Math Fun

For discussing anything related to physics, biology, chemistry, mathematics, and their practical applications.

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Re: Math Fun

Postby Tralix » Fri Jan 18, 2013 1:01 am

Silhouette wrote:
Tralix wrote:It isn't a problem an axiom is not an issue unless you make it one by semantics and then everyone will just think you are a crank.

The problem is solved and successfully disputed according to all science and maths. The fact that you don't understand it, is possibly interesting but not a reason to really discuss your ignorance with you.

I am a crank.
This is not the thread to discuss "my" ignorance, no, whether with or without appeals to authority rather than actual arguments. I'll drop it on the condition that you come up with more lovely puzzles :D I apologise that I know of none.


:)

Sounds like a deal. I will. :D

http://www.cut-the-knot.org/Probability ... ulbs.shtml

Some time ago, Ilia Denotkine has posted the following problem on the CTK Exchange
There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner equally at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world could always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?


Tough one this.

and:

Monkeying around

300 monks live together in a monastery. They have very strict rules which are followed by all of the monks at all times. One of the rules is, that absolutely no communication between monks is allowed. Another is, that mirrors are forbidden. The monks have their three meals a day together in a large hall, the rest of their day is spent with individual contemplation and chores.

One morning, a messenger comes to the monastery and addresses the monks at breakfast. He tells them, that a rare disease is spread throughout the country, and that the monks may have the disease as well. The main symptom of the disease is a large red spot on the head of the afflicted. The disease kills everyone who knows they have it within two hours. The disease was transmitted by a bad shipment of rice, but is not contagious.

On the morning of the eleventh day after the messenger arrives, some of the monks don't turn for breakfast and are found dead in their beds.

Question: How many monks died?


Easier. :)
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"He that cannot obey cannot command."

Benjamin Franklin.

"If you ever actually ask a question about the topic itself, I'll be glad to give it consideration. But it is more than obvious that the topic is not your interest."

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Re: Math Fun

Postby Tralix » Sat Jan 19, 2013 1:34 am

I find it hard to believe that no one can remotely guess at both these problems. They're both intimately related to logic and maths?

Ok clue for both first one parallel advances lead to serial conclusions.

Monks meh this is just straight logic. Not giving a clue to this if you can't solve it, you're not a philosopher.
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Re: Math Fun

Postby Carleas » Sun Jan 20, 2013 5:04 am

I have some questions about the 2nd problem, and also some guesses (more than one because they depend on the answers to the questions).
Does the messenger say that at least one monk definitely has the disease, or only that they "may"? If it's "may," I do not think any monks would die (assuming the question isn't looking for an answer like "however many ate the rice").

If they know that at least one monk has the disease:
Are we counting the day the messenger came as day 1 or day 0? The monks see each other three times a day, so they seen each other either 33 or 30 times depending. And the number of dead monks will be equal to whichever (again, assuming they are told that at least one monk has the disease).

I don't think this as easy a question as you make it out to be.


Still working on the first problem.
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Re: Math Fun

Postby Carleas » Sun Jan 20, 2013 6:55 pm

I don't think I'm equipped to think the first problem through on my own, so I'm going to think here in hopes that someone can make something of my ramblings.

The prisoners can't count the number of people who have been to the room, because they can only transmit 0 or 1 to the next prisoner. They can, however, count the days that pass, and count themselves prior to going in. The days might be useful, espcially since they know that prior to day 100, they won't be getting out. Themselves probably isn't, since they can't really communicate their prisoner number using a bulb with only two states.

The thing I keep getting caught up on is how to deal with multiple repeat visits. The first repeat visit can be passed along by chaning 0 to 1. But the second repeat visit will break that signal, and any visitor that comes after that won't be able to tell if it's been switched and switched back, or if it was never switched in the first place.

I think the best place to start is with a smaller set, say 3 prisoners.
Call the first prisoner to be picked A, and the other two B and C.
After A is picked, either A, B, or C could be picked, and so on ad infinitum.

So we have a tree that looks like this:
________________________________________________________A_________________________________________________________
_________________________/_______________________________|____________________________\_____________________________
________________________A_______________________________B____________________________C____________________________
__________/______________|_______\__________/_____________|____________\____________/__________|____________\__________
_________A______________B_______C________A_____________B___________C__________A___________B___________C_________

In the tree, only the path that ends in the light blue C or the medium green B would be successful in saying that everyone had been to the room.
One thing to note is that this tree can be traversed recursively. If on the second day, the warden picks A again, it's exactly the same situation as it was on the first day.
But if the warden picks A on the third day after picking B or C, it isn't the same as the first day, but the same as the second. Unfortunately, it doesn't seem like this can be communicated. Just using 0 or 1, we can't convey which branch of the tree we're on, because already on the second day there are three possible branches, and only two possible light bulb states.


