How far away is n to the next power of b?

Question

Let n and b be positive integers larger than 1.

Output the distance from n to the next power of b.

For n=5 and b=3, the next power of 3 from 5 is 9 (3^2 = 9), so the output is 9 - 5 = 4.

For n=8 and b=2, the next power of 2 from 8 is 16 (2^4 = 16), so the output is 16 - 8 = 8. Note that n is a power of 2 in this example.

Testcases:

  n b output
212 2 44
563 5 62
491 5 134
424 3 305
469 8 43
343 7 2058
592 7 1809
289 5 336
694 3 35
324 5 301
  2 5 3

This is code-golf. Shortest answer in bytes wins. Standard loopholes apply.

Answer 1

Jelly,  4  3 bytes

ạæċ

A dyadic link taking n on the left and b on the right and returning the result.

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How?

ạæċ - Link: number n, number b | n,b ∈ ℕ
 æċ - ceiling n to the next power of b
ạ   - absolute difference between n and that

Answer 2

x86-64 Assembly (Windows x64 Calling Convention), 14 13 bytes

An inefficient (but svelte!) iterative approach (with credit to @Neil for inspiration):

               HowFarAway PROC
6A 01             push   1
58                pop    rax         ; temp = 1
               Loop:
0F AF C2          imul   eax, edx    ; temp *= b
39 C8             cmp    eax, ecx
72 F9             jb     Loop        ; keep looping (jump) if temp < n
29 C8             sub    eax, ecx    ; temp -= n
C3                ret                ; return temp
               HowFarAway ENDP

The above function takes two integer parameters, n (passed in the ECX register) and b (passed in the EDX register), and returns a single integer result (in the EAX register). To call it from C, you would use the following prototype:

unsigned HowFarAway(unsigned n, unsigned b);

This is limited to the range of a 32-bit integer. It can be easily modified to support 64-bit integers by using the full long registers, but it would cost more bytes to encode those instructions. :-)

Answer 3

Dyalog APL, 10 bytes

2 bytes saved thanks to @ZacharyT

⊢-⍨⊣*1+∘⌊⍟

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Takes n as right argument and b as left argument.

Calculates b⌊logbn + 1⌋ - n.

Answer 4

C (gcc), 39 35 bytes

New undefined behavior thanks to Erik

f(n,b,i){for(i=b;b<=n;b*=i);n=b-n;}

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Answer 5

R, 38 34 bytes

pryr::f({a=b^(0:n)-n;min(a[a>0])})

Anonymous function. Stores all values of b to the power of everything in the range [0,n], subtracts n from each, subsets on positive values, and returns the min.

TIO has a non-pryr version, called as f(n,b); this version needs to be called as f(b,n).

Saved 4 bytes thanks to Jarko Dubbeldam, who then outgolfed me.

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Answer 6

Cubix, 24 20 bytes

-4 bytes thanks to MickyT

Pwp.I|-.;)^0@O?|uq;<

Reads in input like n,b

Fits on a 2x2x2 cube:

    P w
    p .
I | - . ; ) ^ 0
@ O ? | u q ; <
    . .
    . .

Explanation:

I|I0 : read the input, push 0 (counter) to the stack

^w puts the IP to the right place for the loop:

• Pp- : compute b^(counter), move n to top of stack, compute b^(counter) - n
• ? : turn left if negative, straight if 0, right if positive
  • Positive: O@ : output top of stack (distance) and exit.
  • Negative : |? : proceed as if the top of the stack were zero
• <;qu;) : point the IP in the right direction, pop the top of the stack (negative/zero number), move n to the bottom of the stack, u-turn, pop the top of the stack (b^(counter)) and increment the counter
• IP is at ^w and the program continues.

Watch it online!

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Answer 7

Haskell, 20 bytes

n%b=until(>n)(*b)1-n

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until saves the day

Answer 8

05AB1E, 9 8 bytes

sLmʒ‹}α¬

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Explanation

s         # swap order of the inputs
 L        # range [1 ... n]
  m       # raise b to each power
   ʒ‹}    # filter, keep only the elements greater than n
      α   # calculate absolute difference with n for each
       ¬  # get the first (smallest)

Answer 9

MATL, 10 9 bytes

yy:YAn^w-

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Explanation

Consider inputs 694 and 3 as an example.

y    % Implicitly take two inputs. Duplicate from below
     % STACK: 694, 3, 694
y    % Duplicate from below
     % STACK: 694, 3, 694, 3
:    % Range
     % STACK: 694, 3, 694, [1 2 3]
YA   % Base conversion (of 694 with "digits" given by [1 2 3]
     % STACK: 694, 3, [3 3 2 3 1 2]
n    % Number of elements
     % STACK: 694, 3, 6
^    % Power
     % 694, 729
w    % Swap
     % STACK: 729, 694
-    % Subtract. Implicitly display
^    % 35

