Can the number reach 1 by repeatedly subtracting the largest prime less than it?

Question

Challenge:

Given a number, take the largest prime strictly less than it, subtract it from this number, do this again to this new number with the biggest prime less than it, and continue doing this until it's less than 3. If it reaches 1, your program should output a truthy value, else, the program should output a falsey value.

Examples:

All of these should give a truthy value:

3
4
6
8
10
11
12
14
16
17
18
20
22
23
24
26
27
29
30
32
34
35
37
38
40
41
42
44
46
47
48
50

All of these should give falsey values:

5
7
9
13
15
19
21
25
28
31
33
36
39
43
45
49

Rules:

• You can either write a program or function.
• You can assume that the input is bigger than 2.
• Standard loopholes apply
• This is code-golf so the shortest answer wins!

Answer 1

Jelly, 9 8 bytes

’ÆRṪạµ¡Ḃ

Try it online! or verify all test cases.

How it works

’ÆRṪạµ¡Ḃ  Main link. Argument: n

     µ    Combine all atoms to the left into a chain.
’           Decrement; yield n - 1.
 ÆR         Prime range; yield all primes in [2, ..., n -1].
   Ṫ        Tail; yield p, the last prime in the range.
            If the range is empty, this yields p = 0.
    ạ       Compute the absolute difference of p and n.
      ¡   Call the chain to the left n times.
          This suffices since each iteration decreases n, until one of the fixed
          points (1 or 2) is reached.
       Ḃ  Bit; return the parity of the fixed point.

Answer 2

Retina, 31 bytes

.+
$*
+`1(?!(11+)\1+$)11+
1
^1$

Prints 0 (falsy) or 1 (truthy).

Try it online! (The first line enables a linefeed-separated test suite.)

Explanation

.+
$*

Convert input to unary by turning input N into N copies of 1.

+`1(?!(11+)\1+$)11+
1

Repeatedly remove the largest prime less than the input. This is based on the standard primality test with regex.

^1$

Check whether the result is a single 1.

Answer 3

Pyth, 18 15 14 bytes

Thanks to @Maltysen for -1 byte

#=-QefP_TUQ)q1

A program that takes input on STDIN and prints True or False as appropriate.

Try it online

How it works

#=-QefP_TUQ)q1  Program. Input: Q
#          )    Loop until error statement (which occurs when Q<3):
         UQ      Yield [0, 1, 2, 3, ..., Q-1]
     fP_T        Filter that by primality
    e            Yield the last element of that
 =-Q             Q = Q - that
            q1  Q is 1 (implicit variable fill)
                Implicitly print

Old version with reduce, 18 bytes

qu-G*<HGH_fP_TSQQ1

Try it online

How it works

qu-G*<HGH_fP_TSQQ1  Program. Input: Q
              SQ    Yield [1, 2, 3, ..., Q]
          fP_T      Filter that by primality
         _          Reverse it
 u                  Reduce it:
                Q    with base case Q and
                     function G, H -> 
     <HG              H<G
    *   H             *H (yields H if H<G, else 0)
  -G                  Subtract that from G
q                1  The result of that is 1
                    Implicitly print

Answer 4

JavaScript (ES6), 64 63 bytes

Saved 1 byte thanks to @Neil

g=(x,n=x-1)=>n<2?x:x%n?g(x,n-1):g(x-1)
f=x=>x<3?x%2:f(x-g(x-1))

I wrote this in 2 minutes... and it worked perfectly the first time. First user to find the inevitable bug wins....

Try it out

g=(x,n=x-1)=>n<2?x:x%n?g(x,n-1):g(x-1)
f=x=>x<3?x%2:f(x-g(x-1))

<input type="number" value=3 min=3 onchange="A.innerHTML=f(this.value)"><br>
<p id=A>1</p>

How it works

First we define g(x) as the function that finds the first prime number p <= x. This is done using the following process:

1. Start with n = x-1.
2. If n < 2, x is prime; return x.
3. If x is divisible by n, decrement x and go to step 1.
4. Otherwise, decrement n and go to step 2.

