nyuuzyou/wikis huggingface dataset:
Project Euler/16

Project Euler Problem 16 is a problem related to the powers of 2. The task is to find out the "digit sum" (sum of all digits in decimal expansion) of 21000.
==Implementations==
===Aheui===
This pattern works on both AheuiChem[citation needed] and Try It Online. It uses the "decimal multiplication calculator" algorithm which theoretically works for an arbitrarily large number multiplied by a non-zero digit. It also contains some weird parts, for example, a path branches into two and then the two branches joins up immediately, or paths that branches immediately back to itself.
밣발발발따따따밣밣발따따뿌터너벌
ㅇㅇㅇㅇㅇㅇㅇ쑥너벌벌머차바싹볼
ㅇㅇㅇㅇㅇㅇㅇ뿌터너벌벌서
ㅇㅇㅇ아더뭉셕차숙머어ㅇ오쎤너더벊벌아셕어
ㅇㅇㅇㅇㅇ희무선빠박따빠발발다라싼샨오더오
ㅇㅇㅇㅇ먀썩뻐ㅇㅇㅇㅇㅇ요
===C===
#include <stdio.h>
int a[320]={1},k,l,r;
void flow(int d[]){
 int i;
 for(i=0;i<319;++i){
 d[i+1]+=d[i]/10;
 d[i]%=10;
 }
}
int main(){
	for(k=1;k<=1000;++k){
		for(l=0;l<320;++l){
 	 a[l]*=2;
 	}
 	flow(a);
	}
	for(l=0;l<320;++l){
		r+=a[l];
	}
	printf("%d",r);
 return 0;
}
===Gofe===
/1000/&t/10/&q/2/Pt[&sQq>At<#sGqF]>#t.
The language has no interpreter to test whether this program is correct so if you found it incorrect please revise it.
===I fuck, you fuck===
There's a fucker named a
Fuck a 2000 times
There's a fucker named b
Fuck b 2 times
b fucks a over and over
b unfucks b
There's a fucker named c
while a is fucked
c unfucks c
Fuck c 10 times
a unfucks c over and over
c fucks b
c unfucks c
Fuck c 10 times
a unfucks c again
a unfucks a
c fucks a
end fuck
b fucks you
===Python===
#Euler Problem 16
#by Europe2048
def sumdigits(n):
	sum = 0
	for i in str(n):
		sum = sum + int(i)
	return sum
print(sumdigits(2**1000))
Alternate implementation (shorter), by :
 print(sum(map(int,str(2**1000))))
=== Text ===
 1366
==External resources==
* A001370, a related sequence on OEIS. The 1001st term is the solution.
* Problem 16 on Project Euler Official Website (not available)
* Problem 16 on Project Euler Mirror
Project Euler/2

Project Euler Problem 2 is a problem that works with Fibonacci numbers. The task is to work out the sum of all even Fibonacci numbers below four million.
==Implementations==
===Aheui===
This program may need reduction.
바싹바발발나빠뱐주뱔뗘땨뼈뺘뗘뺘변따뗘
삭망희ㅇ루벅포뼈차
싹샥뼈댜챠사빠쑥
ㅇㅇㅇㅇ쌰석쇼뎌
===C===
#include <stdio.h>
int h=0, j=1, j_=0;
int main(){
while(j<4000000){
j_+=j;
j+=j_;
j_=j-j_;
j-=j_;
if(j==(j/2)*2){
h+=j;
}
}
printf("%d",h);
return 0;
}
===Rockstar===
Please note that there are two blank lines before the last line
Henry is gone.
Jack is harkworking.
Jill is gone.
While Jack is lower than 4000000
Let Jill be with Jack.
Let Jack be with Jill.
Let Jill be Jack without Jill.
Let Jack be without Jill.
Let Papa be Jack over 2.
Turn it down.
Let Papa be of 2.
If Papa is Jack
Let Henry be with Jack.
Say Henry.
===Shakespeare===
exeunt at the beginning of a scene sounds absurd, but that's what usually happens in a program with three or more variables. [citation needed] The program can't run without them.
Jack and Jill's affairs.
Hamlet, who adds everything up.
Juliet, who delivers the big things.
Julia, who memorizes the smaller things.
Act I: a.
Scene I: Constant definition.
[ enter Hamlet and Juliet ]
Hamlet: You are a cat.
[ exeunt ]
Scene II: Start loop.
[ exeunt ]
[ enter Julia and Juliet ]
Julia: Are you better than the square of twice the cube of the sum of a big big big cat and a big cat? If so, let us proceed to Scene III.
Juliet: You are the sum of you and me.
Julia: You are the sum of you and me.
Juliet: You are the difference between me and you.
Julia: You are the difference between you and me.
[ exit Julia ]
[ enter Hamlet ]
Hamlet: Are you as good as the product of the quotient between you and a big cat and a big cat?
Juliet: If so, you are the sum of you and me.
Let us return to Scene II.
[ exeunt ]
Scene III: End of affairs.
[ exeunt ]
[ enter Hamlet and Juliet ]
Juliet: Open your heart.
[ exeunt ]
===Wenyan===
有數零。 名之曰「和」。
有數一。 名之曰「大」。
有數零。 名之曰「小」。
恆為是。若「大」大於四百萬者乃止也。
 加「大」以「小」。昔之「小」者。今其是矣。
 加「大」以「小」。昔之「大」者。今其是矣。
 減「大」以「小」。昔之「小」者。今其是矣。
 減「大」以「小」。昔之「大」者。今其是矣。
 
