Complete a Mystery Sequence

Question

Given a sequence of three integers, determine if the sequence is arithmetic (of the form [a, a+d, a+2*d]) or geometric (of the form [a, a*r, a*r^2]) by outputting a fourth term that completes it (a+3*d for arithmetic, a*r^3 for geometric).

Examples:

[1, 2, 3] -> 4 (This is an arithmetic sequence with a difference of 1)
[2, 4, 8] -> 16 (This is a geometric sequence with a ratio 2)
[20, 15, 10] -> 5 (arithmetic sequence, d=-5)
[6, 6, 6] -> 6 (arithmetic with d=0 OR geometric with r=1)
[3, -9, 27] -> -81 (geometric with r=-3)

• The input is guaranteed to be a valid arithmetic and/or geometric sequence (so you won't have to handle something like [10, 4, 99])
• None of the inputted terms will be 0 (both [2, 0, 0] and [1, 0, -1] would not be given)
• For geometric sequences, the ratio between terms is guaranteed to be an integer (ie. [4, 6, 9] would be an invalid input, as the ratio would be 1.5)
• If a sequence could be either arithmetic or geometric, you may output either term that completes it
• You may take input as a 3-term array (or whatever equivalent) or as three separate inputs
• This is code golf, so aim for the lowest byte count possible!

• nyxbird: I think I'll remove any test cases involving 0 (as a term). Someone on the sandbox mentioned [1, 0, 0]
• Stef: Regarding "If a sequence could be either arithmetic or geometric", this happens if and only if the three terms are equal, which corresponds to either (a = 0, d = 0, r = anything) or (a != 0, d = 0, r = 1).

Answer 1

Python, 34 bytes

-1 thanks to @Arnauld

lambda a,b,c:c+(c-b)**2/(b-a or 1)

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How?

Uses the fact that the increments form a geometric sequence in either case. (This avoids the longish (in Python) ternary operator.)

Indeed: \$c+\frac{(c-b)^2}{b-a} = \begin{cases} c+(c-b)=2c-b & \text{if }c-b=b-a \\ c+(c-b)\times \frac c b = \frac{c^2} b & \text{if } \frac c b = \frac b a \end{cases}\$

• pacman256: i accidentally downvoted can you edit so i can change it
• Arnauld: Would it be OK to use / instead of //? Returning integers as floats is fine.
• Stef: With "ternary operator" it's only 2 bytes longer: lambda a,b,c:(c*b/a,c+b-a)[c-b==b-a] (unfortunately cannot compress c-b==b-a into c-2*b+a because that can take values other than 0 or 1)
• Albert.Lang: @Stef you could use c*a<b*b, though, couldn't you?

Answer 2

Raku (Perl 6), 13 bytes

A block taking a sequence of length 3 as a single argument. This task is a perfect match for Raku's sequence operator.

{($_...*)[3]}

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Answer 3

Python 3, 24 bytes

lambda a,b,c:c*(a+c)/b-b

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Note that this fails when \$ a=0 \$ or \$ b=0 \$, but that seems to be OK for this challenge :). We can verify its correctness:

• For an arithmetic sequence \$ b-a = c-b \implies 2b = a+c \$, the formula rightfully produces \$ 2c-b \$: $$ \frac{c*(a+c)}{b}-b = \frac{c*2b}{b}-b = 2c-b $$
• Likewise, for a geometric sequence \$ b/a = c/b \implies a = b^2/c \$, the formula rightfully produces \$ c^2/b \$: $$ \frac{c*(a+c)}{b}-b = \frac{c*(b^2/c+c)}{b}-b = \frac{b^2+c^2}{b}-b = b+\frac{c^2}{b}-b = \frac{c^2}{b} $$

• Neil: Since \$ 2b = a + c \$ and \$ b^2 = ac \$ respectively, you can just say \$ \frac { c ( a + c) } b - b = \frac { 2bc } b - b = 2c - b \$ and \$ \frac { c (a + c) } b - b = \frac { b^2 + c^2 } b - b = b + \frac { c^2 } b - b = \frac { c^2 } b \$.

