How do I scale a cube in OpenGL?

The idea is to make an arm consisting of three segments, each having a cube-like shape. For this I have made the cube inside a function called drawCube() – a set of 6 matrices (glm::mat4) translated and rotated to look like a cube – and would like to rescale this cube in order to make it longer or shorter as I please. I can’t do something of the sort glm::drawCube() cube = glm::scale(drawCube(), glm::vec3(1,2,3)) since drawCube() is not a variable type, so I don’t know how else to approach this.

So basically I created a 3D object and would like to rescale it as a whole instead of rescaling each individual matrix/vector. Is there such an option?

Why browsing cube returns no rows?

I am trying to create a cube in Visual Studio 2019 using two dimensions and one measure, after having defined the datasource and tested over the view that the data is loaded into the project, and after setting the dimensions and the measure to use both at the cube structure as well as the dimension usage levels/tabs (which i suspect i did an error defining relationships in the latter) the cube could be processed and deployed, but when i query over it (using the browser) it simply returns nothing and shows NULL under the measure label i specified.

PS : the deployment required that i skipped the not found key (by specifying ignore the error) in the dimension key errors tab in the advanced settings otherwise it will show error could not process the group of measures properly (i also don’t know why i should do this and how to process the cube succesfully without doing that)

My question is how to get through this and define the relationship between the measure and the dimension at the following interface Cube Relationship Interface (Dimension Usage Tab)(given solely as an example) ?

Is the Gelatinous ice cube familar official?

During the D&D celebration 2020 a new option for a familiar was revealed in the form of the image below.

I assumed it was official material because it was produced directly by wizards event though it does not appear in the the latest book (Tasha’s cauldron).

Is there any official word on if the 2020 celebrations material is legal in; adventure league, ordinary games or essentially homebrew?

enter image description here

Cross-section parts of a Cube and show results graphically

I am trying to visualize a 3D section of a chopped up 1x1x1 Cube. I don’t want to use the Cube[] function to draw this. (unless you think you can still accomplish it that way). Ok, so say you have a simple XY plane, laying flat on the table (lengths 1×1) Now I draw 2 straight lines across it with equations: Y=X+1/2 and Y=X-1/2 and shade the region in between. (it’ll take up 75% of the 1×1 region of course with the top left and bottom right squares being half shaded now, and the other 2 squares fully shaded). Ok simple enough so far.

Now let’s draw up the Z-axis above this (Hold the ruler perpendicular to the table up to a height of 1). Now raise/stretch that same shaded region above to the top, which will still take up the 75% of the now-8 0.5×0.5 cubes). with 2/8 unshaded. Ok now imagine the ZX plane (facing you on the table) and draw up the same 2 equations, now: Z=X+1/2 and Z=X-1/2 so you now have the same regions covering the ZX plane and they now travel to the back of the Cube – and the main point, now intersect the previous region (coming up from XY plane).

I like to visualize both shaded intersection. In other words, how can I draw up both regions, shade them, then put them into 3D for both planes and be able to rotate that final 3D image in any direction I like to visualize that highlighted 3D-intersection.

P.S. Too much in language, for those who like pure equations, in a nutshell this is all I need:

Y=X(+/-)1/2 and Z=X(+/-)1/2 < == 3D Plot and visualize intersection region in between the 2 pairs of lines on each plane.

Assume/hope your solution can take in any Function cutting the cube: Say Y=f(X) and Z=g(X)?

Co-incidentally, assume this is also possible to solve algebraically?

