# Square Root Distance from Integers

Given a decimal number k, find the smallest integer n such that the square root of n is within k of an integer. However, the distance should be nonzero - n cannot be a perfect square.

Given k, a decimal number or a fraction (whichever is easier for you), such that 0 < k < 1, output the smallest positive integer n such that the difference between the square root of n and the closest integer to the square root of n is less than or equal to k but nonzero.

If i is the closest integer to the square root of n, you are looking for the first n where 0 < |i - sqrt(n)| <= k.

# Rules

• You cannot use a language's insufficient implementation of non-integer numbers to trivialize the problem.
• Otherwise, you can assume that k will not cause problems with, for example, floating point rounding.

# Test Cases

.9         > 2
.5         > 2
.4         > 3
.3         > 3
.25        > 5
.2         > 8
.1         > 26
.05        > 101
.03        > 288
.01        > 2501
.005       > 10001
.003       > 27888
.001       > 250001
.0005      > 1000001
.0003      > 2778888
.0001      > 25000001
.0314159   > 255
.00314159  > 25599
.000314159 > 2534463


Comma separated test case inputs:

0.9, 0.5, 0.4, 0.3, 0.25, 0.2, 0.1, 0.05, 0.03, 0.01, 0.005, 0.003, 0.001, 0.0005, 0.0003, 0.0001, 0.0314159, 0.00314159, 0.000314159


This is , so shortest answer in bytes wins.

# Wolfram Language (Mathematica), 34 bytes

Min[⌈.5/#+{-#,#}/2⌉^2+{1,-1}]&


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### Explanation

The result must be of the form $$\m^2 \pm 1\$$ for some $$\m \in \mathbb{N}\$$. Solving the inequations $$\\sqrt{m^2+1} - m \le k\$$ and $$\m - \sqrt{m^2-1} \le k\$$, we get $$\m \ge \frac{1-k^2}{2k}\$$ and $$\m \ge \frac{1+k^2}{2k}\$$ respectively. So the result is $$\\operatorname{min}\left({\left\lceil \frac{1-k^2}{2k} \right\rceil}^2+1, {\left\lceil \frac{1+k^2}{2k} \right\rceil}^2-1\right)\$$.

# Python, 42 bytes

lambda k:((k-1/k)//2)**2+1-2*(k<1/k%2<2-k)


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Based on alephalpha's formula, explicitly checking if we're in the $$\m^2-1\$$ or $$\m^2+1\$$ case via the condition k<1/k%2<2-k.

Python 3.8 can save a byte with an inline assignment.

Python 3.8, 41 bytes

lambda k:((a:=k-1/k)//2)**2-1+2*(a/2%1<k)


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These beat my recursive solution:

50 bytes

f=lambda k,x=1:k>.5-abs(x**.5%1-.5)>0 or-~f(k,x+1)


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# 05AB1E, 16 bytes

nD(‚>I·/înTS·<-ß


Port of @alephalpha's Mathematica answer, with inspiration from @Sok's Pyth answer, so make sure to upvote both of them!

Explanation:

n                 # Take the square of the (implicit) input
#  i.e. 0.05 → 0.0025
D(‚              # Pair it with its negative
#  i.e. 0.0025 → [0.0025,-0.0025]
>             # Increment both by 1
#  i.e. [0.0025,-0.0025] → [1.0025,0.9975]
I·           # Push the input doubled
#  i.e. 0.05 → 0.1
/          # Divide both numbers with this doubled input
#  i.e. [1.0025,0.9975] / 0.1 → [10.025,9.975]
î         # Round both up
#  i.e. [10.025,9.975] → [11.0,10.0]
n        # Take the square of those
#  i.e. [11.0,10.0] → [121.0,100.0]
TS      # Push [1,0]
·     # Double both to [2,0]
<    # Decrease both by 1 to [1,-1]
-   # Decrease the earlier numbers by this
#  i.e. [121.0,100.0] - [1,-1] → [120.0,101.0]
ß  # Pop and push the minimum of the two
#  i.e. [120.0,101.0] → 101.0
# (which is output implicitly)

• Neat, thanks for linking the answer that has the formula used. I was doing mental gymnastics trying to figure out the formula from 05AB1E's ever-odd syntax. – Magic Octopus Urn Feb 26 '19 at 15:40

# JavaScript (ES7),  51  50 bytes

f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n


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(fails for the test cases that require too much recursion)

# Non-recursive version,  57  56 bytes

k=>{for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);return n}


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Or for 55 bytes:

k=>eval(for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n)


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(but this one is significantly slower)

# J, 39 29 bytes

[:<./_1 1++:*:@>.@%~1+(,-)@*:


NB. This shorter version simply uses @alephalpha's formula.

