Henry, Ha ha ! of course you are correct.
Serves me right for tackling an extra problem at 3am, while waiting for a another software compilation to finish. In the light of the next day, it's now obvious that any value between two consecutive primes cannot, by definition, be prime. There is an old adage about carpentry: "measure twice, cut once". The corollary for writing to the forum ought to be: "think twice, write once". — Sent from Mailbox <https://bit.ly/SZvoJe> for iPhone On Sun, May 12, 2013 at 4:51 AM, Henry Rich <[email protected]> wrote: > I don't quite get the problem statement. > > r =. (p+q)%2 is between p and q; so if p and q are CONSECUTIVE primes, r > cannot be prime. > > Henry Rich > > On 5/12/2013 6:07 AM, Alan Stebbens wrote: > > ProgrammingPraxis (at http://programmingpraxis.com/2013/05/10/mindcipher > ) > > offered a problem asking, given p, q as two consecutive pairs of primes, > if > > (p+2)%2 could be prime. > > > > Since both p & q (> 2) are prime, their sum is an even number and not > > prime, but could the half of their sum be a prime? > > > > I'm not much of a mathematician, but I figured I could brute-force an > > approximation with J. > > > > The gist below is my experiment showing that the answer is no, for the > > consecutive pairs of primes in the set of the first million primes. While > > it might be possible for the larger primes, I'm thinking not - just by > > induction. > > > > Probably some of you could show a proof, but it was more fun for me to > > cobble up this in J, and also demonstrate J to the non-J audience at > > Programming Praxis (which has been mostly scheme, Haskell, python, ruby). > > > > https://gist.github.com/aks/5563008 > > > > Here's my experiment: > > > > i. 20 > > 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 > > > > NB. generate the first 20 primes > > p: i. 20 > > 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 > > > > NB. box up consecutive pairs of those primes > > (2 <\ ]) p: i. 20 > > > ┌───┬───┬───┬────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐ > > │2 3│3 5│5 7│7 11│11 13│13 17│17 19│19 23│23 29│29 31│31 37│37 41│41 > 43│43 > > 47│47 53│53 59│59 61│61 67│67 71│ > > > └───┴───┴───┴────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘ > > > > NB. sum up each pair of primes > > +/ each (2 <\ ])p: i. 20 > > ┌─┬─┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬───┬───┬───┬───┬───┐ > > │5│8│12│18│24│30│36│42│52│60│68│78│84│90│100│112│120│128│138│ > > └─┴─┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴───┴───┴───┴───┴───┘ > > > > NB. divide each sum by 2 > > 2 %~ each +/ each (2 <\ ])p: i. > > 20 > > > > ┌───┬─┬─┬─┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┐ > > │2.5│4│6│9│12│15│18│21│26│30│34│39│42│45│50│56│60│64│69│ > > └───┴─┴─┴─┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┘ > > > > NB. now, test each of those results for being prime. 1 p: y -- tests y > > for being prime > > > > 1&p: each 2 %~ each +/ each (2 <\ ])p: i. > > 20 > > > > ┌─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┐ > > │0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│ > > └─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┘ > > > > NB. open the boxed results, so we can add them up > > >1&p: each 2 %~ each +/ each (2 <\ ])p: i. 20 > > 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 > > > > NB. sum/reduce the vector of booleans. If there's a prime, the sum will > > be > 0 > > +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 20 > > 0 > > > > NB. ok. No primes. Let's keep checking for larger groups > > > > +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 1000 > > 0 > > +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 10000 > > 0 > > +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 100000 > > 0 > > > > NB. the previous output took a few seconds. The next will take a few > > minutes > > > > +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 1000000 > > 0 > > ---------------------------------------------------------------------- > > For information about J forums see http://www.jsoftware.com/forums.htm > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
