Game of Cat and Mouse - Numberphile
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- Опубликовано: 27 май 2019
- Ben Sparks discusses the math behind a curious problem....
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Наука
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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)
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
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
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.
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
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?
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!
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.
0:15 - 0:40 how to design a t-shirt 😂😂😂 I think that was the fastest t-shirt design in the world lol
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.
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 ?
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.
"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
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
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!
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.
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.
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.
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?
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
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
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.
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.
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.
One of the better Numerphile videos! The use of Geogebra is a plus too! Thanks!
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
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
Well done. Delivery and graphics with a dash of humor... perfect.
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.
Right now, somewhere, there´s a mouse in a pond holding a smartphone watching this video at x2 speed.
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?
I thoroughly enjoyed this video. Was engaged the whole time. Thanks for posting!!
Really well done! Nice work on the software too!
"Worst mouse ever t-shirt"
Still better than my geometric problem solving
I'm really tempted to use this puzzle for my first "computer AI"
Did you ever do it? What was the result?
update?
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.
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.
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.
Thanks for making original t-shirts and not just super generic ones. It's a small thing but really nice
You're welcome
This video was some vintage numberphile, loved it!
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.
"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.
if the mouse can do maths deserves to escaspe
This was really fun... keep making great videos
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.
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.
That mouse is malnutritioned😂😂
Sciencifier can you swim faster than you can run on land?
@@blackburn3r no
No wonder it swims THAT slow
Very interesting puzzel and simulations. combining the 2 strategies was really a smart idea!
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.
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
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".
In a weird way, the method in which this puzzle was solved reminded me a lot of how certain mechanics work in Kerbal Space program; especially when you gain or lose speed by flying by close enough to a planet/moon, exploiting the gravitational pull and letting yourself be flung off.
The way the cat moved here and how the movements of the mouse dictated how the cat moved in relation; that just gave a very... grativy-ish feel to me, for lack of a better term. So yep, even without actually being able to accelerate in this situation, I thought that the solution would look a bit like this; abuse the "gravity" of the cat to steer it where you want to and at the right moment, change direction and just go the way you need to.
In the end, it didn't play out the way I thought, but still, it was nice to see that I wasn't completely off :D
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
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.
Codeparade made a video about this
Ah, I knew it! Yes in his case it was a goblin, now I remember
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.
It's so nice to see 3.14M subscribers. Kudos!!
Thank you - we marked the milestone: ruclips.net/video/__UlMppZZgs/видео.html
ok that worst mouse ever actually is a tshirt i would buy! Thank you for adding it!
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
first saw a similar puzzle in terrence tao's book. This vid was done beautifully!
Cool video! Looks like a cat-and-mouse Spirograph drawing at the 3:40 mark.
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
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.
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!
this is a really intuitive solution after 1 second of thinking of the problem
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
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?
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 :)
It's really inspiring for me. Thank you
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
Huge fan of the new haircut (possibly I'm years late, but I like it).
That’s probably one of the most interesting math puzzles I ever seen
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.
Nice subtle use of play buttons as background.
Qwertyuoip 123 beat me to it
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.
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!
What's weird is, after all the analysis, once the actual model played out, my adrenaline was pumping like crazy.
My brain started having insights about black holes and orbital mechanics because of this video. Great stuff.
I was gonna say "0:19 new merch?"
But then I saw the pinned comment
and then 0:39
Less than a minute in - I think I have heard this concept before. But I forget how it ended. Will watch further to remember how it goes. Yeah it was a spiral... CodeParade made a video October 2017
This was a really fun one!
The power of math even in such funny things. Fantastic!
Numberphile: Hey can I copy your homework?
CodeParade: Yeah sure just change it a bit so the teacher doesn't notice.
Imagine mice are watching this video to learn the tactics...
I mean this! If this guy had courses to educate on maths, using these simulations, however dry the flavor (like cat and mouse) I would make quite a few adjustments to my lifestyle, just to afford the opportunity to learn form him. It's so intuitive after you've seen it once. Just once and the whole concept, the ratios, the arbitrary number chosen for the cat vs mouse speed, becomes very easy to reproduce. I loved this video. I know it's over a year old, but I'm going to like it, and wish I could love 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.
" that's event horizon " * he said it like it was something easier to understand *
the power of concepts. once you're familiar with it, it IS easier to understand... or at least quicker.
you have done the heavy work beforehand, once, and then can apply it almost instantly, as many times you want.
@@silkwesir1444 yeah but event horizon isn't common knowledge and it is objectively more complicated
@@snowman7514 well they just dropped it in there for those who do know it, they explained it anyway (so no knowledge about event horizons were required to understand anything said in the vid), so I see no harm done.
Actually, it may be the first step of some people learning about what an event horizon is. Just as a drive-by bonus ;)
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.
Sparks is the man. My numberphile favourite by quite a distance.
His Mandelbrot video is top drawer.
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 .
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.
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)
THANK YOU!
I was struggling months ago for a coding challenge, and now I can finally code a solution ❤
Another cool thing about this; if you pick the farthest point on the circle from the closest point he got while using the away tactic, the fraction of the circle the line between those two points make is 3.14/4 . Pi divided by the speed of the cat.
Ohhhh, nice mouse.
The mx speed I mean ofcourse
This is what you would call a bait technique in the fighting game community, loved the explanation though.
My first time watching any of your videos, will become a subscriber now!
Without reading all 857 comments (at this writing) I can think of a solution that would work but requires the mouse be able to hold its breath for a length of time. If it could dive so that cat couldn't see him, it could dive, then swim away from the cat (along the radius line) to a point where it could then safely make a dash for it. But this solution isn't as much fun to solve as yours :)
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.