Brain Teaser No 3
Posted by: Don Atkinson on 11 March 2007
I know I started Brain Teaser No 1 about 5 years ago. ISTR another with No 2 in the title, so hopefully this is not duplicating somebody elses Brain Teaser No 3.....
Flight Around the World
A group of aeroplanes is based on a small island. Each plane holds just enough fuel to take it half way around the world. Any amount of fuel can be transfered from the tank of one aeroplane to another aeroplane whilst the planes are in flight. The ONLY source of fuel available to these aeroplanes is on this small island. Assume that there is no time lost when refueling, either in the air or on the ground.
What is the smallest number of aeroplanes required to ensure the flight of one aeroplane around the world on a great circle, assuming that all areoplanes have the same constant groundspeed and rate of fuel consumption and that all aeroplanes return safely to their island base.
Cheers
Don
Flight Around the World
A group of aeroplanes is based on a small island. Each plane holds just enough fuel to take it half way around the world. Any amount of fuel can be transfered from the tank of one aeroplane to another aeroplane whilst the planes are in flight. The ONLY source of fuel available to these aeroplanes is on this small island. Assume that there is no time lost when refueling, either in the air or on the ground.
What is the smallest number of aeroplanes required to ensure the flight of one aeroplane around the world on a great circle, assuming that all areoplanes have the same constant groundspeed and rate of fuel consumption and that all aeroplanes return safely to their island base.
Cheers
Don
Posted on: 02 April 2007 by Don Atkinson
OOps.....simultaneous post with Ian.
Feel free to provide the explanation, I simply didn't want to spoil Ian's night-shift...
Cheers
Don
Feel free to provide the explanation, I simply didn't want to spoil Ian's night-shift...
Cheers
Don
Posted on: 02 April 2007 by Don Atkinson
quote:messing about.
Surely you mean "structured analysis" ?????
There is a step by step logical solution, but I suppose it could be considered "messing about"...........
Cheers
Don
Posted on: 02 April 2007 by Ian G.
Yep I spent 15mins recalling long division too, but gave up after 10mins more as life intervened. As I'm back amongst the workers this week, there'll be no nightshifts so explain away with pride!
Ian
Ian
Posted on: 02 April 2007 by Don Atkinson
Long division.....
In long division, when two digits are brought down instead of one, there must be a zero in the quotient. This happens twice so we can deduce that the quotient is x080x.
When the divisor is multiplied by the quotient's last digit, the product is a four-digit number, The quotient's last digit must therefore be 9, because we already know that eight times the divisor is a three-digit number. (look at the second subtraction)
The divisor must be less than 125 because eight times 125 is 1 000, a four-digit number. (again, look at the second subtraction)
Its now possible to figure out that the quotient's first digit must be more than 7, because seven times a divisor less than 125 would produce a product that would leave more than two digits after it was subtracted from the first four digits in the dividend. This first digit can't be 9, because that would produce a four-digit number when the divisor is multiplied by it. So it must be 8.
The full quotient is therefore 80809.
The divisor must be more than 123 because 80809 times 123 is a seven-digit number and our dividend has eight digits. The only number between 123 and 125 is........124.
The full reconstruction follows by multiplying 123 x 80809 then doing the long division.....
Cheers
Don
In long division, when two digits are brought down instead of one, there must be a zero in the quotient. This happens twice so we can deduce that the quotient is x080x.
When the divisor is multiplied by the quotient's last digit, the product is a four-digit number, The quotient's last digit must therefore be 9, because we already know that eight times the divisor is a three-digit number. (look at the second subtraction)
The divisor must be less than 125 because eight times 125 is 1 000, a four-digit number. (again, look at the second subtraction)
Its now possible to figure out that the quotient's first digit must be more than 7, because seven times a divisor less than 125 would produce a product that would leave more than two digits after it was subtracted from the first four digits in the dividend. This first digit can't be 9, because that would produce a four-digit number when the divisor is multiplied by it. So it must be 8.
The full quotient is therefore 80809.
The divisor must be more than 123 because 80809 times 123 is a seven-digit number and our dividend has eight digits. The only number between 123 and 125 is........124.
The full reconstruction follows by multiplying 123 x 80809 then doing the long division.....
