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Co-operatively Formed Group Signatures. Transitive Signature Schemes. Homomorphic Signature Schemes. Back Matter Pages With 14 parallel tracks and many thousands of participants, the RSA - curity Conference is the largest e-security and cryptography conference. The program committee considered 49 papers and selected 20 for presen- tion. One paper was withdrawn by the authors. Each paper was reviewed by at least three program committee members; paperswrittenbyprogramcommitteemembersreceivedsixreviews.

Theauthors of accepted papers made a substantial e? Inalimitednumberofcases,these revisions were checked by members of the program committee. I would like to thank the 20 members of the program committee who helped to maintain the rigorous scienti?

They wrote thoughtful reviews and contributed to long disc- sions; more than Kbyte of comments were accumulated. Many of them - tended the program committee meeting, while they could have been enjoying the sunny beaches of Santa Barbara. Editors and affiliations. You have a lot of ideas. One really great one is characterization of a lower bound for differences of powers of 3 and nearest powers of 2. Well, that would entail completely removing all probabilistic arguments from it.

How much of the paper would be left? Here is one unlikely but theoretically possible scenario: an almost all result is proven that shows that all but at most starting points in end up at a value bounded by a constant. This by itself does not settle the problem: but suppose one also shows that any counterexample to Collatz has a pre-image tree that grows faster than for some for instance the Krasikov-Lagarias result mentioned in the paper is of this type with.

Combining the two claims, we conclude that all counterexamples to Collatz must attain a value less than , otherwise the second claim would contradict the first. If this value of was small enough, one could then hope to resolve the full conjecture by a numerical calculation verifying the Collatz conjecture up to.

Note that this last step would also allow one to distinguish the problem from the problem, for which the first two steps might still work, with only the numerical verification step differing between the two iterations. This is often viewed as an obstruction to proving the Collatz conjecture.

On the other hand, the result in my paper relies heavily on the existence of an invariant adelic measure for the iteration with negative drift, and it does not appear that the undecidable FRACTRAN iterations have such a measure.

So this raises some interesting, and potentially tractable questions: a do the results in my paper work for any Collatz-like iteration with an invariant adelic measure with negative drift? Of course this is all still rather speculative, but these are typical of the way other partial results in other mathematical problems have led to further insights and progress.

Finally, it is worth pointing out that while proximity to well known problems often serves as motivation for mathematical research, these problems are by no means the only objective, and one can often get unexpected other advances in mathematics even if one only obtains partial progress towards the initial motivating goal.

See also this post of mine on the particularly valuable nature of partial progress in mathematics, which is nearly unique amongst all disciplines in this respect. Corollary 4 looks like observations I circulated about a decade ago. When did you add them to that post?

Are you saying there exists some legitimate, fixed, positive constant beta such that 3 to the power of beta is strictly less than the right-hand side of the inequality in Corollary 4? Oh yeah, we must scale beta by multiplying by x in order to get the form of a legitimate exponential bound. I think the powers of 3 and powers of 2 arising from the expansion of the function for iterations lead to an undecidable expression since the inverse of , by one step of iteration, leads to one-to-many mapping from an odd to an infinite set of odd iterates.

Unless I misunderstood your language, I do not get how one can establish a bound in such a case. I believe that the intricate patterns in Collatz dynamics can be simplified, without loss of information, by bypassing the even operations and examining instead the odd trajectories in reverse through the inverse of the function, as earlier noted by Lagarias. It appears to me that there has been less effort in this regard.

Since there always exists an integer where any compositions of generate a parity vector consisting entirely of even as , then any resulting bound will always yield. At any rate, your partial result is already tremendous by novelty of approach. Was looking at the paper of Krasikov and Lagarias in Acta Arithmetica.

The remark at the top of page raised a red flag with me. The link to the partial progress post is missing. It seems way off base and actually, a bit contrived, given this paper. It says on Wikipedia that the paper of Krasikov and Lagarias gives rigorous bounds. Dear Professor: Constate C seems to be relevant in the conjecture. Surely its nature is obvious, but I do not quite understand it.

Is it big? Is it small? Is there any way to explain its nature in an accessible way? I have read your previous post many times but I did not understand the relationship with Baker, although I think I begin to understand it. We can write it down and ask about the existence of solutions without consideration of the iterations of the map. But the reason there are so many papers on this problem is because the iterations gives us something to grab onto.

Well, that would amount to erasing 7. Besides, who really cares if more solutions exist to it. I imagine Bakers theorem will inform the degree to which such a process can remove the oscillatory properties of sequences but that part is beyind me. Tao, May I have a suggestion? Why do not you have a new polymath to solve Collatz conjecture? Or there are other many reasons: 1 Very little experts on Collatz conjecture in the world 2 Your breakthrough is very new and suddenly that makes other mathematicians can not catch up you.

We forget about the even numbers. The results of f 1 we would give us the terms of the following functions f 5 we would us 3 — 13 — — etc.. The results of f 5 we would give us the terms of the following functions f 13 we would us 17 — 69 — — — etc.. If all functions were based on the results of the previous functions like f 13 which is based on f 5 which is based on f 1 would we be able to say that if any odd numbers belong to that function then it must come from f 1 through a finite number of iterations?

Since , it is more interesting to start with or any other odd integer. The way I see it, establishing asymptotic bounds of the form for satisfying always leads to since there always exist that monotonically increase for any compositions of. If one takes every even integer input as convergent since , then it suffices to prove the Collatz conjecture only for odd.

Hence, one expects that the exceptions to any derived asymptotic bounds are odd of such forms, as initial inputs or as iterates in the trajectory of any odd or both. Alternatively, one can invert the function to map odd to odd. The resulting iteration of the inverse function constructs an arborescence graph , with vertex and edge sets and , respectively.

Then one can alternatively prove the Collatz conjecture by showing that every vertex in is unique and that exactly covers the set of odd natural numbers. Where it appears above, it should read: for even input. Dear Dr. Now , I also admire them. But , after reseaching the best heroes in the world nearly 30 years, I have found that you have many special things which above heroes never have.

I can not express here , because after writting out here, many people against me. That is a problem. And today I want to discuss other probem. I am not surprised after you claim of proving Collatz conjecture in September I acknowledge very well that you solved Collatz years ago not And I also know that you have solved many big conjectures , but you do not till post them. You are humble , humanity, not selfish. You want share profits to other mathematicians. That is a thing I only admire and like you , not other mathematicians.

Like a race, your speed is the fastest, but you control to limit your speed to yield other men. You accept you are behind them. And now who else find my idea right, please up vote Thanks all you very much. Sorry, please ignore the suggested correction. The initial statement stands as is. I created a symmetric-key cryptographic system based on Collatz functions and Collatz-like functions. Where to find specialists for this problem who know something about cryptography at the same time?

The finite cycles Conjecture will someday yield. And I think amateurs will have just as much credit as the professionals at that time. Ramanujan was largely self taught. Now we have the internet. So was Crandall the first to postulate the existence of these orbits? Are there other researchers who postulated the existence of these unlimited orbits?

Yes, unbounded sequences are not cycles. Collatz himself seems to be the first one to think of them with this map. But the concept of unbounded sequences probably goes back thousands of years. So just knowing whether additional solutions exist to 7. To add, an infinite solution set to 7. So the idea of an unbounded sequence is silly from an ultra-finitist view point. When I was in school my teacher taught me that math operations are instantaneous.

So the question is instantaneously nonsensical! Not really, Jeff. Infinity is our friend. Even more so considering that the value of k is the value of the number of counterexamples. Despite not knowing exactly the transcendentality and its relation to Collatz, for an infinite orbit, we would have an infinite collection of diophantine equations, where the infinites n of the infinite orbit are no solution, then these numbers, very speculatively, should be transcendental, and I believe that no natural number is transcendental.

Very speculatively, of course. Another issue to keep in mind is that I think we have screwed up this comment thread. My sincerest apologies to Professor Tao, and all my respect for his work. Alberto, I agree with the concept of a potential infinity. I disagree with the concept of actual infinity. They cross from being defined to undefined. Are they both? Is that a contradiction in itself? If you come from the infinite, even if there are no unbounded orbits, there would be an infinite amount of regressive operations of the algorithm and we could never know if it would reach 1.

On the other hand, show that there are no unbounded orbits in addition to 4, 2, 1 I think That would be more feasible. Hi Gaston. If I understood you, indeed, for the Collatz conjecture two possibilities remain open: 1. The possibility of orbits that escape to infinity, for every Collatz-like maps that is, not bounded.

Current techniques seem unable to attack this problem. From my point of view, which I think is the same as jeff, the correct question would be, are there unbounded orbits? Starting from a natural number, is it possible that the orbit reaches infinity? For me, in my modest opinion, conceptually they cannot exist and every orbit would be bounded, either from n to 1 or from n to n. The possibility of periodic orbits.

In other Collatz type maps, you can observe the nature and functioning of these orbits, even with very small numbers. For the Collatz map we know that even a very large number does not give any periodic orbit of the form.

