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Offline lightarrow

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« Reply #75 on: 23/07/2008 14:13:44 »
Lightarrow,

Lightarrow,

I like you approach to this not so simple little problem

Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0).

Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5).

Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1.

Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.

Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).
Can you find for me a prove of this theorem? I still haven't found on my books (but there must be somewhere).
Quote

So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.

At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x.

The arc length is given by the formula:

Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2
And of this one also.
I only know this formula: L = ∫0+oo {[x'(t)]2 + [y'(t)]2}1/2dt.
Where x and y are the coordinates of the vector r(t) and t is the parameter of the curve (that is, the polar angle in this case, that you called "x").
Quote
=

Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =

Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 =

Integral from 0 to infinity of e-x =
-e-x from 0 to infinity =
=0 - (-e0) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral.

It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution.
 
Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.
This is very interesting, clever solution!
Quote

Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs.

He also provided the following graphics.
For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)).

So a pentagon will have angles of 108. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines.

In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180)=-0.309
cos(180) = -1
Quote
units towards dog A.

The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.

What do you think?
T = a/v, as I've written in my previous post. If a = 1 and v = 1 unit/minute, then T = 1 minute.
Very interesting problem however, thank you!
« Last Edit: 23/07/2008 14:23:01 by lightarrow »
 

Offline lightarrow

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« Reply #76 on: 23/07/2008 14:33:25 »
BoardChemist,

Quote
I don't know how to hide an answer, but I doubt this is the answer being looked for
.

"" to the mind guess"" of prediction test in  "blue letters", This is post separate from the horse hay one. 

Quote
The puzzle states that the hay is tied down. It does not state that the stake is driven into the ground so maybe the horse just walks over to the hay dragging the stake.(Though my first thought
was that he chews through, or breaks the rope.)

Yessssssss!! The above is the correct answer a little silly I know but many people most often just presume the skate is firmly hamered into the ground

Regards

Alan
What's wrong with LeeE's answer?:
Quote
The horse is standing at the extreme length of the rope, five meters from the stake.  The horse is facing the stake, and five meters further away in the same direction is the hay.  The horse walks the ten meters across the 'circle' and eats the hay
 

Offline lightarrow

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« Reply #77 on: 23/07/2008 14:35:11 »
I suppose it could have been a sawing horse - so no problem, again.
:D
 

Offline lightarrow

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« Reply #78 on: 23/07/2008 14:39:04 »
Two glasses, one nearly filled with liquid A, one with liquid B.  Take a teaspoon of A and put it in B, stir and put a teaspoon of the mixture back in A.   Which now is the purest??
None, if they initially contains the same volume of liquids of the same density (and the density doesn't vary with mixing).
I usually propose to people, at breakfast, a slight variation of this problem:
you take a teaspoon of coffe from a little cup and put it in a large cup of milk [so the initial amounts of coffee and milk can be very different]. Then you mix the milk-coffee, you take a teaspoon of it and put it in the little cup of coffee. Is there more coffee in the cup of milk or milk in the little cup of coffee?

Let be v the volume in the teaspoon, so you take away v of coffee from the little cup and put it in the large cup. Now you take a teaspoon of milk-coffee: this will contain x of coffee and v-x of milk, and you put it in the little cup, so it contains now v-x of milk. Now here you have taken away v of coffee and re-added x of it so you have taken away v-x of coffee from the little cup and so this is the amount of coffee you have put in the cup of milk, so the two amounts are the same.
« Last Edit: 23/07/2008 14:58:23 by lightarrow »
 

Offline Alan McDougall

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« Reply #79 on: 23/07/2008 15:10:51 »
MathProblems.info


Problem 29 Solution
Hard Solution
It is intuitive that at all times the dogs will form the corners of a square, ever decreasing in size and rotating about the center.
Let C be the center of the square at point (0,0). Let the dogs be at initial points (.5,.5), (.5,-.5), (-.5,-.5), and (-.5,.5). Let dog 1 bet at (.5,-.5) who is chasing dog 2 at (.5,.5). Consider the line from C to dog 1. Next consider the line from dog 1 to dog 2. The angle between these lines is 135 degrees.

Theorem: Let C be a fixed point, D a point on a curve, r the distance from C to D, x the polar angle formed by r, and y the angle formed by the angle between CD and the tangent line to the curve at point D. Then tan(y)=r/(dr/dx).