Cheers
Don
Posted on: 02 April 2007 by Alexander
I just started substituting X'es by numbers without converting the problem to a set of equations. "messing about" seems good enough a description. It may be interesting to find out why messing about can work here.
Posted on: 03 April 2007 by Don Atkinson
My Truel Solution
I'm not certain this will print. If it does, I will add an explanation of how it works. If not, I will delete and try again
Cheers
Don
I'm not certain this will print. If it does, I will add an explanation of how it works. If not, I will delete and try again
Cheers
Don
Posted on: 03 April 2007 by Don Atkinson
OK, it seems to have printed out so as to be readable.
The "fault-tree" is read from bottom left to top right.
The nodes represent the output from the previous activity plus the input to the next activity. eg "Ta kills Ts; F aims at Ta"
Each segment represents the route to a possible outcome with its probability expressed as a fraction. So, when "F aims at Ta" there are two possible outcomes, "F survives" (ie "F kills Ta") or "F misses Ta" and each has a probability of 1/2
Where an input could have more than one outcome, there are more than one segments emerging from the node
The sum of all segments emerging from a node must be 1.
The eventual possible outcomes are
"F survives"
"Ta survives"
"Ts survives"
This diagram relies on the fact that F will always "waste" his shots until either Ta is dead or Ts is dead, before taking the first shot in the resulting dual.
To illustrate the use of the fault-tree let's look at the survival chance of Tarquin.
There are two possible ways in which "Ta survives"
Each way is calculated by starting at the "Beginning" and working your way along the route to the final outcome, multiplying the fractions along the way. The results of following both routes are added together
1/2 x 1 x 1/2 x 1 = 1/4 (route 1)
1/2 x 1/5 x 1/2 x 1 = 1/20 (route 2)
1/4 + 1/20 = 6/20 = 3/10 = probability that Ta survives (see diagram below)
I'll post a picture to show these routes and the routes for "F survives" and "Ts survives" together with the arithmetic
Cheers
Don
The "fault-tree" is read from bottom left to top right.
The nodes represent the output from the previous activity plus the input to the next activity. eg "Ta kills Ts; F aims at Ta"
Each segment represents the route to a possible outcome with its probability expressed as a fraction. So, when "F aims at Ta" there are two possible outcomes, "F survives" (ie "F kills Ta") or "F misses Ta" and each has a probability of 1/2
Where an input could have more than one outcome, there are more than one segments emerging from the node
The sum of all segments emerging from a node must be 1.
The eventual possible outcomes are
"F survives"
"Ta survives"
"Ts survives"
This diagram relies on the fact that F will always "waste" his shots until either Ta is dead or Ts is dead, before taking the first shot in the resulting dual.
To illustrate the use of the fault-tree let's look at the survival chance of Tarquin.
There are two possible ways in which "Ta survives"
Each way is calculated by starting at the "Beginning" and working your way along the route to the final outcome, multiplying the fractions along the way. The results of following both routes are added together
1/2 x 1 x 1/2 x 1 = 1/4 (route 1)
1/2 x 1/5 x 1/2 x 1 = 1/20 (route 2)
1/4 + 1/20 = 6/20 = 3/10 = probability that Ta survives (see diagram below)
I'll post a picture to show these routes and the routes for "F survives" and "Ts survives" together with the arithmetic
Cheers
Don
Posted on: 03 April 2007 by Don Atkinson
Probability that Tsunami survives
Remember, you must start at the "Beginning" for each route to survival and this applies even when the routes are superimposed.
The number of "Ts survives" outcomes is infinite, but the probability of each reduces
(1/2 x 4/5 x 1/2 x 4/5) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2 x 4/5) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2 x 1/5 x 1/2 x 4/5) +.....
= 4/25 + (4/25 x 1/10) + (4/25 x 1/100) + (4/25 x 1/1000).....
= 4/25 (1 + 1/10 + 1/100 + 1/1000.....)
= 4/25 (1.11111.....)
= 4/25 x 10/9
= 8/45
Cheers
Don
Remember, you must start at the "Beginning" for each route to survival and this applies even when the routes are superimposed.
The number of "Ts survives" outcomes is infinite, but the probability of each reduces
(1/2 x 4/5 x 1/2 x 4/5) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2 x 4/5) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2 x 1/5 x 1/2 x 4/5) +.....
