But now enters the color photography using color filters problem: We have a monochromatic camera, and we take three photos with it, with a red, a green and a blue filter, and we combine them and we get a color photograph. And violet colors show up just fine! How?
It is my understanding that this is because the frequency of violet is twice that of (low) red, and there's a resonance effect going on
When it receives violet, the blue cone is stimulated because its frequency is not far below violet, and the red one as well, since it's stimulated by the second harmonic
In this model, violet light mostly stimulates the blue cone, compared to the red and green cones. Note that the peak of the blue cone is in the violet range. I have seen some other models where the red cone stimulation actually increases in the violet range, but I don't know if they are accurate or not. Searching the internet reveals that there is a disagreement in this area.
I'm not sure if there should additionally be the requirement that no three floor touch points should be collinear.
This week's Riddler classic puzzle kinda disappointed me, it's a simple Hamiltonian path problem, which can also be solved with dynamic programming.
The newest Numberphile video made me think of this hypothesis: Three non-overlapping spheres (not necessarily of the same radius) are on a floor (ie. a plane). As long as the points where the spheres touch the floor are not collinear, there will exist another plane that also touches all three spheres (in such a way that all three spheres will be on the same side of this second plane). Can this be proven?
Language: cppfor (int i=1; i<99>= 0)
(To solve for minimizing the worst case, replace if(score<=389) with if(max<=6).) The result is:Language: cpp#include <stdio.h> #include <algorithm> using namespace std; int v[4][100]; int main() { for (int i=0; i<=99; i++) { v[0][i] = i; } v[1][0] = v[2][0] = v[3][0] = 0; for (int c1=2; c1<=99; c1++) { for (int i=1; i<=99; i++) { if(i<c1) v[1][i] = v[0][i]; else v[1][i] = min(v[0][i],v[1][i-c1]+1); } for (int c2=c1+1; c2<=99; c2++) { for (int i=1; i<=99; i++) { if(i<c2) v[2][i] = v[1][i]; else v[2][i] = min(v[1][i],v[2][i-c2]+1); } for (int c3=c2+1; c3<=99; c3++) { for (int i=1; i<=99; i++) { if(i<c3) v[3][i] = v[2][i]; else v[3][i] = min(v[2][i],v[3][i-c3]+1); } int score=0, max=0; for(int i=1; i<=99; i++) { score=score+v[3][i]; if(v[3][i]>max) max=v[3][i]; } if(score<=389) { printf("1 %d %d %d max:%d sum:%d\n",c1,c2,c3,max,score); } } } } return 0; }
1 5 18 25 max:6 sum:389
1 5 18 29 max:6 sum:389
At the transition point C, your ship's radial speed is 1 and, with the ship's velocity vector having a magnitude of b, we form right triangle EDC. However, if we call distance BC as r, then the distance of AC is br. Since points A,C,E are collinear as well as B,C,D, the two triangles ABC and EDC are similar, so ABC is a right triangle with legs of 1 and r and hypotenuse br. Note that as your ship travels along AC, the sub's circle grows along BC with the same y-coordinate, so point C is indeed the first point where they meet.
By using the Pythagorean Theorem and rearranging, we get our first identity
1/r = sqrt(b2-1). (*)
Now after your ship meets the edge of the sub's circle, your ship must always maintain a radial speed of 1 while having a speed of b. Thus the tangential speed is sqrt(b2-1), which is 1/r from the identity (*), which depends only on b. (Note that constant radial and tangential velocities at every point is a characteristic of a logarithmic spiral.) Since the tangential speed is distance times angular speed, we get
1/r = x*θ'(x),
where θ'(x) is the angular speed and x is the distance travelled by the sub (alternatively, the radius of the sub's circle). So
θ'(x) = 1/r * 1/x,
so then by integration,
θ2 - θ1 = 1/r * (ln(x2) - ln(x1)).
From the above image, we know that your ship must travel this spiral through the angle CBF, which has an angle of θ2 - θ1 = pi/2. Now the initial distance of the spiral is x1=r and the final distance is x2=1, so this simply gives
pi/2 = 1/r * (-ln(r)).
Then we simply solve for the variables r and b. WolframAlpha gives r≈0.474541, and plugging into (*) and solving gives b≈2.332533, which is our answer.
The following image (created from desmos.com) shows the path your ship must take at speed b≈2.332533 that allows you to reach port safely. The blue part is the straight line and the red part is the logarithmic spiral. Note that the transition between blue and red occurs at the point (0,r) ≈ (0,0.474541).
The answer given to this is that 1 is the result of the empty product. I suppose this makes sense when you consider that every natural number can be expressed as: n = 2p1⋅3p2⋅5p3⋅7p4⋅11p5⋅13p6⋅... as a combination of non-negative integer values of p1, p2, p3, etc. Nothing stops all of the powers from being zero (which is, thus, the unique prime factorization of 1).
