I do.Quote:
Originally posted by solarflare
Yes and yes. But I guess this means nobody watches Futurama?? Not that I expected you would...
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I do.Quote:
Originally posted by solarflare
Yes and yes. But I guess this means nobody watches Futurama?? Not that I expected you would...
Quote:
Originally posted by Simon666
I do.
First, I must say that I have seen alot of the episodes of Futurama, but I don't recognize the reference, so I appologize deeply for my complete ignorance.
That said, I would like to take a crack at the problem at hand. I think the real problem here for me was I may have been trying at first to over analyze the problem. I think I see now what was meant by it being a defintion type problem. Let me just type out my response for one direction of the iff (bipartite to paradoxical) to see if I'm on the right path. I think if it turns out to be right in this direction, the other direction is about as easy.
You are given G, I, and construct the bipartite graph B(X, YI, CI). I assume here that YI was the intention, since I don't think an XI would have any members of Y, but if this wrong let me know! There is also what seems to be a mixing of X standing for the ull set and standing for the "top" set of the bipartite, but if my interpretation is right, it is fairly clear what is meant.
Assume there exists a perfect matching.
Now look at the functions intoTop and intoBottom of the edges of the perfect partition into, respectively, their X and YI vertices.
Each intoBottom is a member of G since each YI is mapped to by a member of CI of the matching.
Each intoTop is a member of G by existence of inverse in a group.
All vertices of the graph B are covered since it is a perfect matching. Also, this shows all are disjoint.
Thus the existence of a perfect matching implies that B is paradoxical. If this way of looking at things is correct, I think it would be fairly easy to show the other implication...
Yes, that is it! There are some notational problems in my definition. For this construction X and Y are really the same set, but formally distinguished. To make it really explicit let G act on X, Identify X with X x {0} and Y with {0} x X then g( <x,0>) = <0,y> iff g(x) = y, then construct the Bipartite graph from there. I just wanted to avoid this extra confusion.
Let me give an example that will hopefully make it clear what is meant and will be useful.
Example: Consider the action of SO(3), the group of rotations, on the unit sphere S2. Let C = {g1,g2,h1,h2} be four distinct elements of SO(3) and I = {g,h} be the partion (I will identify elements of the partition with the index set) g = {g1,g2} and h = {h1,h2}. Form the bipartite graph B(S2 , S2I, CI).
The first vertex set of B is a copy of S2 while the second vertex set is essentially two copies. Suppose I is a perfect matching in B. let Ig1' , Ig2' , Ih1', Ih2' be the edges in I colored g1', g2', h1', h2' respectively. Let
Ag1 = Proj1(Ig1' ),
Ag2 = Proj1(Ig2' ),
Ah1 = Proj1(Ih1' ),
Ah2 = Proj1(Ih2' ),
Here Proj1 is the projection onto the first vertex set (You were calling this the top set)
The {Ag1,Ag2,Ah1,Ah2}
{g1(Ag1),g2(Ag2)}
{h1(Ah1),h2(Ah2)}
Each give a partition of S2, so if B contains a perfect matching then S2 is (2,2)-paradoxical under rotations.
I hope that example makes it clear what is meant by this construction. You are right that the other direction is just as easy.
I will give you a chance for comments before I ask the next question.
No comments necessary. Once I'd written down the definitions and played with them for a bit, the confusions cleared quite nicely.
I hope this series isn't getting boring yet.:) As I have said I am heading for a proof of the Banach-Tarski paradox and one more piece is needed (actually two, but I am going to assume the existence of two independent rotations of the sphere).
DEFINITIONS
The action of a group G on a set X is called locally commutative if any two elements of G, which have a common fixed point, commute as elements of G.
END OF DEFINITIONS
Proposition: Let C be a subset of a group G acting on X and let I be an index set for a partition of C. If <C> (the group generated by C) is a free subgroup of G acting locally commutatively on X, then each connected component of B(X,XI,CI) contains at most one cycle.
I would proceed by contradiction.
I think its clear to me I'm gonna have to think on this one for a little while. I have lots of little pieces I've pulled out of my toolkit but can't seem to fit the pieces coherently together to convince myself a particular line of reasoning might work. My troubles seem to center on the correct meaning of the group action as a free group or possibly what is exactly the points fixed by an action here (I keep seeing it, then I think of something not quite settling about my reasoning), but if I concentrate real hard on my notes, maybe I'll clear the jungle...
Basically here are some of the things that stand out to me:
It seems the contradiction needs something like:
For a bipartite graph with more than one loop and the given action, find a point whose stabilizer is not commutative.
or possibly:
For ..., show that the a particular element must have finite order (so that <C> is not a free subgroup?)...
The only telling analogy that keeps whispering in the back of my mind is that the fundamental group of a circuit is the free group on one generator (and thus abelian since powers of a generator commute), whereas the fundamental group of the figure-8 is the free group on two generators (not abelian). Since any graph with two circuits either shares a path or has a path that connects two of its vertices, it seems a contraction is possible here. But I don't quite see yet the relation of the fundamental group to any of this yet, except the analogy...
I have this feeling that I am missing something more obvious, but the free group thing is throwing me. Although my definition of the group was intimately tied to the action imposed by the fact that the graph was bipartite and had a perfect matching from which I could pull out the functions, I'm having diificulty entering any of it into a proof...
Caffeine is not helping, no matter how hard it tries...
Well the proof of the result is in the above post...
Remember the key about the group being free is that there are no relations...
For completeness let me provide some discussion on the correspondance between group elements and paths, at least as I see it..
DISCUSSION
Given a graph G(V,C), let <C> denote the free group of all reduced words on the elements of C, so that for all c in C, c and its inverse do not appear adjacent in any element of <C>. In general the elements of <C> have only a formal meaning. However, when each element of C is invertible, i.e. is 1-1 onto its range, then an element of <C> determines a partial function from a subset of V to a subset of V. For instance, when G = Gc(V) is a Cayley graph then <C> is just the group generated by C, where many of the elements of <C> could be the identity element. There is a natural correspondance between paths in G(V,C) and elements of <C>. Suppose P(e1,....,en) is a path in G. For i=1,...,n-1, if ei=<x,y,c> then there are two cases:
1. if y is a vertex of ei+1 then let gi = c,
2. if x is an element of ei+1 and x != y, then let gi = c inverse.
for en = <x,y,c>
1. if x is a vertex of en-1 then let gn=c
2. if y is a vertex of en-1 and x != y, then let gn = c inverse
According to this rule one can associate to the path P(e1,...,en) a unique element gp=g1...gn of the free group <C>. In general this "correspondance" is neither 1-1 nor onto. Note that when G = Gc(V) is a Cayley graph, each group element corresponding to a path in G is reduced in the generators C, and that distinct paths starting at the same point yield formally distinct group elements on the generators C.
