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How to use fib method in fMBT

Best Python code snippet using fMBT_python

fibonacci.py

Source:fibonacci.py

1# fibonacci.py2"""3Calculates the Fibonacci sequence using iteration, recursion, memoization,4and a simplified form of Binet's formula5NOTE 1: the iterative, recursive, memoization functions are more accurate than6the Binet's formula function because the Binet formula function  uses floats7NOTE 2: the Binet's formula function is much more limited in the size of inputs8that it can handle due to the size limitations of Python floats9RESULTS: (n = 20)10fib_iterative runtime: 0.0055 ms11fib_recursive runtime: 6.5627 ms12fib_memoization runtime: 0.0107 ms13fib_binet runtime: 0.0174 ms14"""15from math import sqrt16from time import time17def time_func(func, *args, **kwargs):18    """19    Times the execution of a function with parameters20    """21    start = time()22    output = func(*args, **kwargs)23    end = time()24    if int(end - start) > 0:25        print(f"{func.__name__} runtime: {(end - start):0.4f} s")26    else:27        print(f"{func.__name__} runtime: {(end - start) * 1000:0.4f} ms")28    return output29def fib_iterative(n: int) -> list[int]:30    """31    Calculates the first n (0-indexed) Fibonacci numbers using iteration32    >>> fib_iterative(0)33    [0]34    >>> fib_iterative(1)35    [0, 1]36    >>> fib_iterative(5)37    [0, 1, 1, 2, 3, 5]38    >>> fib_iterative(10)39    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]40    >>> fib_iterative(-1)41    Traceback (most recent call last):42    ...43    Exception: n is negative44    """45    if n < 0:46        raise Exception("n is negative")47    if n == 0:48        return [0]49    fib = [0, 1]50    for _ in range(n - 1):51        fib.append(fib[-1] + fib[-2])52    return fib53def fib_recursive(n: int) -> list[int]:54    """55    Calculates the first n (0-indexed) Fibonacci numbers using recursion56    >>> fib_iterative(0)57    [0]58    >>> fib_iterative(1)59    [0, 1]60    >>> fib_iterative(5)61    [0, 1, 1, 2, 3, 5]62    >>> fib_iterative(10)63    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]64    >>> fib_iterative(-1)65    Traceback (most recent call last):66    ...67    Exception: n is negative68    """69    def fib_recursive_term(i: int) -> int:70        """71        Calculates the i-th (0-indexed) Fibonacci number using recursion72        """73        if i < 0:74            raise Exception("n is negative")75        if i < 2:76            return i77        return fib_recursive_term(i - 1) + fib_recursive_term(i - 2)78    if n < 0:79        raise Exception("n is negative")80    return [fib_recursive_term(i) for i in range(n + 1)]81def fib_memoization(n: int) -> list[int]:82    """83    Calculates the first n (0-indexed) Fibonacci numbers using memoization84    >>> fib_memoization(0)85    [0]86    >>> fib_memoization(1)87    [0, 1]88    >>> fib_memoization(5)89    [0, 1, 1, 2, 3, 5]90    >>> fib_memoization(10)91    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]92    >>> fib_iterative(-1)93    Traceback (most recent call last):94    ...95    Exception: n is negative96    """97    if n < 0:98        raise Exception("n is negative")99    # Cache must be outside recursuive function100    # other it will reset every time it calls itself.101    cache: dict[int, int] = {0: 0, 1: 1, 2: 1}  # Prefilled cache102    def rec_fn_memoized(num: int) -> int:103        if num in cache:104            return cache[num]105        value = rec_fn_memoized(num - 1) + rec_fn_memoized(num - 2)106        cache[num] = value107        return value108    return [rec_fn_memoized(i) for i in range(n + 1)]109def fib_binet(n: int) -> list[int]:110    """111    Calculates the first n (0-indexed) Fibonacci numbers using a simplified form112    of Binet's formula:113    https://en.m.wikipedia.org/wiki/Fibonacci_number#Computation_by_rounding114    NOTE 1: this function diverges from fib_iterative at around n = 71, likely115    due to compounding floating-point arithmetic errors116    NOTE 2: this function doesn't accept n >= 1475 because it overflows117    thereafter due to the size limitations of Python floats118    >>> fib_binet(0)119    [0]120    >>> fib_binet(1)121    [0, 1]122    >>> fib_binet(5)123    [0, 1, 1, 2, 3, 5]124    >>> fib_binet(10)125    [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]126    >>> fib_binet(-1)127    Traceback (most recent call last):128    ...129    Exception: n is negative130    >>> fib_binet(1475)131    Traceback (most recent call last):132    ...133    Exception: n is too large134    """135    if n < 0:136        raise Exception("n is negative")137    if n >= 1475:138        raise Exception("n is too large")139    sqrt_5 = sqrt(5)140    phi = (1 + sqrt_5) / 2141    return [round(phi**i / sqrt_5) for i in range(n + 1)]142if __name__ == "__main__":143    num = 20144    time_func(fib_iterative, num)145    time_func(fib_recursive, num)146    time_func(fib_memoization, num)...