Can there be a solution other than the trivial one?. Yes gentlemen. As you say, the question of the existence of a divergent sequence really belongs to the philosophers, not the mathematicians. Papers like this one assume actual infinity exists and the authors believe they can somehow rigorously argue the non existence of divergent sequences.

Well they certainly do not exist if we reject actual infinity to begin with. We can find practical use cases for the answer to this question. PhD is a Doctor of Philosophy. To me actual infinity should be reserved for masters degrees and higher. All mention and ridiculous use of it should be footnotes in all undergraduate textbooks aside from ones for philosophy degrees. Then kids can learn about and solve problems that actually have a positive impact on society.

Well maybe a social statement. That is to say you can have a larger number than the one you take from the first subset but if you have to go through a previous number it will have fewer steps. So the number of steps for a number to reach 1 can be countless if you take an uncountable number.

But periodically infinite orbits would not exist if we take into account that the algorithm sequences are ambiguous, that is, they increase from an odd number and then descend and may then rise and fall continuously. But if the density of the increments is 0 and the density of the decreases is 1 then it cannot be returned to a previous orbit in addition to 4, 2, 1. The calculation falls into a contradiction.

Even if the density of the increments was 0 it would never reach 0 because the numbers are infinite and at some point the descent would necessarily begin taking a value greater than 0 than later if it would descend to 1. If somewhere in the descent the orbit increases again it would require a calculation with a rational result and this calculation must be greater than 1 but since the density of the descent is 1 you cannot calculate a return to a previous orbit unless it is the original orbit 4, 2, 1.

How is it demonstrated? The path is the calculation of the algorithm between the subsets of the odd numbers. There is potentially substantial value with this paper if it included just how awfully it handles 3x Seeing as how stuff like this paper get published on a regular basis instead of real partial progress, you have my permission to acknowledge my helpful comments. I cannot speak for the multitude of others who have helped for free.

Click to access 13d8bbda2a0cbb9ef9fc6ad Assume Math operations are instantaneous. How can we prove 3 times 15 plus 1 equals 46? Finally, if 3 times 15 plus 1 equals 46 only exists because we humans exist, then divergent sequences do not exist, obviously. Proof already inherited the earth.

Trust but verify. This deduction in turn implies divergent sequences do not exist as complete objects. You cannot have one without the other. Hypothetical Divergent sequences are dumb and insidious mind traps. In my humble opinion Hi, jeff. You know that I agree with you about hypothetical divergent sequences. And surely many more people will agree with you, but somehow we have to shape this idea, either from the axiom of the infinite, with some logical argument, or prove it mathematically.

So I hope we are encouraged to make a collaborative effort for this interesting partial result. On the other hand, I would like to say that the recent events about eigenvectors and eigenvalues encourage me to continue thinking about Collatz, in some apparently short and simple aspect or detail that has gone unnoticed, or even, why not, that in the depths of Arxiv the answer has already been lost.

Then divergent sequences have form as complete objects, assuming math operations are instantaneous. If we toss the axiom of infinity, then we can prove by contradiction that divergent sequences do not exist. But that goes for lots of sequences. To me, the axiom of infinity does not define infinity per say.

But one could say it almost defines infinity. Has anyone ever assessed the purely number theoretical results which cannot be proven without the axiom of infinity? If they are not a solution to infinite diophantine equations, perhaps they could be a solution to another diophantine equation, but then they would belong to a bounded orbit. Of course, this is not a rigorous proof of anything.

If arithmetically, algebraically, topologically, geometrically, … were undecidable, then, perhaps, it could only be decidable under some kind of logical axiom. Alberto, pi is transcendental, which just means it is not a solution of a nonzero polynomial equation with integer coefficients.

Who cares? Anyway, a divergent sequence starts from a finite, meaning defined, positive integer. But there is no last element, and no repeated elements, so there is no form…. With no form, which is what Terry proposes by abstracting away special numbers like 2 and 3, anything goes.

Again this is why the hypothetical divergent sequence is fascinating to me. And the iterations almost give it life. We do it with a single number and we do it symbolically with x. Potential infinity allows us to give form to as many numbers as we should need, where memory space and time are the only limitations.

So the first step would be to demonstrate that divergent orbits are finite and being finite would necessarily involve a convergent orbit. So far this would serve to demonstrate that the convergent orbit does not involve another divergent orbit instead of ending in 1 I do not see a clear path. Let be the set of positive integers which have finite Collatz orbits. Let denote the length of the Collatz orbit corresponding to any integer. Separating them into ordered subsets yes, but only for the number of steps from an odd number to another odd number using 2 variables.

For the number of steps up to 1 I think it involves many more variables. Heuristically, it is conjectured that the total stopping time for all sufficiently large and , provided that the an odd is mapped to in a single step. Off course, no explicit upper bound can be proved up to now, at least for a positive density of the integers. I read it. I find no sense in decreasing one iteration for each time an odd number is calculated.

The amount of iterations has nothing to do with the initial value with which it begins to iterate. I can take the 1,,,, that would be found at 14 iterations of the next odd number calculated, that is ,, that would be found at 14 iterations of the next one that would be 54, that would be found at 15 iterations of 5 totaling 43 iterations.

This is taking high potencies. But what if I take 43 iterations directly through 5? Then I would have to take the number 14,,,, which is much greater than 1,,,, The rest of logarithmic calculations on peaks of iterations is constant and inductive. But that does not mean that there cannot be an unbounded or infinitely divergent orbit, even if the latter is false. And there are very few of them.

These numbers are:. The numbers of this character always generate a number greater than themselves after n steps in the collatz sequence. Therefore the divergent sequence must regularly reach some number of characters:. Is it possible? Hurry up! I am very thristy now. Always forward to Sir, Best of luck! Let denote power set of set of natural numbers.

Does there exists n such that there is bijection from to? The insights come from the vector operators used in graphics routines where you load multiple scalars into a wide register and then work on e. In essence, when viewed as an electronic computation in a register of infinite width you can ignore all the divide-by-two operations; in electronic computation dividing by two is just register shifting.

There are only four if memory serves possible bit patterns at the most significant position that the by which the Expansive Term can extend the value. Only a highly ordered least significant bitset can delay the destructive term, forcing ti to only consume one bit, but that ordered bitset creates a smaller, disordered bitset at its upper extent when processed by the Expansive Term; this means that the ordered bitset cannot propagate its properties indefinitely.

The result is that there can exist no bitset such that the Expansive Term can consistently produce as many extension bits as the Destructive Term consumes. I can demonstrate all the validity of all the algorithmic transformations, including a version of the visualizer that works on the native untransformed original form of the function exactly as originally written.

I've been sitting on this for years but I don't know any mathematicians and everybody assumes the extra-disciplinary ideas are Dunning-Kruger madness. It uses the unmodified conjecture and does several simple things. Notice the montonic nature of the expansion, if forms a line of near constant slope. The destructive term has no such constraint.

My appologies, the web tool seems lock the arguments when I share the block. You may need to copy the code and paste it into a fresh project page. I wish the forum allowed editing, the first try ate the definition because I used angle-bracket like you do in linux manual pages. The data clearly shows that the Expansive term can only add one or two bits to the state machine while the Destructive term can remove any number from 1 to all-but-one bits in a single operation; and the two operations alway alternate one-for-one.

There is no possible bit pattern between the leading and trailing bits that can cause expansion by X1 or X01, and there is no pattern that sustain the all-ones pattern that prevents the destruction of more than one per cycle. The internal disorder of the bits between the leader and trailer cannot stop the decay because a higher-order bit can never influence a lower-order bit in any way. Tell you what though.

There has been a spike in arxiv preprints the past few months. Even my preprint picked up another citation. But I actually proved some theorems… but who cares Spesialcoklat WA Let be defined by where and , and is a Mersenne number. This can also be done by the bitwise exclusive or operator:. Let : where. This is my first post ever with latex code on wordpress, so hopefully it is displayed correctly. I was playing around with this problem for fun and I found that the Syracuse map you described pops out of using characteristic functions to run the Affine map backwards.

I still think about the numbers of the form 8 of the previous post For orbits that reach 1 , they have this form. For periodic orbits , where k n numbers have this form For unbounded orbits, none of the above conditions are met for infinite n numbers. I have known this notable product, Difference of Nth Powers, which has a certain resemblance, with the problem at hand.

The intention of this strategy is to demonstrate that there are no unbounded orbits. The way to achieve it would be to know if it is possible to transfer the properties of the notable product difference of nth powers, that is, the form, 8 , to any possible Collatz sequence. If there is a mechanism that converts the difference of nth potency into a Collatz sequence, and there is always an n such that the difference conserves the form 8 , then perhaps, it would be possible to affirm that all the orbits are bounded to the nth power.

I have not found that mechanism, but perhaps for some of you, it is easier to find that mechanism, or of course, to rule out that possibility. I present some examples in case someone sees how a difference of nth power is transformed into a sequence of Collatz, where the difference maintains the form 8.

