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Q: Random walk on the edges of a cube An ant is placed in a corner of a cube and cannot move. A spider starts from the opposite corner, and can move along the cube's edges in any direction $(x,y,z)$ with equal probability $1/3$. On average, how many steps will the spider need to get to the ant?
(This is not homework, it was an interview question.)

A: I have written a short Java program to answer your question numerically. The traversing of the spider is truly random, meaning that it can also traverse in cycles before getting to the ant.
However, you did not defined the term "opposite corner", so I have two different scenarios. Opposite as in across the same plane or as across the cube. In the first scenario, the shortest path is 2 steps, and 3 steps in the second scenario.
I hava used 100 million repeats and the results are the following:
-- First scenario --
Steps sum: 900019866
Repeats: 100000000
Avg. step count: 9.00019866

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

Source code:
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: fixed a typo in the script (and also updated the results)

A: I suggest modeling the problem as a Markov chain where each state represents the distance between the spider and the ant. In this case we have 4 possible states $S_i$ as the distances $i$ can be $\{0,1,2,3\}$.
When the spider is at the opposite corner of the cube, it is at a distance of 3 steps from the ant. It is in state $S_3$.
Building the transition matrix $\mathbf{P}$.

*

*If we draw a cube we see that when we are at state $S_3$, every movement reduces the distance between the spider and the ant to 2 steps. So, when we are at state $S_3$ we move to state $S_2$ with probability 1.

*When we are at state $S_2$, we can go back to state $S_3$ using the edge we arrived from there or we can decrease the distance to only one step if we choose two other edges. So, when we are at state $S_2$ we can move to state $S_1$ with probability 2/3 and to state $S_3$ with probability 1/3.

*When we are at state $S_1$, we can go to state $S_0$ using one of the three possible edges. If we use the other two, we go back to state $S_2$. So, when we are at state $S_1$ we can move to state $S_0$ with probability 1/3 and to state $S_2$ with probability 2/3.

*When we get to state $S_0$, we stay there since it is our goal. $S_0$ is an absorbing state.
\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.

A: I solved your conundrum via Monte Carlo simulations ($n = 10^4$) and obtained $\mathtt{mean(steps)} \approx 10$.

Here is the R code I used:
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))

A: Let $x^*$ be the number of expected steps. Let $x_1$ be the number of expected steps from any corner adjacent to the origin of the spider and $x_0$ ditto for the ant.
Then $x^* = 1 + x_1$ and $x_0 = 1 + \frac{2}{3}x_1$. Since
$$x_1 = 1 + \frac{2}{3}x_0 + \frac{1}{3}x^*= 1 + \frac{2}{3}x_0 + \frac{1}{3} + \frac{1}{3}x_1$$
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
$x_0 = \frac{1}{3}1 + \frac{2}{3}(1 + x_1) = 1 + \frac{2}{3}x_1$.

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

A: 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

A: 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.

A: I believe alesc is on the right track when mentioning "However, you did not defined the term "opposite corner"
Unless I am missing something in the question, there is no correct answer, just answers based on assumptions.
The cube size is not defined I.E. 10 cubic ft, 1000 cubic ft etc.
Ant size is not defined I.E. small garden, carpenter, giant red etc,
Spider type is not defined (to determine step size) I.E small garden, Tarantula etc.
IF you combine all "not defined" variables. the answer could be 0 steps or an undetermined/infinite number of steps.

