How many YHWH are there according to Mark 12:29 and Deuteronomy 6:4? What does the word "one" mean on both verses?

Question: Mark 12:29 of The American Standard Version says, Jesus answered, The first is, Hear, O Israel; The Lord our God, the Lord is one.' " and Deuteronomy 6:4 of Young's Literal Translation says, Hear, O Israel, Jehovah our God is one Jehovah; Is there anything complex about the word "one" in these verses? How many YHWH are there?
Comment: In addition, Mark 12:29 doesn't have any YHWH.
Comment: שְׁמַ֖ע יִשְׂרָאֵ֑ל יְהוָ֥ה אֱלֹהֵ֖ינוּ יְהוָ֥ה׀ אֶחָֽד

Biblia Hebraica Stuttgartensia: with Westminster Hebrew Morphology. (1996). (electronic ed., Dt 6:4). Stuttgart; Glenside PA: German Bible Society; Westminster Seminary.
Comment: Does this answer your question? [Shema and Psalms 110: Is Mark making a case for Jesus' divinity?](https://hermeneutics.stackexchange.com/questions/16867/shema-and-psalms-110-is-mark-making-a-case-for-jesus-divinity) Also see: https://hermeneutics.stackexchange.com/questions/41509/deuteronomy-64-is-the-lord-our-god-is-one-the-same-as-we-only-worship-one-lo
Answer: Great Question!  The word in Deut 6:4 is a complex word, אֶחָֽד׃ ('e·ḥāḏ).  One of the best known and quoted occurrences of this word is in Gen 2:24

That is why a man leaves his father and mother and is united to his
  wife, and they become one flesh.

In Gen 41:26 we see another example of this:

The seven good cows are seven years, and the seven good heads are
  seven years; the dreams are one.

Here is another case where many are referred to as one, Ex 24:3:

When Moses went and told the people all the LORD's words and laws,
  they responded with one voice, "Everything the LORD has said we
  will do."

Thus, the word אֶחָֽד׃ ('e·ḥāḏ) can mean a multiple unity, usually of purpose or action.
In the Old Testament, God is referred to in many places as more than one person such as:

Zech 2:6-12 – the LORD (= YHWH) claims three times that He has been sent by the
LORD.
Isa 48:11-16 – again, the LORD has been sent by the LORD.
Isa 63:7-10 – the LORD (described as a Father) sends His divine servant (the angel of His presence) and His Holy Spirit who is grieved (compare Ps 78:40).
Ex 23:20 – the angel of the LORD's presence has the power to forgive sin (but will not). This and the previous reference clearly make the angel of the presence the preincarnate Jesus.
Hos 1:7 – the LORD saves by the LORD their God.
Prov 30:1-4 – the Son of God is as unfathomable as God Himself

Thus, the Shema (Deut 6:4) does not exclude Jehovah being more than one person, BUT Jehovah is One.  (Gen 2:24 reminds me of the fact that my wife and I (two people) for a single marriage.)
The NT (in Phil 2:10) even quotes passages like Isa 45:23 as specifically applying to Jesus when they were originally written about Jehovah.
Comment: Trinitarians do NOT believe that 1=3 or 3 = 1.
Comment: Linguistic fallacy alert! Words have a range of meaning defined by context. Just because the word "one" can be used differently in various contexts, is not grounds for teaching a plurality of Yahwehs in Mark 12:29, and Deuteronomy 6:4.God is not a plurality of Yahwehs. How many Yahwehs do you believe in?
Comment: Sounds like linguistic fallacy.. 1=3?
Comment: +1 Good post. The term one does not always signify a numerical value.
Comment: It's not whether He's "linguistic fallacy" (whatever that is), or not: it's what the Bible (Gen, Ex, Deut, Prov, Isa, Hos, Zech, Mk, Jn, Philip, etc, Rv) says that counts. In other words, it's not our mind, but He who is God. We didn't invent Him, He invented us. "For My thoughts are not your thoughts, and your ways are not My ways, declares Jehovah. For as the heavens are higher than the earth, so My ways are higher than your ways, and My thoughts higher than your thoughts." That's just His thoughts in Isa 55:8-9. Saying naught of His infinite, origination-less Being
Comment: fallacy alert indeed...lol. The Bible isn't plurality of Jehovahs, plural, but plurality within Jehovah one. Singular. Jn 10:30, 38
Answer: The question about Deut 6:4 involves two words: LORD (YHWH) and God. Let me take them one at a time.
LORD