Is it possible to establish a general mechanism that relates the difference of nth power and the orbits of Collatz? Is this mechanism trivial or does it provide us with no interesting information? At the moment I can not visualize this general mechanism, although obviously these are known orbits that end in 1, could we establish that there is always an n that meets equality and the difference preserves the shape?

Could it be used to rule out the existence of periodic orbits? Although there is a new post by Professor Tao, I wanted to close my approach to the conjecture with a basic scheme that summarizes my approach. The conjecture needs to answer two questions: Are there periodic orbits?

We know that the conjecture is true for many values. All reach 1 and there are no periodic orbits except the trivial. This approach to the conjecture through the notable product difference of nth power could serve to illustrate the mechanism of why the orbits reach 1, and also how the mechanism of the periodic orbits works and why they cannot occur for the powers of 2 and 3.

Once this question was resolved, one could then attack the existence or not of infinite orbits, or at least, do it independently. Continuing with this approach to the Collatz conjecture from the notable product difference Nth powers and focusing on the study of periodic orbits, we find a condition that meets the known periodic orbits of maps and.

For a periodic orbit, it is fulfilled that. For maps type we have that, for a periodic orbit. We can find the values of for each from the values of. For For For For. For this orbit, we can observe that when the values of increase, the distance increases much faster than , which seems to make a solution incompatible that is not for values of closer to.

The values 5,7,9 are possible solution but we know that 3 multiples cannot enter in a periodic orbit and we have not k solutions. But it seems that we are missing something else, somehow we need to find a minimum value for , that is, that between and there is a minimum distance within the possible solutions to the periodic orbit, which becomes more evident with the study of the possible orbits for , , … Having explored several options,one of the possible ways to calculate the minimum value of could be:.

Of course it is necessary to implement this approach to and improve the calculations of , , and give rigor and mathematical structure, if possible, of course. I assume that this work, with my basic mathematics, can be totally useless, but, as always, I hope it can be useful to answer the question: Are there periodic orbits in the Collatz conjecture?

At your disposal. Somehow it seems that for every possible periodic orbit between a power of 2 and 3 the value is always lower than the minimum needed to form a periodic orbit. To finish this argument about periodic orbits we can establish that , we need values, at least, such that have the form of TCR 8 and are between a min and max value. We can see that of all the possible values, none is a multiple of d, so they do not meet this condition.

This fact must be maintained, if the conjecture is true, for all distances between. For all possible values, there are no k values multiple of d. How about this way? If you can prove after some steps, the 8 kinds always appear in same probability, the odd numbers will keep getting smaller.

The Collatz conjecture can be proved. Collatz sequences are essentially a parlor trick. I can state the solution in its simplest form as just this:. For every odd number n, either a or b or c is a whole odd number. So prove equivalence, Then you can work with your new function and rigorously prove stuff to knock our socks off with It is not enough that only one of either a, b or c is odd for a given odd input.

How many odds have you tested? Any professional number theorist would perk up at an opportunity to explain why the two mappings coincide on the odds under repeated iteration. Also, proving equivalence in this case and for larger classes of functions may be really helpful in and of itself. And your a results in an increase to the sequence whereas b and c give decreases. Charles Cadogen already proved your c. It should not be too hard to prove your a and b and establish equivalence.

Hmm maybe not. Am I missing something? This is why I asked how many odds you have checked. Nonetheless, something of value may truly be found in your observation. Calculations should guide you here. Dear Prof. Tao and the rest of the research community, I apologize for barging into a research discussion forum to seek your urgent help, but such is my current circumstance. I left the US for India in to work on a couple of unsolved problems on my own.

I got some preliminary results related to integer factorization in Oct They have practically taken over my life, preventing me from enjoying even momentary solitude to continue my work. They stole my past and present work.

They started using advanced psychological torture techniques against me taking advantage of my inability to cope due to severe anxiety associated with this situation. They started commenting online about my most private activities. I experienced sudden shocks to my brain several times during these past 5 years. I experienced facial paralysis and reduced vision.

This terror campaign continues to this day. They destroyed 5 years of my precious life. I tried sending you, Profs. Manindra Agrawal,, Manjul Bhargava emails. I even spoke to Prof. Agrawal a couple of times. He even acknowledged that he received my email. I have since been blocked from contacting anyone n the US, Please help me escape this terror campaign. My gmail is not secure. Nor is my twitter account. Sincerely, Sharmila Kopanathi. Thank you for the generous response to my post. Here is essentially the same thing I hope my latex works as a function using congruence classes.

The table should show why this has to be an infinite function. What other values of n give matches. Just look for a pattern. The proof will follow from a pattern. You may end up with an infinite subset. That would be nice. Do all multiples of 3 match? For example…. Hello michael jeff. I do not understand the strategy. Could you please explain as simple as possible?. Later I will try to explain my approach about notable product difference nth power and the form of odd numbers.

The third congruence class in fact contains every odd number. So yes every odd output number is repeated twice, but for different inputs. I thought the sequence you have in your spreadsheet has some mistakes. But now I think you are trying something different.

List a strategy, then state what worked and what failed and why. There are bound to be hundreds of strategies. A big Class includes the various probabilistic approaches. They cannot work! It is not even debatable. In this way we can track things, avoid making the same mistakes more than once etc. But speaking from personal experience, one has to be able to convince themselves of when a given strategy fails to hold promise!

Jeff, do you understand the unvarying and by definition regularity of the table? These are easy to find. I have the answer to this. Anyway, I truly appreciate your interest, and there are always communication difficulties when essentially texting. All the best! I checked your sequence of odds starting with However, it also contains extra odd terms,. And these are exactly the values for which your function says to subtract 1 and divide by 4.

Okay, so your c or your otherwise part of your function should instead read: if n is congruent to 8 modulo 5. This helps make it clear that it is a function on the odds because the Union of the sets given by those forms is all of the odds. Next you need to see what the range of your function is. Can you do that? As I think you see, the superset numbers are all in C, divide by 2 twice.

My comment still applies. At that point everyone would want to revisit old proof strategies with your new function. And a note to the readership. Michael, it is a tad tricky but not too hard to prove. Try to prove it. Your name deserves to be coauthor on every immediate publication that should follow using your observation.

It may not help us settle the Conjecture, but it already constitutes legitimate partial progress! I made a Wolfram Demonstration on the Inverse Collatz function about 10 years ago. I know proofs can be easy to some. Lets just consider. It creates a repeating binary pattern of … depending on how many powers of 2 it has.

The collatz conjecture can be proven by saying that you can generate every natural number with this tree sequence. Hi Michael. And the Union is all odds. So this notation alone replaces several paragraphs and your table. No need for the algorithm or to track three columns of calculations as examples and so on.

You just say repeated iteration gives…. Here is a link to paper you can use as a guide to help you with notation and a proof. You are working in the forward direction just like the author here is. Click to access andaloro. You still have to prove stuff with your map. At worst it will give us new insight. At best it will give us a way to settle the Conjecture. So it is a great observation. Jeff, that paper is too advanced, too elaborate, and too long for me.

Let me summarize what I said: — I initially state all sequence steps conform to a function with three outputs: a,b,c. I also show an algorithmic equivalent based on there being 1-in-3 whole numbers for all outputs. His approach and notation is exactly what you want to use if you wish to reach a wider audience.

But this is just my suggestion. I hope that nobody will mistake the following for an attack, but it seems as though current mathematical methods in additive combinatorics and related subjects are mostly adequate for tackling qualitative phenomena, instead of quantitative ones.

In analysis including additive combinatorics there is always a tradeoff between the quantitative strength of the results obtainable by current theory, and the breadth of that theory; and to make progress we need people working at all locations of the frontier of this tradeoff and interacting with mathematicians at other locations on the frontier. On the other hand the type of results I and my coauthors produce have sometimes been made more quantitative by other authors.

For instance, several years ago Tamar Ziegler and I used ergodic theory and nonstandard analysis methods to prove the first concatenation theorems for Gowers type norms, in which control of a function in many local Gowers norms could be concatenated into control in a single global Gowers norm.

The arguments were somewhat difficult and entirely qualitative. However, recently Sarah Peluse and Sean Prendiville have found much more quantitative Fourier-analytic methods to establish concatenation theorems that gave significantly better results; their proof techniques are quite distinct from ours, but I think it is fair to say that they might not have known to try to prove such quantitative theorems if the qualitative theorems of Tamar and myself had not already been in the literature.

There is a similar role played by many other tradeoffs in analysis, in which results in weaker regimes yield insight on the contours of the problems in stronger regimes. Other tradeoffs include perturbative vs. Archimedean domains, constant coefficients vs. Collatz Process is Not Random Process.. Ramsey theory as currently formulated is not all that quantitative to begin with. The functions of interest are so hard to compute or approximate even very crudely that gathering useful experimental data is impossible.