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## 39 bytes, original, brute force

2(>:@])^:((<+.0=])(<.-.)@(-<.)@%:)^:_~]


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Handles all test cases

# Japt, 18 16 bytes

-2 bytes from Shaggy

_=¬u1)©U>½-½aZ}a


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• Might be shorter using Arnauld's solution – ASCII-only Feb 26 '19 at 4:11
• – Shaggy Feb 26 '19 at 11:57
• Oh... of course i could have reversed that :|. Also that %1 && is nasty, not sure if using Arnauld's solution would be shorter (maybe not) – ASCII-only Feb 26 '19 at 11:59
• 16 bytes by reassigning Z¬u1 to Z at the beginning of the function. – Shaggy Feb 26 '19 at 12:00
• The other method appears to be 26: [1,-1]®*U²Ä /U/2 c ²-Z} rm – ASCII-only Feb 27 '19 at 0:18

# Pyth, 22 21 bytes

hSm-^.Ech*d^Q2yQ2d_B1


Try it online here, or verify all the test cases at once here.

Another port of alephalpha's excellent answer, make sure to give them an upvote!

hSm-^.Ech*d^Q2yQ2d_B1   Implicit: Q=eval(input())
_B1   [1,-1]
m                     Map each element of the above, as d, using:
^Q2            Q^2
*d               Multiply by d
h                 Increment
c      yQ          Divide by (2 * Q)
.E                   Round up
^           2         Square
-             d        Subtract d
S                      Sort
h                       Take first element, implicit print


Edit: Saved a byte, thanks to Kevin Cruijssen

• I don't know Pyth, but is it possible to create [-1,1] in 3 bytes as well, or do you need an additional reverse so it becomes 4 bytes? If it's possible in 3 bytes, you could do that, and then change the *_d to *d and the +d to -d. Also, does Pyth not have a Minimum builtin, instead of sort & take first? – Kevin Cruijssen Feb 26 '19 at 13:58
• @KevinCruijssen The order of the two elements isn't important as we're taking the minimum, though I can't think of a way of creating the pair in 3 bytes. A good catch on changing it to - ... d though, that saves me a byte! Thanks – Sok Feb 26 '19 at 14:30
• @KevinCruijssen Also there isn't a single byte minimum or maximum function unfortunately :o( – Sok Feb 26 '19 at 14:31
• Ah, of course. You map over the values, so it doesn't matter if it's [1,-1] or [-1,1]. I was comparing the *d and -d with my 05AB1E answer, where I don't use a map, but can subtract/multiply a 2D array from/with another 2D array, so I don't need a map. Glad I could help to save a byte in that case. :) And thanks for the inspiration for my 05AB1E answer. – Kevin Cruijssen Feb 26 '19 at 14:39