Ps 110 - The Father alone is YHWH (verses 1,2 & 4), as Jesus is called Lord (Adonay) in verses 1 & 5.
Ps 2 confirms that LORD is name for Father only, as "His Anointed One" is called Adonay (verse 4)
Lots of verses from Ps 22 are quoted by NT writers as applying to Jesus since His arrest. (Mt 27:35,39,43,44,46; Mk 15:29,34; Lk 23:27,34; Jhn 19:24,28,34; Heb 2:12; 5:7). Here Jesus calls on God, his YHWH. This Psalm ends: "Future generations will be told about Adonay. They will proclaim his righteousness to a people yet unborn - for he has done it."
YHWH thus belongs to the Father only.
Isa 48:6 YHWH said: "From now on I will tell you of new things, of hidden things unknown to you", and immediately started telling about his arm (being Jesus from Isa 30:30, 40:10, 51:5,9; 52:10, 59:16, 62:8, 63:5, 64:12, 66:14). YHWH said He will bring him (Jesus), and then Jesus himself says that YHWH has sent him - without making himself YHWH.
As nobody has ever seen YHWH (Ex 33:20, Jhn 6:46), Jesus appeared in physical form to Abraham in Gen 18, Him being addressed as Adonay. Scripture do reveal YHWH as speaker here, as early OT people knew nothing of Jesus' existence, but had to know that YHWH is the One and only God over them.
Therefore, also when Isaiah saw Adonay seated on a throne underneath seraphs (Ezekiel's vision was YHWH on a throne high above seraphs)m and had a conversation with Adonay (Isa 6:1-13), the seraphs still worshiped YHWH, unseen through Isaiah's eyes. This is similar to Jesus' announcement that although every knee will bow before him, the praise will still be given to YHWH (Isa 45:23-24), also seen in Phil 1:7 and 2:11. Likewise, the seraphs around the throne in Rev 4:8 never stop worshipping YHWH, although Jesus is there too (Rev 5:6). Jesus provides consistency with the prayer to his Father (Jhn 17:6,26)
See how the NIV has alternative translations of Deut 6:4 in the footnotes as a result of uncertainty about what the original text was:
 "Hear, o Israel:  The LORD our God is one LORD" or
                  "The LORD is our God, the LORD is one" or
                  "The LORD is our God, the LORD alone."

God

'God' as Name is introduced as the Name of more than one Heavenly Being (Gen 1:26), to become the Family Name, like the surname of a human family. Jesus sometimes operated with this Family Name when He visited earth in physical form (Gen 32:24-30, Ex 24:9-11, Deut 10:10, Judg 13:22) or just spoke through a prophet, like Isa 45:21b-23. That happened in the OT only, and was given recognition by Matthew (1:22) and John (1:1-2, 18). Those three occurrences together with Jhn 20:28 deal with the word 'God' as Name for Jesus. All further references to God are applied to the Father only by those two disciples, who received first hand teaching for three and a half years.
The word 'God' as title is attributed to YHWH, with most abundant references in Deuteronomy, as follows:
the LORD, the God of your Fathers - at least 4 times;
the LORD, your God - at least 150 times;
the LORD, our God - at least 20 times.

Scripture consistency opens a door with Rev 19:11-16 to:

see that the One who will rule the nations, who treads the winepress of the fury of the wrath of God Almighty has already started this process as Adonay in the OT. See Lam 1:14-15; 2:1-3,5,7,18-20; 3:31-39,58;  Isa 3:17-18; 7:14,20; 8:7-8; 9:8,13; 10:12; 30:20; Dan 1:2; 9:15-19. Throughout all these passages YHWH's supreme authority is interwoven, for example Lam 2:8,17; 4:11,16; 5:19-22.
know that Jesus' title 'Lord' in the NT is the same Lord (Adonay) of the OT.
Answer: One YHWH having many paniym (faces, facets, operational capacities), each having a different name:

Exo 6:3 - KJV: And I appeared unto Abraham, unto Isaac, and unto Jacob, by the name of God Almighty, but by my name JEHOVAH was I not known to them.
Isa 9:6: - For unto us a child is born, unto us a son is given: and the government shall be upon his shoulder: and his name shall be called Wonderful, Counsellor, The mighty God, The everlasting Father, The Prince of Peace.

Hebrew people knew full well that God Almighty had several definite operative capacities, each having a different name commonly used to describe God Almighty operating within that capacity. Dr. Gary Hedrick of Messianic Perspectives, concerning his podcast entitled, "The Hebrew Names of God, Part 1," explains:

In ancient times, names were much more significant than they are today. They were carefully chosen, and were sometimes conferred by God himself. Names were descriptive of a person's lineage, character, and in some cases even his destiny. Likewise, God applies a variety of Hebrew names and titles to Himself because He wants us to know more about His character, nature, and attributes.

Clearly, the speaker in Psalm 2:7 is the WORD of God--the declarant--the "me"--the Son--the begotten one--who was "made' flesh as the Son of God whose name was prophesied in Isa 9:6, supra, to  be called, "Wonderful, Counsellor, The mighty God, The everlasting Father, The Prince of Peace."

I will declare the decree: the LORD hath said unto me, Thou art my Son; this day have I begotten thee.

