The mouse is thinking: The cat is not fast enough, so I can reach the sweet spot, circle until I'm opposite the cat, and then run straight to the edge. **cat misses by 1 degree** Yay
the mouse just needs to swim at the speed of light because if he does so the cat would go 4x the speed of light and you cant do that or the universe police will show up.
To go above light speed you need infinite power, but let’s suppose the rule of THE GAME YAMI NO GAME are superior. So if the cat goes above lightspeed the cat CANT go slower it always goes faster (cuz you need inf power to slowdown) and then cat ded
Seriously, there was a moment while watching this video when I realized, "This guy is a great teacher!" One of the biggest challenges with teaching is knowing how fast to talk. He's really quite wonderful at it.
2:25 - Initial theory before watching any further: -Starting from the center, the mouse has to cover distance _r_ going directly away from the cat. -The cat has to cover half the circumference of the circle, or (2πr)/2, or πr. -As the cat is going 4x as fast, and 4>π, the cat will get there before the mouse, and at that point the mouse will have a greater distance to go. Conclusion: The mouse is doomed. [resumes video] 12:25 - Took me exactly 10 minutes to realize I was wrong about the mouse being doomed.
As the mouse goes to that narrow sweet spot in an effort to increase the angle between himself and the cat, he could save a large measure of effort by heading toward the center until the angle was great enough, then swim toward the sweet spot again, arriving at just the right time to dash for the escape. This was a fantastic presentation and I learned quite a bit from it. Thank you!
To answer brady's question, the mouse can escape with any ratio less than pi + 1. the formula for the smaller circle is 1 - (pi/x) and the larger circle is just (1/x) (1/x) = 1 - (pi/x) (1 + pi)/x = 1 x = pi + 1
@@LordPelegorn my guess is that once the mouse is at the boundary of the sweet spot, the best move would be to swim directly to the edge. If the mouse were to swim in any direction besides directly perpendicular to the edge, it is 'wasting' time since the cat is travelling much faster than the mouse. You can see this as the cat having a larger angular velocity than the mouse whenever the mouse is outside the sweet spot, hence the cat is closing in in terms of the angle difference.
@@LordPelegorn No. As you get outside the 'circling' ring, the cat is fast enough to be able to start catching up. You need to gain the maximum possible advantage before you try and make a break for it. Spiralling lengthens the path, and the longer you're outside the ring the easier it is for the cat to catch up.
@@LordPelegorn the mouse can only move so fast, the distance covered over time would be the same away from the cat so the outcome would be the same if starting from the critical spot
Wow, mind blown. I kinda forgot about the circling which effectively decreases the distance to the shore. When the video started I immediately thought "When the cat is more than pi-times faster than the mouse the mouse will never make it.". I love this channel and I say that as someone who failed at math.
I loved the moment of realization when he hinted at how we can combine the two strategies to solve the problem. He presented this problem and solution very well. This is truly what maths is all about! P.S.: The number between 4.1 and 4.2 at 16:00 is in fact (pi + 1). It's a fun, easy exercise to work this out.
I was just about to say the same thing :)) The two sweet spot boundaries need to be equal in the worst scenario where the mouse can still escape, so it's 1/x=1-PI/x, so it becomes obvious that x=PI+1. That's fun to think about :)
It's already clear you have the dream job of making money from ridiculing mathematicians. I'll only buy a shirt if you announce officially, that 100% of profits go to Ben Sparks.
Mouse uses circling tactic within sweet spot and gets 180 degrees from the cat. Mouse dashes to the edge of the pond, with the cat close by, but not directly upon it. Mouse shakes himself dry on the edge of the ... oops!
He clearly said that the mouse is faster on land at the beginning of the video. EDIT: I got that it was irony. I was being sarcastic in my correction. a double woosh
It really looks like the mouse could make it if he kept heading to the -180 edge as the cat comes around from 90 degrees, but the mouse turns away. Obviously if it is not possible then it is not possible, but it just looked that way to me.
No actually the mouse runs , while still wet .. the cat pounces, slips on the water skids away, the mouse continues running.. the cat collects itself ... mouse is faster on land the the cat jumps / leaps that is , its as fast in air if not faster .. the mouse dodges , cat lands on water skids again... Its a new mickey mouse story with maths in it.
Hi. Thanks for the interesting topic. As I have calculated, the critical speed ratio (cat/mouse) is Pi + 1. To solve this, you need to consider the angular velocity instead of the velocity itself. Let's note cat and mouse angular velocities, radiuses, and velocities with w_c, r_c, v_c, and w_m, r_m, v_m. Knowing a little bit of physics (or geometry) you can write: w_c = v_c / r_c w_m = v_m / r_m Considering that both cat and mouse are trying their best (maximum speed they can achieve) and the only control options they have are the direction (which both can control) and mouse radius r_m (which only the mouse can control). By dividing the above equations we have: w_c/w_m = (v_c / v_m) * (r_m / r_c) So all variables on the right side are constant except r_m which the mouse can control. In addition, the left side of the equation shows that which of the cat or mouse are controlling the angular difference between them. We already know that the cat is trying to minimize the angular difference while the mouse is trying to maximize it. if w_c is greater than w_m (the left side is greater than 1) then the cat is controlling the angle while if it's not (the left side is less than 1) the mouse has the control on the angle and as mentioned before, r_m is the only thing affects that. So (as mentioned in the video) a specific r_m can be determined which demonstrates which of the cat or mouse has the control on the angle. By assuming the left side equal to 1 we can calculate a specific r_m which is the critical radius of angle control zones (and is noted it as R_m) : R_m = r_c * (v_m / v_c) So if r_m < R_m then the mouse can control the angle but if r_m > R_m then the cat controls it. The mouse can escape if it riches to r_m = r_c and the angular difference is not zero (the cat is not waiting there for the mouse). So the strategy for the mouse is to get to the r_m = R_m with the angular difference of 180 degrees and then swims straight to the edge r_m = r_c. So the mouse has to swim r_c - R_m in the second part of the strategy while the cat must run r_c * Pi around the water. Now if the mouse is fast enough then it can escape and if it's not then it can't. The critical case is that the cat and the mouse gets to their target simultaneously which means: (r_c - R_m) / v_m = r_c * Pi / v_c And by substituting the R_m we have: (r_c - r_c * (v_m / v_c)) / v_m = r_c * Pi / v_c multiplying both sides by v_m / r_c we have (1 - v_m / v_c) = Pi * v_m / v_c (v_m / v_c) * (Pi + 1) = 1 v_c / v_m = Pi + 1 So the critical speed ratio is Pi + 1 which causes the mouse to get to the edge just as the cat arrives and that is 4.1415... which is between the 4.1 and 4.2 that is mentioned in the video.
That was some difficult stuff. Could you not just generalise the inner and outer boundaries? In the video the inner boundary was pi/4 and 4 was the speed, so lets call it pi/v. Thw outer boundary in the video was 1-1/4, so now it is 1-1/v. Lets set these to equal. pi/v=1-1/v. Multiply by v and add 1 -> pi+1=v. The same result...
When doing the circle method, what if the mouse sometimes run on a secant of the circle instead? With this method, the cat should still run in the same direction as the circle method, but the mouse got a chance to reach another point of the circle a bit faster than the circle method (secant line is shorter than arc length).
@@kabirsingh4155 it would so long as the mouse knows when the cat reverses directions and passes the line between it and the mouse along which the center lies
Indeed that is only around 14 km/h (9 mi/h) depending on the breed some domestic cats can sprint at up to 48 km/h (30 mi/h) over short distances. Course that is for a cat in excellent physical condition I suspect that the relatively sedentary lifestyles of many domestic cats would make them quite a bit slower than they could be with better diet and exercise.
I think it's more likely he doesn't have an innate grasp of what 4 m/s looks like. People tend to think of a second as almost instantaneous when it's actually a fair amount of time. But yeah, my immediate reaction was also 'what are you talking about, that's really slow.'
@@xTyphoon51x Now that you mention it, it is true. And I think that is because we generally start counting with 1, and so we associate one with being instantaneous. If you tell someone "Start counting seconds when I tell you... NOW!" The moment you say "NOW!" that person will say one, but if the counting start there he should wait a second (hehe) and then say "one"
This puzzle is about a cat that's a mathematical zero-dimensional point, chasing a mouse that's another zero-dimensional point, around a perfectly circular lake, and *that's* the only thing you find unbelievable?
The mouse can actually escape if the cat is less than approximately 4.6033 times faster than the mouse. Hint: Once the mouse leaves the "safe zone" it doesn't have to dash radially.
@@falmircamion3534 Somehow, I'm not grasping something that I think I should grasp. If not dashing radially from the center, what other way would there be? I'm assuming we're talking about a two-dimensional circle here.
Some of you were saying dashing is the best option. Well outside of the safe circle the cat catches up with the mouse on rotation. So in the moment the mouse leaves the safe zone the best option is to dash
But if the sweetspot is gone, this does not mean that the mouse can't escape, but just that your strategy does not work anymore. Maybe there's a better strategy.
@@TheOneMaddin Rather than dash directly to the edge, dash tangent to the critical circle once you get opposite the cat. The maths gets more complicated but you can escape a faster cat than pi+1
@@digama0 Dash tactic can’t be improved. Outside the outer circumference, cat can ALWAYS decrease the angle. Circling may be able to be improved though.
I loved this puzzle. I instantly went for the low hanging fruit (move to the centre, then dash to the point furthest away from the cat), and found it couldn't be done, which forced me to get inventive. And when I did get it, it felt so satisfying. Will definitely try this one out on my friends.
Taking this a step further: Concidering exhaustion and therefore the necessity to reach the outside as fast as possible we should further ask: what is the fastest way to escape? An animal will most likely swim in more straight lines rather than in circles. so the goal is to reach the point where u can "dash" as fast as possible. As long as you are inside the small radius where u can "circle", the "swim to the opposite point" tactic should be the fastest. but rather than taking the opposite point on the edge, u take the opposide point on the circle that marks the "dash zone". I did not do the maths tho, but i beleave this to be the fastest line, as it suits the behaviour of animals pretty well :)
An improvement would be to spiral out from at or near the center (depending on initial position) to minimize the number of revolutions/laps to reach the angular zenith.
True. Get infinity close to center, take opposition exactly so cat stays stuck, swim out. That's pure geometrical math. My first thought. But with some random noise, your spiral strategy is best!! Beats his strategy by several revolutions. Got my minor in math ages ago, but it looks like that spiral consists of phi over .5pi, radially. Intuitively I just see that constant in there. What do you think?
Specifically, a semi circle path gets you to the angular zenith quite quickly if the cat always runs at full speed toward the mouse. I don't remember the proof of this though
The first phase would indeed be faster if the mouse didn't maintain a constant radius, but aim at the opposite of the cat - not on the shore, but on the circle where it has the same angular velocity as the cat. Because the mouse would take shortcuts closer to the center, it should approach the 180° position a lot faster, esp. if the circle is as thin as in the 4.1 case.