= 4/25 + (4/25 x 1/10) + (4/25 x 1/100) + (4/25 x 1/1000).....
= 4/25 (1 + 1/10 + 1/100 + 1/1000.....)
= 4/25 (1.11111.....)
= 4/25 x 10/9
= 8/45
Cheers
Don
Posted on: 03 April 2007 by Don Atkinson
Probability that F survives
There are two unique outcomes and one infinite outcome
(1/2 x 1 x 1/2) +
(1/2 x 1/5 x 1 x 1/2) +
(1/2 x 4/5 x 1/2) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2 x 1/5 x 1/2) +.....
= 1/4 + 1/20 + 4/20 + (4/20 + 1/10) + (4/20 + 1/100) + (4/20 + 1/1000).....
= 1/4 + 1/20 + [4/20 (1 + 1/10 + 1/100 + 1/1000 +.....)]
= 1/4 + 1/20 + [4/20 (1.11111.....)]
= 1/4 + 1/20 + [4/20 x 10/9]
= 1/4 + 1/20 + 2/9
= 47/90
Cheers
Don
There are two unique outcomes and one infinite outcome
(1/2 x 1 x 1/2) +
(1/2 x 1/5 x 1 x 1/2) +
(1/2 x 4/5 x 1/2) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2) +
(1/2 x 4/5 x 1/2 x 1/5 x 1/2 x 1/5 x 1/2) +.....
= 1/4 + 1/20 + 4/20 + (4/20 + 1/10) + (4/20 + 1/100) + (4/20 + 1/1000).....
= 1/4 + 1/20 + [4/20 (1 + 1/10 + 1/100 + 1/1000 +.....)]
= 1/4 + 1/20 + [4/20 (1.11111.....)]
= 1/4 + 1/20 + [4/20 x 10/9]
= 1/4 + 1/20 + 2/9
= 47/90
Cheers
Don
Posted on: 03 April 2007 by Don Atkinson
The full truel involves all three players drawing lots for the shooting order.
The attached diagram shows the associated fault-tree and probabilities.
Apologies for it being in manuscript, and I hope it is legible when posted.
Cheers
Don
The attached diagram shows the associated fault-tree and probabilities.
Apologies for it being in manuscript, and I hope it is legible when posted.
Cheers
Don
Posted on: 03 April 2007 by Alexander
Truel Variations(calculation-free)
Just for fun I jotted down some cruel variations on the truel theme
- Shootout1: Ts and Ta both are perfect shooters.
First step: Ts and Ta each have 50% chance.
Second step F gets one shot, 50%.
Ts25%/Ta25%/F50%
- Shootout2: Ts hits a fraction better than F, say 51%.
Ts gets 75% in first step and 50% in second step, so 37.5% survival
Ts 25% in first step and something less than 50% in second step so less than 12.5%
So chances for survival Ta37.5%/Ts<12.5/F>50
- Shootout2b, F and Ts work together. Ta strategy doesn't change.
Ta then , depending on getting first, second or third shot, has 1/3+1/6+1/12 chance of surviving first step, and 50% in second step.
So survival becomes 7/24=29%
Because Ts and F have equal chances(50%) at the second step chances there for Ts are 5/12 * 1/2 = 5/24
F also has 50% chance if he's up against Ta in the second stage, because he gets first shot.
Ta29% /Ts21% /F50%
So, once Ts and F are close together you might think Ta might just as well aim at both of them alternatingly.
This would mean though that we end up in the game situation of shootout2b which is very much against the interest of Ta.
So Ta and F have to be convinced that the game is shootout2.
To open up the possibility of adapting to the other's strategy we place the three players much further apart,
so that their chances of hitting each other become much smaller.
- Shootout3: chances of hitting are Ta 0.002/Ts 0.001/F 0.001
Ta can take on Ts while making clear to F to stay out of the game, eg with a TIT for TAT strategy:
if F shoots at Ta, Ta will shoot back at F until F stops shooting.
Ts can try to draw F into the fight by shooting at him if he doesn't participate.
Would it make a difference if we place Ts and F closer together or further apart?
- Shootout4 hitting score Ta 0.001/Ts 0.001/F 0.001, there are 200 bullets each.
Now the game changes again. There is the possibility of everyone surviving.
