Половина, половина и половина

33

Рассмотрим следующую числовую последовательность:

0,12,14,34,18,38,58,78,116,316,516,716,916,1116,1316,1516,132,332,532,

Он перечисляет все двоичные дроби в единичном интервале .[0,1)

(Чтобы облегчить эту задачу, первый элемент является необязательным: вы можете пропустить его и считать, что последовательность начинается с 1/2.)

задача

Напишите программу (полную программу или функцию), которая ...

Выберите один из следующих вариантов поведения:

  • Введите n, выведите n-й элемент последовательности (0-индексированный или 1-индексированный);
  • Введите n, выведите первые n элементов последовательности;
  • Ничего не вводите, выводите бесконечную последовательность чисел, которую вы можете взять один за другим;

правило

  • Ваша программа должна как минимум поддерживать первые 1000 элементов;
  • Вы можете выводить десятичные дроби или дроби (встроенные, целочисленные пары, строки), как вам нравится;
    • Ввод / вывод в виде двоичных цифр в этом вопросе не допускается;
  • Это , самые короткие коды выигрывают;
  • Стандартные лазейки запрещены.

Testcases

input output
1     1/2     0.5
2     1/4     0.25
3     3/4     0.75
4     1/8     0.125
10    5/16    0.3125
100   73/128  0.5703125
511   511/512 0.998046875
512   1/1024  0.0009765625

Эти примеры основаны на 0-индексированной последовательности с включением 0 в начале. Вам нужно будет настроить вход для соответствия вашему решению.

Подробнее

  • OEIS A006257
    • Задача Иосифа: . (Ранее M2216)a2n=2an1,a2n+1=2an+1
    • 0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, ...
  • OEIS A062383
    • : для п > 0 , п = 2 л о г 2 н + 1 или п = 2 пa0=1n>0an=2log2n+1.an=2an2
    • 1, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, ...
  • A006257 (n) / A062383 (n) = (0, 0,1, 0,01, 0,11, 0,001, ...) перечисляет все двоичные дроби в единичном интервале [0, 1). - Фредрик Йоханссон, 14 августа 2006 г.

ТТГ
источник
4
« Ничего не вводить , выводить последовательность бесконечных чисел одну за другой » Должен ли он быть один за другим или нам также разрешено выводить бесконечный список (возможно в Haskell, Elixir, 05AB1E и т. Д.)?
Кевин Круйссен
Могу ли я вывести список строк? напр."1/2" "1/4" "1/8"...
Барранка
@KevinCruijssen Бесконечный список хорош, если вы можете taken элементов из него позже.
18:00
@ Барранка Я думаю, что это приемлемо. Это ничто не отличается для печати фракций на стандартный вывод.
18:00
Когда вы говорите, что ввод / вывод в виде двоичных чисел недопустим , вы имеете в виду, что мы не можем написать функцию, которая возвращает пару, если ints, или a doubleв языке / реализации, где doubleиспользуется формат двоичных символов IEEE ? Надеюсь, вы не имеете в виду, нужно ли было анализировать строку ASCII, если мы хотим получить целочисленный ввод? Обычные целочисленные типы являются двоичными в таких языках, как C. Или вы имеете в виду, что ввод / вывод не может быть массивом или строкой целых или ASCII нулей / единиц?
Питер Кордес

Ответы:

22

Haskell , 25 байт

pred.until(<2)(/2).(+0.5)

Попробуйте онлайн!

Вывод десятичных дробей, индексируемых без начального нулевого члена.

Добавляет 0,5 к входу, затем делит пополам до тех пор, пока результаты не станут меньше 2, затем вычитает 1. Использование выражения без точек экономит на 1 байт более

f n=until(<2)(/2)(n+0.5)-1
XNOR
источник
11

Java 10, 68 64 байта

Сначала попробуйте код гольф!

Вариант 1: найти n-й элемент (1-индексированный)

-4 байта благодаря @Kevin Cruijssen

n->{int x=0;for(;n>>++x!=1;);return((~(1<<x)&n)*2.+1)/(1<<x+1);}

This is an anonymous method that finds the n-th term by removing the most significant bit from n, doubling it and adding one, then dividing by the next highest power of 2.

Try it online!

Code walkthrough:

n->{                      // builds anonymous function with input n
int x=0;                  // stores floor of log(n) (base 2) for most significant digit
for(;n>>++x!=1;);         // calculates floor of log(n) by counting right shifts until 1
return((~(1<<x)&n)        // removes most significant digit of n
*2.+1)                     // multiplies 2 and adds 1 to get the odd numerator
/(1<<x+1);}               // divides by the next highest power of 2 and returns`

Will edit if it's necessary to print the final value instead of returning it.

