Talk:Brilliant numbers: Difference between revisions
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== 100 limit quibble == |
== 100 limit quibble == |
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In my translation of C++, I replaced <code>for j=i to length(primes) do</code> with <code>for j=1 to i do</code>. The reason being I didn't really believe the length()>100 test, whereas with 1 to i it feels (more) like that must be OK. Made no actual difference. Both might be fine. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 22:23, 22 April 2022 (UTC) |
In my translation of C++, I replaced <code>for j=i to length(primes) do</code> with <code>for j=1 to i do</code>. The reason being I didn't really believe the length()>100 test, whereas with 1 to i it feels (more) like that must be OK. Made no actual difference. Both might be fine. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 22:23, 22 April 2022 (UTC) |
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:Both are fine. In one case we get all products p1 x p2 where p1 <= p2, in your case it's p1 >= p2. We can stop when we have at least 100 products because at each iteration of the outer loop we increase the number of digits by 1 (if p1 and p2 both have n digits then p1 x p2 must be greater than the product of any two numbers with fewer than n digits). |
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Revision as of 08:28, 23 April 2022
100 limit quibble
In my translation of C++, I replaced for j=i to length(primes) do with for j=1 to i do. The reason being I didn't really believe the length()>100 test, whereas with 1 to i it feels (more) like that must be OK. Made no actual difference. Both might be fine. --Pete Lomax (talk) 22:23, 22 April 2022 (UTC)
- Both are fine. In one case we get all products p1 x p2 where p1 <= p2, in your case it's p1 >= p2. We can stop when we have at least 100 products because at each iteration of the outer loop we increase the number of digits by 1 (if p1 and p2 both have n digits then p1 x p2 must be greater than the product of any two numbers with fewer than n digits).