Well, although I doubt the speed around a circle is as fast to traverse as a straight line, and cats have very little stamina, they're actually really fast. The domestic cat has a top speed of 45 km/h But we're working in metres per second. So what is 1 m/s in km/h? 3600 seconds in an hour, 1000 metres in a kilometre, so 1 m/s is 3.6 km/h. That's basically the low end of walking speed. (with about 2 m/sec being the very high end of what could be considered 'walkin'g for a human) However, our cat, who has been clocked at 45 km/h is... 45 / 3.6 = 12.5 metres / second. But only in short bursts. Of course, a mouse on land can likely move faster than 1 m/sec too. However, I have seen mice swim, and it'd be generous to say they even hit 1 m/sec in water... XD So... Realistically, a real mouse vs a real cat... The mouse is screwed. (for the purposes of this puzzle anyway)
@@awesomegaming1991 Yes but, again, only in short bursts. Cats can't maintain that speed for very long. Humans are actually one of the best on the planet in terms of endurance running.
@@awesomegaming1991 Yes. But humans are actually pretty slow compared to animals. Humans are endurance optimised, not speed. Meanwhile the fastest land animal is...? That's right. A type of cat. It therefore shouldn't really be a surprise that cats in general are fast.
The boundary between the mouse escaping or not is when the cat is f~=4.603339 times faster than the mouse. This is as opposed to the circle/dash technique from the video that results in f=pi+1. Explanation: Using the circling tactic, the mouse can get to a radius of 1/f at any angle from the cat eventually. In polar coordinates, I'll label the mouse's radial speed towards safety as v_r. "dr/dt = v_r". Because the absolute speed of the mouse is "1", the rate at which the mouse can change angle in radians is obtained from the equation "1=sqrt((dr/dt)^2 + (r*dtheta_mouse/dt)^2)". This boils down to: "dtheta_mouse/dt = sqrt(1 - v_r^2)/r". However, the mouse has to fight against the angular speed of the cat, so we get: "dtheta/dt = sqrt(1 - v_r^2)/r - f". To maximize the distance the mouse travels vs. the rate the angle closes between the cat and the mouse, we can find the right critical point of differentiating dtheta/dr in terms of v_r. "dtheta/dr = (dtheta/dt)/(dr/dt) = (sqrt(1 - v_r^2)/r - f)/v_r = (sqrt(1-v_r^2) - fr)/v_r" I am going to skip the differentiation part, but ultimately after solving for the critical points of d(dtheta/dr)/dv_r; "v_r=sqrt(1 - 1/(fr)^2)" is obtained. I am not a geogebra expert, but I ran a quick simulation in python with the above parameters and it checks out. The radial angle of "pi" is covered between the cat and the mouse when the mouse reaches the edge of the circle when the cat is ~=4.603339 times faster than the mouse. From a more formal math standpoint, given that "dtheta/dr = (sqrt(1-v_r^2) - fr)/v_r", we can substitute for "v_r=sqrt(1 - 1/(fr)^2)" and get "dtheta/dr = (1 - (f*r)^2)/(f*r^2*sqrt(1 - 1/(f*r)^2))" If you paste the below in WolframAlpha, you should see that the following definite integral equates to "-pi" which means that the mouse can just barely squeak by: Integral from 1/4.603339 to 1 of (1 - (4.603339*r)^2)/(4.603339*r^2*sqrt(1 - 1/(4.603339*r)^2))*dr
The math with critical point of differential equation look really complex and I am not able to check it. Also, finally, the optimal path of the mouse is not given. I have put a comment with a different approach leading to the same critical f value.
@@owl94-58 Let r be the distance the mouse is away from the center of the pond. Let the R mean pod radious. Now x = r/R is the relative distance. x ≤ 1 and we consider only the regio where the mouse circles faster. So if the cat is N times faster then we are interested in the region x≥1/N. Now, the best tactics is to run at the angle y away from the line straight t/o the pond edge. By optimisation you find sin(y) = 1 / (x * N). Do you like it a little better, now? So you can see that the mouse starts almost circling but turns gradually towards the bank as it swims. Since N is finite (otherwise the mouse is always lost) sin(y) > 0 and the mouse never really dashes towards the bank if it wants to follow the optimal path
Sorry guys, I am fully aligned with Connor Harriman demonstration but I am still not able to understand the equations of this thread (maybe due to the lack of schematics)
@@owl94-58 Because we gave only the final answers. :-) There is quite some writing to do to explain the solution. But I'll try to explain the "best path" thing. Let's set a notation. The cat is N times faster. Mouse x=r/R is the distance from the center r divided by the pond radious. For x < 1/N the mouse is "angularily faster", as explained in the film. So the mouse can reach a point x=1/N with maximal angular distance from the cat (opposite side of the pond). The question is what to do when x > 1/N (in my previous note I wrongly wrote ≤, no it is corrected). The mouse has fixed goal: to go from x=1/N to x=1. But now the cat is angularily faster, no matter what path the mouse chooses! Fortunately, the cat is π radians (180 degrees) behind. We call the direction of the mouse y. It is a function of x, in general. Now, y=0 is the direction straight to the bank ("dash strategy"). Consider some x. The mouse has to go to x + dx somehow, because it must to finally reach x=1. The question is what direction to choose, so that the angular distance between cat and mouse dß decreses a little as possible. The difference in angular velocities equals N*v/R - v*sin(y)/r=(v/R)*(N - sin(y)/x) and the time mouse needs to go from x to x+dx is R*dx/(v*cos(y))=(R/v)*(dx/cos(y)). Angle change equals angular velocity times time of travel, so dß=(N - sin(y)/x) *(dx/cos(y))=(dx/x)*(N*x - sin(y))/cos(y). We look for a minimum of dß and y is what we can vary. You differentiate with respect to y and ask for which y the differential equals to 0. And the winner is: sin(y) = 1 / (x * N). For x=1/N it is just circling. The closer to the bank, the more the mouse direct itself towards the bank. In the optimal case you have dß=(dx/x)*(N*x - 1/(N*x))/sqrt(1 - 1/(N*x)^2), which integrate from x=1/N to x=1. If this integral is smaller than π, then the mouse manages to escape. You don't normally solve such problems in the comment - it is too confusing.
This puzzle also shows how every persons perspective is different and effects how fast they solve it. I instantly thought of swimming to the point farthest from the cat but figured it was to simple, so I then thought the mouse would have to cicle near the center and then swim to the nearest edge, but couldn't quite visualise whether or not it was possible.
also why is the mouse circulating to increase angle? why not optimise the method to increase the angle from the cat? i might be wrong in the sense that the circle is the most optimal way to do this.
also, before making the final dash, instead dashing straight towards the rim, only take the first step straight, and as soon as the cat breaks symmetry, aim for the part of the rim slightly away from the half the cat is in.
Loved this one. I was kind of into some kind of overly complicated combined version of the dash and circle tactics resulting in an outward spiral before the clue to do them separately. Then everything fell into place. Great problem and great vid.
I thought of a spiral at first too In fact, the solution they present isn’t even the optimal strategy (which stops working at a cat speed of pi + 1). You can increase this to about 4.6 or so by taking a spiral path once you are opposite the cat
Very interesting question; here are two potential alternatives when I first thought about this that I thought might be worth sharing: 1. swim directly away from the cat until you reach the center point, then take a non-circular curve that would outrun the cat. I believe it would be a curve with decreasing radius, but I'm not sure exactly what this curve is and how to calculate this curve; there might be calculus involved. I'm pretty sure this could work when the cat speed is larger than π times the speed of the mouse but there should be a max, and if it's above 4 than this could work. building on this concept, the method in the video might be improved: if the mouse took an optimal spiral inside that 0.25m radius circle it could've gotten out more quickly. 2. this would be an exploit of the coding but is still interesting. swim away from the cat until you reach the center point, then take a zigzag pattern. might be a relatively complex wave. project a line through the cat and the center point, this will be your centerline. make sure you don't go too far before you could turn to the other direction and cross the center line so the cat will turn to the other direction. rinse and repeat. again depends on the speed of the cat, this could go three ways: 1. keep zigzagging until you get out. if you're not fast enough, 2. zigzag until you reach the point that you're no longer able to outrun the center line (further away from the center point you are, the center line moves away from you more quickly), take a straight line to the closest point on the circle. 2.5. if a straight line doesn't work, maybe a spiral mentioned in the first method could help. there will still be a max cat speed for this method to work, but I suspect if this method works, the max cat speed might be higher. in terms of the pattern to take, I believe there is more than one possibility. I think depending on how far away you are from the center, we'll need to figure out how far away you can go from the centerline while also trying to reach the circle, but at the same time consider that the centerline is moving away from you. project the centerline when you turn back and cross it. determine the point of no return based on the projected centerline, and once you reach that point immediately take the straight path perpendicular to the projected centerline and cross it, now the cat will be going the other direction. rinse and repeat. or this pattern could consist of no straight lines but only smooth curves and I think this pattern, if the method works, might be the fastest way out: these curves will likely be non-circular -- so the point of no return will be closer to the projected centerline than before, but on the way back to the centerline you will also be going towards the circle. However I did all this in my head and didn't have a piece of paper nearby to do some numbers, give it a try if you'd like to and let me know if these two methods are feasible or if I'm wrong, I'd love to find out.
A few people have popped in here with the optimal solution of 4.603... and a description of the path the mouse must take, but I wanted to provide a solution with justification for why this is the optimal path. First, when the mouse is infinitesimally far from the border of the lake, the border appears as a straight line. So solve the case of a cat on a line. Assume L is the distance from the mouse to the closest point on the border and x is the distance of the cat from that point. If the mouse moves toward the border at an angle theta from going directly at the closest point, t_m = d_m/v_m = L/(cos(theta)*v_m) and t_c = d_c/v_c = (x+L*tan(theta))/v_c. Now you are looking to maximize t_m - t_c. So differentiate with respect to theta and set equal to zero. d(t_m - t_c)/d(theta) = L*(1/(v_c*cos^2(theta)) - tan(theta)/(v_m*cos(theta))) = 0 You can work that out on your own or just trust me. But when you solve it you get v_m/v_c = sin(theta). That means that the optimal angle to approach the border of the lake is sin^-1(v_m/v_c). Now take the circling tactic from the video. You can't do any better than being exactly opposite the cat inside the circling radius. And once you are outside the cat always has a greater angular velocity that the mouse. This means that the cat will never be able to gain an advantage by changing direction because changing direction will just put them both back on the same diameter with the mouse now slightly further from the center. Now you need to get to the edge as fast as possible, and the shortest path between any two points is a line. So just find the point at which you can aim from the circling radius directly to the border which forms the angle sin^-1(v_m/v_c) with the radius at that point. If the circling radius is r_m and the full radius is r_c, then r_m/v_m = r_c/v_c. Rearrange to get r_m/r_c = v_m/v_c. And we showed that that is equal to sin(theta) earlier. Final step, draw a triangle with one leg from the center of the lake to the border, another leg from that border point tangent to the circleo on which the mouse is able to stay opposite the cat, and the third leg from the tangent point back to the center. The tangent is the path the mouse will follow. It forms an angle theta with the radius (this is the angle offset from the path perpendicular to the border as in the straight line case). And sin(theta) = opposite/hypotenuse = r_m/r_c, therefore this is the optimal angle. That is how you can logically find the best path. But you are going to have to solve the maximum speed of the cat analytically. Calculate the time it takes for the cat to go all the way around to that point and the mouse to get to that point and take the different. Set it to zero and solve. The equation is v_c/v_m = (3*pi/2 - sin^-1(v_m/v_c))/sqrt(1 - (v_m/v_c)^2). Pop that into your graphing calculator or WolframAlpha and you get 4.063338849. Work it out and definitely draw a picture and you should be able to see why that is the optimal path for the mouse to take.
