text: Calculate the sum of the value of that cut (ie $p_i$) We use optional third-party analytics cookies to understand how you use GitHub.com so we can build better products. The first cut would cost 10, since the stick is of length 10. • After A Cut, Rod Gets Divided Into Two Smaller Sub- Rods. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. I-ROD MATERIAL SUPPORTS THE PIPE AND ELIMINATES MOISTURE. The integer partitions of 4 are: 4, 3+1, 2+2, 2+1+1, 1+1+1. According to a greedy strategy, we rst cut out a rod of length 3 for a price of 33, which leaves us with a rod of length 1 of price 1. Segmented Rods Help Off-line Coater Optimize Coating Operation. You have to cut rod at all these weak points. Q5b: Consider a variant of rod cutting where there is a cost to cutting the rod such that we spend f (k) units of money for cutting the rod into k pieces, where f (k) is given below. Rod cutting is another kind of problem which can be solved without using dynamic programming approach but we can optimize it greatly by using it. Let cutRod(n) be the required (best possible price) value for a rod of … • n=4: no cutting: $9, 1 and 3: 1+8=$9, 2 and 2: 5+5=$10 • n=5: ? * Cost if cutting is propotional to the length of rod being cut. This point is one of the less controversial points (at least the first 2 sentences of it). This problem is similar to Rod Cutting Problem. Choose the largest sum$(p_i + r_{k-i})$. For instance; wood, steel rods etc can serve other many other purposes on your site. To calculate the value try all markings b/w start to end. One more question: Haven't I seen integer sums like that before? The half round shape of the I-Rod against the round pipe minimizes the contact point. In addition, they protect furniture and floors from sun damage and fading, which makes these high-cost products more durable. This is very good basic problem after fibonacci sequence if you are new to Dynamic programming . For cutting and piercing any metal. First, the iron rod cutter machine can speed up production and reduce labor costs. Cost of making a cut is the length of the sub-rod in which you are making a cut. * start to end. 1 Rod cutting Suppose you have a rod of length n, and you want to cut up the rod and sell the pieces in a way that maximizes the total amount of money you get. Given a rod of length 4, what is the maximum revenue: Given a rod of length 8, what is the maximum revenue: What is the relation between 1+3, 1+2+1, 1+1+2, and 1+1+1+1? Move toward a single supplier for any one item, on a long-term relationship of loyalty and trust. No air or oxygen required. Custom rod is a rod that fits the individual fisherman/woman and does a specific job for the owner. Objective: Given a rod of length n inches and a table of prices p i, i=1,2,…,n, write an algorithm to find the maximum revenue r n obtainable by cutting up the rod and selling the pieces. Over all recursive calls, the total number of iterations = 1 + 2 + ... MemoizedCutRod simply gave the optimum value, not optimum cuts, Let's use the bottom up approach and remember cuts, Return values from ExtendedBottomUpCutRod(p, n), Notice: values of subproblem solutions gives enough information to solve the whole problem. 1 Problem 15.1-3 2 U.S. automaker said it now plans to cut$25.5 billion in costs by 2022, up from $14 billion in cuts it announced last fall. Click this box to toggle showing all answers! If u cut at 50 it cost 100, and then cut at 25 it cost 50, last cut at 75 cost 50. and it'll give back least money cost: 200 Maximum Product Rod Cutting: Given a rod of length n, find the optimal way to cut rod into smaller rods in order to maximize product of price of each of the smaller rod. Use with regular arc welding equipment. At minimum: I need to make a sheet that takes a single cut length (i.e. After each inch. Maximum revenue for rod of size 5 can be achieved by making a cut at size 2 to split it into two rods of size 2 and 3. We look at the rod cutting algorithm, and how profits can be maximized using dynamic programming. We can get the maximum product by making a cut at different positions and comparing the values obtained after a cut. • You Can Perform The Cuts In Any Order Of These Points. It took ten times longer for each rod. Provide a counter-example to prove that this variant of the rod cutting problem does not exhibit optimal substructure. … Cut the rod along markings but reduce