Note: If <C> is a free group then they are distinct element.
END OF DISCUSSION
Soooo... maybe I was trying to conceptualize too much the functional foundations of the graph and was unable to let the group just be like the normal homotopy groups I am used to playing around with? Edges, paths, and words, oh my!
Then by
you mean my post and particularly the part about the fundamental groups, right?Quote:
Originally posted by souldog
Well the proof of the result is in the above post...
Something like: pick a point between a choice of two cycles (either on the common path they share or on the path that separates them, as the case may be). Then form the path that goes around one cycle then the next (both, say, clockwise). Form the other path that goes around the latter then the first (still clockwise). The two paths are equivalent in their action, but do not commute on the free group (xy != yx) which violates the hypothesis of local commutivity. Thus the graph cannot have two or more cycles because of the contradiction.
It was really that simple, but I just got lost?
That is exactly it.:)
Now let me tie it all toghether..
The Banach-Tarski Paradox : S2 is (2,2)-paradoxical under the action of SO(3).
proof:
I will assume the well known result that there are two independent rotations of the sphere, which generate a free group of rank two. Since every subgroup of a free group is free, there are free groups of any finite rank. In particular there exist four independent rotations of the sphere {g1,g2,h1,h2}.
It is easy to see that the action of any group G of rotations on the sphere is locally commutative. Each rotation of the sphere fixes exactly the two points of ontersection of its axis of rotaion with the sphere. So if two elements of G share a common fixed point, then they must be about the same axis and so must commute as group elements.
Let B(S2, S2I,CI) be as in the example above, where C={g1,g2,h1,h2}. Now by the second proposition it is enough to see that this graph has a perfect matching. Since <C> is a free subgroup of the rotation group and acts locally commutatively on S2, the last proposition shows that each connected component has at most one cycle. Also each vertex in S2 has valency 4 while each vertex in S2I has valency 2. So by the first proposition there is a perfect matching.
QED.
Lets push it a bit further.
An infinite number of spheres from one:
Having assumed the existence of two independent rotations of the sphere g,h the set {g^nh^n: n=1,2,3,4,..} yields countably many independent rotations. So let C={g1,g2,g3,...} be independent rotations of the sphere. Let I={1,3,5,7...} be the odd integers and for each n in I Cn={gn,gn+1}. Then B(S2, S2I,CI) has one copy of the sphere on one side and countably many on the other.The group <C> is a free subgroup of SO(3) and acts locally commutatively, also the valency of each vertex in B is either 2 or infinite. So the graph contains a perfect matching.
For each gn in C let Agn denote the set of vertices in S2 contsained in an element of the perfect matching colored gn' (the extended function). Then {Agn, Agn+1: n=1,3,5,..} is a partion of S2 and for each n = 1,3,5,... {gn(Agn), gn+1(Agn+1)} is a partition of S2.
Note there is nothing special about the sphere here. The above two results apply to any set acted on by a group G, which contains a free subgroup acting locally commutatively. Also since it has been shown that there exist countably many independent rotations of the sphere one can obtain a continumm of spheres from one using the same proof,
Heck lets keep going
The solid Ball in three space is paradoxical:
proof:
Let me denote the unit ball in three space by B3.
A heck I am tired of typing. See if you can prove this.
Well, since I learned the traditional decomposition method to prove Banach-Tarski, I can just use that information here :).Quote:
Originally posted by souldog
The solid Ball in three space is paradoxical:
proof:
Let me denote the unit ball in three space by B3.
A heck I am tired of typing. See if you can prove this.
You just look at the piece of the ball with 0 < radius < 1. These are paradoxical for the exact same argument as the 2-sphere (at each fixed radius). There is also the polar set of mappings on the boundary 2-sphere which map all of the 2-sphere uniquely except for one point to the plane which has the same free group symmetry structure as the 2-sphere except for that point (which can be isolated). Therefore, this boundary is paradoxical in away that we can use to obtain two points (the original center and the non-uniquely mapped "pole" point) to use as centers for the reconstruction of two balls. These two points are just an added component to the bigraph that this construction shows is paradoxical (and in fact the four components and the boundary show the five component decomposition normally used). I could of course give the actual use of the generators in this decomposition and the resulting recomposition partition, but that is not necessary to just show paradox conditions on the ball (and would just be a copy of what is already out there in the books I learned this from originally)...
I've looked over the discussion so far and rewritten everything using my own notations, and I feel much more secure in my understanding of what is going on now. My early confusion was in not recognizing the role of the bipartite graph (and thus making finitary assumptions or mistaking the functional role). I want to thank you, souldog, as these little exercises have made much clearer in my mind the origins of this fascinating little theorem and in the process shown me a nice little bipartite graph construction which seems to be applicable in many situation where one needs to look at mappings and their domain and codomain in general. I am certainly appreciative!
To use the above methods you just tack a couple of edges which use translations into a connected component without cycles of the bipartite graph formed from the unit ball without the origin. These two new edges put a cycle in this connected component and just correspond to translating some points to the origin. You then just proceed as normal, making sure one of these new edges is in the matching.
What I think is nice about these results is they show how "trivial" the Banach-Tarski paradox really is. What one could do now is add some topology. The functions which determine the graph could be partial hemeomorphisms etc..
DEFINITIONS
Let G be a group of homeomrphisms acting on a topological space X. The set X is said to be generically (m,n)-paradoxical under G, if for some m and n there are disjoint subsets of X A1,...,Am,B1,...,Bn and group elements g1,...,gm,h1,...,hn such that each of {A1,...,Am,B1,...,Bn}, {g1(A1),...,gm(Am)} and {h1(B1),...,hn(Bn)} each partition a comeager subset of X.
END OF DEFINITIONS
You might be interested to know about the following two results.
What I am going to call the
Dougherty-Foreman Paradox: S2 is generically (3,3)-paradoxical under the action of SO(3) using open sets.
The above result is entirely constructive.(No Choice)
And
Theorem: S2 is (3,3)-paradoxical under the action of SO(3) using sets with the property of Baire.