The solution to this challenge, f(x), is now fairly straightforward:

1. If x < 3, return x = 1.
2. Otherwise, subtract g(x-1) and try again.

Answer 5

Pyke, 15 11 bytes

WDU#_P)e-Dt

Try it here!

            - stack = input
W           - while continue:
  U#_P)     -     filter(is_prime, range(stack))
       e    -    ^[-1]
 D      -   -   stack-^
         Dt -  continue = ^ != 1

Returns 1 if true and raises an exception if false

Answer 6

Julia, 32 bytes

While it's not going to be the shortest solution among the languages, this might be the shortest of the human-readable ones...

!n=n>2?!(n-primes(n-1)[end]):n<2

Or, to put it in slightly clearer terms

function !(n)
  if n>2
    m=primes(n-1)[end]   # Gets largest prime less than n
    return !(n-m)        # Recurses
  else
    return n<2           # Gives true if n is 1 and false if n is 2
  end
end

Called with, for example, !37.

Answer 7

Mathematica, 32 bytes

2>(#//.x_/;x>2:>x+NextPrime@-x)&

This is an unnamed function which takes an integer and returns a boolean.

Explanation

There's a lot of syntax and funny reading order here, so...

   #                               This is simply the argument of the function.
    //.                            This is the 'ReplaceRepeated' operator, which applies
                                   a substitution until the its left-hand argument stops
                                   changing.
       x_/;x>2                     The substitution pattern. Matches any expression x as
                                   long as that expression is greater than 2.
              :>                   Replace that with...
                  NextPrime@-x     Mathematica has a NextPrime built-in but no
                                   PreviousPrime built-in. Conveniently, NextPrime
                                   works with negative inputs and then gives you the 
                                   next "negative prime" which is basically a
                                   PreviousPrime function (just with an added minus sign).
                x+                 This gets added to x, which subtracts the previous
                                   prime from it.
2>(                           )    Finally, we check whether the result is less than 2.

Answer 8

Python3, 102 92 90 89 88 bytes

f=lambda n:n<2if n<3else f(n-[x for x in range(2,n)if all(x%y for y in range(2,x))][-1])

Golfing suggestions welcome! I see that gmpy contains a function next_prime, but I can't test it yet :(

-2 bytes, thanks to @JonathanAllan!

-1 byte, thanks to @Aaron!

Testcases

f=lambda n:n<2if n<3else f(n-[x for x in range(2,n)if all(x%y for y in range(2,x))][-1])

s="3 4 6 8 10 11 12 14 16 17 18 20 22"
h="5 7 9 13 15 19 21 25 28 31 33 36 39"

for j in s.split(" "):print(f(int(j)))
for j in h.split(" "):print(f(int(j)))

Output is 13 truthy values and 13 falsey values. s contains the truthy cases and h the falseys.

Answer 9

Python, with sympy, 60 bytes

import sympy
f=lambda n:n>2and f(n-sympy.prevprime(n))or n<2

---

My previous method was 83 bytes without sympy using recursion, but I took truthy/falsey to mean distinguishable and consistent, but I have been informed that's an incorrect interpretation. I can't seem to salvage it due to the tail, but I'll leave it here in case someone knows how to do so:

f=lambda n,p=0:n>2and(any(p%x==0for x in range(2,p))and f(n,p-1)or f(n-p,n+~p))or n

Answer 10

Vitsy, 28 26 bytes

This can definitely be shortened.

<]xN0)l1)-1[)/3D-];(pD-1[D

‮

<                    Traverse the code in this direction, rotating on the line.
                     For the sake of reading the code easier, I'm reversing the
                     code on this line. This will be the order executed.

 D[1-Dp(;]-D3/)[1-)1l)0Nx]
 D                         Duplicate the top member of the stack.
  [      ]                 Do the stuff in brackets until break is called.
   1-                      Subtract 1 from the top item of the stack.
     D                     Duplicate the top member of the stack.
      p(                   If the top member is a prime...
        ;                  break;
          -                Pop a, b, push a - b.
           D3/)[         ] If this value is less than 3, do the bracketed code.
                1-         Subtract the top item of the stack by 1.
                  )        If the top item is zero...
                   1       Push 1.
                    l)     If the length of the stack is zero...
                      0    Push 0.
                       N   Output the top member of the stack.
                        x  System.exit(0);

Try it online!

