Repeat average‎

Question

Background

It was a normal presentation that I were in as a audience, until the presenter gave a math problem about repeat taking 2 number out of a list a replacing them with average, claiming that there will be something special about it, and our math teacher, sitting at the end of classroom, exciting rushed to me, and tasked me to code that out.

Task

2 input length and generation

Generate a range, start from 0, with length of length
Randomly choose 2 number (or item) X and Y and replace both X and Y with the average of X and Y
Repeat step 2 for generation times
Output minimum, maximum, average, and the processed list.

Example

I'll let this semi-golfed python code explain itself :)

import random
a,b=map(int,input().split())
l=[*range(a)]
s=lambda:random.randint(0,a-1)
for i in range(b):
    x,y=s(),s();l[x],l[y]=[(l[x]+l[y])/2]*2
print(min(l),max(l),sum(l)/a,l)

Rules

Standard rules applies
Input can be in any convenience form.
As long as you output is clean, understandable, readable, it is allowed.
The two random value X,Y should be uniform statistically independent random by itself(X and Y are not affecting each other no matter what)

what you should do?

Start golfing right now!

Technically the procedure doesn't change the average, it's always (length-1)/2, isn't it? — Neil
– Neil, Commented Dec 23, 2021 at 0:04
@Neil exactly yes, you are smart, you can also prove it with math. — okie
– okie, Commented Dec 23, 2021 at 0:07
"Randomly choose 2 number" - this needs further specification. Does it have to be a uniform random distribution? Must they be independently chosen? — pxeger
– pxeger, Commented Dec 23, 2021 at 1:27

caird coinheringaahing · Accepted Answer · 2021-12-23 00:32:24Z

Jelly, 25 bytes

ḶJX,XƊṬ×ịÆm¥oʋƊ⁹¡µṂ;Ṁ;Æm,

Try it online!

Outputs [[min, max, mean], list]. Takes length as the first argument, and generation as the second

How it works

ḶJX,XƊṬ×ịÆm¥oʋƊ⁹¡µṂ;Ṁ;Æm, - Main link. Takes L on the left, G on the right
Ḷ                         - [0, 1, ..., L-1]
              Ɗ           - Previous 3 links as a monad f(R):
 J                        -   Indices of R
     Ɗ                    -   To the indices of R:
  X                       -     Randomly choose one
    X                     -     Randomly choose one again
   ,                      -     Pair; [i, j]
             ʋ            -   Last 4 links as a dyad g([i, j], R):
      Ṭ                   -     Boolean array with 1s at i and j
           ¥              -     Last 2 links as a dyad h([i, j], R):
        ị                 -       Get R[i] and R[j]
         Æm               -       Take the mean
       ×                  -     Replace the 1s with the mean
            o             -     Or; replace the elements at i and j in R with the mean
               ⁹¡         - Apply f(R) G times
                 µ        - Use this list L' as the new argument
                  Ṃ       - Minimum
                    Ṁ     - Maximum
                      Æm  - Mean
                   ; ;    - [Min, Max, Mean]
                        , - Pair with L'

Graham · Accepted Answer · 2021-12-23 08:09:52Z

3

APL+WIN, 60 bytes

Prompts for length and generation. Index origin = 0

v←⍳l←⎕⋄⍎∊⎕⍴⊂'v[i]←(+/v[i←(?l),?l])÷2⋄'⋄(⌊/v),(⌈/v),(+/v÷l),v

Try it online!Thanks to Dyalog Classic

The results in TIO are output in the order listed in the question. The second example in TIO shows that by generation=1000 all elements of the output have converged to the mean in that particular case.

answered Dec 23, 2021 at 8:09

Graham

7,8061 gold badge15 silver badges13 bronze badges

Add a comment |

lyxal · Accepted Answer · 2021-12-23 00:24:59Z

Vyxal, 25 bytes

ʁ?(₅‹₍℅℅~İṁ£(n¥Ȧ)):₌₍GgṁW

Try it Online!

