If you want to nudge the ketohead narcissist, you have to pander to the ego.

Thanks. My preference is intersectional second ed. I don’t mind conflict as long as it’s not derived from lazy racist tropes.

You can’t claim you climbed a mountain, if each time you fell you just resumed from where you lost grip.

Sure you can; it’s called redpointing.

How do you know it’s a bad idea without knowing what the goal is?

Seems like lots of vpn providers offer port forwarding after a fashion. It surprised me too but there were summary comparisons just a search away.

Assuming your vpn provides a stable remote IP, your client connection needs to use that. Try “whatsmyip” or similar over the vpn. The remote address almost certainly won’t appear in the local output of ip a.

Locally, listen on the “this host”, 0.0.0.0.

You may need to check your firewall locally.

You don’t need to run your http service to troubleshoot - simple tools like netcat can listen for incoming requests - nc -l 0.0.0.0 8000 or what-have-you.

Finally: you might want to look at using a shell host as the client rather than targeting your vpn ip from your local host, just to take hairpin connections out of consideration when troubleshooting.

Generic-ish. It’ll fit any of the input problems I think. You could fool it by using a non-canonical circuit, because it knows nothing about the equivalence of boolean expressions; and it also relies on one swap sufficing to fix an output, so I didn’t go particularly far into turning it into a generic search. Either of those problem extensions would take much more effort from a solver, so my expectation is that they were deliberately avoided.

Haskell part 2, much better solution

Okay, here’s the outline again - this one ran instantly.

Rather than probing with example values, I took a different approach, debugging the structure. I only really care about inputs and outputs, so I wrote something that turns the “wiring diagram” into a map of label -> Expr, where

data Expr = EInput String
          | EAnd Expr Expr
          | EOr Expr Expr
          | EXor Expr Expr
  deriving (Show, Ord)

(the Eq instance is stable in symmatric expressions, eg (==) (EAnd a b) (Eand c d) = a == c && b == d || a == d && b == c)

The expressions are grounded in “inputs” (“x00”…“x44”, “y00”…“y44”) - that is, they just expand out all of the intermediary labelled things.

Then I constructed a circuit that I was after by building a non-swapped 44/45-bit full adder, and produced the same set of expressions for those.

Then: for each output, z00…z45, check the “spec” expression against the actual one. If they’re identical, move on.

Otherwise, find some candidate pairs to swap. For these, I considered all possible labelled outputs except “stable” ones - that is, those that were input depdendencies of z_(i-1) - ie, don’t swap any outputs involved in the computation that’s validated thus far.

searchForSwap :: Exprs -> Layout -> String -> Set.Set String -> [(String, String, Layout, Exprs)]
searchForSwap eSpec actual zz stable =
  let
    vals = Map.keysSet actual & (`Set.difference` stable) & Set.toList
    ds = dependencies actual
  in do
    k1 <- vals
    k2 <- vals
    guard $ k1 < k2
    guard $ k1 `Set.notMember` (ds Map.! k2)    -- don't create any loops
    guard $ k2 `Set.notMember` (ds Map.! k1)
    let actual' = swapPair k1 k2 actual
        eAct' = exprsForLayout actual'
    guard $ eSpec Map.! zz == eAct' Map.! zz
    pure (k1, k2, actual', eAct')

Taking the new layout with swapped outputs and its corresponding set of expressions, carry on searching as before.

A linear scan over the output bits was all that was required - a unique answer poped out without any backtracking.

Anyway, happy Christmas all.

PS. My other version worked (eventually) - it was following this approach that led me to realise that my “spec” full adder was broken too :-D Never skip the unit tests.

(@CameronDev@programming.dev you were asking about alternatives to graphviz-style approaches I recall)

Haskell, programmatic solution

I spent an entire day on this because I didn’t write a unit test to check my “swap outputs” function, which effectively did nothing.

In any case: the approach (which may be more interesting than the code, I know people were interested) involved probing the addition circuit with some example additions - that is, I wrote something that’d let me give alternative inputs from x & y and compute the result using the circuit. I then gave it some simple pairs of values that’d exercise the add and carry bits (ie, pairs chosen from {i << n, n <- {1..43}, i <- {1, 3}}). That gave me some breaking trials.

Because the errors were relatively sparse, I then scanned over pairs of outputs, swapping those that didn’t introduce a data dependency and checking (a) that no new errors were introduced over the trial sets, (b) for any reduction in the number of errors found. I got a bunch fo outputs like this:

swap of ("ccp","mnh") improves matters
bad trial count reduced from 346 to 344

which found the pairs for me. The search could be improved by more carefully tying the probe inputs to the outputs’ dependencies (ie, if the first error comes from the (xi, yi) input bits, then look for swaps of the dependencies introduced by zi) - but in any case, it finds the answer. Phew.