redpajama_set_name: RedPajamaStackExchange


June 14: After eating breakfast at the hotel, Amanda and I packed our stuff into the rental car and checked out of our hotel. We did this for two reasons. First, it was Sunday, and hotel prices were falling all over the region now that it was no longer a weekend--we could afford a hotel a bit closer to the trail! And second, the further I walked along the trail, the longer the drive to get back to it! So we'd be checking out and finding a hotel closer to the trail.
Amanda dropped me off at the Palisades Parkway where I continued the hike from where I left the day before. I didn't see any thru-hikers, but there were plenty of day hikers on the trail! I passed them quickly. The New York City skyline was still visible from various viewpoints, but it was much more hazy today than the day before. I could barely see it with my naked eyes, and couldn't see it at all through my camera. No more photos of NYC today!
Late in the morning, I passed a trail crew building a reroute along the Appalachian Trail. It a minor reroute, replacing a badly eroded section of trail with fancy rock steps. I'd learn later that much of the trail in the Bear Mountain region was being rebuilt with thousands of rock steps for the last decade or so. It's a high-traffic area and they've been working to create a trail that's built to last against the thousands of people who hike the region each year.
Eventually, the trail passed Perkins Tower with its commanding views of the Hudson Valley and HOLY COW there were a lot of people! Cars idled in the road, angling to get a parking spot. Hundreds of people wandered around. It felt positively claustrophobic. Originally, I intended to take a quick snack break, but I rethought that plan. Nope, I needed a little more peace and quiet. I remembered my visit through here in 2003 being relatively nice. I must have walked through then on a weekday or something. Today it was absolutely crazy!
The Hudson River Valley. The NYC skyline is out there somewhere, but it wasn't really visible through all the haze today.
From the Bear Mountain Inn, the trail winds down under a busy road then into the front entrance of the Bear Mountain Zoo. The zoo is filled with troubled animals--injured animals that can no longer survive in the wild. It's a small and kind of sad-looking zoo, but it's a major landmark on the trail for a couple of reasons. One, the trail runs through the zoo! There is a by-pass for hikers who arrive after hours when the zoo is closed, but where's the fun in that? And two, the lowest point of the entire Appalachian Trail happens to be in the zoo, just in front of the bear enclosure. It's all uphill from here! And finally.... for those unfortunate thru-hikers that have managed to make it this far without ever seeing a bear, they're all but guaranteed to finally see them here in the zoo since the trail runs right past that bear enclosure.
When I passed through, there were two bears lounging in the sun in the far back right corner of the enclosure. I could barely see them in their hiding place, and it didn't help that the sun was almost directly behind them. More interestingly, I thought, where the large number of vultures in the trees. They looked ready to swoop down and start attacking all of the pedestrians, but more likely, they were hoping to get some free food from the bears when feeding time came around.
The trail left the zoo through a service entrance in the back of the zoo, almost immediately turning onto the Bear Mountain Bridge crossing the Hudson River. Crossing the bridge on foot is a blast, crossing 150 feet above the Hudson. It's noisy from all of the traffic on the road, but the views are spectacular!
On the other side, the trail climbs steeply, and at the next trail junction, I followed a white blaze to the right. Perhaps ten minutes later, I happened to notice a blue blaze on one of the trees and realized I had taken a wrong turn. ARGH! I know I followed a white blaze at that trail junction, and had I been thinking, I would have realized it was suspicious. I had a mental map of how the trail was laid out and knew there was a trail junction, and the one to the right would lead to Anthony's Nose--a prominent viewpoint immediately above the Bear Mountain Bridge. Obviously, I was on that trail.
Anthony's Nose is about a half-mile off the trail and I'd normally have been inclined to check it out, but I said I'd meet Amanda at a road crossing at a certain time and I was running a bit late so decided to skip Anthony's Nose. Now, I was already halfway there! Crap. Well, in for a penny, in for a pound. I continued on to Anthony's Nose. May as well at this point!
I soon caught up with several young men who were playing music incredibly loud. Seriously? What the heck is wrong with you people?! How do you expect to ever see any wildlife playing a stereo at a volume that others can hear from a quarter of a mile away? I just wanted to hike that much faster to get away from them.
Finally, upon reaching the top at Anthony's Nose, the trail came out to a wonderful viewpoint--perhaps the best of the entire day. Better than Perkins Tower. Better than from the Bear Mountain Bridge. It was just an amazing view. I took a few photos but didn't linger. I was already going to be late for my meet up with Amanda. No time to linger! The young men with the stereo music arrived a few minutes after I did but at least had the common sense to turn it off once they did arrive.
I hurried back down the mountain and along the trail. I arrived at a gravel road (Manitou Road) where Amanda was already waiting for me, arriving about a half hour later than I had projected. The detour up Anthony's Nose didn't set me back so much as the time it took me to get through the zoo because I wanted to read every sign I passed. It was the zoo that put me behind schedule! =) But Amanda must have been reading a really good book or something because she didn't even notice I was late.
We checked into a motel near Killfish, ate dinner at Boston Market (which is actually a restaurant and not a market at all, and it certainly isn't in Boston!), then went to a nearby movie theater where we watched Spy--a hilarious movie starring Melissa McCarthy going undercover for the CIA. It had a lot of really good laughs!
And that was it. It was back to the hotel room for the night, where I tried to get at least a little online work done before I went to sleep!
The fancy new rock steps they've been installing for a decade should weather the test of time!
Amanda met me at the Bear Mountain Inn for a lunch break at a Mexican restaurant further up the road.
The lake behind the Bear Mountain Inn.
The entrance of the Bear Mountain Zoo.
There were two bears in the bear pen, but they were barely visible by hiding themselves in the far back corner of their cage. (That brown mass of fur--that's the bear.) Vultures were everywhere!
This sign is a little misleading. Not only is it a service entrance, but it's also the exit for the Appalachian Trail from the zoo!
View of the Hudson River from on the bridge.
View UP from the deck of the bridge.
And the view from Anthony's Nose looking back down at the Bear Mountain Bridge.
Bear Mountain was one of my favorite parts of living in NY. I found my first letterbox there. It was so pretty, except when the crowds descended.