25 bytes of fun. Takes length then generations and outputs [list, [Max, Min], average]. Can be 24 bytes with the W flag

Explained

ʁ?(₅‹₍℅℅~İṁ£(n¥Ȧ)):₌₍GgṁW
ʁ                          # The range [0, length) - this will be the list we modify
 ?(                        # (generation) times:
   ₅‹                      #   push the top of the stack, and its length + 1
     ₍℅℅                   #   choose two random items from the range [0, length) and place them into a list
        ~İ                 #   without popping anything, get the item at each randomly chosen index from the top list - this leaves [list, indexes, items] on the stack
          ṁ£               #   place the average of those two items into the register
            (...)          #   for each index i in the indexes list:
             n¥Ȧ           #      list[i] = register (the mean)
                 )         # close the main for loop
                  :        # now the top of the stack is the list with all the replaced items. We duplicate it so we can extract the juicy info from it
                   ₌₍Ggṁ   # Push a list of [max, min] and push the mean of that list
                        W  # Wrap everything into a single list and implicitly print

the length, the mighty lyxal god, you forgot about the length!(the list length your code generated is length+1) — okie
– okie, Commented Dec 23, 2021 at 0:13

Arnauld · Accepted Answer · 2021-12-23 00:35:41Z

2

JavaScript (ES6), 145 bytes

Expects (length)(generation). Returns [min, max, avg, list].

with(Math)f=(n,a=[k=n],R=random)=>g=m=>k?g(m,a[--k]=k):m?g(m-1,a[i=R()*n|0]=a[j=R()*n|0]=(a[i]+a[j])/2):[min(...a),max(...a),eval(a.join`+`)/n,a]

Try it online!

edited Dec 23, 2021 at 0:35

answered Dec 23, 2021 at 0:29

Arnauld

206k21 gold badges187 silver badges670 bronze badges

\$\begingroup\$ you forgot the statics about the processed list (average,min,max) \$\endgroup\$

okie
– okie

2021-12-23 00:31:34 +00:00
Commented Dec 23, 2021 at 0:31
\$\begingroup\$ @okie Ugh. I missed that part entirely ... :-/ Edit: now fixed. \$\endgroup\$

Arnauld
– Arnauld

2021-12-23 00:32:11 +00:00
Commented Dec 23, 2021 at 0:32
\$\begingroup\$ Maybe you can simplify eval(a.join`+`)/n to ~-n/2. \$\endgroup\$

tsh
– tsh

2021-12-23 03:08:23 +00:00
Commented Dec 23, 2021 at 3:08

Add a comment |

hyperneutrino · Accepted Answer · 2021-12-23 01:33:23Z

2

Python, 134 bytes

from random import*
R=randrange
def f(l,g):*a,=range(l);exec("a[x]=a[y]=a[x:=R(l)]/2+a[y:=R(l)]/2;"*g);return min(a),max(a),sum(a)/l,a

Attempt This Online!

-13 bytes thanks to pxeger using randrange and :=

edited Dec 23, 2021 at 1:33

answered Dec 23, 2021 at 0:41

hyperneutrino♦

42.8k5 gold badges72 silver badges227 bronze badges

\$\begingroup\$ your loop method looks awesome! I didn't know you could do that if you don't need iter value \$\endgroup\$

okie
– okie

2021-12-23 00:46:37 +00:00
Commented Dec 23, 2021 at 0:46
1

\$\begingroup\$ @okie Ah, yeah, it's a pretty helpful tip. Anyway, if you don't need the iter value, another thing you can do is just repeat the code and exec it :P \$\endgroup\$

hyperneutrino
– hyperneutrino ♦

2021-12-23 01:18:55 +00:00
Commented Dec 23, 2021 at 1:18
1

\$\begingroup\$ 134 bytes \$\endgroup\$

pxeger
– pxeger

2021-12-23 01:23:09 +00:00
Commented Dec 23, 2021 at 1:23

Add a comment |

Neil · Accepted Answer · 2021-12-23 08:54:46Z

Charcoal, 39 bytes

ＦＮ⊞υιＦＥＮＥ²‽ＬυＵＭυ⎇№ιλ⊘ΣＥι§υμκＩ⟦⌊υ⌈υ⊘⊖Ｌυυ

Try it online! Link is to verbose version of code. Explanation:

ＦＮ⊞υι

Generate a range of length length.

ＦＥＮＥ²‽Ｌυ

Generate generation pairs of random integers also in that range, and loop over those pairs.

ＵＭυ⎇№ιλ⊘ΣＥι§υμκ

Update the array at those two indices with the average of the two values that are there.

Ｉ⟦⌊υ⌈υ⊘⊖Ｌυυ

Output the minimum, maximum, average and values.

