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!

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?

Show Spoiler ▼

The book dissapeared on the flight out.  After dinner each night my friend impressed me with an amazing math puzzle.  I haven’t seen the book since.

Serves me right!

This is one of those puzzles, you will understand why I have to do it from memory.

I really like Jeff’s post  A Fun Trick – Guess the Polynomial.  You might want to look at it first.

If you relax the conditions a bit you have a similar sounding puzzle with a very different solution.

So my puzzle is this:

I am thinking of a polynomial.  All of the co-efficients are fractions.   You may use any number as your test number.  When you give me a test number I will tell you the result.

How many test numbers do you need to identify the polynomial?

 I was very intrigued by the recursive sequence you mentioned in the past two episodes–the sequence that begins with 1 and each successive term is the average of all the previous terms times some constant. I have always been fascinated by Pascal’s triangle and all of its surprise appearances in mathematics. Also, my fist encounter with doing mathematics for fun out of my own curiosity was to find a formula for triangle numbers. Like Kyle, I was inspired by bowling pin arrangements. The experience was very rewarding and I have been in love with mathematics ever since.

As you mention in the episode “Stacking Cannonballs,” it is easy to show what the sequence evaluates to through algebra but it is not easy to see what is really going on. This is a view I rarely hear mentioned by math people but I have held to be fundamentally important. It seems to me that geometrical interpretations always shed more light than algebra. Because of the nature of the sequences constructed from this recurrence, I felt there must be a geometric way to look at this problem.

Let us represent numbers by pennies as you suggest on the show. The first case is trivial and does not offer much insight, so we can start by looking at the case with constant k := 2. On the show, you added the terms, took the average, and then multiplied by the constant. I suggest we sum the terms, multiply by the constant, and then take the average because 1) we can and 2) we do not have to think about fractional pennies.

The first two terms are 1 and 2. The next term is found by adding these together. Geometrically, we are adding consecutive numbers together which gives us a triangle number. Multiply this by 2 and arrange as follows:

. \. .
. .\ .

This makes two rows, which conveniently is what needs divided by to give us the average. Look at one row to get the next number: 3. Consider a few more terms written in this way:

.  \. . .
. . \ . .
. . .\  .

.   \. . . .
. .  \ . . .
. . . \  . .
. . . .\   .

.    \. . . . .
. .   \ . . . .
. . .  \  . . .
. . . . \   . .
. . . . .\    .

In each case, we have as many rows as we need to divide by to find the average. Thus we always read across a row to find the next term (all rows sum to the same number). Notice geometrically that the next term is a “vertex” plus an “edge.” This becomes a pattern once we get to later sequences.

Suppose we let the constant k:=3. This generates the sequence of triangle numbers. For each term we sum all the previous triangle numbers, multiply by three and then find the average. The sum of triangle numbers is a tetrahedral number, so let us interpret the sum this way. This is hard to draw with periods via email, but try to image arranging three tetrahedrons in a similar way to the triangles so that we get a three dimensional figure where each plane (instead of row) has the same number of points and the number of planes equals the number of triangle summed. Geometrically, the way to arrange these tetrahedrons is to set the first one flat on the table with one of its faces. Place the next one on an edge. Balance the third one on a vertex. This arrangement will give you an equal number of points on each plane. Thus, summing one plane is our next term. Looking at either the top or bottom plane, notice that this sum is a “face” + “edge” + “vertex”. Although I am unable to visualize the fourth dimension, I image this pattern will continue. Since we would then be multiplying by 4, place one hyper-tetrahedron on a tetrahedral side, one on a triangular face, one on an edge, and one on a vertex. This should divide the total in a convenient number of spaces each with the same number of vertices.

I like this interpretation because the constant represents the dimension we are looking at and the division represents the number of equally distributed levels and the terms of the sequence represent generalizations of the triangle. Sorry this email ran long. I hope this helps you see the sequences in a new way.

This was a very stimulating email!
Here’s a way to reinterpret the 3-D case Trevor discusses:

And indeed, his observation that the pattern should continue is correct. Here’s a typical slice of the 4-D case:

As the “movie” progresses, the little green tetrahedron grows and the orange tetrahedron shrinks; the blue triangular prism widens, and becoming flatter and shorter, and conversely, the green prism becomes longer and narrower. 

