Stand upon the gravesite and you’ll see two great palms towering above all others on the island.  Count paces to the tallest of them and turn 90 degrees clockwise and count the same number of paces and mark the spot with a flag.  Return to the gravesite and count paces to the second-tallest of the trees, turn 90 degrees counter-clockwise and count off that number of paces, marking the spot with a second flag.  You’ll find the treasure at the mid-point between the two flags.

Fortunately, our pirates knew which island the map referred to.  Sadly, upon arriving at the island, the pirates discovered that all evidence of a gravesite had faded.  The captain was preparing to order his men to dig up the entire island to find the fabled treasure when one of the more geometrically inclined pirates walked over to a particular spot and began to dig.  The treasure was quickly unearthed on that very spot.

How did the pirate know where to dig?

Benford’s Law is really quite amazing, at least at first glance: for a wide variety of kinds of data, about 30% of the numbers will begin with a 1, 17% with a 2, on down to just 5% beginning with a 9. Can you spot the fake list of populations of European countries?

 Looking at these lists we have a clue as to when and how Benford’s Law works. Show Spoiler ▼

This is the solution to Morris: Trial/Trual/Whatever.  Please look there before reading the solution.

It turns out the right word is truel, first coined in 1954 by Martin Shubik.

Jeff suggested that they might all choose not to shoot at each other and it would go on forever.  This made me think about the logic.  I’m going to say that someone shoots at someone at some point and everyone knows this.  So everyone knows that if they shoot to miss then one of the others will shoot to hit at some point.

If anyone shoots to hit then they will shoot at the best shot.  If they hit they will be in two-man duel and would prefer to be against the worst shot possible. 

So François knows that no-one will shoot at him.  Xavier knows that one of the others will shoot at him.  

François has two options, he can shoot at Xavier or he can shoot to miss.

If he hits Xavier he will be in a two man dual with JC shooting first, he only has a one in five chance of surviving the first shot so his chance of survival is less than 1/5.

If François shoots to miss he knows that the other two will shoot at each other, if they choose to shoot to hit.  So at some point JC or Xavier will kill the other.  At this point François will be in a two man duel with himself shooting first.  This gives him at least a fifity/fifty chance.

So François shoots to miss.

Xavier and JC know this.  They know they are effectively in a two man duel with each other.  They therefore shoot at each other.

That sorts out the strategy, what about the probabilities?

Let’s consider JC first.  There is a fifty/fifty chance that Xavier or JC will shoot first.  JC’s only chance is to shoot first and hit.  The chance is 1/2 x 4/5 = 2/5.

So 2/5 of the time JC will face François with Francois shooting first.  

Let’s say that the chance that François wins is f.  He has a 1/2 chance of killing JC with his first shot.  There is a 1/2 x 1/5 = 1/10 chance that both miss with their first shot, we are back to the starting position so the probability of François surviving is now f again.

The chance of François surviving is f = 1/2 + 1/10 f.  This gives 9/10 f = 1/2 and f = 5/9.  The chance of François surviving is 5/9.  The chance of JC surviving is 4/9.

This gives JC a total survival chance of 2/5 x 4/9 = 8/45.

Now consider Xavier.  He has a 3/5 chance of facing François with François shooting first.

His chance of surviving is 1/2, the chance that François misses.

His total survival chance is 3/10.

Finally Francois’ chance of surviving is 2/5 x 5/9 + 3/5 x 1/2 = 2/9 + 3/10 = 47/90.

So the probabilities of being the last man standing are:  François 47/90; Xavier 3/10 and JC 8/45.

Amazingly the worst shot has the best chance of survival!

This leaves you in pretty good shape if you need to get from Cincinnati, OH to Destin, FL at 610 Miles, but what if you need to convert some distance that doesn’t happen to be a Fibonacci number?  Just build it up from parts!

100 miles is 89+8+3.  So in kilometers, that’s 144 + 13 + 5 or 162 kilometers.  (160.9344 by conversion…)

OK.  Here’s a puzzle, sort of.  I found this interesting set of numbers recently:

{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 53, 371, 5141, 99481 }

The series doesn’t continue.  That’s all of them.  What’s special about those numbers?

