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Logic Puzzle

post #1 of 17
7/4/09 at 5:27pm
Thread Starter 
Hey guys,

Firstly this post has nothing to do with anything audio.

I heard a logic puzzle a few days ago that has been puzzling me but I can't remember the exact question. Maybe you guys can either solve it, or if my memory is faulty, give me the right question (and answer hopefully).

Line up 100 fish one behind another each with a speed between 1-100(arbitrary units) such that no fish has the same speed as any other. All fish swim in the same direction and slower fish are eaten by faster fish. However, having devoured a slower fish the faster fish will resume at the speed of the slower fish. What is the expected number of surviving fish assuming a normal distribution of the fish (this part is confusing as each fish takes a different speed then surely it should say random distribution)?

For example: if there were an arrangement of 4 fish with speeds 4-2-3-1 swimming to the right then there would only be 1 remaining fish.

As '3' eats '1' so you now have 4-2-1 (remember after eating the '1' the '3' fish resumes at the speed of the fish it just ate). Then '2' eats '1' so you have 4-1 and then '4' eats '1' so just 1 fish remaining.
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post #2 of 17
7/4/09 at 7:12pm
From how I understand the question, it seems like there will always be a faster fish and a slower fish. So wouldn't there end up with only one fish left, with a speed of 1?
I also don't get how there can be a normal distribution if there is one fish for each speed.

Weird. I don't know.
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post #3 of 17
7/4/09 at 7:29pm
i wouldve said opposite, if they all started at the same pint wouldnt none of them catch each other? cause they would just spread out? or am i missing something?

the fastest would start infront and never be causght, the slowest would start at the back and never catch up etc
ie
1234
1.2...3.......4
1...2...............3............................. ......................4
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post #4 of 17
7/4/09 at 8:30pm
Oh, nice catch Al4x. If they are lined up and going in the same direction, that certainly makes sense.

I need to read more carefully, for some reason I thought they were in a pool or pond or something. :P
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post #5 of 17
7/4/09 at 9:14pm
So the answer is binary. All 1s and 0s.
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post #6 of 17
7/4/09 at 9:49pm
Thread Starter 
No I'm guessing that the order of the fish is meant to be random so depending on the distribution there are multiple solutions - remember the full question involves 100 fish

but to simplify things with a 4 fish scenario

if the order was 2-1-3-4 (travelling right) then 3 fish would remain

if the order was 3-4-1-2 (right) then 2 fish would remain

So it comes down to which is the most likely scenario given every possible combination of 100 fishes put in random order

Al4x - you may have misread the question - the fish don't start at the same position - they start in a line (ie like a queue one behind another) - I remember the question saying normally distributed but I think it should have read randomly distributed in which case we don't know where the fastest or slowest fish are or any fish for that matter - there are 100! different ways they could be arranged.
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post #7 of 17
7/4/09 at 9:56pm
By 'normal distribution' maybe they mean (when 9 fishes) 135798642?
I know it's not gaussian, but it is as close as one can get with the given conditions.
Or maybe decimals are permitted? (not just natural numbers)

Although those two orders of distribution would give the same result.
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post #8 of 17
7/4/09 at 10:19pm
Thread Starter 
normal and gaussian are two distinct things - guassian is the classic bell shape - normal should be straight line - I see no way of ordering different speed fish in a straight line in a speed vs frequency graph
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post #9 of 17
7/4/09 at 10:27pm
My mind probably works too simply and I'm missing something, but

1. Since each fish swims at a different speed, there will never be any fish swimming at the same speed.
2. There is no time limit, so eventually the faster fish will catch the slower.
3. Therefore 1 fish remaining.
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post #10 of 17
7/4/09 at 10:31pm
Unless there is a faster fish in front of a slower fish, in which case any number of fish can "get away"
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post #11 of 17
7/4/09 at 11:05pm
Quote:
Originally Posted by xiaoipower View Post
normal and gaussian are two distinct things - guassian is the classic bell shape - normal should be straight line - I see no way of ordering different speed fish in a straight line in a speed vs frequency graph
Thats 'linear'.

Got an oxford or webster lying around? Look it up.

My answer to the puzzle would be 50 fishes (half the original number)
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post #12 of 17
7/4/09 at 11:05pm
I think that the question was recalled incorrectly or we need to know the definition of a "normal" distribution. With out that, it is just speculation.
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post #13 of 17
7/5/09 at 3:15am
so if it was 1-2-3...100 there would still bbe 100
if it was 100-99-98....1 it would be 1

and anything in between
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post #14 of 17
7/5/09 at 3:17am
ah i bet it plots a normsl distribution being random with with number of fish surviving on x-axis and % occurance on y-axis, thatd make a normal curve
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post #15 of 17
7/5/09 at 6:03am
What kind of fish are they
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