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!

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.  

Oh by the way, would you like a cool Math Factor Poster? Click on this to download:

We do it all the time on Math Factor.

Chaim pointed me at the Macalester Problem of the Week.  This led to my making a minor contribution to a published paper.  I can’t claim it’s a world changing paper, or that my contribution amounted to much, but I did get my name in print!  You can read an extract here.  {Just above is a review of a book on symmetry, I’m not sure that is real, one of the authors is called Chaim Goodman-Strauss, clearly a made up name.}

It certainly is a fun paper.  Stan Wagon is a bit of a legend, as you’ll see from the picture.   I’m campaigning for all cycle paths to be built for square wheeled bicycles!

Can you solve some of these problems?

The paper was written by Stan Wagon and Robert Israel.  They kindly added my name when I was able to replace some computer proofs with ‘real proofs’ (that’s a whole other discussion!)

Take a sequence of letters containing no more than ten different characters, for example ABCABCDE.  Can we replace the letters consistantly with digits so that the resulting number is divisible by some target number n?

If n is 2 then it is quite simple, we just replace the last letter with an even digit.  For example can replace ABCABCDE with 12312346.  So we can substitute any sequence of letters to make a number divisible by 2.

Lets try with a target number n=3.  Can we find a method for replacing the letters of any sequence so that the resulting number is divisible by three?

We know that a number is divisible by three if the sum of digits is divisible by three.  ABCABCDE has two As, two Bs, two Cs, one D and an E.  We want 2A + 2B + 2C + D + E to be a multiple of three.  This is quite easy as there are four digits which are divisible by three: 0, 3, 6 and 9.  Use these for A, B and C and then chose D and E so that they add up to a muliple of three.  So we could use 03603612.  It’s okay for numbers to start with zero.

We can always replace any sequence of letters so that the resulting number is divisible by three.  Can you find a method for doing this?

We say that three is ‘attainable’.  

If it is true for a target number, n, then we say that n is attainable.

So far we have said that 2 and 3 are attainable.

What other target numbers are attainable?

Okay so these are my questions:

1.   Is five attainable?

2.   Can you show that three is attainable?

3.  If I tell you that nine is attainable can you show that 18 is attainable?

4.  What about four?

5.  What about six?

6.  What about seven?  I will give you a big clue here, the title of this post.

7.  What other numbers can you show are attainable or not attainable?

Do you think there is a difference between proofs by computer programs or proofs with just the human brain, possibly aided by a pencil and a few sheets of foolscap?

Have fun guys!

In the puzzle you have five different tasks.  On each day one of these tasks is given at random.  How long do you expect it to take to get all five tasks?

First consider a simple case.  Suppose some event has a probability, p, of happening on any one day.  Let’s say that E(p) is the expected number of days we have to wait for the event to happen.  For example if p=1 then the event is guaranteed to happen every day and so E(p)=1.

How can we calculate E(p)? 

Andy does an experiment.  He will wait for the event to happen and record how many days it took.  He will do this several times, for long enough to ensure that he gets an answer that is as accurate as he needs.  He will keep going for N days in total.  Afterwards he will take the average of all of his wait times to get an estimate for E(p).

The average he calculates is the total of all the wait times divided by the number of occurrences of the event.  But we can estimate both of these values and therefore estimate his value for E(p).  The total of all of the wait times is going to be about N.  Since the event has a probability of p of occurring on any particular day the number of occurrences will be about p times N.

So Andy’s average will be about N/(pN) which will be about 1/p.

We can make N as large as we like to make this result as accurate as we like.  So we can confidently say that E(p) = 1/p.

Let’s get back to our puzzle about tasks.  We need to wait for the first task, then the second, then the third and so on.  When there are t tasks left then the chance of getting a new one is t/5.  

So the total waiting time is

    E(5/5) + E(4/5) + E(3/5) + E(2/5) + E(1/5) = 5/5 + 5/4 + 5/3 + 5/2 + 5/1

       = 5(1/5 + 1/4 + 1/3 + 1/2 + 1) = 11 ⁵/₁₂ = 11.4166666… days

You may recognise the harmonic series!

So that’s the proof, now where’s the paradox?

Bob visits Andy when he is able and waits with him until the event occurs.  Of course Andy may well have been waiting for some time.

If Bob turns up just after the event has happened then he will wait for the same time as Andy.  If he turns up just before the event he will wait for one day.  On average he will wait for about half the time that Andy does.

When the event occurs Bob disappears and comes back when he is next able to.

At the end of the experiment Bob averages all of his wait times to get an estimate for E(p).  He gets an answer which is half of Andy’s!

That is the paradox!

Now Carol thinks she understands what is going on here.  The problem is that Andy is distorting his results by always starting the clock straight after an event has occurred.  That guarantees him to get the longest possible wait times.

She thinks the only way to get an accurate answer is to look at the wait time starting from each of the N days and then average these N wait times.

To calculate this she needs to make an assumption.  She knows that the event occurs every 1/p days on average.  She assumes they happen regularly every 1/p days.

 Let’s say they happen every n days where n = 1/p.  Remember Andy’s value for E(p) is 1/p = n.  Bob’s value is half that, so about n/2. 

The wait time will vary between 1 and n.  The average wait time will be n(n+1)/2n = (n+1)/2.

So Carols estimate for E(p) is about n/2 whereas Andy’s was 1/p = n.  So Carol is agreeing with Bob.

I can tell you that Andy had the right value and that the logic I used for Bob and Carol was flawed. 

Can you see where?

“It was a bright, cold day in April, and the clocks were striking thirteen.”

opens George Orwell’s novel ‘Nineteen Eighty-Four’.

1.  By an amazing coincidence thirteen squared is 169 which is the number of times my clock read the right time recently in a single calander day.  Normally it only reads correctly 164 times in a calander day.  This is even more surprising as my clock has been stopped for several years.  How can this be?

My solution combines a number of different techniques.  If you can think of any way a stopped clock can read correctly more than twice a day please post in the comments.  If you can think of something I’ve missed then we may be able to get a bigger answer!

2. I have a second clock which runs slightly fast and I have no way of adjusting it.  How can I make my clock read the right time?

3.  I noticed recently that my third clock was two minutes fast.  It runs at one minute per minute.  It tells me the right time four times a day.  Why?

For inspiration you may want to listen to Peter Sellers (Bluebottle) and Spike Milligan (Eccles) discussing the stopped-clock problem way back in 1957. 

Thanks to New Scientist’s Feedback Column and it’s readers for some of the idea’s here.

This made me smile.  I hope it makes you smile too.

What is

If you think they are the same then why?  If you think they are different then why?

I am in a total immersion game called ‘World of Britain’.

Every day you can take a daily task, if you dare.  There are five tasks and each day one of these tasks is given at random.  You could get the same task each day if you were unlucky.

The tasks are:

·         Cycling in Edinburgh; the best city in the world has plenty of cycle lanes to help you avoid the traffic, some of them are a bit surprising.

·         Cheese Rolling in Gloucestershire; can you beat the cheese? 

·         Bog Snorkelling in Wales; my favourite British sport, bar none!  Watch the action here!  

·         Gurning goes back to at least the thirteenth century.  You may think it is just about pulling funny faces, and you would be right.

·         Mud Racing  How do you know who won?

My math question is:  assuming you win everything you enter how many days would you expect it to take to win all of these competitions?

p.s Gurning is now international.  Here is a US gurner completely oblivious to the exciting belly-flop championship happening behind him.  Apparently “everyone and their butt crack is welcome”