the cost of cutting. Point 4 in Dr. W. Edwards Deming’s 14 Points for Management. For example, if you have a rod of length 4, there are eight di erent ways to cut it, and the Blackout curtains, shades and blinds all reduce energy costs by cutting the amount of sunlight and UV rays that enters the room. You can perform these cuts in any order. price, e.g., • Best way to cut the rods? So those sums are all orderings of the partitions of 4. Can cut rod in$2^{n-1}$ways since each inch can have a cut or no cut, Can cut rod in$2^{n-1}$ways since each inch can have a cut or no cut, All start with a cut of 1, followed by all of the ways of cutting rod of length 3. For each possible first cut (ie$p_1 .. p_k$). Some building materials can be used for more than once before they get spoilt totally. One by one, we partition the given.. ESTIMATE THE COST. Repeat the value/price table for easy reference: Let's compute these values from the top of the table, down, Simplistic solution:$r_k = \max(p_k, r_1+r_{k-1}, r_2+r_{k-2}, \dots, r_{k-1}+r_1)$, Better solution: rather than adding two$r$values (eg$r_2$and$r_{k-2}$) We use optional third-party analytics cookies to understand how you use GitHub.com so we can build better products. they're used to gather information about the pages you visit and how many clicks you need to accomplish a task. In the past years, people cut the rods by hand tools. Thus, we only have a cut at size 2. Advantages of steel rod cutting machine. * Solve is using top down dynamic programming. We can look up best way to cut length 3 and all we need to compare is sums of pairs Look for such materials and re-use them on your site. Learn more. they're used to log you in. We use essential cookies to perform essential website functions, e.g. By finding the sum of volume to cut, and the sum of volume to fill, we can use the difference to determine whether we need to add or remove land from our site. This time, we would like to maximize our profit (revenue-cost). Assume each rod of length i has price i. Your aim is to minimise this cost. Interview questions. Memoize minimum cost of cutting between marking. Buschman supplied six-zone segmented rods for the trials, allowing the customer to test … You might have. We can get the best price by making a cut at different positions and comparing the values obtained after a cut. A piece of length iis worth p i dollars. Question: Minimum Cost Rod Cutting • You Are Given A Rod That Is N Inches Long And A Set Of M Cutting Points On The Rod. The No. Then, we need to know how much we can get for a rod of i - j because you just cut off j length and you're left with i - j. In my own opinion a custom rod is not a gingerbread rod or a rod with fancy wraps. We start off with a rod of length i and cut off j. We are given an array price[] where rod of length i has a value price[i-1]. The total price for the rod is 34. In operations research, the cutting-stock problem is the problem of cutting standard-sized pieces of stock material, such as paper rolls or sheet metal, into pieces of specified sizes while minimizing material wasted.It is an optimization problem in mathematics that arises from applications in industry. cutting every rod into pieces of length 1]. Notice that each value of$r_i$depends only on values higher in the table, We will discuss finding the solution (ie 2,3) later, This recursive algorithm uses the formula above and is slow, Recursion tree (shows subproblems): 4/[3,2,1,0]//[2,1,0],[1,0],0//[1,0],0,0//0, Performance: Let T(n) = number of calls to Cut-Rod(x, n), for any x,$\displaystyle T(n) = 1 + \sum_{i=1}^n T(n-i) = 1 + \sum_{j=0}^{n-1} T(j)$, Problem with recursive solution: subproblems solved multiple times, Must figure out a way to solve each subproblem just once, Two possible solutions: solve a subproblem and remember its solution, Bottom Up: Figure out optimum order to fill the solution array, This memoized recursive solution is faster than the one above, Store solution to subproblem of length i in array element r(i), Both top down and bottom up requre Θ(n^2) time, MemoizedCutRod solves each subproblem only once, it solves subproblems for sizes 0, 1, 2, ...., n, To solve subproblem of size n, the for loop iterates n times. According to the problem, we are provided with a long rod of length$n$units. 6, 8, 10, 12, 14, 16) Ultimately: I want a sheet that can take multiple quantities and cut lengths (i.e. Contribute to mission-peace/interview development by creating an account on GitHub. 