Answer 10

C, 42 40 bytes

Thanks to commenter @Steadybox for the tip

o;p(n,b){for(o=b;n>=b;)b*=o;return b-n;}

Answer 11

Mathematica, 24 bytes

#2^⌊1/#~Log~#2⌋#2-#&

thanks Martin

I/O

[343, 7]

2058

Answer 12

Java (OpenJDK 8), 42 bytes

Based on @GovindParmar's C answer.

n->b->{for(int o=b;n>=b;b*=o);return b-n;}

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Answer 13

JavaScript (ES6), 29 bytes

Very similar to Rick's approach but posted with his permission (and some help saving a byte).

n=>b=>g=(x=b)=>x>n?x-n:g(x*b)

---

Try it

f=
n=>b=>g=(x=b)=>x>n?x-n:g(x*b)
oninput=_=>o.value=f(+i.value)(+j.value)()
o.value=f(i.value=324)(j.value=5)()

*{font-family:sans-serif;}
input{margin:0 5px 0 0;width:50px;}

<label for=i>n: </label><input id=i type=number><label for=j>b: </label><input id=j type=number><label for=o>= </label><input id=o>

Answer 14

R, 30 bytes

pryr::f(b^floor(log(n,b)+1)-n)

Evaluates to the function

function (b, n) 
b^floor(log(n, b) + 1) - n

Which takes the first power greater or equal than n, and then substracts n from that value.

Changed ceiling(power) to floor(power+1) to ensure that if n is a power of b, we take the next power.

Answer 15

JavaScript (ES6), 31 bytes

f=(n,b,i=b)=>b>n?b-n:f(n,b*i,i)

Test cases:

f=(n,b,i=b)=>b>n?b-n:f(n,b*i,i)

console.log(f(212, 2)) // 44
console.log(f(563, 5)) // 62
console.log(f(491, 5)) // 134
console.log(f(424, 3)) // 305
console.log(f(469, 8)) // 43
console.log(f(343, 7)) // 2058
console.log(f(592, 7)) // 1809
console.log(f(289, 5)) // 336
console.log(f(694, 3)) // 35
console.log(f(324, 5)) // 301

Answer 16

Python,  42  41 bytes

f=lambda a,b,v=1:(a<v)*(v-a)or f(a,b,v*b)

A recursive function which, starting with v=1, repeatedly multiplies by b until it strictly exceeds a and then returns the difference.

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Note: The result will never be zero so a>=v and f(a,b,v*b)or v-a may be replaced with (a<v)*(v-a)or f(a,b,v*b) without causing recursion errors.

---

Python 3, 37 bytes?

Using an idea of rici's...

f=lambda n,b:(n<b)*(b-n)or b*f(n/b,b)

which uses floating point arithmetic (hence results may stray from their true distance),
try that here.

Answer 17

Octave, 32 bytes

@(n,b)b^(fix(log(n)/log(b))+1)-n

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Answer 18

Octave, 26 bytes

@(n,b)b^sum(b.^(0:n)<=n)-n

Verify all test cases!

Answer 19

Ruby, 38 bytes

Two different approaches:

->(n,b){p b**(Math.log(n,b).to_i+1)-n}

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->(n,b){p b**(0..n).find{|x|b**x>n}-n}

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Answer 20

Japt, 9 bytes

_q}a@nVpX

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Explanation

_  q}a@  nVpX
Z{Zq}aX{UnVpX}  // Ungolfed
                // Implicit: U, V = input integers
     aX{     }  // For each integer X in [0...1e9), take
          VpX   //   V to the power of X
        Un      //   minus U,
Z{  }           // and return the first one Z where
  Zq            //   Math.sqrt(Z) is truthy.
                //   Math.sqrt returns NaN for negative inputs, and 0 is falsy, so this is
                //   truthy iff Z is positive. Therefore, this returns the first positive
                //   value of V**X - U.
                // Implicit: output result of last expression

Answer 21

Haskell, 31 bytes

f n b=[b^x-n|x<-[1..],b^x>n]!!0

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Answer 22

Perl 6,  31 30  29 bytes

->\n,\b{(b,b* *...*>n).tail -n}

Test it (31)

->\n,\b{b**(log(n,b).Int+1)-n}

Test it (30)

{$^b**(log($^a,$b).Int+1)-$a}

Test it (29)

{($^b,$b* *...*>$^a).tail-$a}

Test it (29)

Answer 23

PARI/GP, 26 24 bytes

f(n,b)=b^logint(b*n,b)-n

Answer 24

Brachylog, 7 bytes

^ʰ↙X>₁-

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Takes input as a list [b, n].

    >₁     n is strictly less than
^ ↙X       some power of
 ʰ         b,
      -    and their difference is
           the output.

