labeling – No difference between TraditionalForm and StandardForm in Graph labels

Traditional form seems to not work properly on Mathematica 11.3 when it is used in Graphic. The next MWE

edges = {1 [UndirectedEdge] 2};
GraphicsRow[{Graph[edges, VertexLabels -> [Rho],
VertexLabelStyle -> Directive["StandardForm", 30]],
Graphic[edges, VertexLabels -> [Rho],
VertexLabelStyle -> Directive["TraditionalForm", 30]],
Graphic[edges, VertexLabels -> [Rho],
VertexLabelStyle -> Directive["DisplayForm", 30]]}]

Produces the next output (I put Standard form in comparison)

graphic

However, when I do

Traditional form[[Rho]]

The character output is that of the left image of the character description.

Then, I started asking myself if it is possible to use characters in Traditional form in the interior Graphics?

javascript – Change how the data is displayed in a graph in height chart

I create graphs with high graphics, but I have trouble understanding the data for users.

I have this query that returns 5 rows for each month:
SELECT * FROM (SELECT *, Row_Number () OVER (PARTITION PER MONTH ORDER BY month ASC) AS rn
FROM (Collaborator SELECT, MONTHNAME, Month,
ROUND (SUM (TIME_TO_SEC (TIMEDIFF / 60), 2) AS Test)
FROM raddb.Table LEFT OUTER JOIN raddb.Type ON raddb.TypeType.Id = raddb.Target.task
WHERE TaskType = & # 39; 1 & # 39;
Group by MONTHNAME, ORDER BY MONTHNAME contributor, ASC test) as Q1) AS Q2
O rn <= 5 ORDER BY 1,2

Results returned:

insert the description of the image here

It brings 5 ‚Äč‚Äčlines of June and May 2.

So now, I'm creating the graph this way:

$ query = "SELECT * FROM (SELECT *, Row_Number () OVER (PARTITION PER MONTH ORDER BY month ASC) AS rn

FROM (Collaborator SELECT, MONTHNAME, Month,

ROUND (SUM (TIME_TO_SEC (TIMEDIFF / 60), 2) AS Test)

FROM raddb.Table LEFT OUTER JOIN raddb.Type ON raddb.TypeType.Id = raddb.Target.task

WHERE TaskType = & # 39; 1 & # 39;

Group by MONTHNAME, ORDER BY MONTHNAME contributor, ASC test) as Q1) AS Q2

Where rn <= 5 ORDER BY 1,2"; 

$viewer = mysqli_query($conn,$query);
$viewer = mysqli_fetch_all($viewer,MYSQLI_ASSOC);
$arrviewer  = $viewer;
$viewer =  json_encode(array_column($viewer, 'Teste'),JSON_NUMERIC_CHECK); 

$axis = '['; 
foreach($arrviewer as $key=>$ view) {
$ axis. = "# $" View['Colaborador'].
if ($ key! = count ($ arr) -1)
$ axis. = & # 39 ;;
}
$ axis. = & # 39;]& # 39 ;;

$ (function () {
var data_viewer = ;

$ (& # 39; container & # 39;) highcharts ({
board: {
renderTo: & # 39; container & # 39;
type: column & # 39;
}
title: {
text: & # 39; Positions & # 39;
}
xAxis: {
categories: 
        }
yAxis: {
title: {
text: hour in hours & # 39;
}
}
series: [{
            name: 'Posicionamentos',
            data: data_viewer
        }]
    });
});

The result is as follows:

insert the description of the image here

In the text box, only the month and time of positioning perished and wanted the collaborator's name to appear as well.

In the XAXIS he wanted the five lines returned to appear each month instead of repeating the same month several times.

For each unweighted undirected graph, each STM is also a SPT

I have the feeling that this could be wrong and that I am looking for a counterexample, however, I could not find one for now .. anyone can I guide myself into the good direction?

unit – How can I create a plexus effect with VFX Graph?

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9.0 pie – Android 9. No graph of total data usage?