So we can only convey one thing to one person at a time. That means the answer is going to involve something like the lightbulb communicating "add 1", or maybe something more veriable like "add today's number." The general idea is that each prisoner is keeping his or her own count, and the lightbulb tells him or her to perform some action on that count. Then there's a rule about turning on the light, something like "turn it on if this is your first visit." But "add 1" won't work with "turn it on if this is your first visit": in the three prisoner case, if it went A, B, C, no matter what happens after that, A's count would be 0, B's would be 1 and C's would be 1. They would never get to three (assuming that's what they're getting to, which it probably isn't but needn't be).

That what I have so far.
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Re: Math Fun

Postby socratus » Fri Feb 22, 2013 8:24 am

Abstract wrote:Rules:

Post a math problem that you have the solution to in bold.
For those who want to answer the problem provide your solution
in a tab so as not to ruin it for others.

Rules subject to change. PM me if you have anything that should be added
or removed or any suggestions for this thread in general.

Euler's Equation and the Reality of Nature.
=.
a)
Euler's Equation as a mathematical reality.

Euler's identity is "the gold standard for mathematical beauty'.
Euler's identity is "the most famous formula in all mathematics".
‘ . . . this equation is the mathematical analogue of Leonardo
da Vinci’s Mona Lisa painting or Michelangelo’s statue of David’
‘It is God’s equation’, ‘our jewel ‘, ‘ It is a mathematical icon’.
. . . . etc.
b)
Euler's Equation as a physical reality.

"it is absolutely paradoxical; we cannot understand it,
and we don't know what it means, . . . . .’
‘ Euler's Equation reaches down into the very depths of existence’
‘ Is Euler's Equation about fundamental matters?’
‘It would be nice to understand Euler's Identity as a physical process
using physics.‘
‘ Is it possible to unite Euler's Identity with physics, quantum physics ?’
==.

My aim is to understand the reality of nature.
Can Euler's equation explain me something about reality?
To give the answer to this. question I need to bind
Euler's equation with an object – particle.
Can it be math- point or string- particle or triangle-particle?
No, Euler's formula has quantity (pi) which says me that
the particle must be only a circle .
Now I want to understand the behavior of circle - particle and
therefore I need to use spatial relativity and quantum theories.
These two theories say me that the reason of circle – particle’s
movement is its own inner impulse (h) or (h*=h/2pi).
a)
Using its own inner impulse (h) circle - particle moves
( as a wheel) in a straight line with constant speed c = 1.
We call such particle - ‘photon’.
From Earth – gravity point of view this speed is maximally.
From Vacuum point of view this speed is minimally.
In this movement quantum of light behave as a corpuscular (no charge).
b)
Using its own inner impulse / intrinsic angular momentum
( h* = h / 2pi ) circle - particle rotates around its axis.
In such movement particle has charge, produce electric waves
( waves property of particle) and its speed ( frequency) is : c>1.
We call such particle - ‘ electron’ and its energy is: E=h*f.
In this way I can understand the reality of nature.
==.
P.S.
' They would play a greater and greater role in mathematics –
and then, with the advent of quantum mechanics in the twentieth
century, in physics and engineering and any field that deals with
cyclical phenomena such as waves that can be represented by
complex numbers. For a complex number allows you to represent
two processes such as phase and wavelenght simultaneously –
and a complex exponential allows you to map a straight line
onto a circle in a complex plane.'
/ Book: The great equations. Chapter four.
The gold standard for mathematical beauty.
Euler’s equation. Page 104. /
#
Euler's e-iPi+1=0 is an amazing equation, not in-and-of itself,
but because it sharply points to our utter ignorance of the
simplest mathematical and scientific fundamentals.
The equation means that in flat Euclidean space, e and Pi happen
to have their particular values to satisfy any equation that relates
their mathematical constructs. In curved space, e and Pi vary.
/ Rasulkhozha S. Sharafiddinov . /
===…
P.S.
Love is the gravity of the soul.
/ by Abstract /
=.
There is gravity attraction in the universe.
And there is love attraction in the world.
Physicists say that the gravity attraction power
is the smallest power.
Then, maybe, the love attraction is stronger than gravity attraction.
=.
The secret of God and Life is hiding in the ' Theory of Light Quanta.'
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Re: Math Fun

Postby Abstract » Thu Feb 28, 2013 2:29 pm

here is a problem for yall:

