Ramsey style theorem with unbounded colors

Question: Let $varepsilon>0$ and $Ninomega$ be sufficiently large (depending on $varepsilon$).
Let $h:subseteq Nrightarrow N$ be such that $h(B)notin B$ for all $Bsubsetneq N$.
Must there be $B_0subsetneq B_1subseteq N$ such that $|B_1|leq varepsilon N$ and $h(B_0)=h(B_1)$?

More generally,

Let $ainomega,varepsilon>0$ and $Ninomega$ be sufficiently large (depending on $a,varepsilon$).
Let $h:subseteq Nrightarrow N$ be such that $h(B)notin B$ for all $Bsubsetneq N$.
Must there be $B_0subsetneq B_1subsetneqcdotssubsetneq B_{a-1}subseteq N$ such that $|B_{a-1}|leq varepsilon N$ and $h(B_0)=cdots= h(B_{a-1})$?

graphics – Interpolating vertex colors with trilinear coordinates instead of barycentric

In 3D graphics, vertex attributes like vertex normals, texture coordinates, and vertex colors are interpolated over the surface of triangles using barycentric coordinates. In actual fact it’s more complex than this and there’s a rabbit hole of perspective correct adjustment and all sorts, but indulge me for a moment. Here’s what this looks like for vertex colors, and Mathematica handles everything about the interpolation behind the scenes:

vtxs = {{0, 0}, {0, 1}, {1, 0}};
cols = {Red, Green, Blue};
tri = Polygon(vtxs, VertexColors -> cols);
pt = {1, 1/2}/2;
  {Yellow, PointSize(Large), Point(pt), Line({pt, #}) & /@ vtxs},
   Text(Style(#1, FontWeight -> Bold, FontSize -> 24), #2) &, 
    {{"R", "G", "B"}, vtxs}),
  Style({Text("u", {.3, .1}), Text("v", {.2, .65}), 
    Text("1-u-v", {.7, .1}), 
    Text(TraditionalForm(p), pt + {1, 1}*0.04)}, White, FontSize -> 24)

barycentric vertex colors

The color at $p$ is a weighted combination $p_c=Rw + Gv + Bu$ where $(u,v,w=1-u-v)$ are the relative areas of the subtriangles such that $u+v+w=1$.

I want to find a reasonably efficient way to interpolate the colors for any triangle without using barycentric coordinates, but using trilinear coordinates instead, and see what images this creates.

That is, we drop a perpendicular to the nearest edge and get a point on that edge, possibly indexing into the $RG,GB,RB$ gradients along the edges, and then weighting the combination of these colors by the trilinear distances or areas of the three regions – whatever is more appropriate as it’s all experimental. Any other suggestions for how to interpolate vertex colors using trilinears is welcome.

trilinear interpolation

This is my code so far.
The next step is to interpolate the corner colors over a pixel grid and make an image, instead of a Graphics with a single point:

poly = RandomPolygon(3);
shell = RegionUnion(MeshPrimitives(poly, 1));
nfs = RegionNearest /@ MeshPrimitives(poly, 1);
rds = RegionDistance /@ MeshPrimitives(poly, 1);
centroid = Mean(First(poly /. Polygon -> List));
 Graphics({FaceForm(None), EdgeForm(Thick), poly, Orange, 
   Line({p, #(p)}) & /@ nfs, Point(centroid)}), {{p, centroid}, 


microsoft word – Creating Text Style with Colors using Pandoc to Docx Converter

I am creating a simple Word document (*.docx) using Pandoc. My input text file is fairly straight forward. I would like to create text with Red/Yellow/Green highlight. I am aware of creating a custom reference document and using that in Pandoc. In fact, I have a custom-reference.docx that I am using, but I am not sure how to add a Text Style for this Text Highlight Colors.

My Pandoc command

pandoc -s word_basics_pandoc.txt -o output.docx –reference-doc=custom-reference.docx

Output in docx, I am trying for

text color highlight

My Sample Text for creating Docx

title: Word report
author: John Doc
date: May 24, 2021
keywords: word,pandoc

This is a sample text document created using Pandoc.

# Text Style with either Text color (Text Highlight color)


## Lists

* one
* two
* three

## Ordered Lists

1. one
2. Two
3. Three

What preset can be used to achieve similar desaturated colors?

What preset can be used to achieve similar desaturated colors as Andrea Dabene? Props if I can also take these pics SOOC (film simulations) on a Fuji. Example pics here.