Richard Neumann · Accepted Answer · 2021-12-31 03:40:33Z

2

Python3, 133 131 128 bytes

from random import*
def f(l,g):
 *r,=range(l)
 while g:shuffle(r);*r,x,y=r;r+=[(x+y)/2]*2;g-=1
 return min(r),max(r),sum(r)/l,r

Try this online

edited Dec 31, 2021 at 3:40

answered Dec 31, 2021 at 3:30

Richard Neumann

2302 silver badges9 bronze badges

Add a comment |

Jonah · Accepted Answer · 2021-12-31 04:05:18Z

2

J, 44 bytes

(<./;>./;];+/%#)@(2&((?#)(2%~1#.{)`[`]}])i.)

Try it online!

Outputs Min / Max / List / Avg

answered Dec 31, 2021 at 4:05

Jonah

34.1k4 gold badges41 silver badges95 bronze badges

Add a comment |

chunes · Accepted Answer · 2022-04-08 07:45:15Z

2

Factor, 87 bytes

[ iota >array [ randomize 2 cut* mean dup 2array append ] repeat dup minmax pick mean ]

Try it online!

answered Apr 8, 2022 at 7:45

chunes

27.8k3 gold badges32 silver badges55 bronze badges

Add a comment |

Razetime · Accepted Answer · 2022-04-08 14:51:35Z

2

BQN (CBQN), 56 bytes

{A←+´÷≠,(⌈´∾⌊´∾A∾⋈){∾˜∘A⌾((2•rand.Range≠𝕩)⊸⊏)𝕩}⍟𝕨↕𝕩}

Attempt This Online!

fixed after ovs' comment.

-1 from ovs.

edited Apr 8, 2022 at 14:51

answered Apr 8, 2022 at 9:00

Razetime

27.6k3 gold badges31 silver badges77 bronze badges

1

\$\begingroup\$ does that recompute 2•rand.Range𝕩 every iteration? \$\endgroup\$

Razetime
– Razetime

2022-04-08 13:19:52 +00:00
Commented Apr 8, 2022 at 13:19
1

\$\begingroup\$ No it didn't. But ∾˜∘A saves a byte over (2⥊A) \$\endgroup\$

ovs
– ovs

2022-04-08 13:41:58 +00:00
Commented Apr 8, 2022 at 13:41

Add a comment |

MarcMush · Accepted Answer · 2022-04-08 11:44:49Z

1

Julia, 81 bytes

l\g=(L=[0.:~-l;];1:g.|>_->(r=rand(1:l,2);L[r].=sum(L[r])/2);(extrema(L),~-l/2,L))

Try it online!

expects length\generation and returns ((min, max), mean, list)

answered Apr 8, 2022 at 11:44

MarcMush

6,88515 silver badges18 bronze badges

Add a comment |

Yousername · Accepted Answer · 2022-04-08 18:52:59Z

1

TI-Basic, 85 bytes

Prompt L,G
seq(I,I,0,L-1→A
For(I,1,G
rand(dim(ʟA→B
SortA(ʟB,ʟA
mean({ʟA(1),ʟA(2→ʟA(1
Ans→ʟA(2
End
Disp min(ʟA),max(ʟA),mean(ʟA),ʟA

^{seq(I,I, can be replaced with randIntNoRep( for -3 bytes if the calculator supports it.}

Output is displayed.

answered Apr 8, 2022 at 18:52

Yousername

4,28011 silver badges24 bronze badges

Add a comment |

Stack Exchange Network

Repeat average‎

Background

Task

Example

Rules

what you should do?

12 Answers 12

Jelly, 25 bytes

How it works

APL+WIN, 60 bytes

Vyxal, 25 bytes

Explained

JavaScript (ES6), 145 bytes

Python, 134 bytes

Charcoal, 39 bytes

Python3, 133 131 128 bytes

J, 44 bytes

Factor, 87 bytes

BQN (CBQN), 56 bytes

Julia, 81 bytes

TI-Basic, 85 bytes

Your Answer

Hot Network Questions

Repeat average‎

Background

Task

Example

Rules

what you should do?

12 Answers 12

Jelly, 25 bytes

How it works

APL+WIN, 60 bytes

Vyxal, 25 bytes

Explained

JavaScript (ES6), 145 bytes

Python, 134 bytes

Charcoal, 39 bytes

Python3, 133 131 128 bytes

J, 44 bytes

Factor, 87 bytes

BQN (CBQN), 56 bytes

Julia, 81 bytes

TI-Basic, 85 bytes

Your Answer

Sign up or log in

Post as a guest

Related

Hot Network Questions