Previous Fibonacci number

Question

The sequence of Fibonacci numbers is defined as follows:

\$ F_0 = 0 \\ F_1 = 1 \\ F_n = F_{n-1} + F_{n-2} \$

Given a Fibonacci number, return the previous Fibonacci number in the sequence. You do not need to handle the inputs \$ 0 \$ or \$ 1 \$, nor any non-Fibonacci numbers.

Errors as a result of floating point inaccuracies are permitted.

This is code-golf, so the shortest code in bytes wins.

Answer 1

Vyxal, 4 bytes

kg/ṙ

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Port of Unmitigated's JavaScript answer

 / # Divide by...
kg # Phi
 ṙ # Round

Answer 2

JavaScript, (削除) 21 (削除ここまで) 17 bytes

n=>n*5**.5-n+1>>1

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-4 bytes thanks to att.

• att: 17 bytes

Answer 3

R, (削除) 23 (削除ここまで) (削除) 22 (削除ここまで) 20 bytes

Edit: -1 byte thanks to pajonk, then -2 bytes thanks to att

\(x)(x*5^.5-x+1)%/%2

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The ratio between consecutive fibonacci numbers famously converges towards the golden ratio phi, equal to (1+5^.5)/2.
This is sufficiently accurate that rounding produces the correct value for all fibonacci numbers, including even the second 1 at the start which we don't need to handle here.

(Edit after reading some other answers: this was independently-derived, but is also the approach taken by Luis Mendo and unmitigated)

• pajonk: Shouldn't this possibly work for -1 byte?
• Dominic van Essen: @pajonk - Thanks - yes! Somehow whenever I'd tried something like that, it always came out longer (probably because I was adding .5 instead of your clever way to add 1)...
• att: 20
• Dominic van Essen: @att - lovely, thanks! Whenever I look at something like this I think 'Doh! Why didn't I think of that...?'

Answer 4

x86-64 assembly, (削除) 13 (削除ここまで) 12 bytes

6a 01 58 99 92 01 c2 39 fa 75 f9 c3

In assembly:

previous:
 push 1
 pop rax
 cdq # zeroes rdx
previous_loop:
 xchg eax,edx
 add edx,eax
 cmp edx,edi
 jne previous_loop
 ret

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Uses the usual calling convention of taking the first argument in rdi and returning in rax. It's also valid x86-32, just with edi and eax instead.

It's pretty simple- it just explicitly computes Fibonacci numbers and returns if the new number is equal to the input. At the end of a loop, rdx contains the next Fibonacci number and rax contains the previous. If rdx is equal to the input rdi, then the loop terminates (and thus rax, the previous number, is returned). Otherwise the two are swapped, and the loop begins again.

This uses 32-bit registers to save a few bytes, so it only works up to the 48th Fibonacci number, 4807526976. This does not fit in a 32-bit integer but its result does so and it is not the same as any previous number modulo 2^32 so the math works out.

It can be expanded to work properly with 64-bit numbers by changing all the registers to 64-bit, at the cost of 2 bytes:

6a 01 58 99 48 0f c1 c2 48 39 fa 75 f7 c3

The added bytes are the REX.W (0x48) prefix that swaps the operand size to 64-bit. xchg rax,rdx; add rdx,rax is also replaced with xadd rdx,rax, which is shorter because it uses one fewer REX.W prefix.

-1 byte thanks to @PeterCordes.

• Peter Cordes: If we start with EAX=1 we can cdq for EDX=0 in 1 byte. That should enable savings even without changing the calling convention, e.g. xchg after add? If necessary, xadd can exchange-and-add between any two registers in one more byte than add. But with EAX involved it's the same size as xchg eax, reg / add. It's faster; only 3 uops on Intel or 2 on AMD, same as xchg, one less than xchg+add. And lower latency. (uops.info).
• Chris: @PeterCordes Good call. Swapping the two just adds one extra loop, since the first loop becomes effectively a no-op of add edx,0. It changes the output for an input of 1 to 0 instead of 1, but those are both equally valid.

Answer 5

R, 27 bytes

\(x){while(x>T)T=F+(F=T);F}

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Similar to this answer, iterates through the Fibonacci sequence until T==x and returns F.