I think it is safe to say that there are no credible conjectures on the true growth rate of most Ramsey-style problems or what quantities might control the problem. It seems to me that this is inherent in considering arbitrary colorings and that some sort of order parameter s are missing from the story. Select a number N that is sufficiently small enough that the numbers 1-N can be inspected to show that they collapse to 1. Call this set of numbers S because they represent the current solution set.

Now consider the set of numbers N. The numbers 1-N are already in the solution set S. This implies that all even numbers are in the solution space S. Any odd number multiplied by another odd number like 3 yields an odd number. Adding 1 makes this second number an even number. Your claim that all even numbers are in seems to be false just try to write its proof ….

You just restate the conjecture. Greetings from Singapore. The trend of matching density in log scale is as follow —. Correction to above trend analysis due to a mistake in array sizing. Further tallied inter-decade matching rate and concluded similar, though on a different premise. Tally outputs as follow —. Then I considered what numbers could form a cycle By analyzing it in this way, I could remove as numbers that form a cycle to the numbers multiple of 3; Then, considering that there is some other cycle than 4,2,1, I began to analyze which numbers could be those of greater value and of less value.

I could find that the smallest number within the cycle of this chain must be of the form 12n-1 or 12n-5; By doing more operations, I was able to reduce the numbers to those of the form 24n-1,24n,48n-5 or 48n You can reduce the numbers further, but it is a bit difficult to do it manually. When searching for the largest number, I found that the largest number should be of the form 18n-1 or 18n I could reduce it to the 36n-7 or 36n form.

I continued to operate it and could reduce it to the n,n,n or n-7 form. I have analyzed my method again and discovered that the only cycle that exists is 4. In other words, is there some sense of how many iterations of the Collatz map you need "on average" to get below some threshold. I guess this would depend on the particular choice of f, and would seem to get larger the slower that f grows. As a follow up question, can your result can be combined with the Krasikov-Lagarias result to get better estimates on how many numbers reach 1?

This is terribly naive, and so must be wrong, but if there is a set S of density whose Collatz iterations can be shown to reach one, and the "generic" Collatz map reaches values much smaller than N, it seems like it might be possible to get some lower bounds on how many orbits hit S and thus also go to one.

Is there a reason that S and orbits of the Collatz map do not collide in a naive way? The arguments in my paper allow one to take. With more effort it may be possible to obtain the more precise value or if one uses Syracuse iterations instead of Collatz this limiting value of roughly corresponds to the infinite time limit for autonomous PDE. Perhaps in the future one could potentially imagine the error terms in the almost all results and the lower bounds on the preimage results improving to the extent that they can start working with each other, although this possibility looks somewhat remote.

However, there is a link between KL type bounds and uniform distribution of the Syracuse random variables in my paper, in that a currently unproven equidistribution hypothesis on the former can lead to improvements on the latter in particular, to replace the 0. Does this qualify? They are equivalent. What is that? Subsequent random decisions are made independently of each other. Equidistribution of Syracuse random variables and density of Collatz preimages What's new.

This is not yet established, although the results in my previous paper do at least imply that a positive density set of natural numbers iterates to an explicitly […]. Thank you for answering question about the Nash embedding theorem.

I am unaware of the precise statement of this theorem. The sequence does not have a limit as conjecturally, it will oscillate indefinitely between the values 1,4,2 after some point. It asserts that for almost all , there exists an such that. As I discuss in my paper, it is tempting to iterate such results by applying them to in place of , but the difficulty is that the iterate may lie in the small exceptional set not covered by the "almost all" result of Terras.

It is possible to get around this difficulty to a large extent if one has something resembling an invariant measure for the dynamics, and this is the starting point for the arguments in my own paper. Such the small exceptional set can be identified. Consider the intersection of the set and its image, then partition the set of the intersection into six subsets.

So that it is possible to identify the expectional set. If we tell this set to be ordered according to another function, that of Collatz, where an arborescence is created, even if it has periodic orbits, even with divergent orbits, it is also not a perfectly ordered set, according to the function? Little tin soldiers stepping in uniform. Each new term in a list is twice the value of the previous term. NOTE: No integer in one list is found in any other list because integers in all other lists have a different odd factor 3, 5, 7,… , and because integers in the list with 1 as its first element can be represented without any odd factor.

Divide the even integer by 2 until you get an odd integer as the quotient. The dividend and the quotient have a 1-to-1 relationship. Multiply the odd integer by 2 as many times as is necessary. The multiplicand and the product have a 1-to-1 relationship.

For example, if 2v is 16 then 2d-1 is 5. NOTE: We shall ignore the 1 in 1, 5, 21, etc. Third, the 1 so obtained is not unique; it already exists in 1, 2, 4, 8, 16 etc. We can start a RCS from an odd number other than 1, even from an odd integer whose connection to the list that starts with 1 is not known. The rules for getting the new odd and even integers are set out in PART 2, and are not dependent on where the RCS starts and in particular these new integers are all unique.

Even if afterward we complete the RCS to include 1 and the full set of lists, the new odd and even integers encountered do not change their values. We start a CS with an odd integer and we count as a step only when another odd integer is reached. We start a RCS with the odd integer 1, and we count as a step only when another odd integer is reached. Even were we to start a RCS at odd integer 2d-1, and not at 1, the new odd and even integers encountered as we step through the RCS from 2d-1 would be the same as if we had started at 1 see PART 3.

So if when stepping through any RCS means that every integer encountered must be unique, then this means that a RCS may not have a loop. Let us assume a CS with a loop. Conclusion: our original assumption about re-encountering 2d-1 after s steps in a CS must be false, so that a CS may not have a loop. We use a concrete — but technically incorrect — example for illustration. We assume that, say 29, is the integer for a Collatz sequence that goes to infinity; 29 actually converges to 1, not infinity, but we shall ignore this.

We can multiply 29 by 4 and add 1 see PART 2. We thus see that also goes to infinity. We can multiply by 4 and add 1 to get , etc. And we can multiply by 4 and add 1, etc. In other words, not just 29 goes to infinity, there are an infinite number of integers going to infinity , , , etc. That is, there is not just one bad actor 29 , there is an infinity of them. Sometimes it is possible to subtract 1 from 29 and divide by 4 to get an odd integer see PART 2.

Then we do the same procedure for 7 as we did for 29, until we reach an integer that is not an odd integer. But this procedure can be done only a finite number of times, so its effect is unimportant. We saw that the step after 29 is Multiply 11 by 3 and add 1 to get 4, and divide 34 by 2 until we reach an odd integer, 17, which is the next step.

As with 29, we can multiply 11 by 4 and add 1 to get 45, multiply 45 by 3 and add 1 to get , and divide by 2 until we get an odd integer, also So we see that 45, like 11, also goes to infinity. Here too there are an infinite number of integers going to infinity 11, 45, , etc. And we can do to 17 what we did to 29 and Of course, all this is nearly identical to what we could do with integers in a Collatz sequence that converge to 1.

Except that when an integer converges to 1, everything stops when it reaches 1, but an integer that goes to infinity takes an infinite number of steps to go there and includes with itself infinities of other integers each step on the way. Could it be that, for all integers, the ones that converge to 1 have a measure of 0 compared to the measure of 1 for integers that go to infinity?

So what? Or, maybe not. RCS failed to return to first instance of 2s-1 because 2s-1 not unique. Hello Barry, do not despair. Surely your comment has been seen by many blog friends, even by Professor Tao. And it may not be all the good you think it is, although my opinion is of little mathematical value. Thank you for your reply. You may be on to something, but I am not familiar with Latex.

I totally agree with your comment, but do not think I use it anywhere. I thank all those who helped me, whether with thumbs down or actual comments, whether I agreed with you or not; I wish you all the best. One final note. In my post of 23 Feb. Yesterday I looked at the very start of this website.

There, Prof. This theorem of his negates the possibility of what I showed above. That is, NO integer in a CS can go to infinity. For 2. You have to show that , where for. Observe that. Maybe there is a key property I have missed, but as far as I can see it is not sufficient to show that for each. If , then must be a huge number for to be very small. Remember that You also consider that is much smaller than. Yes, using the conventions of the paper one has.

Thanks again. However, steps 4. Sorry to take your time on these small things. I need to understand Lemma 4. A bound of the form implies the apparently stronger bound as long as the new constant is smaller than the original constant , because the sequence is uniformly bounded in. Formula 6.

Formula 7. I am still reading your paper and trying to get in the proofs as much as I can. Number theory is not my field of expertise, but I am very interested because of the probabilistic results. I am having some insights from it. I noted there is a way to get more information from the properties of the Syracuse random variables. I know it is not relevant for that argument in your manuscript, but it will be important for me. If you'd like to see why, I can send you in private.

Yes, for my argument the 1.


