Camminata casuale sui bordi di un cubo

Una formica viene posizionata in un angolo di un cubo e non può muoversi. Un ragno inizia dall'angolo opposto e può muoversi lungo i bordi del cubo in qualsiasi direzione con uguale probabilità . In media, di quanti passi avrà bisogno il ragno per arrivare alla formica?$(x,y,z)$$1/3$

(Non si tratta di compiti a casa, è stata una domanda di intervista.)

Suggerisco di modellare il problema come una catena di Markov in cui ogni stato rappresenta la distanza tra il ragno e la formica. In questo caso abbiamo 4 possibili stati $S_i$ le distanze $i$ possono essere$\{0,1,2,3\}$ .

Quando il ragno si trova nell'angolo opposto del cubo, si trova a una distanza di 3 gradini dalla formica. È nello stato $S_3$ .

Costruire la matrice di transizione $\mathbf{P}$ .

• Se disegniamo un cubo, vediamo che quando siamo allo stato $S_3$ , ogni movimento riduce la distanza tra il ragno e la formica a 2 passi. Quindi, quando siamo allo stato $S_3$ ci spostiamo allo stato$S_2$ con probabilità 1.

• Quando siamo allo stato $S_2$ , possiamo tornare allo stato $S_3$ usando il bordo da cui siamo arrivati ​​o possiamo ridurre la distanza a un solo passo se scegliamo altri due bordi. Quindi, quando siamo allo stato $S_2$ , possiamo passare allo stato $S_1$ con probabilità 2/3 e allo stato $S_3$ con probabilità 1/3.

• Quando siamo allo stato , possiamo andare allo stato S 0 usando uno dei tre fronti possibili. Se usiamo gli altri due, torniamo allo stato S 2 . Quindi, quando siamo allo stato S 1 , possiamo passare allo stato $S_1$$S_0$$S_2$$S_1$ con probabilità 1/3 e allo stato S 2 con probabilità 2/3.$S_0$$S_2$

• Quando arriviamo allo stato $S_0$ , restiamo lì poiché è il nostro obiettivo. è uno stato assorbente.$S_0$

$$\begin{equation} \mathbf{P} = \left[\begin{array}{cccc} P_{S_3 \to S_3} & P_{S_3 \to S_2}& P_{S_3 \to S_1} & P_{S_3 \to S_0} \\ P_{S_2 \to S_3} & P_{S_2 \to S_2}& P_{S_2 \to S_1} & P_{S_2 \to S_0} \\ P_{S_1 \to S_3} & P_{S_1 \to S_2}& P_{S_1 \to S_1} & P_{S_1 \to S_0} \\ P_{S_0 \to S_3} & P_{S_0 \to S_2} & P_{S_0 \to S_1}& P_{S_0 \to S_0} \\ \end{array}\right] = \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1/3 & 0 & 2/3 & 0 \\ 0 & 2/3 & 0 & 1/3 \\ 0 & 0 & 0 & 1 \\ \end{array}\right] \end{equation}$$

This is an absorbing Markov chain with three transient states ($S_3$, $S_2$, $S_1$) and one absorbing state ($S_0$).

According to the theory, the transition matrix of a Markov chain with $t$ transient states and $r$ absorbing states can be rewritten as:

$$\begin{equation} \mathbf{P} = \left[\begin{array}{cc} \mathbf{Q}_t &\mathbf{R} \\ \mathbf{0}_{r \times t} & \mathbf{I}_r \\ \end{array}\right] \end{equation}$$

where $\mathbf{Q}_t$ is a $t \times t$ matrix that shows the probability of transitioning from some transient state to another transient state, while $\mathbf{R}$ is a $t \times r$ matrix with the probabilities of transitioning from one of the $t$ transient states to one of the $r$ absorbing states. The identity matrix $\mathbf{I}_r$ shows us that when any of the $r$ absorbing state is reached, there is no transition away from that state. The all zeros matrix $\mathbf{0}_{r \times t}$ can be interpreted as that there is no transition from any of the $r$ absorbing states to any of the $t$ transient states.