Each different operation of God Almighty within His special operative capacity reveals that paniym (or face) to stay within the lane of that capacity while operating as such, as we see in Jesus' obedience to the Father while operating in the capacity of the Son while under the Law of Moses.


Why was my figcaption question migrated?

Question: I had a question here on Webmasters about tables, figures, and figcaptions. Today it was migrated to StackOverflow (https://stackoverflow.com/questions/12783679/a-table-in-a-figure-use-caption-or-figcaption). I'd like to know, why? It's a web development question, so was quite appropriate on this site.
Answer: Webmasters is for questions concerning management and administration of sites, generally as a whole. Questions involving actual code/markup (might, probably) belong on SO. Yes there's plenty of overlap(eg. analytics tends to stay here, and markup questions that're fundamentally about SEO) and threads fall through, but that's the very simple version. Often topics will just get evaluated case by case.
I have no idea why this only got migrated today; Paul will have to address that.
I'm pretty sure I would have filed my vote back when I initially responded.
Comment: Su' has covered it. It only got migrated yesterday, because I'm using the new beta review feature which lets you work through all flags of a particular sort, and this one came up yesterday.


Can you calculate the common volume on excel using Boolean Operation

Question: I would like to calculate the common volume of a solid like the one below but at different angles. I can achieve this using ANSYS Design but I was wondering whether it is possible to calculate this on excel or could I program the Boolean Operation into fortran at all?
How does the Boolean operation work on ANSYS to calculate the common volume?
Thank you for your help!!
Answer: Essentially, the calculation required is:
$$
||\Omega|| := \iiint\limits_{\Omega} 1\,\text{d}x\,\text{d}y\,\text{d}z
$$
with:
$$
\Omega :=
\left\{
(x,\,y,\,z) \in \mathbb{R}^3 :
\begin{aligned}
& x^2+y^2+z^2 - \frac{\left(l_1\,x+m_1\,y+n_1\,z\right)^2}{l_1^2+m_1^2+n_1^2} \le R_1^2\,, \\
& x^2+y^2+z^2 - \frac{\left(l_2\,x+m_2\,y+n_2\,z\right)^2}{l_2^2+m_2^2+n_2^2} \le R_2^2\,, \\
& x^2+y^2+z^2 - \frac{\left(l_3\,x+m_3\,y+n_3\,z\right)^2}{l_3^2+m_3^2+n_3^2} \le R_3^2
\end{aligned}
\right\}
$$
where the integration domain consists of the intersection of three cylinders with axis passing through the origin, of direction $(l_i,\,m_i,\,n_i) \ne (0,\,0,\,0)$ and of radius $R_i > 0$, with $i = 1,\,2,\,3$.
Now, using Wolfram Mathematica 12.0, defining:
{l1, m1, n1, R1} = {1, 0, 0, 1};
{l2, m2, n2, R2} = {0, 1, 0, 1};
{l3, m3, n3, R3} = {0, 0, 1, 1};

cyl1 = (x^2 + y^2 + z^2) - (l1 x + m1 y + n1 z)^2 / (l1^2 + m1^2 + n1^2) - R1^2;
cyl2 = (x^2 + y^2 + z^2) - (l2 x + m2 y + n2 z)^2 / (l2^2 + m2^2 + n2^2) - R2^2;
cyl3 = (x^2 + y^2 + z^2) - (l3 x + m3 y + n3 z)^2 / (l3^2 + m3^2 + n3^2) - R3^2;

Ω = ImplicitRegion[cyl1 <= 0 && cyl2 <= 0 && cyl3 <= 0, {x, y, z}];

and writing:
ContourPlot3D[{cyl1 == 0, cyl2 == 0, cyl3 == 0}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]
RegionPlot3D[cyl1 <= 0 && cyl2 <= 0 && cyl3 <= 0, {x, -1, 1}, {y, -1, 1}, {z, -1, 1}]

we can get an idea about the layout of the cylinders and the shape of $\Omega$:

while writing:
Integrate[1, {x, y, z} ∈ Ω]
NIntegrate[1, {x, y, z} ∈ Ω, PrecisionGoal -> 10, WorkingPrecision -> 10]

it's possible to calculate the $\Omega$ measure:

8 (2 - √2)
4.686291501

where the exact result can only be obtained in very particular cases such as the one shown in this example. Having understood all this, the rest comes by itself, as it's very simple both to change the angles and radii of each cylinder and to add any other cylinders.
That said, right now I would have no idea how to write an algorithm to be implemented, for example, in VBA in Microsoft Excel, so as to calculate the volume numerically.
Answer: This is not an answer which tells you how to compute the volume numerically in any programming language. Instead, I will present some results that may help.

For intersection of any number of infinite long cylinders (not necessarily circular), how to express the volume as a sum of line integrals over the edges.
In the special case where all cylinders are circular, have same radius and their axes intersect at a single point, how to reduce the sum of line integrals to a sum over the vertices.
As a corollary of $(2)$, a closed form expression for the case of three cylinders with common radii is derived.