Code Parade also has a really nice video on this, he also has a link to a playable game for this. (Sorry for all the people I told "Code Bullet" made this video, it was actually Code Parade)
There is a more optimal strategy. If you go the sweet spot before you start circling, you might have to run around for a long time before you get opposite the cat. With such a small angular speed advantage, the mouse might get tired! Instead, run in a small circle until you're opposite the cat, then spiral outward so that you are cancelling the cat's angular speed and staying opposite it, while using the remainder of your speed to move out.
For anybody wondering where the point of no escape for the mouse is, it’s whenever the cat is traveling at faster than π+1 m/s. Anything less than that, and there’s a sweet spot. Anything greater than that, and there’s no escape. Exactly at that and there is a single point-width sweet spot.
Faster solution: 1.) Head to the center. 2.) Go in the opposite direction as the cat. 3.) As soon as the cat moves, keep the angle with the cat as high as possible. 4.) As soon as the mouse can't keep that angle from falling, dash for the edge. This prevents the spinning in circles for several loops before beginning the dash.
Excellent! Though I do think with step 4 as you describe it, the mouse will end up going in infinite circles closer and closer to the circling boundary, and never actually escape. A better step 4 would be to dash out as soon as the mouse hits the dash boundary. I wonder if there is a faster strategy still, it's an interesting problem.
@@jurjenbos228 it does work, it's just a different description of the solution ( eg the mouse can only control the angle of thr cat inside the event horizon where it can rotate faster than the cat)
@@Nomen_Latinum the speed of the strategy depends on how quickly the mouse reaches 180 degrees, and how quickly it reaches the critical zone. Prioritizing either too highly will slow down the escape. Reminds me of the classic calculus example of a dog on the beach retrieving a stick from the water. It has to balance distance on land and in water to minimize time.
After crossing the "event horizon", the cat, assuming perfect behavior, will see that the distance between the mouse and the cat is lesser in the direction it is currently traveling. You can intuit this for yourself by drawing a zig-zag line inside of a circle starting from the center. If you then think of the cat's behavior starting at a point opposite from the line's direction of motion, you can predict the way the cat travels at each turn. Draw a line out from the current direction of motion (the first "zig") to the edge of the pond. Let us call the intersection of this line and the edge of the pond P. Then draw a chord from P to the current position of the cat. Have the cat travel 4x the distance of the "zig" on the side of the chorded circle with the lesser area. Then repeat with the other "zigs". At some point, you will notice that the chord drawn will continuously favor the cat traveling towards the mouse, instead of alternating. The point where this behavior changes is the edge of the "event horizon" shown in the video.
This "escaping the event horizon" and nature of a zig zag relies on discrete time. If you assume continuous time, the mouse can jump the line instantaneously before it or the cat moves a unit step. Edit: I see even at infinitesimal scale, it would come down to speeds dx and 4dx
The maximum speed the cat can travel for the mouse to be able to escape is 1+pi m/s. This is because the sweet spot was greater than (1-pi/4) meters from the boundary and less than (1/4) meters from the boundary. If we assume the mouse's speed to stay as 1 m/s, we can generalise the sweet spot to be an area so that the distance of the mouse in this area from the centre of the pond is x, and the cat's speed being y, to be: 1-(pi/y) < x < 1/y And therefore the largest y can be so that x>=0 (making a singular line the "sweet spot") 1-pi/y = 1/y 1 = pi/y + 1/y 1 = (pi+1)/y y = pi + 1 And therefore the fastest speed the cat can go so that the mouse can still escape is (1 + pi)m/s, which sure enough is ~4.14, between 4.1 and 4.2 .
Somebody else has probably made the observation that the mouse would ideally approach the limit of the circling technique once it reaches 180° distance from the cat, thereby minimizing the total distance to the edge.
@@Finnyke Just run a Google search , it'll probably be much more satisfying that way (since you can go down all the rabbit holes of the problems' history)
A lot of the problems (Hodge conjecture, cough) would be pretty difficult to describe usefully in a RUclips video. If you want an accessible introduction to the problems that cuts an acceptable amount of corners, try Keith Devlin's "The Millennium Problems".
You can do BETTER: after the circle step you should run tangentially (not radially) (epsilon after the cat chose its side you decide the direction of the tangent)
Agree, it would be interesting. Until they do, I highly recommend to read the geogebra wiki, and just fiddle around with it and try to learn it yourself - that's what I did/am doing anyway, and the trial and error was and is super fun. After spending a little time, I was finally able to make a program that draws the logistic map (see the Feigenbaum constant video), and one that draws the Mandelbrot set in color (veeeery slowly), with the desired degree of precision (within GeoGebra's limits). I didn't have any programming experience, but I felt that GeoGebra is pretty user friendly, even if limited in some ways.
In short, it works via modified dash, where you don't go for the closest point on the outline, but diagonally. The exact threshold is hard to compute, but one can see that there's always a small angle away from the cat where the "time to outline" for the mouse doesn't increase as fast as the "time to intercept" for the cat. Therefore, a straight outward dash strat is not optimal. Should the cat try to be "clever" and turn around, it can't win if the mouse reacts instantly. Just dash outward until it turns around again, or if it crosses the 180° line, start the same diagonal dash strat but mirrored.
Trying to understand Brady's question: Is circle+dash really the best strategy? Suppose the cat was some 5 times faster (and never gets tired), is there _nothing_ (not just circle+ dash) that the mouse can do to escape?
That was great. I've been looking for this puzzle for years. The first time I came across it was Martin Gardener in New Scientist (?? Scientific American??) as a puzzle for a gladiator to escape from a lion in a Roman arena. Perhaps that is where it started?
If the sweet spot disappears somewhere between 4.1 and 4.2 does that mean it dissapears at pi+1 aka 1/2 circumference plus radius? Edit: @ironicprayer appears to have hypothesized the same thing.
@@silentobserver3433 I don't think so because the mouse will follow a curved path instead of a straight one and so will take a longer time to get to an edge
@@msolec2000 I meant the boundary where if the mouse starts in the center the dash tactic will always save it, without it having to use to circling tactic first.
If wondering what the biggest possible speed to still find a "sweet spot" is, it's anything less than about 4.142, equal to pi + 1. The reason for this is the equations defining the "sweet area." For the speed 4, it was between 1/4 and 1 - pi / 4. Generalizing, the crossing point can be found with setting the equations equal. So, 1/x= 1 - pi / x. Multiply by x, you get 1 = x - pi. Thus we get, x= pi + 1. Any speed above that will result in no sweet spot, and any speed below that will always have a sweet zone.
I figured it out in about 1 minute on the first pause by skipping all the speed stuff and “extra” calculations, I felt it was easier that way. 1. At what radius can a circling mouse out swim the circling cat to stay on the opposite side - 1/4 the radius, so I made the radius 4 and put the mouse at 1 opposite the cat, who is 4 from the center of course. 2. So the mouse is 3 away from the edge multiplied by the 4 times longer it will take him to swim it straight away, so 12. 3. The cat has to run 1/2 a circle around to the mouse’ exit, so radius*pi, in this case 4*pi which equals about 12.56. 4. When the mouse reaches the edge the cat is .56 away, he’s free! This may bother some because it lacks labels and such, but I like the simplicity of it.
I used a strategy of changing the reference frame, where the cat's position is constant. Instead of having the cat go around the pond, I treated the pond like a disc that the mouse is running on, and the cat is able to spin the disk (like a round table where you spin food toward you. The speed at which the disk can spin is defined in the same way: the outer edge moves four times as fast as the mouse "swims." We can fix the cat now at a certain point. From here, we use very similar technique as the circling + dash, but we do them simlutaneously in a sense. When the mouse is inside r/4 from the center, it can "resist" spinning by swimming in the opposite direction of the disk's movement. However, the mouse has some speed "left over", because it is inside the r/4 region and does not need to use all its speed to resist spinning, or decreasing the angle from the cat. To calculate the left over speed we can use right triangles and the pythagorean theorem, since the dashing direction is perpendicular to the circling direction at any time. It can use the remaining speed to increase its distance from the cat. Eventually, although the left over speed for the mouse decreases as it gets closer to the edge of the r/4 region, it always passes the 1-pi/4 event horizon and dashes away. It's possible to show the the left over speed is always greater than a certain positive speed, so the time it takes the mouse to get to the dashing stage is always less than a finite time. Another fun question is what is the fastest strategy for the mouse to escape, given initial conditions. Starting from the center, I think using the left over speed is the optimal way, but there could be a better way to combine circling and dashing. Very fun puzzle, and I'm happy I saw it eventually!
I've been coming back to some math riddles I thought about a long time ago. This is one of them. I'm glad because now I solve them in much less time then I thought of them then. Interesting how this happens a lot. Happens in video games too, where stuff is much easier after you take a break from it.
Came across this again just now. I'd think that doing "opposite the cat" strategy but with a circle of .23 of the pond size will work a lot better. As it is, the mouse can only improve its angle once it is inside the "dash region" but "improving its angle" is way cheaper when closer to the center of the pond. If the mouse starts its dash exactly at the dash boundary, it will have zero margin. So to escape with the most margin, up until the angle boundary, it can still improve its angle.
I came up with a different solution. From the center - it begins to swim directly away from the cat - the cat must make a decision (randomly?) to start running clockwise or counterclockwise - or the mouse will escape. But the moment the cat decides on a direction - the mouse swims in on a vector that is generally outwards - but biassed slightly in the same clockwise/counter-clockwise direction as the cat is running until the cat realizes that it's running in the wrong direction and won't catch the mouse. By swimming in a direction that continually convinces the cat that it must switch direction, the mouse can get past the event horizon.
I played a flash game about this once when I was a little kid. Spent hours bruteforcing my way to victory. Now I am a videogame programmer and I still remember this game but never understood the math well enough to replicate it. With this I can make tha game that has been in the back of my mind for a lot of time. THANK YOU SO MUCH!
