Legendre
This language is only Turing-complete if and only if Legendre's conjecture is false, i.e. there exists a integer n > 0 such that there are no primes between n^2 and (n+1)^2. This language takes some inspiration from Underload, though in some respects it is very different from it.
Programs in Legendre are made up of series of positive integers (0 is especially banned, because it essentially negates the entire purpose of the language). Each integer corresponds to a base command in Legendre, or a potential user-defined one. Which command it is assigned to is based on the number of primes between its square and the next integer (equivalent to the OEIS sequence A014085).
The language's commands modify a stack, which can hold arbitrarily large positive integers. If the stack ever holds 0, the 0 is immediately removed. In detail, the commands are:
2 (smallest integer producing this command: 1): Push the next integer in the program onto the stack.
3 (smallest producing integer: 4): Pop the top integer on the stack and execute the command associated with it.
4 (smallest: 6): Pop the top integer. If it was 1, increment the top integer on the stack.
5 (10): Swap the top two stack items.
6 (15): Decrement the top integer on the stack. If that results in 0, pop the 0 and discard it.
7 (16): Duplicate the top integer on the stack.
8 (25): Halt execution and print the stack contents.
This is the basic instruction set, which is unable to do anything interesting, let alone loop. However, there is another command, which can be accessed only if Legendre's conjecture proves false.
- 0 (unknown): Remove all items from the stack and combine them into a new function, which will execute all commands starting at the original bottom of the stack and ending at the top, accessible as a command whose "command number" is equal to the one the next integer in the program source corresponds to.
If this command is somehow accessible, the language becomes Turing-complete, as one can simulate a Minsky machine in it.
When the command 8 is execute or the end of the program is reached, the program terminates and the (Unicode) character corresponding to each integer on the stack is printed.
Example programs
1 2 1 3 1 10 4
This simple program pushes the number 2, then 3 and finally a 10, before executing a 4 (command: 3), which causes the 10 (command: 5) to be popped and executed, swapping the 2 and 3.
1 5 3 15 2 1 6 7
This program demonstrates the use of the indirect integer-to-command correspondence. First, a 5 is pushed, then a 15 and a 1, using three different ways of encoding the 2 command. Then, the 1 is popped and as a result, the 15 is incremented to a 16, and finally executed. The program ends with two instances of the number 5 on the stack.
1 1 1 5 ? 24 1 15 1 31 ? 31 24 31
This program demonstrates the use of the 0 command, using ? as a placeholder number. The program first stores '1 5' in the function 9, then '15 31' in 10, before running function 9 (using 24), which pushes 5 onto the stack, and the repeatedly decrements it, until it reaches 0 and is removed. Then, the program halts.
Minsky machine
In order to convert a Minsky machine to Legendre code, the 0 command must be used. Because this command is inaccessible unless Legendre's conjecture is false, I have used a placeholder ? instead.
Note that all Minsky machine instruction line names need to have integers with different A014085 correspondences from each other and the base commands as well as 24 (9) and 31 (10).
Initialization:
1 1 1 1 ? 24
x INC (A/B) y:
A:
1 y 1 24 1 ? 1 6 1 1 16 1 24 ? x
B:
1 y 1 24 1 ? 1 10 1 6 1 1 16 1 10 1 24 ? x
x DEC (A/B) y z:
A:
1 4 1 10 1 15 1 10 1 31 1 1 1 10 1 z 1 1 1 16 1 24 1 31 1 ? 1 24 1 15 1 y 1 6 16 1 24 16 1 ? 1 1 16 1 10 1 1 16 1 24 ? x
B:
1 4 1 10 1 15 1 10 1 31 1 1 1 10 1 z 1 1 1 16 1 24 1 31 1 ? 1 24 1 15 1 10 1 y 1 6 16 1 24 16 1 ? 1 1 16 1 10 1 1 16 1 10 1 24 ? x
x HALT:
1 25 ? x
To create the final program, append all parts (with x,y,z replaced by their counterparts) and add a single integer to start the first instruction in the chain. This should prove the language's Turing-completeness in case Legendre's conjecture is proven false by counterexample.
Interpreter
This interpreter is written in Python (3), and has been tested on all three above examples. Use the -a/--allowZero flags to allow ? to be used, -f/--file to run code directly from a file and -s/--stackOut to output the stack as a Python list instead. If no file is given, the interpreter enters a sort of REPL mode, which is best used with --stackOut.
import sys
import argparse
import io
class I_need_missing(dict): #used to avoid try/except statements. Essentially a dict
def __missing__(self,key):
return None
def appropriate(integer,prev): #returns number of primes between the square of the integer given and the next
return_value = 0
if prev[integer]:
return prev[integer],prev
if integer == "?":
return 0,prev
for i in range(integer ** 2, (integer + 1) ** 2):
t = False
if i > 1:
t = True
for j in range(2,int(i ** 0.5)+1):
t = i/j != round(i/j)
if not t:
break
return_value += t
prev[integer] = return_value
return return_value,prev
def run_command(commandseries,stack,functions,prev): #Runs the appropriate action for each command.
command,prev = appropriate(commandseries.pop(0),prev)
halt = False
if command == 0: #store in given number
functions[appropriate(commandseries.pop(0),prev)[0]] = stack
stack = []
elif command == 2:#push
stack.append(commandseries.pop(0))
elif command == 3:#execute top instruction
commandseries.insert(0,stack.pop())
elif command == 4:#pop, add 1 to new top if popped value was 1
if stack.pop() == 1:
stack[-1] += 1
elif command == 5:#swap top two integers/?
stack[-1],stack[-2] = stack[-2],stack[-1]
elif command == 6:#subtract 1 from top of stack
stack[-1] -= 1
if stack[-1] == 0:
stack.pop()
elif command == 7:#duplicate top of stack
stack.append(stack[-1])
elif command == 8:#halt
halt = True
else:#run custom
try:
commandseries[0:0] = functions[command]
except TypeError:
print("Warning: unassigned function " + str(command) + " is unassigned", file = sys.stderr)
return commandseries,stack,functions,prev,halt
def main(stack,functions,prev):
#Parser for command line options
parser = argparse.ArgumentParser(description = "Interpreter for the Legendre esoteric programming language.")
parser.add_argument("-a","--allowZero", action = "store_true")
parser.add_argument("-f","--file")
parser.add_argument("-s","--stackOut", action = "store_true")
args = parser.parse_args()
allow_zero = bool(args.allowZero)
#Program decoding starts
pre = ""
if not args.file:
pre = input()
if pre == "":
return
else:
pre = open(args.file).read()
mid = pre.split()
final = []
for i in mid:
if i == "?" and allow_zero:
final.append("?")
elif i != 0 or allow_zero: #and allow_zero)
final.append(int(i))
halt = False
#Functional programming at its best
while final and not halt:
final,stack,functions,prev,halt = run_command(final,stack,functions,prev)
#Halting and output
else:
if args.stackOut:
print(stack)
else:
for i in stack:
print(i == "?" and "?" or chr(i),end = "")
print("")
if args.file or halt:
return
else:
main(stack,functions,prev)
if __name__ == '__main__':
main([],I_need_missing(),I_need_missing())