 除「大」以二。所餘幾何。若其等於零者。
 加「大」以「和」。昔之「和」者。今其是矣。
 云云。
云云。
吾有一數。曰「和」。書之。
===Python===
# Project Euler problem 2
# by Europe2048
def fibonacci(n):
 if n < 2:
 return 1
 else:
 return fibonacci(n-1) + fibonacci(n-2)
result = []
for i in range(1, 35):
 if fibonacci(i) % 2 == 0 and fibonacci(i) < 4000000:
 result.append(fibonacci(i))
print(sum(result))
==See also==
==External resources==
* Fibonacci on Wolfram Mathworld
* Problem 2 on Project Euler Official Website (not available)
* Problem 2 on Project Euler Mirror
Project Euler/20

Project Euler Problem 20 is a problem related to factorial. Find the "digit sum" of the factorial of 100. To learn more about "digit sum", click here.
==Implementations==
===Aheui===
The decimal multiplication algorithm has been improved and now it works for an arbitrarily large number multiplied by any number below a million in any integer base below a million.
밤발발따따빠박따뿌터너벌
ㅇㅇㅇ쑥너벌벌머차바싹볼
희ㅇㅇ뿌터너벌벌서
몽다석차바싼숙머오쎤석러더벌벌썩ㅇㅇ너
ㅇㅇㅇㅇ무선빠사빠싹삭따싼산다빠발발도
ㅇㅇ먀썩뻐ㅇㅇㅇ요
===C===
#include <stdio.h>
int a[320]={1},k,l,r;
void flow(int d[]){
 int i;
 for(i=0;i<319;++i){
 d[i+1]+=d[i]/10;
 d[i]%=10;
 }
}
int main(){
	for(k=1;k<=100;++k){
		for(l=0;l<320;++l){
 	 a[l]*=k;
 	}
 	flow(a);
	}
	for(l=0;l<320;++l){
		r+=a[l];
	}
	printf("%d",r);
 return 0;
}
===I fuck, you fuck===
There's a fucker named a
There's a fucker named b
Fuck b 100 times
Fuck a 1 times
while b is fucked
b fucks a again
I unfuck b
end fuck
There's a fucker named c
while a is fucked
c unfucks c
Fuck c 10 times
a unfucks c over and over
c fucks b
c unfucks c
Fuck c 10 times
a unfucks c again
a unfucks a
c fucks a
end fuck
b fucks you
===Python===
# Euler Problem 20
# by Europe2048
import math
def sumdigits(n):
	sum = 0
	for i in str(n):
		sum = sum + int(i)
	return sum
print(sumdigits(math.factorial(100)))
Alternate implenentation (shorter), by :
print(sum(map(int,str(__import__("math").factorial(100)))))
Version by :
print(sum([int(i)for i in __import__("math").factorial(100)]))
==External resources==
* A004152, a related sequence on OEIS. The 101st term is the solution.
* Problem 20 on Project Euler Official Website (not available)
* Problem 20 on Project Euler Mirror
Project Euler/24

