How to better implement recurrence relations in code?

How to better implement recurrence relations in code?

By : Dodi Dodi Saputra Do
Date : November 19 2020, 03:01 PM
To fix the issue you can do Here's a complete solution which is based on the observation that a valid solution for three or more days must begin with one of the following:
code :
cache = {}
modulus = 10**9 + 7

def valid(n):
    if n in cache:
        return cache[n]

    if n == 0:
        v = 1
    elif n == 1:
        v = 2
    elif n == 2:
        v = 4
        v = valid(n-1) + valid(n-2) + valid(n-3)

    v %= modulus

    cache[n] = v
    return v

def main():
    # Preload the cache
    for n in range(1000000):


modulus = 10**9 + 7

def valid(n):
    v0, v1, v2 = 0, 1, 1

    for i in range(n):
        v0, v1, v2 = v1, v2, (v0 + v1 + v2) % modulus

    return v2


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Recurrence Relations

Recurrence Relations

By : Idd
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , The best asymptotic complexity for tribonacci numbers will be using a matrix exponentiation method like the one for Fibonacci numbers. Specifically, written correctly, this is O(log n) integer operations, rather than O(n) (like the dynamic programming method) or O(3n) (like the naive solution).
The matrix of interest is
code :
    [1, 1, 1]
M = [1, 0, 0]
    [0, 1, 0]
    [a, b, c]
M = [1, 0, 0]
    [0, 1, 0]
Recurrence Relations: Proving (n/2) + (n/2) = O(n)

Recurrence Relations: Proving (n/2) + (n/2) = O(n)

By : chris_dev
Date : March 29 2020, 07:55 AM
wish of those help If you are really looking for a proof that O(x) + O(y) = O(x+y), it would work along these lines:
R1 ∈ O(x) ∧ R2 ∈ O(y)
How to do recurrence relations?

How to do recurrence relations?

By : toonishwarhead
Date : March 29 2020, 07:55 AM
will help you Let us count only the multiplies in a call to pow, denoted as M(N), assuming they dominate the cost (a nowadays strongly invalid assumption).
By inspection of the code we see that:
Forming recurrence relations

Forming recurrence relations

By : Alexkl123
Date : March 29 2020, 07:55 AM
I hope this helps . Let a recurrence relation be T(n) = a * T(n/b) + O(n).
This recurrence implies that there is a recursive function which:
code :
int f(int n)
    if(n <= 1)
        return n;
    return f(n-1) + f(n-2);
int f(int n)
    if(n <= 2)
        return n;
    return n * f(n-1); 
Recurrence Relations in Python

Recurrence Relations in Python

By : Vaibhav Chaudhary
Date : March 29 2020, 07:55 AM
I wish this helpful for you This recurrence relation can be answered exactly through the use of the fractions module or with a variable level of precision using the decimal module, which demonstrates the very high level of precision required to find compute 50 iterations accurately.
Jean-François' point about floating point accumulation errors is correct. However the fraction module does not seem to be able to multiply multiple Fraction objects together, so all of the numeric values must be stated within the fraction object. Credit to him for the use of the correct module for this problem.
code :
import math
import fractions

def printRecurrence():
    x = [0]*51 #initialize list of x values
    x[0] = fractions.Fraction(1,1)
    x[1] = fractions.Fraction(1,3)
    for i in range(1, 50):
        x[i+1] = fractions.Fraction(13*x[i]/3) - fractions.Fraction(4*x[i-1]/3)
        print(float(x[i+1]), 3**(-(i+1)), x[i+1]-(3)**(-(i+1)))

import math
from decimal import *
getcontext().prec = 60

def printRecurrence():
    x = [0]*51 #initialize list of x values
    x[0] = Decimal(1)
    x[1] = Decimal(1)/Decimal(3)
    for i in range(1, 50):
        x[i+1] = (Decimal(13)/Decimal(3))*x[i] - (Decimal(4)/Decimal(3))*x[i-1]
        print(float(x[i+1]), 3**(-(i+1)), float(x[i+1]-Decimal(3)**(-(i+1))))