text:
October 1, 2021 October 1, 2021 by Matthew Bennett
The Ascensionist Elusive Target arrives in Hitman 3
Hitman 3 players can now visit Dubai and take a shot at eliminating the newest Elusive Target.
The Ascensionist, otherwise known as Allison Moretta, is an excellent con artist who will stop at nothing to climb the social ladder. The client requesting the end of Moretta is someone who was killed after his usefulness to her was over. However, he had set up a dead man's switch that called in the contract.
Players will have until October 11th to get to Dubai and put a stop to her and her latest scheme.
Later in the Season of Envy, players will be able to explore Colorado for free, unlock new items, and tackle a new set of Featured Contracts.
Hitman 3 is available now on PlayStation 5, Xbox Series X/S, PlayStation 4, Xbox One, Nintendo Switch, PC, and Stadia.
Post tagsHitmanIO Interactive

redpajama_set_name: RedPajamaCommonCrawl


text:
Day 27: Monticello, Albany, Connecting Route, Arbor Ridge
by adventuretoward | Sep 26, 2013 | Ice Age Trail 2013, Trail Journals | 0 comments
Today's run/hike to and through the Arbor Ridge segment was epic for many reasons. First, I ended the day finishing Arbor Ridge, which runs through the Robert O. Cook Memorial Arboretum in Janesville. Did you hear what I said? JANESVILLE! I am now at the end of the second main "segment" of the IAT, and about to embark on my final push north to Sturgeon Bay. WOO HOO!
My friend Terie Cebe was my crew member today, and she picked me up in Albany in a dense fog. We got to the rail-trail a bit north of Monticello and had a little photo shoot for one of the articles I'll be writing. Then I was off. I was going to be on the Badger State and Sugar River Trails for about 16 miles (25.8 km), so I figured this was the time to run – on a flat, crushed limestone path, when it was still cool out and relatively shady. So I ran much of the segment.
Snafu En Route to Arbor Ridge
Only had one snafu. Apparently, at one point the two paths run parallel to one another. But you can't see it, as there's foliage in between. At this point, however, you're supposed to hop from the Badger State to the Sugar River Trail. Either there was no IAT signage or it wasn't prominent, because I didn't see any arrows and so stayed on the Badger State Trail – while Terie waited with my water on the Sugar River Trail.
The two trails split a short bit later, and at that point there's plenty of signage directing you, so all was well. But to ensure she never missed me again, Terie began parking her car across the trail at designated intersections. Ha! There weren't many people on the trail, but one biker said to her, "Well, that's sure an interesting place to park your car."
Oh, saw lots of cranes in one section of the trail. It also passed through a tunnel. Now, when we biked the Elroy-Sparta Trail with our kids years ago, it was a blast to walk through those tunnels, one of which is three-quarters of a mile (1+ km) long. This tunnel was only about a quarter-mile (.4 km), but I got all creeped out.
Creepy Tunnel
First, all of these pigeons were fluttering around the entrance, swooping around me (I hate when birds do that). Then, it was quite dark, despite my phone's flashlight app. And it was kinda creepy, especially with all of the graffiti on the walls and the sounds of water dripping from the ceiling. Since I was alone, my imagination was running wild. So I was glad it was a short tunnel.
Had a 20-mile (32.2 km) connecting route after the nice, shady rail-trails. The temps climbed to about 80 (26.7 C), and despite applying sunscreen several times, I got fried. UGH.