Part I - express volume as a sum of line integrals over the edges.
Let $p = (x,y,z)$ stands for any generic point on $\mathbb{R}^3$.
Let $\Omega_1, \Omega_2, \ldots, \Omega_n$ be $n$ infinite long cylinders whose axes are pointing along directions of unit vectors $a_1,a_2,\ldots, a_n$.  Each $\Omega_i$ is invariant under translation along direction of $a_i$. We will assume the cross-section of $\Omega_i$ cut by any plane perpendicular to $a_i$ is a region bounded by a simple closed piecewise $C^1$ curve.
Let $\Omega = \bigcap_{i = 1}^n \Omega_i$ be their intersection. Its boundary
can be composed into a collection of faces
$$\partial \Omega = \bigcup_{F\in \mathcal{F}} F$$
and each face lies on one of the cylinders.  For any face $F$, let $\Omega_F$ be the cylinder it lies on (i.e $F \subset \partial\Omega_F$). Let $a_F$ be  corresponding unit vector and $A_F(p) = (a_F\cdot p) a_F$. It is easy to verify
$$\nabla \times (p \times A_F(p))  = p - 3 A_F(p)$$
For any $p$ on $F$, $a_F$ is perpendicular to the area element $dS$, this means $A_F(p) \cdot dS = 0$.
Since $\nabla\cdot p = 3$, by divergence theorem and stroke's theorem, we have
$$\begin{align}
V \stackrel{def}{=} \verb/Vol/(\Omega) &=
\int_\Omega dV = \frac13\int_\Omega \nabla\cdot p dV
= \frac13\int_{\partial \Omega} p \cdot dS
= \frac13\sum_{F\in\mathcal{F}}\int_F p \cdot dS\\
&= \frac13\sum_{F\in\mathcal{F}}\int_F (p - 3A_F) \cdot dS
= \frac13\sum_{F\in\mathcal{F}}\int_F \nabla\times( p \times A_F) \cdot dS\\
&= \frac13\sum_{F\in\mathcal{F}}\int_{\partial F} (p \times A_F) \cdot dp
\end{align}
$$
To proceed, decompose each $\partial F$ as a collections of edges.
Let $\mathcal{E}$ be the union of all these edges, we have
$$\bigcup_{e\in \mathcal{E}} e = \bigcup_{F\in \mathcal{F}} \partial F$$
For any edge $e \in \mathcal{E}$, it is in contact with two faces in $\mathcal{F}$. Pick an arbitrary orientation for $e$. Let $F(e)$ be the face
whose induced orientation on $\partial F(e)$ is the same as the one on $e$. Let $G(e)$ be the other face, the induced orientation on $\partial G(e)$ will be opposite to that of $e$.
In terms of them, we can express the volume as a sum of line integral over the edges:
$$\bbox[border:1px solid blue;padding: 1em]{
V = \frac13 \sum_{e\in \mathcal{E}} \int_e (p \times (A_{F(e)} - A_{G(e)}))\cdot dp
}\tag{*1}$$
For general configuration, this is as far as we can go.
When we restrict ourselves to circular cylinders whose central axes intersect at a single point, the situation is slightly better. The line integrals have the form of incomplete elliptic integral of second kind. However, there are many ways the edges can connect to each other (e.g. they can form loops). A generic simplification seems impossible.
Part II - reduction of line integral to a sum over vertices.
However, there is one special case where improvement is possible. Namely, when all the circular cylinder share the same radius. WOLOG, consider the case where the common radius is $1$ and the central axes interest at origin. i.e.
$$\Omega_i = \left\{ p \in \mathbb{R}^3 : |p|^2 - (a_i \cdot p )^2 \right\}\quad\text{ for }\quad 1 \le i \le n$$
