Implement Minceraft

Question

Introduction

The game Minecraft has a 1 in 10000 chance of showing "Minceraft" instead of "Minecraft" on the title screen.

Your challenge

Your challenge is to code a function or program that takes no input, and on average 1 of 10000 times, returns "Minceraft" and the rest of the time returns "Minecraft".

Scoring

This is code-golf, shortest wins!

Answer 1

Minecraft, 358 293 277 276 bytes

Implementing Minceraft in Minecraft

Should be run as a set of commands in game. Should be run on a fresh superflat (redstone ready) world. Byte count is the amount of characters you need to type, including newlines.

/scoreboard objectives add s dummy
/summon bat
/execute as @e[type=bat] store result score o s run data get entity @s UUID[1]
/scoreboard players set c s 2147054151
/execute if score o s < c s run tellraw @s "Minecraft"
/execute if score o s >= c s run tellraw @s "Minceraft"

The UUID of an entity is (mostly) randomly generated and stored as four signed 32 bit integers. One of the integers is compared with 2147054151 which is one ten thousandth of the way between 2^31 and -2^31.

Previous solution because it was so high effort and I'm not deleting it:

Minecraft, 293 bytes

/scoreboard objectives add s dummy
/summon horse
/execute as @e[type=horse] store result score o s run data get entity @s Attributes[1].Base 100000
/scoreboard players set c s 11883
/execute if score o s >= c s run tellraw @s "Minecraft"
/execute if score o s < c s run tellraw @s "Minceraft"

How does it work?

In Minecraft, there are only a limited amount of sources of reliable randomness usable in commands. In Minecraft Functions you can use predicates with specific random chances, but that requires multiple files and submitting that as a zipped data pack would be at least 1 KB.

To avoid this and have it run solely from commands a player can type in, this uses horses as a random source. When a horse is spawned, it is assigned a random speed between 0.1125 and 0.3375 (the units are arbitrary). The random speed calculation is as follows:

$$0.25(0.45 + 0.3x + 0.3y + 0.3z)$$

Where x, y, and z are uniform independent random variables [0, 1). If you find the cumulative density function by convolving the distribution functions (to get the PDF) and then integrating, you will end up with the following piecewise function:

$$ f(x)=\begin{cases}-\frac{9}{16}+15x-\frac{400x^{2}}{3}+\frac{32000x^{3}}{81}&\frac{9}{80}\le x<\frac{3}{16}\\ \frac{29}{4}-110x+\frac{1600x^{2}}{3}-\frac{64000x^{3}}{81}&\frac{3}{16}\le x<\frac{21}{80}\\ -\frac{227}{16}+135x-400x^{2}+\frac{32000x^{3}}{81}&\frac{21}{80}\le x<\frac{27}{80} \end{cases} $$

We need to find the horse speed where the probability of a horse with that speed or lower spawning is one in ten thousand or 0.0001. This can be done by solving the first equation:

$$-\frac{9}{16}+15x-\frac{400x^{2}}{3}+\frac{32000x^{3}}{81} = 0.0001\\ f^{-1}(0.0001)\simeq0.11882574$$

This results in the horse speed being multiplied by 100,000 (because scoreboard values must be integers) and compared with 11883 in the code.

The precision is currently 1.00202 in 10,000. Two bytes can be added or removed for approximately each order of magnitude required. Two additional bytes will make it 1.00012 in 10,000.

Answer 2

Python 3, 63 59 bytes

My idea was to use the time module as a PRNG since the challenge stated

on average 1 of 10000 times, returns "Minceraft" and the rest of the time returns "Minecraft"

and taking the time mod 10000 does exactly that.

Thanks @movatica for -4

import time
print(f"Min{'ceec'[time.time()%1e4>1::2]}raft")

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Alternative, also 59 bytes:

import time
print("Min%sraft"%"ceec"[time.time()%1e4>1::2])

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Answer 3

Python 3, 78 71 69 68 bytes

from random import*
f=lambda:f'Min{randint(0,1e4)and"ec"or"ce"}raft'

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-7 bytes thanks to @Alex bries

-2 bytes thanks to @Alex bries

Answer 4

Python 3 - 45 42 41 Bytes

41 bytes with the constraint of running from a different process every run.

print(f"Min{'ceec'[id(0)%4e4>1::2]}raft")

42 bytes with limited randomness unless running a different process each time [*].

print(f"Min{'ceec'[id({})%4e4>1::2]}raft")

The idea was to use the same wrapper as in here, but a different RNG:
Create a new object (a set dict in this case, because it takes 5 2 characters set() {}), and take the id() of that, which is its memory address.

