fix(learn): Rework Euler Problem 461 (#41333)

* Made instructions more clear + Corrected test cases

* Added highlight for return values

* Adjusted challenge description

* Rework

* Fixed minor linting error

* Address PR Comments

* Adjusted wording in description
This commit is contained in:
Caden Parker
2021-03-10 07:14:27 -08:00
committed by GitHub
parent 11e162cc91
commit 6b9aeff09e

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@ -8,26 +8,46 @@ dashedName: problem-461-almost-pi
# --description--
Let fn(k) = ek/n - 1, for all non-negative integers k.
Let `f(k, n)` = $e^\frac{k}{n} - 1$, for all non-negative integers `k`.
Remarkably, f200(6) + f200(75) + f200(89) + f200(226) = 3.141592644529… ≈ π.
Remarkably, `f(6, 200) + f(75, 200) + f(89, 200) + f(226, 200)` = 3.1415926… ≈ π.
In fact, it is the best approximation of π of the form fn(a) + fn(b) + fn(c) + fn(d) for n = 200.
In fact, it is the best approximation of π of the form `f(a, 200) + f(b, 200) + f(c, 200) + f(d, 200)`.
Let g(n) = a2 + b2 + c2 + d 2 for a, b, c, d that minimize the error: | fn(a) + fn(b) + fn(c) + fn(d) - π|
Let `almostPi(n)` = a<sup>2</sup> + b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup> for a, b, c, d that minimize the error: $\lvert f(a,n) + f(b,n) + f(c,n) + f(d,n) - \Pi\rvert$
(where |x| denotes the absolute value of x).
You are given g(200) = 62 + 752 + 892 + 2262 = 64658.
Find g(10000).
You are given `almostPi(200)` = 6<sup>2</sup> + 75<sup>2</sup> + 89<sup>2</sup> + 226<sup>2</sup> = 64658.
# --hints--
`euler461()` should return 159820276.
`almostPi` should be a function.
```js
assert.strictEqual(euler461(), 159820276);
assert(typeof almostPi === 'function')
```
`almostPi` should return a number.
```js
assert.strictEqual(typeof almostPi(10), 'number');
```
`almostPi(29)` should return `1208`.
```js
assert.strictEqual(almostPi(29), 1208);
```
`almostPi(50)` should return `4152`.
```js
assert.strictEqual(almostPi(50), 4152);
```
`almostPi(200)` should return `64658`.
```js
assert.strictEqual(almostPi(200), 64658);
```
# --seed--
@ -35,16 +55,99 @@ assert.strictEqual(euler461(), 159820276);
## --seed-contents--
```js
function euler461() {
function almostPi(n) {
return true;
}
euler461();
```
# --solutions--
```js
// solution required
function almostPi(n) {
// Find all possible values where f(k, n) <= PI
const f = [];
let max = 0;
while (1) {
let current = Math.exp(max / n) - 1;
if (current > Math.PI) break;
f.push(current);
++max;
}
// Get all pairs where f[i] + f[j] <= PI
const pairs = [];
for (let i = 0; i < max; ++i) {
for (let j = 0; j < max; ++j) {
if (f[i] + f[j] > Math.PI) break;
pairs.push(f[i] + f[j]);
}
}
// Sort all values
pairs.sort((a, b) => a - b);
// Optimal Value for (a + b)
let left = 0;
// Optimal Value for (c + d)
let right = 0;
// minimum error with Math.abs(a + b - Math.PI)
let minError = Math.PI;
// Binary Search for the best match
for (let i = 0; i < pairs.length; ++i) {
let current = pairs[i];
let need = Math.PI - current;
if (need < current) break;
let match;
for (let i = 1; i < pairs.length; ++i) {
if (pairs[i] > need) {
match = i;
break;
}
}
let error = Math.abs(need - pairs[match]);
if (error < minError)
{
minError = error;
left = i;
right = match;
}
--match;
error = Math.abs(need - pairs[match]);
if (error < minError) {
minError = error;
left = i;
right = match;
}
}
let a, b, c, d;
OuterLoop1:
for (a = 0; a < max; ++a) {
for (b = a; b < max; ++b) {
if (pairs[left] == f[a] + f[b]) {
break OuterLoop1;
}
}
}
OuterLoop2:
for (c = 0; c < max; ++c) {
for (d = c; d < max; ++d) {
if (pairs[right] == f[c] + f[d]) {
break OuterLoop2;
}
}
}
return a*a + b*b + c*c + d*d;
}
```