Answer 4

JavaScript (ES6), 28 bytes

(a,b,c)=>c-2*b+a?c*c/b:2*c-b

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Commented

(a, b, c) => // given the 3 integers,
c - 2 * b + a ? // if the sequence is not arithmetic:
 c * c / b // assume it's geometric and return the next geometric term
: // else:
 2 * c - b // return the next arithmetic term

Answer 5

R, 28 bytes

\(a,b,c)c+(c-b)^2/(b-a+!b-a)

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Port of @loopy walt's Python answer.

R, 33 bytes

\(a,b,c)`if`(c-b-b+a,c*c/b,c+c-b)

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Straightforward approach.

Answer 6

Google Sheets, 34 bytes

=IF(A3/A2=A2/A1,A3*A2/A1,A3+A2-A1) 

The formula assumes the sequence is in A1:A3.

Answer 7

Awk, 33 bytes

1,0ドル=1ドル-2*2ドル+3ドル?2ドル*3ドル/1ドル:2ドル+3ドル-1ドル

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You could remove the 1, at the start if you assume that the output will not be 0, saving 2 bytes.

Answer 8

Jelly, 11 bytes

×ばつ÷Ɲ$}E?UḢ

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A monadic link taking a list of three integers and returning an integer.

Explanation

I | Increments (differences between consecutive list members)
 E? | If all equal then:
 + U | - Add the increments to the reversed input list
 $} U | Else, following as a monad applied to the reversed input list:
 ×ばつ | - Multiply by:
 ÷Ɲ | - The result of dividing each pair of neighbouring list members
 Ḣ | Head

Answer 9

Vyxal 3, 13 bytes

 ̄≈[Ṛ+|a÷Ia×ばつ]t

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Port of Nick Kennedy's Jelly answer, sort of?

• Jonathan Allan: Updated with attribution since there are now two Jelly answers. Always best to link regardless in my opinion.

Answer 10

Alice, 29 bytes

3&/O
?+\M@/!]!?-R.n$n?[?-.*~:

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-4 bytes thanks to Nitrodon!

Uses loopy walt's solution, go upvote them!

3&/M\ Read 3 arguments a b c on the stack
 !]! Pop and write c then b on the tape
 ?-R Push b on the stack and calculate b-a
 .n$n if b-a is 0, replace with 1
 ?[? Push b and c on the stack
 -.* Calculate (c-b)^2
 ~: Calculate (c-b)*(c-b)/(b-a or 1)
 ?+ Pop c and calculate c+(c-b)*(c-b)/(b-a or 1)
 \O@ Output the result

• Nitrodon: 29 bytes by using a different code layout.

Answer 11

Funge-98, 46 bytes

<@.+wR-R:VI\OO-R:VIVI(4"FRTH"(4"STRN"
/\OR<@.*

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Answer 12

APL+WIN, (削除) 43 (削除ここまで) 38 bytes

Prompts for integers

↑(=/ ̈a b)/(×ばつ↑a←2÷/v),(↑v+↑b←2-/v←⌽⎕)

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Answer 13

Jelly, 9 bytes

Uses the trick described by loopy walt in their excellent Python answer although the implementation is somewhat different.

IQ2÷\@/+Ṫ

A monadic Link that accepts a triple of integers, [a, b, c], that is either arithmetic or geometric and yields the next integer in the series, d.

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How?

IQ2÷\@/+Ṫ - Link: list of integers, [a, b, c]
I - forward differences -> [b-a, c-b]
 Q - deduplicate
 -> if arithmetic: L=[b-a] (potentially [0])
 else: L=[b-a, c-b] (with no zeros)
 / - reduce L by:
 @ - with swapped arguments:
 \ - last two links as a dyad - f(x, y):
 2 - square -> x^2
 ÷ - divide -> x^2/y
 -> if arithmetic: Z=b-a (only one item so reduce does not perform f)
 else: Z=(c-b)^2/(b-a)
 + - add {[a, b, c]} (vectorises) -> [a+Z,b+Z,c+Z]
 Ṫ - tail -> c+Z (= d)
 -> if arithmetic: d=c+(b-a)
 else: d=c+(c-b)^2/(b-a)

Answer 14

Desmos, 19 bytes

f(a,b,c)=c(a+c)/b-b

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Port of dingledooper's Python 3 solution, so make sure to upvote that one too!

