How many steps does it take from n to 1 by subtracting the greatest divisor?

Question

Inspired by this question over at Mathematics.

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The Problem

Let n be a natural number ≥ 2. Take the biggest divisor of n – which is different from n itself – and subtract it from n. Repeat until you get 1.

The Question

How many steps does it take to reach 1 for a given number n ≥ 2.

Detailed Example

Let n = 30.

The greatest divisor of:

1.   30 is 15  -->  30 - 15 = 15
2.   15 is  5  -->  15 -  5 = 10
3.   10 is  5  -->  10 -  5 =  5
4.    5 is  1  -->   5 -  1 =  4
5.    4 is  2  -->   4 -  2 =  2
6.    2 is  1  -->   2 -  1 =  1

It takes 6 steps to reach 1.

Input

• Input is an integer n, where n ≥ 2.
• Your program should support input up to the language's maximum integer value.

Output

• Simply output the number of steps, like 6.
• Leading/trailing whitespaces or newlines are fine.

Examples

f(5)        --> 3
f(30)       --> 6
f(31)       --> 7
f(32)       --> 5
f(100)      --> 8
f(200)      --> 9
f(2016^155) --> 2015

Requirements

• You can get input from STDIN, command line arguments, as function parameters or from the closest equivalent.
• You can write a program or a function. If it is an anonymous function, please include an example of how to invoke it.
• This is code-golf so shortest answer in bytes wins.
• Standard loopholes are disallowed.

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This series can be found on OEIS as well: A064097

A quasi-logarithm defined inductively by a(1) = 0 and a(p) = 1 + a(p-1) if p is prime and a(n*m) = a(n) + a(m) if m,n > 1.

Answer 1

Jelly, 9 bytes

ÆṪÐĿÆFL€S

Try it online! or verify all test cases.

Background

The definition of sequence A064097 implies that

By Euler's product formula

where φ denotes Euler's totient function and p varies only over prime numbers.

Combining both, we deduce the property

where ω denotes the number of distinct prime factors of n.

Applying the resulting formula k + 1 times, where k is large enough so that φ^k+1(n) = 1, we get

From this property, we obtain the formula

where the last equality holds because ω(1) = 0.

How it works

ÆṪÐĿÆFL€S  Main link. Argument: n

  ÐĿ       Repeatedly apply the link to the left until the results are no longer
           unique, and return the list of unique results.
ÆṪ           Apply Euler's totient function.
           Since φ(1) = 1, This computes φ-towers until 1 is reached.
    ÆF     Break each resulting integer into [prime, exponent] pairs.
      L€   Compute the length of each list.
           This counts the number of distinct prime factors.
        S  Add the results.

Answer 2

05AB1E, 13 11 bytes

Code:

[DÒ¦P-¼D#]¾

Explanation:

[        ]   # An infinite loop and...
       D#        break out of the loop when the value is equal to 1.
 D           # Duplicate top of the stack (or in the beginning: duplicate input).
  Ò          # Get the prime factors, in the form [2, 3, 5]
   ¦         # Remove the first prime factor (the smallest one), in order to get 
               the largest product.
    P        # Take the product, [3, 5] -> 15, [] -> 1.
     -       # Substract from the current value.
      ¼      # Add one to the counting variable.
          ¾  # Push the counting variable and implicitly print that value.

Uses CP-1252 encoding. Try it online!.

Answer 3

Pyth, 11 bytes

fq1=-Q/QhPQ

Test suite

A straightforward repeat-until-true loop.

Explanation:

fq1=-Q/QhPQ
               Implicit: Q = eval(input())
f              Apply the following function until it is truthy,
               incrementing T each time starting at 1:
         PQ    Take the prime factorization of Q
        h      Take its first element, the smallest factor of Q
      /Q       Divide Q by that, giving Q's largest factor
    -Q         Subtract the result from Q
   =           Assign Q to that value
 q1            Check if Q is now 1.

Answer 4

Python 2, 50 49 bytes

f=lambda n,k=1:2/n or n%(n-k)and f(n,k+1)or-~f(k)

This isn't going to finish that last test case any time soon...

