There is a cube and an ant is performing a random walk on the edges where it can select any of the 3 adjoining vertices with equal probability. What is the expected number of steps it needs till it reaches the diagonally opposite vertex?

6Why is this titled and tagged as "logic"? – DaG Mar 20 '11 at 21:03

8@DaG: presumably because the vertices represent the eight distinct truth valuations of three boolean propositional variables, and the edges a change of exactly one of them. – Mitch Mar 20 '11 at 21:57

3[This question](http://mathoverflow.net/questions/39435/expectednumberofpinballstolightupall3channels) is equivalent. – Douglas Zare Mar 21 '11 at 07:36

@Mitch: Right, nice interpretation! So, if I say three false things, but each minute I take back something I had previously said, how long will it take for me to say three true things? – DaG Mar 21 '11 at 08:45
6 Answers
Extending to $d$dimensions
So I had enough time to procrastinate today and I extended this problem to $d$ dimensions. I would appreciate if someone could read through the answer and suggest any simplification to final answer if possible. Do the final numbers $u_n$ constitute a nice well known sequence? Further, what other interesting problems arise out this ant and the cube problem. I think another nice problem is the expected cover time. Thanks
Working out explicitly for $d=3$ gave some nice hints and a good picture of what is going on. We use these to extend to any ddimension.
The first thing is to define a cube in $d$dimensions. A $\mathbf{d}$dimensional cube is a set of vertices of the form $(i_1,i_2,\ldots,i_d)$ where $i_k \in \{0,1\}$. There is an edge between vertex $(i_1,i_2,\ldots,i_d)$ and vertex $(j_1,j_2,\ldots,j_d)$ if $\exists l$ such that $\left  i_l  j_l \right = 1$ and $i_k = j_k$, $\forall k \neq l$.
Two vertices are said to be adjacent if they share an edge. It is easy to note that every vertex shares an edge with $d$ vertices. (Consider a vertex $(i_1,i_2,\ldots,i_d)$. Choose any $i_k$ and replace $i_k$ by $1i_k$. This gives an adjacent vertex and hence there are $d$ adjacent vertices as $k$ can be any number from $1$ to $d$). Hence, the probability that an ant at a vertex will choose an edge is $\frac1{d}$.
For every vertex, $(i_1,i_2,\ldots,i_d)$ let $s((i_1,i_2,\ldots,i_d)) = \displaystyle \sum_{k=1}^d i_k$. Note that $s$ can take values from $0$ to $d$. There are $\binom{d}{r}$ vertices such that $s((i_1,i_2,\ldots,i_d)) = r$.
Let $v_{(i_1,i_2,\ldots,i_d)}$ denote the expected number of steps taken from $(i_1,i_2,\ldots,i_d)$ to reach $(1,1,\ldots,1)$
Let $S_r = \{ (i_1,i_2,\ldots,i_d) \in \{0,1\}^d: s((i_1,i_2,\ldots,i_d)) = r\}$
Claim: Consider two vertices say $a,b \in S_r$. Then $v_a = v_b$. The argument follows easily from symmetry. It can also be seen from writing down the equations and noting that the equations for $a$ and $b$ are symmetrical.
Further note that if $a \in S_r$, with $0 < r < d$, then any adjacent vertex of $a$ must be in $S_{r1}$ or $S_{r+1}$. Any adjacent vertex of $(0,0,\ldots,0)$ belongs to $S_1$ and any adjacent vertex of $(1,1,\ldots,1)$ belongs to $S_{d1}$. In fact, for any $a \in S_r$, $r$ adjacent vertices $\in S_{r1}$ and $dr$ adjacent vertices $\in S_{r+1}$.
Let $u_r$ denote the expected number of steps from any vertex $\in S_r$ to reach $(1,1,\ldots,1)$. For $r \in \{1,2,\ldots,d1\}$, we have \begin{align} u_r & = 1 + \frac{r}{d} u_{r1} + \frac{dr}{d} u_{r+1} & r \in \{1,2,\ldots,d1\}\\\ u_0 & = 1 + u_1 & r = 0\\\ u_d & = 0 & r = d \end{align} Setting up a matrix gives us a taidiagonal system to be solved. Instead, we go about solving this as follows.