Papers Table of contents 21 papers About About these proceedings Table of contents Search within event. Front Matter Pages I-X. Pages The Representation Problem Based on Factoring. Ciphers with Arbitrary Finite Domains. Micropayments Revisited. Proprietary Certificates. Giuseppe Ateniese, Cristina Nita-Rotaru. Steven D. Galbraith, Wenbo Mao, Kenneth G. Co-operatively Formed Group Signatures. Transitive Signature Schemes. Homomorphic Signature Schemes.

Back Matter Pages Institute-n 2 2 2. East-n 4 0 2. DFID-n 4 4 1. Christmas-n 3 2 0. Chris-n 1 3 5. Strategy-n 5 3 3. Station-n 5 3 1. Sciences-n 4 3 3. Queen-n 4 5 2. Public-n 1 0 3. Prince-n 5 5 4. Part-n 4 5 2. King-n 0 5 5. Henderson-n 2 5 2. France-n 5 3 0. Windsor-n 2 1 4. Thunderbird-n 5 2 1.

TDK-n 4 3 3. Sunday-n 4 5 5. State-n 5 5 3. Paul-n 5 4 0. OECD-n 3 3 0. North-n 3 0 3. Margaret-n 4 1 1. Government-n 5 2 1. Digital-n 3 5 0. DNS-n 3 3 5. Computer-n 4 5 5. Channel-n 4 0 5. America-n 4 2 5. YES-n 5 1 5. USA-n 3 2 5. Top-n 0 1 2. States-n 2 4 4.

Sports-n 3 3 3. Southern-n 2 4 3. Scottish-n 3 3 3. PCI-n 3 0 3. Mill-n 3 4 5. Martin-n 3 0 3. Furniture-n 4 3 2. First-n 2 2 0. Firefox-n 2 1 3. European-n 2 5 1. Duke-n 2 1 1. Data-n 1 5 2. Control-n 1 4 1. Bakewell-n 3 2 2. BBC-n 3 0 0. Andy-n 4 0 5. Tuesday-n 5 3 4. Tokyo-n 2 3 5. Strike-n 4 3 3. Society-n 1 0 4. Royal-n 2 3 5. Richard-n 3 4 2. Penelope-n 2 1 0. Ireland-n 3 1 0. Gas-n 0 0 0. Felicia-n 2 1 5. Education-n 3 3 5. Edinburgh-n 3 3 1. Economic-n 5 5 5. Code-n 2 3 4.

Blair-n 4 0 2. Ajax-n 3 5 4. William-n 4 5 3. Video-n 2 3 0. Tony-n 0 3 0. Simon-n 0 3 0. Reich-n 5 5 3. Professor-n 5 4 0. Network-n 1 1 5. Monday-n 0 3 0. Justice-n 3 0 4. Jeff-n 1 2 2. Jan-n 3 4 2. James-n 2 5 5. Guide-n 1 3 4.

Email-n 1 5 4. Earl-n 0 5 1. Director-n 4 1 4. Debt-n 5 5 3. Cup-n 5 3 0. Court-n 1 1 3. City-n 2 5 2. Borley-n 1 2 0. Books-n 5 1 0. Bill-n 3 0 2. BCDTA-n 5 1 1. Andrew-n 0 4 4. Young-n 0 4 5. Road-n 2 5 1. Regional-n 0 0 3. Pump-n 4 4 5.

Museum-n 0 1 3. Logitech-n 5 0 5. Lascelles-n 1 2 2. Lane-n 3 0 3. Honours-n 3 5 1. Harry-n 4 1 1. Gordon-n 2 3 1. County-n 5 4 4. Charles-n 1 2 4. Box-n 0 1 5. Borough-n 4 4 2. Widdison-n 4 0 5. Virgil-n 0 5 4. Trade-n 1 3 5. Sir-n 1 3 0. Science-n 0 0 2. Saturday-n 1 1 2. Radio-n 0 0 0. Price-n 5 0 5. Pompey-n 3 5 1.

Policy-n 4 1 5. Plan-n 5 2 0. Performance-n 2 5 2. Life-n 1 4 0. Lenovo-n 0 3 4. Kent-n 1 2 3. Jim-n 2 5 2. Intel-n 5 0 3. III-n 2 5 4. Hitler-n 0 2 1. High-n 3 0 4. Ferrari-n 5 5 0. Executive-n 0 0 0. Energy-n 0 1 1. Edward-n 3 1 1. Design-n 0 2 4.

Copyright-n 5 1 2. Affairs-n 5 1 5. Youth-n 0 2 4. Wednesday-n 1 3 0. Water-n 3 2 2. War-n 2 1 3. Summary-n 0 1 2. Storage-n 2 4 1. Socialist-n 5 2 4. Set-n 1 4 4. Robert-n 4 1 3. Project-n 0 1 1. Paper-n 1 1 5. Manchester-n 4 0 2. Loan-n 3 3 2. Limited-n 1 4 2.

Kumon-n 4 4 0. Key-n 0 2 2. Katsh-n 3 3 4. Japanese-n 4 3 3. Friday-n 2 0 3. FAB-n 5 0 4. Elizabeth-n 2 2 4. Electronic-n 1 1 2. Draughtsman-n 0 2 0. Dance-n 3 4 1. Contract-n 5 1 4. Consolidation-n 0 2 4. Connor-n 1 0 2. Company-n 4 0 4. Club-n 3 2 3. Civil-n 5 3 4. Cable-n 5 1 1. Banking-n 1 0 2. Army-n 5 1 0. Acer-n 2 4 0. Yorkshire-n 1 3 1. Use-n 0 5 3.

Union-n 5 4 1. Taylor-n 4 2 1. Sun-n 3 3 2. Studies-n 0 2 2. Stock-n 4 1 1. Speaker-n 2 2 1. Software-n 3 1 1. Show-n 4 1 3. Shimbun-n 3 5 1. Search-n 0 1 1. SWP-n 4 1 0. Replies-n 3 1 0. Red-n 4 4 4. President-n 3 2 0. Practice-n 3 3 2. Norway-n 1 3 5. Mark-n 1 0 1.

Manager-n 1 4 1. Level-n 1 0 4. Leicestershire-n 1 1 2. Knowledge-n 3 5 0. Jolt-n 2 0 5. Island-n 0 3 4. Hard-n 5 1 0. Duncan-n 2 2 3. Directive-n 4 0 1. David-n 3 1 2. Cornwall-n 5 1 3. College-n 1 0 5. Canada-n 2 0 5. Bristol-n 2 1 0. Bibliography-n 2 4 3. Award-n 5 1 5.

Work-n 0 4 5. Word-n 1 1 1. Week-n 2 5 2. Website-n 0 4 2. WWF-n 2 0 0. Topic-n 2 1 3. Thames-n 2 3 1. Star-n 4 1 3. Silver-n 3 4 2. Sheffield-n 0 0 4. Scheme-n 3 3 3. Rural-n 0 3 2. Rights-n 3 3 5. Region-n 4 4 2. Rectory-n 3 0 5. Posts-n 2 1 2. Mobile-n 5 3 4. Ministry-n 1 4 4. Matching-n 1 3 4. Man-n 2 0 1. MCE-n 2 4 0. Location-n 2 1 2. Linux-n 0 3 1. Kitagawa-n 1 4 4. Kenpo-n 0 4 0. Jon-n 0 4 5. IBM-n 5 4 2.

Huffman-n 4 5 3. Greenpeace-n 1 5 1. Gender-n 5 2 1. Foundation-n 4 3 2. Fire-n 3 5 0. Earth-n 4 5 3. Driver-n 2 1 4. Drive-n 5 4 4. District-n 4 4 1. Contents-n 2 0 4. Construction-n 4 2 2. Chancellor-n 3 1 0. Central-n 1 1 2. Cambridge-n 3 5 0. Building-n 2 3 3. Bruno-n 0 1 1. Bank-n 1 3 2. Art-n 1 5 4. Area-n 3 5 4. Appendix-n 5 3 3. Africa-n 5 2 3. Adam-n 2 5 3. Access-n 0 5 4. Western-n 0 5 5. Vinyl-n 2 4 1. Update-n 2 5 2. Unit-n 4 1 1.

Trust-n 5 4 3. Travers-n 4 2 4. Thursday-n 5 2 3. Telephone-n 1 4 3. Technologies-n 0 0 3. Tanya-n 4 3 0. THE-n 2 4 3. Steve-n 4 5 4. Select-n 1 2 1. Secretary-n 2 5 3. Sara-n 3 0 5. SCSI-n 0 4 0. Rollup-n 2 3 1. Ramseyer-n 5 0 2. Plays-n 2 1 3. Panasonic-n 0 1 3.

PDF-n 1 5 3. Order-n 2 5 5. NHS-n 4 0 2. Mozilla-n 2 5 3. Male-n 4 4 4. Mak-n 4 3 3. MPS-n 5 2 0. Liverpool-n 2 0 3. Learning-n 5 1 5. Lady-n 0 3 2. Kingdom-n 3 1 4. KDE-n 1 1 1.