Alternatively, here's a 48-byte which returns True instead of 1 for n=2:

f=lambda n,k=1:n<3or n%(n-k)and f(n,k+1)or-~f(k)

Answer 5

Jelly, 10 bytes

ÆfḊPạµÐĿi2

Try it online! or verify most test cases. The last test cases finishes quickly locally.

How it works

ÆfḊPạµÐĿi2  Main link. Argument: n (integer)

Æf          Factorize n, yielding a list of primes, [] for 1, or [0] for 0.
  Ḋ         Dequeue; remove the first (smallest) element.
   P        Take the product.
            This yields the largest proper divisor if n > 1, 1 if n < 2.
    ạ       Yield the abs. value of the difference of the divisor (or 1) and n.
     µ      Convert the chain to the left into a link.
      ÐĿ    Repeatedly execute the link until the results are no longer unique.
            Collect all intermediate results in a list.
            For each starting value of n, the last results are 2 -> 1 -> 0 (-> 1).
        i2  Compute the 1-based index of 2.

Answer 6

Pyth - 15 14 13 bytes

Special casing 1 is really killing me.

tl.u-N/Nh+PN2

Try it online here.

tl                One minus the length of
 .u               Cumulative fixed point operator implicitly on input
  -N              N -
   /N             N /
    h             Smallest prime factor
     +PN2         Prime factorization of lambda var, with two added to work with 1

Answer 7

Retina, 12

• 14 bytes saved thanks to @MartinBüttner

(1+)(?=\1+$)

This assumes input given in unary and output given in decimal. If this is not acceptable then we can do this for 6 bytes more:

Retina, 18

• 8 bytes saved thanks to @MartinBüttner

.+
$*
(1+)(?=\1+$)

Try it online - 1st line added to run all testcases in one go.

Sadly this uses unary for the calculations, so input of 2016¹⁵⁵ is not practical.

• The first stage (2 lines) simply converts the decimal input to unary as a string of 1s
• The second stage (1 line) calculates the largest factor of n using regex matching groups and lookbehinds to and effectively subtracts it from n. This regex will match as many times as is necessary to reduce the number as far as possible. The number of regex matches will be the number of steps, and is output by this stage.

Answer 8

JavaScript (ES6), *44 38

Edit 6 bytes saved thanks @l4m2

(* 4 striked is still 4)

Recursive function

f=(n,d=n)=>n>1?n%--d?f(n,d):f(n-d)+1:0

Less golfed

f=(n, d=n-1)=>{
  if (n>1)
    if(n % d != 0)
      return f(n, d-1) // same number, try a smaller divisor
    else
      return f(n-d)+1  // reduce number, increment step, repeat
  else
    return 0
}

Test

f=(n,d=n)=>n>1?n%--d?f(n,d):f(n-d)+1:0

console.log=x=>O.textContent+=x+'\n';

[5,30,31,32,100,200].forEach(x=>console.log(x+' -> '+f(x)))

<pre id=O></pre>

Answer 9

Mathematica, 36 bytes

f@1=0;f@n_:=f[n-Divisors[n][[-2]]]+1

An unnamed function takes the same bytes:

If[#<2,0,#0[#-Divisors[#][[-2]]]+1]&

This is a very straightforward implementation of the definition as a recursive function.

Answer 10

Octave, 59 58 55 bytes

function r=f(x)r=0;while(x-=x/factor(x)(1));r++;end;end

Updated thanks to Stewie Griffin, saving 1 byte

Further update, saving three more bytes by using result of factorization in while-check.

Sample runs:

octave:41> f(5)
ans =  3
octave:42> f(30)
ans =  6
octave:43> f(31)
ans =  7
octave:44> f(32)
ans =  5
octave:45> f(100)
ans =  8
octave:46> f(200)
ans =  9

Answer 11

Julia, 56 50 45 39 bytes

f(n)=n>1&&f(n-n÷first(factor(n))[1])+1

This is a recursive function that accepts an integer and returns an integer.

Ungolfed:

function f(n)
    if n < 2
        # No decrementing necessary
        return 0
    else
        # As Dennis showed in his Jelly answer, we don't need to
        # divide by the smallest prime factor; any prime factor
        # will do. Since `factor` returns a `Dict` which isn't
        # sorted, `first` doesn't always get the smallest, and
        # that's okay.
        return f(n - n ÷ first(factor(n))[1]) + 1
    end
end

Try it online! (includes all test cases)

Saved 6 bytes thanks to Martin Büttner and 11 thanks to Dennis!