We can do better than this. The mouse can force the cat to travel an extra distance by spiraling outward rather than just traveling directly toward the edge. As long as the extra distance the cat has to travel (divided by how much faster it travels) is greater than the extra distance the mouse must travel by spiraling, the mouse is gaining ground, so to speak. Looking at the sine function, we can see that whenever the angle between the mouse and the cat is close enough to 180 degrees, traveling at some angle diagonally away from the cat will result in the greater distance the cat must travel outweighing the greater distance the mouse must travel. Assume angles are in radians. Let: v = velocity of cat d = distance of mouse from center of pond theta = angle between the mouse's direction of travel and the nearest point on the shore relative to the center of the pond phi = angle between cat and mouse, relative to the center of the pond The mouse is improving its position whenever: (v-sin(theta)/d)/phi < cos(theta)/(1-d) To actually find the optimal path, we need to optimize the path with respect to theta for following equations: Where: dd/dt = cos(theta(t)) dphi/dt = sin(theta(t))/d-v Find theta(t) such that phi is maximized when d(t)=1 If phi(t)>0 until d(t)=1, the mouse escapes. Solving this is left as an exercise for the reader. HINT: Maximize dphi/dd
I can't really follow your maths, but I can't help feeling that the mouse should start from the centre and spiral outward. I'd love to see an animation of that approach.
You're like the guy in my calculus classes who did the exams, in PEN, got perfect scores, and left after 15 minutes. Meanwhile, my paper looked like WWIII re erasures and mistakes. But OTOH, observing folks like you taught me not to pursue a math degree and avoid all that pain.
@@rogergeyer9851 Heh, my comment is a bunch of erasures and mistakes too after someone else somewhere in the comics pointed out that instead of traveling in some sort of spiral, the mouse should travel in a straight line! Shortest path to a given point and all. The principle of forcing the cat to go the long way around still applies.
Damn this was a perfect, long, intense setup for a super satisfying climax. Ben Sparks is a great storyteller :D EDIT: this sounded pretty sexual even though I didn't intend it that way xD
There’s a better strategy for the mouse. Circle around until you are opposite the cat, then go straight out a little bit. Look which way the cat is heading, then go directly perpendicular to your radius in the OTHER direction. You can outrun a cat going 4.6 times your speed.
I haven't finished the video yet, but if the cat always travels strictly towards the mouse, I'm going to guess that the best strategy is some kind of zig-zag path that has the cat just moving back and forth on the opposite side of the pond.
Once the mouse moves out beyond the “circling boundary” it can’t zig-zag to the left or right fast enough to get the cat to change direction. At that point the only escape is to dash
If you are able to dash away from the mouse safely, you should do so. Dashing is the fastest way to reach any point on the edge. Otherwise, you must increase the angle between you and the cat, and you can only do this below the sweet spot marked. Once you have reached an angle sufficient to escape from the cat, you dash again. So, you either dash directly to some point -- not necessarily to the closest point -- or do this circle/dash maneuver. There are no quicker routes. The only open question is whether all dash points are the closest point, or whether you can actually dash at a slight angle to avoid the circle/dash maneuver for some starting configurations.
@@hyreonk Surely it's quicker to just cross the center of the circle, and then alternate directions whilst moving outwards instead of performing the whole circling manouvre near the boundary. Perhaps I'm missing a detail there. It's probably quicker in some circumstances and lesser in others.
I figured out a pretty cool numbers pattern you guys could do a video on. Take any number you want, scramble the digits to create a new number, then subtract them. The answer always result in digits that sum to a multiple of 9. Ex: 351-153= 198. 1+9+8= 18. No matter how large the number or how you rearrange it’s digits this will always work
In the case of ships, both would be 'in the pond', and if the one that follows has a greater speed than the ones that escapes, it will catch up eventually... Also, 'cat' and 'mouse' are points, while ships have size, so while in theory there could be similarities, I believe they would be so theoretic as not to make any real difference
I think that if the cat was faster than ~4.2, then the mouse could try to run in a polygonal shape. It would still outpace the cat and as long as the vertex is outside far enough out to dash, the mouse should be able to time a situation where it could escape. but we'd also have to add some game theory stuff, since the cat might have a better strategy than blindly chasing the mouse. I'll have to think about this.
If the cat much faster (>4.5ish) the cat has a strategy of ‘attempt to stay on the same radius as the mouse’, and since the mouse cannot get off the same radius as the cat when outside the ‘circling region’, and the mouse at any point on the edge of the ‘circling region’ loses to the cat running around to the point of the circle on the same radius as the cat. Note also that the mouse may be able to do a tiny bit better than Pi + 1 via a mixture of moving towards the edge and moving to keep the angle between cat and mouse larger. If you want a more interesting problem, imagine if instead we have many cats moving at the same speed as the mouse. How many cats do we need such that the mouse can not escape?
It took me less than an hour to get this (with no help). You don't even need to wait so long to get the cat at the largest angle. You can just go to the very center at which point the cat is behind you just by going diametrically opposite. You then maintain this by going the necessary amount around the circle to account for the cat's movements and then use the remaining speed you have available (since within the circle you can always match the cat's angle with speed to spare) to progress to the edge of the circle of quarter radius. An interesting generalization of this problem would involve a mouse that could drown if it swims for too long. That would make traveling to the center not entirely equivalent to starting there. It should have been noted that the mouse is aware at all times where the center of the circle is and where the cat is. A blind mouse could never have a perfect winning strategy as a cat can always potentially happen to wait where the mouse ends it's strategy.
Catch a more in-depth interview with Ben on our Numberphile Podcast: ruclips.net/video/-tGni9ObJWk/видео.html
But if the mouse starts in the middle the distance to the edge will scale linearly whereas the cats distance will scale to the square?
@Kire hi
Fun fact: this was question 2 on the 1st ever BMO paper, 1965
wait, this guy is not a pretty, smart redhead???
Critical cat speed would be where 1/x = 1-(pi/x)
You can't run, you can't hide, but you can swim in circle between two event horizons and make a dash to escape.
This comment is perfect.
remember you only have one life
Perfection
party pooper here. technically, only the inner circle is the event horizon.
The mouse is thinking:
The cat is not fast enough, so I can reach the sweet spot, circle until I'm opposite the cat, and then run straight to the edge.
**cat misses by 1 degree**
Yay
the mouse just needs to swim at the speed of light because if he does so the cat would go 4x the speed of light and you cant do that or the universe police will show up.
It just needs to go above 0.25c to get the cat arrested
To go above light speed you need infinite power, but let’s suppose the rule of THE GAME YAMI NO GAME are superior.
So if the cat goes above lightspeed the cat CANT go slower it always goes faster (cuz you need inf power to slowdown) and then cat ded
@@SkKedDy Shrödinger: Are you sure about that?
actually the wouldn't mouse need to swim at the speed of light
only more than a quarter of it
but then they both go to jail
Wow Tom and Jerry is a lot more complex than I remember
ye
Do it on the complex plane
Wow, your right
🤣🤣🤣🤣
@@shambosaha9727 r8
"That would be a tense moment for the mouse"
Yeah i'd also fear death if a gigantic chestnut-shell balloon hybrid came storming toward me
Wallonice Mapper stop commenting
@@alexcerullo3143 bruh he commented twice is that to much?
The mouse has a better character arc than most hollywood action heroes
Clever
Well it’s a circle
@@jayandmatt1517 its better than a line segment
The TVA*
false.
Honestly this guy, as a Numberphile regular, is highly underrated. One of the best if not the best.
Maybe 2nd, but no one can beat cliff stoll
hey vsauce michael here
I'd say he's my favorite right after Matt Parker
ok your name in relation to the comic troubles me
Seriously, there was a moment while watching this video when I realized, "This guy is a great teacher!"
One of the biggest challenges with teaching is knowing how fast to talk. He's really quite wonderful at it.
2:25 - Initial theory before watching any further:
-Starting from the center, the mouse has to cover distance _r_ going directly away from the cat.
-The cat has to cover half the circumference of the circle, or (2πr)/2, or πr.
-As the cat is going 4x as fast, and 4>π, the cat will get there before the mouse, and at that point the mouse will have a greater distance to go.
Conclusion: The mouse is doomed.
[resumes video]
12:25 - Took me exactly 10 minutes to realize I was wrong about the mouse being doomed.
9:36 - "the mouse can't dash from the center. So the obvious question is, where can it dash from?"
I was expecting the ad for dashlane to appear
i see, you are a man who watch too much youtube's videos as well
Brilliant.
@@karolis1 the thread just keeps on giving
Been watching lots of Linus Tech Tips for sure
At first I thought the "press paws" meant my hands. No, it was a pun, and it took me nearly half the video to realise that.
It took me your comment to realise it
my brain is really tiny and I still don't get it
Edit: ok my brain grew by .000002mm and now I realized it's press pause
@@cq.cumber_offishial congrats, you've reached the pun event horizon
paws are legs too
That's what happens when a furry watches a math video
As the mouse goes to that narrow sweet spot in an effort to increase the angle between himself and the cat, he could save a large measure of effort by heading toward the center until the angle was great enough, then swim toward the sweet spot again, arriving at just the right time to dash for the escape.
This was a fantastic presentation and I learned quite a bit from it. Thank you!
To answer brady's question, the mouse can escape with any ratio less than pi + 1.
the formula for the smaller circle is 1 - (pi/x) and the larger circle is just (1/x)
(1/x) = 1 - (pi/x)
(1 + pi)/x = 1
x = pi + 1
That is assuming the method shown is in fact the "best" way for the mouse to try to escape.
Wouldn't some sort of spiral be even more efficient?
@@LordPelegorn my guess is that once the mouse is at the boundary of the sweet spot, the best move would be to swim directly to the edge. If the mouse were to swim in any direction besides directly perpendicular to the edge, it is 'wasting' time since the cat is travelling much faster than the mouse. You can see this as the cat having a larger angular velocity than the mouse whenever the mouse is outside the sweet spot, hence the cat is closing in in terms of the angle difference.
@@LordPelegorn No. As you get outside the 'circling' ring, the cat is fast enough to be able to start catching up. You need to gain the maximum possible advantage before you try and make a break for it. Spiralling lengthens the path, and the longer you're outside the ring the easier it is for the cat to catch up.
@@LordPelegorn the mouse can only move so fast, the distance covered over time would be the same away from the cat so the outcome would be the same if starting from the critical spot
..but could there be a better strategy that lets the mouse escape even at x=pi+1 ?
I know this worst mouse ever is just a distraction to stop us using Parker Squares! You can’t fool me!
No it isnt
Oh yes it is
@@DomenBremecXCVI oh no it isnt. The parker square will never die
Damn... That's not how pantomimes work...
@@DomenBremecXCVI sorry.
Oh no it isn't!
Wow, mind blown. I kinda forgot about the circling which effectively decreases the distance to the shore. When the video started I immediately thought "When the cat is more than pi-times faster than the mouse the mouse will never make it.".
I love this channel and I say that as someone who failed at math.
0:15 - 0:40 how to design a t-shirt 😂😂😂 I think that was the fastest t-shirt design in the world lol
WORST. MOUSE. EVER.
I loved the moment of realization when he hinted at how we can combine the two strategies to solve the problem.
He presented this problem and solution very well. This is truly what maths is all about!
P.S.: The number between 4.1 and 4.2 at 16:00 is in fact (pi + 1). It's a fun, easy exercise to work this out.
I was just about to say the same thing :)) The two sweet spot boundaries need to be equal in the worst scenario where the mouse can still escape, so it's 1/x=1-PI/x, so it becomes obvious that x=PI+1. That's fun to think about :)
Try to beat Cat 4.2 (I did)
@@expioreris How did you do it?