You have a dial that goes from 1 to 9 and the dial steps by one each time ..from 1 to two is one step from two to 3 is one step... without actually setting there and counting all the times it steps and without using divesion or multiplication but only addition if you start at one and take 123459999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999992 steps what number will you land on?
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Re: Math Fun

Postby Flannel Jesus » Thu Feb 28, 2013 4:31 pm

Tralix wrote:
300 monks live together in a monastery. They have very strict rules which are followed by all of the monks at all times. One of the rules is, that absolutely no communication between monks is allowed. Another is, that mirrors are forbidden. The monks have their three meals a day together in a large hall, the rest of their day is spent with individual contemplation and chores.

One morning, a messenger comes to the monastery and addresses the monks at breakfast. He tells them, that a rare disease is spread throughout the country, and that the monks may have the disease as well. The main symptom of the disease is a large red spot on the head of the afflicted. The disease kills everyone who knows they have it within two hours. The disease was transmitted by a bad shipment of rice, but is not contagious.

On the morning of the eleventh day after the messenger arrives, some of the monks don't turn for breakfast and are found dead in their beds.

Question: How many monks died?


Easier. :)

I've seen a similar problem to this before, but as hinted in Carleas's question, you missed an essential detail (one that was given in the similar problem I saw):
The messenger said "the monks may have the disease as well," but in order for the problem to work, what he should have said is "At least one of the monks does, in fact, have the disease." Otherwise none of the monks has to die.

But, assuming he did say that, I think the answer is around 30 (assuming that the period of time between meals is over 2 hours).

[edit]
come to think of it, maybe he doesn't need to say that...I'm not sure.
In any case, my answer remains the same.

[edit2]
I'm still not sure, but I'm pretty confident he does need to say something to that effect...
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Re: Math Fun

Postby Flannel Jesus » Thu Feb 28, 2013 4:50 pm

Abstract wrote:here is a problem for yall:

You have a dial that goes from 1 to 9 and the dial steps by one each time ..from 1 to two is one step from two to 3 is one step... without actually setting there and counting all the times it steps and without using divesion or multiplication but only addition if you start at one and take 123459999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999992 steps what number will you land on?

I think it's 8, assuming that you meant to include the information that it steps from 9 to 1.
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Re: Math Fun

Postby Abstract » Thu Feb 28, 2013 11:01 pm

supposed to tab your answers... but yes your right.
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Re: Math Fun

Postby Silhouette » Thu Feb 28, 2013 11:06 pm

Abstract wrote:here is a problem for yall:

You have a dial that goes from 1 to 9 and the dial steps by one each time ..from 1 to two is one step from two to 3 is one step... without actually setting there and counting all the times it steps and without using divesion or multiplication but only addition if you start at one and take 123459999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999999
9999999999999999999999999999999999999999999999999999999999999999999992 steps what number will you land on?

This requires a conversion of this decimal number to nonal, and observing only the last digit (and adding it to the starting number in nonal).

But there is a simpler way to solve this puzzle as explained here (that achieves the same result):

One can simply rely on a property of decimal numbers and their divisibility by 9.
To illustrate this property: 9 steps would result in no change from the start number, and the same with 9 more steps (18 steps) etc. such that any number of steps divisible by 9 results in no change from the start number. Adding together the digits in decimal numbers divisible by 9 (e.g. 18, 27, 90, 99, 108, 900, 999, 99999999 etc.) always gives another number that is divisible by 9. Continually applying the same addition to the result, in case it is not a single digit result, will always eventually result in 0 if the number is divisible by 9.
If the number is not divisible by 9, the eventual result will be the remainder - or in the case of this puzzle, the offset from the starting point. This property allows us to eliminate the need for division as in the rules for solving this puzzle.

Applying this property to a number with any 9s in it makes any 9s redundant (since adding together any number of 9s with any number will not change the remainder when dividing the number by 9). So we can eliminate all the 9s in the huge beginning number, such that we are essentially dividing 123452 by 9 - or adding 1+2+3+4+5+2 to make 17 and then adding 1+7 to make 8.

Therefore taking the specified huge number of steps around a dial that goes from 1-9 will eventually offset the starting number by 8 steps.
Starting from 1, as the puzzle asks, and advancing 8 steps (to 2,3,4,5,6,7,8,9) will land us on 9.