Is this simpler than the algebra? I don’t think so, really, but it is a neat way to look at things!

I should add a comment about the general case; I think it’s clear from Trevor’s note that he sees how this works, but let’s be explicit: in general, one is chopping up a (k+1)-dimensional prism with k-dimensional tetrahedra as cross-sections. (The movie frame above is one of these cross-sections in the 4-dimensional prism.) The prism, and each slice, is divided into (k+1) tetrahedra, in various reposes. In fact, for each i from 0 to k, there’s a tetrahedron with an i-dimensional facet on the bottom of the prism, and a (k-i) dimensional facet on the top. For example, in the 3-dimensional prism, shown in the first figure, the three tetras are “triangle down” (a two dimensional facet on the bottom of the prism), “edge down” (a one dimensional facet down) and “point down” (a zero dimensional facet down) As we move up, all the roles swap.

For example, suppose we have

-1     -1     3     35     143     399     899 . . . . .
      0     4     32    108     256     500  . . . . .
         4    28    76     148      244  . . . . .
             24    48     72       96   . . . . .
                  24     24    24    . . . . .

What is the polynomial P(n), of degree four, that gives

P(0) = -1 P(1) = -1 P(2) = 3 P(3) = 35 P(4) = 143 , etc.

Can this be expressed simply in terms of the leading values on the left of the table: -1, 0, 4, 24, 24?

And finally, what is the general rule?

Wonderfully, beautifully, the answer is, just as tricycle and prunthaban said in the comments:

24 C(n,4) + 24 C(n,3) + 4 C(n,2) + 0 C(n,1) + (-1) C(n,0)

Where C(n,j) = (n)(n-1)(n-2) … (total of j terms) / j!

For example, C(n,4) = n (n-1) (n-2) (n-3) / 4! In the special case when j=0, we set C(n,0)=1.

So our polynomial here is

(24/4!) (n) (n-1) (n-2) (n-3) + (24/3!) (n) (n-1) (n-2) + (4/2!) (n) (n-1) + (0/1!) (n) + (-1)

That isn’t very “simplified”, but at least we can see the pattern! If we multiply the whole thing out, we get

P(n) = n⁴ – 2 n³ + n² – 1

but it’s hard to see the connection, in this ’simplified’ form, to the original data.

A key property is that C(n,m) + C(n,m+1) = C(n+1,m+1).

This is just a matter of algebra:

C(n,m) + C(n,m+1) = (n) … (n-m+1) / m! + (n) … (n-m+1)(n-m)/ (m+1)!

= you do this part

= (n+1) (n) … (n-m+1) / (m+1)! = C(n+1, m+1).

This is precisely why, these are the numbers that fill in Pascal’s triangle: C(n, m) is the mth entry on the nth row.

Now, how hard is our assertion, that our polynomial is the correct one? Not very, really. Suppose we have a difference table with entries on the left a_k (on the top row), on down to a₀ on the bottom row.

a_k . . . .

 a_k-1 . . . . .

    . . . . .

     a₁ . . . . .

      a₀. . . . .

Then we claim that the jth entry (counting from 0) entries on the ith row (counting up from the bottom 0th row) are given by

P_i(j) = a₀ C(j,i-0) + a₁ C(j,i-1) + . . .  + a_i C(j,i-i)

To prove this, we first show it’s true for the bottom row, then the next row up, etc, etc, clicking along like a typewriter from left to right. (This is just induction on i and j)

For the bottom row, things are pretty easy: all the entries are a₀ C(j,0) = a₀, so check! Also, on the left of each row, the entry, sure enough is P_i(0) = a_i + a₁ 0 + . . .  + a_i 0 = a_i

since (C(0,i) = 0, unless i=0 also– try it!)

Suppose we’ve walked all the way up the spot we care about, on the ith row, jth position. When we get there we will have proven our formula is correct for all the entries below, and for all the entries to the left.

In particular, when we reach the ith row, jth position, we know that (a) the formula is correct for the entry to the left, and (b) the formula is correct for the entry on the row below, to the left. For (a), the entry is P_i(j-1) and for (b), the entry is P_i-1(j-1)

All we have to show is that the value we hope is there, namely P_i(j), is what we get when we add (a) to (b), i.e. would be the correct entry in the difference table. So lets try it!

