Square Root Distance from IntegersCalculate the square root only using ++Sorted Lexical Partition of a...
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Square Root Distance from Integers
Calculate the square root only using ++Sorted Lexical Partition of a NumberReverse and squareThe fastest square root calculatorRobbers - square times square rootCops - square times square rootFermat's factorization helperMiller-Rabin Strong PseudoprimesExact change in fewest bills and coinsApproximate My Squares
$begingroup$
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
kwill 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 code-golf, so shortest answer in bytes wins.
code-golf number integer
asked 1 hour ago
StephenStephen
7,38323395
$endgroup$
add a comment |
$begingroup$
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
kwill 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 code-golf, so shortest answer in bytes wins.
code-golf number integer
asked 1 hour ago
StephenStephen
7,38323395
$endgroup$
add a comment |
$begingroup$
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
kwill 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 code-golf, so shortest answer in bytes wins.
code-golf number integer
asked 1 hour ago
StephenStephen
7,38323395
$endgroup$
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
kwill 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 code-golf, so shortest answer in bytes wins.
code-golf number integer
code-golf number integer
asked 1 hour ago
StephenStephen
7,38323395
asked 1 hour ago
StephenStephen
7,38323395
edited 29 mins ago
Stephen
asked 1 hour ago
StephenStephen
7,38323395
asked 1 hour ago
StephenStephen
7,38323395
asked 1 hour ago
StephenStephen
7,38323395
7,38323395
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
JavaScript (ES7), 51 50 bytes
f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n
Try it online!
(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}
Try it online!
Or for 55 bytes:
k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)
Try it online!
(but this one is significantly slower)
answered 59 mins ago
ArnauldArnauld
76.8k693322
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 36 bytes
Min[⌈(1/#-{1,-1}#)/2⌉^2+{1,-1}]&
Try it online!
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 $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.
answered 31 mins ago
alephalphaalephalpha
21.4k32991
$endgroup$
add a comment |
$begingroup$
Japt, 18 bytes
_¬%1©U>½-Z¬u1 a½}a
Try it online!
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
JavaScript (ES7), 51 50 bytes
f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n
Try it online!
(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}
Try it online!
Or for 55 bytes:
k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)
Try it online!
(but this one is significantly slower)
answered 59 mins ago
ArnauldArnauld
76.8k693322
$endgroup$
add a comment |
$begingroup$
JavaScript (ES7), 51 50 bytes
f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n
Try it online!
(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}
Try it online!
Or for 55 bytes:
k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)
Try it online!
(but this one is significantly slower)
answered 59 mins ago
ArnauldArnauld
76.8k693322
$endgroup$
add a comment |
$begingroup$
JavaScript (ES7), 51 50 bytes
f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n
Try it online!
(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}
Try it online!
Or for 55 bytes:
k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)
Try it online!
(but this one is significantly slower)
answered 59 mins ago
ArnauldArnauld
76.8k693322
$endgroup$
JavaScript (ES7), 51 50 bytes
f=(k,n)=>!(d=(s=n**.5)+~(s-.5))|d*d>k*k?f(k,-~n):n
Try it online!
(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}
Try it online!
Or for 55 bytes:
k=>eval(`for(n=1;!(d=(s=++n**.5)+~(s-.5))|d*d>k*k;);n`)
Try it online!
(but this one is significantly slower)
answered 59 mins ago
ArnauldArnauld
76.8k693322
edited 25 mins ago
answered 59 mins ago
ArnauldArnauld
76.8k693322
answered 59 mins ago
ArnauldArnauld
76.8k693322
answered 59 mins ago
ArnauldArnauld
76.8k693322
76.8k693322
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 36 bytes
Min[⌈(1/#-{1,-1}#)/2⌉^2+{1,-1}]&
Try it online!
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 $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.
answered 31 mins ago
alephalphaalephalpha
21.4k32991
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 36 bytes
Min[⌈(1/#-{1,-1}#)/2⌉^2+{1,-1}]&
Try it online!
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 $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.
answered 31 mins ago
alephalphaalephalpha
21.4k32991
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 36 bytes
Min[⌈(1/#-{1,-1}#)/2⌉^2+{1,-1}]&
Try it online!
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 $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.
answered 31 mins ago
alephalphaalephalpha
21.4k32991
$endgroup$
Wolfram Language (Mathematica), 36 bytes
Min[⌈(1/#-{1,-1}#)/2⌉^2+{1,-1}]&
Try it online!
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 $operatorname{min}left({leftlceil frac{1-k^2}{2k} rightrceil}^2+1, {leftlceil frac{1+k^2}{2k} rightrceil}^2-1right)$.
answered 31 mins ago
alephalphaalephalpha
21.4k32991
edited 56 secs ago
answered 31 mins ago
alephalphaalephalpha
21.4k32991
answered 31 mins ago
alephalphaalephalpha
21.4k32991
answered 31 mins ago
alephalphaalephalpha
21.4k32991
21.4k32991
add a comment |
add a comment |
$begingroup$
Japt, 18 bytes
_¬%1©U>½-Z¬u1 a½}a
Try it online!
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
$endgroup$
add a comment |
$begingroup$
Japt, 18 bytes
_¬%1©U>½-Z¬u1 a½}a
Try it online!
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
$endgroup$
add a comment |
$begingroup$
Japt, 18 bytes
_¬%1©U>½-Z¬u1 a½}a
Try it online!
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
$endgroup$
Japt, 18 bytes
_¬%1©U>½-Z¬u1 a½}a
Try it online!
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
edited 3 mins ago
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
answered 40 mins ago
ASCII-onlyASCII-only
3,3721236
3,3721236
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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