So I'm sorry, FJ but you're wrong! (Due to a condition you forgot, mentioned in the last line of my above tabbed explanation.)
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Re: Math Fun

Postby Abstract » Fri Mar 01, 2013 3:03 pm

silloette definitely got it.
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Re: Math Fun

Postby Flannel Jesus » Fri Mar 01, 2013 5:25 pm

I don't see what makes me wrong in the last line...
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Re: Math Fun

Postby Silhouette » Fri Mar 01, 2013 11:48 pm

Flannel Jesus wrote:I don't see what makes me wrong in the last line...

You'd have been right if the starting number was 9 (plus 8 sets the nonal dial to number 8 ). But it's 1 (plus 8 sets the dial to number 9).

Tralix wrote:
Some time ago, Ilia Denotkine has posted the following problem on the CTK Exchange
There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner equally at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world could always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?

I've finally got this one. Took me over 40 days... sheesh. (Albeit not thinking about it for much of that time by any means! I think I've only sat down and thought about it seriously about 3 or 4 times, but especially counting today this still amounts to a fair while. The one night of planning that the prisoners get probably wouldn't have been enough for me.)

The prisoners must nominate a single "manager" to keep count of the number of times the lightbulb is switched OFF.
It is agreed that this "manager" will be the only prisoner who will ever switch ON the lightbulb. He will never switch it off.
Accordingly, the other 99 prisoners will never switch on the lightbulb, and they will only ever switch it OFF once each.

These are the rules, and things will unfold thus:

Non-manager prisoners can be chosen to visit the central living room any number of times before the manager's first visit, but the lightbulb must remain OFF.
The first time the manager visits the room, the lightbulb will therefore still be in its initial OFF state. He switches ON the lightbulb.
If the manager is chosen two or more days in a row, he will find the lightbulb still ON. He will leave it on.
If a non-manager is chosen after the manager, he will find the lightbulb on and he will turn it OFF. He will never again turn the lightbulb either on or off.
If a non-manager prisoner follows a non-manager prisoner (whether himself or not), he will leave the lightbulb off.
The next time the manager enters the central living room after a non-manager, he will find the lightbulb OFF. He will count +1, and switch the lightbulb ON.

This pattern will continue until the manager has counted 99 non-manager prisoners. The next time he is picked after this point, he can be sure that the other 99 prisoners have been to the room by now, and now that he has (yet again), he can assert with 100% certainty that all 100 prisoners have been to the living room by now. Hello freedom and MENSA.

This works because the binary positions of the lightbulb account for 1: the manager and 0: everyone else. The manager can (and of course must be able to) count the zeroes (and himself, the one 1).

Tralix wrote:Monks meh this is just straight logic. Not giving a clue to this if you can't solve it, you're not a philosopher.

Still flummoxed about this one, so I guess I'm not a philosopher. Maybe your clue about not being a philosopher if you can't solve it is due to the whole lack of mirrors problem and identity issues associated with mirrors. But if that's what you were getting at, I still don't get it.

P.S. In my tabbed explanation for the nonal dial puzzle, I seem to have implied that dividing 9 by 9 gives you 0. Obviously a mistake, I meant 1.
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Re: Math Fun

Postby socratus » Sat Mar 09, 2013 7:50 am

Why Math Works ?
Because math tied with physics.
For example:
I say that there is circle-particle that can change /
transformed into sphere-particle and vice versa
and Euler’s equation cosx + isinx in = e^ix can explain
this transformation / fluctuation of quantum particle.
I try to understand more details.
I have circle- particle with two infinite numbers: (pi) and (e).
I say that this circle-particle that can change into sphere-particle
and vice versa. Then I need third number for these changes.
The third number, in my opinion, is infinite a=1/137
( the fine structure constant = the limited volume coefficient)
This coefficient (a=1/137) is the border between two
conditions of quantum particle. This coefficient (a=1/137) is
responsible for these changes. This coefficient (a=1/137) united
geometry with the physics ( e^2=ah*c)
=..
If physicists use string-particle (particle that has length but
hasn’t thickness -volume) to understand reality
(and have some basic problems to solve this task) then why don’t
use circle-particle for this aim.
It is a pity that I am not physicist or mathematician.
If I were mathematician or physicist I wouldn’t lost the chance
to test this hypothesis.
=..
Best wishes.
Israel Sadovnik Socratus
==…
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Re: Math Fun

Postby Helandhighwater » Wed Mar 20, 2013 11:38 pm

Flannel Jesus wrote:
Tralix wrote:
300 monks live together in a monastery. They have very strict rules which are followed by all of the monks at all times. One of the rules is, that absolutely no communication between monks is allowed. Another is, that mirrors are forbidden. The monks have their three meals a day together in a large hall, the rest of their day is spent with individual contemplation and chores.