Answer 12

Haskell, 59 bytes

f 1=0;f n=1+(f$n-(last$filter(\x->n`mod`x==0)[1..n`div`2]))

Usage:

Prelude> f 30
Prelude> 6

It may be a little inefficient for big numbers because of generating the list.

Answer 13

Japt, 12 bytes (non-competing)

@!(UµUk Å×}a

Test it online! Non-competing because it uses a bunch of features that were added way after the challenge was posted.

How it works

@   !(Uµ Uk Å  ×   }a
XYZ{!(U-=Uk s1 r*1 }a
                       // Implicit: U = input integer
XYZ{               }a  // Return the smallest non-negative integer X which returns
                       // a truthy value when run through this function:
         Uk            //   Take the prime factorization of U.
            s1         //   Slice off the first item.
                       //   Now we have all but the smallest prime factor of U.
               r*1     //   Reduce the result by multiplication, starting at 1.
                       //   This takes the product of the array, which is the
                       //   largest divisor of U.
      U-=              //   Subtract the result from U.
    !(                 //   Return !U (which is basically U == 0).
                       //   Since we started at 0, U == 1 after 1 less iteration than
                       //   the desired result. U == 0 works because the smallest
                       //   divisor of 1 is 1, so the next term after 1 is 0.
                       // Implicit: output result of last expression

This technique was inspired by the 05AB1E answer. A previous version used ²¤ (push a 2, slice off the first two items) in place of Å because it's one byte shorter than s1  (note trailing space); I only realized after the fact that because this appends a 2 to the end of the array and slices from the beginning, it in fact fails on any odd composite number, though it works on all given test cases.

Answer 14

PowerShell v2+, 81 bytes

param($a)for(;$a-gt1){for($i=$a-1;$i-gt0;$i--){if(!($a%$i)){$j++;$a-=$i;$i=0}}}$j

Brutest of brute force.

Takes input $a, enters a for loop until $a is less than or equal to 1. Each loop we go through another for loop that counts down from $a until we find a divisor (!($a%$i). At worst, we'll find $i=1 as a divisor. When we do, increment our counter $j, subtract our divisor $a-=$i and set $i=0 to break out of the inner loop. Eventually, we'll reach a condition where the outer loop is false (i.e., $a has reached 1), so output $j and exit.

Caution: This will take a long time on larger numbers, especially primes. Input of 100,000,000 takes ~35 seconds on my Core i5 laptop. Edit -- just tested with [int]::MaxValue (2^32-1), and it took ~27 minutes. Not too bad, I suppose.

Answer 15

Matlab, 58 bytes

function p=l(a),p=0;if(a-1),p=1+l(a-a/min(factor(a)));end

Answer 16

Clojure, 116 104 bytes

(fn[n](loop[m n t 1](let[s(- m(last(filter #(=(rem m %)0)(range 1 m))))](if(< s 2)t(recur s (inc t))))))

-12 bytes by filtering a range to find multiples, then using last one to get the greatest one

Naïve solution that basically just solves the problem as it's described by the OP. Unfortunately, finding the greatest divisor alone takes up like half the bytes used. At least I should have lots of room to golf from here.

Pregolfed and test:

(defn great-divider [n]
  ; Filter a range to find multiples, then take the last one to get the largest
  (last
     (filter #(= (rem n %) 0)
             (range 1 n))))

(defn sub-great-divide [n]
  (loop [m n
         step 1]
    (let [g-d (great-divider m) ; Find greatest divisor of m
          diff (- m g-d)] ; Find the difference
      (println m " is " g-d " --> " m " - " g-d " = " diff)
      (if (< diff 2)
        step
        (recur diff (inc step))))))

(sub-great-divide 30)

30  is  15  -->  30  -  15  =  15
15  is  5  -->  15  -  5  =  10
10  is  5  -->  10  -  5  =  5
5  is  1  -->  5  -  1  =  4
4  is  2  -->  4  -  2  =  2
2  is  1  -->  2  -  1  =  1
6

Answer 17

Perl 6, 35 bytes

{+({$_ -first $_%%*,[R,] ^$_}...1)}

Try it online!

