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.

If every guest finds their book then the whole group win a trip to Paris.

What is their best strategy?

This puzzle comes from a Peter Winkler book, Mathematical Mind-Benders.   It was called Names in Boxes and was about prisoners trying to escape the death penalty.

This is the guests best strategy:

Associate each box with one of the guests.  Each guest first opens their own box.  If they find their own book they can stop.  Otherwise they open the box associated with the owner of the book.  They keep doing this until they find their own book or have opened 10 boxes.

Incredibly this gives them more than a 33% chance of success.  Even more incredibly this strategy will always have a better than 30% chance of success regardless of the number of guests.

To see how this works lets imagine that Stan repeats the procedure forever.  At some point he will find his book.  Next he will open his own box again.  He will repeat the same loop endlessly.  

If this loop contains up to ten boxes then all of the guests whose boxes lie in the loop will find their book.

The procedure splits the boxes into separate loops.  If no loop is bigger than ten then all of the guests will succeed, otherwise most of them will fail.

It is still true each individual guest has a fifty-fifty chance, the procedure means that they tend to succeed or fail together.

So what is the chance of success?

We have come across these sorts of loops before in Follow Up: Loops and the Harmonic Series. In the comments we noted that on average there will be 1/r loops of length r. As we are interested in loops containing more than half the boxes there can only be one such loop, so we can say the chance of a loop of length r is 1/r.

First we calculate the number of loops of length r if we have n boxes, in the puzzle n =10.  There are n!/r!(n-r)! ways of choosing r boxes and there are r! ways of ordering them.   We could have chosen any box in the loop to be the first one so to avoid double counting we must divide by r.  This gives (n!/r!(n-r)!)r!/r = n!/r(n-r)!

The other n-r boxes could be arranged in (n-r)! ways.  In total there are n! arrangements of all n boxes.  So a random arrangment will have on average (n-r)!/n! occurrences of a particular loop.  

A random arrangement will have on average (n!/r(n-r)!)((n-r)!/n!) = 1/r loops of length r.

Since there can only be one loop larger than 10 this means the chance of success is 1 -1/11 -1/12 -… -1/20 which is about 0.331229…

As the number of guests increases this tends to 1 – integral[ N<x<2N ]( 1/x ) = 1 – ln 2N + ln N = 1 – ln 2 = 0.3068528…

However many guests there are they will always have a better than 30% chance of success!

You have eight batteries and know that four are good and four are dead, but don’t know which are which.  Your only method of testing them is to insert two into a device that will work if you’ve put in two good batteries and not otherwise.  How many such “tests” are required in order to be sure that you’ve located two good batteries?

As of this posting, the answer to this question is not yet on the POTW website, but if you come to this later, the spoiler may be there, so be careful to avoid spoilers if you want to work this through.

Janine is one of twenty guests at a Christmas party.  Each guest is given a book as a present.  Janines’s book is called ‘Living with Crazy Buttocks’.  She isn’t sure what to make of that.

The guests are invited to play a game.  Each book is put into an identical cardboard box.  The boxes can be opened and closed without leaving a mark.  The twenty boxes are piled up around the Christmas Tree.

The guests are told that they will each have the opportunity to open half of the boxes.  Their objective is to find their own book.  If they all succeed the group wins and they will win a trip to Paris.  If any one of them fails then the group fails but they will each get a Twinkie to keep for life.

The guests are taken to another room and then taken to the tree one at a time.  They cannot see what any other guest does at the tree.  They are not able to communicate once  the game starts.  The boxes are put back after each guest, as though they had never been there.

You would think that the chance of the group succeeding was 1/2^20 but they can do much better than that.

The group must come up with a strategy before the game starts.  What is the best strategy to get the group to Paris, and let Janine keep her ’Crazy Buttocks’?

These books are all real.  They will be helpful if you have ever had any of the following thoughts:

We all know the Nazis killed millions of innocent people but what were they like on ecological issues?

I would like to speak Italian but can’t be bothered to learn any Italian words, can you help?

Aubergines are very flushed, just how angry are they?

I think I’m dead, how can I tell for certain?

I am rich but dead.  How should I pimp my coffin?

I am worried about running into large, slow moving objects; can you suggest any strategies to avoid this?

Just how boring was 1587?

I live thousands of miles from Versailles.  Will I get a good view?

I am English, am I human?

My buttocks are insane.  

Combine the four number 1,3,4,and 6 with operators of addition, subtraction, multiplication and division (and parenthesis to indicate order of operation) to yield an expression equal to 24.

I assure you that you can take this in the most straight-forward manner possible.  You aren’t mean to smoosh them together to get “13″ out of 1 and 3.  You aren’t meant to use “1″ as a problem number or something of that sort.  An answer will look something like this:

(4-1)*3/6

except that is equal to 1.5 .  Your expression must equal 24.

I’m interested to hear if this is as difficult for others as it was for me.

   Three Points are taken at random on an infinite Plane.  Find the chance of their being the vertices of an obtuse-angled Triangle.

Note: An obtuse-angled triangle is one that has an angle measuring more than 90 degrees.