TCFP
источник
Welcome to PPCG, good to have you with us :)
Shaggy
Привет, добро пожаловать в PPCG! Отличный первый ответ, +1 от меня. В настоящее время это тот же счетчик байтов, что и в моем ответе на Java, но вы все равно можете использовать некоторые части своего ответа, чтобы сделать его короче моего: {}цикл после может быть ;вместо; Вы можете удалить пробел после return; 2.0может быть 2.; И изменяя n>>x!=1;x++, 1<<xи 1<<x+1к n>>x++!=1;, 1<<x-1, 1<<xсоответственно , также сохраняет байт. Попробуйте онлайн: 64 байта . Приятного пребывания!
Кевин Круйссен
Oh, and if you haven't seen it yet: Tips for golfing in Java and Tips for golfing in <all languages> are both pretty interesting to read through. :)
Kevin Cruijssen
See my 30 bytes answer, originally based on yours but golfed and golfed and golfed.
Olivier Grégoire
9

MathGolf, 5 4 bytes

╫\╨]

Try it online!

How it would look like with the operator working correctly

╫\)╨]   (")" adds 1 to TOS, making rounding behave as expected)

Try it online!

Explanation

╫     Left-rotate all bits in input
 \    Swap top two elements on stack, pushing the input to the top
  ╨   Round up to nearest power of 2
   ]  Wrap in array (just for pretty printing)

I took my inspiration from this question to solve the problem, my "own" solution was around 10-12 bytes I think.

I had intended for the round up to closest power of 2 to return the number itself if it was a number of two, but due to a mistake it rounds to the next power of two (e.g. 4 -> 8 instead of 4 -> 4). This will have to be fixed later, but now it saves me one byte.

maxb
источник
2
I don't know MathGolf but if the ] serves no other purpose than formatting the output, I'd say you don't need to include it in your byte count.
Shaggy
2
I was unsure about it. Since the stack is printed on output as a joined string, it outputs a stack with the numbers 1 and 2 as 12. If that still counts I'll remove a byte
maxb
I think you should leave it in. It sometimes saves a byte to output the stack as one string, sometimes it will cost you a byte.
H.PWiz
@H.PWiz that was my original thinking, since it seems fair to use the strengths of your language. Some languages only print the top of the stack when finished, some print it as a list. Usually it's a 1 byte difference, but it's part of the challenge.
maxb
8

Java 10, 89 85 70 69 68 bytes

v->{for(float j,t=2;;t*=2)for(j=1;j<t;j+=2)System.out.println(j/t);}

Port of @Emigma's 05AB1E answer, so outputs decimals indefinitely as well.
-15 bytes thanks to @Arnauld.

Try it online.

Explanation:

v->{                      // Method with empty unused parameter and no return-type
  for(float j,t=2;;       //  Loop `t` from 2 upwards indefinitely,
                   t*=2)  //  doubling `t` after every iteration
    for(j=1;j<t;          //   Inner loop `j` in the range [1, `t`),
                j+=2)     //   in steps of 2 (so only the odd numbers)
      System.out.println( //    Print with trailing new-line:
        j/t);}            //     `j` divided by `t`
Kevin Cruijssen
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1
How often can I say that I half your byte count? Well, I think this is the first time ;-)
Olivier Grégoire
@OlivierGrégoire Dang, now that is an impressive Java answer. :) I saw your 37-byte version as comment on TCFP's answer, but you even stripped off more bytes. It looks so extremely simple now in your 30-byte version, but it's still ingenious how you've golfed it from the initial version. Well done!
Kevin Cruijssen
7

Java (JDK 10), 30 bytes

n->(n+.5)/n.highestOneBit(n)-1

Try it online!

Returns the nth item in the sequence.