Ok, My Reign Of Terror is over.
Hmmmmmm....
I had heard about an entirely predicative version, but never knew what to look for. Now I have something new to study (and it goes to show another example of why much of modern math doesn't require choice!).Quote:
Originally posted by souldog
Dougherty-Foreman Paradox: S2 is generically (3,3)-paradoxical under the action of SO(3) using open sets.
The above result is entirely constructive.(No Choice)
Ok. Let me change the format a little to something not of this ilk for a little while.
Setting
Its 1983. Seven year old galathaea is sitting in class and listening to the teacher read a story during story-time (I think it was probably the BFG by Dahl). Picture corduroy. Reagan is president. Kids are picking their noses carelessly...
Internal Narrative
A thought is running through galathaea's developing mind, with the autistic self-involvement of youth. Something heard on Mr. Wizard the day before (where, amongst the many wonders of the world, galathaea first learned it was possible to cut a hole in the middle of a 8 1/2 x 11 piece of paper large enough to walk through!). A question is forming...
Action
Suddenly, galathaea raises a hand. The teacher stops reading and looks to galathaea. "Yes?" the teacher asks, expectant.
"If hot air rises and cold air falls, why is it so hot in the deserts and so cold in the mountains?" ignorant little galathaea asks.
Response
The teachers tells galathaea that she will have to go home and ask some friends. When the teacher returns the next day, she tells young galathaea a story so convoluted and full of hand waving, that even young galathaea recognizes it as hand waving and amounted to saying that the air is already cold in high places and gets cooled as it rises.
It took galathaea 6 more years before galathaea finally understood the real reasons, which would have been understandable at the younger age if the teacher had known the real answer.
Repercussions
This was the first time galathaea really noticed that teachers often tell their students false things in school. Throughout the rest of galathaea's scholastic career, this became a hauntingly reoccurring theme that has shaped galathaea to this day. The list of errors conveyed to galathaea through school grew larger and larger each year. Examples of how this problem enter into education are too numerous, but another two examples are indicative.
Recently galathaea was looking over the younger brother's homework (8th grade), and noticed an explanation of tidal forces that mentioned nothing about moments and the fact that the moon was in free fall around the earth (which actually showed why the hand-waving in the book was wrong). Galathaea took the opportunity to explain the idea of forces with moments, and how the forced solid-body movement of the moon around the earth gave rise to pseudo-forces in the well known cross conformation that caused the tide. In the process, galathaea was able to tell bro how this also was one of the two leading reasons for the molten core of the earth(the other being radioactive heating), and how it also gave the moon and all sattelites a force to spin them in the way seen around the solar system (except for the exceptions).
Another example that explores the cause more closely came from a philosophy of science course taken in college. The course had just been entered as an option for education majors to fill one of their requirements, and was thus full of them. The teacher was explaining how certain problems often arise in science that challenge scientist to adjust ore refine their worldview, and an example began to be explained. The choice was Olber's paradox and the challenge it presented to the idea of an infinite, homogeneous, and isotropically populated universe. Suddenly, an education major raised their hand and asked why we were studying actual science (since the teacher went over the math). After a brief recapitulation of the motivation, another ed major chimed in asking if it would be on the test. Suddenly, there were all sorts of objections on the idea that this little excursion didn't fit their list of prerequisites, all (and I mean all) of which came from the education majors. And even though the math was never to be tested in the first place (it was illustrative), the teacher just dropped the example and moved on because too much time had been wasted.
Endgame
This is why galathaea feels so strongly that the future of education lies in interactive computing, most likely on-line, potentially from home. Individual teachers are way too limited in their knowledge, and the lecture/board idiom of teaching is so limited compared to the multimedia and interactive experience available with computers. Things like graph theory and combinatorics should replace things like teaching how to add or subtract at a young age (they are algorithms in number theory after all which are not needed until after the relevant introduction, since we have had things like calculators for a century or so). Kids can manipulate the little sets of objects and learn the basic properties quite easily and intuitively, since many kids already play such games in their heads (I was a maze fanatic, for example). Imagine picking from the best lessons available around the world for a child.
Maybe such an education could finally put an end to incompetent teachers giving their c++ students the void main syndrome...
Reprise
So now I ask you:
If hot air rises and cold air falls, why is it so hot in the deserts and so cold in the mountains?
There is one effect which gives the overwhelmingly largest contribution to this, so it is what I am looking for. Other effects can be mentioned, but focus on the big one.
just a note: the result will be found as Marczewski's problem which was solved by Dougherty and Foreman
Come on guys!!! This one is not a trick question, and I don't expect answers with equations or anything of that sort. I just want an explanation that you could tell a kid that asks this question:
If hot air rises and cold air falls, why is it so hot in the deserts and so cold in the mountains?
I'll give a little hint: This is an on-average stable process. But the "hot air rises" part shows that one part of the process tends equilibrium different than what we see. Driving this stability away from the implied equilibrium must require some other pieces to the explanation. Why is the temperature in mountains cooler than that below? Ther must be some driving source and an explanation for the inversion, so there are at least two small little pieces for this process that give an on-average result different than the idea the hot air above, cool air below you find in, say, a box that is in dynamic equilibrium but hasn't reached thermal equilibrium. Sure its a thermodynamics question, but it can be explained quite simplistically. Where is the heat coming from? Where is it going?
Well in the deserts it's so hot for 2 reasons
1. they are on the equator and the sun ray fall vertivally on them
2. they are low so the air is thick (or dense I dont know the exact word) enough to keep the warmth of the sun whereas on the mauntaing the air is sparse (or someting) and cannot keep the sun's warmth.
Let's look at these points one at a time.Quote:
Originally posted by Seventh Star
Well in the deserts it's so hot for 2 reasons
1. they are on the equator and the sun ray fall vertivally on them
2. they are low so the air is thick (or dense I dont know the exact word) enough to keep the warmth of the sun whereas on the mauntaing the air is sparse (or someting) and cannot keep the sun's warmth.
1. Well, not all deserts are on the equator or even closer to it on average than certain other types of climates. But you have pointed out two important factors: sunlight and latitude. Temperature normally varies on average by latitude due to the amount of sunlight it receives per area. But deserts often have higher temperatures than other climates at the same latitude and elevation, particularly in the summer for that latitude. Where does this higher temperature come from? By explaining this, the explanation for why one commonly finds that deserts in the winter at non-tropical latitudes are often cooler than other climates at that latitude and elevation -- this is not a "counter-example" to the question however, since no such comparison is implied; it is merely another fact to tell a child that asks this...
