A really impossible problem

The problem

Previously we have solved a puzzle that claimed to be impossible but was, in fact, just a bit hard 😀 But in this post we will see a problem that is really impossible to solve, the halting problem:

Given a description of an arbitrary computer program, decide whether the program finishes running or continues to run forever.

For some programs it’s easy to see if they halt or not:

def program_1():
    return 2 + 2

def program_2():
    while True: pass

def program_3():
    from itertools import count
    for i in count():
        if str(2**i)[0] == '7':

def program_4():
    from itertools import count
    for i in count():
        if str(2**i)[-1] == '7':

For others it’s not so simple:

def program_5():
    def z(m, n):
        if m == 0:
            return n + 1
        elif n == 0:
            return z(m - 1, 1)
            return z(m - 1, z(m, n - 1))
    z(4, 4)

And there are simple ones where it’s an open problem to know if they halt or not:

def program_6():
    from itertools import count
    for i in count(1):
        p = 2 ** i
        r = int(''.join(reversed(str(p))))
        for j in count(1):
            q = 5 ** j
            if q > r:
            elif q == r:
The proof

Let’s assume we have a procedure, represented by the function does_halt(), to determine if any given program will halt:

def does_halt(program):
    # magic implementation

We don’t care about the details of the does_halt() function. The only requirements are:

  • Its implementation should be finite.
  • It must accept any Python function that doesn’t take arguments as a program.
  • It must give an answer in an easily bounded time. For example, we should be able to say “if the input program has 1000 bytes, the function should return a boolean value indicating if it halts in no more than NNN seconds”.

The clever point of the proof, though it’s not a new kind of cleverness, is to build a program that “halts only if it doesn’t”:

def paradox():
    if does_halt(paradox):
        while True:

As the paradox() program is finite, including this small amount of code plus the code of does_halt(), does_halt() should return in a finite amount of time. If does_halt() returns False, indicating that paradox() does not halt, the program exits, finishing its execution in a finite time. On the other hand, if does_halt() returns True, indicating that paradox() does halt, it enters an infinite loop.

This proves that some of these propositions must be true:

  • does_halt() must not be finitely implementable.
  • does_halt() must reject some valid programs.
  • does_halt() execution time is unbounded.

One thought on “A really impossible problem

  1. […] the previous post we have seen that the halting problem cannot be solved. But maybe the problem is just in our […]

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