Do we share the prime cluster?

Question

The prime cluster of an integer N higher than 2 is defined as the pair formed by the highest prime strictly lower than N and the lowest prime strictly higher than N.

Note that following the definition above, if the integer is a prime itself, then its prime cluster is the pair of the primes preceding and succeeding it.

Task

Given two integers integers N, M (N, M ≥ 3), output a truthy / falsy value based on whether N and M have the same prime cluster.

This is code-golf, so the aim is to reduce your byte count as much as possible. Thus, the shortest code in every programming language wins.

Test cases / Examples

For instance, the prime cluster of 9 is [7, 11], because:

• 7 is the highest prime strictly lower than 9, and
• 11 is the lowest prime strictly higher than 9.

Similarly, the the prime cluster of 67 is [61, 71] (note that 67 is a prime).

Truthy pairs

8, 10
20, 22
65, 65
73, 73
86, 84
326, 318
513, 518

Falsy pairs

4, 5
6, 8
409, 401
348, 347
419, 418
311, 313
326, 305

Answer 1

Jelly, 6 4 3 5 4 bytes

rÆPE

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How it works

rÆPE    Main link. Arguments: N, M
r       Yield the range of integers between N and M, inclusive.
 ÆP     For each integer, yield 1 if it is prime, 0 otherwise.
   E    Yield 1 if all items are equal (none in the range were prime,
        or there's only one item).

Works because two numbers have different prime clusters iff there is a prime between them, or either number is itself prime; unless both numbers are the same, in which case E returns 1 anyway (all items in a single-item array are equal).

Answer 2

Python 3, 103 95 91 bytes

lambda*z:len({*z})<2or[1for i in range(min(z),max(z)+1)if all(i%k for k in range(2,i))]<[0]

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Answer 3

Perl 6, 52 bytes

{[eqv] @_».&{(($_...0),$_..*)».first(*.is-prime)}}

Test it

Expanded:

{  # bare block lambda with implicit slurpy input ｢@_｣

  [eqv]               # see if each sub list is equivalent

    @_».&{            # for each value in the input

      (

        ( $_ ... 0 ), # decreasing Seq
          $_ ..  *    # Range

      )».first(*.is-prime) # find the first prime from both the Seq and Range

    }
}

Answer 4

Ruby, ₅₇ 54 bytes

->n,m{[*n..m,*m..n].all?{|x|?1*x=~/^(11+)\1+$/}||n==m}

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Uses the horrible regex primality test from my answer (which I had forgotten about until I clicked on it) to the related question Is this number a prime?. Since we have N, M ≥ 3, the check for 1 can be removed from the pattern, making the byte count less than using the built-in.

Note: The regex primality test is pathologically, hilariously inefficient. I believe it's at least O(n!), though I don't have time to figure it right now. It took twelve seconds for it to check 100,001, and was grinding for five or ten minutes on 1,000,001 before I canceled it. Use/abuse at your own risk.

Answer 5

Retina, 58 bytes

\b(.+)¶\1\b

.+
$*
O`
+`\b(1+)¶11\1
$1¶1$&
A`^(11+)\1+$
^$

Try it online! Explanation:

\b(.+)¶\1\b

If both inputs are the same, simply delete everything, and fall through to output 1 at the end.

.+
$*

Convert to unary.

O`

Sort into order.

+`\b(1+)¶11\1
$1¶1$&

Expand to a range of all the numbers.

A`^(11+)\1+$

Delete all composite numbers.

^$

If there are no numbers left, output 1, otherwise 0.

Answer 6

C (gcc), 153 146 bytes

i,B;n(j){for(B=i=2;i<j;)B*=j%i++>0;return!B;}
#define g(l,m,o)for(l=o;n(--l););for(m=o;n(++m););
a;b;c;d;h(e,f){g(a,b,e)g(c,d,f)return!(a-c|b-d);}

-7 from Jonathan Frech

Defines a function h which takes in two ints and returns 1 for truthy and 0 for falsey

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n is a function that returns 1 if its argument is not prime.

g is a macro that sets its first and second arguments to the next prime less than and greater than (respectively) it's third argument

h does g for both inputs and checks whether the outputs are the same.

Answer 7

JavaScript (ES6), 57 56 bytes

Takes input in currying syntax (a)(b). Returns 0 or 1.

a=>b=>a==b|!(g=k=>a%--k?g(k):k<2||a-b&&g(a+=a<b||-1))(a)

Test cases

let f =

a=>b=>a==b|!(g=k=>a%--k?g(k):k<2||a-b&&g(a+=a<b||-1))(a)

console.log('Truthy')
console.log(f(8)(10))
console.log(f(20)(22))
console.log(f(65)(65))
console.log(f(73)(73))
console.log(f(86)(84))
console.log(f(326)(318))
console.log(f(513)(518))

console.log('Falsy')
console.log(f(4)(5))
console.log(f(6)(8))
console.log(f(409)(401))
console.log(f(348)(347))
console.log(f(419)(418))
console.log(f(311)(313))