Let $p_r = \frac{r}{d}$, $\forall r \in \{1,2,\ldots,d1\}$. Then the equations become \begin{align} u_r & = 1 + p_r u_{r1} + (1p_r) u_{r+1} & r \in \{1,2,\ldots,d1\}\\\ u_0 & = 1 + (1p_0) u_1 & r = 0\\\ u_d & = 0 & r = d \end{align} Let $a_{r} = u_{r+1}  u_r$. Then we get \begin{align} p_r a_{r1} & = 1 + (1p_r)a_r & r \in \{1,2,\ldots,d1\}\\\ a_0 & = 1 & r = 0 \end{align} Note that $u_m =  \displaystyle \sum_{k=m}^{d1} a_k$ and $u_d = 0$ \begin{align} a_0 & = 1 & r = 0\\\ a_{r} & = \frac{p_r}{1p_r} a_{r1}  \frac1{1p_r} & r \in \{1,2,\ldots,d1\} \end{align} Let $l_r = \frac{p_r}{1p_r} = \frac{r}{dr}$ \begin{align} a_0 & = 1 & r = 0\\\ a_{r} & = l_r a_{r1}  (1+l_r) & r \in \{1,2,\ldots,d1\} \end{align} \begin{align} a_1 &= l_1 a_0  (1+l_1)\\\ a_2 & = l_2 l_1 a_0  l_2(1+l_1)  (1+l_2)\\\ a_3 & = l_3 l_2 l_1 a_0  l_3 l_2 (1+l_1)  l_3 (1+l_2)  (1+l_3)\\\ a_m & = \left( \prod_{k=1}^{m} l_k \right) a_0  \displaystyle \sum_{k=1}^{m} \left((1+l_k) \left( \prod_{j=k+1}^m l_j \right) \right) \end{align} Since $a_0 = 1$ and $l_0 = 0$, we get \begin{align} a_m & =  \displaystyle \sum_{k=0}^{m} \left((1+l_k) \left( \prod_{j=k+1}^m l_j \right) \right) \end{align} Hence, \begin{align} u_n & =  \displaystyle \sum_{m=n}^{d1} a_m\\\ u_n & = \displaystyle \sum_{m=n}^{d1} \left( \displaystyle \sum_{k=0}^{m} \left((1+l_k) \left( \prod_{j=k+1}^m l_j \right) \right) \right)\\\ u_n & = \displaystyle \sum_{m=n}^{d1} \left( \displaystyle \sum_{k=0}^{m} \left(\frac{d}{dk} \left( \prod_{j=k+1}^m \frac{j}{dj} \right) \right) \right)\\\ u_n & = \displaystyle \sum_{m=n}^{d1} \frac{\displaystyle \sum_{k=0}^{m} \binom{d}{k}}{\binom{d1}{m}} \end{align} Note that \begin{align} u_n & = \frac{\displaystyle \sum_{k=0}^{n} \binom{d}{k}}{\binom{d1}{n}} + u_{n+1} & \forall n \in \{0,1,2,\ldots,d2 \} \end{align} The expected number of steps from one vertex away is when $n = d1$ and hence $u_{d1} = 2^d1$
The expected number of steps from two vertices away is when $n = d2$ and hence $u_{d2} = \frac{2d(2^{d1}  1)}{d1}$
The answers for the expected number of steps from a vertex and two vertices away coincide with Douglas Zhare's comment
Initial Solution
Problems such as these fall in the category of Markov chains and one way to solve this is through first step analysis.
We shall denote the vertices of the cube by numbers from $1$ to $8$ with $1$ and $8$ being the opposite ends of the body diagonal.
Let $v_i$ denote the expected number of steps to reach the vertex numbered $8$ starting at vertex numbered $i$.
$v_1 = 1 + \frac{1}{3}(v_2 + v_4 + v_6)$; $v_2 = 1 + \frac{1}{3}(v_1 + v_3 + v_7)$; $v_3 = 1 + \frac{1}{3}(v_2 + v_4 + v_8)$; $v_4 = 1 + \frac{1}{3}(v_1 + v_3 + v_5)$; $v_5 = 1 + \frac{1}{3}(v_4 + v_6 + v_8)$; $v_6 = 1 + \frac{1}{3}(v_1 + v_5 + v_7)$; $v_7 = 1 + \frac{1}{3}(v_6 + v_2 + v_8)$; $v_8 = 0$;
Note that by symmetry you have $v_2 = v_4 = v_6$ and $v_3 = v_5 = v_7$.
Hence, $v_1 = 1 + v_2$ and $v_2 = 1 + \frac{1}{3}(v_1 + 2v_3)$ and $v_3 = 1 + \frac{2}{3} v_2$.
Solving we get $$\begin{align} v_1 & = 10\\ v_2 = v_4 = v_6 & = 9\\ v_3 = v_5 = v_7 & = 7 \end{align}$$
Hence, the expected number of steps to reach the diagonally opposite vertex is $10$.
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5Beautiful answer! I clicked 3 times quickly on the up arrow but unfortunately it didn't work. – t.b. Mar 20 '11 at 22:36

5The expected number of steps to reach any vertex, in this case, are all integers. This seems surprising to me; does anyone know a nice reason why? – Michael Lugo Mar 21 '11 at 00:33

@Sivaram Hm, Wikipedia doesn't have a dedicated page to "first step analysis." Is this kind of an adhoc thing you came up with or a recognized technique with a kind of theory behind it? – Uticensis Mar 21 '11 at 04:40