This whole thing started (IIRC) with a viral post showing a picture of Portugal leaving a single player inside the field after scoring a goal, otherwise Spain could have resumed the game without waiting for all of them to end the celebration and come back. This rule is obviously fake, but it appeared everywhere... Then Panama and England tried to apply it.
This was a bit depressing to watch though... What were they thinking, LOL? https://www.youtube.com/watch?v=YDHzsO2iE9w https://www.reddit.com/r/soccer/comments/8y7nr8/englands_attempt_to_score_after_croatias_winning/
Find all positive integers a and b, with a > b, such that ab=ba.
You have a gate that requires a passcode to open. There are 10 buttons: the digits 0 to 9. You have forgotten the passcode, but you remember it has four digits. You have no choice but to try them all. Since there are \(10^4\) = 10,000 four-digit passcodes, you might think this would take you 40,000 button presses to guarantee an opened gate. However, this gate’s keypad never resets: The gate opens as soon as the last four buttons you’ve pressed are the correct code, so you can be more efficient. For example, you can try two different codes by pressing just five buttons: The combination “12345” tries both “1234” and “2345.” Of course, pressed for time, you want to press as few buttons as possible while still trying different codes and eventually opening the gate. So the question is: What’s the smallest number of buttons you need to press to make sure you open the gate — i.e., that you’ve tried every possible four-digit combination?
In the game of golf what is the fastest speed a ball can travel over the hole with still ending up in the hole? Assume the ball has been putted, with no elevation applied and is travelling directly over the middle of the hole
You have a gate that requires a passcode to open. There are 10 buttons: the digits 0 to 9. You have forgotten the passcode, but you remember it has four digits. You have no choice but to try them all. Since there are \(10^4\) = 10,000 four-digit passcodes, you might think this would take you 40,000 button presses to guarantee an opened gate. However, this gate’s keypad never resets: The gate opens as soon as the last four buttons you’ve pressed are the correct code, so you can be more efficient. For example, you can try two different codes by pressing just five buttons: The combination “12345” tries both “1234” and “2345.” Of course, pressed for time, you want to press as few buttons as possible while still trying different codes and eventually opening the gate. So the question is: What’s the smallest number of buttons you need to press to make sure you open the gate — i.e., that you’ve tried every possible four-digit combination?
That's true. Normally I wouldn't mind, but the same logic applies to TAS work. We have put a lot of hard work into our run, but our work wasn't respected by many people in the first place, and still hasn't been now. What's even more disappointing is that few people cares like I do, and many people are still blaming me over this and demanding me to shut up.
Consider a circular town of radius 1 mile. You want to build a 1 mile length road inside this town. What is the smallest d for which you can build a road so that every point inside the town is at most d miles away from the street?
Here, d is the distance we want every point in the town to be within the road, r=1-d is the radius of the inner circle, P is a point on the edge of the town (here oriented at the top of the circle), and PA and PB are tangent lines to the inner circle. Then the road is formed by the arc AB (away from P) joined to the line segments from A and B toward P up to a distance of d away from P. The length of PA and PB are each sqrt(1-r2), and so the length of the road is:
r(2π-2θ) + 2(sqrt(1-r2)-(1-r)) = 2r(π-θ+1) + 2sqrt(1-r2) - 2.
Since cos(θ)=r and r=1-d, this becomes:
2(1-d)(π-cos-1(1-d)) + 2sqrt(2d-d2) - 2d.
Using WolframAlpha, when the above is set to 1, you get d≈0.812.
I checked some of the star-like configurations that Aran Jaeger suggested, but I couldn't do any better than this road with them, even with other values for d or length of road.
Now if you use d=0.5, you get 2π/3 - 1 + sqrt(3) ≈ 2.826. So in the case of general d, this method works for any length up to about 45% of the town's outer boundary. Beyond that, the question becomes complicated since now you need the road to cover the center of the town. All I know that, in the limiting case where d→0, parallel lines spaced 2d apart ensure that each point in the town is within d of some line. In this system, a road of length x is in proportion to an area of 2dx, the ratio of road to area being 1/(2d). So, given a circular town of radius 1, the length of the road required for some d close to 0 approaches 1/(2d) * 2π = π/d.
The pentagon case (n=5) is discussed on this site. The solution for n=5 is shown below:
However, it turns out that for n≥6, the minimal Steiner tree is just the polygon on the points minus an edge. (Just because a Steiner tree satisfies all the properties above doesn't mean that it has minimum length!) More information is in this paper. The case n=6 has some local minima, but hexagon minus an edge is the minimum:
As for BrunoVisnadi's problem, assuming the road must be connected, I have so far a value of d≈0.812, where d=1-cos(x) and x is the root of cos(x)(pi-x+1)+sin(x)-1.5 in [0,pi/2]. The road I have is a "broken circle", having the shape of a hairband (I don't know how to describe it better). I'll explain how I got this later.