Write the shortest code, in number of bytes, to display, return, or evaluate to the golden ratio (that is, the positive root of the quadratic equation: \$x^2-x-1=0\$, approximately 1.618033988749895), to at least 15 significant figures. No input will be given to your program.
Sample in Stutsk programming language:
1 100 { 1 + 1 swp / } repeat print
probably other languages too
.5+5**.5/2
.5+5^.5/2 will work in Octave/MATLAB, probably some others too
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tab include in length of program?
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Commented
Nov 19, 2020 at 9:22
GoldenRatio
This is the irrational number itself, not an approximation of it.
Examples (first 2 examples from Mathematica documentation)
FullSimplify[GoldenRatio^4 - GoldenRatio]
FullSimplify[GoldenRatio^20 + 1/GoldenRatio^20]
FullSimplify[GoldenRatio^2 - GoldenRatio - 1]
3 + Sqrt[5]
15127
0
PHP 17 chars
This one is just trolling, but hey.
1.618033988749895
As continued fraction:
{%x%x+1}/1
Or in closed form for 11:
%2%1+sqrt 5
-:1+%:5
some more text for the filter (my first J solution, heh)
-:1+%:5, the result is 1.61803. Is something more needed in the program (or system settings) to get the required "at least 15 significant figures"?
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2÷⍨1+√5
÷2÷1+√5
.5×1+√5
.5+√5÷4
Curses! I can't find a way to do it in less than 7 characters! Dialect is Nars2000.
5**.5/2+.5
This is the same as the Perl & Python submission - thanks to Redwolf Programs for telling me about this.
However, back in 2012, when this answer was originally written, the ** operator did not exist in JavaScript. While almost all browsers and do now support the exponentiation operator, according to Can I Use, as of July 2020, around 9% of users still does not support it, including the latest version of Internet Explorer. Thus, the old version of the answer:
JavaScript (backwards-compatible), 17 chars
Math.sqrt(5)/2+.5
** operator, so 5**.5/2+.5 would work too. Note that this is the same as the python submission.
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5X‚t;O
5 Push 5
X Push 1
‚ Pair: [5, 1]
t Square root: [2.23606797749979, 1.0]
; Halve: [1.118033988749895, 0.5]
O Sum: 1.618033988749895
+++++++[>+++++++<-]>.---.++++++++.-----.+++++++.--------.+++..++++++.-..-.---.+++++.-.+.----.
It works :/ Not very optimised though. I can’t think of any way to optimize this, although I’m sure a way exists.
Explanation:
+++++++ Add seven to cell at 0.
[ Begin a loop.
>+++++++ Add seven to cell at 1.
< Go back to cell 0.
- Decrement the counter there (soon it will reach 0).
] End loop when cell 0 reaches 0 after 7 repetitions.
> Go to cell 1 which is now 49.
. Print it (1).
---. Decrement 3 to get 46 (period) then print it.
Afterwards, not much explanation is needed. Just add some
or subtract some and repeat until everything is printed.
++++++++.-----.+++++++.--------.+++..++++++.-..-.---.+++++.-.+.----.
9227465/5702887
If all you need is enough precision for an IEEE 32 bit float, you can do it in 9 chars:
6765/4181
This will only work for languages that don't treat integer division specially.
9227465/5702887 produces only 13 correct digits - it differs on 14. digit.
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14930352/9227465 is probably the shortest, you can find it using optimal algorithm as advised on math.SE
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Fk5v1+2/
The value is on top of the stack - can be printed by adding p to the end of the program.
F pushes 15 on the stack (trick found here), k sets the precision to 15 digits. The rest is normal postfix notation :-) v is a square root. Trailing p for print was omitted.
p is not needed, because the requirement is to evaluate (not necessarily display) to 15 places.
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J, 10 9 8 chars
p.1,1,_1
(root of polynomial: -x^2+x+1)
>:@%^:_+1
(continued fraction (9 chars))
%:@>:^:_+1
(continued root: (10 chars))
phi
Just a builtin, too short to be compressed. Returns 1.61803398874989484820458683436563811.
A more interesting one. If running in the downloadable version, this will print n digits, given the command:
arn run file.arn -p n
or 25 digits if the -p flag is not provided.