The $(i,j)$ entry of $\mathbf{Q}_t$ represents probability of transitioning from a state $i$ to a state $j$ in exactly one step. To get the probability for $k$ steps we need the $(i,j)$ entry of $\mathbf{Q}_t^k$. Summing for all $k$, we get a matrix that contains in its $(i,j)$ entry the expected number of visits to transient state $j$ after starting from transient state $i$.

$$\begin{equation} \sum_{k=0}^{\infty} \mathbf{Q}_t^k = (\mathbf{I}_t - \mathbf{Q}_t)^{-1} \end{equation}$$

To get the number of steps until being absorbed, just sum the values of each row of $(\mathbf{I}_t - \mathbf{Q}_t)^{-1}$. This can be represented by

$$\begin{equation} \mathbf{t} = (\mathbf{I}_t - \mathbf{Q}_t)^{-1} \mathbf{1} \end{equation}$$

where $\mathbf{1}$ is a column vector with all components equal to 1.

Let us apply this to our case:

As stated above, in our case we have $t$=3 transient states and $r$=1 absorbing state, therefore:

$$\begin{equation} \mathbf{Q}_t = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1/3 & 0 & 2/3\\ 0 & 2/3 & 0 \\ \end{array}\right] \quad \quad \mathbf{R} = \left[\begin{array}{c} 0 \\ 0\\ 1/3 \\ \end{array}\right] \end{equation}$$

The matrix with the expected number of visits is

$$\begin{equation} (\mathbf{I}_t - \mathbf{Q}_t)^{-1} = \left[\begin{array}{ccc} 2.5 & 4.5 & 3 \\ 1.5 & 4.5 & 3\\ 1 & 3 & 3 \\ \end{array}\right] \end{equation}$$

This matrix can be interpreted as follows. Starting from state $S_3$ and before getting absorbed at $S_0$ we visit, on average, $S_3$ 2.5 times, $S_2$ 4.5 times, and $S_1$ 3 times.

The expected number of steps from state $S_3$ to state $S_0$ is given by the first component of the following vector:

$$\begin{equation} \mathbf{t} = \left[\begin{array}{ccc} 2.5 & 4.5 & 3 \\ 1.5 & 4.5 & 3\\ 1 & 3 & 3 \\ \end{array}\right] \left[\begin{array}{c} 1 \\ 1\\ 1\\ \end{array}\right] = \left[\begin{array}{c} 10 \\ 9\\ 7\\ \end{array}\right]. \end{equation}$$

The second and third components of $\mathbf{t}$ are the expected number of steps to $S_0$ if we start from $S_2$ and $S_1$ respectively.

$x^*$$x_1$$x_0$

$x^* = 1 + x_1$$x_0 = 1 + \frac{2}{3}x_1$

$$x_1 = 1 + \frac{2}{3}x_0 + \frac{1}{3}x^*= 1 + \frac{2}{3}x_0 + \frac{1}{3} + \frac{1}{3}x_1$$

we get that $x_1 = x_0 + 2$. So $x_0 = 1 + \frac{2}{3}x_0 + \frac{4}{3}$ implying that $x_0=7$ and $x_1=9$.

We get our answer as $x^*=10$.

Edit:

If we draw the cube with coordinates $(x, y, z)$ then $111$ is the starting position of the spider and $000$ the position of the ant.

The spider can move to either $011$, $101$ or $110$.

By the symmetry of the cube these must have the same number of expected steps to the ant, denoted by $x_1$. From $x_1$, we can either return to the origin (with probability $1/3$) or (with probability $2/3$) we can go to one of the points $001$, $100$, $010$ depending on which state we are in.

Again, by symmetry, these points will have the same number of expected steps which we call $x_0$. From these positions we can reach the goal in one step with probability $1/3$ or go back to one of the $x_1$-positions with probability $2/3$. This means that $x_0 = \frac{1}{3}1 + \frac{2}{3}(1 + x_1) = 1 + \frac{2}{3}x_1$.

One nice abstraction to think of it is this:

Think of the Position of the Ant as $(0,0,0)$ and Spider $(1,1,1)$, now each move the spider can make will essentially switch exactly one of the three components from $1\to0$ or $0\to1$. So the question becomes:

If I randomly switch bits in (1,1,1) after how many steps in average do I get 0,0,0

We see the shortest way is 3 switches. Since it doesn't matter with which bit I start the probability of that happening is 1 * 2/3 * 1/3 = 2/9. If we make 1 mistake (switch one bit back to 1) we will need 5 steps. And the chances of making a mistake are 7/9 - if we want to make only one mistake, we have to get from there back and do everything right again - so the chance of making exactly 1 mistake resulting in 5 steps is 7/9 * 2/9 and the chance of making 2 mistakes aka 7 steps is (7/9)² * 2/9 and so on.