P_i(j-1) + P_i-1(j-1) =

a₀ C(j-1,i-0) + a₁ C(j-1,i-1) + . . .  + a_i-1 C(j-1,1) + a_i C(j-1,0) +

a₀ C(j-1,i-1) + a₁ C(j-1,i-2) + . . .  + a_i-1 C(j-1,0)

Gathering our a’s and using our identity, it falls into place, and we get the desired sum:

a₀ C(j,i-0) + a₁ C(j,i-1) + . . .  + a_i-1 C(j,1) + a_i C(j-1,0)

What about this last term? C(j-1,0) = 1, as does C(j,0), so we are set. Our total is P_i(j) as promised.

1 + 1/2 + 1/3 + 1/4 + . . .
diverges to infinity, growing as large as you please!

If you try adding terms up on a calculator, this scarcely seems possible! By the time you have added the hundredth term, you will have a reached only a whopping 5.187… (and each new term will be less than .01).

After adding up a million terms, you will have made it only to about 14.39272672… — and each new term will be less than .000001. Does the series really diverge?

The eighteenth century mathematician Jacob Bernoulli gave a very elegant proof that it does:

1/2 is at least 1/2

1/3 + 1/4 is at least 1/4 + 1/4 = 1/2

1/5 + 1/6 + 1/7 + 1/8 is at least 1/8 + 1/8 + 1/8 + 1/8 = 1/2

1/9 + . . . + 1/16 is at least 8 x 1/16 = 1/2

etc. So the result of adding up the first 2ⁿ terms 1/2 + 1/3 + . . . + 1/2ⁿ is at least n/2, and in particular, can be as large as we please.

But this does take a long time to get anywhere. To add up to, say, 100, Bernoulli’s proof shows us that 2¹⁹⁸ (about 4×10⁵⁹) terms will suffice. But maybe this is more than we actually need.

A basic fact from calculus is that the area under the curve y = 1/x, from x = 1 to x = N is exactly ln N.

Now the area of a box 1 unit wide and 1/n units tall is 1/n, and boxes of width 1 and heights 1, 1/2, 1/3, . . . altogether have area 1 + 1/2 + 1/3 . . .

Here we see that these boxes can be arranged to show that

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 > ln 8

Shifting the boxes over, we see that

1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 < ln 8

This gives us a much better bound on the harmonic series. Generally, we have that

1 + 1/2 + . . . + 1/n is between ln (n+1) and (ln n) + 1.

So to be sure that the series sums to at least 100, we can be sure that e^100-1 (about 2.7×10⁴³) terms will suffice!

So the sum of the first million terms is about (ln 1,000,000) + γ, and if we want to sum to 100, we need to have n such that ln n + γ is greater than 100; in other words, e ^{(100 – γ)} (about 1.5×10⁴³) terms will do.

The series Σ 1/(n ln n) diverges even more slowly still, taking about e^e^n terms to sum to n (!!) The series Σ 1/(n (ln n) (ln ln n)) takes e^e^e^n terms to sum to n. Etc!!

Before we get into things, we have to be sure to define what we mean by a program or procedure— what specific language or rules or model we are using. But once we’ve done that, all the arguments will be just exactly the same:

We can list out all the programs (in our language or rules or model) in some procedural way. (For example, just start building longer and longer programs up systematically.) Now some of these will run forever and some of these will eventually stop.

We know, because the Halting Problem is not a computable question, that there is no procedure than can tell us whether the Nth program will eventually halt or run forever.

(Sure, one procedure is just to run the Nth program; if the program halts, we’ll eventually learn this. But if it hasn’t halted after a long time, we won’t know if the program is going to run forever, or another century, or just another minute. We won’t know a thing. There is no procedure that will tell us, in a finite amount of time, whether a given arbitrary program will run forever or halt.)

Let’s define a function: B(n). On our list of programs, among the programs that actually do eventually halt (whichever ones they are), how many steps does it take before the nth halting program halts?

Now B(n) cannot be bounded by any function that can be computed! Not n², nor 2ⁿ nor n^n^n^n nor ….

B(n) has to pop above any computable function!!!

I don’t know about you, but this just blows my mind.

Here’s the argument: