Project Euler 44: Find the smallest pair of pentagonal numbers whose sum and difference is pentagona...-优快云博客

本文探讨了如何找到两个五边形数，它们的和与差同样是五边形数，并且差值最小的问题。通过使用C#编程语言，我们实现了一个算法来检查给定数字是否为五边形数，并利用该算法在五边形数中查找符合条件的数对。

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In Problem 42 we dealt with triangular problems, in Problem 44 of Project Euler we deal with pentagonal number, I can only wonder if we have to deal with septagonal numbers in Problem 46. Anyway the problem reads

Pentagonal numbers are generated by the formula, P_n=n(3n-1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …

It can be seen that P₄ + P₇ = 22 + 70 = 92 = P₈. However, their difference, 70 – 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference is pentagonal and D = |P_k – P_j| is minimised; what is the value of D?

I have found a solution which I will present to you in just a few lines, but I must admit that I haven’t been able to justify why it is the right solution it gives us. The solution I have made just finds a solution to the problem which happens to the right solution.

I did not want to generate a list of pentagonal numbers, so I wanted to make a small function which checks if a given number is pentagonal by checking if the inverse function yields an integer, just like in the solution to Problem 42. We could rather easily calculate the inverse function as we did with the inverse function for triangular numbers, or we can cheat and peak at the pentagonal number entry at Wikipedia.

The inverse function is

$\displaystyle n = \frac{\sqrt{24x 1} 1}{6}$

That enables us to make a C# function that checks if a number is pentagonal

 
         private 
         bool 
         isPentagonal( 
         int 
         number) { 
        
         double 
         penTest = (Math.Sqrt(1 + 24 * number) + 1.0) / 6.0; 
        
         return 
         penTest == (( 
         int 
         )penTest); 
        
         }

Once we have this crucial function we can make two nested loops to check pentagonal numbers until we find two where the sum and difference are pentagonal as well. I am frustrated since I can’t figure out why this one is the first. I can prove that it is indeed minimal by testing other numbers until the difference of two numbers reach the result of this problem. However I haven’t done that.

The outer loop of the algorithm counts upwards generating and the inner loop counting downwards testing all pentagonal numbers less than the one generated by the outer loop. The code looks like

 
         int 
         result = 0; 
        
         bool 
         notFound =  
         true 
         ; 
        
         int 
         i = 1; 
        
         while 
         (notFound) { 
        
         i++; 
        
         int 
         n = i * (3 * i - 1) / 2; 
        
         for 
         ( 
         int 
         j = i-1; j > 0; j--) { 
        
         int 
         m = j * (3 * j - 1) / 2; 
        
         if 
         (isPentagonal(n - m) && isPentagonal(n + m)) { 
        
         result = n-m; 
        
         notFound =  
         false 
         ; 
        
         break 
         ; 
        
         } 
        
         } 
        
         }

and gives the result

 
         k = 2167, j = 1020 
        
         The value of D is 5482660 
        
         Solution took 35 ms

Wrapping up

I can see that many other people also can’t give the definitive answer to why the result is correct. If you understand the problem better than I do, please let me know exactly why I find the right solution.

You can as usual find the source code for the problem here.

ref

转载于:https://www.cnblogs.com/forcheryl/p/4899714.html