dnd 5e – Does the Mordenkainen’s Magnificent Mansion spell’s mansion exist on a separate plane of existence?

Yes, there is a difference. They are not the same plane of existence and the Telepathy spell will not work in the situation you have outlined.

The Dungeon Masters Guide (DMG) has this to say on Demiplanes:

Demiplanes are extradimensional spaces that come into being by a variety of means and boast their own physical laws. Some are created by spells. Others exist naturally, as folds of reality pinched off from the rest of the multiverse. Theoretically, a plane shift spell can carry travelers to a demiplane, but the proper frequency required for the tuning fork would be extremely hard to acquire. The gate spell is more reliable, assuming the caster knows of the demiplane.

A demiplane can be as small as a single chamber or large enough to contain an entire realm. For example, a Mordenkainen’s magnificent mansion spell creates a demiplane consisting of a foyer with multiple adjoining rooms, while the land of Barovia (in the Ravenloft setting) exists entirely within a demiplane under the sway of its vampire lord, Strahd von Zarovich. When a demiplane is connected to the Material Plane or some other plane, entering it can be as simple as stepping through a portal or passing through a wall of mist.

From this we can see that the space created by Mordenkainen’s Magnificent Mansion (MMM) is a Demiplane.

The Telepathy spell has this text:

The creature can be anywhere on the same plane of existence as you. The spell ends if you or the target are no longer on the same plane.

From the DMG rules on demiplanes it is clear that the MMM demiplane is not the same plane of existence as the plane where the entrance to MMM was created from. This is made explicit by the fact that the Plane Shift spell requires a tuning fork tuned to the demiplane specifically to access them.

We can see this from Plane Shift’s material component:

a forked, metal rod worth at least 250 gp, attuned to a particular plane of existence

As a result Telepathy will not work between outside MMM and inside MMM.

air travel – Can I bring a single shot of live ammunition onto the plane from US to UK as a souvenir?

As an add on to existing answers consider your question

Can I bring a single shot of live ammo onto the plane from US to UK as a souvenir?

as being seen by security authorities as looking something like

Can I bring a single shot informal firearm onto an aircraft if I also carry the single live ammunition round with me but separately from the firearm ?

Seen in this light, without knowing the specific regulations it’s obvious (or should be 🙂 ) that the answer is probably “NO!!!”, or at very best “Only after very large amounts of questions and paperwork, probably not as carry on, probably not without significant pre-application period and, probably, no!”


If you have Facebook access then this page will be informative Bootleg Gunz – The Bazaar of Bizarre Guns will be informative re what can be achieved.

Genuine: 22 LR, Australia

enter image description here

A “bootleg” / ‘slam-fire’ / … firearm can be implemented in many ways from equipment and material that does not resemble a conventional firearm. The common key component to any such weapon is a live round – if you have that the weapon proper is doable.

Bearing in mind also that if you are able to carry such a round onto an aircraft then Bob & Carol & Ted & Alice (and Uncle Tom Cobbly and all) also can and a number of you can then pass your ammunition to someone who has a/the weapon.

dnd 5e – Can Plane Shift be used to transport from mordenkainen’s magnificent mansion to the same plane the mansion was created?

Sure.

Mordenkainen’s Magnificent Mansion creates what is called a demiplane:

Demiplanes are small extradimensional spaces with their own unique rules. They are pieces of reality that don’t seem to fit anywhere else. Demiplanes come into being by a variety of means. Some are created by spells, such as demiplane, or generated at the desire of a powerful deity or other force. They may exist naturally, as a fold of existing reality that has been pinched off from the rest of the multiverse, or as a baby universe growing in power. A given demiplane can be entered through a single point where it touches another plane. Theoretically, a plane shift spell can also carry travelers to a demiplane, but the proper frequency required for the tuning fork is extremely hard to acquire. The gate spell is more reliable, assuming the caster knows of the demiplane.

So while in the mansion, you are not on the material plane. Plane shift says:

You and up to eight willing creatures who link hands in a circle are transported to a different plane of existence. You can specify a target destination in general terms.