This answer is originally a succession of golfs of TCFP's Java answer. In the end, the golfs didn't look like the original answer anymore (though the math used is the same) so I decided to post the golfs as a separate answer instead of simply commenting on the TCFP's answer. So if you like this answer, go upvote TCFP's answer as well! ;-)

Intermediate golfs were:

n->{int x=0;for(;n>>++x!=1;);return((~(1<<x)&n)*2.+1)/(1<<x+1);} // 64 bytes (TCFP's answer when I started golfing)
n->{int x=0;for(;n>>++x!=1;);x=1<<x;return((~x&n)*2.+1)/x/2;}    // 61 bytes
n->{int x=n.highestOneBit(n);return((~x&n)*2.+1)/x/2;}           // 54 bytes
n->{int x=n.highestOneBit(n);return((~x&n)+.5)/x;}               // 50 bytes
n->((n&~(n=n.highestOneBit(n)))+.5)/n                            // 37 bytes
n->(n-(n=n.highestOneBit(n))+.5)/n                               // 34 bytes
n->(n+.5)/n.highestOneBit(n)-1                                   // 30 bytes, current score
Olivier Grégoire
источник
And I was sitting here thinking my answer was as short as I could make it, you come along and cut it by more than half! Amazing stuff, definitely deserves a +1 from me.
TCFP
@TCFP It was an iterative process over the span of several hours. I actually posted each intermediate golfing as a comment to your answer, but deleted them as I found better golfs. Thanks for the praise ;-)
Olivier Grégoire
6

05AB1E, 11 8 bytes

Saved 3 bytes thanks to Kevin Cruijssen.

∞oDÅÉs/˜

Try it online!

Explanation

∞         # start an infinite list [1...
 o        # calculate 2**N
  D       # duplicate
   ÅÉ     # get a list of odd numbers up to 2**N
     s/   # divide each by 2**N
       ˜  # flatten
Emigna
источник
1
-1 byte by using (infinite list starting at 1): ∞oεDÅÉs/}˜
Kevin Cruijssen
@KevinCruijssen: Cool! That's a command I hadn't seen before. Thanks :)
Emigna
1
Ah, and nice way of saving two more bytes due to the implicit mapping.. Hadn't even thought about that, lol..
Kevin Cruijssen
1
:O how is this possible. You condensed the ~2 page question into 8 bytes.
Cullub
I was thinking using the prime numbers and a list of [1,2,4,4,8,8,8,8,16,16,...,2**n] and prepending the correct indexed prime followed by a /... But that wasn't working so well. Well, but not 8-bytes well. Something like 9LoDÅP)ζ.
Magic Octopus Urn
5

PowerShell, 40 bytes

for($i=2;;$i*=2){1..$i|?{$_%2}|%{$_/$i}}

Try it online!

Outputs the infinite sequence as decimal values. Given language limitations, will eventually run into precision problems, but easily handles the first 1000 entries.

Starts by setting $i=2, then enters a for loop. Each iteration, we construct a range from 1..$i and pull out the odd values with |?{$_%2}. Those are fed into their own inner loop, where we divide each to get the decimal |%{$_/$i}. Those are left on the pipeline and output when the pipeline is flushed after every for iteration. Each iteration we're simply incrementing $i by $i*=2 to get the next go-round.

AdmBorkBork
источник
5

Haskell, 35 32 bytes

Edit: -3 bytes thanks to @Delfad0r.

[(y,2^x)|x<-[1..],y<-[1,3..2^x]]

This is an infinite list of integer pairs.

Try it online!

nimi
источник
5

Haskell, 40 bytes

s=(1,2):[(i*2+u,j*2)|(i,j)<-s,u<-[-1,1]]

Try it online!

Infinite sequence as pairs of integers (starting from (1,2)).

Quite a bit longer than @nimi's answer, but the approach is completely different, so I decided to post it anyway.

This solution is based on the following observation.

Consider the infinite sequence

{12,14,34,18,38,58,78,116,316,}
and apply the following steps.

  • Replace every number ij in the sequence with the list {2i12j,2i+12j}:
    {{14,34},{18,38},{58,78},{116,316},}
  • Join all the lists into a single sequence:
    {14,34,18,38,58,78,116,316,}
  • Add 12 at the beginning of the sequence:
    {12,14,34,18,38,58,78,116,316,}

Notice how you get back to the sequence you started with!

The solution exploits this fact (together with Haskell's laziness) to compute the sequence s.

Delfad0r
источник
4

Python 2 - 68 66 bytes

-2 bytes thanks to Kevin

from math import*
def g(n):a=2**floor(log(n,2));print(n-a)*2+1,2*a

Try it Online!

