; + let steps = 0; + while (x > 1) { + x -= 1; + steps++; + } + // x <= 1 is the smallest value (base case) + // values[0] = f(x) = x + let prev2 = x; + // values[1] = f(x+1) = f(x) + f(x-1) where f(x-1) <= 1 so it's (x - 1) + let prev1 = prev2 + (x - 1); + if (steps === 1) { + return prev1; + } + for (let i = 2; i <= steps; i++) { + const cur = prev1 + prev2; + prev2 = prev1; + prev1 = cur; + } + return prev1; } /**