How it works

{                                 }   # A bare block lambda.
                    [R,] ^$_          # Construct range from arg minus 1, down to 0.
        first $_%%*,                  # Get first element that is a divisor of the arg.
    $_ -                              # Subtract it from the arg.
   {                        }...1     # Do this iteratively, until 1 is reached.
 +(                              )    # Return the number of values generated this way.

Answer 18

Python 3, 75, 70, 67 bytes.

g=lambda x,y=0:y*(x<2)or[g(x-z,y+1)for z in range(1,x)if x%z<1][-1]

This is a pretty straight forward recursive solution. It takes a VERY long time for the high number test cases.

Answer 19

><>, 32 bytes

<\?=2:-$@:$/:
1-$:@@:@%?!\
;/ln

Expects the input number, n, on the stack.

This program builds the complete sequence on the stack. As the only number that can lead to 1 is 2, building the sequence stops when 2 is reached. This also causes the size of the stack to equal the number of steps, rather than the number of steps +1.

Answer 20

C99, 62 61 bytes

1 byte golfed off by @Alchymist.

f(a,c,b)long*c,a,b;{for(*c=0,b=a;a^1;a%--b||(++*c,b=a-=b));}  

Call as f(x,&y), where x is the input and y is the output.

Answer 21

Haskell, 67 bytes

Here's the code:

a&b|b<2=0|a==b=1+2&(b-1)|mod b a<1=1+2&(b-div b a)|1<2=(a+1)&b
(2&)

And here's one reason why Haskell is awesome:

f = (2&)

(-->) :: Eq a => a -> a -> Bool
(-->) = (==)

h=[f(5)        --> 3
  ,f(30)       --> 6
  ,f(31)       --> 7
  ,f(32)       --> 5
  ,f(100)      --> 8
  ,f(200)      --> 9
  ,f(2016^155) --> 2015
  ]

Yes, in Haskell you can define --> to be equivalent to ==.

Answer 22

Matlab, 107 bytes

a=input('');b=factor(a-isprime(a));c=log2(a);while(max(b)>1),b=max(factor(max(b)-1));c=c+1;end,disp(fix(c))

• Non-competing, this is not the iterative translation of my last submission, just another direct algerbraic method, it sums up all binary-logs of all prime factors, kinda ambiguous to illustrate.
• I will golf this more when i have time.

Answer 23

Ruby, 43 bytes

f=->x{x<2?0:1+f[(1..x).find{|i|x%(x-i)<1}]}

Find the smallest number i such that x divides x-i and recurse until we reach 1.

Answer 24

MATL, 17 16 bytes

`tttYfl)/-tq]vnq

Try it Online

Explanation

        % Implicitly grab input
`       % Do while loop
    ttt % Make three copies of top stack element
    Yf  % Compute all prime factors
    l)  % Grab the smallest one
    /   % Divide by this to get the biggest divisor
    -   % Subtract the biggest divisor
    t   % Duplicate the result
    q   % Subtract one (causes loop to terminate when the value is 1). This
        % is functionally equivalent to doing 1> (since the input will always be positive) 
        % with fewer bytes
]       % End do...while loop
v       % Vertically concatenate stack contents (consumes entire stack)
n       % Determine length of the result
q       % Subtract 1 from the length
        % Implicitly display result

Answer 25

Julia, 39 36 bytes

\(n,k=2)=n%k>0?n>1&&n\-~k:\(n-n/k)+1

Try it online!

Answer 26

Clojure, 98 96 bytes

#(loop[n % i -1](if n(recur(first(for[j(range(dec n)0 -1):when(=(mod n j)0)](- n j)))(inc i))i))

uses for :when to find the largest divisor, loops until no such value larger than one is found.

Answer 27

Stax, 9 7 bytes

Åiw^Å♫┌

Run and debug it

These two operations are identical.

1. Take the biggest divisor of n – which is different from n itself – and subtract it from n.
2. Let d be the smallest prime divisor of n. Calculate n * (d - 1) / d.

I like #2 better, so let's count how many times we can do that one instead. So at each step, we decrement the smallest prime factor. Let's work an example starting with 30.