The memory address acts like a uniform hash[*], which has 1/10000 1/40000 (thanks @ovs) chance to end with zeros.

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Update to 42 bytes:
Instead of set(), use {} which is the dict constructor. Can't go shorter for dynamic object creation.

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Update to 41 bytes: - use static object creation!
Thanks to @dingledooper
Works only if the program is run in a different process every run!
Instead of {}, use 0.
0 is an object in Python, so it has an address in virtual memory, and that address is constant for that object during the lifetime of the program. If we were to run that in a for loop, it would not be random.
However, if we run it in a new process every run, we get the desired randomness. This is the case in a stateless server.

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[*] - Objects in Python are freed immediately when they are no longer referenced (reference counting). Thus, running the solution in a loop might always output the same id, as the memory is freed and allocated at the same address. This is OS dependent, and environment dependent, thus may not work as expected, or in a reproducible way.
Running a new process each time solves this issue.

Answer 5

JavaScript, 37 bytes

_=>`Min${new Date%1e4?'ec':'ce'}raft`

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First of all, write Min.

Then get the current time with new Date. Normally it returns something similar to "2021-10-05T10:29:58.432Z". But if you do a mathematical operation on it, this output becomes the epoch time similar to "1633429932422" (a number, basically).

Now divide the number by 10000 (1e4). If the remainder (%) is 0 i.e. false, add ce to Min otherwise add ec.

At last, add raft at the end.

Answer 6

C (gcc), 48 bytes

f(){printf("Min%sraft",rand()%10000?"ec":"ce");}

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Version closer to \$\frac{1}{10000}\$

For a pseudorandomness as close to \$\frac{1}{10000}\$ as possible:

C (gcc), 79 bytes

f(){long p=0,m=1e4,i;for(i=m;i--;)p+=rand();printf("Min%sraft",p%m?"ec":"ce");}

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Answer 7

Taxi, 632 bytes

Go to Heisenberg's:W 1 R 3 R 1 L.Pickup a passenger going to Divide and Conquer.1073741824e-4 is waiting at Starchild Numerology.Go to Starchild Numerology:N 1 L 3 L 1 L 3 L.Pickup a passenger going to Divide and Conquer.Go to Divide and Conquer:W 1 R 3 R 1 R 2 R 1 R.Pickup a passenger going to Trunkers.Go to Trunkers:E 1 R 3 R 1 L.Pickup a passenger going to The Underground.Go to The Underground:E 1 R 2 L.Switch to plan A if no one is waiting.Minecraft is waiting at Writer's Depot.[A]Minceraft is waiting at Writer's Depot.Go to Writer's Depot:N 3 L 2 L.Pickup a passenger going to Post Office.Go to Post Office:N 1 R 2 R 1 L.

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There are two sources of randomness in Taxi: Heisenberg's (which holds a random integer passenger), and Firemouth Grill (which can hold any number of passengers, who are picked up in a random order).

This program uses Heisenberg's to grab a random number, which will be used to choose one of two strings randomly.

Here is the basic control flow of the program:

• Go to Heisenberg's and grab a random integer, in the range of \$[0, 2^{31}-1]\$. (\$2^{31}-1\$ is RAND_MAX in C++, and Taxi uses C++'s random functions.)
• Pick up the value 1073741824e-4 (that's \$\frac{2^{31}}{2 \times 10000}\$) from Starchild Numerology.
• Drop off both passengers at Divide and Conquer, giving us a result of \$\frac{n}{2^{31}} \times 2 \times 10000\$, which is a double in the range \$[0, 2 \times 10000)\$. (We use this range instead of \$[0, 10000)\$ because we will output Minceraft when this number is either 0 or 1.)
• Bring this result to Trunkers, which chops off the fractional part. (The range is still \$[0, 2 \times 10000)\$.)
• Bring this result to The Underground. One of two things will happen:
• • A passenger will be returned, which means that our number was greater than 1 (which has a \$\frac{2 \times 10000 - 2}{2 \times 10000}\$ chance of occurring). Do nothing.
• • No passenger will be returned, which means that our number was either 0 or 1 (which has a \$\frac{2}{2 \times 10000}\$ chance of occurring). Declare the value "Minceraft" at Writer's Depot.
• Declare the value "Minecraft" at Writer's Depot.
• Pick up the first passenger waiting at Writer's Depot. (If both "Minecraft" and "Minceraft" are waiting, "Minceraft" will be picked up, because it was waiting there first.)
• Drop off this passenger at the Post Office, outputting the string and (for purposes of Code Golf) ending the program.