Participants can shoot at the ground to indicate their intent.
- Shootout5: Same as before but if you shoot at someone instead of at the ground you get two extra bullets.
You also get increase your hitting score with a modest fraction.
How can the participants be coaxed into killing each other without anyone being enthusiastic about that outcome?
- Shootout6: same as before but hitting Ta and Ts shoot better : Ta 0.002/Ts 0.002/F 0.001, 200 bullets each.
Poor F, I guess.
Just for fun I jotted down some cruel variations on the truel theme
- Shootout1: Ts and Ta both are perfect shooters.
First step: Ts and Ta each have 50% chance.
Second step F gets one shot, 50%.
Ts25%/Ta25%/F50%
- Shootout2: Ts hits a fraction better than F, say 51%.
Ts gets 75% in first step and 50% in second step, so 37.5% survival
Ts 25% in first step and something less than 50% in second step so less than 12.5%
So chances for survival Ta37.5%/Ts<12.5/F>50
- Shootout2b, F and Ts work together. Ta strategy doesn't change.
Ta then , depending on getting first, second or third shot, has 1/3+1/6+1/12 chance of surviving first step, and 50% in second step.
So survival becomes 7/24=29%
Because Ts and F have equal chances(50%) at the second step chances there for Ts are 5/12 * 1/2 = 5/24
F also has 50% chance if he's up against Ta in the second stage, because he gets first shot.
Ta29% /Ts21% /F50%
So, once Ts and F are close together you might think Ta might just as well aim at both of them alternatingly.
This would mean though that we end up in the game situation of shootout2b which is very much against the interest of Ta.
So Ta and F have to be convinced that the game is shootout2.
To open up the possibility of adapting to the other's strategy we place the three players much further apart,
so that their chances of hitting each other become much smaller.
- Shootout3: chances of hitting are Ta 0.002/Ts 0.001/F 0.001
Ta can take on Ts while making clear to F to stay out of the game, eg with a TIT for TAT strategy:
if F shoots at Ta, Ta will shoot back at F until F stops shooting.
Ts can try to draw F into the fight by shooting at him if he doesn't participate.
Would it make a difference if we place Ts and F closer together or further apart?
- Shootout4 hitting score Ta 0.001/Ts 0.001/F 0.001, there are 200 bullets each.
Now the game changes again. There is the possibility of everyone surviving.
Participants can shoot at the ground to indicate their intent.
- Shootout5: Same as before but if you shoot at someone instead of at the ground you get two extra bullets.
You also get increase your hitting score with a modest fraction.
How can the participants be coaxed into killing each other without anyone being enthusiastic about that outcome?
- Shootout6: same as before but hitting Ta and Ts shoot better : Ta 0.002/Ts 0.002/F 0.001, 200 bullets each.
Poor F, I guess.
Posted on: 03 April 2007 by Alexander
The old coots around here will remember the old self referential sentences. And so do I. Well more or less. This challenge consists of three parts, and it's quite easy really.
Part one: create a self referential row of numbers.
The minimal length of the row is 10 numbers, but it is permitted to make it longer.
The row is n0|n1|n2|n3|n4|n5|n6|n7|n8|n9|...
The first number is the amount of zeroes in the row
The second number is the amount of ones in the row
...
The 10th number is the amount of nines in the row.
Optionally the 11th number is the amount of tens in the row
And so on.
Part two, now there is a pair of rows.
The first row describes the second row and vice versa
Part three, now there are three rows.
The first row describes the second, the second describes the third,
the third describes the first.
Example of a solution for part 2:
row 1: 7/1/0/1/0/0/1/0/0/0
row 2: 6/3/0/0/0/0/0/1/0/0
Part one: create a self referential row of numbers.
The minimal length of the row is 10 numbers, but it is permitted to make it longer.
The row is n0|n1|n2|n3|n4|n5|n6|n7|n8|n9|...
The first number is the amount of zeroes in the row
The second number is the amount of ones in the row
...
The 10th number is the amount of nines in the row.
Optionally the 11th number is the amount of tens in the row
And so on.
Part two, now there is a pair of rows.
The first row describes the second row and vice versa
Part three, now there are three rows.
The first row describes the second, the second describes the third,
the third describes the first.
