Can someone please explain to me what a straight-line distance heuristic would look like while solving the n-puzzle? How would you apply the straight line distance, say for a 8x8 puzzle? Here is an example puzzle:
7 3 4
5 _ 6
8 2 1
Let's recall basic geometry, it is well known that the shortest path between two points is a straight line.
So considering a 8-puzzle, the straight line distance between two tiles is the number of tiles to get from tile A to tile B whether is diagonal, horizontal or vertical line.
Considering the example in your question, let's call d(a,b) the straight line distance between tile a and b:
d(1,_) = 1
d(1,2) = 1
d(1,3) = 2 = d(1,6) + d(6,3) = d(1,_ ) + d(_,3)
d(1,4) = 2
and so on.
We can now generalize that definition to the n-puzzle. Keep in mind that 3 steps are allowed diagonal, horizontal, vertical. In this case the heuristic is usually optimal.
Note: Remember the straight line definition between cities.
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Hi when i am reading about Stanford's Machine Learning materials about autoencoder, found a formula hard to prove by myself. Link to Material
Question is:
" What input image x would cause ai to be maximally activated? "
Screen shot of the Question and Context:
Many thanks to your answers in advance!
While this can be rigorously solved using KLT conditions and Lagrange multipliers, there is a more intuitive way to figure the result out. I assume that f(.) is a monotone increasing, sigmoid type of nonlinearity (ReLU is also valid). So, finding the maximum of w1x1+...+w100x100 + b under the constraint (x1)^2+...+(x100)^2 <= 1 is equivalent to finding the maximum of f(w1x1+...+w100x100 + b) with the same constraint.
Note that g = w1x1+...+w100x100 + b is a linear function of x terms (Name it as g, so later we can refer it by that). So, the direction of largest increase at any point (x1,...,x100) in the domain of that function is the same, which is the gradient. The gradient is simply (w1,w2,...,w100) at any point in the domain, which means if we go in the direction of (w1,w2,...,w100), independent from where we start, we obtain the largest increase in the function. To make things simplier and to allow us to visualize, assume that we are in the R^2 space and the function is w1x1 + w2x2 + b:
The optimum x1 and x2 are constrained to lie in or on the circle C:(x1)^2 + (x2)^2 =1. Assume that we are on the origin (0.0). If we go in the direction of the gradient (blue arrow) (w1,w2), we are going to attain the largest value of the function where the blue arrow intersects with the circle. That intersection has the coordinates c*(w1,w2) and it is c^2(w1^2 + w2^2) = 1, where c is a scalar coefficient. c is easily solved as c= 1 / sqrt(w1^2 + w2^2). Then at the intersection we have x1=w1/sqrt(w1^2 + w2^2) and x2=w2/sqrt(w1^2 + w2^2), which the solution we seek. This can be extended in the same way to 100 dimensional case.
You may ask why we started at the origin and not any other point in the circle. Note that the red line is perpendicular to the gradient vector and the function is constant along that line. Draw that (u1,u2) line, preserving its orientation, arbitrarily with the constraint that it intersects the circle C. Then choose any point on the line, such that it lies within the circle. On the (u1,u2) line, you start at the same value of the function g, wherever you are. Then as you go in the (w1,w2) direction, the longest path taken within the circle always goes through the origin, which means the path you increase the function g the most.
i need to find out contact angle between 2 edges in an image using open cv and python so can anybody suggest me how to find it? if not code please let me know algorithm for the same.
Simplify edge A and B into a line equation (using only the few last pixels)
Get the line equations of the two lines (form y = mx + b)
Get the angle orientations of the two lines θ=atan|1/m|
Subtract the two angles from each other
Make sure to do the special case of infinite slope, and also do some simple math to get Final_Angle = (0;pi)
I've been working on a problem for several weeks and have reached a point that I'd like to make sure I'm not overcomplicating my approach. This is being done in OpenGL ES 2.0 on iOS, but the principles are universal, so I don't mind the answers being purely mathematical in form. Here's the rundown.