Taxi, 691 bytes

1e4 is waiting at Starchild Numerology.Go to Starchild Numerology:W 1 L 2 R 1 L 1 L 2 L.Pickup a passenger going to The Underground.Minceraft is waiting at Writer's Depot.Go to Writer's Depot:E 1 L 2 R.[A]Pickup a passenger going to Cyclone.Go to Cyclone:N.Pickup a passenger going to Firemouth Grill.Pickup a passenger going to Firemouth Grill.Go to Firemouth Grill:S 1 L 2 L 1 R.Go to Fueler Up:E 1 L.Go to The Underground:N.Switch to plan B if no one is waiting.Pickup a passenger going to The Underground.Minecraft is waiting at Writer's Depot.Go to Writer's Depot:N 3 L 2 L.Switch to plan A.[B]Go to Firemouth Grill:S 1 R.Pickup a passenger going to Post Office.Go to Post Office:E 1 R.

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This program uses Firemouth Grill to hold thousands of "Minecraft" and "Minceraft" strings to choose from randomly.

Here is the basic control flow of the program:

• Pick up the value 1e4 from Starchild Numerology to use as a counter.
• Declare the value "Minceraft" at Writer's Depot and go there (but don't pick it up yet).
• Repeat these steps:
• • Pick up the string currently at Writer's Depot, and clone it at Cyclone. (Having two strings along for the ride will earn us extra gas money!)
• • Take both clones, and drop them off at Firemouth Grill.
• • Take our counter, and drop it off at The Underground. One of two things will happen:
• • • A passenger will be left waiting, which was 1 less than our previous counter. Pick up this new counter, and declare the value "Minecraft" at Writer's Depot and go there. At this point, repeat the previous steps.
• • • No passenger will be left waiting, which means that our counter is now 0 and we are done.
• Go to Firemouth Grill, and pick up a passenger. It will most likely be "Minecraft", but there's a 1/10000 chance it will be "Minceraft".
• Drop off this passenger at the Post Office, outputting the string and (for purposes of Code Golf) ending the program.

Taxi, 849 bytes

Go to Fueler Up:W 1 R.Switch to plan B.[A]Go to Fueler Up:E 2 L.[B]Go to Heisenberg's:N 3 R 1 L.Pickup a passenger going to Cyclone.Go to Cyclone:N 1 L 3 L.Pickup a passenger going to Magic Eight.Pickup a passenger going to Cyclone.Go to Cyclone:N 1 R 3 R 1 R.Pickup a passenger going to Equal's Corner.Pickup a passenger going to Equal's Corner.Go to Equal's Corner:N 1 R.2e4 is waiting at Starchild Numerology.Go to Starchild Numerology:N 1 R.Pickup a passenger going to Magic Eight.Go to Magic Eight:W 1 R 2 R 1 R.Switch to plan A if no one is waiting.Pickup a passenger going to The Underground.Go to The Underground:E 2 L.Switch to plan C if no one is waiting.Minecraft is waiting at Writer's Depot.[C]Minceraft is waiting at Writer's Depot.Go to Writer's Depot:N 3 L 2 L.Pickup a passenger going to Post Office.Go to Post Office:N 1 R 2 R 1 L.

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This was my first instinct. It also uses Heisenberg's to generate a random integer within the range \$[0, 2 \times 10000)\$, but it does so by repeatedly generating a random number until it is within this range. The taxi has to stop at Fueler Up to refill its gas tank, and it has to clone passengers at Cyclone and drop the clones off at Equal's Corner to make enough money to re-generate random numbers for as long as necessary, but otherwise, the functionality is the same as the 632-byte program.

It doesn't depend on the specifics of C++, but it does usually take longer to execute, because generating a small enough random number is rarer. If anyone can improve this one, be my guest.