Answer 25

PHP>=7.1, 46 bytes

for([,$n,$b]=$argv;0>=$d=$b**$i++-$n;);echo$d;

PHP Sandbox Online

Answer 26

Python 3, 50 48 bytes

^{Thanks to EriktheOutgolfer for saving 2 bytes!}

lambda n,b:b**-~int(math.log(n,b))-n
import math

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Python doesn't have any fancy log or ceiling builtins, so I just went with the obvious approach with a little bit of golfing flair.

Answer 27

Lua, 74 73 Byte

A straight forward solution, I'm using 10 bytes to ensure that the arguments are treated as numbers, and not strings. Outputs to STDIN.

Edit: forgot to remove the space in w=1 n=n+0, saves one byte

n,b=...b=b+0p=b w=1n=n+0while p<=n do p=math.pow(b,w)w=w+1 end print(p-n)

Explained

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n,b=...           -- unpack the argument into the variable n and b
b=b+0             -- set b's type to number
n=n+0             -- set n's type to number
p=b               -- set a variable to track the current value of the powered b
w=1               -- set the nth power
while p<=n        -- iterate untill the current power is greater or equals to n
do
  p=math.pow(b,w) -- raise b to the power w
  w=w+1           -- increment w
end
print(p-n)        -- outputs p minus the following power of b

Answer 28

Python 2, 48 41 bytes

• @Rod's loop simplification saved 7 bytes!

Full program without recursion or bit twiddling:

i=1;n,b=input()
while n>=i:i*=b
print i-n

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Input format: n, b.

Answer 29

QBIC, 23 bytes

{p=:^q~p>:|_xp-b|\q=q+1

Takes parameter b first, then n.

Explanation

{       DO
p=:^q   SET p to input b (read as 'a' by QBIC fromt he cmd line) ** q (starts as 1)
~p>:    IF p exceeds 'n' (read as 'b' by QBIC fromt he cmd line)
|_xp-b| THEN QUIT, printing p minus b
\q=q+1  ELSE increase q, re-run

Answer 30

Common Lisp, 73 bytes

(defun f(n b)(setq a b)(loop(when(> b n)(return(- b n)))(setq b(* b a))))

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Answer 31

Japt, 16 bytes

This feels longer than it needs to be!

pÂ(MlV /MlU Ä)-V

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Answer 32

J, 17 bytes

[-~]^[:>.]^.[:>:[

Usage

   f =: [-~]^[:>.]^.[:>:[
   424 f 3
305

Explanation

Performs pow(b, ceil(log(n+1))) where log has base b.

Answer 33

dc, 23 bytes

snsb1[lb*dln!<F]dsFxln-

I had hoped to be able to use Z to count digits in a given radix, but it seems to count only in decimal. And we might want a base larger than dc supports. So this takes the obvious approach of successively multiplying by b until a value larger than n is reached, at which point we subtract n and leave the result at top of stack.

Annoyingly, we need !< rather than > comparison because the condition excludes an answer of 0.

Demo

for i in '212 2' '563 5' '491 5' '424 3' '469 8' '343 7'\
         '592 7' '289 5' '694 3' '324 5' '2 5' \
         '12345678901234567890 1000000'
do printf '%s => ' "$i"; dc -e '?rsnsb1[lb*dln!<F]dsFxln-p' <<<"$i"
done

212 2 => 44
563 5 => 62
491 5 => 134
424 3 => 305
469 8 => 43
343 7 => 2058
592 7 => 1809
289 5 => 336
694 3 => 35
324 5 => 301
2 5 => 3
12345678901234567890 1000000 => 999987654321098765432110

Answer 34

AWK, 34+2 bytes

{for(n=$2;n<=$1;n*=$2){}$0=n-$1}1

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Requires the -M option for arbitrary precision to handle the 12345678901234567890 1000000 => 999987654321098765432110 case.

The non-loop version requires the same number of bytes:

$0=$2^(int(log($1)/log($2)+1))-$1

Answer 35

TXR Lisp, 51 bytes

(defun f(n b)(-[(ginterate(op >= n)(op * b)1)-1]n))

Legend: ginterate, op.

Answer 36

APL(NARS), 13 chars, 26 bytes

{-⍺-⍵*1+⌊⍵⍟⍺}

test:

  f←{-⍺-⍵*1+⌊⍵⍟⍺}
  212 f 2
44
  563 f 5 
62
  592 f 7
1809

Answer 37

JavaScript (ES6), 49 bytes

f=(n,b)=>b**(m=Math).floor(m.log(n)/m.log(b)+1)-n

Answer 38

Forth (gforth), 45 bytes

: f >r 1 begin i * 2dup < until rdrop - abs ;

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Code Explanation

: f          \ start a new word definition
  >r         \ store b on the return stack for quick access
  1          \ create a counter to store powers of b
  begin      \ start an indefinite loop
    i *      \ multiply the counter by b
    2 dup <  \ duplicate the counter and n. Check if n is less than the counter
  until      \ if it is, end the loop
  rdrop      \ remove b from the return stack
  - abs      \ subtract and get absolute value (avoid negative)
;            \ end the word definition