One morning, a messenger comes to the monastery and addresses the monks at breakfast. He tells them, that a rare disease is spread throughout the country, and that the monks may have the disease as well. The main symptom of the disease is a large red spot on the head of the afflicted. The disease kills everyone who knows they have it within two hours. The disease was transmitted by a bad shipment of rice, but is not contagious.

On the morning of the eleventh day after the messenger arrives, some of the monks don't turn for breakfast and are found dead in their beds.

Question: How many monks died?


Easier. :)

I've seen a similar problem to this before, but as hinted in Carleas's question, you missed an essential detail (one that was given in the similar problem I saw):
The messenger said "the monks may have the disease as well," but in order for the problem to work, what he should have said is "At least one of the monks does, in fact, have the disease." Otherwise none of the monks has to die.

But, assuming he did say that, I think the answer is around 30 (assuming that the period of time between meals is over 2 hours).

[edit]
come to think of it, maybe he doesn't need to say that...I'm not sure.
In any case, my answer remains the same.

[edit2]
I'm still not sure, but I'm pretty confident he does need to say something to that effect...


No actually that is not necessary in the more difficult version of the problem, it does as you say not matter if you say at least 1 of the monks has the disease as it will become clear if you work through it. Clearly someone has it or the answer would not be how many people die, but why is everyone still alive? We know some monks are found dead in their beds, and we know some monks may have the disease. I know semantics, but it is important.


I'll let this run further before I say what the answer is.
Last edited by Helandhighwater on Wed Mar 20, 2013 11:47 pm, edited 3 times in total.
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Re: Math Fun

Postby Helandhighwater » Wed Mar 20, 2013 11:43 pm

Carleas wrote:I don't think I'm equipped to think the first problem through on my own, so I'm going to think here in hopes that someone can make something of my ramblings.

The prisoners can't count the number of people who have been to the room, because they can only transmit 0 or 1 to the next prisoner. They can, however, count the days that pass, and count themselves prior to going in. The days might be useful, espcially since they know that prior to day 100, they won't be getting out. Themselves probably isn't, since they can't really communicate their prisoner number using a bulb with only two states.

The thing I keep getting caught up on is how to deal with multiple repeat visits. The first repeat visit can be passed along by chaning 0 to 1. But the second repeat visit will break that signal, and any visitor that comes after that won't be able to tell if it's been switched and switched back, or if it was never switched in the first place.

I think the best place to start is with a smaller set, say 3 prisoners.
Call the first prisoner to be picked A, and the other two B and C.
After A is picked, either A, B, or C could be picked, and so on ad infinitum.

So we have a tree that looks like this:
________________________________________________________A_________________________________________________________
_________________________/_______________________________|____________________________\_____________________________
________________________A_______________________________B____________________________C____________________________
__________/______________|_______\__________/_____________|____________\____________/__________|____________\__________
_________A______________B_______C________A_____________B___________C__________A___________B___________C_________

In the tree, only the path that ends in the light blue C or the medium green B would be successful in saying that everyone had been to the room.
One thing to note is that this tree can be traversed recursively. If on the second day, the warden picks A again, it's exactly the same situation as it was on the first day.
But if the warden picks A on the third day after picking B or C, it isn't the same as the first day, but the same as the second. Unfortunately, it doesn't seem like this can be communicated. Just using 0 or 1, we can't convey which branch of the tree we're on, because already on the second day there are three possible branches, and only two possible light bulb states.


So we can only convey one thing to one person at a time. That means the answer is going to involve something like the lightbulb communicating "add 1", or maybe something more veriable like "add today's number." The general idea is that each prisoner is keeping his or her own count, and the lightbulb tells him or her to perform some action on that count. Then there's a rule about turning on the light, something like "turn it on if this is your first visit." But "add 1" won't work with "turn it on if this is your first visit": in the three prisoner case, if it went A, B, C, no matter what happens after that, A's count would be 0, B's would be 1 and C's would be 1. They would never get to three (assuming that's what they're getting to, which it probably isn't but needn't be).

That what I have so far.


That's nice work but you or them or whatever need to work it through a little further. There is a way.

By the way when I say easier their both not easy but it is a matter of realtive ease. I solved both 1 on my own, the other with some help. No man is an island.
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Re: Math Fun

Postby Helandhighwater » Thu Mar 21, 2013 12:35 am

Silhouette wrote:
Flannel Jesus wrote:I don't see what makes me wrong in the last line...

You'd have been right if the starting number was 9 (plus 8 sets the nonal dial to number 8 ). But it's 1 (plus 8 sets the dial to number 9).