Don Thousand
источник
You can golf 1 byte by changing return 2*(n-a) to return(n-a)*2. And you could save an additional byte by using Python 2 instead of 3, so return can be print (with parenthesis).
Kevin Cruijssen
2
@KevinCruijssen For someone who doesn't use Python, you certainly are a better Python programmer than me.
Don Thousand
Hehe. :D Simple things like this comes with experience I guess. I'm on this site for about two years now I think (EDIT: Since April 2016). I sometimes even suggests things to golf for answers that are written in languages I had never seen before.. Some basic things work in most languages. For example, last week I suggested a golf for a T-SQL answer, and I once suggested a golf in a Red answer. xD
Kevin Cruijssen
2
44 bytes using len and bin instead of log.
ovs
@ovs 42 bytes.
Jonathan Frech
4

Python 3, 53 51 bytes

  • Saved two bytes thanks to mypetlion; reusing default parameters to reset n.
def f(m=2,n=1):n<m and print(n/m)&f(m,2+n)or f(m+m)

Try it online!

Jonathan Frech
источник
Swap the parameters to save 2 bytes: def f(m=2,n=1):n<m and print(n/m)&f(m,n+2)or f(m+m)
mypetlion
1
@mypetlion Cool, thank you!
Jonathan Frech
4

R, 42 bytes

function(n)c(y<-2^(log2(n)%/%1)*2,2*n-y+1)

Try it online!

Returns a pair Denominator,Numerator. Uses the formula N=2(n2log2(n))+1 from the Josephus sequence and D=2log2(n)+1 from the other sequence. Happily we are able to re-use the denominator as the two formulas have quite a lot in common!

Giuseppe
источник
3

Racket, 92 91 bytes

(define(f k(m 2)(n 1))(if(> k 0)(if(=(+ n 1)m)(f(- k 1)(+ m m))(f(- k 1)m(+ n 2)))(/ n m)))

Try it online!

  • Saved a byte thanks to Giuseppe -- removing superfluous whitespace.
Jonathan Frech
источник
3

MATL, 8 bytes

BnWGEy-Q

Try it online!

Returns Numerator, then Denominator. Uses the same method as my R answer, although it's a bit more efficient.

Explanation, with input 5:

           # implicit input 5
B          # convert to array of bits
           # STACK: [[1 0 1]]
n          # length (place of Most Significant Bit)
           # STACK: [3]
W          # elementwise raise 2^x
           # STACK: [8]
G          # paste input
           # STACK: [8, 5]
E          # double
           # STACK: [8, 10]
y          # copy from below
           # STACK: [8, 10, 8]
-          # subtract
           # STACK: [8, 2]
Q          # increment
           # STACK: [8, 3]
           # implicit end of program, display stack contents
Giuseppe
источник
2

Shakespeare Programming Language, 426 bytes

,.Ajax,.Ford,.Act I:.Scene I:.[Exeunt][Enter Ajax and Ford]Ajax:You be the sum ofyou a cat.Ford:You cat.Scene V:.Ford:Is twice you nicer I?If solet usScene X.You be twice you.Let usScene V.Scene X:.Ford:Remember twice you.You be the sum oftwice the remainder of the quotient betweenI you a cat.Open heart.You big big big big big cat.Speak thy.Recall.Open heart.You be twice the sum ofa cat a big big cat.Speak thy.Let usAct I.

Try it online!

Outputs the sequence infinitely as both numbers separated by a space, with each item being separated by a newline.

JosiahRyanW
источник
Love it. Lol You be twice the sum of a cat
Cullub
Actually, it's "twice the sum ofa cat a big big cat" (i.e. 10 for some reason).
JosiahRyanW
2

Python 2, 44 bytes

def f(n):m=2**len(bin(n))/4;return 2*n-m+1,m

Try it online!

Function returns a tuple of (numerator, denominator). An input of 0 is not handled (it was optional).

Chas Brown
источник
1
return 2*n-m+1,m can be print-~n+n-m,m to save 2 bytes.
Kevin Cruijssen
2

Excel 48 28 Bytes

Saved 20 bytes (!) thanks to tsh

=(A1+0.5)/2^INT(LOG(A1,2))-1

=MOD(A1+0.5,2^(INT(LOG(A1,2))))/2^INT(LOG(A1,2))

Assumes value in A1, output is in decimal. If you want the output to be in fraction, you can create a custom format for the output cell as "0/###0" and it will show it as fraction.

Explanation: Difficult to explain, since there is a shortcut taken to get to this formula. Basically the numerator is a bit shift left of the input, and the denominator is the next power of 2 higher than the number input.

I originally started with Excel built in functions for BITLSHIFT and BITRSHIFT, but they will shift the entire 48 bits which is not what you want. The functions DEC2BIN (and BIN2DEC) have a limit of -512 to 511 (10 bits) so this wouldn't work. Instead I had to rebuild the number with a modulus of the original number, then times two, then add 1 (since the left digit would always be 1 before a shift).

