Valid term from quadratic sequence?
$begingroup$
You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:
$$T_n=an^2+bn+c$$
You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.
This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.
Test cases
a |b |c |T_n |Y/N
------------------------
1 |1 |1 |1 |Y #n=0
2 |3 |5 |2 |N
0.5 |1 |-2 |-0.5|Y #n=1
0.5 |1 |-2 |15.5|Y #n=5
0.5 |1 |-2 |3 |N
-3.5|2 |-6 |-934|Y #n=-16
0 |1 |4 |7 |Y #n=3
0 |3 |-1 |7 |N
0 |0 |0 |1 |N
0 |0 |6 |6 |Y #n=<anything>
code-golf number decision-problem equation
$endgroup$
add a comment |
$begingroup$
You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:
$$T_n=an^2+bn+c$$
You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.
This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.
Test cases
a |b |c |T_n |Y/N
------------------------
1 |1 |1 |1 |Y #n=0
2 |3 |5 |2 |N
0.5 |1 |-2 |-0.5|Y #n=1
0.5 |1 |-2 |15.5|Y #n=5
0.5 |1 |-2 |3 |N
-3.5|2 |-6 |-934|Y #n=-16
0 |1 |4 |7 |Y #n=3
0 |3 |-1 |7 |N
0 |0 |0 |1 |N
0 |0 |6 |6 |Y #n=<anything>
code-golf number decision-problem equation
$endgroup$
add a comment |
$begingroup$
You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:
$$T_n=an^2+bn+c$$
You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.
This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.
Test cases
a |b |c |T_n |Y/N
------------------------
1 |1 |1 |1 |Y #n=0
2 |3 |5 |2 |N
0.5 |1 |-2 |-0.5|Y #n=1
0.5 |1 |-2 |15.5|Y #n=5
0.5 |1 |-2 |3 |N
-3.5|2 |-6 |-934|Y #n=-16
0 |1 |4 |7 |Y #n=3
0 |3 |-1 |7 |N
0 |0 |0 |1 |N
0 |0 |6 |6 |Y #n=<anything>
code-golf number decision-problem equation
$endgroup$
You are given four numbers. The first three are $a$, $b$, and $c$ respectively, for the sequence:
$$T_n=an^2+bn+c$$
You may take input of these four numbers in any way. The output should be one of two distinct outputs mentioned in your answer, one means that the fourth number is a term in the sequence (the above equation has at least one solution for $n$ which is an integer when $a$, $b$, $c$ and $T_n$ are substituted for the given values), the other means the opposite.
This is code golf, so the shortest answer in bytes wins. Your program should work for any input of $a, b, c, T_n$ where the numbers are negative or positive (or 0), decimal or integer. To avoid problems but keep some complexity, non-integers will always just end in $.5$. Standard loop-holes disallowed.
Test cases
a |b |c |T_n |Y/N
------------------------
1 |1 |1 |1 |Y #n=0
2 |3 |5 |2 |N
0.5 |1 |-2 |-0.5|Y #n=1
0.5 |1 |-2 |15.5|Y #n=5
0.5 |1 |-2 |3 |N
-3.5|2 |-6 |-934|Y #n=-16
0 |1 |4 |7 |Y #n=3
0 |3 |-1 |7 |N
0 |0 |0 |1 |N
0 |0 |6 |6 |Y #n=<anything>
code-golf number decision-problem equation
code-golf number decision-problem equation
edited 11 hours ago
Artemis Fowl
asked 16 hours ago
Artemis FowlArtemis Fowl
1767
1767
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
JavaScript (ES7), 48 bytes
Returns a Boolean value.
(a,b,c,t)=>(t-=c,a?(b*b+4*a*t)**.5%1:b?t%b:t)==0
Try it online!
How?
For sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)
Case $aneq0$
The equation really is quadratic:
$$T_n=an^2+bn+c\
an^2+bn-d=0$$
And its discriminant is:
$$Delta=b^2+4ad$$
It admits an integer root if $Delta$ is non-negative and $sqrt{Delta}$ is an integer.
Case $a=0, bneq0$
The equation is linear:
$$T_n=bn+c\
bn=d\
n=frac{d}{b}$$
It admits an integer root if $dequiv0pmod b$.
Case $a=0, b=0$
The equation is not depending on $n$ anymore:
$$T_n=c\
d=0$$
$endgroup$
add a comment |
$begingroup$
Jelly, 11 10 bytes
_/Ær1Ẹ?%1Ạ
A monadic Link which accepts a list of lists* [[c, b, a], [T_n]]
and yields 0
if T_n
is a valid solution or 1
if not.
* admittedly taking a little liberty with "You may take input of these four numbers in any way".
Try it online! Or see a test-suite.
How?
_/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
/ - reduce by:
_ - subtraction [c-T_n, b, a]
? - if...
Ẹ - ...condition: any?
Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
1 - ...else: literal 1
%1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
- note: (a+bi)%1 yields nan which is truthy
Ạ - all? i.e. all had fractional parts?
- note: all() yields 1
If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ
would also work for 10 (it yields 1
when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)
$endgroup$
1
$begingroup$
That input is perfectly fine.
$endgroup$
– Artemis Fowl
12 hours ago
add a comment |
$begingroup$
Jelly, 15 bytes
_3¦UÆr=Ḟ$;3ị=ɗẸ
Try it online!
Built-in helps here but doesn’t handle a=b=0.
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 38 bytes
Solve[n^2#+n#2+#3==#4,n,Integers]!={}&
Try it online!
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
JavaScript (ES7), 48 bytes
Returns a Boolean value.
(a,b,c,t)=>(t-=c,a?(b*b+4*a*t)**.5%1:b?t%b:t)==0
Try it online!
How?
For sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)
Case $aneq0$
The equation really is quadratic:
$$T_n=an^2+bn+c\
an^2+bn-d=0$$
And its discriminant is:
$$Delta=b^2+4ad$$
It admits an integer root if $Delta$ is non-negative and $sqrt{Delta}$ is an integer.
Case $a=0, bneq0$
The equation is linear:
$$T_n=bn+c\
bn=d\
n=frac{d}{b}$$
It admits an integer root if $dequiv0pmod b$.
Case $a=0, b=0$
The equation is not depending on $n$ anymore:
$$T_n=c\
d=0$$
$endgroup$
add a comment |
$begingroup$
JavaScript (ES7), 48 bytes
Returns a Boolean value.
(a,b,c,t)=>(t-=c,a?(b*b+4*a*t)**.5%1:b?t%b:t)==0
Try it online!
How?
For sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)
Case $aneq0$
The equation really is quadratic:
$$T_n=an^2+bn+c\
an^2+bn-d=0$$
And its discriminant is:
$$Delta=b^2+4ad$$
It admits an integer root if $Delta$ is non-negative and $sqrt{Delta}$ is an integer.
Case $a=0, bneq0$
The equation is linear:
$$T_n=bn+c\
bn=d\
n=frac{d}{b}$$
It admits an integer root if $dequiv0pmod b$.
Case $a=0, b=0$
The equation is not depending on $n$ anymore:
$$T_n=c\
d=0$$
$endgroup$
add a comment |
$begingroup$
JavaScript (ES7), 48 bytes
Returns a Boolean value.
(a,b,c,t)=>(t-=c,a?(b*b+4*a*t)**.5%1:b?t%b:t)==0
Try it online!
How?
For sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)
Case $aneq0$
The equation really is quadratic:
$$T_n=an^2+bn+c\
an^2+bn-d=0$$
And its discriminant is:
$$Delta=b^2+4ad$$
It admits an integer root if $Delta$ is non-negative and $sqrt{Delta}$ is an integer.
Case $a=0, bneq0$
The equation is linear:
$$T_n=bn+c\
bn=d\
n=frac{d}{b}$$
It admits an integer root if $dequiv0pmod b$.
Case $a=0, b=0$
The equation is not depending on $n$ anymore:
$$T_n=c\
d=0$$
$endgroup$
JavaScript (ES7), 48 bytes
Returns a Boolean value.
(a,b,c,t)=>(t-=c,a?(b*b+4*a*t)**.5%1:b?t%b:t)==0
Try it online!
How?
For sake of clarity, we define $d = T_n-c$. (The same variable $t$ is re-used to store this result in the JS code.)
Case $aneq0$
The equation really is quadratic:
$$T_n=an^2+bn+c\
an^2+bn-d=0$$
And its discriminant is:
$$Delta=b^2+4ad$$
It admits an integer root if $Delta$ is non-negative and $sqrt{Delta}$ is an integer.
Case $a=0, bneq0$
The equation is linear:
$$T_n=bn+c\
bn=d\
n=frac{d}{b}$$
It admits an integer root if $dequiv0pmod b$.
Case $a=0, b=0$
The equation is not depending on $n$ anymore:
$$T_n=c\
d=0$$
edited 15 hours ago
answered 15 hours ago
ArnauldArnauld
80.2k797331
80.2k797331
add a comment |
add a comment |
$begingroup$
Jelly, 11 10 bytes
_/Ær1Ẹ?%1Ạ
A monadic Link which accepts a list of lists* [[c, b, a], [T_n]]
and yields 0
if T_n
is a valid solution or 1
if not.
* admittedly taking a little liberty with "You may take input of these four numbers in any way".
Try it online! Or see a test-suite.
How?
_/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
/ - reduce by:
_ - subtraction [c-T_n, b, a]
? - if...
Ẹ - ...condition: any?
Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
1 - ...else: literal 1
%1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
- note: (a+bi)%1 yields nan which is truthy
Ạ - all? i.e. all had fractional parts?
- note: all() yields 1
If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ
would also work for 10 (it yields 1
when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)
$endgroup$
1
$begingroup$
That input is perfectly fine.