Fact Sheet He did not. He is just trolling
@@angelmendez-rivera351 multiple people claimed to have made it possible to do with a cat speed of 4.6
T-Shirts and other merch...
Worst Mouse Ever: teespring.com/worst-mouse-ever-numberphile and Circling Tactic: teespring.com/circling-tactic
Please make sure you pay the original artist. :)
When I saw "NOW AVAILABLE" in the video, I thought you were joking. This is amazing lol
We already have the Parker Square, is this now the Sparks Mouse?
When sales drop off, rebrand it as a "cat and slug" T-shirt.
It's already clear you have the dream job of making money from ridiculing mathematicians. I'll only buy a shirt if you announce officially, that 100% of profits go to Ben Sparks.
Mouse uses circling tactic within sweet spot and gets 180 degrees from the cat.
Mouse dashes to the edge of the pond, with the cat close by, but not directly upon it.
Mouse shakes himself dry on the edge of the ... oops!
He clearly said that the mouse is faster on land at the beginning of the video.
EDIT: I got that it was irony. I was being sarcastic in my correction. a double woosh
Hi Leo. Yeah... true... but... actually, let's see if anybody else gets the point of what I wrote first before I explain it to you.
@@leefisher6366 Some people are just too serious to get irony XD don't be too hard on him.
It really looks like the mouse could make it if he kept heading to the -180 edge as the cat comes around from 90 degrees, but the mouse turns away. Obviously if it is not possible then it is not possible, but it just looked that way to me.
No actually the mouse runs , while still wet .. the cat pounces, slips on the water skids away, the mouse continues running.. the cat collects itself ... mouse is faster on land the the cat jumps / leaps that is , its as fast in air if not faster .. the mouse dodges , cat lands on water skids again... Its a new mickey mouse story with maths in it.
Hi. Thanks for the interesting topic. As I have calculated, the critical speed ratio (cat/mouse) is Pi + 1. To solve this, you need to consider the angular velocity instead of the velocity itself. Let's note cat and mouse angular velocities, radiuses, and velocities with w_c, r_c, v_c, and w_m, r_m, v_m. Knowing a little bit of physics (or geometry) you can write:
w_c = v_c / r_c
w_m = v_m / r_m
Considering that both cat and mouse are trying their best (maximum speed they can achieve) and the only control options they have are the direction (which both can control) and mouse radius r_m (which only the mouse can control). By dividing the above equations we have:
w_c/w_m = (v_c / v_m) * (r_m / r_c)
So all variables on the right side are constant except r_m which the mouse can control.
In addition, the left side of the equation shows that which of the cat or mouse are controlling the angular difference between them. We already know that the cat is trying to minimize the angular difference while the mouse is trying to maximize it. if w_c is greater than w_m (the left side is greater than 1) then the cat is controlling the angle while if it's not (the left side is less than 1) the mouse has the control on the angle and as mentioned before, r_m is the only thing affects that. So (as mentioned in the video) a specific r_m can be determined which demonstrates which of the cat or mouse has the control on the angle. By assuming the left side equal to 1 we can calculate a specific r_m which is the critical radius of angle control zones (and is noted it as R_m) :
R_m = r_c * (v_m / v_c)
So if r_m < R_m then the mouse can control the angle but if r_m > R_m then the cat controls it. The mouse can escape if it riches to r_m = r_c and the angular difference is not zero (the cat is not waiting there for the mouse). So the strategy for the mouse is to get to the r_m = R_m with the angular difference of 180 degrees and then swims straight to the edge r_m = r_c. So the mouse has to swim r_c - R_m in the second part of the strategy while the cat must run r_c * Pi around the water. Now if the mouse is fast enough then it can escape and if it's not then it can't. The critical case is that the cat and the mouse gets to their target simultaneously which means:
(r_c - R_m) / v_m = r_c * Pi / v_c
And by substituting the R_m we have:
(r_c - r_c * (v_m / v_c)) / v_m = r_c * Pi / v_c
multiplying both sides by v_m / r_c we have
(1 - v_m / v_c) = Pi * v_m / v_c
(v_m / v_c) * (Pi + 1) = 1
v_c / v_m = Pi + 1
So the critical speed ratio is Pi + 1 which causes the mouse to get to the edge just as the cat arrives and that is 4.1415... which is between the 4.1 and 4.2 that is mentioned in the video.
My calculations show that even slightly larger than pi+1 it is still possible to escape.
Solution in video is suboptimal;
correct answer is about 4.603.
On top of that, if the ratio is smaller than pi - 1, the mouse is doomed. Plug this equation into desmos for the range. |1-pi/x|
@@DavidSartor0Could you elaborate? Or show source?
That was some difficult stuff. Could you not just generalise the inner and outer boundaries? In the video the inner boundary was pi/4 and 4 was the speed, so lets call it pi/v. Thw outer boundary in the video was 1-1/4, so now it is 1-1/v. Lets set these to equal. pi/v=1-1/v. Multiply by v and add 1 -> pi+1=v. The same result...
When doing the circle method, what if the mouse sometimes run on a secant of the circle instead?
With this method, the cat should still run in the same direction as the circle method, but the mouse got a chance to reach another point of the circle a bit faster than the circle method (secant line is shorter than arc length).
Yes, I did a program with this nethod and it works too
Yes your answer is correct but it will not work in all cases
@@kabirsingh4155 it would so long as the mouse knows when the cat reverses directions and passes the line between it and the mouse along which the center lies
4m/s "That's a quick cat"
Has this guy ever seen an actual cat?
Indeed that is only around 14 km/h (9 mi/h) depending on the breed some domestic cats can sprint at up to 48 km/h (30 mi/h) over short distances. Course that is for a cat in excellent physical condition I suspect that the relatively sedentary lifestyles of many domestic cats would make them quite a bit slower than they could be with better diet and exercise.
I think it's more likely he doesn't have an innate grasp of what 4 m/s looks like. People tend to think of a second as almost instantaneous when it's actually a fair amount of time. But yeah, my immediate reaction was also 'what are you talking about, that's really slow.'
@@xTyphoon51x Now that you mention it, it is true.
And I think that is because we generally start counting with 1, and so we associate one with being instantaneous.
If you tell someone "Start counting seconds when I tell you... NOW!" The moment you say "NOW!" that person will say one, but if the counting start there he should wait a second (hehe) and then say "one"
@@seraphina985 its running in a circle with radius 1m which would make it a lot slower than a straight sprint
This puzzle is about a cat that's a mathematical zero-dimensional point, chasing a mouse that's another zero-dimensional point, around a perfectly circular lake, and *that's* the only thing you find unbelievable?
The mouse can actually escape if the cat is less than approximately 4.6033 times faster than the mouse.
Hint: Once the mouse leaves the "safe zone" it doesn't have to dash radially.
4.14159, not 4.6033... unless, I'm missing something here?
@@jeffo9396 As Maccollo told above, the mouse doesn't need to dash radially, and in fact it is far from being the best option.
@@falmircamion3534 Somehow, I'm not grasping something that I think I should grasp. If not dashing radially from the center, what other way would there be? I'm assuming we're talking about a two-dimensional circle here.
@@falmircamion3534 Actually, thanks, but nm. I see another posting explaining the reasons for it.
Some of you were saying dashing is the best option. Well outside of the safe circle the cat catches up with the mouse on rotation. So in the moment the mouse leaves the safe zone the best option is to dash
If you want more of this kind of stuff, check out the book called Chases and Escapes by Paul Nahin.
Code parade did this first
This question is very simple yet also very hard. Great explanation and i’m gonna say it’s one of the best video on this channel.
"Worst mouse ever"
Me: hold my pencil
Matt Parker: hold my square
Actually give that back I need it
If he holds your pencil? Then how would you draw?
i have 2 pencils
But if the sweetspot is gone, this does not mean that the mouse can't escape, but just that your strategy does not work anymore. Maybe there's a better strategy.
There is
Have seen a different solution on another channel before (less mathematical). Don't remember where. What's the better solution?
@@TheOneMaddin Rather than dash directly to the edge, dash tangent to the critical circle once you get opposite the cat. The maths gets more complicated but you can escape a faster cat than pi+1
@@digama0 Dash tactic can’t be improved. Outside the outer circumference, cat can ALWAYS decrease the angle. Circling may be able to be improved though.
I loved this puzzle. I instantly went for the low hanging fruit (move to the centre, then dash to the point furthest away from the cat), and found it couldn't be done, which forced me to get inventive. And when I did get it, it felt so satisfying. Will definitely try this one out on my friends.
Thinking about the same. How can you prove the optimality of this strategy?
One of the better Numerphile videos! The use of Geogebra is a plus too! Thanks!
Taking this a step further:
Concidering exhaustion and therefore the necessity to reach the outside as fast as possible we should further ask: what is the fastest way to escape?
An animal will most likely swim in more straight lines rather than in circles. so the goal is to reach the point where u can "dash" as fast as possible.
As long as you are inside the small radius where u can "circle", the "swim to the opposite point" tactic should be the fastest. but rather than taking the opposite point on the edge, u take the opposide point on the circle that marks the "dash zone".
I did not do the maths tho, but i beleave this to be the fastest line, as it suits the behaviour of animals pretty well :)
An improvement would be to spiral out from at or near the center (depending on initial position) to minimize the number of revolutions/laps to reach the angular zenith.
True. Get infinity close to center, take opposition exactly so cat stays stuck, swim out. That's pure geometrical math. My first thought.
But with some random noise, your spiral strategy is best!! Beats his strategy by several revolutions.
Got my minor in math ages ago, but it looks like that spiral consists of phi over .5pi, radially. Intuitively I just see that constant in there. What do you think?
Specifically, a semi circle path gets you to the angular zenith quite quickly if the cat always runs at full speed toward the mouse. I don't remember the proof of this though
The first phase would indeed be faster if the mouse didn't maintain a constant radius, but aim at the opposite of the cat - not on the shore, but on the circle where it has the same angular velocity as the cat. Because the mouse would take shortcuts closer to the center, it should approach the 180° position a lot faster, esp. if the circle is as thin as in the 4.1 case.
If you think a 4 m/s cat is quick, you haven't seen my cat when he thinks he's gonna get food :D
You're right. And OTOH, a mouse swimming 1 m/s is pretty darn quick. But hey, it's a puzzle, not an animal behavioral study. :)
Well, although I doubt the speed around a circle is as fast to traverse as a straight line, and cats have very little stamina, they're actually really fast.
The domestic cat has a top speed of 45 km/h
But we're working in metres per second.
So what is 1 m/s in km/h? 3600 seconds in an hour, 1000 metres in a kilometre, so 1 m/s is 3.6 km/h.
That's basically the low end of walking speed. (with about 2 m/sec being the very high end of what could be considered 'walkin'g for a human)
However, our cat, who has been clocked at 45 km/h
is... 45 / 3.6 = 12.5 metres / second.
But only in short bursts.
Of course, a mouse on land can likely move faster than 1 m/sec too.
However, I have seen mice swim, and it'd be generous to say they even hit 1 m/sec in water... XD
So... Realistically, a real mouse vs a real cat... The mouse is screwed. (for the purposes of this puzzle anyway)
Awesome Algodoo Yes. That is not surprising. Cats are genuinely faster than humans. Known fact.