Since the Mansion is on a different plane of existence from the material plane, you can plane shift from the mansion to a location on the material plane, assuming you have a tuning fork properly attuned to the material plane.

air travel – Can I bring a single shot of live ammo onto the plane from US to UK as a souvenir?

It is not permitted to bring ammunition into the UK without a permit. According to this brochure from the UK Border Force (p. 14–15):

There are certain goods you are not allowed to bring into the UK – this is to protect society, animal and public health and the environment.

The following are also banned but in certain cases may be brought into the UK if you have obtained the relevant licence or permit:

Firearms, explosives and ammunition

It is unlikely that it will be worth your while to obtain the proper permits simply to bring one round of ammunition into the UK.

computational geometry – Mahalanobis distance of point to plane algorithm

I am trying to understand the Mahalanobis distance of a point from the plane given by this paper. The algorithm is given below:

  1. Calculate covariance of point $S_{uu}$
  2. Apply a whitening transform to the covariance matrix using SVD where $S_{uu} = RVR^T$. $R$ is the rotation matrix and $V$ is the scale $V = diag(a^2, b^2, c^2)$
  3. The output is transformed into 3 homogenous 4×4 matrices

$Rh = begin{bmatrix} R & 0 \ 0^T & 1 end{bmatrix}, Th = begin{bmatrix} I & 0 \ -u^T & 1 end{bmatrix}, Ah = diag(a,b,c,1)$

where u is the centroid of the plane

4.The transformed plane is given by $q = Ah*Rh*Th*p$ where p is the planar parameters

My first question is regarding the whitening transform. I am not sure how it is being derived (its not given in the paper). If anyone can provide an intuitive explanation, that would be great.

My second question is – I simulated a plane with planar parameters (0, 0, 1) and points on the plane with std error of 0.02. enter image description here We can recover the planar parameters using svd. Now, when I run the algorithm to get the transformed plane using the following code:

# whitening transform of point covariance
cov_pts = np.cov(pts.transpose())
r, v, rt = np.linalg.svd(cov_pts)
abc = np.sqrt(v)
ah = np.diag(v=np.concatenate((abc, np.array((1)))))
rh = np.zeros((4, 4))
rh(0:3, 0:3) = r
rh(3, 3) = 1
th = np.eye(4)
th(3, 0:3) = -center
transformed_plane_fit = np.dot(np.matmul(np.matmul(ah, rh), th),
                                       np.concatenate((plane_fit, np.array((1)))))

the values I get for transformed planar fit is array((-2.99291157e-02, -7.31159923e-04, -1.97310249e-02, 9.11576594e-01)) which doesn’t seem to correspond to the planar parameters. What can be the reason for the discrepancy?

dnd 5e – Are there any elemental plane rules or details?

The 5e information on the planes is, at the time of writing, sparse. There’s some in Dungeon Master’s Guide, but only broad strokes. Mordenkainen’s Tome of Foes alludes to some more details when discussing some extraplanar creatures. Sword Coast Adventure Guide and Eberron: Rising from the Last War discuss some of the cosmology of the Forgotten Realms and Eberron, respectively. But ultimately, there is effectively not much.

Planescape

Previous editions are much, much more detail-rich about the planes. The wider multiverse was, itself, a detailed campaign setting known as Planescape.

Second Edition

Second edition is where the vast, vast majority of Planescape material can be found. Also, whereas 3e tends to avoid making assumptions about other books someone might own, and thus does not really go into planar concerns outside of specific books dedicated to them, in 2e they tended to assume every book was in play and so planar concerns might crop up anywhere and everywhere. This makes providing a list prohibitive.

However, there is a book literally entitled Inner Planes—that is, the elemental ones. That one should definitely be worth a read for you.

Anyway, to offer a bit more detail, 2e introduced the “Great Wheel” as the explanation of the planes, which 5e has restored to its rightful place. The Great Wheel refers to the “Outer Planes,” sixteen aligned planes around the neutral plane known as the Outlands. These are arrayed much like your typical alignment grid, with Lawful-Good Mount Celestia in the “upper left” and the Chaotic-Evil Abyss in the “lower right.” As typically presented, the top part of the wheel is Good (the “Upper Planes”) while the bottom part is Evil (the “Lower Planes”), and the left side is Lawful while the right side is Chaotic. (Of course, since it’s a ring, these designations are fairly arbitrary.) Within the wheel were the Inner Planes (of elements and energy), and within those was the Material Plane (where the vast majority of other campaign settings can be found).

