使用chatgpt(GPT-4)将过程式(的java代码)改成函数式(的elixir代码)

Posted 2023-04-09 yhm138

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天啦噜太可怕了，之前我还嘲笑chatgpt不会小众语言来着。

chatgpt(GPT-4)把过程式(的java代码)改成了函数式(的elixir代码)。chatgpt(GPT-4)在接收2次prompt后，给出的Elixir代码可以不经修改直接AC。
如果你用的是GPT-3.5的版本，它甚至不知道elixir里没有while关键字。
chatgpt对文本的reasoning特别擅长(如翻译，总结全文，根据主题写文章)。
但是chatgpt事实上并不具备创新的能力，如果语料里【过程式代码转函数式代码的例子】足够多，它能完成这一任务我一点不觉得惊讶。
我理解是特别强大的检索整合能力，但是对于原创的新颖的算法题目，它也只能对着题目描述废话一堆给出错误的答案。
update 2023-04-09. 不要尝试将【下标从0开始的编程语言】和【下标从1开始的编程语言】 (如Mathematica,pari/gp等)互相翻译，效果很差。

Problem URL
Accepted elixir code
Prompts and answers
Why do I say elixir is an esoteric programming language

Problem URL

https://leetcode.cn/problems/split-two-strings-to-make-palindrome/description/

Accepted elixir code


defmodule Solution do
  def check_palindrome_formation(a, b) do
    check_concatenation(a, b) || check_concatenation(b, a)
  end

  def check_concatenation(a, b) do
    n = byte_size(a)
    check_concatenation_recursive(a, b, 0, n - 1)
  end

  defp check_concatenation_recursive(a, b, left, right) when left >= right do
    true
  end

  defp check_concatenation_recursive(a, b, left, right) do
    if binary_part(a, left, 1) == binary_part(b, right, 1) do
      check_concatenation_recursive(a, b, left + 1, right - 1)
    else
      check_self_palindrome(a, left, right) || check_self_palindrome(b, left, right)
    end
  end

  def check_self_palindrome(a, left, right) do
    check_self_palindrome_recursive(a, left, right)
  end

  defp check_self_palindrome_recursive(a, left, right) when left >= right do
    true
  end

  defp check_self_palindrome_recursive(a, left, right) do
    if binary_part(a, left, 1) == binary_part(a, right, 1) do
      check_self_palindrome_recursive(a, left + 1, right - 1)
    else
      false
    end
  end
end

Prompts and answers

convert java code to elixir code, ```  format please


class Solution 
    public boolean checkPalindromeFormation(String a, String b) 
        return checkConcatenation(a, b) || checkConcatenation(b, a);
    

    public boolean checkConcatenation(String a, String b) 
        int n = a.length();
        int left = 0, right = n - 1;
        while (left < right && a.charAt(left) == b.charAt(right)) 
            left++;
            right--;
        
        if (left >= right) 
            return true;
        
        return checkSelfPalindrome(a, left, right) || checkSelfPalindrome(b, left, right);
    

    public boolean checkSelfPalindrome(String a, int left, int right) 
        while (left < right && a.charAt(left) == a.charAt(right)) 
            left++;
            right--;
        
        return left >= right;

Here\'s the equivalent Elixir code using pattern matching and recursion:

defmodule Solution do
  def check_palindrome_formation(a, b) do
    check_concatenation(a, b) || check_concatenation(b, a)
  end

  def check_concatenation(a, b) do
    n = String.length(a)
    check_concatenation_recursive(a, b, 0, n - 1)
  end

  defp check_concatenation_recursive(a, b, left, right) when left >= right do
    true
  end

  defp check_concatenation_recursive(a, b, left, right) do
    if String.at(a, left) == String.at(b, right) do
      check_concatenation_recursive(a, b, left + 1, right - 1)
    else
      check_self_palindrome(a, left, right) || check_self_palindrome(b, left, right)
    end
  end

  def check_self_palindrome(a, left, right) do
    check_self_palindrome_recursive(a, left, right)
  end

  defp check_self_palindrome_recursive(a, left, right) when left >= right do
    true
  end

  defp check_self_palindrome_recursive(a, left, right) do
    if String.at(a, left) == String.at(a, right) do
      check_self_palindrome_recursive(a, left + 1, right - 1)
    else
      false
    end
  end
end

This Elixir code follows the same logic as the Java code but uses Elixir\'s pattern matching and recursion capabilities. 
Note that the `check_concatenation_recursive` and `check_self_palindrome_recursive` functions have been defined with 
different patterns to handle the base case and recursive case separately.

yes, the result is true, but how to accelerate it, note that 1 <= a.length, b.length <= 10^5

To improve the performance of the Elixir code, you can leverage the `binary_part/3` function, which is more efficient for working with large binaries (strings in Elixir are binaries) than `String.at/2`. Here\'s the updated code with the optimized functions:

defmodule Solution do
  def check_palindrome_formation(a, b) do
    check_concatenation(a, b) || check_concatenation(b, a)
  end

  def check_concatenation(a, b) do
    n = byte_size(a)
    check_concatenation_recursive(a, b, 0, n - 1)
  end

  defp check_concatenation_recursive(a, b, left, right) when left >= right do
    true
  end

  defp check_concatenation_recursive(a, b, left, right) do
    if binary_part(a, left, 1) == binary_part(b, right, 1) do
      check_concatenation_recursive(a, b, left + 1, right - 1)
    else
      check_self_palindrome(a, left, right) || check_self_palindrome(b, left, right)
    end
  end

  def check_self_palindrome(a, left, right) do
    check_self_palindrome_recursive(a, left, right)
  end

  defp check_self_palindrome_recursive(a, left, right) when left >= right do
    true
  end

  defp check_self_palindrome_recursive(a, left, right) do
    if binary_part(a, left, 1) == binary_part(a, right, 1) do
      check_self_palindrome_recursive(a, left + 1, right - 1)
    else
      false
    end
  end
end


By using `binary_part/3` instead of `String.at/2`, the performance should be improved, especially when dealing with large strings. This is because `binary_part/3` provides direct access to the specified part of the binary without creating a new one, while `String.at/2` creates a new one for each character.