from Hacker News

When the Simplest Concurrent Program Goes Against All Intuition

by simplegeek on 1/14/25, 1:23 PM with 24 comments

• by colonCapitalDee on 1/18/25, 6:30 AM

  It's never actually explained how n=2 is possible. Here's how it works:

  P1 and P2 start executing. P1 sets temp=1, then P2 runs through the loop for i=1..9. Then P1 sets n=temp=1. P2 sets temp=2, then P1 resumes executing for i=2..10, and completes. P2 sets n=temp=2, then completes.

  The intuition is P1 checkpoints, P2 runs through the loop up until the last iteration, P1 resets n to the checkpoint + 1, P2 checkpoints, P1 runs through the loop, then P2 resets n to the checkpoint + 1.

• by tombert on 1/18/25, 2:50 AM

  This was fun to port over to PlusCal to verify it myself:

      EXTENDS Naturals, Sequences
      
      (*--algorithm StateStore {
      
      variables
        n = 0; completions = <<>>; 
      
      define {
          Invar == Len(completions) = 2 => n # 2
      }
      
      fair process (thing \in {"p", "q"})
      variables i = 0, temp = 0; 
        {
         Loop:
              while (i < 10) {
                 First:
                     temp := n + 1;
                 Second:
                     n := temp;
                 Last: 
                     i := i + 1; 
          };
          
          Complete:
              completions := Append(completions, 1)
        }
      }*)
  
  

  (I acknowledge that this isn't the most elegant Pluscal but I think it is a roughly accurate?)

• by Groxx on 1/18/25, 2:39 AM

  To be pedantic (which I feel is fair because it's a concurrency question about possibilities):

  >If we run P and Q concurrently, with ‘n’ initialized to zero, what would the value of ‘n’ be when the two processes finish executing their statements?

  It depends a lot on the observer of `n` and what relationship it has with P and Q. Which isn't defined.

  E.g. a trivial boundary-answer is that it can be 0, if nothing guarantees that P's and Q's threads' memory reaches the `n`-observer's thread. This is somewhat common if you try to synchronize via `sleep`, because that usually doesn't guarantee anything as all.

  We also don't know the computer architecture. We can probably assume at least one byte moves between memory levels atomically, because basically every practical system does that, but that doesn't have to be true. If bits move between memory levels independently, this could observe 31, because it can be any combination of the bits between `00000` and `10100`, which includes `01011` -> there can be a `1` in any position, including all positions, so you can observe a number that was never assigned. (IRL: multi-word values and partial initializations can do this, e.g. it's why double-checked locking is flawed without something to guarantee initialization completes before the pointer is exposed)

• by fjfaase on 1/15/25, 6:19 AM

  This implicitely assumes atomic assignments, meaning that during an assignment all bits representing a value are transfered in one atomic unit. This sounds a bit trivial, but if one would be working with large integers that are stored in multiple memory words, it is less trivial.

  I think it is possible to implement locking with only atomic assigments.

• by hedin_hiervard on 1/18/25, 9:36 AM

  I was once asked a very similar problem during a job interview for a position of a software engineer. I was offered to solve it on a piece of paper or orally. We ran out of time and they offered me to take it home. I did and I spent a few hours thinking about it. Google and ChatGPT were not able to produce better results than the intuitive answer, so I built a test program that confirmed the weird behaviour. After meditating for a few hours I was able to find the correct answer and I was indeed fascinated by this. Do you also think it's too much for a usual interview?

• by sim7c00 on 1/18/25, 7:19 PM

  the best thing i think i read somewhere is to think of modern cpus as a bunch of computers on a network. each task is transmitted in messages over the network potentially, executed and its result sent back in a packet. despite this not being fully accurate it does give an idea that even in a set of 100 operations, due to caches, latencies and also kernel/hardware stuff happening during these 100 operations however much you try to isolate them, the first operation submitted might see its result sent back last.

  this will help to determine or realize better when and where synchronization is needed even though you might not expect it. (or remoddeling of the code).

  its extremely hard to writr concurrent code well. i think i've hardly ever managed lockless code that wasnt super bugy (yet! :D). it takes a lot of focus and patience, and tons details about the platform(s) it might run on.

• by chozang on 1/19/25, 8:34 PM

  Bonus points for mentioning Spin. My "Software Engineering" course in 1987 in grad school would have been more accurately titled, "Using the Spin Model Checker". Except when I was looking for it, I had never came across any mention of Spin since.

• by jpc0 on 1/18/25, 1:15 PM

  As another commentator mentioned, you have violated a precondition, there's no point trying to understand what would happen because its indeterminate.

• by Izikiel43 on 1/18/25, 1:51 AM

  The intuition I developed over the years says that result is unknown, it's a race condition, all bets are off. Specially since doing N+1.

• by dtgriscom on 1/18/25, 2:44 AM

  Seems pretty clear that the result could be anything from 1 to 20.

  (Assumes atomic assignments, as noted by someone else.)