• Dominic van Essen: Realised that this is the shortest R fibonacci function (with F>T) and have cited it here and here...
• Giuseppe: @DominicvanEssen odd, I really thought I had seen this approach somewhere before but can't seem to track it down.
• Dominic van Essen: You might want to post it directly into the fibonacci challenge (as suggested by the author of the otherwise-shortest-but-actually-rather-longer answer...)
• Giuseppe: @DominicvanEssen posted!

Answer 6

Python 3, 41 bytes

f=lambda n,a=0,b=1:n-b and f(n,b,a+b)or a

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Answer 7

C(gcc), (削除) 57,55,53,50,45 (削除ここまで), 36 bytes

x;y;f(a){for(x=y=1;y-a-x;y+=x=y-x);}

Naive approach based on calculating all previous Fibonacci numbers, breaking when we reach the input.

Try it here

Edit: Thanks for Neil, golfing (削除) 2 (削除ここまで) 5 bytes.

Edit2: Thanks Dominic van Essen for golfing an other 5 bytes.

Edit3: Thanks att for golfing 9 bytes.

• rydwolf: Welcome to Code Golf, and nice answer!
• Neil: You already have the previous number, so you can just use y-a as the condition and return x, saving 2 bytes, but you can also use globals and for to save a few more bytes: x;y;c;f(a){for(x=y=1;y-a;c=y,y=x+y,x=c);return x;}.
• Dominic van Essen: 45 bytes with a rather hacky return trick...
• att: 36 bytes

Answer 8

Factor + math.unicode, (削除) 21 (削除ここまで) (削除) 19 (削除ここまで) (削除) 18 (削除ここまで) 14 bytes

[ φ / round ]

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-4 thanks to @DominicvanEssen!

Divide the input by the golden ratio and round the result.

• Dominic van Essen: Interesting fact: 1-phi (the golden ratio) is equal to 1/phi. So you can just divide by the golden ratio instead of multiplying by phi-1...
• Hedge Fleming: Does Factor need the whitespace? If not you could save a few bytes
• noodle person: @HedgeFleming Yes, it does. Each group of characters separated by space is a function.

Answer 9

Python 3, (削除) 29 (削除ここまで) (削除) 28 (削除ここまで) (削除) 27 (削除ここまで) 24 bytes

The footer uses @Noodle9's checker code .

1 byte saved thanks to @Chris!

3 more bytes off thanks to Albert.Lang!

Function that returns an integer-valued float. It fails for large inputs due to floating-point inaccuracies.

lambda n:~n*(1-5**.5)//2

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• Chris: 27 bytes, Try it online!
• Albert.Lang: ~n*(1-5**.5)//2 seems to work just as well.
• Rhaixer: lambda n:round(n/1.618)
• Luis Mendo: @py3programmer Although the challenge allows floating point inaccuracies, that is usually interpreted as inaccuracies inherent to floating-point computations. Defining the golden ratio with only three decimals is not covered by that allowance

Answer 10

Wolfram Language (Mathematica), 17 bytes

Round[2#/++√5]&

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• Greg Martin: haha ++√5 love it!

Answer 11

PowerShell, 31 bytes

[int]("$args"/1.61803398874989)

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The ratio of two adjacent numbers in the Fibonacci series rapidly approaches ((1 + sqrt(5)) / 2). So if N is divided by ((1 + sqrt(5)) / 2) and then rounded, the resultant number will be the previous Fibonacci number...

at least according to this article.

-16 thanks mazzy

• mazzy: 1. The constant 1.61803398874989 is shorter then the expression (1+[math]::sqrt(5))/2) :) 2. $x is redundant. 3. -replace'\..*' is shorter then [math]::round() :) Welcome to the CodeGolf
• mazzy: $args is an array. use "$args" to get a value.
• mazzy: see also Tips for golfing in PowerShell
• mazzy: Try it online! or try &{[int]("$args"/1.61803398874989)} 233 in your PS console
• Ruslan: 1.618033989 is shorter and is sufficient for all 47 possible inputs, here's a test.

Answer 12

Haskell, (削除) 38 (削除ここまで) 37 bytes

This works for an aribrary size of inputs. We iterate through the Fibonacci sequence until we find some number x, that is larger than half the input n. This is works because the previous fibonacci number is roughly 0.618*n. The Fibonacci list f was borrowed from R. Martinho Fernandes.

g n=[x|x<-f,2*x>=n]!!0
f=0:scanl(+)1f

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Answer 13

JavaScript, Node.js, 32 bytes

-1 thanks to Arnauld

n=>(y=1,f=x=>y-n?f(y,y+=x):x)(0)

Explanation:

This is a recursive solution, based off one of the two most common recursive fibonacci algorithms, which involves a function f(x, y) = f(y, x + y) that stops at some point. With this function, x is always a fibonacci number, and y is always the fibonacci number after it, so once y is the fibonacci number we were given as input, we know x is the one before it.