Example of a solution for part 2:
row 1: 7/1/0/1/0/0/1/0/0/0
row 2: 6/3/0/0/0/0/0/1/0/0
Posted on: 08 April 2007 by Alexander
Nothing happened here. The interesting thing about this challenge is that while it looks quite hard if you try to handle it by "structured analysis", a much better approach is very simple, and it is exactly the same approach for parts 1,2 and 3:
Pick any random starting array of 10 or more numbers:
4,2,5,1,1,6,3,8,2,2
and iterate from there, applying the number counting routine again and again. The results are
step1: 0,2,2,1,1,1,1,0,1,0
step2: 3,5,2,0,0,0,0,0,0,0
step3: 7,0,1,1,0,1,0,0,0,0
step4: 6,3,0,0,0,0,0,1,0,0
step5: 7,1,0,1,0,0,1,0,0,0
step6: 6,3,0,0,0,0,0,1,0,0
and we have a pair solution.
You always end up in a loop,but you never know in advance which one, so the rest is a matter of keep trying out different seed arrays. You may end up with a n answer for part1,2,or 3, or even with a 4 step loop.
I tried about 10 seed points after thinking up and posting the challenge and haven't found a second pair solution yet with exactly 10 numbers. I did find a solution to part1: 6,2,1,0,0,0,1,0,0,0.
The general approach for handling such referential arrays/sentences I read in an article by Hofstadter, long ago.
Pick any random starting array of 10 or more numbers:
4,2,5,1,1,6,3,8,2,2
and iterate from there, applying the number counting routine again and again. The results are
step1: 0,2,2,1,1,1,1,0,1,0
step2: 3,5,2,0,0,0,0,0,0,0
step3: 7,0,1,1,0,1,0,0,0,0
step4: 6,3,0,0,0,0,0,1,0,0
step5: 7,1,0,1,0,0,1,0,0,0
step6: 6,3,0,0,0,0,0,1,0,0
and we have a pair solution.
You always end up in a loop,but you never know in advance which one, so the rest is a matter of keep trying out different seed arrays. You may end up with a n answer for part1,2,or 3, or even with a 4 step loop.
I tried about 10 seed points after thinking up and posting the challenge and haven't found a second pair solution yet with exactly 10 numbers. I did find a solution to part1: 6,2,1,0,0,0,1,0,0,0.
The general approach for handling such referential arrays/sentences I read in an article by Hofstadter, long ago.
Posted on: 08 April 2007 by Ian G.
hi,
I tried sticking your iterative method into excel (the countif function makes this very easy) and with random seeds I too only found the solutions you have already given. I couldn't yet find a 3 or 4 step loop. I must have tried about 100 initial guesses.
cheers
Ian
I tried sticking your iterative method into excel (the countif function makes this very easy) and with random seeds I too only found the solutions you have already given. I couldn't yet find a 3 or 4 step loop. I must have tried about 100 initial guesses.
cheers
Ian
Posted on: 08 April 2007 by Don Atkinson
No luck either.
Tried "seeds". Then tried "Googling" and a site from Loughborough University came up with the "seed" approach, but didn't offer much hope of a 10-digit 3-level self-referential number.
Best I have found is 7-digt 3-level solution
4/1/1/0/1/0/0
4/1/0/2/0/0/0
3/3/0/0/1/0/0
Cheers
Don
Tried "seeds". Then tried "Googling" and a site from Loughborough University came up with the "seed" approach, but didn't offer much hope of a 10-digit 3-level self-referential number.
Best I have found is 7-digt 3-level solution
4/1/1/0/1/0/0
4/1/0/2/0/0/0
3/3/0/0/1/0/0
Cheers
Don
Posted on: 09 April 2007 by Ian G.
Alexander: can you confirm whether there is indeed a 10 digit solution to part 3 ??
cheers
Ian
cheers
Ian
Posted on: 09 April 2007 by Alexander
Hello Ian,
I cannot confirm that there is a 10 digit solution to part 3. There is no authority to fall back upon for answers.
In fact when I mentioned the example of a two step cycle I expected there would be more solutions. Which may not be the case. The "sum of numbers is 10" constraint is quite strong and I think a systematic scan is possible. I see several options for handling zeroes. Variants are that they could be ignored or truncated at the last non-zero number. The iteration technique remains the same.