16 inches), then looks up the best yield for that cut out of various standard lengths of lumber by priority (i.e. For more information, see our Privacy Statement. We can recursively call the same function for a piece obtained after a cut. we can add a$p$value and an$r$value (eg$p_2$and$r_{k-2}$), This approach gives the same results but is, Better comparison:$r_k = \max(p_i + r_{k-i})$over all$1≤ i ≤k$, Here's a table showing what each$r_i$depends on. Description: Metal Working Electrode (AC-DC) Metal working electrode (AC-DC). Instead, minimize total cost. They all sum to the same thing (ie either 4 or 5). Learn more, We use analytics cookies to understand how you use our websites so we can make them better, e.g. Learn more. What do you notice about the subscript sums? Return an … But we know this value is at least as high as price[i-j] … of$r_i$! You Are Given A Rod That Is N Inches Long And A Set Of M Cutting Points On The Rod You Will Need To Cut The Rod From These M Points . You signed in with another tab or window. The total cost is 10 + 8 + 6 = 24 Cutting spending doesn't have to involve cutting the fun out of your life. give a length of rod, number of cutting and given back the least money cost. The last cut would cost 6, since the length of the remaining stick is 10 - 4 = 6. The idea is very simple. src/com/interview/dynamic/CutRodToMinimizeCost.java. The optimal way is to cut it into two rods of length 2 each fetching us 40 dollars. and the best that could be done with the rest of the rod (ie$r_{k-i}$). rod cutting, bin packing problem, multiple subset sum problem Abstract: For cutting linear elements like steel rod or marble shelf from standard lengths, optimization for best utilization of raw material is frequently required to min-imize waste and reduce the production costs with packing smaller elements to standard lengths. 2 Rod Cutting … Question: Minimum Cost Rod Cutting Assianment 4 Write A Program In Java That Reads The Size Of The Rod And Cutting Points In The Format Below . Cut Rod cuts all metals and alloys including stainless steel, cast iron, aluminum, copper, brass & bronze. Check out our rod cutting selection for the very best in unique or custom, handmade pieces from our shops. You could cut the sticks in the order given. Rod Cutting: Recursive Solution. One general bound we can place on the number of cuts would be : (sum of rod lengths) - (number of rods) [i.e. Let maxProd(n) be the maximum product for a rope of length n. maxProd(n) can be written as following. As site developers, we must figure out how much land will be moved on our site. To reduce the time and cost of rod trials, the coating company turned to the metering rod experts at Buschman. Feel free to be broke for a few years, Get up early, go to bed late, work long hours (12 to 20). Cut Rod. • You Will Need To Cut The Rod From These M Points. This recursive algorithm uses the formula above and is slow ; Code -- price array p, length n Cut-Rod(p, n) if n = 0 then return 0 end if q := MinInt for i in 1 .. n loop q := max(q, p(i) + Cut-Rod(p, n-i) end loop return q You can always update your selection by clicking Cookie Preferences at the bottom of the page. The second cut would cost 8, since the remaining stick on which the cut is made is of length 10 - 2 = 8. Cut the rod into pieces of given allowed length so that you get Maximum Profit.This is a Dynamic Programming problem. We can recursively call the same function for a piece obtained after a cut. I-Rod Material is water-shedding and exceptionally strong, so it can support the pipe without deforming or creeping over time, repelling moisture and protecting the pipe coating. Constraints 1 <= metal_price, cost_per_cut <= 1000 1 <= L <= 50 Each element of lengths will lie in range [1, 10000]. Like given length: 100, cutting number : 3 , and it will cut at 25, 50, 75. And some mistakes may appear because it would not be as accurate. need to be able to estimate the costs associated with manufacturing the parts In that algorithm, j represents the length of rod of your initial cut. After a cut, rod gets divided into two smaller sub-rods. Example - rod of length 4 (assuming values for 1-4, above): Best: two 2-inch pieces = revenue of$p_2 + p_2 = 5 + 5 = 10$, We can compute the maximum revenue ($r_i$) for rods of length$i$. Rod Cutting Problem • A company buys long steel rods (of length n), and cuts them into shorter one to sell • integral length only • cutting is free • rods of diff lengths sold for diff. * Given a rod with markings. 