Answer 14

05AB1E, 4 bytes

ÅF ̈θ

Try it online or verify some more test cases.

̈θ could be a lot of different alternatives, like Á¤ or `\.

Explanation:

ÅF # Push a list of Fibonacci numbers lower than or equal to the (implicit) input
 ̈ # Remove the last one (the input)
 θ # Keep the next last one
 # (after which it is output implicitly as result)

• Sam: Oof, sorry about accidentally answering it the same way. Guess I hadn't refreshed the page since then. Nice to know my solution was at least somewhat good, haha :)

Answer 15

MathGolf, (削除) 6 (削除ここまで) 4 bytes

)φ/i

-2 bytes porting @emanresuA's Vyxal answer, which is in turn a port of @Unmitigated's JavaScript answer.

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Original 6 bytes answer:

╒fg<┤Þ

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g< could alternatively also be á>: try it online.

Explanation:

) # Increase the (implicit) input-integer by 1
 φ/ # Divide it by the golden ratio 1.618033988749895
 i # Truncate it to an integer
 # (the +1 and truncate act as a round builtin here, which MathGolf lacks)
 # (after which the entire stack is output implicitly as result)

╒ # Push a list in the range [1, (implicit) input-integer]
 f # Get the 0-based n'th Fibonacci number for each value in this list
 g # Filter it by:
 < # Where it's smaller than the (implicit) input-integer
 ┤ # Extract the trailing value
 Þ # Discard the list from the stack, keeping just that value
 # (after which the entire stack is output implicitly)

Answer 16

Python 3, (削除) 50 bytes (削除ここまで) 48 bytes

def f(n,a=0,b=1):
 while b<n:a,b=b,a+b
 return a

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• rydwolf: Welcome to Code Golf, and nice first answer!
• Kevin Cruijssen: Welcome to CGCC! You can save 2 bytes: by changing b!=n to b<n, and by removing the second line and using def f(n,a=0,b=1): instead. Enjoy your stay! Oh, and if you haven't seen it yet Tips for golfing in Python and Tips for golfing in 'all languages' might be interesting to read through. :)
• pxeger: You can also save some bytes by reducing your indentation to only one space per line, because that's all that is required in Python.
• axolotl: @pxeger The code had tabs for indent, so it wouldn't matter

Answer 17

Prolog (SWI), 35 bytes

1+0.
N+M:-nth1(M,_,_),X is N-M,M+X.

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-9 bytes thanks to Jo King

Of course, it's shorter to just use floating point:

Prolog (SWI), 31 bytes

N/M:-M is round(2*N/(1+5^0.5)).

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Answer 18

Retina 0.8.2, 26 bytes

.+
$*
(2円?(1円)|1)+1
1ドル2ドル
1

Try it online! Link includes test cases. Explanation: Loosely based on @MartinEnder's Retina answer to Am I a Fibonacci Number?.

.+
$*

Convert to unary.

(2円?(1円)|1)+1

Match as many Fibonacci numbers as possible. The first iteration only matches the leading 1 while subsequent iterations match iteration of the loop matches the penultimate Fibonacci number (if any), then the previous Fibonacci number, while saving it as the next penultimate Fibonacci number. The match actually sums the Fibonacci sequence, resulting in a sequence one less than the Fibonacci sequence, so a final 1 needs to be matched.

1ドル2ドル

Because the match is summing the sequence, the final capture of the sequence 1ドル is actually the Fibonacci number before the one we want, so we need to add on the penultimate matched Fibonacci number 2ドル.

1

Convert to decimal.

Answer 19

PARI/GP, 16 bytes

n->n*2\/(1+5^.5)

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Port of Unmitigated's JavaScript answer. a\/b is a short way to write round(a/b).

PARI/GP's floating point number is 128-bit by default. This gives the correct result until the 184th Fibonacci number (127127879743834334146972278486287885163).

Answer 20

Jelly, 4 bytes

‘:Øp

Add 1 to the input and integer-divide by the golden ratio.

