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curriculum/challenges/italian/10-coding-interview-prep/project-euler/problem-461-almost-pi.md

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+---
+id: 5900f53a1000cf542c51004c
+title: 'Problem 461: Almost Pi'
+challengeType: 5
+forumTopicId: 302136
+dashedName: problem-461-almost-pi
+---
+
+# --description--
+
+Let `f(k, n)` = $e^\frac{k}{n} - 1$, for all non-negative integers `k`.
+
+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 `f(a, 200) + f(b, 200) + f(c, 200) + f(d, 200)`.
+
+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: $\lvert f(a,n) + f(b,n) + f(c,n) + f(d,n) - \Pi\rvert$
+
+You are given `almostPi(200)` = 6<sup>2</sup> + 75<sup>2</sup> + 89<sup>2</sup> + 226<sup>2</sup> = 64658.
+
+# --hints--
+
+`almostPi` should be a function.
+
+```js
+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--
+
+## --seed-contents--
+
+```js
+function almostPi(n) {
+  
+  return true;
+}
+```
+
+# --solutions--
+
+```js
+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;
+}
+```