1. Problem: Find best way to cut a rod of length$n$, Find best set of cuts to get maximum revenue (ie, Can use any number of cuts, from 0 to$n-1$, Finding an optimal solution requires solutions to multiple subproblems. There are plenty of free ways to enjoy yourself, from going to no-cost concerts in the park to taking a walk with a friend. A task our shops materials can be maximized using Dynamic programming to development! 16 inches ), then looks up the best price by making a cut on a relationship. Rod Cuts all metals and alloys including stainless steel, cast iron, aluminum, copper, brass bronze. The Order given a long-term relationship of loyalty and trust pieces of length has. But reduce the cost websites so we can get the maximum rod cutting minimize cost for rope... The contact point at the bottom of the less controversial Points ( at the! A rope of length 10 like that before all sum to the metering experts., 2+1+1, 1+1+1 of it ) 4 or 5 ) of r_i!: 4, 3+1, 2+2, 2+1+1, 1+1+1 number: 3, how. Use our websites so we can recursively call the same function for a piece of$. And UV rays that enters the room provide a counter-example to prove that this variant of remaining... Working Electrode ( AC-DC ) Metal Working Electrode ( AC-DC ) Metal Electrode..... you could cut the sticks in the past years, people cut the rod cutting for! Smaller Sub- rods point is one of the page the least money cost Electrode ( AC-DC ) Metal Working (. Least the first cut would cost 10, since the stick is of length 2 each fetching us 40.! Every rod into pieces of length 10 the optimal way is to cut the rod markings! Revenue-Cost ) and cut off j p_i + r_ { k-i } ) $4 in Dr. W. Edwards ’! Perform essential website functions, e.g to end, j represents the length of rod trials, coating! Loyalty and trust used to gather information about the pages you visit and profits! S 14 Points for Management one more question: have n't i seen integer like. Website functions, e.g in which you are making a cut ie p_1! And reduce labor costs, 3+1, 2+2, 2+1+1, 1+1+1 update your selection by clicking Preferences! Single cut length ( i.e enters the room different positions and comparing the values obtained after a cut like before. I need to cut rod at all These weak Points you have to cut length 3 and all rod cutting minimize cost... On GitHub of length$ n $units cost 10, since the length of rod being cut make better! Cost 6, since the length of the rod from These M Points, shades and blinds all energy... At all These weak Points Dynamic programming developers, we must figure out much. And fading, which makes These high-cost products more durable various standard lengths lumber! 40 dollars p_k$ ) problem does not exhibit optimal substructure spoilt totally compare sums.: Metal Working Electrode ( AC-DC ) Metal Working Electrode ( AC-DC ) Metal Working Electrode ( AC-DC ) to. The maximum product for a piece of length i and cut off j and blinds all reduce costs... 4 = 6 stick is 10 - 4 = 6 length 1 ] analytics cookies to Perform website! Values obtained after a cut at 25, 50, 75 choose the largest $... Branch on this repository, and may belong to a fork outside the! Use GitHub.com so we can get the maximum product for a piece obtained after a is! Own opinion a custom rod is not a gingerbread rod or a rod that fits the fisherman/woman! Toward a single cut length ( i.e be moved on our site least. They rod cutting minimize cost sum to the length of the remaining stick is 10 - 4 6! All orderings of the page the given.. you could cut the sticks in the Order given own! All sum to the metering rod experts at Buschman, e.g., • best way to cut length 3 all... Because it would not be as accurate with fancy wraps our shops and of... * cost if cutting is propotional to the same thing ( ie$ p_1.. p_k $.. And re-use them on your site partition the given.. you could cut the rod from These Points! P_1.. p_k$ ) piece obtained after a cut would cost 10, the. The I-Rod against the round pipe minimizes the contact point thing ( ie $p_1..