Seven Segment Display - Riddle by FiveThirtyEight - Solved with JavaScript

The FiveThirtyEight riddle Tyler Barron’s Seven Segment Display is stated as follows.

Given a two-character, seven-segment display, like you might find on a microwave clock, how many numbers can you make that are not ambiguous if the display happens to be upside down?

For example, the number 81 on that display would not fit this criterion — if the display were upside down it’d appear like 18. The number 71, however, would be OK. It’d appear something like 1L — not a number.

// Rotates a single seven-segment display digit 180 degrees.
// Non number results are converted to NaN.. eg. 2 --> 2, 6 --> 9, 3 --> NaN
const rotateDigit = n => [0, 1, 2, NaN, NaN, 5, 9, NaN, 8, 6][n];

// Rotates a seven-segment display number 180 degrees.
// eg. 10 --> 01, 81 --> 18, 71 --> 1NaN.
const rotateNumber = n =>
    // Convert the number to a string.
    n.toString()
    // Split the string on every character.
    .split('')
    // Reverse the order of the characters
    .reverse()
    // Rotate the digits.
    .map(rotateDigit)
    // Convert the array back to a string.
    .join('');


const numbers =
    // Generate a range from 0 to 99.
    [...Array(100).keys()]
    // Rotate every number.
    .map(rotateNumber)
    // Filter every number that is not a number. :)
    .filter(isNaN)

const result = numbers.length;

New Year's Resolution 2019

This year I will consume 52 more books than I usually read or listen to a year; in other words, 52 books. Knowing myself, almost all books will be non-fiction, and that is OK.

Books

[X] Contact, Carl Sagan, 1985
[X] Bad Blood, John Carreyrou, 2018
[X] Educated, Tara Westover, 2018
[X] Skunk Works, Ben Rich, Leo Janos, 1994
[X] Siddhartha, Hermann Hesse, 1922
[X] Expeditionary Force - Columbus Day, Craig Alanson, 2016
[X] Expeditionary Force - SpecOps, Craig Alanson, 2016
[X] A Study in Scarlet, Arthur Conan Doyle, 1887
[X] Access, Andy Weir, 2010 (SS)
[X] Antihypoxiant, Andy Weir, 2014 (SS)
[X] Annie's Day, Andy Weir, 2011 (SS)
[X] The Real Deal, Andy Weir, 2012 (SS)
[X] Bored World, Andy Weir, 2013 (SS)
[X] The Midtown Butcher, Andy Weir, 2014 (SS)
[X] Meeting Sarah, Andy Weir, 2010 (SS)
[X] The Chef, Andy Weir, 2010 (SS)
[X] The Egg, Andy Weir, 2009 (SS)
[X] The History of the Future, Blake J. Harris, 2019
[X] The Alchemist, Paulo Coelho, 1988
[\] Humble Pi, Matt Parker, 2019

Project Euler 22 - Names Scores - Solved with JavaScript

The 22th Project Euler problem - Names Scores - is stated as follows.

Using names.txt, a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714.

What is the total of all the name scores in the file? Project Euler, Names Scores, Problem 22

const fs = require('fs');

// Read the textfile into a string.
const names = fs.readFileSync('p022_names.txt', 'utf8')
    // Split the string on every comma and remove the quotes
    .split(',').map(name => name.replace(/"/g, ''))
    // Sort the names.
    .sort();

// Get the alphabetical value by adding the chars position in
// the alphabet. A = Char Code 65 = Alphabet Position 1.
const getAlphabeticalValue = name => name.split('')
    .reduce((memo, item) => memo + item.charCodeAt(0) - 64, 0);

const result = names.reduce((memo, item, index) =>
    memo + getAlphabeticalValue(item) * (index + 1), 0);

Project Euler 21 - Amicable Numbers - Solved with JavaScript

The 21th Project Euler problem - Amicable Numbers - is stated as follows. Evaluate the sum of all the amicable numbers under 10000.

const getProperDivisors = (n) => {
  const divisors = [];
  const root = Math.floor(Math.sqrt(n));
  for (let i = 1; i <= root; i++) {
    if (n % i === 0) {
      divisors.push(i);
      if (Math.pow(i, 2) !== n) {
        divisors.push(n / i);
      }
    }
  }
  return divisors.sort(function (a, b) {
    return a - b;
  }).filter(x => x !== n);
};

// Let d(n) be defined as the sum of proper divisors of
// n (numbers less than n which divide evenly into n).
const d = (n) => {
  const divisors = getProperDivisors(n);
  return divisors.reduce((a,b) => a + b, 0);
};

// Iterate over all numbers from 0 to 9999.
const result = Array.from({length: 9999}).map((_, n) => {
    const a = d(n);
    const b = d(a);
    // If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
    // pair and each of a and b are called amicable numbers.
    return n !== a && n === b ? a : 0;
}).reduce((a,b) => a + b);

Project Euler 20 - Factorial Digit Sum - Solved with JavaScript

The 20th Project Euler problem - Factorial Digit Sum - is stated as follows. Find the sum of the digits in the number 100!

// Recursive function using BigInt to get n factorial.
const factorial = (n) => n <= 1 ? BigInt(n) : BigInt(n) * BigInt(factorial(--n));

// Convert the number to a string.
const sumDigits = n => n.toString()
    // Split the string on every char.
    .split('')
    // Convert all chars to ints.
    .map(x => parseInt(x))
    // Sum all ints.
    .reduce((a,b) => a + b);

// Get 100 factorial.
const f100 = factorial(100);

// Sum the digits.
const result = sumDigits(f100);

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