The above circle is a lake. Point G is a girl in a boat. Point B is a brute who wants to catch her. The brute can only run along the shoreline, while the girl can float with her boat all over the lake. While the girl goes through a distance of a radius, the brute may cover a semi-circumference. Once she reaches a point on the shore before the brute, she must be considered to have escaped (she runs more quickly than him on land). Find out how the girl escapes.

- from Dynamic Chess Strategy, Mihai Suba, Pergamon 1991.

As it happens, Mihai Suba was in town last weekend, third on tiebreak while sharing the top score in a fifteen-minute tournament in Huesca. Your correspondent was there, purely as a spectator, sharing as I do Botvinnik's dislike of rapidplay chess: partly for reasons of health and partly because my results in that discipline resemble those achieved by Fort William.

Anyway, I didn't know Mr Suba was going to be there and if I had, I might have brought along my copy of Dynamic Chess Strategy and asked him what the solution to the puzzle was, seeing as I've never been able to work it out. Never mind: I shall ask our readers instead. Perhaps they can also tell me what the Xs on the diagram are all about, which go entirely unmentioned in Suba's text. I never even noticed that before this week. Perhaps that goes some way to explaining why I never solved it.

## 19 comments:

Those Xs are a mystery all right. The Brute isn't blowing kisses at the Girl in the Boat, I take it?

seani

What if G head in the opposite direction to B, then which ever direction B takes, G arcs around in the opposite direction, still tending towards the far edge, but also gaining on the half-way position opposite the B -which B might be out of position for.

I believe the solution is that she makes sure she is looking at him directly whilst rowing, but I can't quite prove it.

What if G head in the opposite direction to B, then which ever direction B takes, G arcs around in the opposite direction, still tending towards the far edge, but also gaining on the half-way position opposite the B -which B might be out of position for.You couldn't do that as a graphic?

Tom's solution sounds interesting.

1. Is it necessary to assume that G's boatspeed when making the adjustments required by keeping an eye on B is unaffected?

2. Can anyone calculate the revised distance that G will travel (previously R) thanks to these adjustments, and whether this remains proportionate to that travelled by B (2 Pie R)

In short, I feel an attack of calculus coming on and the need for an accredited mathematician.

seani

Aren't the Xs the two protagonists themselves? I think so - from the diagram the labels are a bit confusing. G is mid way between the shore and the centre and B is at the shoreline.

If G begins by heading east directly away from B then at some point she will notice in which direction he's following and adjust her course accordingly. ie she heads east and as B begins to head clockwise or north at first to catch her she heads south east... and so on. At least that's what intuition says.

A colleague of mine solved it like this.

"First, go to the centre. Then, wherever the brute is at that time, row directly away from him for a short way, say a quarter of a radius. Finally, wherever the brute is at that time, row to the point on the shore directly opposite him (that is, picturing a diameter from the brute, through the centre, to the opposite side).

"The reason it works is, once you have rowed a quarter radius (anything strictly between zero and half a radius will do), away from the brute, he will be somewhere in the quadrant behind you, with the centre of the lake between you and that quadrant. So you will be closer than one radius to the point opposite him, and therefore you will get there before he does."

Sort of a triangulation strategy?

If this bears any resemblance to real life the girl will row her boat towards the brute as fast as possible, when she will at the very least have sex with him, and later probably marry him as well.

But not your real life, presumably?

Real life is alas shared by us all.

I scoured the internet for a solution to this puzzle. My search led me here to, among other things, a bizarre true story about a wild rabbit attacking former U.S. President Jimmy Carter.

In the linked puzzle, a rabbit (our girl in a boat) is in the center of a lake trying to evade the secret service agent (our brute). The secret service agent is a bit faster than our brute, but that shouldn't change much.

As far as I can tell, you'll need a math degree to fully understand the solution to this problem, which is described in excrutiating detail in the comments to the above post. For those lacking that degree, the girl (rabbit) can escape through a sort of spiral (see here), which actually makes a lot of sense once I thought about it, the idea perhaps being to move at all times in a direction opposite the brute.

Watch out for killer rabbits.

But the bigger question which no one has addressed is: what on earth has all this got to do with chess?!

Adam B.

You'll have to read Suba to find out, Adam. (I recommend it: it's a good book.)

There are many possible solutions but here is one ...

Assume (without loss of generality) that the circle has a radius = 1 and take the (x,y) coordinates of the centre as (0,0).

1. G starts due E and B can take either the S or the N route - let's assume S route. G continues E until B has reached a point exactly due SW of the centre. (B is now at coordinates (-0.71,-0.71) and G at coordinates (0.39,0) - the 0.39 is actually pi/8 i.e. half the distance B has travelled.)

2. G now heads due N. B has to change strategy because the quickest route to catch G is clockwise travel rather than anticlockwise. When B has returned to his original point (-1,0) G stops at the point (0.39,0.39).

3. G now changes direction again and heads for the point (1,0), i.e. the eastmost point on the pond. Pythagoras shows the distance to get there is 0.72 and B can't get there in time because the distance around half the circle is 1.57 (pi/2).

It's all about feinting, innit?

Aaargh! The general logic above was correct, but the co-ordinates I gave were wrong. G follows the path (0,0) to (0.25,0) to (0.25,0.25) to (1,0) and B still cannot catch her. Above I assumed ratio of velocities was 2 whereas it is pi, of course.

So without having read Suba, but adding "feinting" to "rabbits", it all is suggestive of how to offer a draw to a stronger chess player with halitosis.

Although I'm not sure which side needs the breath problem.

seani

I've played some guy with a similar problem, forget his name. Blackpool and Doncaster. Got a draw and a loss. Can't remember how I offered the draw. (I assume it was me, because about 90% of draw offers in my games are mine rather than the opponent's.)

Let the inital position of the brute be B,that of the girl G, the centre of the lake C,the point on the shore opposite B be A, a point on AC one third of a radius from C be D, the mid-point of AC be E, a point on the SE shore (at 5 o'clock) equidistant from A and C be F.

The girl sets off towards A. If when she reaches C the brute has not moved, she reaches A before him. If he sets off round the shore she tacks towards the point on the shore opposite him. If she notices that he has gone clockwise once she reaches D, he will have travelled one-sixth of a circumference to a point opposite F. If we take CF=6,by Pythagoras' theorem EF=root27 as CE=3, and so DF=root28, which is smaller than 6, so she beats the brute to the shore.

If the brute sets off before her,she tacks away from his position. It should be plain from observation that a loop from G to B not passing C will be shorter than two radii, and so she can make this journey before the brute can make a complete circumference.

I think Martin Gardner may discuss this somewhere.

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