recomb.js

Source:recomb.js

...11		12		// define the standard factorial function to compare with13		fact = function(n){ return n <= 1 ? 1 : n * fact(n - 1); },14		// define the standard fibonacci function to compare with15		fib  = function(n){ return n <= 1 ? 1 : fib(n - 1) + fib(n - 2); },16		17		// prepare the sequence of arguments for comparison18		seq = df.listcomp("i for(i = 0; i < 15; ++i)"),19		20		// build a set of results for our argument list using the standard factorial function21		factTable = df.map(seq, fact),22		// build a set of results for our argument list using the standard fibonacci function23		fibTable  = df.map(seq, fib);24	25	tests.register("dojox.lang.tests.recomb", [26		function testFactLinrec1(t){27			var fact = df.linrec("<= 1", "1", "[n - 1]", "a * b[0]");28			t.assertEqual(df.map(seq, fact), factTable);29		},...

1029.js

Source:1029.js

1/**2 * FUNC   |   CALLS   | VALUE |3 * fib( 0) =   0 calls =   04 * fib( 1) =   0 calls =   15 * fib( 2) =   2 calls =   16 * fib( 3) =   4 calls =   27 * fib( 4) =   8 calls =   38 * fib( 5) =  14 calls =   59 * fib( 6) =  24 calls =   810 * fib( 7) =  40 calls =   1311 * fib( 8) =  66 calls =   2112 * fib( 9) = 108 calls =   3413 * fib(10) = 176 calls =   5514 */15/**16 * Call FIB function:17 * * C(n) = [ (n - 2 > 0) ? ( C(n - 2) + 1 ) : (return 0) ) ] + [ (n - 1 > 0) ? C(n - 1) + 1 : (return 0) ]18 * * F(n) = F(n - 2) + F(n - 1)19 */20const { format } = require("util")21const { readFileSync } = require("fs")22const [numLines, ...input] = readFileSync("/dev/stdin", "utf8").split("\n")23/** @typedef {{ calls: number, value: number }} countableFibProp */24/** @type { Map<number, countableFibProp> } */25const FibonacciList = new Map([26	[0, { calls: 0, value: 0 }],27	[1, { calls: 0, value: 1 }],28	[2, { calls: 2, value: 1 }]29])30function recusiveFibonacciWithCallsCounter(nth = 0) {31	nth = Math.floor(Math.max(0, nth)) // is nth leq 0? So set nth to 032	if (FibonacciList.has(nth) === false) {33		// If not included, set it34		const fibA = recusiveFibonacciWithCallsCounter(nth - 2)35		const fibB = recusiveFibonacciWithCallsCounter(nth - 1)36		FibonacciList.set(nth, {37			calls: (fibA.calls + 1) + (fibB.calls + 1),38			value: fibA.value + fibB.value39		})40	}41	return FibonacciList.get(nth)42}43function main() {44	const output = new Array(Number.parseInt(numLines, 10))45	for (let index = 0; index < output.length; index += 1) {46		const nth = Number.parseInt(input[index])47		const { calls, value } = recusiveFibonacciWithCallsCounter(nth)48		output[index] = format("fib(%d) = %d calls = %d", nth, calls, value)49	}50	console.log(output.join("\n"))51}...

RecursiveD5.py

Source:RecursiveD5.py

1#  Fibonacci sequence calculations23# n:      0  1  2  3  4  5  6  7  8  94# fib(n): 0  1  1  2  3  5  8  13 21 3456# A non-recursive function for calculating the nth Fibonacci number:7def fib(n):8    if n < 2:9        return n10    prev = 011    curr = 112    for i in range(2,n+1):13        new_curr = prev + curr14        prev = curr15        curr = new_curr16    return curr1718# Calculating the ratios of pairs of Fibonacci numbers -- like fib(6)/fib(5)19def CalcFibRatios(low,high):20    for i in range(low,high+1):21        print(timeit(i)/timeit(i-1))22        23# The recursive function for Fibonacci24def fib_r(n):25    if n < 2:26        return n27    return fib_r(n-1) + fib_r(n-2)2829# Timing how long it takes to calculate fib_r(n)30def timeit(n):31    import time32    start = time.time()33    fibr2(n)34    end = time.time()35    return end - start3637# Recursive Fib:38# 1 = fib(2)[2 calculation] = fib(1)[1 calculation] + fib(0)[1 calculation]39# 2 = fib(3)[3 c] = fib(2)[2 c] + fib(1)[1 c]40# 3 = fib(4)[5 c] = fib(3)[3 c] + fib(2)[2 c]41# 5 = fib(5)[8 p] = fib(4)[5 p] + fib(3)[3 c]42# So, for Recursive Fib, Fib(n) requires fib(n) calculations to occur43#44# Explicit Fib:45# 1 = fib(2)[3 c]46# 2 = fib(3)[6 c]47# 3 = fib(4)[9 c]48# 5 = fib(5)[12 c]49# So, for Explicit Fib, Fib(n) required 3n calculations5051def fibr2(n):52    return fibr2h(0, 1, 1, n)5354def fibr2h(pv, cv, i, t):55    if i == t:56        return cv57    if i - 1 == t:58        return pv59    hc = pv + cv60    hi = i + 161    hp = cv62    return fibr2h(hp, hc, hi, t)
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