Let $\mathcal{V}$ be the collection of vertices of $\Omega$.
For any $v \in \mathcal{V}$, Let $E(v)$ be the collection of edges connect to $v$.
For any $e \in E(v)$, let $u$ be the other end-point of $e$.
Choose the orientation of $e$ so that it is pointing from $u$ to $v$. Let $F = F(e)$, $G = G(e)$ to the two faces in contact to $e$ and $a = a_F$, $c = a_G$
be corresponding vectors.
For any $p \in e \subset \Omega_F \cap \Omega_G$, we have
$$|p|^2 - (a\cdot p)^2 = |p|^2 - (c\cdot p)^2 = 1
\implies (c\cdot p)^2 = (a\cdot p)^2$$
Flipping the sign of $c$ is needed, we can assume $a\cdot p = c \cdot p$ along $e$. i.e. $e$ is lying on the plane $p\cdot( a - c) = 0$.
Choose a coordinate system so that $a = (0,0,1)$ and $c = (\sin\beta,0,\cos\beta)$.
In this coordinate system,
$$\begin{array}{crcl} & |p|^2 - (a\cdot p)^2 = 1 &\iff& x^2 + y^2 = 1\\
\text{ and }\quad & p \cdot a = p \cdot c &\iff& z = \cos\theta z + \sin\beta y
\end{array}$$
We can parameterize $e$ by a single $\theta$.
$$p = (x,y,z) = (\cos\theta,\sin\theta,k\sin\theta)
\quad\text{ where } \quad k = \frac{\sin\beta}{1 - \cos\beta}
$$
Notice
$$\begin{align}(p \times (a - c))\cdot dp
&= \left|
\begin{matrix}
x & y & z\\
0 &-\sin\beta & 1 - \cos\beta\\
dx & dy & dz
\end{matrix}
\right|
= (1-\cos\beta) \left|
\begin{matrix}
x & y & ky\\
0 &-k & 1\\
dx & dy & dz
\end{matrix}
\right|\\
&= 2(ydx - xdy) = -2d\theta\end{align}$$
The line integral over $e$ can be evaluated as
$$\begin{align}\int_e (p \times (A_F - A_G))\cdot dp
&= \int_e (a\cdot p) (p \times (a - c))\cdot dp\\
&= -2k \int_e \sin\theta d\theta
= 2k \int_e d\cos\theta
= 2k \int_u^v dx\\
&= \frac{2 c \times a }{1 - c\cdot a}\cdot (v - u)
\end{align}
$$
Using these, we can decompose the line integrals in $(*1)$ into sum of contribution from end points of the edges. Regroup them according to the vertices, we convert the sum over edges to a sum over the vertices.
$$\bbox[border:1px solid blue;padding: 1em]{
V = \frac23 \sum_{v\in \mathcal{V}} v \cdot \sum_{e \in E(v)}\frac{a_{G(e)} \times a_{F(e)}}{1 - a_{F(e)}\cdot a_{G(e)}}
}\tag{*2}$$
Part III - closed form expression for volume of three cylinder, same radii.
Back to our problem of three cylinders and consider the special case where
all radii are $1$.
Let $a = a_1,b = a_2,c = a_3$  be the three unit vectors. Define
$$\begin{cases}
A = b\cdot c\\
B = c\cdot a\\
C = a \cdot b\end{cases}
\quad\text{ and }\quad
\begin{cases}
\tilde{a} = b \times c\\
\tilde{b} = c \times a\\
\tilde{c} = a\times b\end{cases}
$$
It is easy to verify
$$
\begin{cases}
|\tilde{a}|^2 = 1 - A^2,\\
|\tilde{b}|^2 = 1 - B^2,\\
|\tilde{c}|^2 = 1 - C^2,
\end{cases}
\quad\text{ and }\quad
\begin{cases}
\tilde{a}\cdot\tilde{b} = AB-C\\
\tilde{b}\cdot\tilde{c} = BC-A\\
\tilde{c}\cdot\tilde{a} = CA-B\\
\end{cases}
$$
In general, when $a,b,c$ are linear independent, $\Omega$ will have $14$ vertices. $6$ of them has degree $4$ and the remaining $8$ has degree $3$.
