Sunday, February 12, 2023
HomeSoftware EngineeringThe right way to discover the Product of Consecutive Fib Numbers in...

The right way 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 job is to seek out the n_th quantity in Van Eck’s Sequence. (1-based)

The Answer in Python

Choice 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__

Choice 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 circumstances to validate the answer

from resolution import seq
import check

from random import randint

@check.describe("Pattern exams:")
def exams():
    @check.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)):
            check.assert_equals(seq(i+1), s[i])
    @check.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)):
            check.assert_equals(seq(i+50), s[i])

@check.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]
    
    @check.it('200 random exams:')
    def _():
        for _ in vary (200):
            a = randint(100, 1000)
            exp = ref_sol(a)
            check.assert_equals(seq(a), exp)



Supply hyperlink

RELATED ARTICLES

LEAVE A REPLY

Please enter your comment!
Please enter your name here

- Advertisment -
Google search engine

Most Popular

Recent Comments