2. Here, there is the start of something, but it still does not have explanatory power. You are correct that the air at low elevations is on average denser than that at high elevations, but this does not prevent the hot air from rising. In fact, there are regions of the atmosphere that do have this temperature profile where hot air is above cool air, even though density does on average decrease even in those regions. And there are regions in the atmosphere that again go from hotter temperature to lower as one goes higher. So these must have something in common with our layer of the atmosphere; ie. they have a source of heat and a reason for the temperature to decrease as elevation increases. A hint here is that the decrease in temperature does indeed have to do with density, but not for the reason given.
So, it is a start at working towards the solution. But it is not a correct line of reasoning as is...
Here's a shot in the dark: countercurrent diffusion?
Hot air going up passes cold air going down in the opposite direction, allowing for maximum diffusion... but I have a feeling the correct answer has nothing to do with this. Probably because there's less pressure higher up, the warm air can expand, causing it to cool.
1. well the denser air can keep the warmth stored during the day. The ground also. (the sand doesnt do this really good and hense the really cold temperatures in deserts at night). Really high places do not have much ground to keep the warmth (only rocks) and the less dense air also
The ground of the deserts reflects the suns rays and the air manages to keep it. In higher grounds it does not.
2. Oceans and seas (even big rivers) keep the temperature stable in areas near to them (only horisontally but not vertically)
But anyway i think that the ground temperatyre plays the most important role followef by the density of air.
Quote:
Originally posted by solarflare
but I have a feeling the correct answer has nothing to do with this. Probably because there's less pressure higher up, the warm air can expand, causing it to cool.
And here we have the outline for the basic answer to the question. Now I'll try to toss it together in an explanation that could be given to a kid who asks this that would be useful to get a start at understanding what is going on:Quote:
Originally posted by SeventhStar
The ground also. (the sand doesnt do this really good and hense the really cold temperatures in deserts at night). Really high places do not have much ground to keep the warmth (only rocks) and the less dense air also
The ground of the deserts reflects the suns rays and the air manages to keep it. In higher grounds it does not.
2. Oceans and seas (even big rivers) keep the temperature stable in areas near to them (only horisontally but not vertically)
But anyway i think that the ground temperatyre plays the most important role followef by the density of air.
Where does the heat come from?
Its warmer during the day than it is at night. So its pretty clear that the heat around us is because of the sun. It might be thought (by this hypothetical kid) that because the sun is really hot, it makes the Earth warm just like when you stand close to a hot stove you get warm. The sun is really hot, but for something really hot to heat another thing the way a stove does requires there to be something in between the two things to carry the heat. A stove does this through the air, but space doesn't have as much stuff in it, and not much heat can move from the sun to the Earth this way. However, the sun is also very bright, and the sun makes the Earth warm by shining light on it. Light and heat are two different forms of the same thing: energy. When light shines on things, some of it gets bounced off of it (like a mirror), some of it goes right through it, and some of it gets absorbed. A window, for example, lets you see alot of the light that shines through it. It is the absorption that is the key here.
When light gets absorbed by something, what is actually happening is that the light energy changes into a type of energy that makes the small pieces of the thing (the atoms, if the kid knows about them) start to move around alot, since moving is a kind of energy too. These atoms wiggle around faster as they absorb the light, and this wiggling is what we call heat. The hotter an object is, the more these things wiggle.
So, alot of the light from the sun just passes through the air and hits the ground (some of it gets absorbed and some reflected -- and this can be explained to the child as well). You can't see through ground very well (:)) so it must be absorbing or reflecting much of the light. The light that is absorbed heats the ground, and that is why the ground gets warm and where a lot of the heat on the ground comes from (of course, there is internal heating from the tidal forces and radioactive decay as I think I mentioned in my question backgrounder, but it is not the main source of heat -- this too can be told to the kid).
Why is the desert ground hotter than the ground on a forest?
Things are held together tightly by these things called forces. Even though all the pieces (atoms) of stone are wiggling around really fast, the stone doesn't fly apart because these forces hold it together. But different types of things have different amounts of force holding it together. (An explanation to the kid could point out that atoms and molecules are different for different things, and each different type of atom or molecule pulls the other atoms and molecules to it in different strengths). Since these pieces are held in place with different strengths, some things need more energy than others to cause them to wiggle around as much as other things (a small introduction to momentum and inertia could fit in here as well, but it is not as big a factor as the binding forces). This is a big point, because the thing we call temperature is not the amount of energy they have absorbed (which we call heat) but how fast they wiggle around. So some things get hot much faster than other things (a great place to start an introduction to specific heat).
Water gets heated slower than most other things around us. Although it is a liguid and moves around a lot, and it sure seems easier to pull it apart than a rock, water has some really strong forces in it keeping it together (hydrogen bonding could eventually be explained, but that should wait), and rocks normally let their atoms or molecules wiggle around alot faster without letting them break out of their place. So places that have a lot of water tend to warm up slower as they absorb the sun's light. Deserts are places that don't have a lot of water (eventually again the kid can be taught why deserts don't have a lot of water, telling them about mountains, rain shielding, and all). So the ground gets warm very quickly in the desert. In forests, and any place where there is a lot of plants, there is a lot of water (since plants need the water, they grow best where there is a lot of it).
The temperature of the ground warms the air that touches it, and if they are touching for a long time, they tend to reach the same temperature (this is the equilibrium law of thermodynamics). So the air around the ground in deserts tends to be around the same temperature as the hot ground, and the air around the ground in wetter places tends to be the same somewhat cooler temperature as that place's ground. It is the temperature hear that gets shared, not the amount of energy or heat there.
But if hot air rises, why is it so much cooler the higher up you go?
Yes, hot air does rise (an explanation could be given). Its not exactly true that the higher up you go the cooler it gets, though. If you go up really high you will see that the temperature gets cooler then warmer then cooler again and there are actually different levels to the atmosphere (an introduction to the atmosphere's stratification could be given here). But those other levels are much higher than even the highest mountains, and here at the lowest level, things do get cooler as you go up. Why? Its because of the way that things behave as they expand.