The 10 number example was the simplest example of the technique I could come up with but you can just as well apply the technique when creating a self referential sentence that goes like "this sentence contains four a's, three b's, ... and one z." or "No one in his right mind would deny that this sentence consists of seven a's ...and twenty six apostrophes.".
I cannot confirm that there is a 10 digit solution to part 3. There is no authority to fall back upon for answers.
In fact when I mentioned the example of a two step cycle I expected there would be more solutions. Which may not be the case. The "sum of numbers is 10" constraint is quite strong and I think a systematic scan is possible. I see several options for handling zeroes. Variants are that they could be ignored or truncated at the last non-zero number. The iteration technique remains the same.
The 10 number example was the simplest example of the technique I could come up with but you can just as well apply the technique when creating a self referential sentence that goes like "this sentence contains four a's, three b's, ... and one z." or "No one in his right mind would deny that this sentence consists of seven a's ...and twenty six apostrophes.".
Posted on: 09 April 2007 by Ian G.
Thanks Alex - a follow-on teaser is what is the fewest attempts in a systematic scan for the three step, 10-digit solution? The 'sum of the numbers is 10' constraint makes this an intesting challenge me-thinks.
And no, I don't know yet if it can be done nor what the answer is.
Ian
And no, I don't know yet if it can be done nor what the answer is.
Ian
Posted on: 09 April 2007 by Alexander
Once in a final cycle, the previous step is seen by the current step as one of the following,
(because permutations do not matter), so an exhaustive search is done with the following seed arrays(pad with zeroes)
(10)(9,1),(8,1,1),(8,2),(7,3),(7,2,1),(7,1,1,1),
(6,4),(6,3,1),(6,2,2),(6,2,1,1),(6,1,1,1,1),
(5,5),(5,4,1),(5,3,2),(5,3,1,1),(5,2,2,1),(5,2,1,1,1),(5,1,1,1,1,1)
(4,4,2),(4,3,3),(4,3,2,1),(4,3,1,1,1),(4,2,2,2),(4,2,2,1,1),(4,2,1,1,1,1),skip (4,1,1,1,1,1,1) because it yields count 6
(3,3,3,1)(3,3,2,2)(3,3,2,1,1),skip(3,3,1,1,1,1),(3,2,2,2,1),(3,2,2,1,1,1),skip(3,2,1,1,1,1),skip(3,1,1,1,1,1,1)
combinations of twos and ones will lead to any of the above so we can skip them too.
(because permutations do not matter), so an exhaustive search is done with the following seed arrays(pad with zeroes)
(10)(9,1),(8,1,1),(8,2),(7,3),(7,2,1),(7,1,1,1),
(6,4),(6,3,1),(6,2,2),(6,2,1,1),(6,1,1,1,1),
(5,5),(5,4,1),(5,3,2),(5,3,1,1),(5,2,2,1),(5,2,1,1,1),(5,1,1,1,1,1)
(4,4,2),(4,3,3),(4,3,2,1),(4,3,1,1,1),(4,2,2,2),(4,2,2,1,1),(4,2,1,1,1,1),skip (4,1,1,1,1,1,1) because it yields count 6
(3,3,3,1)(3,3,2,2)(3,3,2,1,1),skip(3,3,1,1,1,1),(3,2,2,2,1),(3,2,2,1,1,1),skip(3,2,1,1,1,1),skip(3,1,1,1,1,1,1)
combinations of twos and ones will lead to any of the above so we can skip them too.
Posted on: 09 April 2007 by Ian G.
Yep Alex, nicely done.
I just whanged those attempts into my excel spreadsheet and found no three step solutions.
ian
I just whanged those attempts into my excel spreadsheet and found no three step solutions.
ian
Posted on: 09 April 2007 by Alexander
Ian, I don't remember ever having created anything with Excel but you're making fine use of it.
There is one open question for volunteers with powertools:
Don found a new solution by shortening the array till the number of zeroes started to interfere with the other numbers.
The 8 and 9 number arrays didn't appear terribly interesting - but who knows. Less than seven may give some new patterns.
Will longer arrays of 20 or 10.000 yield more interesting solutions, or just the small tweak on the two 10-number solutions?