@@awesomegaming1991 Yes but, again, only in short bursts. Cats can't maintain that speed for very long. Humans are actually one of the best on the planet in terms of endurance running.
@@awesomegaming1991 Yes. But humans are actually pretty slow compared to animals.
Humans are endurance optimised, not speed.
Meanwhile the fastest land animal is...?
That's right. A type of cat.
It therefore shouldn't really be a surprise that cats in general are fast.
Then an owl came by and ruined the day for both cat and mouse.
What if there was a owl-safe zone for the mouse near the cat?
And what if the cat was an scp called the friendly cat
Did it bore the two to death by pontificating pointless pieces of preposterous wisdom?
Stop uploading while I’m trying to do homework
Numberphile 😱
Actually that was what exactly happened to me
Jokes on you, I've got holidays 😎
my homework IS numberphile
Ha. I'm out of school =P
@@kodekiwi9203nearly out myself. just finished physics and maths and now im kinda sad about that
The boundary between the mouse escaping or not is when the cat is f~=4.603339 times faster than the mouse.
This is as opposed to the circle/dash technique from the video that results in f=pi+1.
Explanation:
Using the circling tactic, the mouse can get to a radius of 1/f at any angle from the cat eventually.
In polar coordinates, I'll label the mouse's radial speed towards safety as v_r. "dr/dt = v_r".
Because the absolute speed of the mouse is "1", the rate at which the mouse can change angle in radians is obtained from the equation "1=sqrt((dr/dt)^2 + (r*dtheta_mouse/dt)^2)".
This boils down to: "dtheta_mouse/dt = sqrt(1 - v_r^2)/r".
However, the mouse has to fight against the angular speed of the cat, so we get: "dtheta/dt = sqrt(1 - v_r^2)/r - f".
To maximize the distance the mouse travels vs. the rate the angle closes between the cat and the mouse, we can find the right critical point of differentiating dtheta/dr in terms of v_r.
"dtheta/dr = (dtheta/dt)/(dr/dt) = (sqrt(1 - v_r^2)/r - f)/v_r = (sqrt(1-v_r^2) - fr)/v_r"
I am going to skip the differentiation part, but ultimately after solving for the critical points of d(dtheta/dr)/dv_r; "v_r=sqrt(1 - 1/(fr)^2)" is obtained.
I am not a geogebra expert, but I ran a quick simulation in python with the above parameters and it checks out.
The radial angle of "pi" is covered between the cat and the mouse when the mouse reaches the edge of the circle when the cat is ~=4.603339 times faster than the mouse.
From a more formal math standpoint, given that "dtheta/dr = (sqrt(1-v_r^2) - fr)/v_r", we can substitute for "v_r=sqrt(1 - 1/(fr)^2)" and get "dtheta/dr = (1 - (f*r)^2)/(f*r^2*sqrt(1 - 1/(f*r)^2))"
If you paste the below in WolframAlpha, you should see that the following definite integral equates to "-pi" which means that the mouse can just barely squeak by:
Integral from 1/4.603339 to 1 of (1 - (4.603339*r)^2)/(4.603339*r^2*sqrt(1 - 1/(4.603339*r)^2))*dr
The math with critical point of differential equation look really complex and I am not able to check it. Also, finally, the optimal path of the mouse is not given. I have put a comment with a different approach leading to the same critical f value.
Actually you can integrate the equation by hand. You then get an equation for the critical speed u: pi = sqrt(u^2 - 1) - atan(sqrt(u^2 - 1)).
@@owl94-58 Let r be the distance the mouse is away from the center of the pond. Let the R mean pod radious. Now x = r/R is the relative distance. x ≤ 1 and we consider only the regio where the mouse circles faster. So if the cat is N times faster then we are interested in the region x≥1/N. Now, the best tactics is to run at the angle y away from the line straight t/o the pond edge. By optimisation you find sin(y) = 1 / (x * N). Do you like it a little better, now?
So you can see that the mouse starts almost circling but turns gradually towards the bank as it swims. Since N is finite (otherwise the mouse is always lost) sin(y) > 0 and the mouse never really dashes towards the bank if it wants to follow the optimal path
Sorry guys, I am fully aligned with Connor Harriman demonstration but I am still not able to understand the equations of this thread (maybe due to the lack of schematics)
@@owl94-58 Because we gave only the final answers. :-) There is quite some writing to do to explain the solution. But I'll try to explain the "best path" thing. Let's set a notation. The cat is N times faster. Mouse x=r/R is the distance from the center r divided by the pond radious. For x < 1/N the mouse is "angularily faster", as explained in the film. So the mouse can reach a point x=1/N with maximal angular distance from the cat (opposite side of the pond). The question is what to do when x > 1/N (in my previous note I wrongly wrote ≤, no it is corrected). The mouse has fixed goal: to go from x=1/N to x=1. But now the cat is angularily faster, no matter what path the mouse chooses! Fortunately, the cat is π radians (180 degrees) behind. We call the direction of the mouse y. It is a function of x, in general. Now, y=0 is the direction straight to the bank ("dash strategy"). Consider some x. The mouse has to go to x + dx somehow, because it must to finally reach x=1. The question is what direction to choose, so that the angular distance between cat and mouse dß decreses a little as possible. The difference in angular velocities equals N*v/R - v*sin(y)/r=(v/R)*(N - sin(y)/x) and the time mouse needs to go from x to x+dx is R*dx/(v*cos(y))=(R/v)*(dx/cos(y)). Angle change equals angular velocity times time of travel, so dß=(N - sin(y)/x) *(dx/cos(y))=(dx/x)*(N*x - sin(y))/cos(y). We look for a minimum of dß and y is what we can vary. You differentiate with respect to y and ask for which y the differential equals to 0. And the winner is: sin(y) = 1 / (x * N). For x=1/N it is just circling. The closer to the bank, the more the mouse direct itself towards the bank. In the optimal case you have dß=(dx/x)*(N*x - 1/(N*x))/sqrt(1 - 1/(N*x)^2), which integrate from x=1/N to x=1. If this integral is smaller than π, then the mouse manages to escape.
You don't normally solve such problems in the comment - it is too confusing.
This puzzle also shows how every persons perspective is different and effects how fast they solve it. I instantly thought of swimming to the point farthest from the cat but figured it was to simple, so I then thought the mouse would have to cicle near the center and then swim to the nearest edge, but couldn't quite visualise whether or not it was possible.
I'm really tempted to use this puzzle for my first "computer AI"
Did you ever do it? What was the result?
update?
A smarter mouse would zig-zag out to the sweet spot, maintaining the maximum angle for as long as possible, and thus escape much quicker.
also why is the mouse circulating to increase angle? why not optimise the method to increase the angle from the cat? i might be wrong in the sense that the circle is the most optimal way to do this.
also, before making the final dash, instead dashing straight towards the rim, only take the first step straight, and as soon as the cat breaks symmetry, aim for the part of the rim slightly away from the half the cat is in.
@@kaidatong1704 That would increase the distance the mouse would have to swim, therefore giving the cat advantage.
Spinning Leaf it also increases the distance the cat has to run
@@kaidatong1704 The cat moves faster than the mouse. Any increased distance outside of the threshold will allow the cat to catch up.
Loved this one. I was kind of into some kind of overly complicated combined version of the dash and circle tactics resulting in an outward spiral before the clue to do them separately. Then everything fell into place. Great problem and great vid.
I thought of a spiral at first too
In fact, the solution they present isn’t even the optimal strategy (which stops working at a cat speed of pi + 1). You can increase this to about 4.6 or so by taking a spiral path once you are opposite the cat
@@Muhahahahaz It's not a spiral, but a straight line.
A spiral seems more intuitive. I don't remember why a straight line is better.
Very interesting question; here are two potential alternatives when I first thought about this that I thought might be worth sharing:
1. swim directly away from the cat until you reach the center point, then take a non-circular curve that would outrun the cat. I believe it would be a curve with decreasing radius, but I'm not sure exactly what this curve is and how to calculate this curve; there might be calculus involved. I'm pretty sure this could work when the cat speed is larger than π times the speed of the mouse but there should be a max, and if it's above 4 than this could work. building on this concept, the method in the video might be improved: if the mouse took an optimal spiral inside that 0.25m radius circle it could've gotten out more quickly.
2. this would be an exploit of the coding but is still interesting. swim away from the cat until you reach the center point, then take a zigzag pattern. might be a relatively complex wave. project a line through the cat and the center point, this will be your centerline. make sure you don't go too far before you could turn to the other direction and cross the center line so the cat will turn to the other direction. rinse and repeat. again depends on the speed of the cat, this could go three ways: 1. keep zigzagging until you get out. if you're not fast enough, 2. zigzag until you reach the point that you're no longer able to outrun the center line (further away from the center point you are, the center line moves away from you more quickly), take a straight line to the closest point on the circle. 2.5. if a straight line doesn't work, maybe a spiral mentioned in the first method could help. there will still be a max cat speed for this method to work, but I suspect if this method works, the max cat speed might be higher. in terms of the pattern to take, I believe there is more than one possibility. I think depending on how far away you are from the center, we'll need to figure out how far away you can go from the centerline while also trying to reach the circle, but at the same time consider that the centerline is moving away from you. project the centerline when you turn back and cross it. determine the point of no return based on the projected centerline, and once you reach that point immediately take the straight path perpendicular to the projected centerline and cross it, now the cat will be going the other direction. rinse and repeat. or this pattern could consist of no straight lines but only smooth curves and I think this pattern, if the method works, might be the fastest way out: these curves will likely be non-circular -- so the point of no return will be closer to the projected centerline than before, but on the way back to the centerline you will also be going towards the circle.
However I did all this in my head and didn't have a piece of paper nearby to do some numbers, give it a try if you'd like to and let me know if these two methods are feasible or if I'm wrong, I'd love to find out.
Smart!
A few people have popped in here with the optimal solution of 4.603... and a description of the path the mouse must take, but I wanted to provide a solution with justification for why this is the optimal path.
First, when the mouse is infinitesimally far from the border of the lake, the border appears as a straight line. So solve the case of a cat on a line. Assume L is the distance from the mouse to the closest point on the border and x is the distance of the cat from that point. If the mouse moves toward the border at an angle theta from going directly at the closest point, t_m = d_m/v_m = L/(cos(theta)*v_m) and t_c = d_c/v_c = (x+L*tan(theta))/v_c.
Now you are looking to maximize t_m - t_c. So differentiate with respect to theta and set equal to zero.
d(t_m - t_c)/d(theta) = L*(1/(v_c*cos^2(theta)) - tan(theta)/(v_m*cos(theta))) = 0
You can work that out on your own or just trust me. But when you solve it you get v_m/v_c = sin(theta). That means that the optimal angle to approach the border of the lake is sin^-1(v_m/v_c).
