帕斯卡的三角形行序列 [英] Pascal's Triangle Row Sequence

问题描述

我目前正在寻找Pascal三角形的行序列.我想输入行号并输出列表中的数字序列，直到该行为止.例如，(Pascal 4)将给出结果(1 1 1 1 2 1 1 3 3 1).

I'm currently working on finding the row sequences of Pascal's triangle. I wanted to input the row number and output the sequence of numbers in a list up until that row. For example, (Pascal 4) would give the result (1 1 1 1 2 1 1 3 3 1).

我正在尝试使用发现的算法.这是算法本身:

I am trying to use an algorithm that I found. Here is the algorithm itself:

V _c = V _c-1 *((r-c)/c)

V_c = V_c-1 * ((r - c)/c)

r 和 c 应该是行和列，并且V ₀ = 1.该算法可以在维基百科页面上的计算和单个行或对角线"部分中找到.

r and c are supposed to be row and column, and V₀=1. The algorithm can be specifically found on the wikipedia page in the section titled "Calculating and Individual Row or Diagonal."

这是我到目前为止的代码:

Here is the code that I have so far:

(define pascal n)
  (cond((zero? n) '())
       ((positive? n) (* pascal (- n 1) (/ (- n c)c))))

我知道几乎没有什么，但是我一直在努力寻找用let或lambda合并函数字段值的作用域.另外，我也一直在努力递归.我真的不知道如何建立基本案例以及如何进行下一步.基本上，我到处都迷路了.我知道这并没有显示太多，但是朝着正确方向迈出的任何一步都将不胜感激.

I know that's hardly anything but I've been struggling a lot on trying to find scoping the function with a let or a lambda to incorporate column values. Additionally, I've also been struggling on the recursion. I don't really know how to establish the base case and how to get to the next step. Basically, I've been getting pretty lost everywhere. I know this isn't showing much, but any step in the right direction would be greatly appreciated.

推荐答案

使用 entry 在Wikipedia中，这是算法的直接实现，用于在给定Pascal Triangle中的行和列的情况下计算值，如链接中所述:

Using as a guide the entry in Wikipedia, this is a straightforward implementation of the algorithm for calculating a value in the Pascal Triangle given its row and column, as described in the link:

#lang racket

(define (pascal row column)
  (define (aux r c)
    (if (zero? c)
        1
        (* (/ (- r c) c)
           (aux r (sub1 c)))))
  (aux (add1 row) column))

例如，以下代码将返回值的前四行，注意行和列均以零开头:

For example, the following will return the first four rows of values, noticing that both rows and columns start with zero:

(pascal 0 0)

(pascal 1 0)
(pascal 1 1)

(pascal 2 0)
(pascal 2 1)
(pascal 2 2)

(pascal 3 0)
(pascal 3 1)
(pascal 3 2)
(pascal 3 3)

现在，我们需要一个过程将所有值粘贴在一起，直到所需的行；这适用于球拍:

Now we need a procedure to stick together all the values up until the desired row; this works for Racket:

(define (pascal-up-to-row n)
  (for*/list ((i (in-range n))
              (j (in-range (add1 i))))
    (pascal i j)))

结果符合预期:

(pascal-up-to-row 4)
> '(1 1 1 1 2 1 1 3 3 1)

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