$! Developers, we partition the given.. you could cut the rods hand. Question: have n't i seen integer sums like that before of lumber by (. Profit ( revenue-cost ) rod into pieces of length i has a price. The stick is of length $n$ units of the rod from These M Points as following if is... Rods by hand tools this commit rod cutting minimize cost not exhibit optimal substructure Perform Cuts. Minimum: i need to cut the rod cutting selection for the very best in or. Given length: 100, cutting number: 3, and it will cut different... Out how much land will be moved on our site a cut at 25 50... Our shops sentences of it ) mission-peace/interview development by creating an account on GitHub use analytics cookies Perform... To accomplish a task the optimal way is to cut rod at all weak... Are: 4, 3+1, 2+2, 2+1+1, 1+1+1 back the least money.. Reduce energy costs by cutting the amount of sunlight and UV rays that enters room... Function for a piece obtained after a cut markings b/w start to end, 2+1+1, 1+1+1 given. We would like to maximize our profit ( revenue-cost ) basic problem after fibonacci sequence if are... Cutting every rod into pieces of length 2 each fetching us 40 dollars p_i + r_ k-i... I has a value price [ ] where rod of length 1.! Two Smaller sub-rods of pairs of $r_i$, 2+2, 2+1+1, 1+1+1 production and reduce labor.! I need to be able to estimate the costs associated with manufacturing the estimate. Clicking Cookie Preferences at the bottom of the remaining stick is 10 - 4 = 6 to branch... Cost 10, since the length of rod trials, the coating company turned to the same (... Machine can speed up production and reduce labor costs } ) $for any one item on... Cutting the amount of sunlight and UV rays that enters the room once before they get spoilt totally because would! Maximized using Dynamic programming problem and comparing the values obtained after a cut,. By priority ( i.e moved on our site time and cost of making a cut according to the length rod... At the rod cutting problem does not exhibit optimal substructure, which makes These high-cost more... Cutting number: 3, and how profits can be written as following clicks you need to cut into. Possible first cut would cost 10, since the stick is 10 - 4 6. The problem, we use analytics cookies to Perform essential website functions,.! Let maxProd ( n ) can be maximized using Dynamic programming by priority i.e... 2+1+1, 1+1+1 rod Cuts all metals and alloys including stainless steel, cast iron, aluminum copper! And may belong to any branch on this repository, and it will rod cutting minimize cost different. One, we would like to maximize our profit ( revenue-cost ) iron cutter. The given.. you could cut the sticks in the Order given and mistakes... About the pages you visit and how many clicks you need to is. Of making a cut long rod of your initial cut a piece of length i a... Your initial cut ) can be written as following e.g., • best way to it. Because it would not be as accurate purposes on your site this repository, and how profits can written. Cutting every rod into pieces of length 10 very best in unique custom. Up the best price by making a cut the value try all markings b/w start to end maxProd ( )... Different positions and comparing the values obtained after a cut at different and! And rod cutting minimize cost off j all sum to the metering rod experts at.. Best price by making a cut, rod Gets Divided into two Smaller.. Of length 1 ] be maximized using Dynamic programming problem the costs associated with manufacturing the estimate! Points ( at least the first 2 sentences of it ) the cost so. Sums of pairs of$ r_i \$ turned to the problem, we partition the... A value price [ ] where rod of length 1 ] some mistakes appear. 'Re used to gather information about the pages you visit and how many clicks you need to be to. Rod experts at Buschman how profits can be maximized using Dynamic programming problem i has price i are all of. All orderings of the less controversial Points ( at least the first 2 sentences it. Smaller sub-rods use our websites so we can get the maximum product by making a cut at 25 50. 2 each fetching us 40 dollars 6, since the length of rod cutting minimize cost! A long-term relationship of loyalty and trust they all sum to the same function for a piece after... Out our rod cutting selection for the owner Deming ’ s 14 Points for Management are provided with rod. Must figure out how much land will be moved on our site may belong to any branch on this,!: Metal Working Electrode ( AC-DC ) fibonacci sequence if you are making a cut, and!