Now take the circling tactic from the video. You can't do any better than being exactly opposite the cat inside the circling radius. And once you are outside the cat always has a greater angular velocity that the mouse. This means that the cat will never be able to gain an advantage by changing direction because changing direction will just put them both back on the same diameter with the mouse now slightly further from the center. Now you need to get to the edge as fast as possible, and the shortest path between any two points is a line. So just find the point at which you can aim from the circling radius directly to the border which forms the angle sin^-1(v_m/v_c) with the radius at that point.
If the circling radius is r_m and the full radius is r_c, then r_m/v_m = r_c/v_c. Rearrange to get r_m/r_c = v_m/v_c. And we showed that that is equal to sin(theta) earlier.
Final step, draw a triangle with one leg from the center of the lake to the border, another leg from that border point tangent to the circleo on which the mouse is able to stay opposite the cat, and the third leg from the tangent point back to the center. The tangent is the path the mouse will follow. It forms an angle theta with the radius (this is the angle offset from the path perpendicular to the border as in the straight line case). And sin(theta) = opposite/hypotenuse = r_m/r_c, therefore this is the optimal angle.
That is how you can logically find the best path. But you are going to have to solve the maximum speed of the cat analytically. Calculate the time it takes for the cat to go all the way around to that point and the mouse to get to that point and take the different. Set it to zero and solve. The equation is v_c/v_m = (3*pi/2 - sin^-1(v_m/v_c))/sqrt(1 - (v_m/v_c)^2). Pop that into your graphing calculator or WolframAlpha and you get 4.063338849.
Work it out and definitely draw a picture and you should be able to see why that is the optimal path for the mouse to take.
We need a folow-up video on this
Imagine mice are watching this video to learn the tactics...
Right now, somewhere, there´s a mouse in a pond holding a smartphone watching this video at x2 speed.
Code Parade also has a really nice video on this, he also has a link to a playable game for this.
(Sorry for all the people I told "Code Bullet" made this video, it was actually Code Parade)
Thaaaat's where I recognized the puzzle
Which video is it? I am subscribed to him but I don't remember this.
@@jmcbresilfrPardon me, i meant Code Parade
@@FinetalPies btw it's codeparade not code bullet oops
There is a more optimal strategy. If you go the sweet spot before you start circling, you might have to run around for a long time before you get opposite the cat. With such a small angular speed advantage, the mouse might get tired! Instead, run in a small circle until you're opposite the cat, then spiral outward so that you are cancelling the cat's angular speed and staying opposite it, while using the remainder of your speed to move out.
For anybody wondering where the point of no escape for the mouse is, it’s whenever the cat is traveling at faster than π+1 m/s. Anything less than that, and there’s a sweet spot. Anything greater than that, and there’s no escape. Exactly at that and there is a single point-width sweet spot.
Faster solution:
1.) Head to the center.
2.) Go in the opposite direction as the cat.
3.) As soon as the cat moves, keep the angle with the cat as high as possible.
4.) As soon as the mouse can't keep that angle from falling, dash for the edge.
This prevents the spinning in circles for several loops before beginning the dash.
The interesting point of this puzzle is that this strategy doesn't work!
Excellent! Though I do think with step 4 as you describe it, the mouse will end up going in infinite circles closer and closer to the circling boundary, and never actually escape. A better step 4 would be to dash out as soon as the mouse hits the dash boundary. I wonder if there is a faster strategy still, it's an interesting problem.
@@jurjenbos228 it does work, it's just a different description of the solution ( eg the mouse can only control the angle of thr cat inside the event horizon where it can rotate faster than the cat)
@@Nomen_Latinum the speed of the strategy depends on how quickly the mouse reaches 180 degrees, and how quickly it reaches the critical zone. Prioritizing either too highly will slow down the escape. Reminds me of the classic calculus example of a dog on the beach retrieving a stick from the water. It has to balance distance on land and in water to minimize time.
Codeparade made a video about this
Ah, I knew it! Yes in his case it was a goblin, now I remember
you can zigzag away "vibrate away" starting from the center would work. it will keep the cat balanced in same position
Woulnt that only work while in the "event horizon"? As soon as you leave it you can't zig-zag quick enough away from the cat to escape, I believe.
After crossing the "event horizon", the cat, assuming perfect behavior, will see that the distance between the mouse and the cat is lesser in the direction it is currently traveling.
You can intuit this for yourself by drawing a zig-zag line inside of a circle starting from the center. If you then think of the cat's behavior starting at a point opposite from the line's direction of motion, you can predict the way the cat travels at each turn. Draw a line out from the current direction of motion (the first "zig") to the edge of the pond. Let us call the intersection of this line and the edge of the pond P. Then draw a chord from P to the current position of the cat. Have the cat travel 4x the distance of the "zig" on the side of the chorded circle with the lesser area. Then repeat with the other "zigs". At some point, you will notice that the chord drawn will continuously favor the cat traveling towards the mouse, instead of alternating. The point where this behavior changes is the edge of the "event horizon" shown in the video.
This "escaping the event horizon" and nature of a zig zag relies on discrete time. If you assume continuous time, the mouse can jump the line instantaneously before it or the cat moves a unit step.
Edit: I see even at infinitesimal scale, it would come down to speeds dx and 4dx
The maximum speed the cat can travel for the mouse to be able to escape is 1+pi m/s. This is because the sweet spot was greater than (1-pi/4) meters from the boundary and less than (1/4) meters from the boundary. If we assume the mouse's speed to stay as 1 m/s, we can generalise the sweet spot to be an area so that the distance of the mouse in this area from the centre of the pond is x, and the cat's speed being y, to be:
1-(pi/y) < x < 1/y
And therefore the largest y can be so that x>=0 (making a singular line the "sweet spot")
1-pi/y = 1/y
1 = pi/y + 1/y
1 = (pi+1)/y
y = pi + 1
And therefore the fastest speed the cat can go so that the mouse can still escape is (1 + pi)m/s, which sure enough is ~4.14, between 4.1 and 4.2 .
Somebody else has probably made the observation that the mouse would ideally approach the limit of the circling technique once it reaches 180° distance from the cat, thereby minimizing the total distance to the edge.
"Worst mouse ever t-shirt"
Still better than my geometric problem solving
Give us more info about all 7 millennium prize problems.
They already did about the Riemann Hypothesis.
@@randomdude9135 Yeah, and he is asking for more. So do I
@@Finnyke Just run a Google search , it'll probably be much more satisfying that way (since you can go down all the rabbit holes of the problems' history)
@@rewrose2838 but kraft paper and poorly drawn geometry and equations is so much more entertaining :/
A lot of the problems (Hodge conjecture, cough) would be pretty difficult to describe usefully in a RUclips video. If you want an accessible introduction to the problems that cuts an acceptable amount of corners, try Keith Devlin's "The Millennium Problems".
"Assume the cat can go 4 meters a second - that's a quick cat" Sounds like mine when he's got a case of the midnight zoomies.
I think swimming 1 meter a second is really fast for a mouse.
You can do BETTER: after the circle step you should run tangentially (not radially) (epsilon after the cat chose its side you decide the direction of the tangent)
Really proud of myself as someone who is really bad at math that i figured it out in the first minute
Can you make a Geogebra video to learn how to program in it???
hear hear!
any programming would be cool
And if nothing else, make the source code available so we can play with them!
Agree, it would be interesting. Until they do, I highly recommend to read the geogebra wiki, and just fiddle around with it and try to learn it yourself - that's what I did/am doing anyway, and the trial and error was and is super fun. After spending a little time, I was finally able to make a program that draws the logistic map (see the Feigenbaum constant video), and one that draws the Mandelbrot set in color (veeeery slowly), with the desired degree of precision (within GeoGebra's limits). I didn't have any programming experience, but I felt that GeoGebra is pretty user friendly, even if limited in some ways.
Real question is where can we get the brown paper bitmap
That mouse is malnutritioned😂😂
Sciencifier can you swim faster than you can run on land?
@@blackburn3r no
No wonder it swims THAT slow
Even when the sweet spot is gone there still may be a more complex working tactic.
As others have pointed out: there is.
Some claim to have found tactics for the mouse to win against a cat that's 4.6 times as fast as the mouse.
In short, it works via modified dash, where you don't go for the closest point on the outline, but diagonally. The exact threshold is hard to compute, but one can see that there's always a small angle away from the cat where the "time to outline" for the mouse doesn't increase as fast as the "time to intercept" for the cat. Therefore, a straight outward dash strat is not optimal.
Should the cat try to be "clever" and turn around, it can't win if the mouse reacts instantly. Just dash outward until it turns around again, or if it crosses the 180° line, start the same diagonal dash strat but mirrored.
@@achtsekundenfurz7876 I really don’t think that works because I don’t know how you would get the cat to actually change direction in time.
Well done. Delivery and graphics with a dash of humor... perfect.
Sparks is the man. My numberphile favourite by quite a distance.
His Mandelbrot video is top drawer.
1:28 PRESS PAWS
im done
All cats I've known hate it when you press their paws.
@@Omnifarious0 I recently learned that cats have whiskers on their ankles. They are touch sensitive there.
Now you've told your tail.
Trying to understand Brady's question: Is circle+dash really the best strategy? Suppose the cat was some 5 times faster (and never gets tired), is there _nothing_ (not just circle+ dash) that the mouse can do to escape?
This is interesting, how does one even understand the "space" (for lack of better word) of strategies for the mouse?
5 times faster means 6 times as fast.
ok that worst mouse ever actually is a tshirt i would buy! Thank you for adding it!
It's so nice to see 3.14M subscribers. Kudos!!
Thank you - we marked the milestone: ruclips.net/video/__UlMppZZgs/видео.html
If the mouse swims at the speed of light, the cat will run faster than light, breaking the universe and dying. Easy
That was great. I've been looking for this puzzle for years. The first time I came across it was Martin Gardener in New Scientist (?? Scientific American??) as a puzzle for a gladiator to escape from a lion in a Roman arena. Perhaps that is where it started?
If the sweet spot disappears somewhere between 4.1 and 4.2 does that mean it dissapears at pi+1 aka 1/2 circumference plus radius?
Edit: @ironicprayer appears to have hypothesized the same thing.
Yes it does. Another commenter showed all the maths
@smoov22 There a better strategy. The value is 4.6 times faster (see my other comment).
For this solution, it is generally (r/v)(π+v)
So if pi m/s is the upper boundary for the dash tactic to always work, what is the upper boundary for the first two tactics you mentioned?
Exercise for the viewer.
Actually if the mouse starts from the center, then these tactics are both the same as dash, so it is also π m/s
@@silentobserver3433 I don't think so because the mouse will follow a curved path instead of a straight one and so will take a longer time to get to an edge
@@robo3007 It's actually 4.14159... ie, 1+pi.
@@msolec2000 I meant the boundary where if the mouse starts in the center the dash tactic will always save it, without it having to use to circling tactic first.
If wondering what the biggest possible speed to still find a "sweet spot" is, it's anything less than about 4.142, equal to pi + 1. The reason for this is the equations defining the "sweet area." For the speed 4, it was between 1/4 and 1 - pi / 4. Generalizing, the crossing point can be found with setting the equations equal. So, 1/x= 1 - pi / x. Multiply by x, you get 1 = x - pi. Thus we get, x= pi + 1. Any speed above that will result in no sweet spot, and any speed below that will always have a sweet zone.