Answer 15

C (gcc), 24 bytes

f(a,b,c){a=c*(a+c)/b-b;}

Another port of dingledooper's answer!

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Answer 16

Vyxal, 6 bytes

+∇/*0-

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Port of dingledooper's clever formula, go look at that for an explanation of why this works.

 # input c, a, b
+ # c+a
 ∇ # rotate stack to b, c, c+a
 / # c/b
 * # (c+a) * c/b
 0 # last input (b)
 - # (c+a) * c/b - b

This uses four necessary operations (+/-*) and two stack-manipulation operations (0∇). I'm fairly sure that a solution with only one stack-manipulation op is impossible but I'd love to be proven wrong.

• Kevin Cruijssen: Your explanation is incorrect. It's c/b not b/c (e.g. (c+a) * b/c - b for the first test case would be (3+1) * 2/3 - 2 = 2⁄3 instead of (3+1) * 3/2 - 2 = 4). Your code and TIO are fine.

Answer 17

Perl 5 -ap, 41 bytes

$_=2*($b=<>)-$_-($c=<>)?$c*$c/$b:$c+$b-$_

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Answer 18

MathGolf, 6 bytes

+╠*k;,

Port of dingledooper's Python 3 answer, so make sure to upvote that answer as well!

I/O as floats, and inputs are in the order \$c,a,b\$.

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Explanation:

+ # Add the first two (implicit) inputs together: c+a
 ╠ # Divide it by the third (implicit) input: (c+a)/b
 * # Multiply it by the first (implicit) input: (c+a)/b*c
 k; # Push the second input, and discard it
 , # Subtract it by the third (implicit) input: (c+a)/b*c-b
 # (after which the entire stack is output implicitly as result)

The k;, could also be ?-Þ for the same byte-count: Try it online.

 ? # Triple-swap the top three values: a,(c+a)/b,b
 - # Subtract the top two: a,(c+a)/b-b
 Þ # Only keep the top value of the stack: (c+a)/b-b
 # (after which the entire stack is output implicitly as result)

Answer 19

Retina 0.8.2, 65 bytes

.+
$*
¶1+¶
$'$&
1(?=1*¶1+¶(1+))
1ドル
(?=1+¶(1+)¶)1円
1
(1+)¶1円¶1+
1

Try it online! Takes input on separate lines but link is to test suite that splits on non-digits for convenience. Limited to positive integers due to use of unary arithmetic. Explanation: Uses @dingledooper's formula.

.+
$*

Convert to unary.

¶1+¶
$'$&

Add c to a.

1(?=1*¶1+¶(1+))
1ドル

Multiply a+c by c.

(?=1+¶(1+)¶)1円
1

Divide that by b.

(1+)¶1円¶1+

Subtract b and remove b and c.

1

Convert to decimal.

Answer 20

Charcoal, 16 bytes

×ばつ+θζζηη

Try it online! Link is to verbose version of code. Explanation: Trivial implementation of @dingledooper's formula.

35 bytes to validate the input:

NθNηNζ¿=⊗η+θζI−+ζηθ¿∧θ=×ばつζηθ

Try it online! Link is to verbose version of code. Works with edge cases such as 0 0 0, 1 0 0, 1 0 -1. Doesn't output anything if the input is not an arithmetic or geometric sequence.

Notes for both solutions:

• 6 bytes (NθNηNζ) could be saved by requiring the input to be in JSON format.
• Floating-point arithmetic could be used by replacing ÷ with ∕.

Answer 21

SAKO, 41 bytes

PODPROGRAM:S(A,K,O)
S()=×ばつ(A+O)/B-B
WROC