DEV Community πŸ‘©β€πŸ’»πŸ‘¨β€πŸ’»

DEV Community πŸ‘©β€πŸ’»πŸ‘¨β€πŸ’» is a community of 963,864 amazing developers

We're a place where coders share, stay up-to-date and grow their careers.

Create account Log in
dev.to staff
dev.to staff

Posted on

Daily Challenge #14 - Square into Squares

Good morning, all.

This challenge finishes our second week of the Daily Challenge series. It'll be our most difficult challenge yet!

We are grateful to user g964, who posted this challenge and many others on CodeWars.

Given a positive integer n, return a strictly increasing sequence of numbers so that the sum of the squares is equal to nΒ².

If there are multiple solutions, return the result with the largest possible value:

For example: decompose(11) must return [1,2,4,10].

Note: there are actually two ways to decompose 11Β², 11Β² = 121 = 1 + 4 + 16 + 100 = 1Β² + 2Β² + 4Β² + 10Β² but you shouldn't return [2,6,9], since 9 is smaller than 10.

Hint: Very often will xk be n-1.

Good luck and happy coding!


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

Top comments (12)

Collapse
 
aoneill01 profile image
Andy O'Neill

I think this works for JavaScript:

function squaresTotaling(max, goal) {
  if (max ** 2 === goal) return [max];
  if (goal <= 0 || max <= 0) return null;
  const subProblem = squaresTotaling(max - 1, goal - max ** 2);
  if (subProblem) return [...subProblem, max];
  return squaresTotaling(max - 1, goal);
}

function decompose(val) {
  return squaresTotaling(val - 1, val ** 2);
}
Collapse
 
alvaromontoro profile image
Alvaro Montoro

JavaScript

const decompose = number => {
  let components = [];
  let modifier = 1;
  do {
    // initialize the values before the loop
    components = [];
    let remaining = number ** 2;

    // from number - 1 until 1 check if the square of the number is less than the remaining
    for (let number = Math.floor(Math.sqrt(remaining)) - modifier; remaining > 0 && number > 0; number--) {
      if (number**2 <= remaining) {
        components.push(number);
        remaining -= number**2;
        // if a squared component is found, continue from the square root of the remaining
        number = Math.floor(Math.sqrt(remaining)) + 1;
      }
    }
    // the modifier is used to check for the next set of
    modifier++;
  // do this while there are duplicates (duplicates mean that one number -in particular 1- was repeated)
  // this code is a variation of https://stackoverflow.com/a/34192063/3695983
  } while (components.length !== new Set(components).size);
  return components.reverse();
}

And a live demo on CodePen.

Collapse
 
alvaromontoro profile image
Alvaro Montoro

Not really inspired lately, my code is getting bigger and messier... and I'm running behind on these challenges :-/

Collapse
 
coreyja profile image
Corey Alexander

I feel ya! Me tooooo

I got this one started but ran into a tougher example that I haven't worked through yet! Definitely want to get it wrapped up but I'm definitely not doing 1 a day anymore

Collapse
 
choroba profile image
E. Choroba • Edited on

Perl solution, using recursion:

#!/usr/bin/perl
use warnings;
use strict;

sub _decompose {
    my ($n, $target, $decomposition, $sum) = @_;

    return $decomposition if $target == $sum;

    my $smaller = int sqrt $n - 1;
    while ($smaller) {
        my $small_square = $smaller ** 2;
        if ($sum + $small_square <= $target) {
            my $d = _decompose(
                $smaller ** 2, $target,
                [ $smaller, @$decomposition ], $sum + $smaller ** 2
            );
            return $d if @$d;
        }
        --$smaller;
    }
    return []
}

sub decompose {
    my ($n) = @_;
    my $square = $n ** 2;
    return _decompose($square, $square, [], 0)
}

use Test::More tests => 2;
is_deeply decompose(11), [1, 2, 4, 10], 'eleven';
is_deeply decompose(12), [1, 2, 3, 7, 9], 'twelve';

I added 12 as a test, too, because for 12, 11 is not part of its decomposition.

Collapse
 
coreyja profile image
Corey Alexander

Here is my Rust version!

I started with an iterative solution, but I was working off a simplifying assumption that turned out to be false once I starting testing larger examples.

After I realized my simplification wasn't going to work I went for a recursive solution. I think it turned out pretty well! I also got to try out my first Macro in Rust so that was cool!

fn decompose_recursive(goal: u32, i: u32, curr: Vec<u32>) -> Option<Vec<u32>> {
    macro_rules! try_recusing {
        ( $goal:expr, $i:expr, $next:expr ) => {{
            let attempt = decompose_recursive($goal, $i, $next);

            if attempt.is_some() {
                return attempt;
            }
        }};
    }

    if goal == 0 {
        return Some(curr);
    }

    if i <= 0 {
        return None;
    }

    if goal >= i.pow(2) {
        let mut next = curr.clone();
        next.push(i);

        try_recusing!(goal - i.pow(2), i - 1, next);
    }

    try_recusing!(goal, i - 1, curr);

    None
}

pub fn decompose(n: u32) -> Option<Vec<u32>> {
    let reverse_vec = decompose_recursive(n.pow(2), n - 1, vec![])?;
    let vec = reverse_vec.iter().rev().cloned().collect();
    Some(vec)
}

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

    #[test]
    fn it_works_for_a_nonexistant_example() {
        assert_eq!(decompose(4), None);
    }

    #[test]
    fn it_works_for_the_dev_to_example() {
        assert_eq!(decompose(11), Some(vec![1, 2, 4, 10]));
    }

    #[test]
    fn it_works_for_50() {
        assert_eq!(decompose(50), Some(vec![1, 3, 5, 8, 49]));
    }
}
Collapse
 
matrossuch profile image
Mat-R-Such

Python solution:

def decompose(n):
    if n == 0:
        return -1
    elif n == 1:
        return [1]

    array_power=[pow(i,2) for i in range(1,n)]          #array with powers [1^2,2^2,3^2...,(n-1)^2]
    array_power=array_power[::-1]
    array_help,power,i,j=[],pow(n,2),0,n-1              # power= n^2, i=element(array), j= base of the power

    while power > 0:
        if power >= array_power[i]:
            power-=array_power[i]
            array_help.append(j)
        i+=1
        j-=1
        if i == n-1 and power > 0:
            return [n]
    return array_help[::-1]


print(decompose(1)) 
Enter fullscreen mode Exit fullscreen mode
Collapse
 
yagnesh_raja profile image
YAGNESH RAJA

Once try for 50 You should get[1,3,5,8,49],

Collapse
 
meave9786 profile image
meave9786

This is the use full blog here most people have to seen it and save the batter update of euchre free online really thanks for the share me it is the amazing.

Update Your DEV Experience Level:

Settings

Go to your customization settings to nudge your home feed to show content more relevant to your developer experience level. πŸ›