* fix: clean-up Project Euler 241-260 * fix: typo * Update curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-255-rounded-square-roots.md Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
		
			
				
	
	
		
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			68 lines
		
	
	
		
			2.1 KiB
		
	
	
	
		
			Markdown
		
	
	
	
	
	
| ---
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| id: 5900f46d1000cf542c50ff7f
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| title: 'Problem 255: Rounded Square Roots'
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| challengeType: 5
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| forumTopicId: 301903
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| dashedName: problem-255-rounded-square-roots
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| ---
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| 
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| # --description--
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| 
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| We define the rounded-square-root of a positive integer $n$ as the square root of $n$ rounded to the nearest integer.
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| 
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| The following procedure (essentially Heron's method adapted to integer arithmetic) finds the rounded-square-root of $n$:
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| 
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| Let $d$ be the number of digits of the number $n$.
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| 
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| If $d$ is odd, set $x_0 = 2 × {10}^{\frac{d - 1}{2}}$.
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| 
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| If $d$ is even, set $x_0 = 7 × {10}^{\frac{d - 2}{2}}$.
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| 
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| Repeat:
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| 
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| $$x_{k + 1} = \left\lfloor\frac{x_k + \left\lceil\frac{n}{x_k}\right\rceil}{2}\right\rfloor$$
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| 
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| until $x_{k + 1} = x_k$.
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| 
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| As an example, let us find the rounded-square-root of $n = 4321$.
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| 
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| $n$ has 4 digits, so $x_0 = 7 × {10}^{\frac{4-2}{2}} = 70$.
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| 
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| $$x_1 = \left\lfloor\frac{70 + \left\lceil\frac{4321}{70}\right\rceil}{2}\right\rfloor = 66 \\\\
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| x_2 = \left\lfloor\frac{66 + \left\lceil\frac{4321}{66}\right\rceil}{2}\right\rfloor = 66$$
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| 
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| Since $x_2 = x_1$, we stop here. So, after just two iterations, we have found that the rounded-square-root of 4321 is 66 (the actual square root is 65.7343137…).
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| 
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| The number of iterations required when using this method is surprisingly low. For example, we can find the rounded-square-root of a 5-digit integer ($10\\,000 ≤ n ≤ 99\\,999$) with an average of 3.2102888889 iterations (the average value was rounded to 10 decimal places).
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| 
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| Using the procedure described above, what is the average number of iterations required to find the rounded-square-root of a 14-digit number (${10}^{13} ≤ n < {10}^{14}$)? Give your answer rounded to 10 decimal places.
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| 
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| **Note:** The symbols $⌊x⌋$ and $⌈x⌉$ represent the floor function and ceiling function respectively.
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| 
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| # --hints--
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| 
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| `roundedSquareRoots()` should return `4.447401118`.
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| 
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| ```js
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| assert.strictEqual(roundedSquareRoots(), 4.447401118);
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| ```
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| 
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| # --seed--
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| 
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| ## --seed-contents--
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| 
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| ```js
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| function roundedSquareRoots() {
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| 
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|   return true;
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| }
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| 
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| roundedSquareRoots();
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| ```
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| 
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| # --solutions--
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| 
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| ```js
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| // solution required
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| ```
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