FizzBu​​zz Ruby一线 [英] FizzBuzz Ruby one-liner

问题描述

Rosettacode.org在Ruby中具有出色的单行FizzBu​​zz解决方案.

Rosettacode.org has this excellent one-line FizzBuzz solution in Ruby.

1.upto(100){|n|puts'FizzBuzz '[i=n**4%-15,i+13]||n}

麻烦是，我不明白.使我感到困惑的部分是"n以4模-15的幂".有人有解释或对解释的引用吗?我想在其他问题中使用这种选择子字符串的方式. 有关FizzBu​​zz的详细信息，请参阅[ https://rosettacode.org/wiki/FizzBu​​zz ]

The trouble is, I don’t understand it. The part that puzzles me is the "n to the power of 4 modulo -15". Does anyone have an explanation or a reference to an explanation? I want to use this way of selecting substrings in other problems. For more information on FizzBuzz, see [https://rosettacode.org/wiki/FizzBuzz]

推荐答案

我不知道他们是如何发现提高到第四幂的，但是-15是因为FizzBu​​zz处理的是3的倍数或5的倍数或5的倍数. 3和5(即15的倍数)中的一个...然后取反，最终可以很好地处理负指数.我们可以看到它与模块化幂一起使用.此处的内存有效方法部分显示:

I don't know how they discovered to raise to the fourth power, but the -15 is because FizzBuzz deals with multiples of 3 or multiples of 5 or multiples of both 3 and 5 (ie, multiples of 15)...then negating it ends up working with negative indices quite well. We can see that it works with Modular Exponentiation. The Memory-efficient method section there says:

c mod m =(a⋅b)mod m
c mod m = [(a mod m)⋅(b mod m)] mod m

c mod m = (a ⋅ b) mod m
c mod m = [(a mod m) ⋅ (b mod m)] mod m

在我们的例子中， c 是我们的 n ，所以我们有

In our case, the c is our n, so we have

c ** 4 % m

使用指数定律，我们知道(c ** e1) * (c ** e2) = c ** (e1 + e2) ，所以c ** 4 = (c ** 2) * (c ** 2)，所以我们现在有一个a和b，它们都是c ** 2.因此:

using the law of exponents, we know that (c ** e1) * (c ** e2) = c ** (e1 + e2), so c ** 4 = (c ** 2) * (c ** 2), so we now have an a and a b, which are both c ** 2. Thus:

(c ** 4) % m = ((c ** 2) * (c ** 2)) % m
             = (((c ** 2) % m) * ((c ** 2) % m)) % m
             = (((c ** 2) % m) ** 2) % m

并再次执行相同的步骤:

and following the same steps, again:

(c ** 2) % m = (c * c) % m
             = ((c % m) * (c % m)) % m
             = ((c % m) ** 2) % m

最后:

(c ** 4) % m = ((((c % m) ** 2) % m) ** 2) % m

当m = -15时，c % m的唯一值是(-14..0)，我们可以构建一个简单的表进行查看.由于我们只对模的结果进行运算，因此我们只需要证明这15个数字是有效的:

When m = -15, the only values for c % m are (-14..0) and we can build a simple table to look at. Since we only ever operate on the result of a modulo, we only need to be able to prove these 15 numbers work:

c%m    **2     %m    **2     %m
-14 => 196 => -14 => 196 => -14
-13 => 169 => -11 => 121 => -14
-12 => 144 => -06 =>  36 => -09
-11 => 121 => -14 => 196 => -14
-10 => 100 => -05 =>  25 => -05
-09 =>  81 => -09 =>  81 => -09
-08 =>  64 => -11 => 121 => -14
-07 =>  49 => -11 => 121 => -14
-06 =>  36 => -09 =>  81 => -09
-05 =>  25 => -05 =>  25 => -05
-04 =>  16 => -14 => 196 => -14
-03 =>   9 => -06 =>  36 => -09
-02 =>   4 => -11 => 121 => -14
-01 =>   1 => -14 => 196 => -14
 00 =>   0 =>  00 =>   0 =>  00

现在，看我们的表，3的所有倍数的值均为-09，5的所有倍数的值均为-05，3和5的倍数的值设置为00；其他所有内容都是-14(如果我们使用15而不是-15，则我们将分别具有6、10、0和1，并且需要查找才能将其转换为字符串索引).将它们插入 String#[] 的开始参数中字符串'FizzBuzz '可以给我们:

Now, looking at our table, the values for all multiples of 3 are -09, the values for all multiples of 5 are -05, and things that are multiples of 3 and 5 are set to 00; everything else is -14 (If we had used 15 instead of -15, we'd have 6, 10, 0, and 1, respectively, and would need a look up to turn that into string indices). Plugging those in for the start parameter of String#[] with the string 'FizzBuzz ' gives us:

'FizzBuzz '[-9] # => 'F'
'FizzBuzz '[-5] # => 'B'
'FizzBuzz '[0]  # => 'F'
'FizzBuzz '[-14]# => nil

并在这些数字上加13以得到长度:

and adding 13 to those numbers to get the length:

'FizzBuzz '[-9, 4]   # => "Fizz"
'FizzBuzz '[-5, 8]   # => "Buzz "
'FizzBuzz '[0, 13]   # => "FizzBuzz "
'FizzBuzz '[-14, -1] # => nil

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