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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.
Edit: -1 byte thanks to pajonk, then -2 bytes thanks to att
\(x)(x*5^.5-x+1)%/%2The 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)
.5 instead of your clever way to add 1)...
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May 26, 2022 at 10:43
\(x){while(x>T)T=F+(F=T);F}Similar to this answer, iterates through the Fibonacci sequence until T==x and returns F.
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
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 47th Fibonacci number, 2971215073. It can be expanded to work 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.
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).
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May 26, 2022 at 21:10
add edx,0. It changes the output for an input of 1 to 0 instead of 1, but those are both equally valid.
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math.unicode, [ φ / round ]
-4 thanks to @DominicvanEssen!
Divide the input by the golden ratio and round the result.
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.
Edit: Thanks for Neil, golfing 2 5 bytes.
Edit2: Thanks Dominic van Essen for golfing an other 5 bytes.
Edit3: Thanks att for golfing 9 bytes.
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;}.
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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
~n*(1-5**.5)//2 seems to work just as well.
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May 26, 2022 at 17:46
[int]("$args"/1.61803398874989)
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
1.61803398874989 is shorter then the expression (1+[math]::sqrt(5))/2) :) 2. $x is redundant. 3. -replace'\..*' is shorter then [math]::round() :) Welcome to the CodeGolf
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$args is an array. use "$args" to get a value.
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&{[int]("$args"/1.61803398874989)} 233 in your PS console
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1.618033989 is shorter and is sufficient for all 47 possible inputs, here's a test.
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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
-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.
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)
)φ/i
-2 bytes porting @emanresuA's Vyxal answer, which is in turn a port of @Unmitigated's JavaScript answer.
Original 6 bytes answer:
╒fg<┤Þ
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)
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. :)
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May 27, 2022 at 6:42
1+0.
N+M:-nth1(M,_,_),X is N-M,M+X.
-9 bytes thanks to Jo King
Of course, it's shorter to just use floating point:
N/M:-M is round(2*N/(1+5^0.5)).
.+
$*
(\2?(\1)|1)+1
$1$2
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?(\1)|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$2
Because the match is summing the sequence, the final capture of the sequence $1 is actually the Fibonacci number before the one we want, so we need to add on the penultimate matched Fibonacci number $2.
1
Convert to decimal.
n->n*2\/(1+5^.5)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).
→←xİf
Makes use of the built-in list of Fibonacci numbers.
→←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
--2!(1-%5)*
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)
_\mG}
No round builtins, so I use the old increment and floor trick.
_\mG}
} # The input number incremented
\ # Divided by
mG # The golden ratio
_ # Floor
import."math"
func f(n float64)float64{return Round(n/Phi)}
wow built-in constants
import."math"
func f(n float64)float64{return Round(2*n/(1+Pow(5,0.5)))}
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)
Expects input in edx and outputs to eax.
b8 dd bc 1b cf f7 f0 d1 e8 c3
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^{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.