Janesville's arboretum was pretty cool. Hate to say it, Madisonians, but I think it's better than ours. Ended the day with a trail record 36 miles (57.9 km) under my belt. That felt great, although I think I might need different meds – these antibiotics don't seem to be working. Rats.
Tonight I'm at a gorgeous B&B in Elkhorn, Ye Olde Manor House. Rooms are stylish and comfy, there's an indoor pool and the food is fantastic. Definitely one to keep in mind.


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Submissions/Hackathon goes Hogwarts: Creating learning environments at tech events
Submission no. 2014 Subject - C9 and C5
Hackathon goes Hogwarts: Creating learning environments for tech events
Claudia Garád, Magdalena Reiter
claudia.garad wikimedia.at, mail magdalenareiter.at
--Claudia.Garad (talk)
MaxdalenaReiter
Wikimedia Österreich, Jugend hackt Österreich, OPEN COMMONS Linz (Austria)
https://www.wikimedia.at https://jugendhackt.org
"Coding for a better world" is the credo for a series of hackathons that started in 2013 in Berlin and is meanwhile a program that takes place regularly in different cities in Germany, Switzerland and Austria. "Jugend hackt" (which means "young people are hacking") is not only a weekend with the possibility for people from 12 to 18 years to think about socially relevant issues and create helpful prototypes. But it is also an format with an integrated educational concept behind it that has been constantly refined and improved. Its aim is to address a diversity of people, not only those that are already associated with technology. Therefore one of the main functions of the organisation team and the mentors is to create a space that is comfortable and leaves no one behind.
The organizing team from Wikimedia Österreich applied many concepts and learnings from "Jugend hackt" to the Wikimedia Hackathon 2017. Especially their mentoring approach to welcome and integrate newcomers seemed to be an idea with a lot of potential which hasn't been systematically applied to Wikimedia tech events before. In our presentation we will showcase the concepts behind the youth hackathons and how they can relate to tech community building in the Wikiverse. We will also share first experiences and learnings from the Hackathon, which will take place in May 2017 in Vienna.
The Wikimedia movement still struggles to integrate newcomers in a sustainable way. We will present our learnings from the work with young people at tech events: On the one hand the strategies of youngsters differ from those of adults and are worth to look at. On the other hand having educational concepts in the background can also help when working with adults. We will show how turning Wikimedia (tech) events into learning environments for newcomers can contribute to a positive welcome culture, reduce entry barriers for new volunteers, and foster sustainable motivation.
WikiCulture & Community
Hackathon goes Hogwarts
Plani (talk) 10:05, 27 March 2017 (UTC)
Shikeishu (talk) 09:31, 30 March 2017 (UTC)
Schiste (talk) 13:17, 3 April 2017 (UTC)
Daniel Mietchen (talk) 03:01, 9 April 2017 (UTC)
Susannaanas (talk) 04:50, 11 April 2017 (UTC)
Raimund Liebert (WMAT) (talk) 09:27, 11 April 2017 (UTC)
Saguaromelee (talk) 19:44, 11 April 2017 (UTC)
Birgit Müller (WMDE) (talk) 18:13, 12 April 2017 (UTC)
--Cornelius Kibelka (WMDE) (talk) 08:05, 17 July 2017 (UTC)
Retrieved from "https://wikimania2017.wikimedia.org/w/index.php?title=Submissions/Hackathon_goes_Hogwarts:_Creating_learning_environments_at_tech_events&oldid=40977"


