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A super-prime is a prime number whose index in the list of all primes is also prime. The sequence looks like this:
3, 5, 11, 17, 31, 41, 59, 67, 83, 109, 127, 157, 179, 191, ...
This is sequence A006450 in the OEIS.
Given a positive integer, determine whether it is a super-prime.
2: false 3: true 4: false 5: true 7: false 11: true 13: false 17: true 709: true 851: false 991: true
This is code-golf, so the shortest answer in each language wins.
ÆRÆNċ
ÆRÆNċ Main link. Argument: n
ÆR Prime range; yield the array of all primes up to n.
ÆN N-th prime; for each p in the result, yield the p-th prime.
ċ Count the occurrences of n.
Thanks to user202729 for saving 3 bytes.
PrimeQ/@(#&&PrimePi@#)&
This makes use of the fact that Mathematica leaves most nonsensical expressions unevaluated (in this case, the logical And of two numbers) and Map can be applied to any expression, not just lists. So we compute the And of the input and its prime index, which just remains like that, and then we Map the primality test over this expression which turns the two operands of the And into booleans, such that the And can then be evaluated.
ÆRi³ÆP
Uses the same technique as my Japt answer: Generate the primes up to n, get the index of n in that list, and check that for primality. If n itself is not prime, the index is 0, which is also not prime, so 0 is returned anyway.
õ fj bU Ä j
This is actually very straight-forward, unlike my original submission:
õ fj bU Ä j
Uõ fj bU +1 j Ungolfed
Implicit: U = input integer
Uõ Generate the range [1..U].
fj Take only the items that are prime.
bU Take the (0-indexed) index of U in this list (-1 if it doesn't exist).
+1 j Add 1 and check for primality.
This is true iff U is at a prime index in the infinite list of primes.
Implicit: output result of last expression
p=lambda m:(m>1)*all(m%x for x in range(2,m))
f=lambda n:p(n)*p(len([*filter(p,range(n+1))]))
Returns 0/1, at most 4 bytes longer if it has to be True/False.
f, you can reformat your code like this and exclude it from the byte count.
\$\endgroup\$
– musicman523
Jul 7 '17 at 21:57
ÆCÆPaÆP
ÆC counts the number of primes less than or equal to the input (so, if the input is the nth prime, it returns n). Then ÆP tests this index for primality. Finally, a does a logical AND between this result and ÆP (primality test) of the original input.
p x=2==sum[1|0<-mod x<$>[1..x]]
f x=p$sum[1|y<-[1..x],p y,p x]
Try it online! Usage: f 991 yields True.
n=input()
s={2};p=i=1
while i<n:
p*=i;i+=1
if p*p%i:s|={i}
print{n,len(s)}<=s
^(?3)(?=((?=(\2?+x*?((?!(xx+)\4+$)))xx)x)*(x*))\5(?3)xx
Try it online! - Perl
Try it online! - PCRE
Takes its input in unary, as a sequence of x characters whose length represents the number. Based on Prime counting function.
^ # tail = N = input value
(?3) # Assert N is not composite
(?=
# Calculate π(N) = the number of primes <= N, by counting
# from the largest to the smallest prime.
( # J = 0
(?=
# \2 starts at zero, and on each subsequent iteration, contains the difference
# N-P-(J-1) where P is the previously found prime, and J is the running total of
# our prime count.
(
\2?+ # Start from the previous value of \2, atomically so that it
# can't be backtracked and started again from zero if the
# following fails to match. This will make tail = P-1, where
# P is the previously found prime.
x*? # Advance as little as necessary to make the following match,
# and add this to \2, while subtracting it from tail.
((?!(xx+)\4+$)) # Define subroutine (?3): Assert tail is not composite.
# Note that this needs to be inside group \2 for it to work
# in PCRE1 and older versions of PCRE2, which atomicize
# groups that have nested backreferences.
)
xx # Assert tail is prime by eliminating the false positives 0, 1
)
x # J += 1; tail -= 1
)* # Iterate zero or more times, until there are no smaller primes remaining
# At this point, head = π(N), and tail = N - π(N)
(x*) # \5 = tail = N - head = tool to make tail = head
)
\5 # tail = π(N)
(?3)xx # Assert tail is prime
Alternative 55 bytes:
^(?3)(?=((?=(\2?+x*?((?!(xx+)\4+$|x?$))))x)*(x*))\5(?3)
Try it online! - Perl
Try it online! - PCRE
This eliminates the false negatives \$0\$ and \$1\$ in the definition of the primality subroutine instead of patching its second and third uses.
^(?!(xx+)\1+$)(?=(x*?(?!(xx+)\3+$)x)*x)(?>(?<-2>x)*)(?<!^x?|^\4+(x+x))
This uses the .NET feature of balanced groups to do the prime counting. Since .NET regex has no subroutines, this has three copies of the primality test.