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Topological Graph Theory – Layer Thickness

A book incorporating a G-chart consists of placing the vertices of G on a back and assigning graphic edges to the pages so that the edges of the same page do not intersect. The page number is a measure of the quality of an embedded book that corresponds to the minimum number of pages in which chart G can be incorporated.

If a graph $ G $ is a finite overlay graph $ B $,
is there a relationship between their page number?

I think the cover chart is more complicated than the basic chart. Similarly $ pn (G) geq pn (B) $ hold in general?

Graph Theory – Online Algorithm Searching for a Clique of Size K

I'm trying to write an online algorithm capable of detecting cliques of size k. I start with a set of vertices first. For each iteration, I add an edge. The algorithm will detect the first time that an edge would create a clique of size k. What is an efficient algorithm that can carry out this task and what is the temporal complexity?

co.combinatorics – A transitive vertex graph has an almost perfect match missing from an independent set of vertices

Consider a graph of the power of the cycle $ C_n ^ k $, represented by a graph of Cayley with generator $ {1,2, ldots, k, n-k, ldots, n-1 } $ on the group $ mathbb {Z} _n $. Suppose I delete an independent set of vertices of the form $ {i, i + k + 1, ldots, lfloor frac {n} {k + 1} rfloor + i $ or a single vertex. Then, is it possible to get a perfect / nearly perfect match when I always delete the independent vertex set? If no, then is it possible in case the graph is equal cycle power?

I hope so, because we can couple the vertices between two independent sets of the form above or between the independent set and the single vertex to obtain a nearly perfect maximum match (in the case where the # The order of the induced subgraph is odd) or perfect (in case the order of the induced subgraph is even). We can also see by observing that when the maximal independent set of vertices (as indicated above) is suppressed, we have a Hamiltonian cycle in the induced subgraph (the proof of this point does not seem clear to me) . Counterexamples? Also, can we generalize this, if it is true, to any transitive vertex graph, that is, is there an independent set of vertices (non singleton), so that the removal of this set induces a perfect match / almost perfect? Thank you in advance.

Randomized algorithms – How can a maximum number of minimum cuts in a graph be exactly $ n choose $ 2?

According to my instructor, $ n choose $ 2 is the maximum number of minimum cuts that we can have on a graph. To prove it, he showed the lower limit using an n cycle chart. To prove the upper limit, he drew the argument of two facts:

  • Probability of finding $ i ^ {th} $ min cut $ geq frac {2} {n (n-1)} = frac {1} {n choose 2} $
  • Event to find $ i ^ {th} $ min cut is disjointed.

So, by adding the probabilities, he proved the upper limit of $ n choose $ 2.

Now, if we look at a tree, graphically, with $ n $ nodes, then we will be able to conclude $ (n-1) $ cuts min which is less than $ n choose $ 2 cuts ($ n geq3) $. Did I miss something?

tracing – Changing the graph to become an animation

p1 = ParametricPlot3D[{(3 + Cos[v]) Peach[u], (3 + Cos[v]) Cos[u],
Peach[v]}, {u, 0, 2 Pi}, {v, 0, 2 Pi},
Plot Style -> Opacity[.5], Mesh -> None];

Erase everything[p2, p3]
p2[tmax_] : = ParametricPlot3D[{Cos[Sqrt[2] t](3 + Cos
Peach[Sqrt[2] t](3 + Cos
p3[tmax_] : = ParametricPlot3D[{Cos[Sqrt[2] t](2 + Cos
Peach[Sqrt[2] t](2 + Cos
Line -> ({CapForm[None], FaceForm[Green, White], Tube[#, .25]} &);

Manipulate[Show[p1, p2[tmax], p3[tmax], PlotRange -> All, Axes -> False], {tmax, 1, 50}]

enter the description of the image here

To export as GIF animation:

frames = Table[Show[p1, p2[tmax], p3[tmax], PlotRange -> All], {tmax, 1, 50}];
Export["frames.gif", frames]

enter the description of the image here