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# Daily Challenge #13 - Twice Linear

Hey, everyone. If yesterday was our rest day, today is the warm-up.

We have another great challenge from user g964 on CodeWars:

Consider a sequence `u`, where u is defined as follows:

1. The number `u(0) = 1` is the first one in `u`.
2. For each `x` in `u`, `y = 2 * x + 1` and `z = 3 * x + 1` must also be in `u`.
3. There are no other numbers in `u`.

Ex: `u = [1, 3, 4, 7, 9, 10, 13, 15, 19, 21, 22, 27, ...]`

1 gives 3 and 4. 3 gives 7 and 10 while 4 gives 9 and 13. This pattern continues as far you as allow it to.

Your task is to create a function `dbl_linear` with a parameter of `n` that returns the element `u(n)` in an ordered sequence (excluding duplicates).

Good luck! Let us know if you like these mathematical challenges and we'll stick with the trend.

Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge for a future post? Email yo+challenge@dev.to with your suggestions!

## Discussion (17)

Corey Alexander

Here is my Rust solution! I think my answer is pretty good except I should be using a better data structure than a `Vec` here, since I really want to push/pop from both ends. If I did that I would be able to avoid needing to sort and reverse my vec, which would help on the processing speed!

``````fn twice_linear(size: usize) -> Vec<u32> {
let mut processed: Vec<u32> = vec![];
let mut unprocessed: Vec<u32> = vec![];

let mut current = 1;
while processed.len() < size {
processed.push(current);

// This block of code is NOT the most effiecient
// I should switch to a data store that can push/pop
// from both ends effieciently, as to avoid needing
// the sort AND the revserse
unprocessed.push(current * 2 + 1);
unprocessed.push(current * 3 + 1);
unprocessed.sort();
unprocessed = unprocessed.iter().rev().cloned().collect();

current = unprocessed.pop().unwrap();
}

processed
}

pub fn twice_linear_at(u: usize) -> u32 {
twice_linear(u + 1).pop().unwrap()
}

#[cfg(test)]
mod tests {
use crate::*;

#[test]
fn it_works_for_the_base_cases() {
assert_eq!(twice_linear_at(0), 1);
assert_eq!(twice_linear_at(1), 3);
assert_eq!(twice_linear_at(2), 4);
}

#[test]
fn it_works_for_the_example() {
assert_eq!(twice_linear_at(3), 7);
assert_eq!(twice_linear_at(4), 9);
assert_eq!(twice_linear_at(5), 10);
assert_eq!(twice_linear_at(6), 13);
assert_eq!(twice_linear_at(7), 15);
assert_eq!(twice_linear_at(8), 19);
assert_eq!(twice_linear_at(9), 21);
assert_eq!(twice_linear_at(10), 22);
assert_eq!(twice_linear_at(11), 27);
}

#[test]
fn it_can_also_return_the_vec() {
assert_eq!(
twice_linear(12),
vec![1, 3, 4, 7, 9, 10, 13, 15, 19, 21, 22, 27]
);
}
}
``````
Oleksii Filonenko

Take a look at std::collections::VecDeque :)

Corey Alexander

Thank you! I knew there was one that fit the bill better!

Yozen Hernandez

Here's my solution in Perl. I actually think its a bit too slow, even when using `state` variables to cache the results in subsequent calls. Maybe I'll look into optimizing it later.

``````#!/usr/bin/env perl

use v5.24;
use strict;
use warnings;
use feature qw(signatures);
no warnings "experimental::signatures";
use List::Util qw(first uniq);

sub dbl_linear (\$n) {
state @u = (1);
state \$last_n = 0;

return \$u[\$n] if \$u[\$n];

for my \$i ( \$last_n .. \$n ) {
my \$val = \$u[\$i];
@u = sort { \$a <=> \$b } uniq (@u, map {\$_ * \$val + 1} (2, 3));
}

\$last_n = \$n;
return \$u[\$n];
}

use Test::More tests => 5;
is( dbl_linear(0), 1,  "u(0) == 1" );
is( dbl_linear(1), 3,  "u(1) == 3" );
is( dbl_linear(6), 13, "u(6) == 13" );
is( dbl_linear(100), 447, "u(100) == 447" );
is( dbl_linear(1000), 8488, "u(1000) == 8488" );
``````

I'll write an explainer for it when I get a few minutes later.

edh_developer

For n = 100,000 , on an ok laptop, it runs in roughly 45 seconds

``````import time

n = 100000

print "n = %d" % n

def insert(l,val):
index = len(l) - 1
done = False

while not done and index >= 0:
if l[index] == val:
done = True
elif l[index] < val:
l.insert(index + 1,val)
done = True
else:
index -= 1

if not done:
l.insert(0,val)

list = [1]
index = 0

start = time.time()

while index < n:
x = list[0]
y = (2 * x) + 1
z = (3 * x) + 1

insert(list,y)
insert(list,z)

# Remove unnecessary entries from the list, which otherwise
# grows to unreasonable size for large values of n.
list.remove(x)
while (index + len(list)) > n :
list.pop()

index += 1

end = time.time()

print "result = %d" % list[0]
print "elapsed time (seconds) = %3.6f" % (end - start)
``````
Corey Alexander

Let us know if you like these mathematical challenges and we'll stick with the trend.