@Billare: I didn't come up with this. It is usually discussed in an introductory stochastic/Markov chain class. Of course you need not use this. You can write an infinite expression for $v_i$'s and note the repeating pattern to get the answer... – Mar 21 '11 at 04:45

@Michael Lugo: For higher dimensional cubes, the values are not necessarily integers. For the $4$cube, the values are $64/3$, $61/3$, $56/3$, and $15$. – Douglas Zare Mar 21 '11 at 07:34

@Douglas: I haven't worked it out for 4cube but I am wondering how the denominator has a factor of $3$... My intuition says the denominator should have a factor of $4$ – Mar 21 '11 at 07:42

@Sivaram: You can write the system of equations as $M \vec v = \vec 1$ where $\vec 1$ is the vector with each entry $1$ and $M$ is a $4\times 4$ tridiagonal matrix. You can clear the denominators of $M$ by multiplying by $4$, but that does not work for $M^{1}$. – Douglas Zare Mar 21 '11 at 08:21

By the way, the expect time to get to a vertex starting from an adjacent vertex is $2^d1$, or $15$ for $d=4$. That lets you check that the expected number of steps from a distance of $2$ is $\frac{d}{d1} (2^d1)  1$, or 56/3 for $d=4$. – Douglas Zare Mar 21 '11 at 08:30

@Douglas: $d=4$, then $\frac{d}{d1}(2^d1)1 = \frac{4}{3}(2^41)1 = \frac{4}{3}(15)1 = 19$. Further I am not sure if the formula is right since $d=3$ gives $\frac{3}{2}(2^31)1 = \frac{3}{2} \times 7  1 \neq 9$. I think you mean $\frac{d}{d1}(2^d2)$? – Mar 21 '11 at 08:45


what is the 1 added for here? i thought that if ant starts from v1, it has 3 options to choose v2,v4 or v6, so i understand how 1/3(v2+v4+v6) is coming. Can someone explain the 1? – huy Sep 08 '20 at 04:11

