How does Dominated work? Can it really make me attack allies or walk off a cliff?

In one of the games I was in, a creature was able to Dominate my friend and he was getting angry that the DM forced his character to attack and knock prone an NPC that we were trying to protect. They got into a long argument trying to figure out if the DM can do that or not. We finally resolved the problem by deciding that the DM can only choose the action the dominated character can do, but not force him to move off a cliff or attack with an at-will attack of the DM’s choice on an NPC.

So I ask: how does Dominated really work, can a DM force a player’s character go walk off a cliff or attack an NPC?

Does Air Walk Require a Fly Action?

The end of the Fly action states:

If you’re airborne at the end of your turn and didn’t use a Fly action this round, you fall.

However, it also requires you to have a fly speed. While this makes perfect sense for the Fly spell, the Air Walk spell says:

The target can walk on air as if it were solid ground.

This gives the target the ability to move through the air, but it doesn’t specifically say it’s a move speed. Additionally, if the target treats the air as solid ground, they wouldn’t normally need to spend an action to not fall through the “floor.”

My concern is that these two spells are the same level, but seem to function very differently. Does the Air Walk spell necessitate the usage of the Fly action, or does it literally allow you to stand on air?

Personagem volta pra trás no walk cycle unity 3d

Olá, a minha dúvida é a seguinte:

Sou iniciante na Unity3d e estou com um problema nas animações, o personagem caminha, e volta pra trás, e assim repetidamente. Segue a imagem do animation controller: Meu animation ontroller

Meu script C#:

using UnityEngine; public class playerMove : MonoBehaviour {     private Animator anim;     public float speed = 2.0f;     void Start()     {        anim = GetComponent<Animator>();     }     void Update()     {        if (Input.GetKey(KeyCode.W))         {             transform.Translate(Vector3.forward * Time.deltaTime);             anim.SetBool("isWalking", true);             anim.SetBool("idle", false);         }         else if (Input.GetKey(KeyCode.S)) {         transform.Translate(Vector3.back * Time.deltaTime);         anim.SetBool("isWalking", true);         anim.SetBool("idle", false);     }     else {         anim.SetBool("isWalking", false);         anim.SetBool("idle", true);     } } } 

Transition matrix for shortest path walk

Consider a directed, weighted graph $ G$ . Let $ s$ and $ t$ be two distinct vertices of $ G$ and consider a walker that starts at $ s$ and traverses a random shortest path from $ s$ to $ t$ , chosen uniformly at random from all shortest paths from $ s$ to $ t$ .

Question: What is the transition matrix $ P$ that characterises this “random walk” (specifically, choosing a random shortest path) that starts at $ s$ and ends at $ t$ ?

Intuitively, it seems clear that $ P$ is given by \begin{equation} P_{ij} = \frac{\omega_{ij}}{\sum_k \omega_{ik}}\,, \end{equation} where $ \omega_{ij}$ is the number of shortest paths from $ s$ to $ t$ that traverse the directed edge $ (i,j)$ .

To see this, consider the $ k$ shortest paths from $ s$ to $ t$ . If we let $ k$ walkers walk each of the $ k$ shortest paths, their empirical transition matrix is given by $ P$ .

However, this is not a rigorous proof. Can someone provide a rigorous proof or fill in the missing step(s)?

Sampling in large graph using simple random walk

I’m studying sampling techniques in online social networks. The assumption is we don’t have full access to the network(i.e, we don’t know the size of the network). However crawling is supported, i.e, starting with any arbitrary node we can access its neighbors. Hence we go for random walk(Markov Chain) based crawling techniques as it requires only neighborhood information for any node.

Let’s assume we choose simple random walk(SRW) for crawling the network.

In SRW we start with an arbitrary seed node then choose one of the neighbors as new node. At the new node we choose one of its neighbors and this process continues.

Let G be our graph show in above figure, assume graph is large and size of the graph is not known. We go for SRW with node ‘A’ as initial node.

enter image description here

Since the size of the graph is not known how do we select initial distribution $ X_0$ (at time 0), and initial transition probabilities $ P_0$ (at time 0). After time 0 if I choose one of the neighbors(say B), how the transition probabilities $ P_1$ (at time 1) and distribution $ X_1$ (at time 1) are calculated.

WIll a proliferating 3D random walk a.s. revisit the origin?

The concept of a “proliferating random walk” on a lattice is that at any time $ t \in \Bbb N \cup 0$ , there is some set consisting of at least one particle, each of which is on its own lattice point. When taking a time step, each particle also randomly decides whether to split into two particles, each of which moves to a different lattice point. If two particles would reach the same lattice point at some step, they coalesce back into one particle.

To be precise for the problem of interest, define for any real constant $ 0 \leq \rho \leq 1$ a $ \rho$ -proliferating walk in $ d$ dimensions as a Markov process evolving in discrete integer time steps, with the following properties:

  • The state $ S_t$ at any time $ t$ is described by a finite subset of $ \Bbb Z^d$ .

  • The “evolution component” attributed to any point $ x \in \Bbb Z^d$ is the set $ $ e_{t+1}(x) = \left\{ \begin{array}{cl} \emptyset & x \not\in S_t \ \left\{\begin{array}{l} r_2(x) \mbox{ with probability }\rho \r_1(x) \mbox{ with probability }1-\rho \end{array}\right.& x \in S_t \end{array}\right. $ $ where $ r_1(x)$ is set containing a single point one lattice step from $ x$ with all such sets chosen with equal probability, and $ r_2(x)$ is set containing two points each one lattice step from $ x$ with all such sets chosen with equal probability.

  • The state evolves as $ $ S_{t+1} = \bigcup_{x\in \Bbb R^d} e_{t+1}(x) = \bigcup_{x\in S_t} e_{t+1}(x) $ $

  • The initial state is $ S_0 = \{ 0^d \}$ , that is, one particle at the origin.

Further, for some $ d$ and $ \rho$ , define the probability $ RV(d,\rho)$ of revisiting the origin as the probability that at some time $ t>0$ the origin will be an element of $ S_t$ . I don’t use the term “return to origin” because that ought to be reserved to mean that $ S_t = S_0$ . The walk can revisit the origin even if $ S_t$ has many elements. If the $ d$ -dimensional random walk with proliferation $ \rho$ revists the origin a.s, we say that $ RV(d,\rho) = 1$ .

For $ \rho = 0$ , the proliferating random walk is an ordinary symmetrical random walk, and will return to the origin with probability of Polya’s random walk constant, roughly 34%. But for $ \rho > 0$ the number of particles in $ S_t$ tends to increase with $ t$ (and this increase is rapid for $ d>2$ ), so it is plausible that for sufficiently large $ \rho$ the Markov process for $ S_t$ revisits the origin almost surely, at least in $ 3$ dimensions.

It is easy to prove that $ RV(d,\rho) \geq RV(d,0)$ , and not much harder to show that if $ \xi > \rho$ then $ RV(d,\xi) \geq RV(d,\rho)$ . I strongly suspect that $ RV(3,1) = 1$ , that is, the 3-D random walk with particle doubling at every time step revisits the origin a.s. — but I can’t prove that.

The generic question that arises is to characterise, for any given $ d$ , the range of $ \rho$ such that $ RV(d,\rho) < 1$ . In particular:

Is $ RV(3,1) = 1$ , and if so, what can be said about the minimum value of $ \rho$ such that $ RV(3,\rho) = 1$ ?

Entropy of endpoints of a random walk in a dense graph

Let $ p\in[0,1]$ be a constant and let $ G$ be a graph with $ n$ vertices and $ \approx p\binom{n}{2}$ edges. If you’d like, consider $ p=1/2$ .

Let $ X$ be a random vertex of $ G$ chosen proportional to its degree (i.e. the probability that $ X=v$ is $ d(v)/2|E(G)|$ ). That is, $ X$ is chosen according to the stationary distribution of a random walk. Now, generate two standard random walks starting at $ X$ denoted $ A_0,A_1,\dots$ and $ B_0,B_1,\dots$ where $ A_0=B_0=X$ .

For $ k\geq1$ , define $ \alpha_k$ so that $ H(A_k,B_k) =\log_2(\alpha_kn^2)$ where $ H$ denotes the binary entropy. That is, $ \alpha_k:=2^{H(A_k,B_k)}/n^2$ . The way that entropy works is that $ \alpha_k\approx1$ if and only if the distribution on $ (A_k,B_k)$ is roughly uniformly on $ V(G)^2$ .

Question: Given the value of $ \alpha_i$ , what kind of lower bound can one prove on $ \alpha_j$ for $ j>i$ ?

I am mainly interested in constant values of $ i,j$ . For example, I would be quite interested to know whether all such graphs satisfy any (non-trivial) bound of the form $ $ \alpha_2\geq f(\alpha_1)$ $ for some function $ f$ . I’d be very happy to know if there are any existing results related to this; please excuse my lack of knowledge on random walks…

Just how gaseous is the “gaseous form” of Wind Walk?

The Wind Walk spell allows creatures to assume a “gaseous form” that enables them to fly very quickly:

You and up to ten willing creatures you can see within range assume a gaseous form for the duration, appearing as wisps of cloud. While in this cloud form, a creature has a flying speed of 300 feet and has resistance to damage from nonmagical weapons. The only actions a creature can take in this form are the Dash action or to revert to its normal form.

Unlike the spell whose actual name is Gaseous Form, the gaseous form of Wind Walk doesn’t say anything about various abilities that might be enabled by being non-solid, such as entering another creature’s space or squeezing through small cracks. However, many of these abilities are implied by the term “gaseous form”, even if you don’t read it as a reference to the spell of the same name.

So, does Wind Walk actually make an affected creature non-solid and allow them to do anything a gas could, similar to Gaseous Form? Other than a very fast fly speed, what additional abilities, if any, are granted to a creature by Wind Walk?