Integrating multidimensional PDFs on Boolean cube separated by hyperplane

Problem Statement: Let $ f:[0,1]^d \rightarrow [0,1]$ be a $ d$ -dimensional probability density function. Further, for $ c\in [0, 1]$ and $ \mathbf{w}\in \mathbb{R}^d$ , let $ H = \{\mathbf{x} \in [0,1]^d : \mathbf{w}\bullet\mathbf{x} = c\}$ be a hyperplane. This separating hyperplane partitions partitions the unit hypercube $ [0, 1]^d$ , into two regions,

$ $ U = \{\mathbf{x} \in [0,1]^d :\mathbf{w} \bullet \mathbf{x} > c\} \quad \text{ and }\quad L = \{ \mathbf{x} \in [0, 1]^d: \mathbf{w} \bullet \mathbf{x} \leq c\}$ $

Under what conditions on $ f$ can one efficiently compute (approximately or exactly) the cumulative probability of $ f$ over $ L$ (or equivalently over $ U$ ), i.e. $ \int_{L} f(\mathbf{x})d\mathbf{x}$ ?

Preliminary Ideas: If $ f$ is something like the uniform distribution then the integral can be computed exactly. For sake of example, suppose that $ \mathbf{0}\in L$ and $ H$ represents the upper limits of integration. Then we are integrating

$ $ \int_{0}^c\int_{0}^{\frac{c – w_dx_d}{w_{d-1}}}\dots\int_{0}^{\frac{c – w_2x_2 – \dots w_dx_d}{w_1}} 1 dx_1 \dots dx_{d-1} dx_d$ $

Which each integral has a closed form solution derivable in polynomial time, and there are only $ d$ number of integrals. Assuming this is correct, then if $ f$ is any polynomial it can also be solved exactly.

Question Are there any characteristics, or common families of PDFs in which this problem is polynomial time solvable? In practice, PDFs such as a Gaussian are computed numerically, is there any literature on how well of a numerical approximation can be achieved in this setting, for $ d$ -dimensional PDFs which are computed numerically?

Passive check and Gelatinous Cube Transparent traits

Since Gelatinous Cube have the transparent trait, do you need to do an action perception check to spot it or you could use your passive perception?

Transparent trait: Even when the cube is in plain sight, it takes a successful DC 15 Wisdom (Perception) check to spot a cube that has neither moved nor attacked. A creature that tries to enter the cube’s space while unaware of the cube is surprised by the cube.

Why is targeting an adjacent attacker with a 5 foot cube area attack considered a ranged attack?

An opponent has moved adjacent up into a character’s face and swung at them. On their turn, in retaliation, the character would like to attack back with their favoured cube area attack, made at the size of a 5 foot sized cube to be ergonomic. Oddly the rules as written (see below) seems to qualify this attack as a ranged attack even though the target is adjacent and every other area attack also containing the attacker would not. Is this an oversight, an intentional design decision, or is there anything I’m overlooking that makes this ruling invalid?

The rules leading me to this conclusion appears here:

Ranged Attacks in Melee

Any time you make a ranged attack and there is an enemy within melee reach of you, you have disadvantage 1 on your attack roll. Area attacks are considered ranged attacks if the area does not include at least one space adjacent to the attacker.

The 5 foot cube placed on the attacker’s square does not include at least one space adjacent to the attacker but it does include attacker’s square itself which intuitively feels like it shouldn’t be a ranged attack as well as other area attacks. RAW however, this means it’s a ranged attack and imposes disadvantage 1. To me a more intuitive ruling and writing of it would be:

Ranged Attacks in Melee

Any time you make a ranged attack and there is an enemy within melee reach of you, you have disadvantage 1 on your attack roll. Area attacks are considered ranged attacks if the area does not include the attacker or at least one space adjacent to the attacker. (changes italicized)

Are there existing rules or other evidence the designer’s intention was for this scenario to be a ranged attack? If so, why only 5 foot cubes and not every other area effect (they have to include a square adjacent to the opponent as well)? Is there perhaps another mechanical reason I can’t find that this attack should be considered ranged? Is the attack simply supposed to impose disadvantage 1 and being considered ranged is simply a byproduct?

In the case that it shouldn’t be considered ranged (or only considered ranged for the purpose of disadvantage 1), I would like to revise this confusing wording. I have found the Open Legends repository and my intention is to submit a pull request if I understand the rules correctly and this ruling is against the RAI. However, I’m asking my question here first to gain assurance, as I know that I am very new to the system and may be overlooking something.