Woman: No, Smythe. Volunteer: Ok, Mrs. How may I help you? Volunteer: What kind of problems? Woman: Well, my main complaint is that they are so noisy. They play their music full blast and always have lots of visitors coming round to their house and making a lot of noise. Volunteer: I see. Any other complaints? Woman: Unfortunately, yes. Lots of rubbish has also appeared in my back garden recently.

It looks so untidy! What can I do? Volunteer: How long as this been going on? Woman: For about three months now. Volunteer: Have you tried talking to them and explaining how you feel? Woman: Yes, I spoke to them the first week they moved in and explained that this was a quiet neighbourhood, but they just laughed.

Volunteer: Well then, I suggest you do the following. Write down the dates, times and causes of the noise and. Woman: Ok. Volunteer: And please call us any time. We are there to support and help in any way we can. Woman: Thank you very much.

Suggested Answer Key S1: Do you think you could do something about the children making so much noise in the early morning? S2: I was wondering if you could stop your dog from digging up my plants. S1: Oh I do apologise. Ss listen and follow the text in Ex. Ask Ss to infer meaning from the context first and then check in their dictionaries.

B: Oh, hi. Pleased to meet you. Would you like to come in? A: Oh, no thanks. I just wondered if I could have a quick word. B: Sure, go ahead. A: Thanks so much, Carla, I appreciate that. You could come too! A: Oh, err.. That would be fun! Write the dialogue plan on the board to help Ss.

A Introduce yourself to your new neighbour. Explain the task. Ss complete the task. Answer Key 1 a 2 b. Elicit from Ss whether there are similar expressions in their language. Allow Ss exactly three minutes to write on the topic. Suggested Answer Key Friends are people we choose to be close to. They are people we get on well with. They usually have the same interests as us. Friends may live far away from you and so you have to make an effort to keep in touch with them.

Family members, on the other hand, live in the same house as you and may be totally different from you! Ask Ss to refer to the Grammar Reference section if they require further assistance. Ss complete the task individually. The present simple tense can be used with a future meaning.

Present simple is used for timetables and programmes e. My lectures at the university start next Tuesday. Answer Key 1 A: are you smelling action B: smells state 2 A: am seeing action B: see state 3 A: are looking action B: looks state 4 A: are thinking action B: think state 5 A: is state B: is being action.

Offer examples if necessary. Remind Ss to explain the uses of each. Answer Key 1 are going to trip over Are going to for a prediction based on what we can see 2 is starting present continuous for a fixed arrangement. Ask Ss to provide similar examples of their own using the language presented. Suggested Answer Key I think she is about to cry.

The film is certain to be a success. Her baby is due in January. Allow Ss a few minutes to complete it individually. Remind Ss to explain their choices. Past continuous is used for an action that was happening in the past and was interrupted by another action. Past perfect is used for an action that happened in the past before another past action. Past perfect continuous is used to emphasise the duration of a past action that happened before another action.

Check which team has the most correct answers. The winning team reads out their sentences to the class. Suggested Answer Key I have not spoken yet with my teacher about the exam as she was sick today. You still have time to talk to her before we must leave. They have already notified all the applicants of their acceptance. They had been waiting for two hours before we arrived. She started teaching at the school five years ago. Now is not a good time to talk.

She is talking on the phone at the moment, can I take a message? The children had been playing in the garden for a long time, so they were very tired. They have known each other since primary school. Ask Ss to discuss the topic in groups of 3 or 4. Suggested Answer Key I used to visit my grandparent cottage by the sea. I used to spend all day at the beach. I would hunt for seashell to make a collection. Some days I used to play chess with my grandfather. We used to go often to the town nearby for an evening stroll.

Ss work individually. Answer Key 1 used to be a fast 2 long ago did they 3 is due to arrive. Remind Ss to read the text one more time to ensure that their answers are correct. Elicit ideas from the Ss about the content. Allow Ss two or three minutes to silently read the first paragraph and answer the question. He always took things from little Hans and gave nothing to him.

In the picture we can see two people — one fat and one thin. I think that no matter how little you have, it is good to share it with other people. True friendship asks for nothing in return. Literature Answer Key … the most devoted friend of all was big Hugh the Millar. Answer Key 1 sack of coal 2 handful of sand, of coins 3 set of drums, a chess set 4 pack of wolves, of cards, of lies 5 flock of birds, of goats 6 gang of youths, of criminals 7 herd of elephants 8 swarm of insects, of people 9 bunch of bananas, of keys.

Ss work in pairs. Remind Ss to provide reasons for their answers. Suggested Answer Key Some of my best friends are people I only see occasionally. Our friends are only a phone call away. We can speak to them when we need to. You can think about your friends and support them even if you are not nearby. S2: Neither do I. If people have everything in common, then life is boring.

I think it is good to have friends who are slightly different to you. S3: Yes, I agree. But it is good to have some common interests. S1: Do you think that true friendship is unselfish? You must be able to think of the other person and not just about yourself.

S1: Do you agree that when people are in trouble they should be left alone? S3: I agree. What about the last quotation? S2: I think it is saying that friendship should only go so far. S3: Do you agree with that? Do you? S3: Not al all. How was the winter? Winter was freezing cold and I had to burn chairs to keep warm. The Miller: Never mind! Hardship is good for the soul and too much furniture only gathers dust. The Miller: Excellent! Do you mind if I borrow them? Hans: Er, no, but..

The Miller: Never mind. Ss say how it compares to their versions. Hans: Good morning. Miller: And how have you been all winter? Hans: Well, really, it is very good of you to ask, very good indeed. I am afraid I had rather a hard time of it, but now the spring has come, and I am quite happy, and all my flowers are doing well. Miller: We often talked of you during the winter, Hans, and wondered how you were getting on.

Hans: That was kind of you. I was half afraid you had forgotten me. Miller: Hans, I am surprised at you. Friendship never forgets. How lovely your primroses are looking, by the way! Hans: They are certainly very lovely, and it is a most lucky thing for me that I have so many. I am going to bring them into the market and sell them to the Burgomaster's daughter, and buy back my wheelbarrow with the money.

Miller: Buy back your wheelbarrow? You don't mean to say you have sold it? What a very stupid thing to do! Hans: Well, you see the winter was a very bad time for me, and I really had no money at all to buy bread with. So I first sold the silver buttons off my Sunday coat, and then I sold my silver chain and lastly I sold my wheelbarrow.

But I am going to buy them all back again now. Miller: Hans, I will give you my wheelbarrow. It is not in very good repair; indeed, one side is gone, and there is something wrong with the wheel-spokes; but in spite of that I will give it to you. I think that generosity is the essence of friendship, and, besides, I have got a new wheelbarrow for myself. Hans: Well, really, that is generous of you.

I can easily repair it, as I have a plank of wood in the house. Miller: A plank of wood! Why, that is just what I want for the roof of my barn. There is a very large hole in it, and the corn will all get damp if I don't stop it up. How lucky you mentioned it!

I have given you my wheelbarrow, and now you are going to give me your plank. Pray get it at once, and I will set to work at my barn this very day. Hans: Certainly. Miller: It is not a very big plank, and I am afraid that after I have mended my barn-roof there won't be any left for you to mend the wheelbarrow with; but, of course, that is not my fault.

And now, as I have given you my wheelbarrow, I am sure you would like to give me some flowers in return. Here is the basket, and mind you fill it quite full. Hans: Quite full? Miller: Well, really, as I have given you my wheelbarrow, I don't think that it is much to ask you for a few flowers. I may be wrong, but I should have thought that friendship, true friendship, was quite free from selfishness of any kind.

Hans: My dear friend, my best friend, you are welcome to all the flowers in my garden. I would much sooner have your good opinion than my silver buttons, any day. Miller: Good-bye, little Hans. Hans: Good-bye. Suggested Answer Key para 1: School was boring until a new girl with a strange name joined the class. Ask Ss what they would put in an article describing someone.

Then ask Ss how they would organise their writing. Allow Ss a few minutes to reread the article individually and match the headings and paragraphs. Explain to Ss that they must complete the description of her physical appearance using vocabulary from Ex. Invite Ss to come up with their own examples. She always looks well-dressed in fashionable clothes. He is in his mid-thirties with a freckled face and a long nose. He loves to dress in formal clothes. She has a heart-shaped face and sad eyes.

He shaves his head because he is going bald but he keeps a goatee because he says his chin is too pointy. She has an oval face with a cheeky grin that is framed by her long brown hair usually in a ponytail. Allow Ss some time to complete the task. Remind Ss why it is important when we write to describe negative qualities using mild language so as not to offend anyone.

Suggested Answer Key 1 Andy can sometimes be rather proud. Answer Key 1 confident 2 shy 3 sensitive. Suggested Answer Key 1 Although she is outgoing, occasionally she can be moody. However, he can sometimes be stubborn.

