## How to turn three two-dimensional plots into one 3D-plot

I am trying to make a complex function grapher (code below), and have made real and imaginary output 3d graphs, and have created functions to "slice" the graph into two dimensional contour plots for any given x, y or z, (leaving the other two variables as the axes for this). Often seeing layers of a graph is a lot easier to understand than the 3d planes the first function creates. I would like to create a "skeleton graph", in which you can see the x y and z intercept plots on the 3d graph so that you can understand the general shape of the function. I am unable to do this, and am asking for help with this. Code left below.

complexfunctiongraph[f_] :=   Show[Plot3D[    z = Re[f[x + (y*I)]], {x, Xrange[[1]], Xrange[[2]]}, {y,      Yrange[[1]], Yrange[[2]]},     PlotRange -> {Zrange[[1]], Zrange[[2]]}, Boxed -> False,     AxesStyle -> {Red, Green, Blue},     AxesLabel -> {"Re Input", "Im Input", "Output"},     PlotStyle -> {Green, Thick}],    Plot3D[z = Im[f[x + (y*I)]], {x, Xrange[[1]], Xrange[[2]]}, {y,      Yrange[[1]], Yrange[[2]]},     PlotRange -> {Zrange[[1]], Zrange[[2]]}, Boxed -> False,     AxesStyle -> {Red, Green, Blue},     AxesLabel -> {"Re Input", "Im Input", "Output"},     PlotStyle -> {Blue, Opacity[.5]}]] 
complexfunctionzeroX[f_, n_] :=  Show[ContourPlot[    Re[f[n + (y*I)]] == z, {y, Yrange[[1]], Yrange[[2]]}, {z,      Zrange[[1]], Zrange[[2]]}, ContourStyle -> Green,     FrameStyle -> {Blue, Green}, FrameLabel -> {"IM input", "Output"}],    ContourPlot[    Im[f[n + (y*I)]] == z, {y, Yrange[[1]], Yrange[[2]]}, {z,      Zrange[[1]], Zrange[[2]]}, ContourStyle -> {Blue, Opacity[.5]},     AxesStyle -> {Blue, Green}, AxesLabel -> {"IM imput", "Output"}]] complexfunctionzeroY[f_, n_] :=  Show[ContourPlot[    Re[f[x + (n*I)]] == z, {x, Xrange[[1]], Xrange[[2]]}, {z,      Zrange[[1]], Zrange[[2]]}, ContourStyle -> Green,     FrameStyle -> {Blue, Red}, FrameLabel -> {"RE input", "Output"}],    ContourPlot[    Im[f[x + (n*I)]] == z, {x, Xrange[[1]], Xrange[[2]]}, {z,      Zrange[[1]], Zrange[[2]]}, ContourStyle -> {Blue, Opacity[.5]},     FrameStyle -> {Blue, Red}, FrameLabel -> {"RE input", "Output"}]] complexfunctionzeroZ[f_, n_] :=   Show[ContourPlot[    Re[f[x + (y*I)]] == n, {x, Xrange[[1]], Xrange[[2]]}, {y,      Yrange[[1]], Yrange[[2]]}, ContourStyle -> Green,     FrameStyle -> {Red, Green}, FrameLabel -> {"RE input", "IM input"}],   ContourPlot[    Im[f[x + (y*I)]] == n, {x, Xrange[[1]], Xrange[[2]]}, {y,      Yrange[[1]], Yrange[[2]]}, ContourStyle -> {Blue, Opacity[.5]},     FrameStyle -> {Red, Green},     FrameLabel -> {"RE input", "IM input"}]] 
Xrange = {-10, 10}; Yrange = {-10, 10}; Zrange = {-3, 3}; 

## Can a ballista be fired once per round with a crew of three?

The rules for the ballista on p.255 of the DMG state that to make an attack the weapon must be loaded, aimed and fired. Each of these takes one action. However, could each of these actions be performed by a crew of three, with one character to perform each step, thus allowing the ballista to be fired once per round?

My assumption is yes, but there’s nothing textual to support this. Am I missing something?

## Do Rokugan’s Dragons have three or four toes?

I couldn’t find it in what my collection gives me (or I am blind), but: is there a mention in the sources – or an official depiction – that details if Rokugani Dragons have three or four toes on their hands?

Or… might it be totally different and have five as depicted in some of the TCG cards?