The problem 24 of Project Euler is to find out the 1000000th lexicographic permutation of 0~9.
==Implementations==
===C++===
#include<iostream>
#include<algorithm>
using namespace std;
int a[10]={0,1,2,3,4,5,6,7,8,9};
int main(){
 for(int i=1;i<1000000;i++) next_permutation(a,a+10);
 for(int i=0;i<10;i++) cout<<a[i];
 return 0;
}
Unfortunately, this does not work on the answer checking in Project Euler Mirror, but this is technically correct, isn't it?
Project Euler/28

Project Euler Problem 8 is a problem related to the well known number spiral. The spiral expands indefinitely in all four directions, as shown below. The task is to find the sum of the numbers on the diagonals of the 1001 by 1001 spiral formed the same way, starting from the "1" at the center. However, many of the programs listed here utilizes the relationship between the numbers in the spiral, instead of the actual position of the numbers.
43	44	45	46	47	48	49
42	21	22	23	24	25	26
41	20	7 	8 	9 	10	27
40	19	6 	1 	2 	11	28
39	18	5 	4 	3 	12	29
38	17	16	15	14	13	30
37	36	35	34	33	32	31
==Implementations==
===Aheui===
Please note that the variable that functions like Ajax in the Shakespeare program counts down from 8, rather than 4 in other programs. This is done to shorten the program a little.
발발나싹박싼발발나산뺘투뱗텨뱛볎뱕뗘뱍뗘떠
싹삭다삳밧타뿌섣썯도볋처삭망희
뽀더서써뻐선처마산본
===Rockstar===
Heiss is hardworking
Juda is hardworking.
Jei is hard-working.
While Jei is weaker than 1002
Arr is blow.
While Arr is not gone
Let Juda be with Jei.
Let Heiss be with Juda.
Knock Arr down.
Build Jei up, up....
Say Heiss.
===Shakespeare===
Note at Problem 2.
Scotchgate conspiracy.
Hamlet, who adds everything up.
Juliet, who sends messages to Hamlet.
Julia, who sends messages to Juliet.
Ajax, who can only count to four.
Act I: x.
Scene I: Initialization.
[ enter Hamlet and Juliet ]
Hamlet: you cat.
Juliet: you cat.
[ exit Juliet ]
[ enter Julia ]
Hamlet: you big cat.
[ exeunt ]
Scene II: Start outer loop.
[exeunt]
[ enter Hamlet and Julia ]
Hamlet: Are you as good as the sum of a big cat and the cube of twice the sum of a big big cat and a cat?
If so, let us proceed to Scene V.
[ exit Julia ]
[ enter Ajax ]
Hamlet: You big big cat!
[ exeunt ]
Scene III: Start inner loop.
[ exeunt ]
[ enter Julia and Ajax ]
Julia: Are you as good as nothing?
If so, Let us proceed to Scene IV.
[ exit Ajax ]
[ enter Juliet ]
Julia: You are the sum of you and me.
[ exit Julia ]
[ enter Hamlet ]
Juliet: you are the sum of you and me.
[ exit Juliet ]
[ enter Ajax ]
Hamlet : you are the sum of you and a pig. Let us proceed to Scene III.
Scene IV: end inner loop.
[ exeunt ]
[ enter Hamlet and Julia ]
Hamlet: You are the sum of you and a big cat. Let us proceed to Scene II.
Scene V: end outer loop.
[ exeunt ]
[ enter Hamlet and Julia ]
Julia: Open your heart.
[ exeunt ]
===Wenyan===
有數一。 名之曰「和」。
有數一。 名之曰「上」。
有數二。 名之曰「下」。
恆為是。若「下」大於千者乃止也。
 為是四遍。
 加「下」以「上」。昔之「上」者。今其是矣。
	加「上」以「和」。昔之「和」者。今其是矣。
 云云。
 加「下」以二。昔之「下」者。今其是矣。
云云。
吾有一數。曰「和」。書之。
==External resources==
* Problem 28 on Project Euler Official Website (not available)
* Problem 28 on Project Euler Mirror
Project Euler/3