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9 Page 9 de 13
Adam Ferrie and the Beginnings of Unitarianism in Canada
Ferrie, Helke (Author)
The Scottish-born businessman Adam Ferrie (1777-1863) immigrated to Montreal in 1829 after a successful but turbulent business life in Glasgow. In Montreal he began business in dry goods, hardware, groceries, and stationery, later opening branches run by his sons in Upper Canada. An outspoken political reformer in Scotland, in Lower Canada Ferrie counselled firm but balanced loyalism. He opposed both the radical Patriotes and Doric Club in the 1830s and 1840s, and the "pretended liberals" Hincks and Lafontaine in the 1850s. He helped organize Montreal's Board of Trade in 1842, sat on the city's municipal council from 1840 to 1843, and was appointed to the Legislative Council in 1843. He was involved in several philanthropic undertakings in Montreal, including helping cholera-stricken immigrants in 1832 and 1834, heading the committee managing the Montreal Public Bakery from 1837 to 1840, and a woodyard where the poor could purchase fuel at reasonable prices. He retired to Hamilton in 1853 to be closer to his children. The author, the wife of one of Adam Ferrie descendants, describes Adam Ferrie's tenuous links to the establishment of Unitarianism in Montreal in the 1840s.
The Canadian Unitarian and Universalist Historical Society
http://www.canadianuuhistoricalsociety.ca/Helke%20Ferrie%20%20(1999)%20Adam%20Ferrie.pdf
Ferrie, Helke. "Adam Ferrie and the Beginnings of Unitarianism in Canada." 1–15. [S.l.]: [s.n.], 1999. http://www.canadianuuhistoricalsociety.ca/Helke%20Ferrie%20%20(1999)%20Adam%20Ferrie.pdf.
Search by title and author on WorldCat
http://quescren.concordia.ca/fr/resource/3YJ3YBFJ
Unitarians


text:
require 'test_helper'

class ProjectTest < ActiveSupport::TestCase
  test "should provide name" do
    p = Project.new
    assert ! p.save
  end
end

redpajama_set_name: RedPajamaGithub


3 Faculty of Law Faculty of Law, Centenary Buildings, Chepauk, Triplicane PO, Madras-600 005.
4 Government Law College Gandhi Nagar, Katpadi, Vellore -632 006.
10 Sriram Law Academy Plot No: 1690, H Block, 15th Street, Anna Nagar, Chennai 600 040.


text:
"""Tests for the format specification classes."""

import unittest

from dfvfs.analyzer import specification

from tests import test_lib as shared_test_lib

class FormatSpecificationStoreTest(shared_test_lib.BaseTestCase):
  """Class to test the specification store."""

  def testAddSpecification(self):
    """Function to test the AddSpecification function."""
    store = specification.FormatSpecificationStore()

    format_regf = specification.FormatSpecification('REGF')
    format_regf.AddNewSignature(b'regf', offset=0)

    format_esedb = specification.FormatSpecification('ESEDB')
    format_esedb.AddNewSignature(b'\xef\xcd\xab\x89', offset=4)

    store.AddSpecification(format_regf)
    store.AddSpecification(format_esedb)

    with self.assertRaises(KeyError):
      store.AddSpecification(format_regf)

if __name__ == '__main__':
  unittest.main()