So the formula for the expected average number of steps is:

$$\mathbb E(\mathrm{steps}) = \sum_{n=0}^{\infty} (3 + 2n) \cdot \frac{2}{9} \cdot \left ( \frac{7}{9} \right ) ^{n} = 10$$

Just to compliment tiagotvv's answer:

I don't naturally think of these kinds of problems as matrices (even though they are). I have to draw it out, which I've done below. You can see that there are 3 places to move from S, all of which are As. From any A, you can either return to the S, or move to one of two Bs. From any B, you can move to the E, or to one of two As. This all translates to the transition matrix given by tiagotvv, which can also be drawn in graph form.

Because I am terrible at math, I would just try to simulate your problem. You can do this with the markovchain package in R.

  library(markovchain)
  library(ggplot2)

  # Create a markovchain object, given the states and their transition matrix

  mcCube <- new("markovchain", 
                states = c("S", "A", "B", "E"),
                transitionMatrix = matrix(data = c(0,   1,   0,   0,
                                                   1/3, 0,   2/3, 0,
                                                   0,   2/3, 0,   1/3,
                                                   0,   0,   0,   1), 
                                          byrow = T, nrow = 4),
                name = "cube")

  # The following code calcuates the probability of landing on E after taking
  # between 1 and 100 steps from the start, given the above set of transition
  # probabilities.

  start <- c(1, 0, 0, 0)

  list <- list()

  for (i in 1:100){

    list[[i]] <- (start * mcCube^i)[4] 

  }

   a <- do.call(rbind, list)

   data <- data.frame(propE = a, 
                      steps = c(1:100))

   ggplot(data, aes(x = steps, y = propE)) +
    geom_line(size = 1) +
    ylab("Probability you reached the spider") +
    xlab("Number of steps taken") +
    theme_bw() +
    theme(panel.grid.minor = element_blank())

  # This code simulates 1000 different applications of the markov chain where you 
  # take 1000 steps, and records the step at which you landed on E

  list <- list()
  for (i in 1:1000) {


    b <- rmarkovchain(n = 1000, object = mcCube, t0 = "S", include.t0 = T)

    list[[i]] <- 1001 - length(b[b == "E"])

  }

  data <- as.data.frame(do.call(rbind, list))

  ggplot(data, aes(x = V1)) +
    geom_density(fill = "grey50", alpha = 0.5) +
    geom_vline(aes(xintercept = mean(V1))) +
    ylab("Density") +
    xlab("Number of steps to reach E") +
    theme_bw() +
    theme(panel.grid.minor = element_blank())

  mean(data$V1)  # ~10 is the average number of steps to reach E in this set of
                 # simulations

tiagotvv's answer can be calcuated in R as:

q = matrix(c(0,   1,   0,   
             1/3, 0,   2/3, 
             0,   2/3, 0), byrow = T, nrow = 3)


(solve(diag(3) - q) %*% c(1, 1, 1))[1] # = 10

Parity considerations give a very clean solution, using surprisingly simple machinery: no Markov chains, no iterated expectations, and only high school level summations. The basic idea is that if the spider has moved an even number of times in the $x$ direction, it has returned to its original $x$ coordinate so can't be at the ant's position. If it has moved an odd number of times in the $x$ direction, then its $x$ coordinate matches the ant's. Only if it has moved an odd number of times in all three directions will it match the $x$, $y$ and $z$ coordinates of the ant.

Initially the spider has made zero moves in any of the three directions, so the parity for each direction is even. All three parities need to be flipped to reach the ant.

