FIBonacci sequence

Question

For this code golf, you will receive an input of a fibonacci sequence, that is, a normal Fibonacci sequence but with one number incorrect. _{See, the sequence is fibbing! Get it? :D}

Your job is to find out which number is incorrect, and print the index (0-based) of that number.

For example:

Input : 1 1 2 9 5 8 13
Output: 3
Input : 8 13 21 34 55 80
Output: 5
Input : 2 3 5 5 13 21
Output: 3

Specifications:

• The sequence may start at any number.
• The first two numbers of the input will always be correct.
• Shortest code (character count) wins.

• Volatility: Does the input have to be space-delimited or can commas be used as well?
• Doorknob: @Volatility Input is space-delimited.
• Luis Mendo: The job is to find only the first such number, right? For example, if you started from the right in the first sequence you could think that 8 is incorrect because it doesn't equal 9+5
• Doorknob: @LuisMendo There will always be only one such number.
• Doorknob: @LuisMendo Okay, let me reword that: There will always be exactly one way to change a single number that causes the sequence to be correct.

Answer 1

GolfScript (18 chars)

~]:^,,{^>3<~-+}?2+

The key to keeping this short is ? (find).

Answer 2

J, (削除) 30 (削除ここまで) 23

(2+0 i.~2&}.=[:}:}:+}.)

Answer 3

Golfscript, (削除) 31 (削除ここまで) (削除) 28 (削除ここまで) (削除) 26 (削除ここまで) (削除) 25 (削除ここまで) 23

~]-1%~1{)3円3ドル$-=}do])\;

Answer 4

APL (19)

1+1⍳⍨(1↓1⌽k)≠2+/k←⎕

Explanation:

• k←⎕: store user input in k
• 2+/k: sum each pair of elements in k (i.e. 1 1 2 3 -> 1+1 1+2 2+3 -> 2 3 5)
• 1↓1⌽k: rotate k to the right by 1 and then drop the first element (i.e. 1 1 2 3 -> 2 3 1)
• ≠: find the place where these lists are not equal
• 1⍳⍨: find the location of the first 1 in this list (location of the incorrect number)
• 1+: add 1 to compensate for the dropped element

Answer 5

K, 32

{2+*&~(n@n@x)=x+(n:{1_x,x 0})@x}

Answer 6

dc, (削除) 36 (削除ここまで) 32

?zszsasb[lalbdsa+dsb=x]dsxxlzz-p

dc is a reverse-Polish calculator, so obviously you need to input the numbers in reverse order ;)

$ dc fib.dc <<< "999 13 8 5 3 2 1 1"
7
$ dc fib.dc <<< "999 1 1"
2

Answer 7

JavaScript, 70

for(n=prompt().split(' '),i=n.length;i---2;)if(n[i-2]- -n[i-1]!=n[i])i

Answer 8

Javascript ((削除) 69 (削除ここまで) (削除) 68 (削除ここまで) (削除) 61 (削除ここまで) (削除) 60 (削除ここまで) 55)

for(s=prompt(i=2).split(' ');s[i]-s[i-1]==s[i-2];i++);i

---

(60)

s=prompt(i=2).split(' ');for(;s[i]==+s[i-1]+ +s[i++-2];);--i

---

(61)

s=prompt(i=1).split(' ');for(k=+s[1];k+=+s[i-1],k==s[++i];);i

---

(68)

s=prompt(i=1).split(' ');for(k=+s[1];k+=+s[i-1],k==s[++i];);alert(i)

---

(69)

s=prompt(i=1).split(' ');k=+s[1];for(;k+=+s[i-1],k==s[++i];);alert(i)

Answer 9

Awk: 55

{for(i=3;i<=NF;i++)if($i+$(i-1)!=$(i+1)){print i;exit}}

Answer 10

Ruby, 66

My first attempt at an (somewhat) complicated Ruby program:

p gets.split.map(&:to_i).each_cons(3).find_index{|a,b,c|a+b!=c}+2

• britishtea: You can save quite a few characters if you replace gets.split with $* (ARGV) to take input as command line arguments instead of on the standard input stream. The space between p and $* can then also be safely removed.

Answer 11

Python, 74

a=map(int,raw_input().split())
i=2
while a[i-2]+a[i-1]==a[i]:i+=1
print i

I had this solution first, but Doorknob answered the question about the format of input right before I had time to post it:

Python, 66

a,b=input(),input()
i=2
while input()==a+b:a,b=b,a+b;i+=1
print i

Assumes newline separated input.

Answer 12

VB.net (77)

Assuming the numbers are already in a IEnumerable(Of Integer).