I figured it out in about 1 minute on the first pause by skipping all the speed stuff and “extra” calculations, I felt it was easier that way.
1. At what radius can a circling mouse out swim the circling cat to stay on the opposite side - 1/4 the radius, so I made the radius 4 and put the mouse at 1 opposite the cat, who is 4 from the center of course.
2. So the mouse is 3 away from the edge multiplied by the 4 times longer it will take him to swim it straight away, so 12.
3. The cat has to run 1/2 a circle around to the mouse’ exit, so radius*pi, in this case 4*pi which equals about 12.56.
4. When the mouse reaches the edge the cat is .56 away, he’s free!
This may bother some because it lacks labels and such, but I like the simplicity of it.
Nice subtle use of play buttons as background.
Qwertyuoip 123 beat me to it
if the mouse can do maths deserves to escaspe
Numberphile: Hey can I copy your homework?
CodeParade: Yeah sure just change it a bit so the teacher doesn't notice.
I used a strategy of changing the reference frame, where the cat's position is constant. Instead of having the cat go around the pond, I treated the pond like a disc that the mouse is running on, and the cat is able to spin the disk (like a round table where you spin food toward you. The speed at which the disk can spin is defined in the same way: the outer edge moves four times as fast as the mouse "swims." We can fix the cat now at a certain point.
From here, we use very similar technique as the circling + dash, but we do them simlutaneously in a sense. When the mouse is inside r/4 from the center, it can "resist" spinning by swimming in the opposite direction of the disk's movement. However, the mouse has some speed "left over", because it is inside the r/4 region and does not need to use all its speed to resist spinning, or decreasing the angle from the cat. To calculate the left over speed we can use right triangles and the pythagorean theorem, since the dashing direction is perpendicular to the circling direction at any time. It can use the remaining speed to increase its distance from the cat. Eventually, although the left over speed for the mouse decreases as it gets closer to the edge of the r/4 region, it always passes the 1-pi/4 event horizon and dashes away. It's possible to show the the left over speed is always greater than a certain positive speed, so the time it takes the mouse to get to the dashing stage is always less than a finite time.
Another fun question is what is the fastest strategy for the mouse to escape, given initial conditions. Starting from the center, I think using the left over speed is the optimal way, but there could be a better way to combine circling and dashing.
Very fun puzzle, and I'm happy I saw it eventually!
I've been coming back to some math riddles I thought about a long time ago. This is one of them. I'm glad because now I solve them in much less time then I thought of them then. Interesting how this happens a lot. Happens in video games too, where stuff is much easier after you take a break from it.
This is what you would call a bait technique in the fighting game community, loved the explanation though.
I was gonna say "0:19 new merch?"
But then I saw the pinned comment
and then 0:39
A new game : Is there an actually mouse, which Is smart enough to use this to escape?
I think so. This is like catching a baseball. Doing it is easier than calculating it.
A mouse running away from my cat jumped in my pool once while my cat did circles. The mouse got away.
Cool video! Looks like a cat-and-mouse Spirograph drawing at the 3:40 mark.
Came across this again just now. I'd think that doing "opposite the cat" strategy but with a circle of .23 of the pond size will work a lot better. As it is, the mouse can only improve its angle once it is inside the "dash region" but "improving its angle" is way cheaper when closer to the center of the pond. If the mouse starts its dash exactly at the dash boundary, it will have zero margin. So to escape with the most margin, up until the angle boundary, it can still improve its angle.
The mouse can escape. Code Parade did an example of this problem using goblins with his video on it.
You mean Code Parade?
@@ariesunderground5936 Yeah that's it
Actually, the mouse could do better if it dashed in a diagonal before reaching 180deg instead of taking a 90deg turn when it reaches 180.
Right, but to be fair maximizing the distance to the cat (once on land) was not stated as a requirement.
@@TruthNerds Well maximizing that distance and finding the maximum speed ratio is the same.
Let's call it a Parker-Mouse
The kitty?
*P A R K E R P U S S Y* 😂
but it's a Sparks Mouse. Ben Sparks drew it.
@@narrativium Sparker mouse I guess
I came up with a different solution. From the center - it begins to swim directly away from the cat - the cat must make a decision (randomly?) to start running clockwise or counterclockwise - or the mouse will escape. But the moment the cat decides on a direction - the mouse swims in on a vector that is generally outwards - but biassed slightly in the same clockwise/counter-clockwise direction as the cat is running until the cat realizes that it's running in the wrong direction and won't catch the mouse. By swimming in a direction that continually convinces the cat that it must switch direction, the mouse can get past the event horizon.
I played a flash game about this once when I was a little kid. Spent hours bruteforcing my way to victory.
Now I am a videogame programmer and I still remember this game but never understood the math well enough to replicate it.
With this I can make tha game that has been in the back of my mind for a lot of time. THANK YOU SO MUCH!
We can do better than this.
The mouse can force the cat to travel an extra distance by spiraling outward rather than just traveling directly toward the edge. As long as the extra distance the cat has to travel (divided by how much faster it travels) is greater than the extra distance the mouse must travel by spiraling, the mouse is gaining ground, so to speak.
Looking at the sine function, we can see that whenever the angle between the mouse and the cat is close enough to 180 degrees, traveling at some angle diagonally away from the cat will result in the greater distance the cat must travel outweighing the greater distance the mouse must travel.
Assume angles are in radians.
Let:
v = velocity of cat
d = distance of mouse from center of pond
theta = angle between the mouse's direction of travel and the nearest point on the shore relative to the center of the pond
phi = angle between cat and mouse, relative to the center of the pond
The mouse is improving its position whenever:
(v-sin(theta)/d)/phi < cos(theta)/(1-d)
To actually find the optimal path, we need to optimize the path with respect to theta for following equations:
Where:
dd/dt = cos(theta(t))
dphi/dt = sin(theta(t))/d-v
Find theta(t) such that phi is maximized when d(t)=1
If phi(t)>0 until d(t)=1, the mouse escapes.
Solving this is left as an exercise for the reader.
HINT: Maximize dphi/dd
It's the perfect slope circled.
I can't really follow your maths, but I can't help feeling that the mouse should start from the centre and spiral outward. I'd love to see an animation of that approach.
You're like the guy in my calculus classes who did the exams, in PEN, got perfect scores, and left after 15 minutes. Meanwhile, my paper looked like WWIII re erasures and mistakes.
But OTOH, observing folks like you taught me not to pursue a math degree and avoid all that pain.
@@rogergeyer9851 Heh, my comment is a bunch of erasures and mistakes too after someone else somewhere in the comics pointed out that instead of traveling in some sort of spiral, the mouse should travel in a straight line! Shortest path to a given point and all. The principle of forcing the cat to go the long way around still applies.
Damn this was a perfect, long, intense setup for a super satisfying climax. Ben Sparks is a great storyteller :D
EDIT: this sounded pretty sexual even though I didn't intend it that way xD
Didn’t occur to me until you added the edit.
You fools. It's not between-two-event-horizons. It's inside the goldilock zone!
I dont know why i liked this comment, since i dont understand it, but im 100th
Very interesting puzzel and simulations. combining the 2 strategies was really a smart idea!
There’s a better strategy for the mouse. Circle around until you are opposite the cat, then go straight out a little bit. Look which way the cat is heading, then go directly perpendicular to your radius in the OTHER direction. You can outrun a cat going 4.6 times your speed.
I haven't finished the video yet, but if the cat always travels strictly towards the mouse, I'm going to guess that the best strategy is some kind of zig-zag path that has the cat just moving back and forth on the opposite side of the pond.
Once the mouse moves out beyond the “circling boundary” it can’t zig-zag to the left or right fast enough to get the cat to change direction. At that point the only escape is to dash
The next challenge would be to find the quickest escape route for any given point
If you are able to dash away from the mouse safely, you should do so. Dashing is the fastest way to reach any point on the edge. Otherwise, you must increase the angle between you and the cat, and you can only do this below the sweet spot marked. Once you have reached an angle sufficient to escape from the cat, you dash again.
So, you either dash directly to some point -- not necessarily to the closest point -- or do this circle/dash maneuver. There are no quicker routes. The only open question is whether all dash points are the closest point, or whether you can actually dash at a slight angle to avoid the circle/dash maneuver for some starting configurations.
@@hyreonk Surely it's quicker to just cross the center of the circle, and then alternate directions whilst moving outwards instead of performing the whole circling manouvre near the boundary. Perhaps I'm missing a detail there. It's probably quicker in some circumstances and lesser in others.
@@vertxxyz Yes! I was surprised that a zig-zag technique was not mentioned!
Huge fan of the new haircut (possibly I'm years late, but I like it).
This video was some vintage numberphile, loved it!
I figured out a pretty cool numbers pattern you guys could do a video on. Take any number you want, scramble the digits to create a new number, then subtract them. The answer always result in digits that sum to a multiple of 9.
Ex: 351-153= 198. 1+9+8= 18.
No matter how large the number or how you rearrange it’s digits this will always work
*1. 3 open buttons?* _That's bold._
2. Did you put your RUclips play button on the ground so they would be in the video?
3. Rule of threes.
He's flexing obviously
I wonder if this tactic issue would ever come up with ships.
In the case of ships, both would be 'in the pond', and if the one that follows has a greater speed than the ones that escapes, it will catch up eventually... Also, 'cat' and 'mouse' are points, while ships have size, so while in theory there could be similarities, I believe they would be so theoretic as not to make any real difference
I think that if the cat was faster than ~4.2, then the mouse could try to run in a polygonal shape.
It would still outpace the cat and as long as the vertex is outside far enough out to dash, the mouse should be able to time a situation where it could escape.
but we'd also have to add some game theory stuff, since the cat might have a better strategy than blindly chasing the mouse. I'll have to think about this.
If the cat much faster (>4.5ish) the cat has a strategy of ‘attempt to stay on the same radius as the mouse’, and since the mouse cannot get off the same radius as the cat when outside the ‘circling region’, and the mouse at any point on the edge of the ‘circling region’ loses to the cat running around to the point of the circle on the same radius as the cat. Note also that the mouse may be able to do a tiny bit better than Pi + 1 via a mixture of moving towards the edge and moving to keep the angle between cat and mouse larger. If you want a more interesting problem, imagine if instead we have many cats moving at the same speed as the mouse. How many cats do we need such that the mouse can not escape?
this is a really intuitive solution after 1 second of thinking of the problem
It took me less than an hour to get this (with no help).
You don't even need to wait so long to get the cat at the largest angle. You can just go to the very center at which point the cat is behind you just by going diametrically opposite. You then maintain this by going the necessary amount around the circle to account for the cat's movements and then use the remaining speed you have available (since within the circle you can always match the cat's angle with speed to spare) to progress to the edge of the circle of quarter radius.
An interesting generalization of this problem would involve a mouse that could drown if it swims for too long. That would make traveling to the center not entirely equivalent to starting there.
It should have been noted that the mouse is aware at all times where the center of the circle is and where the cat is. A blind mouse could never have a perfect winning strategy as a cat can always potentially happen to wait where the mouse ends it's strategy.