So the line of sight from the center to the initial point of dog 1 would be be at a 270 degree angle. The line of sight from dog 1 to dog 2 would be a vertical line of 90 degrees. The angle between these lines is 270 degrees. The tangent of 270 is -1. So we have r/(dr/dx) = -1. Thus r(x)=c*e-x.

At the moment the dogs start walking x equals 0. At this moment the distance from the center to any of the dogs is 2-1/2. So r(x)=2-1/2*e-x.

The arc length is given by the formula:
Integral from 0 to infinity of ((f(x))2+(f'(x))2)1/2 =
Integral from 0 to infinity of ((2-1/2*e-x)2 + (-2-1/2*e-x)2)1/2 =
Integral from 0 to infinity of (2-1*e-2x + 2-1*e-2x)1/2 =
Integral from 0 to infinity of e-x =
-e-x from 0 to infinity =
=0 - (-e0) =
0 -(-1) = 1 The path the dogs take is called a logarithmic spiral. It is an interesting paradox that the dogs will make an infinite number of circles, yet the total distance is constant. Every revolution the size of the square will to e-2*pi/21/2 = 0.03055663 times it's size before the revolution.



--------------------------------------------------------------------------------


Easy Solution
David Wilson suggested the more simple solution which follows...

Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

Antonio Molins went further to state that as dog A is heading direcctly towards dog B, dog B is moving towards from dog A at rate equal to cosine of any of the interior angles of the shape made by the dogs. He also provided the following graphics.

For example consider a pentagon. Remember that for a n-gon of n sides each angle will be 180*(1-(2/n)). So a pentagon will have angles of 108. Let's assume each dog walks at a rate of 1 unit per minute, where the initial shape has sides of 1 unit. As pointed out by David Wilson we can forget about the spiral paths but instead think of the dogs walking in straight lines. In the case of the penagon for every one unit dog A draw towards dog B's initial position, dog B will draw cos(180)=-0.309 units towards dog A. The question is, how long will it take for them to meet? Let the answer to that question be t. In t seconds dog A will cover t units, and dog B will cover .309*t units. The total distance traveled by A equals the distance traveled by B plus 1 (the initial separation). So solving for t: t=0.309+1 --> t=1/(1-.309) --> t=1.4472.

Thus the general formula is 1/(1+cos(t)), where t is an interior angle of the shape.



Number of sides = 3
Lengh of initial side = 1
Each angle = 180*(1-(2/3)) = 60
Length of path = 1/(1+cos(60)) = 1/(1+0.5) = 2/3



Number of sides = 4
Lengh of initial side = 1
Each angle = 180*(1-(2/4)) = 90
Length of path = 1/(1+cos(90)) = 1/(1+0) = 1



Number of sides = 5
Lengh of initial side = 1
Each angle = 180*(1-(2/5)) = 108
Length of path = 1/(1+cos(108)) = 1/(1-.309) = 1.4472



Number of sides = 6
Lengh of initial side = 1
Each angle = 180*(1-(2/6)) = 120
Length of path = 1/(1+cos(120)) = 1/(1-.5) = 2



Number of sides = 10
Lengh of initial side = 1
Each angle = 180*(1-(2/10)) = 144
Length of path = 1/(1+cos(144)) = 1/(1-.809) = 5.2361



Number of sides = 15
Lengh of initial side = 1
Each angle = 180*(1-(2/15)) = 156
Length of path = 1/(1+cos(156)) = 1/(1-.9135) = 11.5668

Michael Shackleford, A.S.A.

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Offline Alan McDougall

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« Reply #80 on: 23/07/2008 15:15:50 »
LightArrow

I am impressed by your solutions to these difficult problems!

I could give you the link by PM, your maths is better than mine.

perhaps you could then post some math problems from this site

Regards

Alan

 

Offline lightarrow

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« Reply #81 on: 23/07/2008 17:31:58 »
LightArrow

I am impressed by your solutions to these difficult problems!

I could give you the link by PM, your maths is better than mine.

perhaps you could then post some math problems from this site

Regards

Alan


Yes, I'm interested, thank you.
 

lyner

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« Reply #82 on: 24/07/2008 11:37:50 »
Quote
Easy Solution
David Wilson suggested the more simple solution which follows...
Suppose dog A is pursuing dog B who is pursuing dog C. During the entire pursuit, the dogs remain at the corners of a square, and angle ABC is a constant 90 degrees. That is, B's path towards C is always perpendicular to A's path towards B, and B has zero velocity in the direction along A's path. Since B neither approaches nor recedes from A during the walk, A simply covers the intial separation of 1.