After the spider's first move (let's label that direction $x$), exactly one direction has odd parity and the other two ($y$ and $z$) are even. To catch the ant, only those two parities need to be reversed. Since that can't be achieved in an odd number of subsequent moves, from now on we consider pairs of moves. There are nine possible combinations for the first paired move:

$$(x,x), \,(x,y), \,(x,z), \,(y,x), \,(y,y), \,(y,z), \,(z,x), \,(z,y), \text{or} \,(z,z)$$

We need to move in the $y$ and $z$ directions to reach the ant after one paired move, and two out of nine combinations will achieve this: $(y,z)$ and $(z,y)$ would ensure all three parities are odd.

The other seven combinations leave one odd and two even parities. The three repeated moves, $(x,x)$, $(y,y)$ or $(z,z)$, leave all parities unchanged so we still require one $y$ and one $z$ movement to reach the ant. The other pairs contain two distinct moves, including one in the $x$ direction. This switches the parity of $x$ and one of the other parities (either $y$ or $z$) so we are still left with one odd and two even parities. For instance the pair $(x,z)$ leaves us needing one more $x$ and one more $y$ to reach the ant: an equivalent situation (after relabelling of axes) to where we were before. We can then analyse the next paired move in the same way.

In general paired moves start with one odd and two even parities, and will either end with three odd parities (with probability $\frac{2}{9}$) and the immediate capture of the ant, or with one odd and two even parities (with probability $\frac{7}{9}$) which returns us to the same situation.

Let $M$ be the number of paired moves required to reach the ant. Clearly $M$ follows the geometric distribution on the support $\{1, 2, 3, \dots\}$ with probability of success $p = \frac{2}{9}$ so has mean $\mathbb{E}(M) = p^{-1} = \frac{9}{2} = 4.5$. Let $N$ be the total number of moves required, including the initial move and the $M$ subsequent paired moves. Then $N = 2M + 1$ so, applying linearity of expectations, $\mathbb{E}(N) = 2\mathbb{E}(M) + 1 = 2 \times 4.5 + 1 = 10$.

Alternatively you might note $P(M \geq m) = (\frac{7}{9})^{m-1}$ and apply the well-known formula for the mean of a discrete distribution taking only non-negative integer values, $\mathbb{E}(M)=\sum_{m=1}^\infty P(M\geq m)$. This gives $\mathbb{E}(M)=\sum_{m=1}^\infty (\frac{7}{9})^{m-1}$ which is a geometric series with first term $a=1$ and common ratio $r=\frac{7}{9}$ so has sum $\frac{a}{1-r} = \frac{1}{1-7/9}=\frac{1}{2/9}=\frac{9}{2}$. We can then take $\mathbb{E}(N)$ as before.

Comparison to Markov chain solutions

How might I have spotted this from the Markov chain transition matrix? Using @DLDahly's notation, the states in the transition matrix correspond to my description of the number of the number of directions with odd parity.

The one-step transition matrix is

$$\begin{equation} \mathbf{P} = \left[\begin{array}{cccc} P_{S \to S} & P_{S \to A}& P_{S \to B} & P_{S \to E} \\ P_{A \to S} & P_{A \to A}& P_{A \to B} & P_{A \to E} \\ P_{B \to S} & P_{B \to A}& P_{B \to B} & P_{B \to E} \\ P_{E \to S} & P_{E \to A} & P_{E \to B}& P_{E \to E} \\ \end{array}\right] = \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1/3 & 0 & 2/3 & 0 \\ 0 & 2/3 & 0 & 1/3 \\ 0 & 0 & 0 & 1 \\ \end{array}\right] \end{equation}$$

The first row show us that after one movement, the spider is guaranteed to be in state A (one odd and two even parities). The two-step transition matrix is:

$$\begin{equation} \mathbf{P}^{(2)} = \mathbf{P}^{2} = \left[\begin{array}{cccc} 1/3 & 0 & 2/3 & 0 \\ 0 & 7/9 & 0 & 2/9 \\ 2/9 & 0 & 4/9 & 1/3 \\ 0 & 0 & 0 & 1 \\ \end{array}\right] \end{equation}$$

The second row shows us that once the spider has entered state A, in two moves time it has either returned to state A with probability $7/9$ or has reached state E (all odd parities) and captured the ant, with probabilty $2/9$. So having reached state A, we see from the two-step transition matrix that the number of two-step moves required can be analysed using the geometric distribution as above. This isn't how I found my solution, but it is sometimes worth calculating the first few powers of the transition matrix to see if a useful pattern like this can be exploited. I have occasionally found this to give simpler solutions than having to invert a matrix or perform an eigendecomposition by hand - admittedly something that is only really relevant in an exam or interview situation.