I like em, but I think it's fun to have a good mix of different types of challenges!

One suggestion is to provide more examples/test cases. Especially for the mathy ones, where the correct answers might not be super obvious to everyone, having many examples/test cases can definitely make it easier! Just a thought!

Dylan Paulus • Edited on

Nim

Time: `./main 0.04s user 0.01s system 39% cpu 0.113 total`

``````import sequtils, algorithm

proc dblLinear(u: int): seq[int] =

for i in 0..<u:
let y = 2 * result[i] + 1
let z = 3 * result[i] + 1

result.sort()

echo \$dblLinear(100000)
``````
edh_developer

Not familiar with nim. It looks like result is a list, and that duplicate values are being added to it. Is that not the case?

Dylan Paulus • Edited on

I think you're right! Was testing with too small of an input-set to see any duplicates. :)

Cheap fix (./main 341.16s user 0.52s system 99% cpu 5:42.40 total):

``````import sequtils, algorithm

proc dblLinear(u: int): seq[int] =

for i in 0..<u:
let y = 2 * result[i] + 1
let z = 3 * result[i] + 1

if not (y in result):
if not (z in result):

result.sort()

echo \$dblLinear(100000)
``````
Nans Dumortier • Edited on

Not sure I have well understood this one ... Am I correct with this JS function ? 🙈

``````const doubleLinear = (n) => {
let u = [1];
let i = 0;
while(i < n) {
u = [...u, u[i] * 2 + 1, u[i] * 3 + 1];
u = Array.from(new Set(u))
u.sort((a, b) => a - b);
i++;
}
return u;
}
``````

EDIT: added the line `u = Array.from(new Set(u))` to get unique values !

Alvaro Montoro

JavaScript

``````const twiceLinear = number => {
let series = { 1: 1 };
let keys = Object.keys(series);
let index = 0;

while (index < number) {
series[ keys[index] * 2 + 1 ] = 1;
series[ keys[index] * 3 + 1 ] = 1;
index++;
keys = Object.keys(series);
}

return keys;
}
``````

I've been running late lately... but here is a live demo on CodePen

Mat-R-Such

Python

``````def dbl_linear(n):
u=[1]
if n == 0:
return u
else:
for i in range(n):
y,z=2*u[i]+1,3*u[i]+1
u.append(y)
u.append(z)
u=sorted(u)
return u
print(dbl_linear(10))
``````
Kerri Shotts • Edited on

A little late to the party (3am my time), but oh well! Here's my submission:

``````function dblLinear(n) {
const series = [1];

const calc = x => ({
y: 2 * x + 1,
z: 3 * x + 1
});

const ascendingOrder = (a, b) => a - b;

for (let idx = 0; idx <= n; idx++) {
let x = series[idx];
const { y, z } = calc(x);
for (let v of [y, z]) {
if (series.indexOf(v) < 0) {
series.push(v);
series.sort(ascendingOrder);
series.splice(n+1);
}
}
}
return series[n];
}
``````

Full code w/ some tests: gist.github.com/kerrishotts/029c8f...

This is one where it would have been super helpful to have some answers to compare against. Beyond the initial few digits, I'm just assuming things are correct, which may not be true.

ListNUX

Not too good with maths, hope I understood well. Here is a (very inefficient, so I added a constraint to the max num to give) attempt in bash:

``````#!/bin/bash

declare -a numbersarr
numbersarr[0]=1

echo "Array position \"n\""
# make sure we are given a number and the number is not too large
re="^[0-9]+\$"
if ! [[ \$myn =~ \$re ]] || [ \$myn -gt 100 ]; then
echo "not a number or number too big" >&2; exit 1
fi

arrl=1

while [ \$arrl -le \$myn ]; do
for i in "\${numbersarr[@]}"; do
let x=2\*i+1
let y=3\*i+1
numbersarr+=( \$x \$y )
done
sorted_unique=(\$(echo "\${numbersarr[@]}" |xargs -n1 | sort -gu | xargs))
arrl="\${#sorted_unique[@]}"
done

# declare -p sorted_unique
echo " >> item in pos \${myn}: "\${sorted_unique[\$myn]}
``````
Timothy Foster

Haskell, featuring tail recursion!

``````import Data.Set (Set)
import qualified Data.Set as Set

u :: Int -> Set Int
u layers = u' layers (Set.singleton 1)
where
u' 0 s = s
u' layer s = u' (layer - 1) (Set.unions [s, Set.map (\x -> 2*x+1) s, Set.map (\x -> 3*x+1) s])

dbl_linear :: Int -> Int
dbl_linear n = Set.elemAt n (u layers)
where layers = ceiling \$ logBase 2 \$ fromIntegral (n + 1)
``````
Abel Mihailescu

Nice challenge.

edh_developer

This fails for larger values of n. For n = 100, for instance, it returns 463 rather than 447.