In order to visualize the configuration, let's imagine $a,b,c$ lies close to the +ve $z$, $x$ and $y$ axes. In this case, one of the $6$ vertices with degree $4$ will lie close to +ve $x$-axis. It is $v_{100} = \frac{\tilde{b}}{|\tilde{b}|}$. The $(a_{F(e)},a_{G(e)})$ for the 4 edges connect to $v_{100}$ are $(a,c), (-c, a), (-a,-c),(c,-a)$. The contribution of $v_{100}$ to $V$ in formula $(*2)$ equals to
$$\frac43 \frac{c \times a}{|c\times a|}
\cdot \left(\frac{c \times a}{1 - c\cdot a} + \frac{a \times -c}{1 + c\cdot a}\right)
= \frac83 \frac{1}{\sqrt{1-B^2}}
$$
The contribution from other $5$ vertices has similar form, with $B$ may be replaced by $A$ or $C$.
An example of a vertex will degree $3$ is located at
$$v_{111} = \frac{\tilde{a} + \tilde{b} + \tilde{c}}{\lambda}
\quad\text{ where }\quad \lambda^2 = 2(1-A)(1-B)(1-C)$$
The $(a_{F(e)},a_{G(e)})$ for the 3 edges connect to it are $(a,b), (b, c), (c,a)$. The contribution to $V$ from the first edge is
$$\frac23\frac{\tilde{a} + \tilde{b} + \tilde{c}}{\lambda}\cdot
\frac{b \times a}{1 - a\cdot b}
= -\frac{2}{3\lambda(1-C)}(\tilde{a} + \tilde{b} + \tilde{c})\cdot \tilde{c}\\
= -\frac{2}{3\lambda(1-C)}(AC-B + BC - A + 1 - C^2)
= - \frac{2}{3\lambda}(1 + C - A - B)
$$
The contribution from other two edges is similar. The total contribution of $v_{111}$ to $v$ is
$$-\frac{2}{3\lambda}(3 - A-B-C) =
-\frac{2}{3}\sum_{(u,v,w) \in \Lambda} \sqrt{\frac{1-u}{2(1-v)(1-w)}}$$
where $\Lambda$ is the set of cyclic permutations of $(A,B,C)$
The contribution to the other $7$ vertices of degree $3$ is similar. The end result is
$$\bbox[border: 1px solid blue;padding: 1em]{
V = \frac{4}{3}\sum_{(u,v,w)\in \Lambda'} \left[\frac{1}{\sqrt{1-u^2}} - \sqrt{\frac{1-u}{2(1-v)(1-w)}}\right]
}\tag{*3a}$$
where $\Lambda'$ is following set of twelve $3$-tuples.
$$\begin{align}
&(A,B,C), (-A,-C,B), (A,-B,-C), (-A,C,-B),\\
& (B,C,A), (-B,-A,C), (B,-C,-A), (-B,A,-C),\\
&(C,A,B), (-C,-B,A), (C,-A,-B), (-C,B,-A)
\end{align}$$
In the special case when $a,b,c$ are orthogonal to each other, $A = B = C = 0$ and the volume reduces to the known result:
$$V = \frac{4}{3} (12)\left[ 1 - \frac{1}{\sqrt{2}}\right] = (16-8\sqrt{2})$$
When $a, b, c$ are linear dependent, above arguments cease to work. $\Omega$ now has $2$ vertices and $6$ edges. However, since RHS of $(*3a)$ is well behaved as long as $|A|,|B|,|C| < 1$. By continuity, $(*3a)$ remains valid even when $a, b, c$ are linear dependent.
Actually, in this special case, there is a simpler expression for $V$. Notice $a,b,c$ are determined up to a sign. Let
$$\alpha = \cos^{-1}A, \beta = \cos^{-1}B, \gamma = \cos^{-1}C \quad\in (0,\pi)$$
It is always possible to choose the signs of $b$ and $c$ to make
$\alpha + \beta + \gamma = 2\pi$. If one do that, one can simplifies $(*3a)$  to
$$\bbox[border: 1px solid blue;padding: 1em]{
\begin{align}
V &= \frac83 \left(\sqrt{\frac{1+A}{1-A}} + \sqrt{\frac{1+B}{1-B}} + \sqrt{\frac{1+C}{1-C}}\right)\\
&= \frac83 \left(\cot\frac{\alpha}{2} + \cot\frac{\beta}{2} + \cot\frac{\gamma}{2}\right)
\end{align}
}\tag{*3b}
$$
As an example, consider the case $a, b, c$ lies on the plane $z = 0$ with angle $120^\circ$ among them. Above formula gives $V = \frac{8}{\sqrt{3}}$ which can be proved with elementary means.