Project Euler Problem 3 is a problem related to factorization. The problem asks for the largest prime factor of a particular number.
==Implementations==
===Aheui===
This program takes the 12-digit constant as input. It works on Try It Online but unfortunately fails on AheuiChem.
방싹박빠삭빠싸사투빠사빠싹삭ㅇ루
ㅇㅇㅇㅇㅇ귀희멍차속누석썩뻐서차수
ㅇㅇㅇ오ㅇㅇㅇㅇㅇㅇ서ㅇㅇ더너벌벌
===C===
If the program generates an error on your computer, you may try replacing #define lnt long long with #define lnt __int64 or #define lnt __int64_t, and/or replacing %lld with %I64d. [citation needed]
#include <stdio.h>
#define lnt long long
lnt a=600851475143;
lnt i=2;
int main(){
while(a!=i){
if(a%i==0){
a/=i;
}else{
i+=1;
}
}
printf("%lld",i);
return 0;
}
===Rockstar===
The king was hiding everything, weathering randomly after I went several short millimeters over him.
The state is me
while the king isn't the state.
Let the rule be the king over the state.
Turn it down
if nothing is the king without the rule of the state.
Let the king be over the state,
else
build the state up.
Say the state.
===Python===
# Euler Problem 3
# by Europe2048
# Fixed by None1
import math
test = int(input("What number do you want to check for? "))
sqrt = math.floor(math.sqrt(test))
prev_primes = []
factors = []
i=2
while i<sqrt:
 flag=1
 while 1:
 flag=0
 for j in prev_primes:
 if i % j == 0:
 flag=1
 break
 if not flag:
 break
 i+=1
 if (test / i) % 1 == 0:
 factors.append(i)
 prev_primes.append(i)
print(factors[-1])
==See also==
Other pages related to prime factors and factorization:
3	12	21	23
Adjacent numbered pages:
==External resources==
* Prime factor on Wolfram Mathworld
* Prime Factorization on Wolfram Mathworld
* A page on xkcd related to factorization
* Problem 3 on Project Euler Official Website (not available)
* Problem 3 on Project Euler Mirror
Project Euler/4

Project Euler Problem 4 is a problem related to palindromic numbers. The goal is to find out the largest palindromic number (in base 10) that is the product of two 3-digit numbers.
==Implementations==
===Aheui===
This program uses a palindrome detector that would theoretically work for all positive whole numbers below 1 billion[citation needed]. The palindrome detector also contains a code snippet for integer logarithm (designed for determining how many digits there are in a number). This program costs approximately half an hour to run at full speed on AheuiChem. It may still need reduction.
바싿밟밣밝반따따따밟타빠싹빠삭빠싸사따빠바쌀뿌서더너벌벌
삭다사박발따나살발발나타뿌소쎡버너벗뻐설뻐머차박발따나솔
쏙뚜번벌석러떠번벌뻐서야쳐오마빠박루쀼서뻐섣차사마빠삭뿌
소떠삭발발나타발ㅇㅇㅇ초저박뱔셔땨쳐나사싹삭토초어쏟수써
아아초삭무밝북따붍싹샥샫오망희ㅇㅇㅇ싿삳자아야먀오소타아
발냐별토밟봃따또탸뽀뺘셛샤뗘뺘셕썩
===Rockstar===
Rhythm is bam-bam-bam.
A feather is up.
The party is perfection.
Music wants emotion.
If emotion is as weak as a feather,
give back silence.
Put rhythm in life.
While life is lower than emotion without rhythm.
Rock the world with emotion at life.
Build life up.
Join the world.
Give the world back.
Healing takes time.
Shatter time.
If time is smaller than a feather
give right back
else
let first be time at the party.
Let last be time without rhythm.
Let last be time at last.
Send first is last and healing taking music taking time.
This is for you, mom.
You are you.
Let this be of you.
Let just be this.
Glass is shattering
until gone.
Let the story be just of this
if the story is as weak as glass.
Break it down.
While the story is greater than glass
if healing taking the story,
let glass be the story.
Break it down.
Knock just down.
Let the story be just of this.
Knock this down.
Let just be this.
Say glass.
===Python===
# Euler Problem 4
# by Europe2048
def isPalindrome(s):
 return s == s[::-1]
maxnum = 0
for i in range(10000, 998001):
 if isPalindrome(i) == False:
 i += 1
 continue
 
 prev_primes = []
 factors = []
 
 for n in range(100, 999):
 if (test / n) % 1 == 0:
 factors.append(n)
 
 if factors != []:
 if i > maxnum:
 maxnum = i
 i += 1
 continue
==External resources==
* A308803 and A327897, two related sequences on the OEIS. "308803" is a palindrome as well.
* Problem 4 on Project Euler Official Website (not available)
* Problem 4 on Project Euler Mirror
Project Euler/5