Tralix wrote:
Some time ago, Ilia Denotkine has posted the following problem on the CTK Exchange
There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner equally at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world could always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?

I've finally got this one. Took me over 40 days... sheesh. (Albeit not thinking about it for much of that time by any means! I think I've only sat down and thought about it seriously about 3 or 4 times, but especially counting today this still amounts to a fair while. The one night of planning that the prisoners get probably wouldn't have been enough for me.)

The prisoners must nominate a single "manager" to keep count of the number of times the lightbulb is switched OFF.
It is agreed that this "manager" will be the only prisoner who will ever switch ON the lightbulb. He will never switch it off.
Accordingly, the other 99 prisoners will never switch on the lightbulb, and they will only ever switch it OFF once each.

These are the rules, and things will unfold thus:

Non-manager prisoners can be chosen to visit the central living room any number of times before the manager's first visit, but the lightbulb must remain OFF.
The first time the manager visits the room, the lightbulb will therefore still be in its initial OFF state. He switches ON the lightbulb.
If the manager is chosen two or more days in a row, he will find the lightbulb still ON. He will leave it on.
If a non-manager is chosen after the manager, he will find the lightbulb on and he will turn it OFF. He will never again turn the lightbulb either on or off.
If a non-manager prisoner follows a non-manager prisoner (whether himself or not), he will leave the lightbulb off.
The next time the manager enters the central living room after a non-manager, he will find the lightbulb OFF. He will count +1, and switch the lightbulb ON.

This pattern will continue until the manager has counted 99 non-manager prisoners. The next time he is picked after this point, he can be sure that the other 99 prisoners have been to the room by now, and now that he has (yet again), he can assert with 100% certainty that all 100 prisoners have been to the living room by now. Hello freedom and MENSA.

This works because the binary positions of the lightbulb account for 1: the manager and 0: everyone else. The manager can (and of course must be able to) count the zeroes (and himself, the one 1).

Tralix wrote:Monks meh this is just straight logic. Not giving a clue to this if you can't solve it, you're not a philosopher.

Still flummoxed about this one, so I guess I'm not a philosopher. Maybe your clue about not being a philosopher if you can't solve it is due to the whole lack of mirrors problem and identity issues associated with mirrors. But if that's what you were getting at, I still don't get it.

P.S. In my tabbed explanation for the nonal dial puzzle, I seem to have implied that dividing 9 by 9 gives you 0. Obviously a mistake, I meant 1.


Well you are right about one of those, but let's let it run. And the other one, excuse me saying you are not a philosopher, cause I am not, and that one I needed help on. Although I am told it is easier. It is a good exercise in logic, something I admit freely I was and nor still am one now. Maths I can do, the painstaking methodology of the solution is once you have an idea easy. :)

And by the way you don't need a manager, but nonetheless your solution is correct. Might be interesting to think why if you can be bothered but frankly after spending all that time I'd probably not. I kinda had an epiphany when I did it about the switches, in fact I stumbled on the solution almost straight away, although the finer points needed some work, (one of those rare moments you just get inspired) and from then on it was a matter of 1,0 without needing necessarilly to know who had been in the room exactly. :)


These sorts of things do not have just 1 answer. Which is why I like them.
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Re: Math Fun

Postby Silhouette » Thu Mar 21, 2013 11:26 am

Helandhighwater wrote:Well you are right about one of those, but let's let it run. And the other one, excuse me saying you are not a philosopher, cause I am not, and that one I needed help on. Although I am told it is easier. It is a good exercise in logic, something I admit freely I was and nor still am one now. Maths I can do, the painstaking methodology of the solution is once you have an idea easy. :)

And by the way you don't need a manager, but nonetheless your solution is correct. Might be interesting to think why if you can be bothered but frankly after spending all that time I'd probably not. I kinda had an epiphany when I did it about the switches, in fact I stumbled on the solution almost straight away, although the finer points needed some work, (one of those rare moments you just get inspired) and from then on it was a matter of 1,0 without needing necessarilly to know who had been in the room exactly. :)


These sorts of things do not have just 1 answer. Which is why I like them.

Hmm. I went through quite a few alternative potential solutions before I found the one I did, and they all turned out to be flawed except the one I posted. Are you sure there is at least one other way? I cannot think of a way to do it aside from what I came up with because of that last line in my tabbed solution.