What are good builds for 'Roland' in Borderlands

Question: I'm having trouble finding information for some cookie-cutter type builds for Roland, so I put this question out there in the same vein as this one: What are good builds for 'Brick'?
I mostly play solo (heavy Combat Rifle user), but It'd be great to get some info on a good co-op build, too.
Answer: Let's organize the skills, and not by tree:
Offense: Impact, Metal Storm, Overload, Scattershot, Assault, Sentry, Guided Missile, Grenadier, Deploy, Refire, and Barrage.
Defense: Defense, Quick Charge, Fitness, Grit, Stat, Cauterize, Stockpile, Aid Station, Supply Drop, and Revive.
Turret offense: Sentry, Guided Missile, Deploy, Refire, and Barrage.
Turret defense: Stockpile, Aid Station, Supply Drop, and Revive.
Altruistic: Cauterize and Revive. These skills provide no benefit to yourself directly. You cannot Cauterize yourself with an explosive and the turret despawns when you're downed.
Any other skill, I'll call either personal offense or personal defense. The Scorpio Turret action skill is a given. Scattershot or Assault depends on whether you prefer shotguns or rifles, respectively.
I ordered them in a specific way and split turret skills into their own sub-sets because, grenades and the turret are an afterthought.
Grenadier: Grenades are iffy. Grenades are iffy for several reasons: (1) They can backfire on you, particularly the bouncing betties, and this can prevent you from advancing further until it's done exploding as well as hindering your vision, (2) there's a bug where terrain will cause the grenade to disappear, (3) if it's caught in a bad spot, its blast radius may be ineffective, (4) their damage is delayed, which is bad in comparison to how you probably could've killed the enemy sooner by focusing on gunning, and (5) considering how the Grenadier skill and the Warmonger class mod specifically up grenade damage, this is the only thing that Roland has that boosts and resonates with grenade damage except for shock with the Shock Trooper COM.
On the other hand, Roland has the only directly grenade-oriented skill, and not to denounce grenades completely, grenades still have their place...in mobs. Grenadier depends on how often that you use grenades. And having at least one point in it means there's reprieve for the obsessive ("must have full ammo"), while being more hands-free and offensive than Supply Drop.
Turret skills: The turret is similar to grenades. It's delayed, it's location is confined, its damage isn't easy to boost, and if you've developed an attachment to its resupplying ability, limits your movement. Furthermore, it interrupts your ability to aim and shoot your weapon to deploy it most of the time, it only fires in spurts, and its cooldown is long, being the only one of all classes that cannot be easily redeployed in seconds, not to mention at the lower levels.
On the other hand, it's an extra gun and it's useful when there are many enemies in one spot. Even if all personal offensive and personal defensive skills are filled out (which comes to a total of 11-1 for preference of shotgun or rifle), that leaves 19 skill points that you'll have to spend on something non-personal.
Cauterize: You only need one point in this to heal your team-mates by shooting them. If you play with others, put at least one point in this.
What are the offensive COMs? Heavy Gunner, Rifleman, Commando, Shock Trooper, and the more exclusive Patriot Vladof, Gunman S and S, Champion Atlas, and Marine. I have little experience with the loyalty enhancement mods and Marine. From my testing, Commando and Rifleman outdo Heavy Gunner and Shock Trooper. I followed Gearbox Software Forum's DaCheat's Shock Trooper guide, but shock Bulldogs just don't proc enough. I've also tried a shock Destroyer and Orion.
Patriot Vladof loyalty enhancement mod: Let's see what I have for that: Vladof Hammer, Vladof Revolution, and The Chopper. Vladof Hammer is better used by Brick and its ammunition feeding device size is annoyingly small, Vladof Revolution is outdone by Atlas rifles, and The Chopper is for PvP. +61% Weapon Damage, Cauterize, Defense, Grenadier, and +56% Weapon Reload Speed. The Rifleman COM adds more damage.
Gunman S and S loyalty enhancement mod: Serpens = rifle. Draco doesn't have enhanced proccing like normal legendary weapons. +61% Weapon Damage, Metal Storm, Overload, Assault, +56% Weapon Reload Speed. Each bonus is a personal offense bonus, which is lovely. The only other major weapon is the Crux, but again, that's Brick's bag.
Champion Atlas loyalty enhancement mod: Havoc, Ogre (elemental challenge), and Ajax's Spear. +61% Weapon Damage, Fitness, Grit, Impact, +84% Weapon Accuracy. C'mere, sweetheart!
Marine: Launchers (rockets). "I wanna be like you, Brick! (Except worse.)"
In PvP, Shock Trooper is viable, because shock is the king element in PvP. This is because a player's total hit-points are mostly composed of shield.
Support: Let me put it like I put so long ago: Tactician? No one's gonna say, "Wow, +47% Shield Recharge Rate and +19% Team Shield Capacity," cocking their head with an affectionate smile. "Thanks, [insert your username here]!" You know what's noticeable? Ammo regeneration via the Support Gunner COM. You're the only one who can. You love it.
But other than for the open admiration, Tactician's cool. Leader for XP-boosting. You're playing support when you're playing in a lower level person's game, where you're too powerful. You're not if you're equally levelled. Not to forget that errbody loves Support Gunner for general play.
Caveat: If you don't have the good enough gear, one spec may stand above the rest in what is available to you.
That's when the Heavy Gunner COM shines: it applies to all.
Comment: One other one year old late note. Roland is exceptional because of his ammo regen mod, something BL2 doesn't even have, and that it regenerates Rocket Ammo, albeit very slowly.  So imagine, you have a high damage burst weapon, with effectively unlimited ammo.  And this is for every member in your team. How is this not bad at all?
Answer: Here's a good resource that I used when deciding on what to do with my Roland. It covers a lot of different builds and would be far too long to post here without having to quote a lot of text :) http://forums.gearboxsoftware.com/showthread.php?t=108033


Build Dynamic Search page for Custom Object ''Building''

Question: I am a solo admin with no dev skills. We have a requirement to create a searchable page for our custom object ''Building'' they would like this page to allow user to search for building records, select the appropriate one, and make edits to fields. I was going to use Salesforce sites.com feature to host a VF page, to allow access to everyone without a SF account. Looking for any help in the steps I need to take to create this VF page for this custom object. I do not know what it will involve, ie. class creation?, triggers needed? vf page?
Comment: Is your page public? Why use sites?
Comment: Any requirements above and beyond search? If not, does the SF global search not meet your needs?
Comment: Sorry I shouldve clarified, we need the user to be able to search for the building records, ability to select the record view the details and also edit the fields they need on that record.
Answer: Yes, ofcourse you need to use VF pages and apex class!
Your VF page should have an <apex:inputText> component.  This is the text box where the users enter the text to search and an <apex:commandButton>'s action to call the method in Apex class, that does the searching.
You should use SOQL query in the method for searching the text entered. And the returned results should be sent to another VF page which displays the results.  If the searching is to be done in multiple objects, you should be using SOSL.  But in your case, SOQL should be enough. The 2nd VF page will have <apex:pageblocktable> or <apex:repeat> to show the results returned by the apex method.
You need to learn about the attributes/components I highlighted in the above paragraphs to proceed with your task.
Do try and post your code next time, so we can help further when you are struck.  But this will let you start.
Here is list of visualforce components for your reference
https://developer.salesforce.com/docs/atlas.en-us.pages.meta/pages/pages_compref.htm


What is the difference between homomorphic encryption and homomorphic signature?