Ask Ss why it is a good idea to use the senses when we describe something it brings the description to life and makes it more interesting to read. Allow Ss a minute or so to find an example of this technique in the model in Ex. Confirm the correct answer. Check Ss answers. Answer Key 2 Dave is clean-shaven and dresses in a suit and tie with neatly-combed hair. Answer Key You could describe a family member, a neighbour, someone in your local area, a famous person alive or dead.

For someone who is alive, you would use present tenses. For someone who is dead or someone you no longer see, you would use past tenses. If necessary, Ss can complete the task as homework. I always thought she was different and admired her for not caring about what people thought of her. Suddenly, we found ourselves spending a lot of time together and we became really close. I have always admired her individuality and her sense of style.

She is a qualified hairdresser, so she is always experimenting with her hair! You never know from one day to the next what she will look like! She is far from boring! Another thing I admire about Alison is her calm personality. She never seems to be in a hurry. She is both patient and caring. Moreover, she is good at giving advice.

However, she does tend to take ages to make decisions and in that way she is very different from me. But, I think that she is wise not to rush into things. In her free time she likes to take up new hobbies. She is far more adventurous than I am and I often wish I was more like her. She is very creative. But the most important thing of all is that she makes an effort to keep our friendship going.

She regularly writes, calls and comes to visit. I think we will be very good friends for a long time to come. I sincerely hope so. Culture Corner Objectives Answer Key The pie chart tells us about the variety of races that have found a home in Britain.

Explain that a pie chart is used to illustrate the relative proportions of a group of things. Ss justify their answers. Answer Key 1 Jerome 2 Li. Encourage Ss to try to guess the meaning of the word from the context before using their dictionaries to check their guesses.

Allocate roles. Suggested Answer Key Interviewer: How would you describe your nationality? Interviewer: Why did your family move to Britain? Rupa: To work in a factory. Interviewer: When did they come to Britain? Rupa: In Interviewer: What language do you speak at home? Rupa: Gujarati. Interviewer: Do you spend a lot of time in the Indian community? Rupa: Yes, I do. But I also have friends that are not Indian.

Interviewer: Where do you live? Interviewer: What do you like about living in Newham? Li: It is culturally diverse. Jerome: I was born in Birmingham, England. But my parents are from The Caribbean, from Jamaica. Interviewer: How would you describe the Caribbean community? Jerome: I would say it is well integrated. A lot has changed since the s when there was a lot of racism.

Go through the questions and elicit answers from Ss around the class. Suggested Answer Key Russia is a diverse multicultural society. More then ethnic groups, many with their own national territories make up the population of Russia. Post-soviet Russia has evolved with three distinct minority ethnic groups in the country.

Germans are the largest of these minority groups with a population of one million. Germans first came to Russia in and settled along the Volga River. The Germans came to Russia to provide essentials skills as craftsman and as traders. They became an autonomous republic that was dissolved in World War II. The North Koreans are a recently new minority group to Russia.

In North Korea allowed many Koreans to migrate to Russia due to poor economic conditions in their own country. They immigrated to Russia and concentrated working in commercial activities. There is a history of racism suffered by the North Koreans due mainly to their threat to local merchants.

The Roma people are a very detached minority group in Russia whose origins date back to the s when they migrated into Russia from Europe. They live in small separate communities and tend to sell items in street markets. They have yet to integrate into Russian society and are often discriminated against. Russia struggles with racial discrimination due to its vast ethnic diversity. However, this diversity helps create a culturally rich society with much to offer for everyone.

Explain that it was taken during the Victorian era in Britain. Accept any reasonable answers. Answer Key 1 An upper class family as they are wearing good clothes and appear to be posing in front of their own house. Answer Key 1 having 2 politely 3 running 4 their 5 heard.

Suggested Answer Key Upper, middle class and working class families have large families but working class families were usually larger. In upper and middle class families the children were looked after by a nanny, whereas in working class families it was the older siblings who looked after the younger ones. Working class fathers usually worked in factories, whereas fathers from the upper and middle class worked in banking or insurance. Upper and middle class families lived comfortably and did not have to carry out dirty jobs but working class families did.

In working class families the mother was responsible for the household chores but in upper and middle class families they had servants. For entertainment, working class families went to cheap music halls, watched sports matches or went to see firework displays whereas upper and middle class families visited museums. Alternatively, explain the task and set it as homework.

In the next lesson, Ss read their projects to the class. Suggested Answer Key My family has only two children whereas the Victorian family has twelve children. The Victorian family is dressed very formally while my family is dressed casually. My family is smiling and appears happy in the photo whereas the Victorian family is very serious and they are not smiling.

Both my family and the Victorian family had their picture taken in front of their house. Suggested Answer Key I see children rushing to the bus stop, deliveries being dropped off at he local stores and people rushing about to get to work. I smell bread baking in the local bakery. I hear lots of noise from the traffic in the streets.

I feel alert and tense as the busy day is about to begin. Suggested Answer Key My neighbourhood is busy as many people live in the area. It is somewhat polluted and noisy because of all the traffic. There is however a quiet green park nearby that we can go to when we want to escape the crowed street. I like my neighbourhood as it is lively and never boring.

Eyes closed, Ss listen and imagine the scene. Remind Ss that we can understand the purpose of a leaflet by 1 looking at the pictures and 2 reading the headings. Elicit from Ss what the purpose of the leaflet is. Suggested Answer Key set up: arrange for sth to happen take turns: alternate with someone limited raw materials: few basic resources preserve landfill space: not use up land in rubbish dumps reduce: make smaller greenhouse gases: substances that cause the atmosphere to get hotter time-consuming: taking a long time councils: local governing bodies remove: take away wastes: does not use well services: public assistance provided by governments sick and tired: extremely bored concrete jungle: aggressive urban environment filtering out pollutants: reducing poisonous substances in the air 1 My friends and I set up a study group to help us prepare for the exams.

Ss should make a poster similar to the one in the unit. Invite Ss to rephrase it into their own words and discuss to what extent they agree with it. Suggested Answer Key I think the quote is talking about not being so selfish and thinking only about ourselves and what we want and rather to think about what we must do in our daily lives to help others and the world around us.

Answer Key 1 D 2 A. Play the recording. Select individual Ss to read out the correct answer. Rep: Certainly, Sir. Which area do you live in? Man: I live in East Swinton. That area is included in the project. Can I have your name, please? Man: Mike Browne. Rep: That sounds like a great idea, Mr Browne. Oh, and perhaps some daffodils and snow drops? Rep: Yes, that should be fine. We have plenty of all of those types. And to which address should we send the bulbs? Customer: You can send them to 51, Green Road.

You should be receiving the bulbs within the next week. We have 1 million bulbs to give away altogether! Man: Wow! Thank you for signing up and happy planting! Man: Thank you! Ask a S to read out the extract. I miss you a lot too!

But, you must have made some new friends by now! What are they like? Do your new friends love going to the cinema as much as you do? Do they go to the same school as you? Best wishes, Francine. Remind Ss to include information about all the prompts. Suggested Answer Key My family has five members and we are all very close.

My mother, father, and grandmother take good care of my little sister and myself. Even though my parents work long hours, they always find some time to spend with my sister and me. We try to eat together every night when it is possible.

I am proud that I can talk about anything with my parents and grandmother. I really think my grandmother is great, as she has lived such an interesting life and always gives great advice. We all think family is very important and enjoy spending time together. We often go to the beach or for a picnic on Sundays.

We also play card games together or spend an evening talking and laughing. Are you closer to a particular family member? What is your relationship with your sister like? Explain that in this module they will learn about stressful situations, peer pressure, how to express negative feelings, how to sympathise, encourage and persuade, etc. Look at Module 2 In order to stimulate discussion and interest, ask Ss which page each picture is from. Two teenage girls talking to each other. What do you think they are talking about?

They might be talking about someone in their class at school. What celebration is it for? Do you bake things at home? What other things do people eat and drink at that time of the year? What is your favourite celebration? Where is it? Do you like eating fast food? Why not? What are the good things about living in a big city? Are there any disadvantages? Which country is it from? Have you ever seen anything like this before? Find the page number s for Allow Ss time to browse through the module and find the relevant information.

Then ask them to explain what each one is and elicit simple information about each item. When did she live? Where was she born? Which parts of the body does it explain? What do you know about how they work? What does it mean? Have you ever experienced this? What are they about? What style are they written in? Suggested Answer Key The Literature section is about an author and a literature extract.

This one p. Ask Ss why the text was written and who it was written for. In other words, to name the purpose of the text. Answer Key The purpose of the text is to give advice to teenagers about how to handle stress.

Ask Ss to predict some of the vocabulary they expect to find in the poem. Write words on the board. Ss follow the text and listen. Suggested Answer Key S1: I agree with the poem that stress is a natural reaction and that it is a normal part of our lives. S2: I agree that change is what usually brings on stress, but I do think that it is unique that some people handle it better then others. S1: Stress is the way our bodies cope with the environment.