If running in the online version, this will print the first 50 digits.
l[├Qn0
Unpacked: :-1+:/5
:- Halve
1 Literal one
+ Plus
:/ Square root of
5 Literal five
++. This increments variables too, which gives it a unique purpose.
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Commented
Nov 19, 2020 at 16:55
(floating-point; accurate to 15 significant figures)
½→√5
This seems (to me) to be surprisingly readable for a golfing language...
√5 # square root of 5
→ # increment
½ # halve
(calculation as arbitrary precision rational number)
!Ẋ/İf!5İ⁰
Try it online! (TIO header converts the rational number [expressed in Husk as a fraction] to its first 1000 decimal digits)
/ # get the ratio of
Ẋ # every pair of elements of
İf # the fibonacci sequence;
! # now select the ratio at position
!5İ⁰ # 10^5 (change to !9İ⁰ for more accuracy with same byte-count)
(1+1%)/1
Adapted from the built-in docs.
(...)/1 set up a converge-reduce, seeded with 1 and run until two successive iterations return the same result
(1+1%) add one to the inverse of the current value, and feed that into the next iteration
(5**0.5)/2+0.5
Based on the Javascript Perl answer above.
5**0.5 is shorter than Math.sqrt(5).
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Commented
Sep 27, 2012 at 14:14
sqrt was....
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Commented
Sep 27, 2012 at 14:21
.5+5^.5/2
In R, evaluates full double precision. More digits can be seen by setting options(digits=99). The question says "evaluate", so that goes with the rules.
= sign before this.
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Hexdump:
b1 7f d9 e8 d9 e8 de c1 d9 fa e2 f8 c3
Disassembly:
100139D0 B1 7F mov cl,7Fh
100139D2 D9 E8 fld1
again:
100139D4 D9 E8 fld1
100139D6 DE C1 faddp st(1),st
100139D8 D9 FA fsqrt
100139DA E2 F8 loop again (100139D4h)
100139DC C3 ret
Uses the converging sequence an = sqrt(an-1 + 1).
The number of iterations is determined by the contents of ecx, which is mostly garbage. The minimal number of iterations is 127, which guarantees good precision (actually, 30 iterations should be enough). In the worst case, the calculation will take a few minutes (232-1 iterations), and in the best case, it's instantaneous (127 iterations).
(1+5**.5)/2
= 1.618033988749895
Saved 4 bytes!
After posting this answer I noticed the example Stutsk program and realized that I could probably save a few bytes. My new answer is based off the example given in the question. This program works because the golden ration can be expressed as a continued fraction.
golden_ratio = 1+1/(1+1/(1+1/...))
ff*101.;n~<
$1$,1+$:?!^1-
f201.;n,2+1~<
$:5$,+2,$:?!^1-
The second line approximates the sqrt of 5. After looping 15 times, this value is used to calculate the golden ratio.
15k5v2/.5+p
The most character-consuming task is setting the decimal precision..
N[x/.Solve[x^2-x-1==0][[2]],16]
1.618033988749895
(It's going to be the longest code, I expect...:)
φ
Builtins ftw ¯\_(ツ)_/¯
Without builtins it's 6 bytes:
51α√½Σ
Explanation:
φ # Push golden ratio builtin 1.618033988749895
# (output the entire stack joined together implicitly as result)
5 # Push 5
1 # Push 1
α # Wrap the last two values into a list: [5,1]
√ # Take the square-root of each value: [2.23606797749979,1.0]
½ # Halve each: [1.118033988749895,0.5]
Σ # And sum this list: 1.618033988749895
# (after which the entire stack joined together is output implicitly as result)
X5mq+2/
Just one more byte than 05AB1E! Pretty simple stack-based translation of the equation on the Wikipedia page:
X5 Push 1 and 5 on to the stack
mq Square root the top number
+ Add the top two numbers on the stack
2/ Divide by two
(implicit) output the stack
w 5**.5+1/2
Output: 1.618033988749894849
This highlights a quick about MUMPS: order of operations is evaluated left to right. Something like .5+5**.5/2 would give us (5.5**.5)/2 (1.172603939955857389).