Project Euler Problem 5 is related to the concept of Lowest Common Multiple. The task is to work out the lowest common multiple of all integers from 1 through 20.
==Implementations==
===Aheui===
Warning: tall stack
This program searches for the smallest multiple of 20, then searches for the smallest multiple of 19 among the multiples of 20, and then searches for the smallest multiple of 18 among the multiples of 20×19, and so on.
발발나싹발밤따뿌망희
뿌석뻐서썬뻐석처속
사싹샥도루발나툐
빠쌱션속쳐볼서
===Python===
Python has a library with a LCM function, so we don't need to manually define it this time.
# Euler Problem 5
# by Europe2048
import math
for i in range(1, 21):
 for j in range(1, 21):
 print(math.lcm(i, j), end = " ")
 print("\n")
==External resources==
* Problem 5 on Project Euler Official Website (not available)
* Problem 5 on Project Euler Mirror
Project Euler/6

Project Euler Problem 6 is related to the sum of integer sequences. The task is to work out the difference between the square of the sum of integers from 1 through 100, and the sum of the squared integers from 1 through 100. However, many of the programs below calculates the sum of n×n×(n-1) for the integers n from 1 through 100 instead, and still get the correct result.
==Implementations==
===Aheui===
바싹밤발발따뚜
투너벌벌뻐뻐뻐
따따싹삭두빠초
뭉뱔벌냐셔토삭
희
===C===
#include <stdio.h>
int h=0,j=100;
int main(){
while(j>0){h+=j*j*(j-1);--j;}
printf("%d",h);
return 0;
}
===Rockstar===
Jesus is a scratching rockplayer.
Henry is gone.
While Jesus isn't nothing
Let Es be Jesus without 1.
Let El be Jesus of Jesus of Es.
Let Henry be with El.
Knock Jesus down.
Say Henry.
===Shakespeare===
The delivery of the broken cubes.
Hamlet, who adds everything up.
Juliet, who always counts down.
Act I:a.
Scene I: define constant.
[ enter hamlet and juliet ]
Hamlet: you are the square of the sum of a big big big cat and a big cat.
Scene II: start loop.
Juliet: you are the sum of you and the product of me and the product of me and the sum of me and a pig.
Hamlet: you are the sum of you and a pig.
Are you as good as nothing?
If not, let us proceed to Scene II.
Juliet: open your heart.
[ exeunt ]
===Python===
Note: The end='' in the last line is added to eliminate the trailing newline.
r=0
j=0
for j in range(1,101) :
	r+=j*j*(j-1)
print(r,end=)
===Wenyan===
有數零。名之曰「H」。
有數百。名之曰「J」。
恆為是。若「J」等於零者乃止也。
 減「J」以一。乘其以「J」。乘其以「J」。加其以「H」。昔之「H」者。今其是矣。
	減「J」以一。昔之「J」者。今其是矣。
云云。
吾有一數。曰「H」。書之。
==External resources==
* A045991, a related sequence on OEIS.
* A052149, another related sequence on OEIS. The 100th term is the solution to this problem.
* Problem 6 on Project Euler Official Website (not available)
* Problem 6 on Project Euler Mirror
Project Euler/7