$endgroup$
– Artemis Fowl
12 hours ago
add a comment |
$begingroup$
Jelly, 11 10 bytes
_/Ær1Ẹ?%1Ạ
A monadic Link which accepts a list of lists* [[c, b, a], [T_n]]
and yields 0
if T_n
is a valid solution or 1
if not.
* admittedly taking a little liberty with "You may take input of these four numbers in any way".
Try it online! Or see a test-suite.
How?
_/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
/ - reduce by:
_ - subtraction [c-T_n, b, a]
? - if...
Ẹ - ...condition: any?
Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
1 - ...else: literal 1
%1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
- note: (a+bi)%1 yields nan which is truthy
Ạ - all? i.e. all had fractional parts?
- note: all() yields 1
If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ
would also work for 10 (it yields 1
when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)
$endgroup$
1
$begingroup$
That input is perfectly fine.
$endgroup$
– Artemis Fowl
12 hours ago
add a comment |
$begingroup$
Jelly, 11 10 bytes
_/Ær1Ẹ?%1Ạ
A monadic Link which accepts a list of lists* [[c, b, a], [T_n]]
and yields 0
if T_n
is a valid solution or 1
if not.
* admittedly taking a little liberty with "You may take input of these four numbers in any way".
Try it online! Or see a test-suite.
How?
_/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
/ - reduce by:
_ - subtraction [c-T_n, b, a]
? - if...
Ẹ - ...condition: any?
Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
1 - ...else: literal 1
%1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
- note: (a+bi)%1 yields nan which is truthy
Ạ - all? i.e. all had fractional parts?
- note: all() yields 1
If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ
would also work for 10 (it yields 1
when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)
$endgroup$
Jelly, 11 10 bytes
_/Ær1Ẹ?%1Ạ
A monadic Link which accepts a list of lists* [[c, b, a], [T_n]]
and yields 0
if T_n
is a valid solution or 1
if not.
* admittedly taking a little liberty with "You may take input of these four numbers in any way".
Try it online! Or see a test-suite.
How?
_/Ær1Ẹ?%1Ạ - Link: list of lists of integers, [[c, b, a], [T_n]]
/ - reduce by:
_ - subtraction [c-T_n, b, a]
? - if...
Ẹ - ...condition: any?
Ær - ...then: roots of polynomial i.e. roots of a²x+bx+(c-T_n)=0
1 - ...else: literal 1
%1 - modulo 1 (vectorises) i.e. for each: keep any fractional part
- note: (a+bi)%1 yields nan which is truthy
Ạ - all? i.e. all had fractional parts?
- note: all() yields 1
If we could yield non-distinct results then _/Ær1Ẹ?ḞƑƇ
would also work for 10 (it yields 1
when all values are solutions, otherwise a list of the distinct solutions and hence always an empty list when no solutions - this would also meet the standard Truthy vs Falsey definition)
edited 11 hours ago
answered 13 hours ago
Jonathan AllanJonathan Allan
53.6k535173
53.6k535173
1
$begingroup$
That input is perfectly fine.
$endgroup$
– Artemis Fowl
12 hours ago
add a comment |
1
$begingroup$
That input is perfectly fine.
$endgroup$
– Artemis Fowl
12 hours ago
1
1
$begingroup$
That input is perfectly fine.
$endgroup$
– Artemis Fowl
12 hours ago
$begingroup$
That input is perfectly fine.
$endgroup$
– Artemis Fowl
12 hours ago
add a comment |
$begingroup$
Jelly, 15 bytes
_3¦UÆr=Ḟ$;3ị=ɗẸ
Try it online!
Built-in helps here but doesn’t handle a=b=0.
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
_3¦UÆr=Ḟ$;3ị=ɗẸ
Try it online!
Built-in helps here but doesn’t handle a=b=0.
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
_3¦UÆr=Ḟ$;3ị=ɗẸ
Try it online!
Built-in helps here but doesn’t handle a=b=0.
$endgroup$
Jelly, 15 bytes
_3¦UÆr=Ḟ$;3ị=ɗẸ
Try it online!
Built-in helps here but doesn’t handle a=b=0.
answered 13 hours ago
Nick KennedyNick Kennedy
1,29649
1,29649
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 38 bytes
Solve[n^2#+n#2+#3==#4,n,Integers]!={}&
Try it online!
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 38 bytes
Solve[n^2#+n#2+#3==#4,n,Integers]!={}&
Try it online!
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 38 bytes
Solve[n^2#+n#2+#3==#4,n,Integers]!={}&
Try it online!
$endgroup$
Wolfram Language (Mathematica), 38 bytes
Solve[n^2#+n#2+#3==#4,n,Integers]!={}&
Try it online!
answered 9 hours ago
J42161217J42161217
13.7k21253
13.7k21253
add a comment |
add a comment |
If this is an answer to a challenge…
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…Avoid asking for help, clarification or responding to other answers (use comments instead).
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