Finding the power sandwich

Question

Introduction

Finding the closest power to a number is a common enough problem. But what if you need both the next-highest and next-lowest power? In this challenge you must find the closest powers to a given number - the 'power sandwich' if you will, where the given number is the filling and the powers are the bread. Mmm, tasty.

Challenge

Given a power P >0 and a number N >0, output the largest integer x^P that is smaller or equal to N, and the smallest integer y^P that is greater or equal to N.

Input should be taken as a list of two positive (>0) integers, first the power P and then the number N. Output should be a list of two integers, the first being smaller or equal to N, the second being greater or equal to N, and both being a power of P.

If N is a power of P already, the output should be the list [N, N].

This is code-golf, so the shortest code (as measured in bytes) wins.

Example Input and Output

Input:

[2, 17]

Output:

[16, 25]

Explanation: 16 is the biggest square number (power of 2) less than or equal to 17, and 25 is the smallest square number greater or equal to 17.

Test cases

[2, 24] -> [16, 25]
[2, 50] -> [49, 64]
[3, 8] -> [8, 8]
[1, 25] -> [25, 25]
[3, 25] -> [8, 27]
[4, 4097] -> [4096, 6561]
[2, 10081] -> [10000, 10201]
[11, 2814661] -> [177147, 4194304]
[6, 1679616] -> [1000000, 1771561]

Answer 1

JavaScript (ES7), 39 bytes

Expects (p)(n).

p=>n=>[x=(q=n**(1/p)|0)**p,(q+=x<n)**p]

Try it online!

Alternate version (same size)

p=>n=>[0,1].map(i=>(n--**(1/p)+i|0)**p)

Try it online!

Answer 2

R, 36 35 33 bytes

\(P,N)abs(c(r<-N^(1/P),-r)%/%1)^P

Attempt This Online!

Answer 3

Vyxal, 7 bytes

Ėe₍⌊⌈$e

Try it Online!

Trivial implementation.

Ėe₍⌊⌈$e­⁡​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁣‏⁠‎⁡⁠⁤‏⁠‎⁡⁠⁢⁡‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏‏​⁡⁠⁡‌­
Ėe       # ‎⁡Exponentiate N with the reciprocal of P
  ₍⌊⌈    # ‎⁢Pair results of floor, ceil in a list
     $e  # ‎⁣Swap (returns first input) and exponentiate with P
💎

Answer 4

Python, 46 bytes

based on port of Arnauld's JavaScript answer

lambda p,n:[int(q:=n**(1/p))**p,(-(-q//1))**p]

Attempt This Online!

floor and ceil of p-th root to the power of p

Answer 5

BQN, 10 bytes

(⌊⋈⌈)∘√´⋆⊑

Try it here.

Explanation:

(⌊⋈⌈)∘√´⋆⊑ # tacit function which takes a list [P,N]
 ⌊⋈⌈       # take the floor paired with the ceil
(   )∘√´   # of the P-th root of N
        ⋆⊑ # and raise to the power P

Answer 6

Desmos, 51 bytes

f(l)=[floor(l[2]^{1/l})^l[1],ceil(l[2]^{1/l})^l[1]]

Try it on Desmos! Expects list [p,n] and returns list [x^p,y^p] as per specifications.

Explanation:

f(l)=[                      ,                     ]    Given a list, return a list
      floor(l[2]^{1/l})^l[1]                           floor(n^1/p)^p
                             ceil(l[2]^{1/l})^l[1]     ceil(n^1/p)^p

Answer 7

Jelly, 8 bytes

*İ}Ḟ,ĊƊ*

Try it online!

Profoundly boring direct floor-ceil solution. However, if the N-is-a-power-of-P case can be relaxed to permit outputting just [N]...

Jelly, 7 bytes

*İ}¹ị*€

Try it online!

...we actually have a builtin for floor-ceil. Namely, the index-into dyad ị.

*İ}        Pth root of N.
   ¹       (Break undesired initial 2,2,2 chain.)
    ị      Index that into
     *€    the list of x^P for all x <- 1..N.

Answer 8

Desmos, 37 bytes

f(p,n)=[floor(k),ceil(k)]^p
k=n^{1/p}

Try It On Desmos!

Try It On Desmos! - Prettified

The best I could think of for now.

Answer 9

05AB1E, 8 bytes

zmDî‚ï¹m

Port of @AidenChow's Desmos answer.

Inputs in the order \$P,N\$.

Try it online or verify all test cases.

Explanation:

$$[N_a,N_b]=\left[\left\lceil N^\frac{1}{P}\right\rceil,\left\lfloor N^\frac{1}{P}\right\rfloor\right]$$

z         # Push 1/P, where P is the first (implicit) input-integer
 m        # Take N to the power this 1/P, where N is the second (implicit) input-integer
  D       # Duplicate it
   î      # Ceil the copy
    ‚     # Pair the two together
     ï    # Floor/cast both to integers
      ¹m  # Take both values in the pair to the power P
          # (after which the resulting pair is output implicitly)

Answer 10

Charcoal, 29 bytes

ＮθＮη≔¹ζＷ‹Ｘζθη≦⊕ζＩＸ⟦⁻ζ›Ｘζθηζ⟧θ

Try it online! Link is to verbose version of code. Explanation:

ＮθＮη

Input P and N.