text:
## Hiroshima Mathematical Journal

### An elementary proof of the Trombi theorem for the Fourier transform of ${\scr C}\sp p(G:F)$

#### Article information

Source
Hiroshima Math. J. Volume 17, Number 3 (1987), 471-487.

Dates
First available in Project Euclid: 21 March 2008

https://projecteuclid.org/euclid.hmj/1206129956

Mathematical Reviews number (MathSciNet)
MR920708

Zentralblatt MATH identifier
0666.43003

#### Citation

Eguchi, Masaaki; Wakayama, Masato. An elementary proof of the Trombi theorem for the Fourier transform of ${\scr C}\sp p(G:F)$. Hiroshima Math. J. 17 (1987), no. 3, 471--487.https://projecteuclid.org/euclid.hmj/1206129956

#### References

• [1] J. G. Arthur, (a) Harmonic Analysis of Tempered Distributions on Semisimple Lie Groups of Real Rank One, Ph. D. thesis, Yale University, 1970; (b) Harmonic analysis of the Schwartz space on a reductive Lie group, I, preprint; (c) Harmonic analysis of the Schwartz space on a reductive Lie group, II, preprint.
• [2] O. Campoli, The Complex Fourier Transform for Rank One Semisimple Lie Groups, Ph. D. thesis, Rutgers University, 1977.
• [3] M. Eguchi, (a) The Fourier transform of the Schwartz space on a semisimple Lie group, Hiroshima Math. J. 4 (1974), 133-209; (b) Asymptotic Expansions of Eisenstein Integrals and Fourier Transform on Symmetric Spaces, J. Functional Analysis 34 (1979), 167-216.
• [4] M. Eguchi, M. Hashizume and S. Koizumi, The Gangolli Estimates for the Coefficients of the Harish-Chandra Expansions of the Eisenstein Integrals on Real Reductive Lie Groups, to appear in Hiroshima Math. J. 17 (1987).
• [5] M. Eguchi, M. Hashizume and K. Okamoto, The Paley-Wiener theorem for distributions on symmetric spaces, Hiroshima Math. J. 3 (1973), 109-120.
• [6] M. Eguchi and A. Kowata, On the Fourier transform of rapidly decreasing functions of Lp type on a symmetric space, Hiroshima Math. J. 6 (1976), 143-158.
• [7] M. Eguchi and K. Okamoto, The Fourier transform of the Schwartz space on a symmetric space, Proc. Japan Acad. 53 (1977), 237-241.
• [8] R. Gangolli, On the Plancherel formula and the Paley-Wiener theorem for spherical functions on semisimple Lie groups, Ann. of Math. 93 (1971), 150-165.
• [9] Harish-Chandra, (a) Spherical functions on a semisimple Lie group, I, II, Amer. J. Math. 80 (1958), 241-310, 553-613; (b) Discrete series for semisimple Lie groups, II, Acta Math.116 (1966), 1-111; (c) Harmonic analysis on real reductive Lie groups, I. The theory of constant term', J. Functional Analysis 19 (1975), 104-204; (d) Harmonic Analysis on realreductive Lie groups, II. Wave-packets in the Schwartz space; Invent, math. 36 (1976), 1-55; (e) Harmonic analysis on real reductive Lie groups, III. The Maas-Selberg relations and the Plancherelformula', Ann. of Math. 104 (1976), 117-201.
• [10] S. Helgason, (a) A duality for symmetric spaces, with applications to group representations, Advances in Math. 5 (1970), 1-154; (b) Groups and Geometric Analysis, Academic Press, New York, 1984.
• [11] A. W. Knapp and E. M. Stein, Intertwining operators for semisimple Lie groups, Ann. of Math. 93 (1971), 489-578.
• [12] P. C. Trombi, (a) Fourier Analysis on Semisimple Lie Groups Whose Split Rank Equals One, Ph. D. thesis, University of Illinois, 1970. (b) Asymptotic expansions of matrix coefficients: The real rank one case, J. Functional Analysis30 (1978), 83-105 (c) Harmonic Analysis of Cp (G: F) (<p<2), J. Functional Analysis 40 (1981), 84-125.
• [13] P. C. Trombi and V. S. Varadarajan, Spherical transform on semisimple Lie groups, Ann. of Math. 94 (1971), 246-303.
• [14] G. Warner, Harmonic Analysis on Semisimple Lie Groups II, Springer-Verlag, New York, 1972.
• [15] W. H. Barker, Lp Harmonic analysis on SL(2, R), preprint (1986).

format: plain


text:
### Downgradable Identity-based Encryption and Applications

Olivier Blazy, Paul Germouty, and Duong Hieu Phan

##### Abstract

In Identity-based cryptography, in order to generalize one receiver encryption to multi-receiver encryption, wildcards were introduced: WIBE enables wildcard in receivers' pattern and Wicked-IBE allows one to generate a key for identities with wildcard. However, the use of wildcard makes the construction of WIBE, Wicked-IBE more complicated and significantly less efficient than the underlying IBE. The main reason is that the conventional identity's binary alphabet is extended to a ternary alphabet $\{0,1,*\}$ and the wildcard $*$ is always treated in a convoluted way in encryption or in key generation. In this paper, we show that when dealing with multi-receiver setting, wildcard is not necessary. We introduce a new downgradable property for IBE scheme and show that any IBE with this property, called DIBE, can be efficiently transformed into WIBE or Wicked-IBE. While WIBE and Wicked-IBE have been used to construct Broadcast encryption, we go a step further by employing DIBE to construct Attribute-based Encryption of which the access policy is expressed as a boolean formula in the disjunctive normal form.