+1, the bunch of equations in the end are exact the same equations that appears when one uses markov chain, with start at 1 and absorption state as 8. – Dhruv Joshi Aug 03 '21 at 13:59
Call the set containing only the starting vertex $A$. You can move to any of three vertices next. Call the set of them $B$. For the next step, you can go back to $A$, or you can move on to any of three new vertices. Call the set of those vertices $C$. Finally, call the set containing the goal vertex $D$.
Call the expected number of steps from $A$ to $D$ $E(AD)$ etc.
Consider $E(BD)$. We can write an equation for $E(BD)$ by considering what happens if you start at $B$ and take two steps. You could go to $C$ and then to $D$. The probability of this is $2/3$ for the first step and $1/3$ for the second, or $2/9$ over all.
You could also go to $C$ and back, or to $A$ and back. Either way, your new expected number of steps to $D$ is the same as it was before, because you're back where you started. The probability of this is $7/9$ because probabilities add to one.
This gives
$E(BD) = 2/9(2) + 7/9(2 + E(BD))$
which means
$E(BD) = 9$
It takes one step to go from $A$ to $B$, so
$E(AD) = 10$
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Hint: there are three kinds of vertices besides the target itself: 1) those one step from the target 2) those two steps from the target 3) the vertex opposite to the target Let u_i be the expected number of steps, starting at a vertex of type i, until it reaches the target. Considering the possibilities for the first step from a vertex of type i, write three equations expressing u_i in terms of the other u_j. Then solve.
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This is a beautiful idea that can be used to solve problems of this nature. Could you please share some resources to solve this type of problems for any polyhedron? – Gray_Rhino Sep 19 '20 at 13:45
Here is what I thought. It is a Markov Chain Problem.
Mark the start point as $E_3$, here we define $E_n$ which means that it takes $n$ sides to reach the final point. we have such relationship: take one side, it will definitely go to point as $E_2$:
$$E_3 = E_2 + 1$$
the probability is 2/3 to $E_1$ and 1/3 back to $E_3$:
$$E_2 = 2/3 (E_1 + 1) + 1/3 (E_3 + 1)$$
similar to $E_1$:
$$E_1 = 1/3 + 2/3 (1 + E_2)$$
solve these equations, you will get $E_3$ = 10.
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The other answers are great and probably the best way to solve this problem, but here's another approach just for fun.
cube with labeled vertices for reference
The goal is to determine the expected number of steps required to move from vertex A to vertex H.
We will parameterize each vertex by the minimum number of steps $d$ away from vertex H. So for vertex A, $d=3$. Vertices B, C, and E have $d=2$. Vertices D, G, and F have $d=1$, and vertex H has $d=0$.
We will denote the probability of moving from a vertex with $d=i$ to one with $d=j$ $p_{ij}$.
By looking at the cube figure,
$p_{32}=1$ since you can't move any further from the target vertex
$p_{23}=1/3$ , $p_{21}=2/3$
$p_{12}=2/3$ , $p_{10}=1/3$
ALet $x_{ij}$ be the number of times moved from a vertex with $d = i$ to one with $d=j$.
Since we stop at the condition of reaching vertex H ($d=0$), $x_{10}=1$, as we only want to move there once. In order to have a complete path from vertex A to vertex H, $x_{21} \geq 1$, $x_{32} \geq 1$. Furthermore, the number of steps backward from a vertex must be one less than the number of forward steps. Thus $x_{23} = x_{32}1$, $x_{12} = x_{21}1$.
Therefore the probability of a path with $x_{32}$ transitions from $d=3$ to $d=2$ and $x_{21}$ transitions from $d=2$ to $d=1$ is just the probability of each transition raised to the number of those transitions.
$Pr(x_{32},x_{21}) = a p_{32}^{x_{32}}p_{21}^{x_{21}}p_{12}^{x_{21}1}p_{23}^{x_{32}1} p_{10}$
Where $a$ is the number of possible arrangements of those transitions.
Using $p_{32}=1$, $p_{10}=p_{23}$, and $p_{12}=p_{21}$, this can be rewritten as
$Pr(x_{32},x_{21}) = a p_{21}^{2x_{21}1}p_{23}^{x_{32}}$.
Determining $a$ is somewhat tricky. The first transition and last transition are fixed. All the transitions from a vertex of $d=2$ to $d=1$ must be followed by the reverse except for the last one. Otherwise we would have reached the target vertex prematurely. Likewise all the transitions from a vertex of $d=3$ to $d=2$ must be followed by the reverse, as you can't be further away. This means transitions from $d=1$ to $d=2$ and from $d=2$ to $d=3$ can not be freely arranged. Also the last transition from $d=2$ to $d=1$ must be the second to last transition, occurring right before the transition to the target vertex. This leaves the number of transitions to be freely arranged in any order is
$x_{23}+x_{21}1$.
The number of ways we can arrange $x_{23}$ transitions in $x_{23}+x_{21}1$ total transitions is
${x_{23}+x_{21}1}\choose x_{23}$ or, by writing $x_{23}$ in terms of $x_{32}$, ${x_{32}+x_{21}2}\choose {x_{32}1}$.
By substituting a for this we arrive at the probability we are looking for.
$ Pr(x_{32},x_{21}) = {{x_{32}+x_{21}2}\choose{x_{32}1}} p_{21}^{2x_{21}1} p_{23}^{x_{32}}$.
The total number of transitions for a path with $x_{32}$ transitions from $d=3$ to $d=2$ and $x_{21}$ transitions from $d=2$ to $d=1$ is $2x_{32} + 2x_{21} 1$.
Therefore to determine the expected number of transitions, $n_{trans}$, we multiply this by the probability and sum over $x_{32}$ and $x_{21}$ from 1 to $\infty$.
$\left<n_{trans}\right> = \sum\limits_{x_{21}=1}^\infty\sum\limits_{x_{32}=1}^\infty \left(2x_{32} + 2x_{21} 1\right){{x_{32}+x_{21}2}\choose{x_{32}1}} p_{21}^{2x_{21}1} p_{23}^{x_{32}} $
The result of this summation is $\left<n_{trans}\right> = 10$
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I've been challenged whith this exercice but with a slight difference.
The difference is that if I get vertex 8 (beginning from 1) then the ant dies. And the question is "What probabilities I have to die?"
I know from the above answers that the number of expected steps from 1 to 8 is 10, then it means that is more difficult to die on vertex 8 than in vertex for example 7, due to the expected number of steps to reach vertex 7 is smaller than in vertex 8 and that means that the probability to die is higher on vertex 7 because of is easier to reach. But...I don't know exactly how to express it in terms of PROBABILITY.
Anyone could open my mind? Thanks



Is there any other vertex where you can die? I understood you were walking on the edges of the cube until you met your fate at vertex 8. Which will happen almost surely. – Did Mar 30 '11 at 13:13

Yes, there is another vertex where can die. It is vertex number 5. The other question is if the ant has some probabilty to survive. I'm totally sure the answer is not. – Mar 30 '11 at 13:44

Didier, I already know the answer. Probability to die on vertex 8 is 3/7 and in vertex 5 is 4/7. The solution has just been released by the teacher who ask for answers. The solution are in the following link http://www.elpais.com/articulo/sociedad/hormiga/salvacion/posible/elpepusoc/20110330elpepusoc_6/Tes – Mar 30 '11 at 13:53

It's spanish, use the google translate option. Thanks for your interest to help me, if you need some help in the translation don't hesitate to say it to me – Mar 30 '11 at 13:56