≔¹ζ

Start with y=1.

Ｗ‹Ｘζθη≦⊕ζ

Increment y until it is large enough.

ＩＸ⟦⁻ζ›Ｘζθηζ⟧θ

Output the Pth powers of x and y.

61 bytes for a more efficient version:

ＮθＮη≔¹ζ≔¹εＷ‹Ｘεθη≦⊗εＷ›⁻εζ¹«≔÷⁺εζ²ιＦ¬‹Ｘιθη≔ιεＦ¬›Ｘιθη≔ιζ»ＩＸ⟦ζε⟧θ

Try it online! Link is to verbose version of code. Explanation:

ＮθＮη

Input P and N.

≔¹ζ≔¹ε

Start with both x and y equal to 1.

Ｗ‹Ｘεθη≦⊗ε

Double y until it is large enough.

Ｗ›⁻εζ¹«≔÷⁺εζ²ιＦ¬‹Ｘιθη≔ιεＦ¬›Ｘιθη≔ιζ»

Perform a binary search to narrow down the range of x and y.

ＩＸ⟦ζε⟧θ

Output the Pth powers of x and y.

19 bytes using floating-point arithmetic:

Ｎθ≔ＸＩη∕¹θηＩＸ⟦⌊η⌈η⟧θ

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input P.

≔ＸＩη∕¹θη

Take the Pth root of N.

ＩＸ⟦⌊η⌈η⟧θ

Output the Pth powers of the floor and ceiling.

Answer 11

GolfScript, 25 24 23 bytes

{0\{@@).2$?.-1$<}do@;@}

Try it online!

I’m pretty sure this method is optimal, considering GolfScript doesn’t have floats and thus no min/max. It’s possible that there’s a shorter way to do the stack manipulation, but that seems pretty unlikely to me at this point. I stand corrected (by myself)—I was able to get rid of the variable i. Now I think this is optimal…but then I’ll probably find another 1-byte save somewhere else 😂

This is a block (function) taking N then P on the stack and leaving the two values on top of the stack.

Explanation:

Code	Stack (bottom to top)
{…} Block	2 17
0\	2 0 17
{…}do Run until result is 0…	Showing first pass over:
@@ Stack shift	17 2 0
) Increment	17 2 1
: Duplicate	17 2 1 1
2$ Push third from stack	17 2 1 1 2
? Exponentiate	17 2 1 2
: Duplicate	17 2 1 2 2
-1$ Push bottom of stack	17 2 1 2 2 17
< Less than?	-1 2 1 2 0
(pop and either repeat or stop)	17 2 1 2

At the end of the loop, the stack looks like 17 1 4 9 16 2 5 25 so we do @;@ to make the top of the stack be 25 16.

It took like twenty minutes of fiddling around to get the stack manipulation to work right.

Answer 12

C#, 79 bytes

(p,n)=>(Math.Pow(Math.Pow(n,1d/p)is{}x?(int)x:x,p),Math.Pow(Math.Ceiling(x),p))

Alternate approach that uses 87 bytes

(p,n)=>{var(i,x)=(0,0d);for(;(x=Math.Pow(++i,p))<n;);return(Math.Pow(x==n?i:~-i,p),x);}

Try it online!

Answer 13

APL+ WIN, 18 bytes

Prompts for power and integer:

((⌊p),⌈p←⎕*÷n)*n←⎕

Try it online! Thanks to Dyalog Classic

Answer 14

MATL, 14 bytes

w1Y\^tkwXkh1G^

Try on MATL Online or Test all cases

w    % Get both inputs and bring P to the top of the stack
1Y\  % Take the inverse of P
^    % Raise N to that power i.e. take N's P-th root
tk   % Take a copy of that root and floor it
wXk  % And ceil the original copy
h1G^ % Raise both those values to the power of P

Answer 15

Vyxal 3, 7 bytes

ė*∦⌊⌈$*

Try it Online!

Takes two integers as its argument and returns two integers. Works the same as various other answers including @mathscat’s Vyxal one and @UnrelatedString’s Jelly one.

Explanation

ė*∦⌊⌈$*
ė       | Reciprocal of n
 *      | To the power of p
  ∦⌊⌈   | Pair ceiling with floor and wrap in a list
     $* | To the power of p

```

Answer 16

Rust, 67 66 bytes

|p,n:f64|{let x=n.powf(1./p);(x.floor().powf(p),x.ceil().powf(p))}

Try it online!

Trivial implementation: takes n to the power of 1/p, and then calculates floor and ceil to the power of p.

Takes the integer inputs as 64-bit floats and returns its output as a tuple of 64-bit floats.

-1 byte thanks to ceilingcat