Project Euler Problem 7 is a problem related to prime numbers. It requires you to find the 10001st prime number.
==Implementations==
===Aheui===
This may theoretically work, but it is very ineffective. It even failed on Try It Online.
박싹밤발발따따빠따발발나다뿌서더너벌벌석터너벌벌머차삭빠사빠싹삭루
수ㅇㅇㅇㅇㅇ희멍터너벌벌석차박빠삭빠싸사타야ㅇ야ㅇ오두ㅇ너벌벌서처
마ㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇ야오야여요뼈ㅇㅇㅇㅇㅇ어
Second try that also fails:
방박싹사뿌서
누벌벌석차박싼산빠빠따삭빠산싹삭주
타망희ㅇㅇ요나도ㅇ루석썩뻐선뻐석처산마수
ㅇㅇㅇㅇㅇㅇ볼벌선처산마숙ㅇㅇㅇ터너벌벌
ㅇㅇㅇㅇㅇ오ㅇㅇㅇ더너벌벌
Third attempt based on the C program below, and it still fails:
방박싹사뿌서
너벌벌석차ㅣ아순머희멍터
ㅇㅇㅇㅇㅇㅇ우빠쏞
ㅇㅇㅇ우ㅇㅇ삲뺘아빠따삭빠삲싹삭주
ㅇㅇㅇ어ㅇㅇ쏜ㅇㅇ루석썩뻐섢뻐석처수
ㅇㅇㅇㅇㅇㅇ솒ㅇㅇ처숙어마숞터너벌벌
ㅇㅇㅇㅇㅇ숀ㅇ더너벌벌ㅇ쏜뻐숙
ㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇㅇ쏜ㅇㅇ뻐
There may be an alternative that uses the queue.
===C===
This program works on Try It Online.
#include <stdio.h>
int e[10002]={0},i=0,ei=0;
int a=2;
int c=10001;
int main(){
for(a=2;c>0;++a){
 if(ei==0){e[ei]=a;++ei;--c;}
 else{
 for(i=0;e[i]*e[i]<=a;++i){
 if(a%e[i]==0){break;}
 }
 if(a%e[i]!=0){e[ei]=a;++ei;--c;}
 }
}
printf("%d",a-1);
return 0;
}
==External resources==
* Prime number on Wolfram Mathworld
* A000040, a related sequence on OEIS. The 10001st term is the solution.
* Problem 7 on Project Euler Official Website (not available)
* Problem 7 on Project Euler Mirror
Project Euler/8

Project Euler Problem 8 is a problem related to searching on a 1D strip. It requires finding the largest product of 13 adjacent digits in a 1000-digit constant. The constant can be found on the talk page.
==Implementations==
===Aheui===
This program takes the 1000-digit constant as input. Unfortunately, the program works on Try it Online but fails on AheuiChem.
밯발발나다뿌
코터떠벍벍처마바바바바바바바바바바바바수머
싸싼순희쑨버버버버버버버버버버버버버버빠쏙
뽀섣떠몽숙ㅇㅇㅇㅇ머선ㅇ쳐펴선
사순ㅇ손빠싸마샤숟빠싿며
쑌퍄녀뺘별쌷별퍄며빠싿삳조
==External resources==
* Problem 8 on Project Euler Official Website (not available)
* Problem 8 on Project Euler Mirror
Project Euler/9

Project Euler Problem 9 is a problem related to Pythagorean triples. The task is to find out the value of a×b×c where the sum of the natural numbers a, b, and c is equal to 1000 and the three numbers form a Pythagorean triple.
==Implementations==
===Aheui===
This program doesn't work out the actual values of a, b, and c. Instead, it uses three medium values x (stores in main stack), m (stored in "n" stack) and n (stored in "nh" stack) acting as a "pythagorean triples generator" where a=x(m2-n2), b=2xmn, c=x(m2+n2). We can easily find out that a+b+c=2mx(m+n) and abc=2x3mn(m+n)(m-n)(m2+n2), and the program uses that shortcut to work out the answer.
발발나반싼빠싾삲빠산빠싾빠반따삲다싼산뚜
뿌썩뿌머투떠떠오뗘벓벌벌벌뻐떠선썬뻐서뻐
쑨뽀섢쏙차빠뱛오발발발따따따주사불다산오
산투두섢썮뻐너벌벌머서뻐머머처모발노
뿌썩빠싼산타우요어어쳐녀뱔별뱔별냐며다셚
따삲빠따싼샨야댜야쌱오사빠따싹삭떠망희
===Rockstar===
The line marked "See note above" is technically useless[citation needed], but unluckily it couldn't be removed.
The palace is gone.
The stable is gone.
Xenocrates is hardworking
The monk is hard-working
The nun is hardworking
While the palace is gone
Let the stable be the monk with the nun.
Let the stable be of the monk.
Let the stable be with the stable.
If the stable of Xenocrates is 1000
Let my tower be the monk of the monk
Let your tower be the nun of the nun
Let my tower be of my tower
Let your tower be of your tower
Let the palace be my tower without your tower
Let the palace be of the monk
Let the palace be of the nun
Let the palace be of Xenocrates
Let the palace be of Xenocrates
Let the palace be of Xenocrates
Let the palace be with the palace.
Else
If the stable of Xenocrates is higher than 1000
 Xenocrates is hardworking.
 Build the nun up.
 If the monk is the nun
 Build the monk up.
 The nun is hardworking.
 Else
 Xenocrates is hardworking. (See note above)
Else
 Build Xenocrates up.
Say the palace.
==External resources==
* Problem 9 on Project Euler Official Website (not available)
* Problem 9 on Project Euler Mirror
Project Euler (language)

