# RogerBW's Blog

 Perl Weekly Challenge 139: Jort Primes 18 November 2021 I’ve been doing the Weekly Challenges. The latest involved date calculations and numerical decompositions. (Note that this is open until 21 November 2021.) Task 1: Jort Sort You are given a list of numbers. Write a script to implement JortSort. It should return true/false depending if the given list of numbers are already sorted. Finally I'm allowed to return a Boolean rather than 1/0! Hurrah! This is of course Jenn Schiffer's Javascript framework. It must be good, it's got a logo and everything! (More history here.) What's more, with my massive manly and testosterone-soaked brain, I have improved the algorithm. Which of course isn't the point, and arguably a sort followed by an array comparison would be more in the spirit of the original. But I did get to learn some new language features, so there's that. The basic idea of this approach is to compare pairs of entries: if an earlier one is higher than a later one, the list isn't sorted and we can return without doing the rest of the comparisons. In Perl (the only language here without explicit Boolean constants) and PostScript I lay that out by hand using array indices: ``````sub jortsort { my \$a=shift; foreach my \$i (1..\$#{\$a}) { if (\$a->[\$i-1] > \$a->[\$i]) { return 0; } } return 1; } /jortsort { /a exch def /ret true def 1 1 a length 1 sub { dup 1 sub a exch get exch a exch get gt { /ret false def exit } if } for ret } bind def `````` In Python I have `zip`, which lets me merge two iterables into one. ``````def jortsort(a): `````` So I make two separately-iterable copies of the list `````` j, k = tee(a) `````` step once through one of them `````` next(k,None) `````` then iterate them in parallel `````` for i in zip(j,k): if i[0] > i[1]: return False return True `````` And in Raku, Ruby and Rust, as well as `zip` as above I have explicit methods (with wildly different names) for taking overlapping N-sized chunks off a single iterable. (To be fair, Python gets `pairwise` to do the same thing in 3.10, at least for the case of 2-sized chunks, but I'm running 3.7 here on Debian/oldstable.) Probably one could wrap something like `any` round them to get rid of the explicit looping entirely, but eh. ``````sub jortsort(@a) { for @a.rotor(2 => -1) -> @i { if (@i[0] > @i[1]) { return False; } } return True; } def jortsort(a) a.each_cons(2) { |i| if i[0] > i[1] then return false end } return true end fn jortsort(a: Vec) -> bool { for i in a.windows(2) { if i[0] > i[1] { return false; } } true } `````` Task 2: Long Primes Write a script to generate first 5 Long Primes. A prime number (p) is called Long Prime if (1/p) has an infinite decimal expansion repeating every (p-1) digits. OK, there are clearly two steps here: generate primes, then check the decimal expansion. There's some disagreement in the more general case over whether 2 should count; it and 5 are the only two primes where (1/p) has a non-repeating expansion, but the example sets us right. So it's "full reptend primes" rather than "long-period primes". (By inspection, 3 and 5 won't qualify under either condition, and therefore I start at 7.) Step one, a reasonably efficient primality tester, since I haven't got round to writing one for these things before. Shown in Python, for compactness, working on the basis that after 2 and 3 every prime number has the form (6k+1) or (6k-1) (because 6k+(0, 2, 3 or 4) will always be divisible by 2 or 3). (Raku has a built-in probabilistic tester, but with numbers these small I want to be deterministic.) ``````def is_prime(n): if n>2 and n%2==0: return 0 if n>3 and n%3==0: return 0 lim=int(sqrt(n)) k6=0 while 1: k6+=6 for t in [k6-1,k6+1]: if t <= lim: if n % t == 0: return False else: return True `````` Then it's a matter of finding the length of the decimal expansion. I found some fascinating blog posts on ways of doing this, and ended up using the simple one, optimising it further because I'm always dealing with fractions of the form (1/prime). I use the 6k±1 technique again to find prime candidates, test them, then get the length of the decimal expansion using Brent's approach of repeated modulus calculation. Python again, taking a parameter for the number to produce because a mere 5 was boring. ``````def longprime(n): nn=n ba=[7] k6=6 while nn>0: if len(ba)==0: k6+=6 ba=[k6+1,k6-1] b=ba.pop() if is_prime(b): k=1 l=0 while 1: k*=10 l+=1 k %= b if k==1: break if l==b-1: o.append(b) nn-=1 return o `````` The other languages all work basically the same way. Full code on github. Comments on this post are now closed. If you have particular grounds for adding a late comment, comment on a more recent post quoting the URL of this one. 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