Answer 11

CJam, 21 16 bytes

Thanks to Dennis for saving 4 bytes.

ri{_1|{mp},W=-}h

Try it online!

Explanation

ri       e# Read input and convert to integer N.
{        e# Run this block as long as N is positive (or until the program aborts
         e# with an error)...
  _1|    e#   Duplicate and OR 1. This rounds up to an odd number. For N > 2, this
         e#   will never affect the greatest prime less than N.
  {mp},  e#   Get all primes from 0 to (N|1)-1.
         e#   For N > 2, this will contain all primes less than N.
         e#   For N = 2, this will contain only 2.
         e#   For N = 1, this will be empty.
  W=     e#   Select the last element (largest prime up to (N|1)-1).
         e#   For N = 1, this will result in an error and terminate the program, which
         e#   still prints the stack contents though (which are 1, the desired output).
  -      e#   Subtract from N. Note that this gives us 0 for N = 2, which terminates the 
         e#   loop.
}h

Answer 12

MATL, 13 bytes

`tqZq0)-t2>}o

Try it online! Or verify all test cases at once.

Explanation

`        % Do...while
  t      %   Duplicate. Takes input implicitly in the first iteration
  qZq    %   All primes less than that
  0)     %   Get last one
  -      %   Subtract (this result will be used in the next iteration, if any)
  t      %   Duplicate
  2>     %   Does it exceed 2? If so: next iteration. Else: execute the "finally" 
         %   block and exit do...while loop
}        % Finally
  o      %   Parity. Transforms 2 into 0 and 1 into 1
         % End do...while implicitly
         % Display implicitly

Answer 13

Perl, 42 bytes

Includes +1 for -p

Run with input on STDIN

reach1.pl:

#!/usr/bin/perl -p
$_=1x$_;$_=$`while/\B(?!(11+)\1+$|$)|11$/

Uses the classic primality regex

Answer 14

.NET Regex, 38 bytes

Just to show that it can be checked in a single regex.

^(?>(?<=(.*))..+(?<!^\1\2+(.+.)|$))+.$

Input is assumed to be in unary.

Explanation

It simply implements the requirement to the word, repeatedly removing the biggest prime and check whether there remainder is 1.

• (?>(?<=(.*))..+(?<!^\1\2+(.+.)|$))+: Non-backtracking group makes sure the biggest prime we found is not overriden, and + simply repeat the process of matching the biggest prime.

  • (?<=(.*))..+(?<!^\1\2+(.+.)|$): Match the biggest prime less than the remaining number

    • (?<=(.*)): Record how much we have subtracted to establish an "anchor" point for assertion.

    • ..+: Look for the biggest number...

    • (?<!^\1\2+(.+.)|$): ... which is prime and less than the remaining number.
      • (?<!^\1\2+(.+.)): The usual prime test routine, with ^\1 tacked in front to make sure we are checking the amount matched by ..+
      • (?!<$): Assert less than the remaining number

Answer 15

Clojure, 125 bytes

#(loop[x %](if(> x 2)(recur(- x(loop[y(dec x)](if(some zero?(vec(for[z(range 2 y)](mod y z))))(recur(dec y))y))))(quot 1 x)))

Yikes, that is one long piece of code. The most verbose language strikes again!

Ungolfed:

(defn subprime [n]
  (loop [x n]
    (if (> x 2)
      (recur
        (- x
          (loop [y (dec x)]
            (if (some zero? (vec (for [z (range 2 y)] (mod y z))))
              (recur (dec y)) y))))
      (quot 1 x))))

Answer 16

Floroid, 45 30 29 bytes

f=Bb:b<2Fb<3Gf(b-en(b-1)[-1])

Answer 17

Perl 6,  54 53 52  51 bytes

{($_,{$_-($_-1...2).first: *.is-prime}...3>*)[*-1]==1}
{($_,{$_-($_-1...2).first: *.is-prime}...3>*).any==1}
{any($_,{$_-($_-1...2).first: *.is-prime}...3>*)==1}

