Hard Question!

[john321]

This is how I see the breeding at any given ratio.

 

D= Dominators, S= Sharers

 

D=(10), S-S=(5,5), D-S-S=(10,3,3), D-D=(10,0)

 

Which should add up to 34 Dominators and 20 Sharer breeds at any given ratio. But, I still cannot induce a correct answer!

[EL]

Guarantee no one will solve it!

Consider two breeding strategies of the fictional Furble. Dominator Furbles can fight for a breeding territory' date=' and if they win, will be able to rear 10 offspring. An alternative is to share territory with another Furble which will allow each to rear 5 offspring. Sharers who attempt to share with dominators will be forced out of the territory, although they will be able to find a new territory. Assume sharers become extra cautious after encountering a dominator and so will always find another territory to share the next time around, but due to lost time will only be able to produce 3 offspring. Dominators are always able to force sharers out of the territory and rear 10 young. Dominators who meet dominators will win 50% of the time. When they lose, they are not able to reproduce that season due to sustained injuries. Individual Furbles cannot switch strategies.

With a total population of 2000 dominator and sharer Furbles, how many would you expect to be dominators?[/quote']

 

Let us build a legend.

Let {T} denote one season

Let {D} denote Dominator

Let {S} denote Sharer

Let {.} denote "encounters"

D.D = (50/100) * (10 + 0) = 5D/T

S.S = 5S/T

D.S = 10D/T + 3S/T = (10D + 3S)/T

The big picture here is that D are central, while S are peripheral and the population is always expanding.

From the equations above, we notice that 2/3 of the D will successfully breed, 20D/6D (assuming heterosexual breeding) because territorial behaviour is only harboured by such type of animals.

Be very careful with the terms of constant breeding, where Sharers will "always find another territory to share the next time around"; In other words New Sharers will never confront Dominators and will breed at a constant rate of 9S/6S while older Sharers are driven off and will also breed at 9S/6S consequently.

 

The colony should have two radii, r, which is the central dominance zone radius and R, which is the total population colony radius. If we do not care about calculating confrontation encounters, then we calculate for a population of 12 Furbles breeding 20D and 9S. If we do, then the peripheral D-couples will breed 10D outwards while the central D will have 50% only breeding 10D per couple.

The dominators dominate pi r^2 territorial breeding area while the sharers share ((pi R^2) - (pi r^2))territorial breeding area. R^2 must be directly proportional to the whole population, while r^2 is only directly proportional to the dominators. Notice that (2pi r) is the factor of the mixed confrontation border at which dominators expand their territory at the expense of the sharers, while (2pi R) represents the new territorial possibilities.

Having 2000 Furbles implies that 1000 Furble-couples are directly proportional to (pi R^2).

Use the rules to extract rate values for r and R per season then build an area relation with breeding ground to estimate the stable rate of growth of each type, and finally calculate the precise percentage of each.

[robotochan]

I think all of you have over simplified this. I took it as an equal probability of the following meetings (d=dominator s=sharer)

1/4 D+D, 1/4 D+S, 1/4 S+S, 1/4 S+D. I then took the number of offspring that would be produced D+D = 12D (dont think of it mathematically) D+S = 12D+8S S+S=12S and S+D = 8S+12D, (i have added into the equations the partners they might meet but only done it for the meetings of each group.) therefore when you add up the amount of each you have 28Sharers and 36Dominators, which using ratios then for a total of 2000 will give you the answer that for some reason i cant give, so wont. but i can pm it to you to check.

 

P.S what the hell is that furbles program for, i played with it for a while you cant even make them explode or fight or anything.

[Karl1986]

Guarantee no one will solve it!

 

 

 

Consider two breeding strategies of the fictional Furble. Dominator Furbles can fight for a breeding territory' date=' and if they win, will be able to rear 10 offspring. An alternative is to share territory with another Furble which will allow each to rear 5 offspring. Sharers who attempt to share with dominators will be forced out of the territory, although they will be able to find a new territory. Assume sharers become extra cautious after encountering a dominator and so will always find another territory to share the next time around, but due to lost time will only be able to produce 3 offspring. Dominators are always able to force sharers out of the territory and rear 10 young. Dominators who meet dominators will win 50% of the time. When they lose, they are not able to reproduce that season due to sustained injuries. Individual Furbles cannot switch strategies.

 

With a total population of 2000 dominator and sharer Furbles, how many would you expect to be dominators?[/quote']

 

 

The way I see it... is there is no indication of the amount of dominators/sharers. It doesn't say every dominator has a dominator offspring, and same with sharers. Therefore, there is no way to say? :/

[astafford]

The problem does have a solution, I found it after worrying about the interpretation of the question for some time, even made a breve simulation. I won't give the answer because the test will loose its value if the answer is disclosed.

 

You have to realise something about Dominators which is not explicitly stated, but is required for there to be a solution.

 

Regards

 

Alan

[johny321]

Is EL's Solution correct?

[Jaan T]

I've come up with three alternate theories depending on interpretation, and none of them work. I'm sure that I'm on the right track.

 

None of my theories look like EL's though, and if I uderstand what he has done, then his theory is incorrect.

 

I'm happy to submit Ideas if anyone would like?

[K9-47G]

This problem reminds me of the many examples that Richard Dawkins gave in his book, The Selfish Gene.

[serg]

100$ reward to first person who send me answer for question 7 or 8 from HASELBAUER - DICKHEISER test for exceptional intelligence.

 

Answer and strong hint on full solution will be sufficient. After confirmation I’ll get back to you for details of money transfer. After you receive money I’ll expect full solution.

 

If you don’t care about 200$ I can offer answers 2-3 questions from this test except (5, 12, 14).

Depends on which questions you ask. (I can’t promise that all of them have perfect solutions.)

 

I promise never to publish any part of solutions anywhere.

I promise not to reveal any part of identity of the winner or any person with right solution.

 

Public discussion of questions will be ignored.

Lame excuses “I have answer, but I not gonna send you because … ” will be ignored.

Tricks - “send me first, I’ll send you later” will be ignored.

 

Serg.

[Sisyphus]

"People who care about their IQs are dorks." - Stephen Hawking

 

Now just imagine what he'd say about someone willing to pay hundreds of dollars to cheat on an IQ test and get into a society for people with high IQs.

 

EDIT: Fun questions, though! Apparently I have an astonishly high IQ! Either that, or the patience to figure out every question on an untimed test...

[TwistOfCain]

No.

 

 

Well' date=' yes. It matters for a few reasons: -

 

1) You have presented the copyrighted property of a company that has expressly asked that the answer not be given. It would be polite to respect that wish.

 

2) If the answer is given, then the Admins may get a request to delete the thread from the institute when the thread propagates.

 

3) Providing the answer will undermine the effectiveness of the test for people who google for answers during quizzes.

 

4) Because the question is copyrighted material, posting it without the thread citing reference is in breach of the forum rules. My post showing the source protects the site.

 

 

I have no idea. It's possible, I suppose. It's not my request, you would have to take it up with the International High IQ society. I'm guessing it's a preventative measure rather than an infallible system.

 

I don't really mind either way, I'm just mentioning it.[/quote']

 

 

IF you had at IQ of 96, would you be intelligent enough to do that in the first place?

[serg]

100$ reward to first person who send me answer for question 7 or 8 from HASELBAUER - DICKHEISER test for exceptional intelligence.

 

...

 

Serg.

 

30 days are over. I got what I wanted.

I withdraw 100$ reward and the rest of promises.

 

Serg.