To establish lore, let’s assume the following hierarchy, top to bottom:

• text from Official RPG books for the Legends of the 5 Rings RPGs (Priority to the variant that has most mentions among different editions)
• text from Oriental Adventures material not under the L5R rules
• text Official Adventures not in the print products (aka official campaign)
• Official L5R Novels text
• Official RPG Artwork
• Official TCG descriptions
• Official TCG artwork

## What happens when the feat Born of the Three Thunders modifies thunderhead?

The feat Born of the Three Thunders (CAr 76) reads, in part,

When you cast a spell with either the electricity descriptor or the sonic descriptor that deals hit point damage, you can declare that spell to be a spell of the three thunders, with half its damage dealt as electricity damage and half dealt as sonic damage. In addition, the spell concludes with a mighty thunderclap that stuns all creatures that take damage from the spell for 1 round unless they succeed on a Fortitude save, then knocks stunned creatures prone unless they succeed on a Reflex save (both saves at the same DC as the base spell).

It’s clear the feat is intended to work on instantaneous spells, but a few electricity spells and sonic spells that inflict damage have durations other than instantaneous. The one I’m interested in is the 1st-level Sor/Wiz spell thunderhead [evoc] (SpC 219), which has the electricity descriptor, has a 1-round-per-level duration, and says that it

creates a small thundercloud over the subject’s head. The cloud moves with the subject, following it unerringly even if he becomes invisible or leaves the region. In every round of the spell’s duration, a miniature bolt of lightning leaps from the thundercloud to strike the subject. Each bolt deals 1 point of electricity damage that is negated by a successful Reflex save.

Here’re the questions:

1. Which of the following is true: When the miniature lightning bolts from spell thunderhead, when modified by the feat Born of the Three Thunders, strike a foe…
• only the first miniature lightning bolt will trigger the effects of the feat Born of the Three Thunders.
• each miniature lightning bolt will trigger the effects of the feat Born of the Three Thunders.
2. Which of the following is true: At least one miniature lightning bolt from spell thunderhead, when modified by the feat Born of the Three Thunders, will…
• have an equal chance (a coin toss) of inflicting either 1 point of electricity damage or point of sonic damage.
• inflict 1 point of electricity damage and 1 point of sonic damage (because even if halved minimum damage is 1).
• inflict 0 points of damage (because half of 1 rounded down is 0).

Note: There’s probably an argument that says the spell thunderhead, when modified by the feat Born of the Three Thunders, doesn’t do anything as the miniature lightning bolts inflict the damage and those come from the thundercloud which is created by the spell, and the thundercloud does no damage. I’d rather not entertain that possibility as it gets into the differences between the conjuration and evocation schools of magic. To avoid such arguments, assume that the feat Born of the Three Thunders actually has an effect when applied to the spell thunderhead. Thanks.

“Who Cares?”
A buddy wants to play an electricity-chucking dude. And while dragonfire adept and warlock are interesting, a wizard with the feat Storm Bolt (CM 47) serves this purpose, too, and is… um… a wizard. So, yeah. In looking for ways to make electricity better, the feat Born of the Three Thunders makes electricity attacks more versatile, which is awesome. The feat Mark of the Dauntless (Dra 142) later, and the Wiz6 can do pretty well, shocking folks all day long if he keeps at least 1 Born of the Three Thunders lightning bolt prepared. The thunderhead thing arose because I wanted an interesting 1st-level electricity spell that could be Born of the Three Thunders. (By the way, I know the Storm Bolt feat and the Born of the Three Thunders feat don’t interact. It’s just a theme thing.)

## One Histogram or three in “hspec” option

I have a data set that naturally breaks into three subsets that I want to display as stacked histograms. If I use an hspec option like "Probability", it seems apply to each separately. I would prefer to see it applied to the original, union of the three. Is this possible?

## Segmenting a binary image into three parts

I am trying to segment a binary image into three parts, head, body, and tail. The binary image is a stack tiff file of a moving fruit fly larvae. I was able to do this initially by removing points within a certain distance of the centroid:

However, this broke down if the larvae’s head moved too close to the centroid, because the head points would then be removed:

I have been looking for creative solutions to this problem. Trying to fit it to a curved ellipsoid and get the vertice points that make up the tail and the head. However, I have not been very successful! I was curious if anyone had any insight into this problem or a suggestion for a creative solution? Thank you! 🙂

Here are the raw binary images:

1)

2)

## Is there any compiler IR paradigms other than Three Address Code and Static Single Assignment (SSA) Form?

Specifically, I have tried boiling down the operations of a VM into the smallest possible standardized units and arrived at this:

allocate_stack(size) fetch(offset_from_stack_pointer) store(offset_from_stack_pointer) call(index) 

So you might have:

allocate_stack 6 fetch -2 store 0 fetch -1 store 1 call 72 

Basically, the allocate_stack would tell you how much space you are allocating for fields in the stack. Then the fetch would queue up something relative to the stack to be stored (or fetch_a would fetch an absolute address). Then store would store the fetched value in the space relative to the stack (or store_a for absolute position). And finally, call <index> would call that specific function.