Mrs Smith has two children.  The eldest is a boy.  What is the chance that both are boys?

Mrs Jones has two children.  One is a boy.  What is the chance that both are boys?

Mrs Brown has two children.  One is a boy born on a Tuesday.  What is the chance that both are boys?

From Gary Foshee published on Ed Pegg’s http://www.mathpuzzle.com/ 

When two men get up ridiculously early to fire pistols at each other we call it a duel. Personally I prefer to lie in.

But what is the right term when three men skip breakfast to fire pistols at each other?

In a cold, misty field near the outskirts of Paris the sun peers over the horizon to see three men face each other with pistols.  Xavier is an expert shot, he never misses. Jean-Christophe is a very good shot, he will get you four times out of five. Francois only has a fifty/fifity chance of hitting his target.

They each take turns to fire their pistol.  

What is the best strategy for each of them and what odds would you give for the last man standing?

After the bodies had been cleared away I started to wander home only to hear the referee say ‘Maintenant, M. Galois et ami’

After winning their trip to Paris, the guests became elated and celebrated with the consumption of some adult beverages. Ever responsible, the host confiscated the keys to all cars to ensure that no one drove home drunk. Later on, when things started to calm down, party-goers started to request the return of their keys claiming to be sober enough for the drive home.

Having once been out-done by the guests, our host took another whack. He distributed all of the keys, but did so randomly. He then presented a challenge he felt sure they’d only be able to satisfy if they were indeed sober enough to drive. They were allowed to exchange keys, but only in rounds. During each round, each party-goer could either do nothing or pair up with another party-goer and exchange the sets of keys each was holding. (Each party-goer could be part of at most one pairing per round.) No one would be allowed to drive home unless everyone recovered their own keys.

The host wished to allow only a fixed number of rounds. To be fair, he wanted to be sure that it would indeed be possible to make the change. However, he also wanted to make it as difficult as possible for the party-goers. What is the minimum number of rounds must allow them to ensure that an exchange would be possible?

For clarity, all key recipients can discuss, share information such as who has the keys of whom, and agree upon a strategy. Also, careful readers will realize that there were 20 guests at the party originally. Sadly, it was a rather disorderly party and some guests did leave early, but many more appeared. Everyone present at the key ceremony had a key confiscated, and everyone with a key confiscated received a key for this challenge, but neither you nor the host knows just how many there are.

In Yoak: Batteries, and the Problem-of-the Week Jeff posed a great problem from Stan Wagon’s Problem of the Week.

You have eight batteries, four good and four dead. You need two good batteries to work the device; if either battery is dead then the device shows no sign of life. How many tests using two batteries do you need to make the device work?

In some similar sounding puzzles we might change the tests depending on the results we get as we go along. That doesn’t work here. We will stop if any test succeeds, we only carry on as long as the tests fail. We never get any information that we could use to choose our next test. That means we can choose the combinations we are going to use before we start our tests.

There are two ways of posing this problem. The one we have used is to make the device work. The other is to identify two working batteries which requires one less test.

Another problem is to find all four working batteries (which laciermaths neatly dealt with in the comments) and yet another is to find the most efficient approach which minimises the average number of tests.

The following seven combinations are guaranteed to make the device work.

(1,2),(1,3),(2,3),(4,5),(4,6),(5,6),(7,8)

Any six of these tests will identify two working batteries, for example our tests could be:

(1,2),(1,3),(2,3),(4,5),(4,6),(5,6)

If all of these tests fail then at most one battery from {1,2,3} can work and at most one battery from {4,5,6} can work. As we know there are four good batteries we can deduce that both 7 and 8 are working batteries.

So we have an answer to the problem, six tests are sufficient to identify two good batteries and seven tests are sufficient to work the device.

But hold on a minute! We haven’t yet shown that there isn’t an answer with fewer tests.

I came up with a proof but I won’t repeat it here because Jim Schmerl gave a much better one to POW. The better proof works for any number of batteries with any number being good.

It says: With n batteries and r good batteries then the best solution is to split the batteries into r-1 groups of roughly equal size and then test each pair within each group.

Let’s consider our seven tests again.

Here n is 8 and r is 4. We split the batteries into r-1 = 3 groups and test each pair in each group. At least one group has two good batteries so this will definitely give a solution.