It cannot be as simple as that; if I am chasing someone and, eventually, I catch them, you can't say that the distance I have run is just our initial separation; I may have to run miles and miles before catching them.
 

Offline Alan McDougall

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« Reply #83 on: 24/07/2008 11:41:39 »
Sophie correct it is not as simple as that, read light arrows postshe made a valiant effort.

The easy solution just gives an approximation
 

lyner

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« Reply #84 on: 24/07/2008 13:08:59 »
It does give a sort of lower limit, I grant you, but that's about all.
The shortest way for them to get together (the actual lower limit) is if they all aim at the centre - then they travel Root 2 of their initial separation.
« Last Edit: 24/07/2008 13:19:26 by sophiecentaur »
 

Offline lightarrow

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« Reply #85 on: 29/07/2008 20:26:58 »
Do you know this problem:

There is a solid sphere (full inside) and you make a hole in it with a drill which go straight through the centre of the sphere to the opposite side (you probably have in english a better way to describe such a kind of hole).

The hole is 6 cm long.
Which is the remaining (residual) volume of the sphere?

"But, wait a moment, you didn't tell us the sphere's radius!" Exactly!
« Last Edit: 29/07/2008 23:35:05 by lightarrow »
 

lyner

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« Reply #86 on: 29/07/2008 20:54:38 »
Are you sure there's nothing else you need to tell us?
Why couldn't you drill a 6cm diameter hole through the Earth or through a football?

 

Offline LeeE

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« Reply #87 on: 29/07/2008 23:10:49 »
I think this is the problem lightarrow's thinking of:

Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.

Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!

This can be worked out mathematically or logically (I solved it via logic).
 

Offline lightarrow

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« Reply #88 on: 29/07/2008 23:34:17 »
Are you sure there's nothing else you need to tell us?
Why couldn't you drill a 6cm diameter hole through the Earth or through a football?
Sorry, it's 6 cm long, not in diametre. I've correct the post.
 

Offline lightarrow

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« Reply #89 on: 29/07/2008 23:36:03 »
I think this is the problem lightarrow's thinking of:

Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.

Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!

This can be worked out mathematically or logically (I solved it via logic).
Exactly that problem, right.
 

lyner

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« Reply #90 on: 30/07/2008 07:46:09 »
So we can assume the answer is always the same so an infinitely thin drill through a 6cm diameter sphere would leave 4XPiX33/4 material behind - i.e. taking nothing out of the sphere.
Proving that it works for a range of sphere sizes is a bit harder - I will try it sometime. The question implies that original sphere could be ANY radius greater than 3cm. Wierd.
 

Offline lightarrow

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« Reply #91 on: 30/07/2008 13:24:09 »
So we can assume the answer is always the same so an infinitely thin drill through a 6cm diameter sphere would leave 4XPiX33/4 material behind - i.e. taking nothing out of the sphere.
Proving that it works for a range of sphere sizes is a bit harder - I will try it sometime. The question implies that original sphere could be ANY radius greater than 3cm. Wierd.
(4/3)π33 = 36π.
Apart the mistake in the formula for a sphere's volume ( :)) you answered correctly. Clearly with more computations you can show that the remaining volume doesn't depend on the radius (if ≥ 3). Weird, yes.
 

lyner

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« Reply #92 on: 30/07/2008 13:44:19 »
I was so pre-occupied with putting in the supercripts that I missed that!
Durrr!
It could work, approximately, for olives too!
 

Offline BenV

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« Reply #93 on: 30/07/2008 14:53:29 »
Hold on, you guys have confused me...

If you were to drill a hole through a sphere, right through the centre, and the hole was 6cm long, then the sphere must have a diameter of 6cm , no?

In which case, surely we need to know the diameter of the bore hole to determine how much volume is lost - i.e. volume of (6cm diameter) sphere, minus volume of (6cm length, n diameter) cylinder?

Edit - I'm basing this on the idea that drilling through something involves coming out of the other side - drilling 6cm into something is different.

Edit2 - And going back to the horse thing - if 'the stake wasn't hammered into the ground' is a valid answer, then so is 'he chewed through the rope', 'the stake was between him and the hay' and 'a kindly old man was passing and moved the hay closer to him'.

Edit3 - I've just realised that I may be a pedant...
« Last Edit: 30/07/2008 16:08:44 by BenV »
 

Offline lightarrow

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« Reply #94 on: 30/07/2008 17:39:41 »
Hold on, you guys have confused me...