 Dim p = xs.Skip(2).TakeWhile(Function(c, i) c = xs.Skip(i).Take(2).Sum).Count + 2

Answer 13

JS, 52B

for(s=prompt(i=2).split` `;s[i]-s[i-1]==s[i++-2];);i

Answer 14

Matlab / Octave, 39 bytes

Thanks to Stewie Griffin for saving a byte! (- instread of ~=)

@(x)find(diff(x(2:end))-x(1:end-2),1)+1

This is an anonymous function that inputs an array and outputs a number.

Try it online!

Answer 15

Kotlin, 77 bytes

{val r=it.split(' ').map{it.toInt()}
var i=2
while(r[i]==r[i-1]+r[i-2])i++
i}

Beautified

{
 val r = it.split(' ').map { it.toInt() }
 var i=2
 while(r[i] == r[i-1] + r[i-2]) i++
 i
}

Test

var f:(String)->Int =
{val r=it.split(' ').map{it.toInt()}
var i=2
while(r[i]==r[i-1]+r[i-2])i++
i}
data class Test(val input: String, val output: Int)
val TESTS = listOf(
 Test("1 1 2 9 5 8 13", 3),
 Test("8 13 21 34 55 80", 5),
 Test("2 3 5 5 13 21", 3)
)
fun main(args: Array<String>) {
 val fails = TESTS
 .asSequence()
 .map { it to f(it.input) }
 .filter { (test, res) -> test.output != res }
 .toList()
 if (fails.isEmpty()) {
 println("Test Passed")
 } else {
 fails.forEach{ println(it)}
 }
}

Answer 16

Raku, 45 bytes

{first :k,?*,(@($/=.words)Z-(0,ドル1,ドル*+*...*))}

Try it online!

$/ = .words stores the words (numbers) of the input string into the pattern-match variable $/. Conveniently, the elements of that variable can be accessed with 0ドル, 1ドル, etc, so 0,ドル 1,ドル * + * ... * forms the Fibonacci sequence starting with the first two input numbers. The input numbers and the Fibonacci sequence are then zipped together using the subtraction operator (Z-), forming a new sequence which is nonzero only at the place where the two series are different. Then first :k, ?*, ... returns the index (thanks to the :k parameter) of that location; ?* is an anonymous predicate function that coerces its argument to a boolean value.

Answer 17

Python (90)

a=map(int,raw_input().split())
print min(i for i in range(2,len(a))if a[i-2]+a[i-1]!=a[i])

Answer 18

Mathematica 59

Because space-delimited input is required, StringSplit needs to be employed. The following assumes that the input is in the form of a string i.

s = StringSplit@i;
p = 3; While[s[[p - 1]] + s[[p - 2]] == s[[p]], p++]; p - 1

Answer 19

Haskell, 48

f l=length$fst$span id$zipWith(==)l1ドル:scanl(+)1l

Answer 20

Jelly, 11 bytes

ÆḞ€i$€In1i1

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Answer 21

QBIC, 31 bytes

_!_!{_!~a+b=c|a=b┘b=c┘s=s+1\_Xs

Explanation

_!_! Ask the user for two umbers, assign them to 'a' and 'b'
{ DO
 _! Ask for a third number (this will be assigned to 'c' on every iteration)
 ~a+b=c IF the previous two terms add up to the third
 |a=b THEN shift b into a, 
 ┘b=c and c into b
 ┘s=s+1 increment s (starts as 3 in QBIC)
 \_Xs ELSE quit, printing the step counter

I'm not quite sure if this is allowed; the sequence is entered one term at a time, and the program aborts on error, not after entering the entire sequence.

Answer 22

Pxem, 0 bytes (content) + 52 bytes (filename).

• Filename (escaped): 002円._._X.w.c.t.v.m.v.+._.c.t.v.m.v.-001円.r.x.n.d.a001円.+.vX.a

Usage

• From stdin
• Must be an actual FIBonacci sequence
• Each integer are separated with blank characters

With comments

XX.z
# Initial stack: F(i-1), F(i-2), i
# i is initially 2
.a002円._._XX.z
# while :; do
.aX.wXX.z
 # to: (F(i-1)+F(i-2)), i, F(i-1)
 .a.c.t.v.m.v.+XX.z
 # input is F(i)
 # to: F(i), (F(i-1)+F(i-2)), i, F(i-1), F(i)
 .a._.c.t.v.m.vXX.z
 # to 0, abs(F(i)-F(i-1)-F(i-2)), i, F(i-1), F(i)
 # if (pop!=pop); then
 .a.-001円.r.xXX.z
 # print pop; exit; fi
 .a.n.d.aXX.z
 # to F(i), F(i-1), (i+1)
 .a001円.+.vXX.z
# done
.aX.a

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Answer 23

Husk, 5 bytes

δV≠İf

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You know it's a good day when you get to use decorV in an answer.