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Cake day: June 12th, 2023
  • Kogasa@programming.devtoFuck AI@lemmy.world•When billion-dollar AIs break down over puzzles a child can do, it's time to rethink the hype
    1·
    2 months ago

    1000 or so moves

    Ok, this nerd-snipes me a bit. The real point of the Towers of Hanoi problem, at least if you’re doing it in a discrete math class, is to count exactly the number of moves for n discs. Call this number f(n).

    To make this post easier to read, we will label the discs from 1 to n, with 1 being the bottom disc and n being the top. We will also label the pegs A, B, and C, with A being the starting peg and C being the target peg. This lets us describe moves in a compact notation: 3: A -> C means disc 3 moves from A to C.

    For n=0, we vacuously need 0 moves to win.

    For n=1, we need just 1 move: 1:A->C.

    For n=2, we can do it in 3 moves, so f(2) = 3. The sequence is:

    • 2: A->B # unblock the next move
    • 1: A->C # move 1 to its final location
    • 2: B->C # move 2 to its final location

    Now suppose we know f(n) for n between 1 and N, with N >= 2. For n=N+1, we can move N+1: A -> C and attempt to use our strategy for moving the remaining N discs from A to C, shuffling N+1 around as needed. Since it’s the smallest disc, we can always move it to any pillar, and any time we want to move something to the pillar it’s on, it must be moved first. Let’s see how that plays out for N=2:

    • 3: A->C # move the N+1 disc
    • 2: A->B # start the N-disc strategy
    • 3: C->B # unblock the next step
    • 1: A->C # next step of N-disc strategy
    • 3: B->A # unblock the next step
    • 2: B->C # last step of N-disc strategy
    • 3: A->C # finish the N+1-disc problem

    We hazard a guess that every step of the N-disc strategy is preceded by a move of the N+1 disc, plus one final move. In other words, f(N+1) = 2f(N) + 1. Some careful analysis will justify this guess.

    Now we have a recurrence relation for f(n), but how to solve it? A classical technique is “guess the formula magically, then prove it by induction.” It’s certainly doable here if you compute a few values by hand:

    f(2) = 3

    f(3) = 2(3) + 1 = 7

    f(4) = 2(7) + 1 = 15

    f(5) = 2(15) + 1 = 31

    f(6) = 2(31) + 1 = 63

    You’ll probably see it right away: f(n) = 2^n - 1. Indeed, we can prove this by induction: the base case n=2 holds as f(2) = 3 = 2^(2) - 1, and if f(n) = 2^n - 1 then f(n+1) = 2f(n) + 1 = 2(2^n - 1) + 1 = (2^(n+1) - 2) + 1 = 2^(n+1) - 1 as desired.

    We may conclude f(12) = 2^(12) - 1 = 4095. In other words, with 12 discs, exactly 4095 moves are required.

    –

    Bonus: an alternative to the “guess and prove” technique that is generalizable to a broad class of recurrence relations. The technique is called “generating functions.” Given the sequence f(n), we consider the formal power series

    F(x) = f(0) + f(1)x + f(2)x^(2) + … + f(n)x^(n) + …

    This F(x) is the “ordinary generating function” for the sequence f(n). Strictly speaking, it may not be a well-defined function of x since we haven’t made any assumptions about convergence, but we will assume (for now, and prove later) that the set of such formal power series behaves algebraically much like we expect. Namely, given the recurrence relation above, we can write:

    F(x) - f(0) - f(1)x = f(2)x^(2) + f(3)x^(3) + f(4)x^(4) + … + f(n)x^n + …

    = (2f(1) + 1)x^(2) + (2f(2) + 1)x^(3) + … + (2f(n-1) + 1)x^(n) + …

    = 2x(f(1)x + f(2)x^(2) + … + f(n)x^(n)) + (x^(2) + x^(3) + … + x^(n) + …)

    = 2x(F(x) - f(0)) + x^(2)/(1-x)

    In our case, we have f(0) = 0, f(1) = 1, f(2) = 3 so we can write more succinctly:

    F(x) - x = 2xF(x) + x^(2)/(1-x)

    Solving for F,

    F(x)(1 - 2x) = x + x^(2)/(1-x)

    = x(1 + x/(1-x))

    F(x) = x(1 + x/(1-x))/(1 - 2x)

    = x/(2x^(2) - 3x + 1)

    Ok, great. We’ve found that our generating function, convergence notwithstanding, is that rational function. We can use partial fraction decomposition to write it as

    F(x) = 1/(1 - 2x) - 1/(1-x)

    which has the advantage of telling us exactly how to compute the coefficients of the Taylor series for F(x). Namely,

    1/(1-x) = 1 + x + x^(2) + … + x^(n) + …

    1/(1 - 2x) = 1 + 2x + 4x^(2) + … + 2^(n) x^(n) + …

    So F(x) = (1-1) + (2-1)x + (4-1)x^(2) + … + (2(n)-1)x(n) + …

    The nth coefficient of the Taylor series for F about 0 is 2^(n)-1, and by the definition of F as the ordinary generating function for f(n), we have f(n) = 2^(n) - 1. (The rigorous justification for ignoring convergence here still needs to be done; for now, this can be seen as a useful magic trick.)

  • Kogasa@programming.devtoFuck AI@lemmy.world•When billion-dollar AIs break down over puzzles a child can do, it's time to rethink the hype
    3·
    2 months ago

    Not necessarily do logic, but mimic it, like it can mimic coherent writing and basic conversation despite only being a statistical token muncher. The hope is that there’s sufficient information in the syntax to model the semantics, in which case a sufficiently complex and well-trained model of the syntax is also an effective model of the semantics. This apparently holds up well for general language tasks, meaning “what we mean” is well-modeled by “how we say it.” It’s plausible, at face value, that rigorous argumentation is also a good candidate, which would give language models some way of mimicking logic by talking through a problem. It’s just not very good in practice right now. Maybe a better language model could do better, maybe not for a reasonable cost.