Kyurinagas Revenge has been developed and published under the banner of Recotechnology S.L. This game was released on 25th October, 2016. You can also download Sonic Adventure 2 Battle.
Kyurinagas Revenge is a 2D game with loads of platforming, puzzles and actions. In this game you will fight alongside two heroes, a street fighter Broccoli Joe and samurai Kaoru Tamanegi. They will fight all along Feudal Japan in their latest adventure. During the game you can easily switch between the characters at any point during the game. You need to defeat your enemies and also need to solve the puzzles which the Kyurinaga will throw at you. The game has got a single player mode as well as a co-op mode. This game has got some stellar visuals and the sounds are also quite impressive. You can also download Minecraft Story Mode Episode 5.


Are you craving for some lip-smacking Italian food at Wynwood? Many are of the opinion that Italian cuisine is about only pasta and pizza and they turn to generic pizza stores where pizza doesn't taste anywhere close to authentic Italian ones. Start by identifying the restaurants that serve authentic Italian food.
One way to search for Wynwood restaurants and bars is by using Yellow Pages. Else you can look up online for information on local eat-outs. Proceed then by checking the individual websites of the restaurants and read through their online menu.
Are there bread slices in the starter menu list? Usually Italian restaurants serve complimentary bread slices and don't charge for it. They would have it as a dish in their menu only if the bread is too good.
Does it serve pizza with chicken, ketchup or pineapple toppings? Italians don't top their pizza with meat like chicken breast or roast-beef, pineapple, chorizo and cedar cheese. Keep in mind that ketchup is never added on an Italian pizza.
Do they spread tomato or pesto sauce on bread? Pesto is a condiment for pasta in Italy. They aren't used to dress salads and aren't spread on breads. Italians serve bread with a dash of extra olive oil and chopped tomatoes.
Is pepperoni pizza included in the menu? No authentic Italian restaurant would mention pepperoni in their menu which means pepper in their language. They would use 'salami' instead.
Now that you have a list of restaurants in hand, you have to find the best one in the area. You need to check out for food's quality, price and ambience. Go through the online reviews to know more on the taste and quality of the food. Though restaurants serving authentic Italian food may be costly, there are some which are affordable. Ask the locals for recommendations as they would have a better idea. After finalizing on a restaurant, get to know the dishes they specialize in as you don't want to miss their best dishes.
Wynwood has some of the coolest restaurants. Enjoy your stay with sumptuous Italian food.


It has been more than ten years since Ryan Nyquist last tasted victory at X Games, but he will be hungrier than ever before when he travels to Austin next week in search of a fifth gold medal.
The 36-year-old last celebrated first place when he doubled up in BMX Park and Dirt at X Games 2003 but has made the podium six times since then – including four second-place finishes.
But even with 20 events under his belt since first rocking up to X Games back in 1997, Nyquist is still targeting a return to the top step.
"Anytime I go to an event there's not a doubt in my mind I want to go there and ride well and try to get the top spot," he said. "It would be weird to me to go to an event and not care about doing well. I feel like that would be a sign for me that maybe I'm done.
"To me those medals still mean a lot. Now that a lot of years have gone past they almost mean more, because it's not as frequent.
He is renowned for a mixing old-school style with modern bangers and Nyquist believes this unique approach is key to both his longevity in the sport and success at X Games 2015.
"My style of riding is a little bit different to most people so my progression has always been a little bit branched out from the typical rider," he added.
"I think that kind of helps me stand out too because I'll do some tricks that a lot of guys won't. I'm fortunate because I feel like I've taken a path in my riding that not many have and I've kind of found my niche in the sport.
A stacked field is expected in Austin, particularly for BMX Park, but Nyquist isn't phased by the many up-and-coming riders who are making their names on the contest scene and online with big new tricks.
That's because Nyquist, spurred on by the relentless progression in the sport, might have a few surprises of his own to unveil in Texas and even further down the road.
"I have some stuff I'm working on but I'll just see how it goes," he said. "I'm a very methodical rider in the fact that I need to know I'm going to pull something before I just send it. It's very rare you'll see me just through a Hail Mary and hope it works.
"But sometimes things change. When you see people dropping in and it's all hammers then it's like well, maybe it's time.
Almost 20 years into his X Games career, Ryan Nyquist is far from calling it a day and gold medals could still be on the menu in Austin.
Check out how Ryan Nyquist gets on this year with daily reports from X Games Austin 2015 on Extreme.com!


