## How to plot the solutions of $f(z,a)=0$ on the complex plain of $z$?

I have an equation $$f(z,a)=0$$ where $$z$$ is a complex variable $$\frac {56}{10} and $$a$$ is a real variable $$0.

I want to plot solutions of this equation on the complex plain $$z$$ as $$a$$ increases from $$0$$ to $$\pi$$, and show the behavior of $$z$$ by changing the color of the curve (as $$a$$ increases, the curve changes from blue to red), something like this plot

f[z_,a_]:=9 + 4 Cos[a - (273 z)/50] - 4 Cos[a - 2 z] - 2 Cos[2 z] -   3 Cos[(173 z)/50] + Cos[4 z] - 2 Cos[(273 z)/50] -   3 Cos[(373 z)/50] - 4 Cos[a + 2 z] + 4 Cos[a + (273 z)/50] -   4 I Sin[a - (273 z)/50] + 4 I Sin[a - 2 z] - 2 I Sin[2 z] -   3 I Sin[(173 z)/50] + I Sin[4 z] - 2 I Sin[(273 z)/50] -   3 I Sin[(373 z)/50] - 4 I Sin[a + 2 z] + 4 I Sin[a + (273 z)/50];   f[z,a]==0  56/10 <z<\[Pi] 0<a<\[Pi]  

## Wich solutions is better for translate some data?

I’ve a little mobile app who need some translation. Currently, I just want to translate my app in one language, I don’t think I’ll need more. So I’m facing a problem, should I translate stuff in database, or in translations files on client side ?

The user will be able to select his language, but first i need to determine it with the locale variable for the registration (because he must select his country at registration).

First I tought about creating one translation table for each table who need it.

But it seems to not be the good solution if I need to add more language, I’ll have to alter all translations tables.

So, my second solution is to create a language table, keep the translation table for each table but as one-to-many relationship.

I think this solution is better, but I’m not sure about performance. I know it’s a little app and the question does not really arise, but we never know.

Now, I’m stuck because It means that every time I need translation, I have to query the database. There are tons of contries, and I’ll like to have auto completion. Fruits will be displayed in several pages and used in others tables.

So I thought about keep translations on client side within json files. For exemple translations/fr_FR.json

"fruits": [   {     "id": 1,     "original": "banana",     "translation": "banane"   },   {     "id": 2,     "original": "apple",     "translation": "pomme"   } ],   "countries": [   {     "id": "ES",     "original": "spain",     "translation": "espagne"   } ] 

Everytime I need translation, I can use a function to get the translation from the file.

It means more calculations, and user will have to update his app if translation has changed.

Wich solution is better ?

## [ GSA SER Emails Mega Thread ] – GSA SER catchall Emails / Service Providers / Solutions

I made to create this thread for everyone.   Since we use GSA SER we need quality emails.  I know there are many catchall Email providers /  Major Email providers such as Yahoo , Google etc which is kinda expenses but best.
What kind of catchall services you guys use ?
What kind of solutions ?
How to create catchalls if you do your own ( If then what hosting / providers allows that )
Any tutorials etc ?
So far i use mixed emails.
I purchase catchalls from market places (where i am not soo happy and many catchall sellers domains are black list)
I purchase gmail / Hotmail / some private providers i found
Feel free to share yours and feel free to ask anything from mine..
LETS TALK ABOUT GSA Search engine ranker and Emails…
It’s all about emails..
Let’s make this thread a GSA SER Email solution mega thread .
Note for admins : @Sven @S4nt0s&nbsp;  : I just wanted to create this thread to share knowledge . If mega threads or these kind of threads not allowed in this community please remove this.

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## Real solutions of third and fourth degree equations

A few hours ago I "discovered" that if a third or fourth degree equation has distinct real solutions, it’s possible to calculate them avoiding complex numbers.

In particular, we have:

poly = (x - 1)(x - 2)(x - 3); {c, b, a} = CoefficientList[poly, x][[1 ;; 3]]; d = a^2/3 - b; e = 2 a^3/27 - a b/3 + c; f = ArcCos[-3 Sqrt[3] e/(2 d Sqrt[d])]; N[{-a/3 + 2 Sqrt[3 d] Cos[(f - 4 Pi)/3]/3,    -a/3 + 2 Sqrt[3 d] Cos[(f - 2 Pi)/3]/3,    -a/3 + 2 Sqrt[3 d] Cos[(f - 0 Pi)/3]/3}] 

{1., 2., 3.}

poly = (x - 1)(x - 2)(x - 3)(x - 4); {d, c, b, a} = CoefficientList[poly, x][[1 ;; 4]]; e = 3 a^2/4 - 2 b; f = 2 c - a b + a^3/4; g = b^2 + 12 d - 3 a c; h = 27 a^2 d - 9 a b c + 2 b^3 - 72 b d + 27 c^2; i = ArcCos[h/(2 g Sqrt[g])]; j = (e + 2 Sqrt[g] Cos[i/3])/3; N[{-a/4 - (Sqrt[j] + Sqrt[e - j + f/Sqrt[j]])/2,    -a/4 - (Sqrt[j] - Sqrt[e - j + f/Sqrt[j]])/2,    -a/4 + (Sqrt[j] - Sqrt[e - j - f/Sqrt[j]])/2,    -a/4 + (Sqrt[j] + Sqrt[e - j - f/Sqrt[j]])/2}] 

{1., 2., 3., 4.}

After a few moments of joy, however, I noticed that, for example, if I write (x - 10^-15) instead of (x - 1), I get respectively 6.66134*10^-16 and 8.88178*10^-16 instead of 10^-15 as the solution.