How?

a => b =>                 // given a and b
  a == b |                // if a equals b, force success right away
  !(g = k =>              // g = recursive function taking k
    a % --k ?             //   decrement k; if k doesn't divide a:
      g(k)                //     recursive calls until it does
    :                     //   else:
      k < 2 ||            //     if k = 1: a is prime -> return true (failure)
      a - b &&            //     if a equals b: neither the original input integers nor
                          //     any integer between them are prime -> return 0 (success)
      g(a += a < b || -1) //     else: recursive call with a moving towards b
  )(a)                    // initial call to g()

Answer 8

R, 63 46 bytes

-17 by Giuseppe

function(a,b)!sd(range(numbers::isPrime(a:b)))

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Pretty simple application of ETHProductions' Jelly solution. Main interesting takeaway is was that with R boolean vectors any(x)==all(x) is equivalent to min(x)==max(x).

Answer 9

PARI/GP, 28 bytes

v->s=Set(v);#s<2||!primes(s)

Try it online with all test cases!

Returns 0 or 1 (usual PARI/GP "Boolean" values).

Explanation:

v must be a vector (or a column vector, or a list) with the two numbers N and M as coordinates. For example [8, 10]. Then s will be the "set" made from these numbers, which is either a one-coordinate vector (if N==M), or a two-coordinate vector with sorted entries otherwise.

Then if the number #s of coordinates in s is just one, we get 1 (truthy). Otherwise, primes will return a vector of all primes in the closed interval from s[1] to s[2]. Negation ! of that will give 1 if the vector is empty, while negation of a vector of one or more non-zero entries (here one or more primes) will give 0.

Answer 10

Jelly, 6 bytes

ÆpżÆnE

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-2 thanks to Dennis.

Answer 11

APL (Dyalog Unicode), 18+16 = 34 24 bytes

⎕CY'dfns'
∧/=/4 ¯4∘.pco⎕

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Thanks to Adám for 10 bytes.

The line ⎕CY'dfns' (COPY) is needed to import the dfns (dynamic functions) collection, included with default Dyalog APL installs.

How it works:

∧/=/4 ¯4∘.pco⎕ ⍝ Main function. This is a tradfn body.
             ⎕ ⍝ The 'quad' takes the input (in this case, 2 integers separated by a comma.
          pco  ⍝ The 'p-colon' function, based on p: in J. Used to work with primes.
    4 ¯4∘.     ⍝ Applies 4pco (first prime greater than) and ¯4pco (first prime smaller than) to each argument.
  =/           ⍝ Compares the two items on each row
∧/             ⍝ Applies the logical AND between the results.
               ⍝ This yields 1 iff the prime clusters are equal.

Answer 12

Python 2, 87 86 bytes

lambda*v:v[0]==v[1]or{1}-{all(v%i for i in range(2,v))for v in range(min(v),max(v)+1)}

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Answer 13

C (gcc), 103 bytes 100 bytes

i,j,p,s;f(m,n){s=1;for(i=m>n?i=n,n=m,m=i:m;i<=n;i++,p?s=m==n:0)for(p=j=2;j<i;)p=i%j++?p:0;return s;}

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Answer 14

Mathematica, 39 27 26 bytes

Equal@@#~NextPrime~{-1,1}&

Expanded:

                         &  # pure function, takes 2-member list as input
       #~NextPrime~{-1,1}   # infix version of NextPrime[#,{-1,1}], which
                            # finds the upper and lower bounds of each
                              argument's prime clusters
Equal@@                     # are those bounds pairs equal?

Usage:

Equal@@#~NextPrime~{-1,1}& [{8, 10}]
(*  True  *)

Equal@@#~NextPrime~{-1,1}& [{6, 8}]
(*  False  *)

Equal@@#~NextPrime~{-1,1}& /@ {{8, 10}, {20, 22}, {65, 65}, 
    {73, 73}, {86, 84}, {326, 318}, {513, 518}}
(*  {True, True, True, True, True, True, True}  *)

Equal@@#~NextPrime~{-1,1}& /@ {{4, 5}, {6, 8}, {409, 401}, 
    {348, 347}, {419, 418}, {311, 313}}
(*  {False, False, False, False, False, False}  *)

Contributions: -12 bytes by Jenny_mathy, -1 byte by Martin Ender

Answer 15

Haskell, 81 bytes

A straightforward solution:

p z=[x|x<-z,all((0/=).mod x)[2..x-1]]!!0
c x=(p[x-1,x-2..],p[x+1..])
x!y=c x==c y

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Answer 16

J, 15 bytes

-:&(_4&p:,4&p:)

How it works:

   &(           ) - applies the verb in the brackets to both arguments
            4&p:  - The smallest prime larger than y
      _4&p:       - The largest prime smaller than y
           ,      - append
 -:               - matches the pairs of the primes

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