使用scanl和自引用列表的斐波那契数列

时间:2018-07-02 08:26:42

标签: haskell

我正在通过Chris Allen Haskell编程书来学习Haskell。我被困在使用scanl和自引用列表定义Fibonacci系列的一部分上。 我了解对以下代码的评估:

scanl (+) 1 [1..3]
1 : (scanl (+) ((+) 1 1) [2,3])

我读到(+)在两个参数上都是严格的,因此下一步将((+)1 1)替换为2。

1 : (2 : (scanl (+) ((+) 2 2) [3]))
1 : (2 : (4 : (scanl (+) ((+) 4 3) [])))
1 : (2 : (4 : (7 : (scanl (+) 7 []))))
1 : (2 : (4 : (7 : [])))

我无法理解以下代码的评估以及它如何产生正确的斐波那契数列。

fibs = 1 : scanl (+) 1 fibs.

我很难找到非空列表(x:xs)的模式匹配如何用于自引用列表。 感谢帮助和任何资源,以更好地了解自引用/无限列表。

1 个答案:

答案 0 :(得分:2)

我们可以使用由1:引起的相位差来仅使用scanl进行递归后向引用,如您在上一个示例中所示。

let fib = 1 : scanl (+) 1 fib in take 5 fib

以这种形式替换scanl的定义时,我们首先看到(x:xs)与现有的:相匹配:

1 : 1 : scanl (+) 2 (drop 1 fibs)

此时,第二个元素存在,因此我们可以提取另一个x

1 : 1 : 2 : scanl (+) 3 (drop 2 fibs)
1 : 1 : 2 : 3 : scanl (+) 5 (drop 3 fibs)
1 : 1 : 2 : 3 : 5 : scanl (+) 8 (drop 4 fibs)

我们并不是真正使用drop,而是通过为每个元素匹配:来沿着我们正在构建的相同列表前进。我必须给其余列表命名,以递归地引用它,并发现drop形式可以方便地表示生成和起源。实际上,像这样使用drop会使整个列表保留在内存中,并从头开始遍历不必要的遍历以查找下一个元素。实际上,只有小块形状像递归let表达式那样作为尾巴,直到我们对其求值:

fibsfrom a b = let morefibs = a : scanl (+) b morefibs in morefibs

如果将此函数以1 1或其他任意位置作为种子,则可以通过连续的一对斐波那契数对种子来获得整个列表。尝试使用例如2 3或5 8.尾巴的公式先回到列表的前一个元素,因此可以始终将其拆分为头尾。

有时将引用作为图形查看会更清楚。我们将递归视为一个循环,但是仍然可以进行评估,因为我们只需要head list即可执行下一步。这是前几个步骤的graphviz点渲染(为回旋处的“运行代码段-全页”方法表示歉意,堆栈溢出无法整齐地处理点或svg)。

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