Question: I want to apply homomorphic signature instead of homomorphic encryption in Provable data possession.
So I want to know about homomorphic signature and homomorphic encryption.
Comment: Your question is way too broad and will likely be closed. I did my best, but you need to be more specific in what you are looking for. A general introduction (like I gave), example crypto systems, implementations, how to apply it to your problem, or something else. Your question as stated is like saying, "I want to ride a scooter instead of a motorcycle. So tell me about scooters and motorcycles."
Answer: Concerning your very broad question: Wikipedia can tell you about homomorphic signatures.
However, its application is quite specific, and I have no idea if it fits your scenario/requirements. The homomorphic property has direct consequences for the signatures: It can not achieve existential unforgeability, because this contradicts the homomorphic property (with signature $S(m)$ you can create signature $S(m+m)$, in additive notation). Furthermore, common signatures use cryptographic hash functions to reduce the size of the signature to a fixed length. For homomorphic signatures you can not do this, because the cryptographic hash function does not preserve any structure.
Additionally, the proposed scheme on the wiki page is really not practical: At first, you need to fix the input size at key generation and you need a pairing friendly group (okay, we can do that). The signing algorithm is quite fast (just normal elliptic curve operations). But the signature verification is REALLY slow since it requires $D$ pairing operations, and $D$ can be quite large (The "hash" input is a vector space with $D$ dimensions). Even though the signature itself is relatively short, the verification key is also quite large ($2D+1$ points on the elliptic curve).
Answer: First start with some notation. Say we have a plaintext space $P$ which forms a group. And an encryption function which goes from the plaintext space to the ciphertext space, say $E : P\to C$.
$E$ is homomorphic if $E$ forms a group homomorphism, i.e. given $E(x)$ and $E(y)$ for $x,y\in P$ we can efficiently construct $E(x\cdot y)$ without the private key, where $\cdot$ is the group operation in $P$. That would be homomorphic encryption.
Homomorphic signatures are very close, but instead we have a signature function, $S$. Then, given two valid signatures $S(x)$ and $S(y)$, we should be able to construct a signature of $S(x\cdot y)$ efficiently without the private key.
Comment: @user11236 advantage compared to what? to each other? that is like comparing apples and oranges.
Comment: What is the advantage and disadvantage of homomorphic encryption and homomorphic signature?


Is there a keyword that can make a class variable readonly from outside the class but not on the inside?

Question: Basically, the readonly keyword doesn't let me modify a field after I first create the class instance. I could use a property but in this case its just extra overhead. Is there a keyword to make a class field readonly from only outside the class?
Comment: The property is the way to go. The overhead is minimal.
Answer: make the field private, provide getter and setter for it.
Make the setter private.
This way the value can be seen from outside the class by the getter,but, cannot be set/written  from outside the class.
this makes the property read-only from outside the class.
Answer: As others have stated, use a property. If you don't want to split the property into one Getter and one Setter then make the setter private.
Public Class Foo

    Public Property Abc() As Object
        Get
            Return Me.m_Abc
        End Get
        Private Set(value As Object)
            Me.m_Abc = value
        End Set
    End Property

    Private m_Abc As Object

End Class

However: The common way is to set the access level of the field to Friend making it accessible within the same assembly, but not from outside the assembly.
Public Class Foo

    Public ReadOnly Property Abc() As Object
        Get
            Return Me.m_Abc
        End Get
    End Property

    Friend m_Abc As Object

End Class

Answer: No there isn't. This type is scenario is precisely why properties are provided in the first place. You get a whole lot of flexibility.
However, if you insist you want to use a read only field, you can use reflection to change the value:-
Public Class TestClass

    Public ReadOnly MyNumber As Integer

    Public Sub New()
        'Readonly fields can only be changed this way
        'in the constructor
        Me.MyNumber = 900
    End Sub

    Public Sub ChangeNumber(ByVal num As Integer)
        SetNumber(num)
    End Sub

    Private Sub SetNumber(ByVal num As Integer)

        Dim fi = Me.GetType.GetField("MyNumber")

        'Reflection can change the value of
        'a read only field after construction
        fi.SetValue(Me, num)
    End Sub

End Class

Note that this is a very terrible thing. Reflection shouldn't be used for this sort of thing as you're going to take a performance hit. Just use properties and save yourself the trouble.