As for the monk one, I simply don't know where to start:
If they are not allowed to communicate in any way then they will have to find out whether they have the large red spot on their head by themselves.
Yet if no mirrors are allowed, and they are not able to directly see their head by themselves, then they cannot find out whether they have the large red spot on their head by themselves either.
It doesn't seem right to get past this by using any accidental or unconscious communication in your solution, because that is still disallowed.
Nor does it seem right to get past this by using any alternative reflective surface in your solution, since it would at least act like a mirror, and therefore be disallowed. Though perhaps that technicality could be forgiven, it still seems wrong.
And that doesn't leave any other room for them to find out about a solely visual indication. And then there's something special about 11 days? I just don't get it.

I just need a hint on where to start, really.
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Re: Math Fun

Postby Helandhighwater » Thu Mar 21, 2013 11:08 pm

Silhouette wrote:
Helandhighwater wrote:Well you are right about one of those, but let's let it run. And the other one, excuse me saying you are not a philosopher, cause I am not, and that one I needed help on. Although I am told it is easier. It is a good exercise in logic, something I admit freely I was and nor still am one now. Maths I can do, the painstaking methodology of the solution is once you have an idea easy. :)

And by the way you don't need a manager, but nonetheless your solution is correct. Might be interesting to think why if you can be bothered but frankly after spending all that time I'd probably not. I kinda had an epiphany when I did it about the switches, in fact I stumbled on the solution almost straight away, although the finer points needed some work, (one of those rare moments you just get inspired) and from then on it was a matter of 1,0 without needing necessarilly to know who had been in the room exactly. :)


These sorts of things do not have just 1 answer. Which is why I like them.

Hmm. I went through quite a few alternative potential solutions before I found the one I did, and they all turned out to be flawed except the one I posted. Are you sure there is at least one other way? I cannot think of a way to do it aside from what I came up with because of that last line in my tabbed solution.

As for the monk one, I simply don't know where to start:
If they are not allowed to communicate in any way then they will have to find out whether they have the large red spot on their head by themselves.
Yet if no mirrors are allowed, and they are not able to directly see their head by themselves, then they cannot find out whether they have the large red spot on their head by themselves either.
It doesn't seem right to get past this by using any accidental or unconscious communication in your solution, because that is still disallowed.
Nor does it seem right to get past this by using any alternative reflective surface in your solution, since it would at least act like a mirror, and therefore be disallowed. Though perhaps that technicality could be forgiven, it still seems wrong.
And that doesn't leave any other room for them to find out about a solely visual indication. And then there's something special about 11 days? I just don't get it.

I just need a hint on where to start, really.


The monk thing drove me nuts at first until someone started me off on the right track. Ok so we know at least 1 monk has it, so start there, then work through what would happen if say 2 had it and so on. It'll surprise you how it figures. Start with the simplest scenario, then use more complicated ones is the only hint I can give without giving it all away.

There is a non manager necessary I can show by which I mean just once does there have to be a person in the room who knows a condition, it's just a different way of doing what you did, to which I say kudos, it's basically about position of switches according to the first occurrence, rather than who is in the room, it's entirely a solution that is no better or worse than that solution. An an optimal solution that then only relies on one time a situation is found, I found out about after I had solved it, that frankly introduced me to laws of probability I was completely unaware of, very interesting but I think if you want that solution, you should probably pm me. Suffice to say they should not be inducted into MENSA but hired by any think thank. Your solution is correct and the best given what you know, but is not optimal. :)
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Re: Math Fun

Postby Helandhighwater » Sat Mar 23, 2013 1:16 am

The monk one starts with how you know you have it without mirrors or magic. It finishes with the only possible solution to those that do. I can say no more than that. But how would anyone know they have it given the original problem. Sight is a clue! :)

How does one perceive anything? Start with the most simplest of assumptions, someone must have it, who or how many have it and why? it's a serial problem not an outright solution after one day as the problem states.

Edit: to say how many.
Last edited by Helandhighwater on Sat Mar 23, 2013 10:13 pm, edited 2 times in total.
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Re: Math Fun

Postby Helandhighwater » Sat Mar 23, 2013 10:11 pm

Ok another clue, some of the monks are found dead, it's more than one. So you can rule out at least that the herald of doom had it, and none of the monks did at least on day 1 you have to know you have it to die, so 1 monk at least must have it from the start however we are neutral about the messenger. If one monk has it and everyone can see he has and that monk can see that, and people are dying day after day... You have to work it through each day from the "assumption" that at least 1 person has it who is a monk. Some monks die, it is not likely they will all die at once? It does only say some monks are found dead on the 11th day which only says that some people died on the final day, it does not say anything about the previous days, it might mean some had already died it is unlikely it means no one did... Tis a demon of language and assumption this one. It's a pain in the ass because the only way some things can happen is if they are determined by previous days. So you have to be working all over then time period. :evilfun:
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Re: Math Fun

Postby Helandhighwater » Sun Mar 24, 2013 12:12 am

And one more thing although I think most have given up: if people die on the last day... we can only assume that they were because they are... and? It's a clever little bugger because when you have the solution you realise why the monks die, but it takes a little logic to determine when and where, and you need the final solution to see it clearly. It's kinda like that puzzle where you have all the corner pieces when you start, but the edges and more importantly the filler piece details and how they are filled in are sometimes little more than intuition and guesswork until you can see the whole picture or at least part of it so you have to start filling it in like you might a jigsaw puzzle, jogging forward and backward across all parts of the puzzle.