Answer 8

Clojure, 42 bytes

#(str"Min"(if(<(rand)1e-4)"ce""ec")"raft")

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Answer 9

R, 45 44 43 bytes

`if`(runif(1)<1e-4,"Minceraft","Minecraft")

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Boring, but shorter than any more-interesting attempt that I've tried so far...

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R, 42 bytes (p=0.000101) p=1 in 10000.5

Edit: much closer to exactly 1 in 10000 (for the same bytes) thanks to Robin Ryder

`if`(rexp(1)>1e-4,"Minecraft","Minceraft")

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Outputs "Minecraft" if an exponentially-distributed random variable is greater than 1e-4.

The exponential distribution is governed by a parameter - lambda - that defaults to 1 in the R rexp() function. This gives cumulative p(up to x)=1-exp(-lambda*x), equal to 9.9995e-05 with x=1e-4 and lamda=1, which is pretty close to 1 in 10000 (actually 1 in 10000.5).

Answer 10

Charcoal, 17 bytes

MinＰceraft¿‽×χφec

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

Min

Print Min.

Ｐceraft

Print ceraft without moving the cursor.

¿‽×χφ

If a random number in the range [0..10000) is nonzero, then...

ec

... correct it to ec.

Answer 11

Retina, 29 bytes

ce9999*$(ec
L@$`..
Min$&raft

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ce9999*$(ec

Insert ce and 9999 copies of ec.

L@`..

Select one of the pairs of letters at random....

L$`
Min$&raft

... and wrap it between Min and raft.

Answer 12

05AB1E, 17 bytes

’£Ì³±’œ4°LΩΘ5!*è™

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’£Ì³±’                -- "minecraft"
      œ               -- permutations: ["minecraft", "minecratf", ...]
       4°LΩ           -- random integer in the range 1..10000
           Θ          -- 1 if it’s equal to 1, 0 otherwise
            5!*       -- multiply by 5! (120)
               è      -- index into the list of permutations
                ™     -- titlecase

Answer 13

APL, 26 bytes

'Min','raft',⍨'ec'⌽⍨1=?1E4

Saved 1 byte thanks to Adam

Answer 14

Minecraft, 226 bytes.

/scoreboard objectives add d dummy
/scoreboard players set c d 2147054151
/execute store result score s d run seed
/execute if score s d < c d run tellraw @s "Minecraft"
/execute if score s d >= c d run tellraw @s "Minceraft"

(Includes a trailing newline, to run the final command)

I saw the other Minecraft answer by Okx and had some suggestions. I would've put them in the comments of the other response, but I figured I changed it enough for its own submission, and another CGSE member said it's fine so whatever.

I'm only just now noticing the only change I made was using /seed. Ah, well. The things you learn when you have five ideas for improvements and four of them don't work.

Answer 15

PowerShell, 38 bytes

"Min$(('ec','ce')[!(random 1e4)])raft"

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Answer 16

V (vim), 80 bytes

iMin<Esc>:r!echo $RANDOM
C<c-r>=<c-r>"%10000
<esc>:s/^0\n/ce
:s/\d\+/ec
kgJAraft

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Generate a random no. in 0-9999, and then replace 0 with ce and anything else with ec.

Answer 17

Jelly, 18 17 bytes

“ẹ?ŀɼƲṬ`Ỵȧ»ȷ4XỊịḲ

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Full program.

-1 byte thanks to Jonathan Allan!

How it works

“ẹ?ŀɼƲṬ`Ỵȧ»ȷ4XỊịḲ - Main link. Takes no arguments
“ẹ?ŀɼƲṬ`Ỵȧ»       - Compressed string "Minceraft Minecraft"; Call that S
           ȷ4     - 10000
             X    - Random integer between 1 and 10000
              Ị   - Is that equal to 1?
                Ḳ - Split S on spaces
               ị  - Index into the split S

Answer 18

J, ²⁸ 25 bytes

'Minecraft'120&A.~0=?@1e4

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^{-3 thanks to Bubbler}

Note: J does not allow 0-argument functions, but this function ignores it's argument, which is the J equivalent

To see this work, change 1e4 to 2 in the TIO, which will make the letter swap happen 50% of the time.