text:
Q: What's the max size of a Xively datapoint value I would like to write JSON to a datapoint value in Xively (so I can store a value dictionary in a single datapoint). What is the max size (e.g. string length) I can write to a datapoint?

A: 255 bytes, so basically 255 utf-8 characters.


text:
Being human, we all have shortcomings, and we all make mistakes, and after marriage, our mistakes and sins do hurt our spouse.
Living Islam
Husband and Wife on the Path to Allah
When I first started studying the Deen of Islam in-depth, I harbored an almost 'awestruck' respect and reverence of practicing, religious Muslims; the way a teenaged, star-struck fan admires a popular celebrity in modern-day pop culture.
The idyllic picture of their personal lives that I'd painted in my naive young brain was based on my admiration of their lofty morals.
I assumed that the more righteous a person was, the more perfect and discord-free their personal and family life would be. I presumed that they would never have any disagreements with other Muslims, much less with their close family members.
More than a decade later, today, I know better, both because of my personal life experience and as a result of gaining some knowledge of Islam since those bygone days of youthful naiveté.
Below I'd like to share some of the truthful facts about the marriages and family lives of righteous Muslims that Allah has allowed me to glean through the study of His Book and of the seerah of Prophet Muhammad (peace be upon him), including the lives of some of his companions.
Fighting & Differences
I have learnt that, even if both a husband and wife are very righteous and Allah-fearing, it is not possible for their married life to be devoid of differences and the occasional fighting.
A happy marriage between two righteous people does not mean a complete absence of fighting and disagreements between them, nor is it true that neither spouse will never dislike something about the other.
After marriage, every husband-wife duo will have to live through some days in which they are very angry at their partner; ticked off about something the latter did or said.
Being human, we all have shortcomings, and we all make mistakes, and after marriage, our mistakes and sins do hurt our spouse – the one person who lives with us day in and day out.
Several examples of this glaring reality of marriage viz. the presence of disagreements between righteous husbands and wives are present in the Quran and sunnah.
The first example is that of Prophet Muhammad (peace be upon him) and his wives, when he once became so angry at them all, that he vowed to not speak to or visit any of them for a month. This incident created quite a furor in Madinah and shows us that, even though the Prophet and his wives were the most righteous of this ummah, and all of them were destined for Jannah, their married lives were not without the occasional fights and undercurrents of anger.
There were even days when the Prophet's wives got upset with him, as proved by another relevant hadith that mentions how, during such times, when Aisha daughter of Abu Bakr would be upset with her husband, she'd swear by Allah by saying, "By the Lord of Ibrahim", instead of, "By the Lord of Muhammad", the latter being her normal preference. Another narration mentions how his wife Hafsah bint `Umar admitted to not talking to him sometimes during the day, in front of her father `Umar (may Allah be pleased with them all).
Another famous incident that is quoted in sahih ahadith, mentions how Ali ibn Abi Talib got his nickname 'Abu Turab' from his father-in-law, the Prophet. Ali once left his home in a huff after a disagreement with his wife, Fatimah daughter of Muhammad (may Allah be pleased with them both), and lay down on the dusty masjid floor so that some of his clothes/body got covered in dust.
This is indicative of how the happy marriage of Ali and Fatimah, one of the most righteous married couples in Islamic history, was not devoid of disagreements either.
An important point to note about this incident is that even though the Prophet clearly saw that Ali was angry, he lovingly called him by a nickname (perhaps to appease him?), which means that the older, close family members of the quarreling married couple (especially their parents and parents-in-law) should not make a big hullabaloo about their occasional trivial tiffs, unless the fighting gets serious in nature.
Hardship & Scarcity of Resources