Ho scritto un breve programma Java per rispondere numericamente alla tua domanda. La traversata del ragno è veramente casuale, il che significa che può anche attraversare in cicli prima di arrivare alla formica.

Tuttavia, non hai definito il termine "angolo opposto", quindi ho due diversi scenari. Di fronte come attraverso lo stesso piano o attraverso il cubo. Nel primo scenario, il percorso più breve è di 2 passaggi e 3 passaggi nel secondo scenario.

Ho usato 100 milioni di ripetizioni e i risultati sono i seguenti:

-- First scenario --
Steps sum: 900019866
Repeats: 100000000
Avg. step count: 9.00019866

-- Second scenario --
Steps sum: 1000000836
Repeats: 100000000
Avg. step count: 10.00000836

Codice sorgente:

import java.util.Random;
import java.util.concurrent.atomic.AtomicLong;
import java.util.stream.IntStream;

public class ProbabilityQuizSpider {

    // Edges of the cube
    private static final int[][] EDGES = new int[][] {
            {1, 3, 7}, // corner 0
            {0, 2, 4}, // corner 1
            {1, 3, 5}, // corner 2
            {0, 2, 6}, // corner 3
            {1, 5, 7}, // corner 4
            {2, 4, 6}, // corner 5
            {3, 5, 7}, // corner 6
            {0, 4, 6}  // corner 7
    };

    private static final int START = 0; // Spider
    private static final int FINISH = 5; // Ant
    private static final int REPEATS = (int) Math.pow(10, 8);

    public static void main(String[] args) {

        final Random r = new Random();
        final AtomicLong stepsSum = new AtomicLong();

        IntStream.range(0, REPEATS).parallel().forEach(i -> {

            int currentPoint = START;
            int steps = 0;

            do {

                // Randomly traverse to next point
                currentPoint = EDGES[currentPoint][r.nextInt(3)];

                // Increase number of steps
                steps++;

            } while(currentPoint != FINISH);

            stepsSum.addAndGet(steps);

        });

        // Results
        System.out.println("Steps sum: " + stepsSum.get());
        System.out.println("Repeats: " + REPEATS);
        System.out.println("Avg. step count: " + (((double) stepsSum.get()) / ((double) REPEATS)));

    }

}

EDIT: corretto un refuso nello script (e anche aggiornato i risultati)

Ho risolto il tuo enigma tramite simulazioni Monte Carlo ($n = 10^4$) e ottenuto $\mathtt{mean(steps)} \approx 10$.

Ecco il codice R che ho usato:

ant = c(0,0,0) # ant's coordinates 

sim = 1e4 # number of MC simulations
steps = numeric() # initialize array of steps

for (i in 1:sim)
{
  spider = c(1,1,1) # spider's coordinates
  count = 0 # initialize step counter

  # while ant's coordinates == spider's coordinates
  while (!isTRUE(all.equal(ant, spider)))
  {

  # random walk in one of three dimensions
  xyz = trunc(runif(1,1,4))

  # let the spider move
  if (spider[xyz] == 1) 
    {
    spider[xyz] = 0
    } else if (spider[xyz] == 0) 
    {
    spider[xyz] = 1
    }

  # add one step
  count = count + 1
  }

# add the number of step occurred in the ith iteration
steps = c(steps, count)

# print i and number of steps occurred
cat("\n", i, " ::: ", count)
}

# print the mean of steps
(mean(steps))

Credo che Alesc sia sulla buona strada quando menziona "Tuttavia, non hai definito il termine" angolo opposto "A meno che non manchi qualcosa nella domanda, non esiste una risposta corretta, solo risposte basate su ipotesi. La dimensione del cubo non è definita IE 10 piedi cubi, 1000 piedi cubi ecc. La dimensione della formica non è definita IE piccolo giardino, carpentiere, rosso gigante ecc., Il tipo di ragno non è definito (per determinare la dimensione del passo) IE piccolo giardino, Tarantola ecc. SE si combinano tutti "non definito "variabili. la risposta potrebbe essere 0 passi o un numero indeterminato / infinito di passi.