A single counterexample would be enough so it shouldn't be too hard. Heh heh
There is one open question for volunteers with powertools:
Don found a new solution by shortening the array till the number of zeroes started to interfere with the other numbers.
The 8 and 9 number arrays didn't appear terribly interesting - but who knows. Less than seven may give some new patterns.
Will longer arrays of 20 or 10.000 yield more interesting solutions, or just the small tweak on the two 10-number solutions?
A single counterexample would be enough so it shouldn't be too hard. Heh heh
Posted on: 09 April 2007 by Alexander
The following real world issue is in line with Don's shootout math above:
Several asian countries have a bit of a demographic problem because there are more boys being born than girls, boys being more valuable for the family.
China's "one child" policy didn't exactly help here, it may even have exacerbated the problem.
At this site I read a proposal for a more balanced birth control policy for China.
The proposal is to instate a "one boy" policy.
Realistic calculations would drive us too far (the one child policy is not that straightforward either)but let's cast this into a simple math problem that can help us understand the principle at work.
What would the maximum ratio of girls/boys be if people would follow the rule but always strive for a boy? I haven't calculated anything yet. Bedtime. ttfn.
Several asian countries have a bit of a demographic problem because there are more boys being born than girls, boys being more valuable for the family.
China's "one child" policy didn't exactly help here, it may even have exacerbated the problem.
At this site I read a proposal for a more balanced birth control policy for China.
The proposal is to instate a "one boy" policy.
Realistic calculations would drive us too far (the one child policy is not that straightforward either)but let's cast this into a simple math problem that can help us understand the principle at work.
What would the maximum ratio of girls/boys be if people would follow the rule but always strive for a boy? I haven't calculated anything yet. Bedtime. ttfn.
Posted on: 10 April 2007 by Don Atkinson
quote:What would the maximum ratio of girls/boys be if people would follow the rule but always strive for a boy?
Just to be clear.....
We assume that the natural birth rate is 50% boys and 50% girls. Parents must produce off-spring until they get one boy, then must stop.
You could have g/b or g/g/b or b (for example)
You are not allowed to have b/g or g/b/g or g or g/g or b/b
Cheers
Don
Posted on: 10 April 2007 by Ian G.
It will be nifty if this is right...
The sequence always ends when a boy is born into the family so the number of boys is always 1.
If a girl is born another attempt is made to have a boy so after 2 tries there is a 1/4 chance of ending the sequence, resulting in 1 girl and 1 boy.
If a second girl is born there is again another attempt giving a 1/8 chance of producing a boy resulting in then a family of 2 girls and 1 boy.
etc.
So the average number of girls is then the probabilty of each outcome times the number of girls in that outcome, so
1/4 * (1) gb
+ 1/8 * (2) ggb
+ 1/16 * (3) gggb
+ 1/32 * (4) ggggb
+ ....
Which miraculously sums to exactly 1 !
So this policy does not change the ratio of boys to girls - just the average size of families.
It seems odd at first but on reflection (I think) the insistence on continuing until a boy is produced is exactly matched by the extra girls produced in the process.
Ian
The sequence always ends when a boy is born into the family so the number of boys is always 1.
If a girl is born another attempt is made to have a boy so after 2 tries there is a 1/4 chance of ending the sequence, resulting in 1 girl and 1 boy.
If a second girl is born there is again another attempt giving a 1/8 chance of producing a boy resulting in then a family of 2 girls and 1 boy.
etc.
So the average number of girls is then the probabilty of each outcome times the number of girls in that outcome, so
1/4 * (1) gb
+ 1/8 * (2) ggb
+ 1/16 * (3) gggb
+ 1/32 * (4) ggggb
+ ....
Which miraculously sums to exactly 1 !
So this policy does not change the ratio of boys to girls - just the average size of families.
It seems odd at first but on reflection (I think) the insistence on continuing until a boy is produced is exactly matched by the extra girls produced in the process.
Ian
Posted on: 10 April 2007 by Don Atkinson
quote:It seems odd at first but on reflection (I think) the insistence on continuing until a boy is produced is exactly matched by the extra girls produced in the process.
The usual "fault tree" confirms your probabilities, ie the overall ratio of boys/girls remains 50/50.
Also the average number of children per family doubles from 1 child to 2 children (but pity those with 6 girls and a boy.........)
Cheers
Don