Instagram for more pics: https://www.instagram.com/andreadabene/?hl=en

plotting – Colors in RegionPlot legend

I’m trying to put a common legend for two-row RegionPlot. However, I haven’t been able to get the colors in the legend right:

ra = RegionPlot[{y >= x^2 - 4 , y >= x^2 - 4 && y <= 0 }, {x, -10, 
   10}, {y, -10, +10}]

rb = RegionPlot[{y >= x^2 - 16 , y >= x^2 - 4 && y <= 0 }, {x, -10, 
   10}, {y, -10, +10}]

Legended[GraphicsRow[{ra, rb}, Spacings -> Scaled[0.4], 
  Epilog -> Inset["Plot Title", Scaled[{0.5, 0.95}]]], 
 SwatchLegend[{Blue, Orange}, {"label1", "label2"}]]

Legended[GraphicsRow[{ra, ra}, Spacings -> Scaled[0.4], 
  Epilog -> Inset["Plot Title", Scaled[{0.5, 0.95}]]], 
 SwatchLegend[{LightBlue, LightOrange}, {"label1", "label2"}]]

I’ve also tried to change the colors of the regions. However, as far as I could tell, the declared color applied to the boundaries; the interior was a shade lighter, something I couldn’t reproduce in the legend. Any ideas on how to address the problem?

plotting – How can I assign mixed colors to each point in ListPlot?

I have this data which is in the form {{xi,yi,{ri,bi,gi}}} where xi and yi are the data and {ri,bi,gi} are the weight of red, black and green colors that coincide with each point {xi,yi}. To visualize this I used Blend to give the assigned color for each point as follows

colorbar(rd_, bk_, gr_) := Blend({Red, Black, Blue}, Rescale({rd, bk, gr}, {0, 1})) 

and then I tried this

    ListPlot(data((All, 1 ;; 2)), ColorFunction -> (colorbar(#((1)), #((2)), #((3)))) & /@ 
  data((All, 3 ;; 3))((All, 1)), ColorFunctionScaling -> False)

Now, this should show each point with the assigned color, but all points appear with the same color

enter image description here

How can I solve this, please?

Julia: ImplictPlots in a loop: setting colors

How can I assign colors with implicit_plot?

using ImplicitPlots, Plots
v = ((x,y) -> (x+3)*(y^3-7)+18, (x,y) -> sin(y*exp(x)-1))
p = plot()
for f in v 
    implicit_plot!(f; xlims = (-1, 3), ylims = (-1, 2), framestyle = :origin, lc = (:red, :blue))


Would it be easy to pass a gradient of blue colors for the second function?

Bonus question: Can I pass a vector or tuple to implicit_plot?

using ImplicitPlots, Plots
f(x,y) = (((x+3)*(y^3-7)+18, sin(y*exp(x)-1)))
implicit_plot(f; xlims = (-1, 1), ylims = (-1, 1))

ERROR: LoadError: MethodError: no method matching isless(::Float64, ::Tuple{Float64, Float64})

I guess it’s not supported, but just wanted to double-check. I tried to put together a wrapper, but I’m not sufficiently familiar with how to pass functions around.

Proof: Graph Coloring algorithm needs at most twice as many colors to color a graph as the optimal solution

Given an algorithm

(1) create the complement $bar{G}$ of input graph $G$

(2) calculate a maximum matching $M$ on $bar{G}$

(3) color the two vertices of every edge $e_i in M$ with color $i$

(4) color remaining nodes on the original graph with a greedy approach (every vertices receives the smallest color possible)

Proof that this algorithm needs at most twice as many colors as $chi(G)$.

Can someone give me a hint how to start this?

How to find and replace colors in Google Docs

I have a google docs and there is a bunch of repeating words miscolered and I would like to know how to find and replace colors please.

I’ve already searched this type of topic a few days ago right here: How can I automate Find & Replace to change font colours in Google Docs?

What I didn’t understand about it is that It included Google Scripts but I can’t for some reason access Google App Scripts through my Google Docs Document. The only thing that is stopping me from answering my own question is that I can’t use Google Scripts on my Document and I am not using a Google Workspace account.

Why color-wheel websites (Adobe…) claim these colors are complementary

I just learn about the different color harmony styles models, and I was playing with Adobe color wheel website using the complementary style, but I’m realizing that the proposed colors are not really on the other side of the Hue circle:

enter image description here

Here, you see that the blueish line has hue 174° and the brown color has hue 19°, and 174°-19°=155°, while I would expect their distance to be 180° since they are complementary. And if I ask to Krita, they are indeed not on opposite parts of the Hue circle:

enter image description here

enter image description here

So what’s wrong with Adobe color wheel? I also tried with https://paletton.com, and the Paletton gives similar results compared to Adobe color. Are all color wheels crazy, or am I missing something? I also tried to check how footage was behaving, and they seem to agree with my definition, for instance this photo uses colors #27AB9E (Hue=174°) and #ff4473 (hue=344°), which are much more aligned, and we have 344-174 = 170° which is way closer to 180°.

enter image description here

enter image description here