Yeah, I remember when I saw this for the first time. It’s astonishing how powerful lazy evaluation can be at times.

Haskell bits and pieces

The nice thing about Haskell’s laziness (assuming you use Data.Map rather than Data.Map.Strict) is that the laziness can do a ton of the work for you - you might’ve spotted a few Haskell solutions in earlier days’ threads that use this kind of trick (eg for tabling/memoisation). Here’s my evaluation function:

eval l =
  let
    v = l & Map.map (\case
                       Const x -> x
                       And a b -> v Map.! a && v Map.! b
                       Or a b  -> v Map.! a || v Map.! b
                       Xor a b -> v Map.! a /= v Map.! b)
  in v

For part 2, we know what the graph should look like (it’s just a binary adder); I think this is a maximal common subgraph problem, but I’m still reading around that at the mo. I’d love to know if there’s a trick to this.

Gotcha, thanks. I just re-read the problem statement and looked at the input and my input has the strongest possible version of that constraint: the path is unbranching and has start and end at the extremes. Thank-you!

(I ask because everyone’s solution seems to make the same assumption - that is, that you’re finding a shortcut onto the same path, as opposed to onto a different path.)

Hey - I’ve a question about this. Why is it correct? (Or is it?)

If you have two maps for positions in the maze that give (distance to end) and (distance from start), then you can select for points p1, p2 such that

d(p1, p2) + distance-to-end(p1) + distance-to-start(p2) <= best - 100

however, your version seems to assume that distance-to-end(p) = best - distance-to-start(p) - surely this isn’t always the case?

More like because you’re eating broccoli.

Thanks. It was the third thing I tried - began by looking for mostly-symmetrical, then asked myself “what does a christmas tree look like?” and wiring together some rudimentary heuristics. When those both failed (and I’d stopped for a coffee) the alternative struck me. It seems like a new avenue into the same diophantine fonisher that’s pretty popular in these puzzles - quite an interesting one.

This day’s puzzle is clearly begging for some inventive viaualisations.

I should add - it’s perfectly possible to draw pictures which won’t be spotted by this test, but in this case as it happens the distributions are exceedingly nonuniform at the critical point.

Haskell, alternative approach

The x and y coordinates of robots are independent. 101 and 103 are prime. So, the pattern of x coordinates will repeat every 101 ticks, and the pattern of y coordinates every 103 ticks.

For the first 101 ticks, take the histogram of x-coordinates and test it to see if it’s roughly randomly scattered by performing a chi-squared test using a uniform distrobution as the basis. [That code’s not given below, but it’s a trivial transliteration of the formula on wikipedia, for instance.] In my case I found a massive peak at t=99.

Same for the first 103 ticks and y coordinates. Mine showed up at t=58.

You’re then just looking for solutions of t = 101m + 99, t = 103n + 58 [in this case]. I’ve a library function, maybeCombineDiophantine, which computes the intersection of these things if any exist; again, this is basic wikipedia stuff.

day14b ls =
  let
    rs = parse ls
    size = (101, 103)
    positions = map (\t -> process size t rs) [0..]

    -- analyse x coordinates. These should have period 101
    xs = zip [0..(fst size)] $ map (\rs -> map (\(p,_) -> fst p) rs & C.count & chi_squared (fst size)) positions
    xMax = xs & sortOn snd & last & fst

    -- analyse y coordinates. These should have period 103
    ys = zip [0..(snd size)] $ map (\rs -> map (\(p,_) -> snd p) rs & C.count & chi_squared (snd size)) positions
    yMax = ys & sortOn snd & last & fst

    -- Find intersections of: t = 101 m + xMax, t = 103 n + yMax
    ans = do
      (s,t) <- maybeCombineDiophantine (fromIntegral (fst size), fromIntegral xMax)
                                       (fromIntegral (snd size), fromIntegral yMax)
      pure $ minNonNegative s t
  in
    trace ("xs distributions: " ++ show (sortOn snd xs)) $
    trace ("ys distributions: " ++ show (sortOn snd ys)) $
    trace ("xMax = " ++ show xMax ++ ", yMax = " ++ show yMax) $
    trace ("answer could be " ++ show ans) $
    ans

C++ is one if those languages where writing a library feels hugely different from using it. Boost is a case in point here: there are brilliant peiple behind it, but (error messages aside) the ergonomics of using thise libs in an application are usually pretty good.

(Scala felt similar to me. There are other languages where it feels much less like I’m swapping hats as I flip between parts of a codebase.)

He doesn’t want to be co-president. He’s a founder of the USA.