So basically these are 2-address codes instead of 3-address codes. Does anything like this exist out there where they’ve expanded upon the idea to find compiler optimizations and such, like 3-address-code and SSA form? If not like this, does anything else exist outside of 3-address-code and SSA form?

## CEO wants to terminate fifty developers (three developers each month with the interval of ten days) based on the lowest test scores

XYZ Soft is facing a big loss in the industry. Therefore, the company has started to offload its software developers. It started to conduct a test of hundred developers on daily basis and store their test scores in a data structure. After ten days, its CEO wants to terminate fifty developers (three developers each month with the interval of ten days) based on the lowest test scores.

Suppose you are working as a most senior developer in the company then which of the following data structure you will recommend to the CEO in the above given scenario.

1. 1.  AVL Tree 
2. 2.  Heap 

## Greedy Probabilistic Algorithm for Exact Three Cover

I have a probabilistic greedy algorithm for Exact Three Cover. I doubt it’ll work on all inputs in polytime. Because the algorithm does not run $$2^n$$ time. I will assume that it works for some but not all inputs.

Our inputs are $$S$$ and $$B$$

$$S$$ is just a set of integers

$$B$$ is a list of 3-element {}s

## Algorithm

1. Input validation functions are used to ensure that sets of 3 are $$elements$$$$S$$.

2. A simple $$if~~statement$$ makes sure that $$|S|$$ % $$3$$ = $$0$$

3. I treat the sets like lists in my algorithm. So, I will sort all my sets from smallest to largest magnitudes (eg {3,2,1} now will be sorted to {1,2,3}

4. I will also sort my list of sets called $$B$$ in an ordering where I can find all {1,2,x}s with all other {1,2,x}s. (eg, sorted list {1,2,3},{1,2,4},{4,5,6},{4,5,9},{7,6,5} )

5. I will also genereate a new list of sets containing elements where a {1,2,x} only occurs one time in $$B$$.

6. Use brute force on small inputs and on both sides of the list $$B$$ up to $$|S|$$ / $$3$$ * $$2$$ sets. (eg. use brute force to check for exact covers on left and right side of the list B[0:length(s)//3*2] and reversed B[0:length(s)//3*2])

## Seed the PRNG with a Quantum Random Number Generator

for a in range(0, length(B)):     o = quantumrandom.randint(0, length(B))     random.seed(int(o))  # I will create a function to shuffle B later  def shuff(B, n):     for i in range(n-1,0,-1):         random.seed()         j = random.randint(0,i+1)         B[i],B[j] = B[j],B[i] 

## Define the number of times while loop will run

n = length(s)  # This is a large constant. No instances # are impractical to solve.  while_loop_steps = n*241*((n*241)-1)*((n*241)-2)//6 

## While loop

stop = 0 Potential_solution = [] opps = 0 failed_lists = 0 ss = s  while stop <= while_loop_steps:      opps = opps + 1     stop = stop + 1      shuff(B,length(B))          if length(Potential_solution) == length(ss)//3:         # break if Exact         # three cover is         # found.         OUTPUT YES         failed_lists = failed_lists + 1         HALT      # opps helps     # me see     # if I miss a correct     # list           if opps > length(B):         if failed_lists < 1:             s = set()             opps = 0               # Keep B[0]     # and append to     # end of list     # del B[0]     # to push >>     # in list.       B.append(B[0])     del [B[0]]     Potential_solution = []     s = set()          for l in B:         if not any(v in s for v in l):             Potential_solution.append(l)             s.update(l) 

Run a second while loop for new_list if Step 5 meets the condition of there being only ONE {1,2,x}s )eg. {7,6,5} shown in step 4

## Two Questions

How expensive would my algorithm be as an approximation for Three Cover?

And, what is the probability that my algorithm fails to find an Exact Three Cover when one exists?

## What are the three points of view in Kolmogorov Complexity?

I was reviewing for my finals and find this question that I have totally no clue.

Compare the following to statements from three points of view:

1. There exists a constant $$c > 0$$ such that for all palindromes $$x \in \{0, 1\}^*$$ we have $$K(x) \leq \lfloor x / 2 \rfloor + c$$.

2. There exists a constant $$c > 0$$ such that for all $$x \in \{0, 1\}^*$$ we have $$K(\overline{x}) \leq K(x) + c$$ where $$\overline{x}$$ is the complement of $$x$$.

So what are the three points of view am I suppose to use and where should I start?