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Answer 21

Husk, 5 bytes

→←xİf

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Makes use of the built-in list of Fibonacci numbers.

Explanation

→←xİf
 İf Take the list of Fibonacci numbers
 x Split it on occurrences of the input
 ← Get the first sub-list
→ Get the last element of that

Answer 22

Stax, 10 bytes

Çâ'Ç,1⁄4)í"─

Run and debug it

Approach

• Start with an infinite Fibonacci generator
• Keep values while they are less than the input, resulting in array e.g. [1, 1, 2, 3, 5].
• Extract last element.

• Razetime: 9 bytes with |5

Answer 23

K (ngn/k), (削除) 12 (削除ここまで) 11 bytes

--2!(1-%5)*

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Applies @Albert.Lang's approach from this comment.

• (1-%5)* calculate 1 minus the square root of 5 (i.e. -1.2360679774997898) and multiply it by the (implicit) input
• -2! integer divide by 2 (rounding towards 0, e.g. -8!-7 returns -1)
• - negate (and implicitly return)

Answer 24

APL(Dyalog Unicode), ^{(削除) (削除ここまで)}17 bytes ^SBCS

×ばつ1+5*.5}

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5

• pxeger: 5឵឵឵឵឵឵឵឵឵឵឵឵឵឵឵

Answer 25

Pip, 10 bytes

Ua*DRT5//2

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Answer 26

Fig, \5ドル\log_{256}(96)\approx\$ 4.116 bytes

_\mG}

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No round builtins, so I use the old increment and floor trick.

_\mG}
 } # The input number incremented
 \ # Divided by
 mG # The golden ratio
_ # Floor

Answer 27

Go, 59 bytes

import."math"
func f(n float64)float64{return Round(n/Phi)}

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wow built-in constants

Calculated version, 72 bytes

import."math"
func f(n float64)float64{return Round(2*n/(1+Pow(5,0.5)))}

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Answer 28

m68k assembly (GNU), 16 bytes

70 00 72 01 d0 81 c1 41 b2 af 00 04 66 f6 4e 75

Or in assembly:

begin:
 moveq.l #0, %d0;
 moveq.l #1, %d1;
loop:
 add.l %d1, %d0;
 exg.l %d0, %d1;
 cmp.l 4(%sp), %d1;
 bne loop;
 rts;

Works with both the Linux and the SysV calling conventions (parameters on stack, return in %d0, %d0-%d1 are scratch registers).

The c function definition is:

unsigned long prev_fibonacci(unsigned long num);

All calculations are done in 32 bit mode. 16bit is the same size, only faster.

Code execution takes 22 cycles plus 46 cycles per loop iteration. (On a 68000 w/ a 16bit bus and no waitstates)

Answer 29

x86-64 machine code, 10 bytes

Expects input in edx and outputs to eax.

b8 dd bc 1b cf f7 f0 d1 e8 c3 

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Explanation

Disassembly:

b8 dd bc 1b cf mov eax,0xcf1bbcdd
f7 f0 div eax
d1 e8 shr eax,1
c3 ret

The function approximates \$\operatorname{round}(\frac{n}{\phi})=\lfloor\frac{n+\frac{\phi}{2}}{\phi}\rfloor=\lfloor\frac{1}{2}\cdot\frac{2^{32}n+2^{31}\phi}{2^{31}\phi}\rfloor\$, where \$\phi=\frac{1+\sqrt{5}}{2}\$. \2ドル^{31}\phi\approx3474701533\$, which fits nicely into a 32-bit integer as 0xcf1bbcdd. Since div r32 works on edx:eax, n can be loaded into edx and 0xcf1bbcdd can be loaded into eax to create the numerator of the fraction, while also having eax as the denominator of the fraction after div eax. Afterwards, all that is left is to divide by two, which is handled by shr eax, 1.

Answer 30

Regex (ECMAScript), 153 bytes

(?=(x*).*(?=x{4}(x{5}(1円{5}))(?=2円*$)3円+$)(|x{4})(?=xx(x*)5円)4円(x(x*))(?=(6円*)7円+$)(?=6円*$8円)6円*(?=x1円7円+$)(x*)(9円x?)|)((5円10円)(9円10円12円?)?|x{5}|xx?x?)\B

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Takes its input in unary, as a string of x characters whose length represents the number. Returns its output as the length of the match.