=MOD(A1                        Use MOD for finding what the right digits are
       +0.5                    this will later add the left "1" to the right digits
           ,2^INT(LOG(A1,2)))) Take Log base 2 number of digits on the right
                               this creates the numerator divided by 2 (explained later)
/ 2^INT(LOG(A1,2))             The denominator should be 2^ (Log2 + 1) but instead of 
                               adding a 1 here, we cause the numerator to be divided by 2 instead
                               This gives us a fraction.  In the numerator, we also added .5
                               instead of 1 so that we wouldn't need to divide it in both the
                               numerator and denominator
Then tsh showed how I could take the int/log out of the mod and remove it from numerator/denominator. 

Examples: enter image description here

Keeta
источник
What about =(A1+0.5)/2^INT(LOG(A1,2))-1?
tsh
2

C++, 97 75 71 bytes

-26 bytes thanks to tsh, ceilingcat, Zacharý

float f(int i){float d=2,n=1;while(--i)n=d-n==1?d*=2,1:n+2;return n/d;}

Testing code :

std::cout << "1\t:\t" << f(1) << '\n';
std::cout << "2\t:\t" << f(2) << '\n';
std::cout << "3\t:\t" << f(3) << '\n';
std::cout << "4\t:\t" << f(4) << '\n';
std::cout << "10\t:\t" << f(10) << '\n';
std::cout << "100\t:\t" << f(100) << '\n';
std::cout << "511\t:\t" << f(511) << '\n';
std::cout << "512\t:\t" << f(512) << '\n';
HatsuPointerKun
источник
You may just omit if(!i)return 0; since 0 is not required in the challenge.
tsh
1
When trying to golf C-like language. You should avoid using while but try for. for(;exp;) is as same as while(exp) but you can write two more other statement into it. Prefer ?: instead of if else, which would be shorter in most cases.
tsh
1
I don't think you need the (...) around d-n-1.
Zacharý
1

JavaScript (ES6), 44 bytes

Returns the n-th term, 1-indexed.

f=(n,p=q=1)=>n?f(n-1,p<q-2?p+2:!!(q*=2)):p/q

Try it online!

Arnauld
источник
1

><>, 19 18 bytes

Using xnor's idea, fixed by Jo King, -1 byte by making better use of the mirrors and another -2 bytes by Jo King because the ! was superfluous and ; is not required.

2*1+\1-n
2:,2/?(

Try it online!

PidgeyUsedGust
источник
You should be checking if it is smaller than 2 first, otherwise the first element is -0.25. Fix for the same amount of bytes
Jo King
Thanks! I also managed to remove another byte by reusing the mirrors.
PidgeyUsedGust
Why did you invert the condition? 16 bytes
Jo King
Didn't notice that it would continue the loop. Are we allowed to not properly finish?
PidgeyUsedGust
Yeah, terminating with an error is fine as long as the OP doesn't specify otherwise
Jo King
1

APL (Dyalog Unicode), 15 bytes

1-⍨.5∘+÷2*∘⌊2⍟⊢

Try it online!

Anonymous prefix lambda.

Thanks to Adám for 4 bytes and to Cows quack for 2 bytes.

How:

1-⍨.5∘+÷2*∘⌊2⍟⊢  Anonymous lambda, argument   10
            2⍟⊢  Log (⍟) of  in base 2. 210  3.32192809489...
                 Floor. 3.32192809489...  3
        2*∘       Take that power of 2. 2³  8
       ÷          Use that as denominator
   .5∘+            + 0.5  10.5. Using that as numerator: 10.5÷8  1.3125
1-⍨               Swap the arguments (⍨), then subtract. 1-⍨1.3125  1.3125-1  0.3125
J. Sallé
источник
1

C# (.NET Core), 69 bytes

a=>{int b=1,c=2;while(a-->1){b+=2;if(b>c){b=1;c*=2;}}return b+"/"+c;}

Try it online!

Ungolfed:

a=> {
    int b = 1, c = 2;   // initialize numerator (b) and denominator (c)
    while (a-- > 1)     // while a decrements to 1
    {
        b += 2;         // add 2 to b
        if (b > c)      // if b is greater than c:
        {
            b = 1;      // reset numerator to 1
            c *= 2;     // double denominator
        }
    }
    return b + "/" + c; // return fraction as string
}
Meerkat
источник