The air above us is held on to the Earth by a force called gravity which causes things to have weight depending only on how much of the thing (mass, not number, which could be introduced briefly) there is. If you look at any height, the air at that height is pressed down towards the ground by the gravity of the earth (which has alot of mass) as well as by the weight of all of the air above it. Since there is less air above as you go higher (and since gravity gets weaker as you move away from the earth), air at higher places is forced down on less and less as you go up higher. This causes the air to be "squished" more really close to the eath and less squished as you move up (an exploration of pressure and density would go here).
Now the important thing here is that the lower pressure of the air as you go higher causes any denser air from below that might be rising to expand (adiabatic expansion), and also any air from above that may be falling gets squished as it goes down. And the interesting thing is that when a gas expands it cools down and when a gas get squished, its temperature rises (toss in an explanation of this basic gas law here, keeping it simple but the explanation correct in terms of kinetic energy and such). The heat is still there, it is just not the same temperature. This cooling occurs faster than any equilibrium of temperatures might happen, and so as you go up, the temperature of the air decreases.
At very high elevations, there are places where the atmosphere actually absorbs alot of the sun's radiation due to the presence of things like ozone and ions (which are also created by such absorptions), and it is the fact that these species are stratified (because of which radiation frequency actually passes to which level), that causes the different temperature profiles in the various atmospheric strata. It also explains why only certain intervals of light are able to pass through the atmosphere (mainly the visible spectrum and some extremes like certain radio) and many others are absorbed at various levels. Of course, that explanation is a bit technical and wouldrequire some elaboration for use in telling it to kids. One could also explain to kids the "coincidence" of why the light that is transmitted is the light that animals, plants, and photoreceptive bacteria "see" is precisely because that is the light that is transmitted and they could then evolve to receive it, if one wanted to go there.
But why isn't the ground of mountains warmed like the rest of the ground?
Well, it is. But the light is hitting at an angle for most of the mountain, and this spreads the light out more (insert explanation of how this is seen for different latitudes as well). But also, the air here is also cooler, so equilibrium is reached at a lower temperature. And also, it tends to rain or snow more at these higher elevatin pieces of land (which continues the rain shield discussion earlier), and snow tends to reflect more light than rocks and sand, so less of the light is absorbed as heat as well. All of these reasons add up to why the mountains are cooler than the land lower down.
So....
So, this is the kind of explanation for this question you can give to a kid. I didn't expect this kind of detail as a response to my question, I just wanted to show how one would wrap it all up to turn that kind of question from a kid into an opportunity to give the kid a deeper insight and physical inuition about the world and how things work. Notice that the responses that were given to the question implied alot of background and would not have clarified the situation at all to a child that didn't already posess such a background. If you add alot of visuals to the presentation, maybe break up the discussion into a few days worth of explanations that went into the bits and pieces in a systematic way, and introduced the question at the very beginning as the goal, you then have a very nice lesson plan that could be given to a kid. There is no need to introduce the equations, or even give the kid the expectation that they must memorize or learn every piece of information, because long term potentiation requires repeated exposure to material before it is learned (to paraphrase Natural Born Killers, "repitition works!"). Every year (or whatever interval might be found optimal to learning) similar materials can be introduced, slowly adding on the technical details as the child's ability to understand and use mathematical representations for physical events matures. It's kind of like souldog's questions leading up to the Banach-Tarski paradox (which is what inspired me to ask this question). Of course, I am sure that everyone who read the question probably already knew most if not all of this already, which is why it may have seemed like I was trying to ask a trickier question, or why it wasn't worth answering. But I sometimes think that some of the teachers out there who do know this stuff still would not have taken such a question as the educational opportunity I believe it is. I think they probably would have two different reasons for that as well; first, it might seem too difficult to explain in words and concepts that the child understands (which really underestimates children, or even animals, for that matter, who have highly developed physical intuitions), or because they do not realize that succinct answers are often more confusing to the child than detailed ones and may not realize all of the details being assumed by their answer, because most of the pieces are left unexplained. Of course, in galathaea's case, the teacher just didn't know the answer....
So really, the main answer was: its warmer on the ground because that is where the light gets absorbed (and hotter in deserts because of specific heat) and cools as one goes higher in the atmosphere because of adiabatic cooling. But that sounds like an answer that a kid memorizes for a test and then promptly forgets...
So I guess the points would go to both solarflare and SeventhStar! The thread is now open again for questions!
I hope you didn't expect us to give THAT answer!!!
Anyway I have a question:
Suppose you have an unlimited number of cells that are linked together like this
() cell, - link
...-()-()-()-()-...
Each cell has two conditions on -(1)- and off -(0)-
Each cell checks it's neighbours at every priod of time T and:
if BOTH neighbours are on or both are off it turns (stays) off
if only ONE of the nighbours is on it turns (stays) on
you have only One cell turned on at the beggining:
...all off...-(0)-(0)-(1)-(0)-(0)-...all off...
how many periods of time T would pass in order to have X - cells ON
Well, each round will always have only a power of two cells on, and each power of two will occur first exactly on the same number round (1 on round 1, 2 on round 2, 4 on round 4, etc...). So if the question is asking for the round where exactly X are ON, then it is equal to X iff X is a power of two, otherwise it is never. If it is not asking for exact number on, just enough, then it will occur on the round of the first power of two greater than or equal to X. It looks like an inductive proof suffices here.
The pattern is that the cells tend to switch between two states:
10000...00001
and
101010...010101
From the first position, say there are 2^(n)-1 zeros between the ones. They will each form a separate pattern of 2^(n-1) ones, stretching 2^(n-1)-1 cells in each direction from the initial one, and taking 2^(n-1)-1 turns to do so. This means that the first pattern will change into two small versions of the second pattern - but guess what, they're side by side, so it's really now one version of length 2^(n+1)-1, containing 2^(n) ones. I think it is clear enough that the second pattern will then change into the first pattern (twice as long) after just one time interval.
Notice that to get 2^(n) ones from 2^(n-1) ones, it takes 2^(n-1)-1+1 turns. Simple manipulation shows that (since the first time interval doesn't count) to get X ones from ...000010000... takes X-1 time intervals.
This thread seems to have lost its head, so I decided to figure out some other basic stuff about the past problem to fill time...
First start with a notation. Write 1-1 for the sequence 1, 1-3 for the sequence 101, and generically 1-(2n-1) for the alternating sequence 1010...01 with n 1's. Write 0-1 for the sequence 0, and generally 0-n for the sequence of n 0's. So the first few iterations are therefore:
0-Inf 1-1 0-Inf (Inf is short here for infinite)
0-Inf 1-3 0-Inf
0-Inf 1-1 0-3 1-1 0-Inf
...
Basically there are three basic relations that come in handy. There are two basic transformation types, where we focus on a 1-x group and see how it alters its surrounding zero group and how it changes form "internally". Obviously the neighboring 0-x's can potentially be altered by two 1-x groups. Using S to indicate the successor function, we get the two transformations
S(0-m | 1-n | 0-m) -> 0-(m-1) | 1-1 0-n 1-1 | 0-(m-1) (for m>2, n>1)
S(0-m | 1-1 | 0-m) -> 0-(m-1) | 1-3 | 0-(m-1) (for m>2)
and the one direct equivalence relation (I call it a collapse :))
0-x | 1-n 0-1 1-n | 0-x = 0-x | 1-(2n+1) | 0-x
that settles all the case of m, n.
Using these relations, its fairly easy to prove some simple statements:
- 1 sequences are the same length throughout the string on a given round
- # of 1 sequences is always a power of 2
- collapse occurs on steps after the existence of 0-3 in a string, therefore every four rounds (though complete collapse occurs only on the powers of 2 that are used for the proof of the original problem)
Okay, that's enough for that completely useless exploration... I was curious about the bipartite graph problem series. As I've looked over my notes and tried to use it to study along with some other graph problems (the book by Lovasz, if it edifies), I came across some other proofs of the perfect matching and they all seem to focus in on the proof of what is called the Konig-Hall theorem. I was just curious, souldog (if you are still peeking at this thread), if my interpretation is correct that the first problem was indeed a version of this theorem (well, I mean it is fairly simple to set the problem so that the Konig-Hall theorem gives the perfect matching existence immediately, so maybe I really am asking if it was intentional or known this connection)?
Ahh... maybe I'll go bug some other thread now...
Yes, I am aware of it. I usually relate it to the Hall-Rado theorem and am not aware off the top of my head what the distinction between the two is. I didn't want to bring in any theorems at that point in the discussion.
Ok, I am going to use my power as granted me by the United Association of Solipsists to declare the previous question dead, since as far as I can tell it has been answered correctly. I use my power to grant the point to solarflare who gave the inductive proof (though he may have forgotten to clarify that X was only powers of two since it was already clarified previously). Solarflare also correctly answered the question in terms of intervals instead of the rounds I used.
But I'll ask the next question. Its about one of my favorite areas of recreational math: series manipulations. Now I would like to use the S notation, but I don't want to go to all the trouble of putting the indices so I will leave them blank and use these two rules:
- All series are infinite from 0 on up.
- The series variable will be "i".
Definitions
We will denote by (not the standard notation, vBulletin ain't TEX)
(a; n)
with a e C and n e Z >= 0, to be the shifted or rising factorial which is defined as
a (a + 1) (a + 2) ... (a + n - 1) or P (a + i)
where the product is over i from i = 0 to i = n - 1. By defintion we take (a; 0) = 1 for all a. Notice that unlike the normal factorial, a may be any complex number.
Now you can build infinite series with these things. One defines a hypergeometric series pFq(a1, a2, ..., ap; b1, b2, ..., bq; x) as
S ( [(a1; i) (a2; i) ... (ap; i)] / [(b1; i) (b2; i) ... (bq; 1)] x^i / i! ).
In other words, it is a sum of terms each with p rising factorials in the numerator and q rising factorials in the denominator all multiplied by x to the ith power divided by the normal factorial of i (which can also be represented as (1; i)). All the a1, ..., bq are parameters to the function just like x. There are many common functions that have representations as hypergeometric functions. For example, the simplest hypergeometric function, 0F0(x), is just the common exponential function (to Euler's base).
On a different line of definitions, say you have a given series
Q = S f(i, ...)
over a function f that takes the series variable i as one of its parameters (it may have other parameters). Now I will make an exception to my previous series rules and give the range of the series variable i explicitly after the sum.
We define the (m, n) - multisection of the series Q over the basis f as the sum
S f(i, ...) (where i = m (mod n) and i >= 0).
In other words, the sum is over all nonnegative i that have remainder m when divided by n. For example, look again at the exponential series with f(i, x) = x^i / i!. The (0,2) - multisection of the exponential function over f is then
1 + x^2/2! + x^4/4! + x^6/6! + ...
which is just the hyperbolic cosine. Similarly, the (1, 2) - multisection is just the hyperbolic sine.
Problem
Express the general (m, n) - multisection of the exponential function over the basis f given above as a hypergeometric function. The hypergeometric function's parameters may vary with m, n and even the numbers p and q of rising factorial terms can vary with m, n.
I will let you play around with it for a little while. Its really not a difficult problem once you realize that you need a couple of relations for (a; n). I'll give a hint in a day unless it is solved or somebody says they don't want the hint because they are making progress and don't want it spoiled.
Bravo Galathaea for taking the initiative. SeventhStar has been very lax in his responsibilities.
Nice Question. How do you create the mathematical notation with vB? Could you direct me to a link with the needed keywords?
I actually just use [font=symbol]S[/font] for the sigma and with capital P for pi. I also increase the size. To check out how people do things that I haven't seen before, I usually will hit the quote button to see the whole vBulletin commands used. There's a few that aren't documented in the help, which is annoying. Sam Hobbs has a thread in the testing forum where I recently saw another way to get extra characters.
ooops I'm really sory guy. I forgot abou this thread :rolleyes:
(NOW WHEN THE SUBJECTS OF THE CODEGURU MAILS ARE OFF I AM I LITTLE CONFUSED :mad: )
But galathaea's got it (first). So he got the point and the right to ask (which he did).
I've been trying to perform a general counting of the points :) why dindt you do it before me?...
So if you'd let me ask I have a question that would be next on the thread:
Who has how many points in this thread ( a tough one :D ) ?
oh and think that we should stop with this strictly mathematical questions... They are not nice (:D)
They scare people!
And most importantly I, personally dont have time to think over them. Try simpler (as a condition) questions. If it hard but it's said in a simple way I would remeber it and think over it. I may be too lazy to write down these LOOOONG questions :rolleyes:
If anyone dissagrees let me here his point.
You will need to figure out formulas for:
(a; x + y) = ? (something in terms of rising factorials without sums)
(a; xy) = ? (something in terms of rising factorials without products)
If you look at the (m, n) - multisection sum given in a slightly different way, you get to a point where you have to apply these two transformations and right away a little touch up gives you the hypergeometric series. Its really quite a short manipulation once you figure out how to apply the the two formulas above.