This is based on Am I a Fibonacci Number? (161 bytes), with some modifications that take advantage of the difference between that problem and this one.

 # tail = N = input number; no need for an anchor, because all
 # Fibonacci inputs except for 1 will return a match
(?=
 (x*) # 1円+1 = potential number for which 5*(1円+1)^2 ± 4 is a
 # perfect square; this is true iff 1円+1 is a Fibonacci number,
 # which we shall call F_n. Outside the surrounding lookahead
 # block, 1円+1 is guaranteed to be the largest number for which
 # this is true such that 1円 + 5*(1円+1)^2 + 4 <= N.
 .*
 (?= # tail = (1円+1) * (1円+1) * 5 + 4
 x{4}
 ( # 2円 = (1円+1) * 5
 x{5}
 (1円{5}) # 3円 = 1円 * 5
 )
 (?=2円*$)
 3円+$
 )
 (|x{4}) # 4円 = parity - determined by whether the index of Fibonacci
 # number 1円+1 is odd or even;
 (?=
 xx # tail = arithmetic mean of (1円+1) * (1円+1) * 5 and 6円 * 6円
 # = ((F_n * F_n * 5) + (L_n * L_n)) / 2 = L_{2n}
 (x*)5円 # 5円 = floor(tail / 2) = floor(L_{2n} / 2)
 )
 4円
 # require that the current tail is a perfect square
 (x(x*)) # 6円 = potential square root, which will be the square root
 # after the following two lookaheads; 7円 = 6円-1
 (?=(6円*)7円+$) # 8円 = must be zero for 6円 to be a valid square root
 (?=6円*$8円)
 # 6円 is now the Lucas number L_n corresponding to the Fibonacci number F_n.
 6円*
 (?=x1円7円+$) # tail = F_{2n} = L_n * F_n = 6円 * (1円+1), where 6円 is larger
 (x*)(9円x?) # 9円 = floor(tail / 2);
 # 10円 = ceil(tail / 2); the remainder tail % 2 will always be
 # the same as the remainder discarded by 5,円 because
 # F_{2n} is odd iff L_{2n} is odd, thus this ceil()
 # can complement the floor() of 5円 when adding 5円 + 10円
| # Allow everything above to be skipped, resulting in all
 # capture groups being unset.
)
(
 # Note that if the above was skipped using the empty alternative in the lookahead,
 # the following will evaluate to 0. This relies on ECMAScript NPCG behavior.
 (5円10円) # 12円 = F_{2n+1} = (L_{2n} + F_{2n})/2 = 5円 + 10円; head=12円
 (
 9円10円 # head = F_{2n+2} = F_{2n} + F_{2n+1}
 # = 9円+10円 + 12円
 12円? # head = F_{2n+3} = F_{2n+2} + F_{2n+1} (optionally)
 # = head + 12円;
 # This is needed only for an input of N = 21, which is an
 # exception because 2^2*5+4 = 24 > 21, thus the algorithm
 # chooses 1^2*5+4 = 9 as its square instead. This won't break
 # any subsequent Fibonacci inputs, because executing this
 # optional will in those cases set tail = 0, breaking the
 # final assertion.
 )? # Do the above optionally
|
 x{5}|xx?x? # The cases 2→1, 3→2, 5→3, and 8→5 cannot be handled by our
 # main algorithm, so match them here.
)
\B # Assert head ≠ 0 and tail ≠ 0

Regex (ECMAScript), 162 bytes

\B(?=(x*).*(?=x{4}(x{5}(1円{5}))(?=2円*$)3円+$)(|x{4})(?=xx(x*)5円)4円(x(x*))(?=(6円*)7円+$)(?=6円*$8円)6円*(?=x1円7円+$)(x*)(9円x?)|)(9円10円(5円10円)12円?|x{21}|x{8}|x{5}|xx?x?)$

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This is an exact clone of Am I a Fibonacci Number? (161 bytes) with the ^ anchor replaced with \B, which is a shorthand for (?!^|$). So the regex engine goes forward one character at a time looking for a match, thus matching the largest Fibonacci number less than the input. As such this is much slower than the 153 byte version.

(What we'd really like is (?!^) instead of \B, so we could return an output of \0ドル\$ from an input of \1ドル\$, but that would take 3 more bytes, and the problem doesn't require handling \1ドル\$.)