These sorts of diagrams are called graphs. A graph is just a set of vertices connected by edges. A graph with v vertices and all possible edges is called K_v. Our solution consists of K₃, K₃ and K₂.

Now how do we show that this is minimal. Well we have some clever tricks.

The figure above is certainly a solution but it isn’t minimal.

7 has three edges and 8 has just one edge. We can reduce the number of edges by a procedure I call ‘Make 7 like 8′. The procedure involves:

1.    Remove all edges from 7.

2.    Connect 7 to the same vertices as 8.

3.    Connect 7 to 8.

That procedure gives:

We now have 11 edges instead of 12.

If we start with a solution then the ‘Make a like b’ procedure will give another solution. Why is this?

• If the four good batteries include both 7 and 8 then they will be tested as we test the pair 7 and 8.
  
• If the four good batteries do not include 7 then they are tested in the original solution and so will be tested in the new solution. We have only changed tests that include 7.
  
• If the four good batteries include 7, but not 8, then they will be tested. The equivalent group, replacing 7 with 8, is tested. For example consider {1,4,6,7}. The equivalent group is {1,4,6,8}. We know this is tested as it doesn’t involve 7 and so is tested in the original solution. But 7 connects to the same vertices as 8 so {1,4,6,7} is tested in the new solution.
  

We can repeat this trick as often as we like. You might like to play around with it, start with a solution with a ridiculous number of edges and see how far you can reduce it by repeating this trick.

The number of edges connected to a vertex, a, is called the degree of a which is written d(a). The procedure ‘Make a like b’ will reduce the number of edges if d(a) > d(b) + 1.

We can therefore make the following claim:

Claim 1: In an optimal solution the degree of any two vertices will differ by no more than one.

So far so good but we need to go a bit further. We can use the same trick to show the following:

Claim 2: In an optimal solution if a-b and b-c then a-c (where a-b means there is an edge between a and b).

This means the solution splits the batteries into groups where each pair within a group is tested.

Proof of Claim 2:

Suppose there is a chain a-b-c but there is no edge between a and c (so a-c is not true). We can always reduce the number of edges.

Case 1: If d(b) > d(a) we ‘Make b like a’. This reduces the number of edges by at least one, we remove the b-c edge.

Case 2: Similarly if d(b) > d(c) we ‘Make b like c’.

Case 3: Otherwise we have d(b) <= d(a) and d(b) <= d(c). We ‘Make a like b’ and then ‘Make c like b’. These two procedures will also reduce the number of edges by at least one.

To illustrate this lets go back to our example. We left it like this.

We have the chain 1-3-5 but not 1-5. d(3) > d(1) so we ‘Make 3 like 1′. That gives:

Now we have the chain 4-5-6 but not 4-6. As d(4) <= d(5) and d(6) <= d(5) we have the third case. First ‘Make 4 like 5′ and then ‘Make 6 like 5′.

What our two claims show is that an optimal solution will split the batteries into roughly equal groups (the degree of any two vertices can differ by no more than one), each pair within a group is tested.

It only remains to show that we should use three groups.

More than three groups is not a solution, we could have one good battery in each group.

The other possibilities are to have one group, K₈, which has 28 edges or to have two groups, K₄ and K₄, which has 12 edges.

So having three groups, with 7 edges, is optimal.

In the general solution we should have as many groups as possible which is r-1.

So this proves the general solution we have before: With n batteries and r good batteries then the best solution is to split the batteries into r-1 groups of roughly equal size and then test each pair within each group.

This is equivalent to Turan’s theorem.

For an optimal solution we need to consider the ‘complement graph’. This just means the graph with an edge precisely where our solution graph doesn’t have one. As our solution has 7 edges and there are 28 possible edges the complement graph has 21 edges. Such a graph is called a Turan graph.

The Turan graph can be got by splitting the vertices into r-1 groups and then connecting each pair which are not in the same group.

Since each group of four vertices in the solution graph contains at least one edge the same four vertices in the complement graph are missing an edge. In other words the complement graph does not contain K₄ , it is K₄-free.

Turan’s theorem says the graph with n vertices with the most edges which is K_r-free is the Turan graph, exactly the complement of our solution graph.