Good thing about this is if you google it you will only find the simple solutions to the simpler problems. This one is by no means either simple nor does it have an absolute answer done in an absolute way, as are all the best puzzles, but there is one.

So we know because some monks die, that there is a monk, 1 at least, that has a dot on his head by deduction if not induction... If no monks had a dot no one would die and the last part of the problem would be a paradox, it's just obviously logical, no one has a dot in the community it's not contagious even if the original messenger had one, it is not certain anyone would die in fact it's certain no one would.

The easier puzzle tells you this, but this is not the easier puzzle. So when someone dies... we know... what? Look closely at the language some of the monks may have it, not 1 of the monks may have it. Why would he not say some of the monks may have it if he knew only one by sight had it, since it is non contagious that would be illogical... why would he say some or 1 monks have it when he can see no one has it, so hence no one would know they had it, no one would die, and so on... And there's more as well, and it goes for quite some time before you work your way to the one of the monks definitely has it, or more... It's not easy, it's a pain in the ass, it's only virtue is it is logical.
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Re: Math Fun

Postby Helandhighwater » Mon Mar 25, 2013 1:30 am

Damn no one will even try it. I must admit the first time I postulated this, no one would even try it, given to me I was much the same. Is there no answer then?

Clue: sometimes the best answer is the most inconvenient.

And clue 2 probably why I was called a bad philosopher was because of my assumptions, my mistakes and my lack of ability to reason. The person who invented this kinda got me thinking, that sometimes there is an answer, and the only way to find it is to try it out. And then... No more clues you can either get it or you can't. :)

You can't see any way of solving this well then why, how does one exhaust all logical possibilities by claiming you cannot see? I am certainly not clever enough to make a puzzle this good, but the man who did, was of course more clever than I or perhaps more wise.
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Re: Math Fun

Postby peterpan » Mon Mar 25, 2013 6:13 pm

Helandhighwater wrote:Damn no one will even try it. I must admit the first time I postulated this, no one would even try it, given to me I was much the same. Is there no answer then?

I'm trying! I'm trying!
I get the start of process to find the solution but my mind is going around itself in circles.....
I put myself in the position of one of the monks.
1st meal goes as follows -
If I know that 1 monk is sick but I don't see a red dot, I must me the sick one and therefore die in 2 hours.
If I see a red dot, and everyone else(apart from sick monk) does too then I am not sick and the monk with the dot dies.
The same assumption goes for there being 1,2 or more monks if we know that there are that number sick......
But I don't understand how we get to know from how many are sick originally to how many die after X days or X meals.


Now I have a head-ache and suspect that I may even have a red dot on my fore-head.....
And I had rice for dinner....
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Re: Math Fun

Postby Flannel Jesus » Mon Mar 25, 2013 6:22 pm

peterpan wrote:If I know that 1 monk is sick but I don't see a red dot, I must me the sick one and therefore die in 2 hours.

But you don't know that 1 monk is sick.

If there was only 1 guy with a red dot on his head, he would see no red dots, and remember that the guy said, verbatim, "the monks may have the disease as well."
He sees no red dots, and there's no reason he necessarily has to have a red dot -- the monks may have the disease, and they may not -- so he doesn't die at all.

Now, what if there are 2 guys? Guy 1 and Guy 2 -- they each see one other person with a red dot, right? Guy 1 sees Guy 2, and vice versa. And they see that everyone else doesn't have a red dot. So they know that there's either 1 or 2 people with the disease. If there's 1 person with the disease, it's the other guy, but because of the logic above, he knows that the other guy doesn't necessarily have to die, so when he sees that the other guy lives...well, nothing. Still each one thinks that he may have it, or he may not have it, so no one has to die.

And you can do similar logic for 3 guys, 4 guys, etc.

The only way this riddle really works is if you change the wording -- it doesn't work with 'may'. The guy needs to say, 'at least one of you DOES have the disease' for it to work.
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