{any($_,{$_-(^$_).grep(*.is-prime)[*-1]}...3>*)==1}

Explanation:

# bare block lambda with implicit parameter ｢$_｣
# used to generate all of the rest of the elements of the sequence
{
  # create an any Junction of the following list
  any(
    $_, # initialize sequence with the inner block's argument

    # bare block lambda with implicit parameter ｢$_｣
    {
      # take this inner block's argument and subtract
      $_ -

      ( ^$_ )            # Range up-to and excluding ｢$_｣
      .grep(*.is-prime)\ # find the primes
      [ * - 1 ]          # return the last value
    }

    ...   # keep doing that until

    3 > * # the result is less than 3

  # test that Junction against ｢1｣
  # ( returns an ｢any｣ Junction like ｢any(False, False, True)｣ )
  ) == 1
}

Example:

# show what is returned and if it is truthy
sub show ($_) {
  # ｢.&{…}｣ uses the block as a method and implicitly against ｢$_｣
  my $value = .&{any($_,{$_-(^$_).grep(*.is-prime)[*-1]}...3>*)==1}
  say join "\t", $_, ?$value, $value.gist;
}

show 3;  # 3    True    any(False, True)
show 4;  # 4    True    any(False, True)
show 5;  # 5    False   any(False, False)
show 10; # 10   True    any(False, False, True)
show 28; # 28   False   any(False, False, False)
show 49; # 49   False   any(False, False)
show 50; # 50   True    any(False, False, True)

Answer 18

Irregular, 63 bytes

p~?1_$-1p:;
n=i(0)?1_$-1p:;
_~
N=n
1(?!(11+)\1+$)11+~1
^11$~0
N

I created this language two days ago, and the basic premises are that there are no built in loops, the only features are basic arithmetic and decision making, and program evaluation is based on regular expressions.

Explanation

p~?1_$-1p:;
n=i(0)?1_$-1p:;
_~
N=n

This part converts the input into unary. It repeatedly subtracts 1 from the input until it equals 0, prepending 1_ each time. It then removes all of the _s. If I hadn't forgotten a break in my code it could be written as so:

p~?1_$-1p:;
_~
n=i(0)?1_$-1p:;

The next part repeatedly removes the largest prime from the input until it is equal to 1 or 11, with 11 being replaced with 0.

1(?!(11+)\1+$)11+~1
^11$~0
N

I used the regex from Martin Ender's answer.

Answer 19

Haskell, 79 bytes

Not really short but pointfree :)

(<2).until(<3)(until(flip(`until`(+1))2.(.)(<1).mod>>=(==))pred.pred>>=flip(-))

Answer 20

PowerShell v2+, 81 bytes

param($n)while($n-gt2){$n-=(($n-1)..2|?{'1'*$_-match'^(?!(..+)\1+$)..'})[0]}!--$n

Takes input $n. Enters a while loop so long as $n is still 3 or greater. Each iteration, subtracts a number from $n. The number is the results of the regex primality test applied against a range ($n-1)..2 via the Where-Object (?) operator, then the first [0] of the results (since the range is decreasing, this results in the largest one being selected). After concluding the loop, $n is either going to be 1 or 2, by definition, so we pre-decrement $n (turning it into either 0 or 1), and take the Boolean-not ! thereof. That's left on the pipeline and output is implicit.

Examples

PS C:\Tools\Scripts\golfing> 3..20|%{"$_ --> "+(.\can-the-number-reach-one.ps1 $_)}
3 --> True
4 --> True
5 --> False
6 --> True
7 --> False
8 --> True
9 --> False
10 --> True
11 --> True
12 --> True
13 --> False
14 --> True
15 --> False
16 --> True
17 --> True
18 --> True
19 --> False
20 --> True

Answer 21

Matlab, 51 bytes

v=@(x)x-max(primes(x-1));while(x>=3)x=v(x);end;x==1

This is VERY similar to the JS6 solution by ETHProductions, but needs the variable to be in the workspace.

Answer 22

Python 2.7: 88 87 Bytes

r=lambda n:n>2and r(n-[a for a in range(2,n)if all(a%b for b in range(2,a))][-1])or n<2

Thx @TuukkaX for -1 more byte!