^ # tail = N = input value
(?!(xx+)\1+$) # Assert N is not composite
# Calculate π(N) = the number of primes <= N, by counting
# from the largest to the smallest prime.
(?=
(
x*? # Advance as little as necessary to make the following match
(?!(xx+)\3+$) # Assert tail is not composite
x # Eliminate the false primality positive of 0, and advance forward
# so that the next prime can be found (if we didn't do this, the
# regex engine would exit the loop due to a zero-width match)
)* # Every time this loop matches an iteration, the capture group 1
# match is pushed onto the stack. This (balanced groups) is how we
# count the number of primes.
x # Eliminate the false primality positive of 1
)
(?>(?<-2>x)*) # Pop all group 2 captures off the stack, doing head += 1 for each one,
# atomically so it won't backtrack to force it to match the following:
(?<!^x?|^\4+(x+x)) # Assert head is prime
ÝØ<Øså
ÝØ<Øså
Ý # Push range from 0 to input
Ø # Push nth prime number (vectorized over the array)
< # Decrement each element by one (vectorized)
Ø # Push nth prime number again
s # swap top items of stack (gets input)
å # Is the input in the list?
ÅP can be used instead of ÝØ (not that it saves any bytes, but hey, it's a built-in that does the trick).
\$\endgroup\$
– Makonede
Mar 30 at 0:37
&P_QP_smP_dS
&P_QP_smP_dS
Implicit input
mP_dS Primality of all numbers from 1 to N
s Sum of terms (equal to number of primes ≤ N)
P_ Are both that number
&P_Q and N prime?
P=Prime;!P@P@Range@#~FreeQ~#&
-6 bytes from @MartinEnder
sI~p>@hs
s - is_prime(input)
I~p>@hs - if ^:
~p> - first_n_primes(input)
@ - ^.index(input)
h - ^+1
s - is_prime(^)
~µ:||\_x0]{p=p-µq|~q=a|_xµp]q=q+1
~ | IF .... THEN (do nothing)
: the number 'a' (read from cmd line)
µ | is Prime
\_x0 ELSE (non-primes) quit, printing 0
] END IF
{ DO
In this next bit, q is raised by 1 every loop, and tested for primality.
p keeps track of how may primes we've seen (but does so negatively)
µq| test q for primality (-1 if so, 0 if not)
p=p- and subtract that result from p (at the start of QBIC: q = 1, p = 0)
~q=a| IF q == a
_xµp QUIT, and print the prime-test over p (note that -3 is as prime as 3 is)
] END IF
q=q+1 Reaise q and run again.
f=filter
p x=2==(length$f(\a->mod(x)a==0)[1..x])
s=map(\(_,x)->x)$f(\(x,_)->p x)$zip[1..]$f(p)[2..]
r x=x`elem`(take x s)
(\(_,x)->x) is snd, (\(x,_)->p x) is (p.fst). Both fst and snd are in Prelude, so no need for imports.
\$\endgroup\$
– Laikoni
Jul 7 '17 at 23:52
r x=elem x$take x s. However, in this case you can go pointfree (introducing backticks again) and omit the function name: elem<*>(`take`s).
\$\endgroup\$
– nimi
Jul 8 '17 at 12:32
x=#(input@@)a=function{p=1;k=$1==2;f=2;while(f<$1)do{p=p*$1%f;k=1;f=f+1};return p*k}r=1;i=2;while(i<x)do{if(a@i)then{r=r+1}i=i+1}print@((a@x*a@r)>0)
Saved 2 bytes thanks to Tom Carpenter.
A very naive implementation using built-in functions:
isprime(x)&isprime(nzz(primes(x)))
(p=@isprime)(x)&p(nnz(primes(x)))
\$\endgroup\$
– Tom Carpenter
Jul 8 '17 at 17:28
def a(n):
r=[2];x=2
while x<n:x+=1;r+=[x]*all(x%i for i in r)
return{n,len(r)}<=set(r)
Constructs r, the list of primes <= n; if n is prime, then n is the len(r)'th prime. So n is a super prime iff n in r and len(r) in r.
return 1 if x is a super-prime, 0 otherwise.
without using an isprime-kind function.
x->a=[0,1];for i=3:x push!(a,0∉i%(2:i-1))end;a[sum(a)]&a[x]
ṗṗ
One of the cases when the behaviours of functions returning non booleans is really helpful.
user's solution.
ṗ£İp
⁰ due to implicit input, plus, putting it on a seperate line just declares it as a new function, which doesn't affect the first line
\$\endgroup\$
– Jo King♦
Oct 11 '20 at 8:56
P_hx.fP_Z
Uses a fairly different approach to the existing Pyth solution. Returns True for super-primes, False for non-super-primes, and [] (falsy) for composite numbers.
Tired of all the extremely obscure golfing languages in this thread? Here's the practical language solution you're looking for! Let me know if you see a way to shave off a few bytes.
require'prime';n=gets.to_i;p n.prime?&&(Prime.take(n).index(n)+1).prime?
math.primes, Saved 13 bytes thanks to @Bubbler! Saved 4 bytes thanks to @chunes!
[ dup nprimes index ?1+ prime? ]
[ dup nprimes index ?1+ prime? ]
dup nprimes ! First n primes (generates extra primes)
index ! Find the index of the input in that
! when the index is not found, an f is outputted
! If the index exists,
?1+ ! If it's a number, increment
! If it's f, output 0 (which is not prime)
prime? ! Is that prime?
[ dup primes-upto index ?1+ prime? ]. ?1+ increments an integer or consumes an f and gives 0.
\$\endgroup\$
– Bubbler
Mar 29 at 23:28
nprimes instead of primes-upto. It's less efficient of course, but small price to pay.
\$\endgroup\$
– chunes
Mar 30 at 0:46