# Perl 6, 3433 29 bytes

-1 byte thanks to Grimy

{+(1...$_>*.sqrt*(1|-1)%1>0)}  Try it online! • -1 byte by replacing >= with >. Square roots of integers are either integer or irrational, so the equality case provably cannot happen. – Grimmy Feb 26 '19 at 13:03 • @Grimy Thanks, this seems to be allowed according to the challenge rules. (Though floating-point numbers are always rational, of course.) – nwellnhof Feb 26 '19 at 13:16 # APL (Dyalog Unicode), 27 bytesSBCS ⌊/0~⍨¯1 1+2*⍨∘⌈+⍨÷⍨1(+,-)×⍨  Try it online! Monadic train taking one argument. This is a port of alephalpha's answer. ### How: ⌊/0~⍨¯1 1+2*⍨∘⌈+⍨÷⍨1(+,-)×⍨ ⍝ Monadic train ×⍨ ⍝ Square of the argument 1(+,-) ⍝ 1 ± that (returns 1+k^2, 1-k^2) ÷⍨ ⍝ divided by +⍨ ⍝ twice the argument ∘⌈ ⍝ Ceiling 2*⍨ ⍝ Squared ¯1 1+ ⍝ -1 to the first, +1 to the second 0~⍨ ⍝ Removing the zeroes ⌊/ ⍝ Return the smallest  # C# (Visual C# Interactive Compiler), 8985 71 bytes k=>{double n=2,p;for(;!((p=Math.Sqrt(n)%1)>0&p<k|1-p<k);n++);return n;}  Try it online! -4 bytes thanks to Kevin Cruijssen! • You can save a byte by putting the n++ in the loop, so the -1 can be removed from the return: k=>{double n=1,p;for(;Math.Abs(Math.Round(p=Math.Sqrt(0d+n))-p)>k|p%1==0;n++);return n;} – Kevin Cruijssen Feb 26 '19 at 12:39 • Also, the 0d+ can be removed, can it not? – Kevin Cruijssen Feb 26 '19 at 12:59 • @KevinCruijssen Yes it can, I just forgot the n was already a double – Embodiment of Ignorance Feb 26 '19 at 15:56 # Java (JDK), 73 70 bytes k->{double i=1,j;for(;(j=Math.sqrt(++i)%1)==0|j>=k&1-j>=k;);return i;}  Try it online! -3 bytes thanks to @ceilingcat • @ceilingcat Neat, thanks. – Sara J Jul 23 '19 at 20:40 # Java 8, 85 bytes n->{double i=1,p;for(;Math.abs(Math.round(p=Math.sqrt(i))-p)>n|p%1==0;i++);return i;}  Port of EmbodimentOfIgnorance's C# .NET answer. Try it online. The Math.round can alternatively be this, but unfortunately it's the same byte-count: n->{double i=1,p;for(;Math.abs((int)((p=Math.sqrt(i))+.5)-p)>n|p%1==0;i++);return i;}  Try it online. # MathGolf, 16 bytes ²_b*α)½╠ü²1bαm,╓  Try it online! Not a huge fan of this solution. It is a port of the 05AB1E solution, which is based on the same formula most answers are using. ## Explanation ² pop a : push(a*a) _ duplicate TOS b push -1 * pop a, b : push(a*b) α wrap last two elements in array ) increment ½ halve ╠ pop a, b, push b/a ü ceiling with implicit map ² pop a : push(a*a) 1 push 1 b push -1 α wrap last two elements in array m explicit map , pop a, b, push b-a ╓ min of list  • Is every symbol considered a byte in code golfing? Because some of your characters require more than a single byte. I don't mean to nit-pick, I'm genuinely wondering :) – schroffl Feb 27 '19 at 12:05 • Good question! A "byte" in golfing relates to the minimum file size required to store a program. The text used to visualize those bytes can be any bytes. I have chosen Code Page 437 to visualize my scripts, but the important part is the actual bytes that define the source code. – maxb Feb 27 '19 at 13:07 • A good example of the number of characters and number of bytes being different is this answer. Here, the 'ԓ' character is actually 2 bytes, but the rest are 1 byte characters. – maxb Feb 27 '19 at 13:08 # Forth (gforth), 76 bytes : f 1 begin 1+ dup s>f fsqrt fdup fround f- fabs fdup f0> fover f< * until ;  Try it online! ### Explanation Starts a counter at 1 and Increments it in a loop. Each iteration it checks if the absolute value of the counter's square root - the closest integer is less than k ### Code Explanation : f \ start a new word definition 1 \ place a counter on the stack, start it at 1 begin \ start and indefinite loop 1+ \ add 1 to the counter dup s>f \ convert a copy of the counter to a float fsqrt \ get the square root of the counter fdup fround f- \ get the difference between the square root and the next closes integer fabs fdup \ get the absolute value of the result and duplicate f0> \ check if the result is greater than 0 (not perfect square) fover f< \ bring k to the top of the float stack and check if the sqrt is less than k * \ multiply the two results (shorter "and" in this case) until \ end loop if result ("and" of both conditions) is true ; \ end word definition  # Jelly, 13 bytes I have not managed to get anything terser than the same approach as alephalpha - go upvote his Mathematica answer! ²;N$‘÷ḤĊ²_Ø+Ṃ


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### How?

²;N$‘÷ḤĊ²_Ø+Ṃ - Link: number, n (in (0,1)) ² - square n -> n²$          - last two links as a monad:
N           -   negate        -> -(n²)
;            -   concatenate   -> [n², -(n²)]
‘         - increment       -> [1+n², 1-(n²)]
Ḥ       - double n        -> 2n
÷        - divide          -> [(1+n²)/n/2, (1-(n²))/n/2]
Ċ      - ceiling         -> [⌈(1+n²)/n/2⌉, ⌈(1-(n²))/n/2⌉]
²     - square          -> [⌈(1+n²)/n/2⌉², ⌈(1-(n²))/n/2⌉²]
Ø+  - literal         -> [1,-1]
_    - subtract        -> [⌈(1+n²)/n/2⌉²-1, ⌈(1-(n²))/n/2⌉²+1]
Ṃ - minimum         -> min(⌈(1+n²)/n/2⌉²-1, ⌈(1-(n²))/n/2⌉²+1)


# Japt, 14 bytes

_=¬aZ¬r¹©U¨Z}a


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_=¬aZ¬r¹©U¨Z}a     :Implicit input of integer U
_                  :Function taking an integer Z as an argument
=                 :  Reassign to Z
¬                :    Square root of Z
a               :    Absolute difference with
Z¬             :      Square root of Z
r            :      Round to the nearest integer
¹           :  End reassignment

# Perl 5-p, 42 bytes
$t=sqrt++$\while($p=abs$t-int$t)>$_||!\$p}{