BBC Radio 4 - Puzzles for Today

[detuned]

So the Today programme on radio 4 have been broadcasting brainteasers for a while, which you can have a look at here:

http://www.bbc.co.uk/programmes/p057wxwl

It's an eclectic mix to say the least, but every now and again there's something interesting. #90 (November 3rd) has caught my attention today. The problem goes as follows:

Alien bees invade Earth. On the third and fourth days of its life, each bee gives birth to a live clone, then dies at the end of its fourth day. The invasion begins with one bee. How many bees are there at the end of the twentieth day?

Today’s #PuzzleForToday has been set by Dr Gihan Marasingha, Senior Lecturer in mathematics at the University of Exeter

(click this link for the published solution)

Ignoring the fact this puzzle is not well-defined - you need to make some assumption about the age of the initial invading bee - I cannot get the published solution under any permutation. I'll post my logic later, but I'd be interested to hear if anyone had any thoughts...

[puzzlemad]

I'm with you Tom, I can't get a number anywhere near.

[detuned]

In case anyone doesn't want this spoiled, select the following text with your mouse...

So there are 4 possible ages of bees: 0, 1, 2, or 3 days old - 4 years old would mark the end of a bee's 4th year, which is when we're told it dies. At the end of day t lets denote the numbers of those bees by a(t), b(t), c(t) and d(t). The total number of bees is then just the sum of these 4 sequences.

The setup of the problem determines the following relations between the sequences:

d(t) = c(t-1)
c(t) = b(t-1)
b(t) = a(t-1)
a(t) = c(t-1) + d(t-1)

The first three are simply ageing, the last is the cloning condition.

To obtain an answer, clearly you need to give an initial condition for a(0), b(0), c(0), d(0) - which is why this problem is not well defined. Presumably one of these is equal to 1, and the other three are equal to 0.

Anyhow, running these 4 iterations through, I get either: 48, 59, 70 or 38. None of these is remotely close to 200...

[Nilz]

My interpretation was that a(t)=c(t)+d(t) (not t-1 on the RHS). However my maths still doesn't give me the quoted answer (I get 237).

[puzzlemad]

In case anyone doesn't want this spoiled, select the following text with your mouse...

That's odd Tom - I couldn't see anything to select - but when I hit the quote tab I could see your text. I'd got the same as you.

[Fred76]

In case anyone doesn't want this spoiled, select the following text with your mouse...

So there are 4 possible ages of bees: 0, 1, 2, or 3 days old - 4 years old would mark the end of a bee's 4th year, which is when we're told it dies. At the end of day t lets denote the numbers of those bees by a(t), b(t), c(t) and d(t). The total number of bees is then just the sum of these 4 sequences.

The setup of the problem determines the following relations between the sequences:

d(t) = c(t-1)
c(t) = b(t-1)
b(t) = a(t-1)
a(t) = c(t-1) + d(t-1)

The first three are simply ageing, the last is the cloning condition.

To obtain an answer, clearly you need to give an initial condition for a(0), b(0), c(0), d(0) - which is why this problem is not well defined. Presumably one of these is equal to 1, and the other three are equal to 0.

Anyhow, running these 4 iterations through, I get either: 48, 59, 70 or 38. None of these is remotely close to 200...

At the end of the day, the 3 days old bees die (if at the end of the first day of its life, it is 0 day old, then if it dies at the end of its fourth day, it's 3 days old), then d(t)=0 for any t.

c(t) = b(t-1)
b(t) = a(t-1)
a(t)=b(t-1)+c(t-1) should fix your issues.

It works then for intial conditions a(1)=1, b(1)=0 c(1)=0 (the invasion begins with one bee, but it's not specified how old it is).

Fred

[PuzzleScot]

If anyone reads this far, spoilers really needn't be necessary.

OK, I can see where the 200 comes from.
If:
1) No bees die!
2) and his counting is out.

At the end of the 19th day, there are 179 bees <=4 days, and 21 > 4 days.
I suppose technically he says "how many bees" rather than "how many live bees"
But yeah, poor on many levels.

Answers to questions like this really should include workings. Just saying "It's related to the Fibonacci sequence" doesn't count as an explanation.

[Fred76]

No, the dead bees are not counted.

Here is a table of alive bees population at the end of the 20 first days of invasion:



Fred

[detuned]

In case anyone doesn't want this spoiled, select the following text with your mouse...

That's odd Tom - I couldn't see anything to select - but when I hit the quote tab I could see your text. I'd got the same as you.

It's definitely there - you can't immediately see it as I deliberately changed the font colour to match the background colour.

[detuned]

Thanks for the explanation Fred, that makes sense

[PuzzleScot]

Thanks for the explanation Fred, that makes sense

Yeah, that makes more sense! At the end of the 4th day, the bee is already dead, but we were all counting it anyway.

[puzzlemad]

Thanks for the explanation Fred, that makes sense

Yeah, that makes more sense! At the end of the 4th day, the bee is already dead, but we were all counting it anyway.

I wasn't. My mistake was interperating the English. Thanks to the non-native, not first language English speaker for putting me right . I was counting as days old, not days of life so had nowhere near the right number.

[kiwijam]

The solution briefly mentioned Fibonacci.
That was referring to the observation that the the number born today equals the number born 2 days ago plus the number born 3 days ago.
So that gives a Fibonacci-style sequence, and from that sequence t18 + t19 + t20 = 200.