Aside from the planes themselves, major points of interest in the Great Wheel include Sigil, which is the great City of Doors atop the infinite Spire in the middle of the Outlands, and is considered the exact center of the multiverse (about which the “wheel” revolves). Sigil is ruled by the enigmatic Lady of Pain, who enforces very few rules but among them is that she is not to be worshiped as a goddess; violations of this rule are punishable by flaying, a punishment which the Lady delivers personally. Another rule is a bar on gods and other so-called “powers”—the Lady summarily curb-stomped the last one who challenged that rule, and the complete and utter lack of any attempt by others to challenge it suggests she could do so again.

The other notable thing are the exemplars, the creatures of raw belief and alignment that spawn naturally on the Outer Planes. They are (left-to-right, top-to-bottom) the LG archons, the NG guardinals, the CG eladrin, the LN modrons, the N rilmani, the CN slaad, the LE devils, the NE yugoloths, and the CE demons. I’ve written more about the exemplars here.

Third Edition

Third edition, or more specifically “v.3.5 revised edition,” had two dedicated Planescape books, Manual of the Planes and Planar Handbook. Other books such as Fiendish Codex I and II, Lords of Madness, and Fiend Folio also touch on the planes to a fair degree.

Third edition expanded upon 2e’s Planescape, but for the most part it just translated various 2e things into 3e terms. There is novel stuff in there (particularly in the Fiendish Codexes), but 2e is still the primary source.

Fourth Edition

Fourth edition had a number of unexplained, and unpopular, departures from the canon of prior editions, and 5e has seemed to be quietly ignoring a lot of them. Nonetheless, this is where you would look for various planar features that originated in 4e (its novel content tended to be more popular than its alterations to existing content), like the Feywild. Like 2e, 4e explicitly assumes all books are “core,” which does again tend to cause things to be a bit scattered all over the place.

Also, 4e just didn’t do a lot of fluff—where 2e and 3e included books with hundreds of pages of near-solely fluff, going into the setting in great detail, 4e tended to limit fluff to a sentence or three in the introduction to some mechanical option (whether it be a class, a feat, or a monster). So there just isn’t as much to find in 4e anyway.

You are probably safe ignoring 4e unless you are particularly interested in the Feywild or the Elemental Chaos (and much of the detail of the latter seems to have changed in 5e, since we have the elemental planes again that 4e replaced with the Elemental Chaos).

diophantine equations – Finding Pythagorean quadruples on a given plane?

In 2D one cannot construct Pythagorean triples $x^2+y^2=m^2$ ($x,y,minmathbb{Z}$) that lie on every line through the origin (e.g., a Pythagorean triple with $x=y$ would require $sqrt{2}$ to be rational).

What happens when moving to planes in 3D?

Given $a,b,cinmathbb{Z}$ can one find $x,y,z,minmathbb{Z}$ such that $mne 0$, $x^2+y^2+z^2=m^2$, and $ax + by + cz = 0$?

I would be happy with a counterexample (as in the 2D case) but happier with a construction, since it would lead to a nifty algorithm for approximating a 3D model with one that has only rational-coordinate unit-length normals.

What I have noted so far:

If $a,b,c,||(a,b,c)||$ is itself a Pythagorean quadruple the answer is clearly “yes”, and the construction involves using $(a,b,c)$‘s perpendiculars(1) to transform Pythagorean triples from the 2D plane.

(1) All Pythagorean quadruples have (at least two) perpendiculars owing to the form of their parameterization — https://en.wikipedia.org/wiki/Pythagorean_quadruple

graphics – Find point along x axis where yz plane flips to back facing after perspective projection. Have a solution but don’t know why it works

I am trying to find the point along the x axis where a plane with normal pointing along the x axis would flip to back facing after perspective projection. Essentially the red line in this image.

My first attempt to find it I just sampled along the x axis checking if a triangle there was back facing after projection. This only gave me an approximation but I could make the approximation as accurate as I wanted by just sampling more planes so I used this as a baseline to verify other solutions.

I then tried the following:

vec4 center = projection.inverse()*vec4(0,0,0,1);
center /= center.w;
float d = dot(center.xyz, vec3(1,0,0));
vec3 point = vec3(1,0,0)*d;

But this did not work, it was close but not right. See this image, the red line is the sampled point and the blue is the one calculated from that.

I then tried changing the calculation for center to projection.inverse()*vec4(0,0,-1,1) to see what that would look like. With the the sampled point in red, the blue line being the previous version and the green the new version. I got the following image which as you can see is still not correct.

But I noticed that the green was always directly between the blue and the red so I tried the following.

vec4 c1 = projection.inverse()*vec4(0,0,0,1);
c1 /= c1.w;
float d1 = dot(c1.xyz, vec3(1,0,0));
vec4 c2 = projection.inverse()*vec4(0,0,-1,1);
c2 /= c2.w;
float d2 = dot(c2.xyz, vec3(1,0,0));
float d = (d2-d1)*2 + d1;
vec3 point = vec3(1,0,0)*d;

Which seems to be correct but I have no idea why that works. It is always extremely close to the sampled point so I assume it is correct.

My question is why does this work?

real analysis – On some property of the zeros of $zeta(s)$ in the complex plane

This property is rather elementary, and not at all specific to $zeta$, so I am wondering if it has any value in studying the zeros of the Riemann zeta function in the critical strip. It is a well known result? I can provide a proof sketch if you are interested, and it has been checked numerically.

If $zeta(s)=0$, with $s=sigma +it$ and $0<sigma<1$ then for all real $theta$, we have

$$sum_{n=1}^{infty}(-1)^{n+1}frac{cos(theta+tlog n)}{n^sigma}=0.$$

This would be true even if by chance, one of the zeroes is outside the critical line $sigma=frac{1}{2}$. In particular let $t_0$ be the imaginary part of a zero of $zeta(s)$. Then in order to find all $sigma$‘s such that $zeta(sigma +it_0)=0$, we only need to focus on the $sigma$‘s that satisfy the following equation for every $theta$:

$$sum_{n=1}^{infty}(-1)^{n+1}frac{cos(theta+t_0log n)}{n^sigma}=0.$$

Of course the Riemann Hypothesis (RH) implies that $sigma$ must be equal to $frac{1}{2}$, but my assertion is true even if RH is not true. So if you prove that my equality can only happen if $sigma=frac{1}{2}$, then you would have proved RH. I have several strong reasons to believe that my equality does not lead to a proof of RH, yet I am wondering if it has any value.

reference request – Does every 5-celled animal tile the plane?

An animal in the plane is a finite set of grid-aligned unit squares in $mathbb{R}^2$. (The definition is the same as a polyomino, but where we relax the connectivity requirement.) One may equivalently think of them as finite subsets of $mathbb{Z}^2$.

In Erich Friedman’s investigation of tilings of rectangles by 1D animals, it is mentioned offhandedly that Coppersmith showed every 4-celled animal tiles the plane in 1985 (and can do so without reflections); I was able to locate the paper Coppersmith – Each four-celled animal tiles the plane (link to PDF). 6-celled animals do not, in general, since they can have a hole:

6-celled animal with hole

Has any progress been made on the 5-cell case since 1985 (either with or without reflections permitted)? I’d be interested in computational results as well, e.g. “every size-5 subset of a $7times 7$ grid tiles the plane”. (I have confirmed that every size-5 subset of a $3times 3$ grid tiles the plane, and every size-5 subset of a $4times 4$ grid at least covers a $10times 10$ square.)

Since I realize the answer to this question may well be “no further progress has been made and the problem is difficult”, I’m interested in any statements that can be made about restrictions of this problem to natural subcases. For instance:

  • Can every size-5 subset of $mathbb{Z}$ tile the plane? Of $mathbb{Z}times {0,1}$?

  • I didn’t specify above, but one can consider whether reflections are allowed in a tiling or not. If so, a proof may be easier (it seems Coppersmith restricted himself to the rotation case), and if not, a counterexample might be more tractable.

  • Is there expert consensus on what the likely answer is?

Previously on math.SE here, without any progress on the question.