Counter example to Zooko’s triangle? [on hold]

So Zokoo’s triangle is a conjecture that is often explained by this trilemma:





pick two

But what if we hash the public key of an user and turn it into a meaningful identifier?

For example, we can use the hash as an input to get an unique profile picture with Gravatar (the default profile picture generator of Stack Exchange). We can also pass the hash as a seed to a neural network that generates profile pictures with people, animals, landscapes, etc.

But that’s not good, we can’t fetch an user identified by their profile picture, so we have to create meaningful character strings. I call it the “exquisite corpse” method: Let’s say the hash contains 6 alphanumerical characters, we can divide it into 3 parts of 2 characters each. The first part identifies an adverb in a dictionary of 3844 words ((26*2+10)²), the second part identifies an adjective is a dictionary of 3844 words and the third part a noun in a dictionary of 3844 words. This way, we obtain usernames like Extremely Metaphorical Chicken.

Maybe you would say that relying on a common arbitrary dictionary of words makes it not distributed, but in this case the arbitrary choice of the hash function would also make it not distributed. The ‘distributed’ criterion means that every peer on the network can do name resolve by themselves, not that we can’t rely on a common standard. Every peer can download the dictionary and/or the Gravatar generator (although I concede installing a whole neural network would be too much).

Is it a proper counter-example to Zooko’s triangle or I am mistaken? If so, should we change “human-meaningful” to “human-choosable”?

Thank you for your help.

to do a triangle [on hold]

this is my code

class Pattern {     public static void main (String [] args)         {         int STAR = "*";         int SPACE = " ";         int HEIGHT = 5;          {              Display();              console.ReadLine();              {               static public void  Display()                  {                      for (int i = 0; i < HEIGHT; i++)                      {                          for (int j = 0; j < i; j++)                          console.write(SPACE);                           for (int j = 0; j < HEIGHT; j++)                          console.Write(STAR);                          console.WriteLine();                      }                  }              }          }     } } 

my errors are this error: illegal start of expression static public void Display() error: class, interface, or enum expected }

Intersection of sphere with triangle containing moving vertices

First off, apologies if I cannot properly articulate my question in the most formal way. However, I believe my question should be simple enough to grasp anyhow.

In $ \mathbb{R}^3$ , I would like to determine the time of contact, if any, between:

  • An unmoving unit sphere, whose center is at the origin
  • A triangle, each of whose vertices follow independent, linear motion from $ t = 0$ to $ t = 1$ . In other words, each triangle vertex has a start and end position, which are linearly interpolated by $ t$ .

The sphere may touch the triangle on a vertex, an edge, or on the triangle’s surface. Vertex testing is simple enough as it’s analogous to static line segment-sphere intersection.

For edge testing, parameterizing a line-sphere intersection test by $ t$ appaers to lead to solving a degree 4-polynomial, which isn’t ideal. I believe that doing the same for triangle surface testing (parameterizing a sphere-plane intersection with $ t$ ) would lead to solving a 6-degree polynomial.

Would there be any applicable non-analytical methods to approximate the intersection (other than directly approximating the polynomial roots)? Or maybe there is an analytical method that I’m not thinking of. In addition, would further constraining the motion of the triangle potentially simplify a solution?

divide and conquer algorithm for finding a 3-colored triangle in an undirected graph with the following properties?

In an undirected Graph G=(V,E) the vertices are colored either red, yellow or green. Furthermore there exist a way to partition the graph into two subsets so that |V1|=|V2| or |V1|=|V2|+1 where the following conditions apply: either every vertex of V1 is connected to every vertex of V2 or no Vertex of V1 is connected to a vertex of V2 . This applies recursively to all induced subgraphs of V1 and V2

I can find all triangles in the Graph by multiplying the adjacency matrix with itself three times and step up the nodes corresponding to the non zero entries of the main diagonal. Then I can see if the nodes of the triangle are colored the right way. O(n^~2,8)! But given the unique properties of the graph I want to find a solution using divide and conquer to find the colored triangle. this is an example graph with the given properties. I need to find the bold triangle: this is an example graph with the given properties. I need to find the bold triangle

Blue boxes symbolize the partitions are fully connected, purple boxes mean no connection between the partitions There is no connection between the coloring of nodes and the partitions, also the partitions have to be computed on the fly

Printing Pascal’s triangle for n number of rows in Python

I have written a program that takes a number n and prints Pascal’s triangle having n number of rows.