Project Euler is a joke language by User:None1. It is made to complete the Project Euler problems with the shortest code possible.
There are only 28 valid programs, A-Z, 0 and 1, which solves the Project Euler problem 1-28 respectively.
==Examples==
===Project Euler problem 1===
 1
===Smallest 1000 digit fibonacci number===
 25
===digit sum of 100!===
 20
 Category:2023
Promo

Promo (Procedural Model) is a programming language based on Brainfuck created by Alex Costea, meant to be an easy and powerful model of computation. It uses 6 characters and 8 commands.
Like Brainfuck, it operates on a list, however the list is infinite in both directions and all numbers on it can get arbitrary big. Unlike Brainfuck, it takes only one (or zero) natural numbers as input and returns one number as output. The input is written in the initial cell, and the value of the final cell is considered the output. Also unlike Brainfuck, every program is valid (calls to non-existing functions are simply ignored).
Note: please read the official specification here: . This page is not necessarily up to date.
The first 4 characters have the same meaning as in Brainfuck. Every character not in "+-<>#@" is ignored.
Command	Meaning
+ 	add 1
- 	remove 1
> 	move right by 1
< 	move left by 1
# and @	when started with #, call the function corresponding to the number in binary written by # as 1 and @ as 0. Example: #@#@ calls function 10. Calls are not made if the current cell is 0.
@ 	when followed by a string containing only "@", do a dynamic call. This means: 
# let k = current cell
# move to the right as many times as there are @'s
# if the current cell is not 0, call function k
# move left by the same amount
+- 	separates functions between then, starting with function 0 (which is called automatically at the beginning), 1, 2, 3...
>< 	Only when written as the first 2 characters, take a number as an input. If omitted, input will automatically be considered 0.
== Examples ==
This program calculates 2^n of a given number (or loops forever for negative inputs).
 ><>+<#>+--#>#@<+--#@++
As the only output is a number, a traditional Hello, World! is impossible (unless encoded as a Base 256 number).
Cat program
 ><
Quine (kinda cheating?)
 0
== Busy Beaver lower bounds ==
BusyBeaver(n) is defined as the biggest number that can be outputted by a program that takes no output with n characters. Those are the current lower bounds for it, and some examples:
n 	BusyBeaver(n)	Example
0 	0 	[the empty program]
1 	1 	
2 	2 	
3 	3 	
4 	4 	
5 	5 	
6 	6 	
7 	7 	
8 	8 	
9 	9 	
10	10 	++++++++++
11	12 	+#+#+--#+++
12	21 	
13	39 	
14	84 	
15	155 	
16	340 	
17	780 	
18	1554 	
19	9330 	
20	2391483 	
21	2^256-2 	++#+#+#+--#+#@+--#@++
22	> 10^1000000 	
23	> 10^10^76 	
== Compiler ==
https://alex-costea.github.io/Promo/
Propositio

Propositio is an esolang based on logic theory.
==Commands==
 A -> B A implies B, if A then B
 !A Returns the opposite of A, can replace A in other commands
 A Sets A to true. !A sets A to false
 A. Outputs A.
 A? Only does the next instruction if A is true.
 :A Sets A's value to the user's input.
 1. Sets jump point 1. Can be any number of digits as long as it's only numbers.
 > 1 Goes to jump point 1.
 # Ends program.
 A !> B Breaks all implications between A and B (including ones with !A or !B, or reversed ones from B -> A). Otherwise, any time A is true 
 B will be reset to true.
==Programs==
====Truth Machine====
 :A
 1.
 A.
 A?
 > 1
 #
====Cycler====
This program takes A and B, then swaps their values.
 :A
 :B
 A -> C
 !A -> !C
 A !> C
 !A
 B -> A
 !B -> !A
 B !> A
 !B
 C -> B
 !C -> !B
 C !> B
 !C
 A.
 B.
Category:LanguagesCategory:2024Category:Unknown computational class