The harder forrmula to figure out is the product, but I'll give a graphical hint for multiplying by 2 to get you started on that one.
Code:............
||
\/
. . . . . . . . . . . .
(and shrink the spaces by extracting a coefficient)
I'll just wrap this one up so we can move on to a different question. First the addition formula follows from breaking up the multiplication chain. Its fairly easy to see visually:
...........
or written out:
(a; i + m) = (a; m) (a + m; i)
The multiplication formula takes a little more consideration, but it follows when you realize a series of facts on the multiplication chain. First, since the chain is of length a multiple of the multiplier, you can break the chain apart into separate multiplier number of chains. The particular choice one wants to make is to keep the distance between elements the same for each chain, and picking the chains in an interlocked fashion like the visual clue given accomplishes this constant size difference (for the clue, it is 2). Notice how this decomposition of the chain is similar to how one may take a buffer and view it as an matrix (vector of vectors) in a column decomposition. Now notice that if you were to take a normal rising factorial chain (which has common difference 1) and multiply each of the elements by some number n, you get a multiplication chain with common difference n. So you can do this reversed and extract a coefficient from each term of a chain with some common difference not 1 and you get a standard rising factorial multiplied by a coefficient. Doing this for the separate chains when you break apart the multiplication factorial gives you this formula:
(a; n i) = n^(n i) P ((a + j) / n; i)
(for j from 0 to n - 1)
With thos two formulas, its basically two steps to the solution, with a little cleaning up each time. I'll break it up somewhat to make it clear what is happening each step.
S x^i / (1; i) (with i = m (mod n), i >= 0)
||
|| (addition formula and switch summation range)
\/
S x^(i + m) / (1; i + m)
= x^m / (1; m) S x^i / (1 + m; i)
(with i = 0 (mod n), i >= 0)
||
|| (multiplication formula and final range switch)
\/
x^m / (1; m) S x^(n i) / (1 + m; n i)
= x^m / (1; m) S (x^n)^i / [n^(n i) P ((1 + m + j) / n; i)]
= x^m / (1; m) S ((1; i) / P ((1 + m + j) / n; i)) ((x / n)^n)^i / (1; i)
= x^m / (1; m) 1Fn(1; {(1 + m + j) / n}j; (x / n)^n)
(with i >= 0; 0 <= j < n and {...}j meaning the set over j )
Notice how the two manipulations were just what was needed to get the right hand side to simplify to i when we switched the bounds over to that of standard hypergeometric form.
Now this is a simple little series of transformations that can be applied to any hypergeometric series to produce its (m, n) - multisection. And since there is an alternate form for expressing multisections in terms of the function's value on the n nth-roots of unity when the basis is equivalent to the exponential basis x^i (as the hypergeometric functions are), you have an entire class of hypergeometric identities which form a particular basis of identities on which one can classify a lot of information on the standard identities. Additionally, the (m, n) - multisection of the exponential series over the given basis obeys the function d^n / (dx)^n = id, so we have a particular basis over differential equations of order n much like one uses normal fourier analysis over the trignometric decomposition to get the two element symmetries and related information. Also, one has almost identical ability to do these transformations on basic (or q-) hypergeometric series, and filling in the details gives all sorts of relations in group theory over this really nice basis of relations. Finally, because the rising factorial and hypergeometric functions in general have formulations in terms of integrals, you get integral relationships over this entire basis.
So you can see that this little problem can be used as the starting point for an exposition into many pieces of analysis, combinatorics, and group theory. It also has the appearance of an identity probably found by thousands of undergrads or grads each year when they first start messing around with hypergeometric series. However, although I haven't made a detailed search yet, I have yet to see it actually mentioned in any of the standard texts like the various Dover textbooks, Andrews/Askey/Roy, the various works of Gauss, Jacobi, etc. An ex-professor of mine had mentioned that she thought she had seen an exposition of these bases under multisections and the theory, but she was quite busy and I have yet to find any references, though many special cases of the results are quite classical and it seems likely Gauss would have known of these results since the multiplication theorem is due to him (though I haven't seen explicitly the addition formula, it is obviously used in many proofs, often too obvious to be spelled out as a theorem). I spent a summer working out the various directions this could go a few years ago because I felt I could really use this basis well as a stepping stone into learning about all the areas mentioned. There are some very strange looking inversion requirements as well, and this whole area still interests me intensely. I now use alot of the experience gained from studying this one problem and its ramifications to work out many of the entires in Ramanujan's Notebooks as published by Berndt as exercises and try to figure out related problems for myself. I thought it would be nice to introduce here and think that anyone following this thread that wants to get some really nice exercise in the fields mentioned should follow through with the calcuations. I really think it can be quite an educational experience, and it is quite interesting once you get started.
Well, SeventhStar mentioned that he would enjoy it better if the questions weren't quite so in-depth so everyone can work on them more easily without needing to get all the definitions straight. I've been working more towards the in-depth questions with the possibility of extending the knowledge learned over various fields of knowledge, as I though that would be a better way to spend my time. Maybe there is a point somewhere in the middle that we can all work on and enjoy, or maybe some other format entirely would be better? I'll just leave this thread open, and the next question can be proposed in the format that the posting person desires, and we'll see where we can meet :)?
I will give this thread a brief resuscitation attempt. I'll just throw out a problem and let it sit. No time limits. No hints unless someone is actually interested and asks. Just a little problem. Its a mathematical problem, but it probably has no need for definitions to anyone who has followed this thread as it is covered usually at the high-school level, so maybe it falls within SeventhStar's criteria. I just don't like to see this thread die without an open problem...
Lets say you are given two vectors (3 dimensional) A and B and a constant b. Now, you are given the information that
A x X = B (x is cross-product)
A . X = b (. is vector dot product)
Solve for X in terms of A, B, b, and the magnitude of A.
Try it out. There are nice little pieces to this puzzle, and it doesn't involve any higher mathematics. If you understand the problem, you'd understand the solution.
Well I may have not understood your problem because I only needed the first equation (3 equation for 3 unknow variable)..
Without calcul error solution may looks like...(and assuming cross product is vectorial product)
x1=(b1a1+b2a2+b3a3)/(a1a3-a2a3);
x2=(b1a1+b2a2+b3a3)/(a2a1-a3a1);
x3=(b1a1+b2a2+b3a3)/(a3a2-a1a2);
where 1 2 3 are indices for x y z axes...