Here is my code –

n = int(input("Enter number of rows: ")) a = []  for i in range(n):     a.append([])     a[i].append(1)      for j in range(1, i):         a[i].append(a[i - 1][j - 1] + a[i - 1][j])     if (n != 0):         a[i].append(1)  for i in range(n):     print("   " * (n - i), end = " ", sep = " ")      for j in range(0, i + 1):         print('{0:5}'.format(a[i][j]), end = " ", sep = " ")     print() 

Here are some example outputs –

Enter number of rows: 3               1             1     1          1     2     1  

Enter number of rows: 6                        1                      1     1                   1     2     1                1     3     3     1             1     4     6     4     1          1     5    10    10     5     1  

Enter number of rows: 10                                    1                                  1     1                               1     2     1                            1     3     3     1                         1     4     6     4     1                      1     5    10    10     5     1                   1     6    15    20    15     6     1                1     7    21    35    35    21     7     1             1     8    28    56    70    56    28     8     1          1     9    36    84   126   126    84    36     9     1  

Therefore, I would like to know whether I could make this program shorter and more efficient.

Any help would be highly appreciated.

Tangent line of a function in 1st quadrant forms a triangle. Find the Area.

So, i have this particular question, that is Calculus based. Here is the question below:

Consider graph of f(x) = exp(-x), for x>= 0 . The tangent line to the graph of f(x) at x = a, and intersects the x-axis at point A, and the y-axis at point B. Determine the area of triangle AOB in terms of a.

Hence the coordinate on the x-axis is (A,0) and coordinate on the y-axis is (0,B).

I did some work on this. The slope of the tangent line would be : m = -exp(-a) So my equation of the tangent line would be : y = [-exp(-a)]x + B, next i plug in the point : (A,0) Then get the relation B = A*exp(-a), therefore the full equation of the tangent line is: y = -exp(-a)[x-A]

Now the area is of a right triangle with vertices AOB, that area is : Area = AB/2. Plugging in the relation of B = Aexp(-a). Then Area = [(A^2)*exp(-a)]/2

BUT my formula depends on the x-axis coordinate A. SO not sure if this A can be replace with it being related to ‘a’, the point of tangent.

Hope some one can give some input on this.

Yau’s problem: Construct a triangle given a side, an angle, and an angle bisector

In Shing-Tung Yau’s autobiography The Shape of a Life, he mentions a problem that he came up with as a teenager.

Suppose you know the length of one side of a triangle, one angle, and the length of one angle bisector. Can you construct the corresponding triangle, using just a compass and ruler? I worked on this problem for the better part of a year and made little headway. …

One day, I found a book that discussed [this problem]. I learned that it could not be solved, which came as quite a relief. The book cited a recent argument that proved you could not construct one, and only one, triangle that satisfied three of these conditions.

I was excited to see that “my problem” had stumped other people and was only recently shown to be insoluble. I further realized that this same problem was similar to one that dated back many centuries: Could you trisect an angle if you had only a ruler and compass? No, you could not. Nor could you solve another long-standing problem, “squaring a circle.” … I was proud to find out that my problem was in the same category as these two classic problems.

I’m curious to learn more about the history and literature of this problem. I tried doing some searching but when I use the obvious keywords, I get too many hits on unrelated problems.

Possibly this question belongs on some other stackexchange site; I’m willing to migrate it if people think it should be.