Step    n   Factors             Operation
0       30  [2, 3, 5]           Replace 2 with 1
1       15  [1, 3, 5] = [3, 5]  Replace 3 with 2
2       10  [2, 5]              Replace 2 with 1
3       5   [1, 5] = [5]        Replace 5 with 4
4       4   [4] = [2, 2]        Replace 2 with 1
5       2   [1, 2] = [2]        Replace 2 with 1
6       1   [1] = []            End of the line

From the example we can see a recursive definition for steps(x) that solves the problem. Presented here in python-ish pseudo-code.

def steps(n):
    pf = primeFactors(n)
    return sum(steps(d - 1) for d in pf) + 1

The stax program, unpacked into ascii looks like this.

Z       push a zero under the input
|fF     for each prime factor of the input, run the rest of the program
  v     decrement the prime factor
  G     run the *whole* program from the beginning
  +^    add to the running total, then increment

Run this one

Answer 28

x86, 24 23 bytes

Straightforward implementation of problem. Input in ecx, output in edi.

I wonder if there's a way to test if n > 1 in less than 3 bytes...

-1 by using cdq to zero edx. This requires the input to be a positive signed number (<=2^31-1), which I think is reasonable.

.section .text
.globl main
main:
        mov     $100, %ecx      # n = 100

start:
        xor     %edi, %edi      # result = 0
loop1:
        mov     %ecx, %ebx      # d = n
loop2: 
        dec     %ebx            # --d
        mov     %ecx, %eax      
        cdq     
        div     %ebx            # n % d 
        test    %edx, %edx
        jnz     loop2           # do while (n % d) 

        inc     %edi            # ++result
        sub     %ebx, %ecx      # n -= d 
        cmp     $1, %ecx
        ja      loop1           # do while (n > 1)

        ret

Hexdump:

00000039  31 ff 89 cb 4b 89 c8 99  f7 f3 85 d2 75 f6 47 29  |1...K.......u.G)|
00000049  d9 83 f9 01 77 ec c3                              |....w..|

Answer 29

Pyth, 17 16 bytes

L?tbhy-b*F+1tPb0

Try it online! (The y.v at the end is for function calling)

---

Original 17 bytes:

L?tb+1y-b*F+1tPb0

Try it online! (The y.v at the end is for function calling)

(I actually answered that question with this Pyth program.)

Answer 30

JavaScript (ES6), 70 54 bytes

f=(n,i=2)=>n<i?0:n%i?f(n,i+1):n>i?f(i)+f(n/i):1+f(n-1)

Implementation of the provided recursive formula, but now updated to use recursion to find the divisor too.

Answer 31

Java (JDK 10), 60 bytes

n->{int c=0,d;for(;n>1;c++,n-=d)for(d=n;n%--d>0;);return c;}

Try it online!

Answer 32

Pyke, 11 bytes (noncompeting)

D3Phf-oRr;o

This uses a new behaviour where if there is an exception raised after a goto, it restores the state from before the goto (except variable definitions) and continues. In this case it is equivalent to the following python code:

# Implicit input and variable setup
inp = input()
o = 0
# End implicit
try:
    while 1:
        inp -= factors(inp)[0] # If factors is called on the value 1, it returns an empty
                               # list which when the first element tries to be accessed
                               # raises an exception
        o += 1 # Using `o` returns the current value of `o` and increments it
except:
    print o # This in effect gets the number of times the loop went

This is all possible using Pyke without a while loop construction - yay goto!

Try it here!

Answer 33

Perl, 57 + 1 (-p flag) = 58 bytes

$n=$_;$n-=$n/(grep!($n%$_),2..$n/2,$n)[0],$\++while$n>1}{

Usage:

> echo 31 | perl -pe '$n=$_;$n-=$n/(grep!($n%$_),2..$n/2,$n)[0],$\++while$n>1}{'

Ungolfed:

while (<>) {
# code above added by -p
    # $_ has input value
    # $\ has undef (or 0)
    my $n = $_;
    while ($n > 1) {
        my $d = 1;
        for (2 .. ($n / 2)) {
            if ($n % $_ == 0) {
                $d = $n / $_;
                last;
            }
        }
        $n -= $d;
        $\++;
    }
} {
# code below added by -p
    print;  # prints $_ (undef here) and $\
}