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You are currently viewing Easy methods to discover the Product of Consecutive Fib Numbers in Python

Easy methods to discover the Product of Consecutive Fib Numbers in Python


0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, 4, 0, 5, 3, 0, 3, …

That is the Van Eck’s Sequence.

Let’s undergo it step-by-step.

Time period 1: The primary time period is 0.
Time period 2: Since we haven’t seen 0 earlier than, the second time period is 0.
Time period 3: Since we had seen a 0 earlier than, one step again, the third time period is 1
Time period 4: Since we haven’t seen a 1 earlier than, the fourth time period is 0
Time period 5: Since we had seen a 0 earlier than, two steps again, the fifth time period is 2.
And so forth…

Your activity is to search out the n_th quantity in Van Eck’s Sequence. (1-based)

The Answer in Python

Possibility 1

from collections import Counter

c=Counter()
SEQ = [0]
for i in vary(1000):
    n = SEQ[-1]
    if not c[n]: c[n]=i
    SEQ.append(i-c[n])
    c[n]=i
    
seq=SEQ.__getitem__

Possibility 2

def dist(arr):
    for i in vary (1, len(arr)):
        if arr[-1-i] == arr[-1]:
            return i
    return 0

def seq(n):
    s = [0, 0]
    for _ in vary (n):
        s.append(dist(s))
    return s[n-1]

def seq(n):
    van, eck = [0], 0
    whereas n := n - 1:
        van.insert(0, eck := van.index(eck, 1) if eck in van[1:] else 0)
    return eck

Take a look at instances to validate the answer

from resolution import seq
import take a look at

from random import randint

@take a look at.describe("Pattern exams:")
def exams():
    @take a look at.it("Small numbers")
    def _():
        s = [0, 0, 1, 0, 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6]
        for i in vary (len(s)):
            take a look at.assert_equals(seq(i+1), s[i])
    @take a look at.it('Bigger numbers')
    def __():
        s = [3, 1, 42, 0, 5, 15, 20, 0, 4, 32, 0, 3, 11,
             18, 0, 4, 7, 0, 3, 7, 3, 2, 31, 0, 6, 31, 3,
             6, 3, 2, 8, 33, 0, 9, 56, 0, 3, 8, 7, 19, 0,
             5, 37, 0, 3, 8, 8, 1, 46, 0, 6, 23, 0]
        for i in vary (len(s)):
            take a look at.assert_equals(seq(i+50), s[i])

@take a look at.describe('Random exams:')
def r():

    def dist(arr):
        for i in vary (1, len(arr)):
            if arr[-1-i] == arr[-1]:
                return i
        return 0

    def ref_sol(n):
        s = [0, 0]
        for _ in vary (n):
            s.append(dist(s))
        return s[n-1]
    
    @take a look at.it('200 random exams:')
    def _():
        for _ in vary (200):
            a = randint(100, 1000)
            exp = ref_sol(a)
            take a look at.assert_equals(seq(a), exp)

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