A x X = B (x is cross-product)
A . X = b (. is vector dot product)
Solve for X in terms of A, B, b, and the magnitude of A.
Now
||A x X|| = ||A||*||X||*sina (alfa is angle formed between A and X)
||A . X|| = ||A||*||X||*cosa (alfa is angle formed between A and X)
Thus:
||A . X||² + ||A x X||² = ||A||²*||X||²*(cos²a + sin²a) = ||A||²*||X||²
or
sqrt(B . B + b*b)/(||A||) = ||X||
The direction of X is given by B x A. So we have the direction and the size, which should be sufficient.
False.Quote:
Originally posted by Simon666
The direction of X is given by B x A. So we have the direction and the size, which should be sufficient.
Let A=[1 0 0] and X=[1 1 0]; Then B=[0 0 1] and BxA is not in the direction of X.
Use the Vector Triple Product rule.
Ax(BxC)=B(A.C)-C(A.B)
Using AxX=B and A.X=b .
Ax(AxX)=A(A.X)-X(A.A)
Ax(B)=A(b )-X||A||^2
so X=(b *A-AxB)/||A||^2
I'm soooo sorry I didn't get back earlier! I didn't think anyone would respond so quickly (normally this thread lies dormant for quite some time) and I've been busy with work, and I didn't feel like it, and my chi was all bunched up, and... Anyway, Tom Frohman got the full answer, but I actually had solved it pretty much along the lines of Simon666 (but with the correct direction calculated by rotating the frame so that A and X were in the (x,y) plane -- thus B is along the z axis -- and then re-rotating on completion). But I really like Tom's answer, since its so much more direct and elegant (using the triple product rule seems to bring out the reasoning for the direction of X much better than the rotation).
Pierre31, notice that there are actually an entire family of solutions to each equation (the tranformation in each equation is not invertible).
Note that both Tom's and Simon's answers are the same for length. Since A and B are perpendicular, AxB has length ||A|| ||B|| along the direction perpendicular to both A and B (using the right-hand rule). b A is perpendicular to this last vector (since its in the direction of A), so the sum (or difference as this case is) is just that of two edges of a right-triangle and the theorem of Pythagoras applies. You can then factor out a length of ||A|| which cancels a term in the denominator.
Anyway, I think the point belongs to Tom, but either Tom Frohman or Simon666 can post the next question if they like.
Well I'm not much on constructing puzzles so I'll give you a checker problem. Sorry its either that or a chess problem.
This is a stroke problem. White is moving up the board and red is moving down.
White has a sequence of moves that takes all of the red pieces by force. White will make 7 moves all told(multiple jumps still count as 1 move in this tally).
White moves first.
Checker notation. If you look at the board and position in the enclosed file you will see the squares of the board are numbered 1 to 32 starting at the top left green square and moving left to right and then down.
If white moves the piece at the lower left out of the corner the move is recorded as 29-25 (not the right first move.) For jumps and multiple jumps just name the squares that the piece lands on. Omit the square jumped over.
I was able to remove all red pieces by force but I moved the red pieces in such a way that if red was actually trying to win then they would never do this. Here goes.
W: 6->2
R: 13->17
W: 21->14
R: 24->27
W: 14->10
R: 20->24
W: 10->3->12->19->28
R: 27->31
W: 22->18
R: 31->26
W: 30->23
That's only 6 moves by White. One assumption that I made was the White piece that made it to spot 3 it was kinged and allowed to keep going with the jumping. If not then it would be 7 moves. Just keep the white there and continue the jump on the next turn while red moves continue in the order listed. It’s been a while since I played checkers so I might be way off.
-Ben
Not the solution.
In the actual solution red can only make one choice at each move.
That is a huge hint.
On each turn, red will only have one choice as to what move he may legally make. For this to be true either there is only 1 move available to red as all the other pieces are gone or tied up or else red has to take a jump (if a jump is available you are required to take it.)
Tom I'm not really good familiar with the rules of this game so tell me:
If I can jump over to take a piece of my opponent am I obliged to by these rules, or can I make another move?
If a jump is available, you must take it.
If during a jump you land on the last rank and are "kinged", the move ends there.
A piece that is alreay a king keeps jumping if it can.
Forcing your opponent to jump is the most important principle of this combination.
I play by the rules of English Draughts. I think there are 14 million variants of checkers played around the world. I guess I really didn't think about this. In the US almost everybody has played checkers when they were young and almost everybody knows the rules.
Ok I got it. I'll write it out in columns, 1st white, 2nd red, with all jump locations for clarity.
I didn't know that one must jump if available. Are those tournament rules? Also, the big thing that kept sticking me was I really had this desire to go 06 - 02 instead of 06 - 01, and I have no idea why. It wasn't until I saw this that I was able to solve it today (I had the set up ready last night and it took me that long to see that step -- I think its the first step where white has a choice on how to force black). Anyway, it was a nice problem indeed!Code:22 - 17 13 - 22
29 - 25 22 - 29
30 - 25 29 - 22
21 - 17 22 - 13
06 - 01 13 - 06
01 - 10 - 03 - 12 - 19 - 28 20 - 24
28 - 19 win
Well done, we have a winner.
Having to take a jump is a rule in most checker variants. It is the basis of most checker combinations.
ok I have a riddle now. Just wait a sec for me to post it ;)
there are four men that need to pass a bridge through the night. They are soldiers and two of them are injured, one - seriously.
they have a flashlight and the bridge cannot be passed without a flashlight and it can carry no more than two men.
the first onw can pass the bridge for 1 minute, the second for 2 the third, slightly injured man, for 5 and the fourth for 10.
In 17 minutes the birdge will be blown to pieces so they need to pass it within that time.
How will they do it?
Let me name these 4 soldiers for A( 1minute needed to pass the bridge) , B(2 min), C(5 mins), D(10mins).
First, A and B pass the bridge with the flashlight.
Then A comeback with flashlight.
Time Now: 3 mins.
Second, C and D pass the bridge withe the flashlight carried back by A.
B carry the flashlight and pass the bridge to get A.
Time Now: 3+10+2 = 15 min.
Last, A and B pass the bridge in the left 2 min..
Time Now: 15+2 = 17 min..
The bridge is blown.. No enemy is coming..