I guess there is also emotional stress as well as physical stress. Check answers around the class. Answer Key tense: to tighten shallow: to take small amount of air in with each breath rationally: based on reason nutritious: healthy and nourishing essential: Important, necessary positive: hopeful, confident. Preparing short notes using the categories given, practising their talk and timing it. Remind Ss that they should refer back to the text for ideas and that they should not write every word down.

When we are stressed, our muscles tense, our breathing becomes shallow and adrenaline is released into our bloodstream. This helps us focus and gives us the strength to survive in difficult situations. Stress is sometimes good because it keeps us alert, but it is bad for the heart and our general well-being if we have a lot.

Thankfully, we can quite often avoid stress and we can also learn how to cope with it. When I feel stressed, for example, when I have too much work to do, am sitting exams or am organising an important event, I try to do a lot of aerobic exercise or listen to some calm music. That usually helps. Elicit what part of speech they are. Allow Ss a few minutes to complete the task. Answer Key 1 tired 2 getting 3 take 4 proportion.

Allow Ss time to read through the items, filling in the gaps. Answer Key 1 snarled 2 stammer. Ask the questions to the class. Invite Ss to justify their answers. Listening and Speaking Skills Suggested Answer Key In the picture there are two teenage girls whispering to themselves while another teenage girl stands, arms crossed, in the background.

It seems that the two girls are talking about the other girl. The picture is ripped in two to show separation. The girl in the background is obviously unhappy and might even be angry because the other girls are teasing her or excluding her. Allow Ss a few minutes to complete the rest of the task. Answer Key 1 influence 2 persuade 3 dissuade. Ss listen and complete the task.

Play the recording again for Ss to check their answers. Speaker 1: I wanted to just get a job last year when I finished school, but all my friends were going to university and I started to feel like the odd one out. I ended up applying to go to university too. It was the best thing I ever did! Speaker 3: I gave up playing the violin when I started secondary school because it was taking up too much time and I was missing chances to go out with my friends.

Speaker 4: There was a new girl in my class and all my friends were horrible to her. She tried to be my friend but I ignored her. I know she was upset but I was scared of what the others would say if I was nice to her. Speaker 5: Last year I started hanging out with a new group of friends. They were into punk music and crazy hairstyles.

So, I decided to get my hair dyed purple! I looked absolutely ridiculous! Answer Key 1 hundred and one 2 two 3 million. Answer Key 1 E 2 A. A: I have just found out that Suzie has been lying to me. I thought she was my best friend. A: Are you alright?

You look a bit worried. B: Well, I could be better. You see, Lucy and Hannah want me to go shopping with them this afternoon and miss Art class. A: I know what you mean. I think you should do what you think is right for you. B: Thanks for listening. B: Cheer up! Ask various pairs to repeat the exchanges following the intonation patterns in the recording. Suggested Answer Key A: Can you help me with my homework? Ss close their books and discuss with a partner new words and phrases they have learned in this unit.

Ask Ss to make sentences using them. Grammar: relative clauses; clauses of purpose; clauses of result; clauses of reason Vocabulary: phrasal verbs with put; dependent prepositions. Explain that it is more informal to end a sentence with a preposition. Allow Ss some time to complete the task individually.

Answer Key 1 Do you know the company which she works for? Remind Ss that they provide extra information, some of it necessary defining and some of it unnecessary non-defining. Ask Ss to say which is defining sentence 1 and nondefining sentence 2 : The woman who lives at number 14 has been arrested.

Mr Jenkins, who is very friendly, lives near the park. Answer Key 1 Buckingham Palace, where the royal family live, is a popular tourist attraction in London. Who, which, where and that can be omitted when they are the object of a relative clause.

Answer Key 1 so 3 in case 2 to 4 with a view to. Check answers with the class. Ss work individually to fill in the gaps. S3: There were so many people there that there was nowhere to stand and people kept bumping into me. S4: I was so bored that I kept looking at my watch. S2: The music was so loud that the police came to tell us to turn it down. S3: I had such a bad time that I never want to go to a party ever again! S4: On my way home I felt so cold that I wanted to cry.

Select a S to read out the extract. Ss answer individually. The road was closed due to the fact that there had been an accident. He gave up his job due to the fact that he was suffering from bad health. Allow Ss time to complete the task. Ss make sentences about themselves using the phrases. Play in teams. Ss from each team show their drawings in turn.

The other team guesses what the phrasal verb is. Answer Key 1 is the house in which 2 a good time 3 such a heavy suitcase 4 in case it gets 5 put off because of the. Explain that they will be reading an extract from Jane Eyre. Answer Key Jane Eyre is about an orphaned girl who is brought up by her unkind aunt and then sent to an awful school. Elicit their meanings. Suggested Answer Key threats: statements of intent to hurt inflictions: harm or damage obedient: following orders tottered: almost fell over dependant: someone who relies on others for financial support predominated: became most important.

Answer Key he spent some three minutes in thrusting out his tongue at me as far as he could, without damaging the roots. Invite individual Ss to describe it. It refers to the book that Jane was reading and in the picture she is handing it to John. He used to grab my mobile phone and hide it somewhere. He hated sharing his house with me as much as I hated being there. He wanted all of the attention to himself.

He spent it all on clothes. He liked to wear designer stuff whereas I had to make do with second-hand. Many years later, when I got married to my true love and had my own family, I vowed to treat my children equally and make our home full of happiness.

I am glad to say things are working out really well for me now. Suggested Answer Key I often remember relaxing on the sofa and my step-brother Danny would storm in and disturb me. How I hated living in that house! Ever since mum died and dad married Sharon my life had been hell. Danny always found new ways to. Explain that each piece of writing usually has a purpose. Answer Key 1 semi-formal Looking forward to meeting you 2 informal Hi Sally!

Writing Skills 3 semi-formal Regarding our recent telephone conversation … 4 informal Lots of love 5 semi-formal Dear Mrs Carter, I am writing on behalf … 6 informal Hi Megan. With Ss, place the people in the right position on the scale e. Elicit the purpose of each and invite comments from the class about how they differ stylistically. Answer Key A giving advice to a friend informal B asking for advice from an agony aunt semiformal.

Paragraphs 2 and 3 contain the main subject s of the letter. Paragraph 4 contains the closing remarks. The style is different because of the person the letter is being written to. We use an informal style for letters to friends and a semi-formal style for letters to people we do not know well or when we want to be more polite or respectful. Answer Key 1 I hope my advice helps you. Answer Key No, it is not written in an appropriate style because it uses too many short forms, colloquial language and the greeting and ending are inappropriate for a semi-formal email.

I would be happy to help you out. I would be pleased to send you a list of all their email addresses. It would be a good idea to have the party on a Friday or Saturday night because it would be difficult for everyone to come on a Thursday.

I would not mind coming earlier to help you. Regards, Matthew Summers. I would like to express our gratitude for what we felt was an informative and enjoyable afternoon. In particular, the members of my class were impressed by the useful advice that you gave regarding how to deal with peer pressure.

I would also like to ask you is it would be possible for you to come again in the future and talk to another class. Several teachers expressed a desire to share your good advice with other classes in the school as the problem is one that all pupils may face.

We would be most grateful if you could find the time to visit us again in the near future and look forward to your reply Regards, Peter Quinn. Great to hear that your exams are over. How did they go? Before I go, who else have you invited? Email me soon! Suggested Answer Key 1 When can I call? Ss use dictionaries as necessary. I would like to help them put an end to bullying and also help young people help each other.

Young people can learn important skills such as listening to each other, empathy and problemsolving skills. I think it is an important organisation. There are many ways to get involved. You can find out how to stop bullying by ordering an anti-bullying pack, or you can do some fundraising during antibullying week.

Or you can simply make a donation or volunteer to become a childline counsellor. Alternatively, ask Ss to bring information to the next lesson and complete the project then. It helps children and young people in Moscow who have all kinds of special needs: physically and mentally disabled children and the homeless. Read the text aloud with Ss. Ss listen and read. Suggested Answer Key 1 The nervous system consists of the brain, the spinal cord and a network of nerves.

Objectives weight: heaviness solid waste: rubbish that is not liquid dumped: placed somewhere carelessly landfills: areas where rubbish is buried manufacture: mass production cut back: reduce in number minimal: the least biodegradable: breaks down naturally when buried concentrated: made stronger by removing water from it delivered: sent to a particular address.

Elicit from Ss what they think the text is about. Ss listen to and read the text again. Suggested Answer Key A: What should people have in mind when choosing a product? B: Well, many people look at its appearance. Skip to main content Skip to table of contents. Advertisement Hide. This service is more advanced with JavaScript available.

Papers Table of contents 21 papers About About these proceedings Table of contents Search within event. Front Matter Pages I-X. Pages The Representation Problem Based on Factoring. Ciphers with Arbitrary Finite Domains. Micropayments Revisited. Proprietary Certificates. Giuseppe Ateniese, Cristina Nita-Rotaru.

Steven D. Galbraith, Wenbo Mao, Kenneth G. Co-operatively Formed Group Signatures.

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