0.111111111111 0.111111111111 -1e-60
0.037037037037 0.037037037037 -4e-60
0.0123456790123 0.0123456790123 -1.57e-59
0.00411522633745 0.00411522633745 -6.243e-59
0.00137174211248 0.00137174211248 -2.4961e-58
0.000457247370828 0.000457247370828 -9.984e-58
0.000152415790276 0.000152415790276 -3.993577e-57
5.08052634253e-05 5.08052634253e-05 -1.59742978e-56
1.69350878084e-05 1.69350878084e-05 -6.38971879e-56
5.64502926948e-06 5.64502926948e-06 -2.5558875048e-55
1.88167642316e-06 1.88167642316e-06 -1.02235500146e-54
6.27225474386e-07 6.27225474386e-07 -4.08942000567e-54
2.09075158129e-07 2.09075158129e-07 -1.63576800226e-53
6.96917193763e-08 6.96917193763e-08 -6.54307200905e-53
2.32305731254e-08 2.32305731254e-08 -2.61722880362e-52
7.74352437514e-09 7.74352437514e-09 -1.04689152145e-51
2.58117479171e-09 2.58117479171e-09 -4.18756608579e-51
8.60391597238e-10 8.60391597238e-10 -1.67502643432e-50
2.86797199079e-10 2.86797199079e-10 -6.70010573727e-50
9.55990663597e-11 9.55990663597e-11 -2.68004229491e-49
3.18663554532e-11 3.18663554532e-11 -1.07201691796e-48
1.06221184844e-11 1.06221184844e-11 -4.28806767185e-48
3.54070616147e-12 3.54070616147e-12 -1.71522706874e-47
1.18023538716e-12 1.18023538716e-12 -6.86090827496e-47
3.93411795719e-13 3.93411795719e-13 -2.74436330998e-46
1.3113726524e-13 1.3113726524e-13 -1.09774532399e-45
4.37124217466e-14 4.37124217466e-14 -4.39098129597e-45
1.45708072489e-14 1.45708072489e-14 -1.75639251839e-44
4.85693574962e-15 4.85693574962e-15 -7.02557007356e-44
1.61897858321e-15 1.61897858321e-15 -2.81022802942e-43
5.39659527735e-16 5.39659527735e-16 -1.12409121177e-42
1.79886509245e-16 1.79886509245e-16 -4.49636484708e-42
5.99621697484e-17 5.99621697484e-17 -1.79854593883e-41
1.99873899161e-17 1.99873899161e-17 -7.19418375532e-41
6.66246330538e-18 6.66246330538e-18 -2.87767350213e-40
2.22082110179e-18 2.22082110179e-18 -1.15106940085e-39
7.40273700597e-19 7.40273700597e-19 -4.60427760341e-39
2.46757900199e-19 2.46757900199e-19 -1.84171104136e-38
8.22526333997e-20 8.22526333997e-20 -7.36684416545e-38
2.74175444666e-20 2.74175444666e-20 -2.94673766618e-37
9.13918148886e-21 9.13918148886e-21 -1.17869506647e-36
3.04639382962e-21 3.04639382962e-21 -4.71478026589e-36
1.01546460987e-21 1.01546460987e-21 -1.88591210636e-35
3.38488203291e-22 3.38488203291e-22 -7.54364842542e-35
1.12829401097e-22 1.12829401097e-22 -3.01745937017e-34
3.76098003645e-23 3.76098003657e-23 -1.20698374807e-33
1.25366001171e-23 1.25366001219e-23 -4.82793499227e-33
4.17886668798e-24 4.1788667073e-24 -1.93117399691e-32
1.39295549185e-24 1.3929555691e-24 -7.72469598763e-32
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