The next fact based on real life, which my studies of the Quran and sunnah have taught me over the years, about marriages between righteous couples, is that they are almost always tested by the afflictions of hunger, hardship and/or straitened economic circumstances.
It is true that Prophet Muhammad deliberately chose to live a life of poverty, despite having full access to worldly resources that could allow him to live like a rich king (especially later on in his life), and all his wives supported him in this choice.
Allah had actually revealed Quranic verses offering his wives a choice: either they choose Allah and (staying married to) His Messenger whilst living in hardship, or they go for the luxuries of this worldly life. They all unanimously chose the latter, without hesitation.
However, there are other examples too. The Quran has made praiseworthy mention of a Muslim couple during the time of the Prophet, who honored and cared for their unexpected overnight guest to such a degree, that they both agreed to sacrifice the only single meal in their house that night, which was reserved just for their children, in order to serve it to their guest. They turned off the lamp before eating so that their guest would not find out that they were not eating anything.
Sahih Muslim has also recorded a narration about this incident. The Arabic word that Allah uses in the Quran to describe their situation is "khasasah", which means, poverty, or dire need.
Whichever the era in time since Islam came to this world, one constant thing that I have observed in the biographical accounts of the lives of our pious predecessors (al-salaf al-salih) is that they endured poverty and hunger.
Young singles of this ummah who wish to marry someone righteous and henceforth live a life upon the Deen of Islam, should therefore be forewarned that the trial of hunger and poverty might come from Allah as a test of their marriage, but to recall that, as it is the sunnah of Prophet Muhammad, it definitely has a lot of good in it.
If the husband and wife are able to weather this trial with patience and righteousness, they will emerge from it stronger and closer to Allah, both as a couple as well as individuals, Insha'Allah.
Separation by Long Distance
Lastly, yet another trying experience that many real life marriages endure, especially those between righteous couples, is being separated for a long time in the path/for the sake of Allah.
Many a time, if her husband is a da'ee who is actively involved in the propagation of Allah's Deen, a wife will have to endure days, weeks if not months or years without him at home, living either with his extended family or just with her children, alone, busy raising their next generation.
However, it is not just du'at whose marriages are tested by the separation of husband and wife for long times. Sometimes, when the husband loses his job (which happens a lot, in many marriages, cue point number two above) and the family is direly in need of an income, the only job offer that Allah might send his way could be in another city or country.
However, modern government-enforced visa and immigration restrictions as well as other factors related to the schooling/upbringing of their children, sometimes prevent a wife from joining her husband in the other country where he works, leading to weeks if not months of separation, which takes a toll on their marriage.
The prime example of this in Islamic history, whence a righteous couple endured a trying separation for the sake of Allah, is that of Prophet Ibrahim (peace be upon him) and his wife Hajar. He left her and their infant son Ismael in the barren and desolate valley of Makkah, at the command of Allah, so that it could eventually become inhabited and the house of Allah could be built there.
Any modern-day Muslim couple who wishes to marry a righteous person and eventually live a married life according to superlative levels of taqwa, piety, and faith, should recall the marital challenges faced by our pious predecessors and how they faced the difficulties after getting married with exemplary patience (sabr).
The path to Jannah is thorny, but with a righteous spouse by one's side, it becomes easier to tread than when trodden alone.
happy married life Allah heart spouse Couple weds peaceful
About Sadaf Farooqi
Sadaf Farooqi is an author, blogger and freelance writer based in Karachi, Pakistan. To date, Sadaf has authored over 300 original articles, most of which can be accessed on her blog, "Sadaf's Space" (sadaffarooqi.wordpress.com). She has recently started self-publishing her past articles as non-fiction Islamic books, which are available on Amazon and Kindle (www.amazon.com/author/sadaffarooqi)