If you were to drill a hole through a sphere, right through the centre, and the hole was 6cm long, then the sphere must have a diameter of 6cm , no?
Make a draw of the bored sphere seen from a lateral view. How do you  measure the hole's lenght?
« Last Edit: 30/07/2008 17:50:53 by lightarrow »
 

Offline LeeE

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« Reply #95 on: 30/07/2008 22:52:19 »
The question specifically doesn't mention the diameter of the hole, so assuming that it's not a trick question, which it isn't, the significance of not specifying the hole diameter is that the answer is the same, regardless of the diameter of the hole.

Thus, sophiecentaur's approach is the correct logical solution.

As long as the length of the hole remains the same, any diameter will give the same answer.
 

Offline Alan McDougall

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« Reply #96 on: 31/07/2008 11:45:49 »
Guys,
I decided to post the solution,





The Hardest Logic Puzzle Ever

Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.

Boolos provides the following clarifications:[1]

It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
Random will answer 'da' or 'ja' when asked any yes-no question.[1]






Hence whenever you ask them a yes-no question, they reply 'Bal' or 'Da' - one of which means yes and the other no. The trouble is that we do not know which of 'Bal' or 'Da' means yes and which means no". There are other related puzzles in The Riddle of Scheherazade (see especially, p. 114).

More generally this puzzle is based on Smullyan's famous Knights and Knaves puzzles (e.g. on a fictional island, all inhabitants are either knights, who always tell the truth, or knaves, who always lie. The puzzles involve a visitor to the island who must ask a number of yes/no questions in order to discover what he needs to know).

A version of these puzzles was popularized by a scene in the 1980's fantasy film, Labyrinth. There are two doors with two guards. One guard lies and one guard doesn't. One door leads to the castle and the other leads to "certain death". The puzzle is to find out which door leads to the castle by asking one of the guards one question. In the movie Sarah does this by asking "Would he tell me that this door leads to the castle?".


[edit] The solution
Boolos provided his solution in the same article in which he introduced the puzzle. Boolos states that the "first move is to find a god that you can be certain is not Random, and hence is either True or False".[1] There are many different questions that will achieve this result. One strategy is to use complicated logical connectives in your questions (either biconditionals or some equivalent construction).

Boolos' question was:

Does 'da' mean yes if and only if you are True if and only if B is Random?[1]
Equivalently:

Are an odd number of the following statements true: you are False, 'ja' means yes, B is Random?
The puzzle's solution can be simplified by using counterfactuals.[2][3] The key to this solution is that, for any yes/no question Q, asking either True or False the question

If I asked you Q, would you say 'ja'?
results in the answer 'ja' if the truthful answer to Q is yes, and the answer 'da' if the truthful answer to Q is no. The reason this works can be seen by looking at the eight possible cases.

Assume that 'ja' means yes and 'da' means no.
(i) True is asked and responds with 'ja'. Since he is telling the truth the truthful answer to Q is 'ja', which means yes.

(ii) True is asked and responds with 'da'. Since he is telling the truth the truthful answer to Q is 'da', which means no.

(iii) False is asked and responds with 'ja'. Since he is lying it follows that if you asked him Q he would instead answer 'da'. He would be lying, so the truthful answer to Q is 'ja', which means yes.

(iv) False is asked and responds with 'da'. Since he is lying it follows that if you asked him Q he would in fact answer 'ja'. He would be lying, so the truthful answer to Q is 'da', which means no.

Assume 'ja' means no and 'da' means yes.
(v) True is asked and responds with 'ja'. Since he is telling the truth the truthful answer to Q is 'da', which means yes.

(vi) True is asked and responds with 'da'. Since he is telling the truth the truthful answer to Q is 'ja', which means no.

(vii) False is asked and responds with 'ja'. Since he is lying it follows that if you asked him Q he would in fact answer 'ja'. He would be lying, so the truthful answer to Q 'da', which means yes.

(viii) False is asked and responds with 'da'. Since he is lying it follows that if you asked him Q he would instead answer 'da'. He would be lying, so the truthful answer to Q is 'ja', which means no.

Using this fact, one may proceed as follows.[2]

Ask god B, "If I asked you 'Is A Random?', would you say 'ja'?". If B answers 'ja', then either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers 'da', then either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, A is not Random.


Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you True?', would you say 'ja'?". Since he is not Random, an answer of 'ja' indicates that he is True and an answer of 'da' indicates that he is False.
Ask the same god the question: "If I asked you 'Is B Random?', would you say 'ja'?". If the answer is 'ja' then B is Random; if the answer is 'da' then the god you have not yet spoken to is Random. The remaining god can be identified by elimination.

