A Cat and Mouse Game
The Numberphile video above talks through an investigation in which a mouse is swimming in a pond, with a hungry cat prowling around the edge. The cat can’t swim, but can run at a speed of 4m/s. The mouse can swim at a speed of 1m/s and can run faster than the cat on the land. The question is, can the mouse ever escape and run to safety?
Modelling the cat
The cat, C travels the circumference of the circle radius r at a speed of 4 m/s. The mouse is at point M at time t.
We can then calculate how long it would take the cat to make one rotation of the circle:
If C is vertically above O at time 0 then we can measure the horizontal and vertical position over time with the vector:
Can the mouse dash to safety?
When the Mouse starts at the centre and is directly beneath the Cat (CM perpendicular to OB), if the mouse heads due south it will have a position vector of:
The cat can reach the opposite side of the circle in pi x r /4 seconds and the mouse will reach the same point in r seconds. Therefore because:
The mouse will always be caught by the cat.
When will the dash strategy work?
So, how far does the mouse have to be away from the centre for this strategy to work? The Cat takes pi x r/4 seconds to reach the vertically opposite point of the circle, so the mouse must take less than pi x r/4 seconds. As the mouse travels at 1 m/s it must be at least pi x r/4 metres from the edge. This means it must be at least r – pi x r/4 from the centre. We can show this on the diagram:
So the dash strategy will work if the mouse is vertically opposite and greater than a distance of r – pi x r/4 from the centre. This gives the following lower bound for the mouse radius:
Next – is it possible for the mouse to get vertically opposite the cat? For this we want the angle MOC to be pi radians. This will be possible when the angular rotation of the mouse is faster than the cat.
If the cat is 4 times as fast as the cat then the angular rotation is the same when the circle the mouse makes is 4 times smaller. This is a circle from the centre with radius r/4 . If the circle is less than r/4 then the mouse will be able to complete rotations faster than the cat and so progressively increase MOC. This gives:
We can see this on a diagram:
We can see that this makes a narrow strip between 2 circles. Therefore if the mouse heads into this region then they will first be able to increase MOC until it is pi radians and then when it has done this, it will then be able to successfully dash to the vertically opposite side of the circle and escape.
Maximum escape times
I can now work out the upper bounds of escape times for this strategy. We can express the bound for the mouse radius as:
So, let’s take a = 0.1:
And let’s say the cat is already vertically above the mouse. We can now create a strategy that always leads to escape and find an upper bound for the time this will take.
First the mouse heads straight upwards to the edge of the mouse radius (and the cat doesn’t move). This will take (in seconds):
Next the mouse starts to rotate around the edge of the circle with radius:
As the mouse does this, the cat also rotates to try and keep up. The cat travels at:
radians per second.
The mouse will travel 2pi radians in (seconds):
Therefore this is
radian per second.
So, every second the mouse will gain an angular rotation of (radians):
The mouse can dash for the edge when the angular rotation gain is pi radians (i.e the mouse is on the opposite side of the circle defined by the mouse radius), so this gives:
Solving for t gives:
Lastly the mouse then dashes for safety at a point vertically opposite the cat. This is a distance of:
which the mouse covers in the same number of seconds.
Putting this all together, the total time to escape is:
This is the upper bound for the time taken to escape when the mouse starts in the centre because attempting to reach the mouse radius by first heading directly to the cat and then rotating when vertically underneath is the worst possible scenario in terms of time taken (it would be clearly quicker to set out away from the cat).
So, if the large circle has a radius of 1 and the mouse starts at the centre, if we head to a point:
from the centre then this strategy guarantees we will escape within (seconds)
What mouse radius should the mouse head to?
So for a radius of 1, when we head to a point on the mouse radius:
what is the optimum value of a to minimise the upper bound of time taken to escape? We have the upper bound for the time given by:
We can plot this on a graph:
Because this is an increasing function for the given domain then the upper bound for the minimum time will be when a approaches 0. So the mouse should aim for a point as close as possible to:
for all r. As the mouse radius approaches r/4 we see that the time approaches infinity – this is because when r = pi/4 the mouse and the cat circle at the same angular speed – and so the mouse is never able to get vertically opposite.