I intuitively believe that it’s a numerical problem and that the main cause is ArcCos, but I’m not too sure and also I’m not able to establish if something can be done to solve the issue, or if I have to give up and I must necessarily rely on it to the good Newton method.

Thank you!

## Hosting solutions supporting SSLv3 [closed]

What hosting companies/services still offer SSLv3 in their free or paid SSL certificates?

(The outdated SSLv3 is supported by default in very old browsers).

## Interpreting the accuracy of solutions to the correspondence problem

I have two pictures of the same object, taken by a car travelling down the road like shown on the right side of the image below. I want to find pixels of the object in each frame that correspond to each other.

Now, in the description of the Middlebury Stereo Evaluation v.3 dataset it says

Maximum disparities range from 200 to 800 pixels at full resolution.

This leads me to my two questions:

1. Do I understand correctly that algorithms working with the Middlebury dataset had to match pixels that were a distance of 200 to 800 pixels apart, like shown on the left side of the image?
2. Consider the leaderboard for the Middlebury Stereo Evaluation. Does an average absolute error metric of 1.4 mean, that above problem could be solved for the images in the dataset with an average accuracy of 1.4 pixels?

## CLRS 22.3-1, How Come Solutions Online State There Can’t Be Edges From WHITE to GRAY nodes “at any point… during search”?

The exercise (from the book Introduction To Algorithms) states

Make a 3-by-3 chart with row and column labels WHITE, GRAY,and BLACK. In each cell (i, j) indicate whether, at any point during a depth-first search of a di- rected graph, there can be an edge from a vertex of color i to a vertex of color j . For each possible edge, indicate what edge types it can be. Make a second such chart for depth-first search of an undirected graph.

The colors WHITE, GRAY, BLACK correspond to Undiscovered, discovered but not finished, and finished. The following solution is what multiple sites & universities have posted(such as: walkccc, Rutgers University):

 |       | WHITE         | GRAY                | BLACK                | |-------|---------------|---------------------|----------------------| | WHITE | All kinds     | Cross, Back         | Cross                | | GRAY  | Tree, Forward | Tree, Forward, Back | Tree, Forward, Cross | | BLACK | -             | Back                | All kinds            | 

I will draw a minimal counter example as it helps understand my conflict:

• Start at node 0: 0 is GRAY
• PAUSE
• At this point, 3 is still white and has an edge to 0
• Resume and keep going, eventually the edge from 3 to 0 will be discovered as a tree edge

This contradicts the solutions saying you can only have Cross/Back edges going form WHITE->GRAY. This example can be easily modified to contradict many of the elements in the table. I think the solutions are doing one of the following:

• Assuming that the graph is a tree and that we start at its root. (Wrong as DFS doesn’t need a tree graph and any node can be started from).
• More likely (Just thought of this), interpreting the question of "can there be an edge" as "can there be an edge that we have discovered". In which case, the solutions work, as although the edge from 3->0 was a WHITE->GRAY edge at one point, we hadn’t discovered it yet.

## NMinimize: How to avoid solutions that do not satisfy constraints within a certain tolerance?

I just started to use Mathematica a few weeks ago. Using NMinimze, I would like to avoid solutions that do not satisfy certain constraints (although they "almost" satisfy them). Do you know how to change the following command to find a solution satisfying "completely" all the constraints, solving the same minimization problem?

NMinimize[{((e*(1 - Sqrt[(g - e)^2 + (f - h)^2]) + (g - e)*(1 -        Sqrt[f^2 + e^2])) + (h*(1 -        Sqrt[(g - e)^2 + (f - h)^2]) + (f - h)*(1 -        Sqrt[g^2 + h^2])))/((g + f)*  Max[1 - Sqrt[(g - e)^2 + (f - h)^2], 1 - Sqrt[g^2 + h^2]]), 0 <= e <= 1, 0 <= f <= 1, e^2 + f^2 == 1, e <= g <= 1, 0 <= h <= f, Sqrt[(g - e)^2 + (f - h)^2] <= 1, g^2 + h^2 <= 1}, {e, f, g, h}]  

## finding the combinatorial solutions of series and parallel nodes

I have n nodes, and I want to find the (non duplicate) number of possible ways in which these nodes can be combined in series and parallel, and also enumerate all the solutions. For example, for n=3, there are 19 possible combinations.

 0 (0, 1, 2)  1 (0, 2, 1)  2 (1, 2, 0)  3 (1, 0, 2)  4 (2, 0, 1)  5 (2, 1, 0)  6 [0, 1, 2]  7 [0, (1, 2)]  8 [0, (2, 1)]  9 (0, [1, 2]) 10 ([1, 2], 0) 11 [1, (0, 2)] 12 [1, (2, 0)] 13 (1, [0, 2]) 14 ([0, 2], 1) 15 [2, (0, 1)] 16 [2, (1, 0)] 17 (2, [0, 1]) 18 ([0, 1], 2) 

In the notation above, a series combination is denoted by (..) and a parallel combination is denoted by [..]. Duplicates are removed, for example [0,1,2] is the same as [1,2,0] since all of them are happening in parallel so the order does not matter here.

Can you give me an algorithm for this, or if any such algorithm already exists, then point me to it?

(I tried googling for a solution, but did not hit any relevant answer, maybe I was entering the wrong keywords.)

Note: for a sequential-only solution, the answer is easy, it is n!, and the enumeration of the solutions is also easy. But when parallelism (especially non duplicates) is added to the problem, it gets very complex.