diff --git a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md index 5f3d2bb361..09c3e10628 100644 --- a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md +++ b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-70-totient-permutation.md @@ -8,24 +8,42 @@ dashedName: problem-70-totient-permutation # --description-- -Euler's Totient function, φ(`n`) \[sometimes called the phi function], is used to determine the number of positive numbers less than or equal to `n` which are relatively prime to `n`. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6. The number 1 is considered to be relatively prime to every positive number, so φ(1)=1. +Euler's Totient function, ${\phi}(n)$ (sometimes called the phi function), is used to determine the number of positive numbers less than or equal to `n` which are relatively prime to `n`. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, ${\phi}(9) = 6$. The number 1 is considered to be relatively prime to every positive number, so ${\phi}(1) = 1$. -Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180. +Interestingly, ${\phi}(87109) = 79180$, and it can be seen that 87109 is a permutation of 79180. -Find the value of `n`, 1 < `n` < 107, for which φ(`n`) is a permutation of `n` and the ratio `n`/φ(`n`) produces a minimum. +Find the value of `n`, 1 < `n` < `limit`, for which ${\phi}(n)$ is a permutation of `n` and the ratio $\displaystyle\frac{n}{{\phi}(n)}$ produces a minimum. # --hints-- -`totientPermutation()` should return a number. +`totientPermutation(10000)` should return a number. ```js -assert(typeof totientPermutation() === 'number'); +assert(typeof totientPermutation(10000) === 'number'); ``` -`totientPermutation()` should return 8319823. +`totientPermutation(10000)` should return `4435`. ```js -assert.strictEqual(totientPermutation(), 8319823); +assert.strictEqual(totientPermutation(10000), 4435); +``` + +`totientPermutation(100000)` should return `75841`. + +```js +assert.strictEqual(totientPermutation(100000), 75841); +``` + +`totientPermutation(500000)` should return `474883`. + +```js +assert.strictEqual(totientPermutation(500000), 474883); +``` + +`totientPermutation(10000000)` should return `8319823`. + +```js +assert.strictEqual(totientPermutation(10000000), 8319823); ``` # --seed-- @@ -33,16 +51,68 @@ assert.strictEqual(totientPermutation(), 8319823); ## --seed-contents-- ```js -function totientPermutation() { +function totientPermutation(limit) { return true; } -totientPermutation(); +totientPermutation(10000); ``` # --solutions-- ```js -// solution required +function totientPermutation(limit) { + function getSievePrimes(max) { + const primes = []; + const primesMap = new Array(max).fill(true); + primesMap[0] = false; + primesMap[1] = false; + + for (let i = 2; i < max; i += 2) { + if (primesMap[i]) { + primes.push(i); + for (let j = i * i; j < max; j += i) { + primesMap[j] = false; + } + } + if (i === 2) { + i = 1; + } + } + return primes; + } + + function sortDigits(number) { + return number.toString().split('').sort().join(''); + } + + function isPermutation(numberA, numberB) { + return sortDigits(numberA) === sortDigits(numberB); + } + + const MAX_PRIME = 4000; + const primes = getSievePrimes(MAX_PRIME); + + let nValue = 1; + let minRatio = Infinity; + + for (let i = 1; i < primes.length; i++) { + for (let j = i + 1; j < primes.length; j++) { + const num = primes[i] * primes[j]; + if (num > limit) { + break; + } + + const phi = (primes[i] - 1) * (primes[j] - 1); + const ratio = num / phi; + + if (minRatio > ratio && isPermutation(num, phi)) { + nValue = num; + minRatio = ratio; + } + } + } + return nValue; +} ```