I have 2 points in 3D space along with a control point that I am using to produce a bezier curve with the following equation:
B(t) = (1 - t)2P0 + 2(1 - t)tP1 + t2P2
The start/end points are being positioned at dynamic coordinates on a fairly large sphere, so x/y/z varies greatly, making a static solution not so practical. I'm currently rendering the points using GL_LINE_STRIP. The next step is to render the curve using GL_TRIANGLE_STRIP and control the width relative to height.
According to this quick discussion, a good way to solve my problem would be to find points that are parallel to the curve along both sides taking account the direction of it. I'd like to create 3 curves in total, pass in the indices to create a bezier curve of varying width, and then draw it.
There's also talk of interpolation and using a Loop-Blinn technique that seem to solve the specific problems of their respective questions. I believe that the solutions, however, might be too complex for what I'm going after. I'm also not interested bringing textures into the mix. I prefer that the triangles are just drawn using the colors I'll calculate later on in my shaders.
So, before I go into more reading on Trilinear Interpolation, Catmull-Rom splines, the Loop-Blinn paper, or explore sampling further, I'd like to make sure what direction is most likely to be the best bet. I think I can say the problem in its most basic form is to take a point in 3D space and find two parallel points along side it that take into account the direction the next point will be plotted.
Thank you for your time and if I can provide anything further, let me know and I'll do my best to add it.
This answer does not (as far as I see) favor one of the methods you mentioned in your question, but is what I would do in this situation.
I would calculate the normalized normal (or binormal) of the curve. Let's say I take the normalized normal and have it as a function of t (N(t)). With this I would write a helper function to calculate the offset point P:
P(t, o) = B(t) + o * N(t)
Where o means the signed offset of the curve in normal direction.
Given this function one would simply calculate the points to the left and right of the curve by:
Points = [P(t, -w), P(t, w), P(t + s, -w), P(t + s, w)]
Where w is the width of the curve you want to achieve.
Then connect these points via two triangles.
For use in a triangle strip this would mean the indices:
0 1 2 3
Edit
To do some work with the curve one would generally calculate the Frenet frame.
This is a set of 3 vectors (Tangent, Normal, Binormal) that gives the orientation in a curve at a given parameter value (t).
The Frenet frame is given by:
unit tangent = B'(t) / || B'(t) ||
unit binormal = (B'(t) x B''(t)) / || B'(t) x B''(t) ||
unit normal = unit binormal x unit tangent
In this example x denotes the cross product of two vectors and || v || means the length (or norm) of the enclosed vector v.
As you can see you need the first (B'(t)) and the second (B''(t)) derivative of the curve.
In my game I draw the line using two points. I want to calculate the points between the line. Please give any formula for finding points between the two points.
If your two points are A and B then
r(t) = A + t(B - A) , where t is greater than or equal to 0 and less than or equal to 1
is the equation of the line joining A to B, and consequently allows you to find any point lying on the line between them (by using an appropriate value for t).
Does this help??
(This is a follow-up from this previous question).
I was able to successfully use OpenCV / Hough transforms to detect lines in pictures (scanned text); at first it would detect many many lines (at least one line per line of text), but by adjusting the 'threshold' parameter via trial-and-error, it now only detects "real" lines.
(The 'threshold' parameter is dependant on image size, which is a bit of a problem if one has to deal with images of different resolutions, but that's another story).
My problem is that the Hough transform sometimes detects two lines where there is only one; those two lines are very near one another and (apparently) parallel.
=> How can I identify that two lines are almost parallel and very near one another? (so that I can keep only one).
If you use the standard or multiscale hough, you will end up with the rho and theta coordinates of the lines in polar coordinates. Rho is the distance to the origin, and theta is normally the angle between the detected line and the Y axis. Without looking into the details of the hough transform in opencv, this is a general rule in those coordinates: two lines will be almost parallel and very near one another when:
- their thetas are nearly identical AND their rhos are nearly identical
OR
- their thetas are near 180 degrees apart AND their rhos are near each other's negative
I hope that makes sense.
That's interesting about the theta being the angle between the line and the y-axis.
Generally, the rho and theta values are visualized as being the angle from the x-axis to the line perpendicular to the line in question. The rho is then the length of this perpendicular line. Thus, a theta = 90 and rho = 20 would mean a horizontal line 20 pixels up from the origin.
A nice image is shown on Hough Transform question