• 0=?@1e4 Does a random number between 0 and 9999 inclusive equal 0? Returns 1 one in 10000 times.
• 120&A. Transposes the requires letters.
• Conditional execution at rate according to step 1 happens as a result of the way & works when invoked dyadically.

alternative 1, 27 bytes

4|.'raftMin','ec'|.~0=?@1e4

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Rotate |. the string 'ec' with probability 1/10k, then append it to the string 'raftMin, then rotate the result by 4.

alternative 2, 28 bytes

'Minecraft'C.~3<@,4#~0=?@1e4

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Answer 19

Python 2, 62 61 bytes

-1 byte thanks to xnor!

from random import*
print'Min%sraft'%'ceec'[random()>1e-4::2]

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If a probability of \$\left({92 \over 256}\right)^9 \approx 0.0000999841\$ to print Minceraft is fine, 59 bytes is possible:

import os
print'Min%sraft'%'ceec'[max(os.urandom(9))>91::2]

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Python 2, 63 bytes

from random import*
print'Min%xraft'%(236-30*0**randrange(1e4))

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Answer 20

Swift, 55 53 bytes

print(.random(in:1...1e4)<2 ?"Minceraft":"Minecraft")

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This is also my first Code Golf post.

Thanks to @Dingus for getting taking off 2 bytes!

Answer 21

05AB1E, 19 18 bytes

„ecтnиćRªΩ”–·ÿraft

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Run 10000 times

„ec      push string "ec"
т        push 100
n        squared, 10000
и        repeat, push ["ec"]*10000
ć        head extract
R        reverse
ª        append
Ω        choose random element
”–·ÿraft push "Min" + tos + "raft", which is implicitly outputed

Answer 22

Mathematica, 48 44 43 41 bytes

If[Random[]<.1^4,"Minceraft","Minecraft"]

Here's a simple unit test which found a bug in my first version.

Table[If[Random[]<.1^4,"Minceraft","Minecraft"], {ix, 10^6}] // Counts

And the result:

<|"Minecraft" -> 999909, "Minceraft" -> 91|>

Answer 23

Befunge-93, 100 97 81 bytes

"tfarec"v
77+:v>0#<
v*2\_v
?1v0:#
^#< ^-1\$#+
+55:$<  |`-1*:*:
>:#,_@ |<
^ "Min"<

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Uses rejection sampling to generate a random integer between 0 and 9999 inclusive, and then swaps the letters if that number is 0.

I abused some coincidences that arose in my code to golf it.

Answer 24

Vyxal, 27 20 18 bytes

`↔ṅ`₴k2ṁċ[`ec`|`ce`]₴`r…ż`,

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Edit: this is actually 7 less bytes, at only 20:

k2ṁ[`↓∑ġ¬`|`↔ṅ⇩ṡ…ż`]

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Edit 2: Thanks to @emanresu A, this has been shortened to 18 bytes.

k2ṁ[`↓∑ġ¬`|`↔ṅ⇩ṡ…ż

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Answer 25

Bash, 43 bytes

x=(ce ec);echo Min${x[RANDOM%9999?1:0]}raft

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RANDOM%9999 has a range of [0,9999], and in bash 0 evaluates to false, so the ternary works as expected. Supposedly, the $RANDOM function has an "approximately linear distribution", but I haven't proved that with a million iterations. Hat-tip to the Zsh solution.

Answer 26

JavaScript, 51 44 bytes

_=>`Min${Math.random()*1e4>1?'ec':'ce'}raft`

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That's a lotta bytes for such a simple task...

-7 thanks to the OP

Answer 27

Bash, 59 bytes

printf "%0.sMinecraft\n" {0..9999}|sed '1s/ec/ce/'|shuf -n1

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Change 9999 to 1 to see it work with 50% probability.

Repeats the line "Minecraft" 10,000 times. Changes the first line only to "Minceraft". Shuffles the lines randomly and returns the first.

Answer 28

Factor, 40 38 bytes

1e-4 [ "Minceraft"] [ "Minecraft"] ifp

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-2 bytes thanks to @Bubbler!

Factor has a version of if called ifp that calls either quotation depending on a probability.

There is also an interpolating version that is slightly longer:

1e4 random 1 < "ce""ec"? [I Min${}raftI]

Answer 29

Ruby, 35 34 bytes

$><<"Min#{rand<1e-4?:ce: :ec}raft"

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Thanks to Dingus for a saved byte.

Answer 30

Excel, 41 40 bytes

-1 byte thanks to MarcMush

="Min"&IF(RAND()<0.1^4,"ce","ec")&"raft"

I tried typing 0.0001 as 1E-4 to save two bytes. Excel changes it to 0.0001.