[edit] Random's behaviour
Most readers of the puzzle assume that Random will provide completely random answers to any question asked of him; however, the puzzle does not actually state this. In fact, Boolos' third clarifying remark explicitly refutes this assumption.[2]

Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
This says that Random randomly acts as a liar or a truth-teller, not that Random answers randomly.

A small change to the question above yields a question which will always elicit a meaningful answer from Random. The change is as follows:

If I asked you Q in your current mental state, would you say 'ja'?[2]
We have effectively extracted the truth-teller and liar personalities from Random and forced him to be only one of them. This completely trivializes the puzzle since we can now get truthful answers to any questions we please.

1. Ask god A, "If I asked you 'Are you Random?' in your current mental state, would you say 'ja'?"
If A answers 'ja', then A is Random:


2a. Ask god B, "If I asked you 'Are you True?', would you say 'ja'?"
If B answers 'ja', then B is True and C is False.

If B answers 'da', then B is False and C is True. In both cases, the puzzle is solved.

If A answers 'da', then A is not Random:


2b. Ask god A, "If I asked you 'Are you True?', would you say 'ja'?"
If A answers 'ja', then A is True.

If A answers 'da', then A is False.


3. Ask god A, "If I asked you 'Is B Random?', would you say 'ja'?"
If A answers 'ja', then B is Random, and C is the opposite of A.

If A answers 'da', then C is Random, and B is the opposite of A.

We can modify Boolos' puzzle so that Random is actually random by replacing Boolos' third clarifying remark with the following.

Whether Random says 'ja' or 'da' should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he says 'ja'; if tails, he says 'da'.
With this modification, the puzzle's solution demands the more careful god-interrogation given at the end of the The Solution section.


[edit] Exploding god-heads
In A simple solution to the hardest logic puzzle ever,[2] B. Rabern and L. Rabern develop the puzzle further by pointing out that it is not the case that 'ja' and 'da' are the only possible answers a god can give. It is also possible for a god to be unable to answer at all. For example, if the question "Are you going to answer this question with the word that means no in your language?" is put to True, he cannot answer truthfully. (The paper represents this as his head exploding, "...they are infallible gods! They have but one recourse their heads explode") Allowing the "exploding head" case gives yet another solution of the modified puzzle (modified so that Random is actually random) and introduces the possibility of solving the original puzzle (unmodified) in just two questions rather than three. In support of a two-question solution to the puzzle, the authors solve a similar simpler puzzle using just two questions.

Three gods A, B, and C are called, in some order, Zephyr, Eurus, and Aeolus. The gods always speak truly. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English and will answer in English.
Note that this puzzle is trivially solved with three questions (just ask away!). To solve the puzzle in two questions, the following lemma is proved.

Tempered Liar Lemma. If we ask A "Is it the case that {[(you are going to answer 'no' to this question) AND (B is Zephyr)] OR (B is Eurus) }?", a response of 'yes' indicates that B is Eurus, a response of 'no' indicates that B is Aeolus, and an exploding head indicates that B is Zephyr. Hence we can determine the identity of B in one question.

Using this lemma it is simple to solve the puzzle in two questions. A similar trick (tempering the liar's paradox) can be used to solve the original puzzle in two questions.


 

Offline lightarrow

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Try solving these easy problems by deductive thinking
« Reply #97 on: 01/08/2008 13:01:22 »
I decided to post the solution,
...
The Hardest Logic Puzzle Ever
...
GULP!

Ehm...
Something a bit more simple:

A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
 

Offline lightarrow

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« Reply #98 on: 02/08/2008 14:30:36 »
A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?
I know it's simple, but the result is quite amazing.

Anyway, if you don't like that problem, try this:

you have two fuses which burn exactly in one hour each; they don't burn in a regular way, however, so you can't say that half lenght will burn in half an hour, and so on.
The problem is that you have to measure exactly 45 minutes. How will you do it?
 

Offline Chemistry4me

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« Reply #99 on: 20/11/2008 06:52:08 »
[quote author=Alan McDougall link=topic=15562.msg188243#msg188243 date=1217501149

Ehm...

A sample of 100 kg of potatoes has a water content of 99%. It is put under the sun to dry a little, and the water content at the end becomes 98%. How does the potatoes weight now?

Is the water content 97.02%? Do the potatoes weigh 98kgs?
 

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Try solving these easy problems by deductive thinking
« Reply #99 on: 20/11/2008 06:52:08 »

 

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