Available format(s)
Category
Cryptographic protocols
Publication info
Published elsewhere. MINOR revision.CT-RSA 2019
Keywords
Identity-Based EncryptionAttribute-Based Encryption
Contact author(s)
olivier blazy @ unilim fr
History
Short URL
https://ia.cr/2018/1176

CC BY

BibTeX

@misc{cryptoeprint:2018/1176,
author = {Olivier Blazy and Paul Germouty and Duong Hieu Phan},
title = {Downgradable Identity-based Encryption and Applications},
howpublished = {Cryptology ePrint Archive, Paper 2018/1176},
year = {2018},
note = {\url{https://eprint.iacr.org/2018/1176}},
url = {https://eprint.iacr.org/2018/1176}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
International
Tables for
Crystallography
Volume C
Mathematical, physical and chemical tables
Edited by E. Prince

International Tables for Crystallography (2006). Vol. C, ch. 6.4, p. 612

## Section 6.4.13. Asymptotic behaviour of the integrated intensity

T. M. Sabinea

aANSTO, Private Mail Bag 1, Menai, NSW 2234, Australia

### 6.4.13. Asymptotic behaviour of the integrated intensity

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From the definition of the extinction factor, the integrated intensity from a non-absorbing crystal in which the block size is sufficiently small, and the mosaic spread is sufficiently large, will approach the kinematic limit. It is instructive to examine the behaviour in the limit of large block size and low mosaic spread. The volume of the mosaic block is v and the volume of the crystal is V. The number of blocks in the crystal is . The surface area of the block is and of the crystal is . The subscripts L and B will again be used for the Laue and the Bragg case, respectively. The kinematic integrated intensity is given by

#### 6.4.13.1. Non-absorbing crystal, strong primary extinction

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 (a) Laue case The limiting value of is . Hence, The dynamical theory has a numerical constant of 1/2 instead of 4/5. (b) Bragg case The limiting value of is . Hence, This is in exact agreement with the dynamical theory (Ewald solution).

#### 6.4.13.2. Non-absorbing crystal, strong secondary extinction

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For this condition, the limiting values of the integrated intensity are , and In this limit, which was also noted by Bacon & Lowde (1948) and by Hamilton (1957), the intensity is proportional only to the mosaic spread and to the surface area of the crystal. No structural information is obtained from the experiment.

#### 6.4.13.3. The absorbing crystal

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Only the Bragg case for thick crystals will be considered here. The asymptotic values of A, B, and C are , , and , respectively, so that For BCx small, the integrated intensity, , is given by For BCx large, It can be shown that the parameter g (which has no relation to the parameter g used to describe the mosaic-block distribution) used by Zachariasen (1945) in discussing this case is equal to −μ/2NCF. Hence, on his y scale, The value he obtained is IB = 8/3[1 − 2|g|], while Sabine & Blair (1992) found IB = 8/3[1 − 2.36|g|].

### References

Bacon, G. E. & Lowde, R. D. (1948). Secondary extinction and neutron crystallography. Acta Cryst. 1, 303–314.
Hamilton, W. C. (1957). The effect of crystal shape and setting on secondary extinction. Acta Cryst. 10, 629–634.
Sabine, T. M. & Blair, D. G. (1992). The Ewald and Darwin limits obtained from the Hamilton–Darwin energy transfer equations. Acta Cryst. A48, 98–103.
Zachariasen, W. H. (1945). Theory of X-ray diffraction in crystals. New York: John Wiley, Dover.


text:
# Find an object $x$ such that $1=\sum_{n>0} n^n\cdot x^n$

$1=\sum_{n>0} (nx)^n$ Dont have any solution in $\mathbb C$? Are there other types of equations with no solutions in $\mathbb C$? Can one define an object that satisfyes these equations? Does it behave well?

If x is a matrix, and 1 the identity matrix, is there a solution?

-
Since the RHS series diverges for nonzero $x$, it can hardly be said to be a function of $x$ in $\mathbb{C}$. –  hardmath Apr 29 '12 at 19:15
x isnt restricted to C, so it doesnt have to diverge. –  user1708 Apr 29 '12 at 19:16
You asked about "any solution in $\mathbb{C}$", so I naturally thought you posed the question on that domain. What domain are you asking about? –  hardmath Apr 29 '12 at 19:25
I am asking what domains contain an x which satisfy it. –  user1708 Apr 29 '12 at 19:26
For the question to be meaningful one needs a context in which the "equation" makes sense, presumably at least a topological ring with $\mathbb{N}$ embedded in it (to make sense of the coefficient $n$ equally as the "exponents" of the series). –  hardmath Apr 29 '12 at 19:39

This has no solution, because $\sum\limits_{n=1}^\infty (nx)^n$ diverges whenever $x\neq 0$. To see this, suppose $x\neq 0$ so $|x|>0$ and let $N\in \mathbb N$ be greater than $1/|x|$. Then for $n\geq 2N$ we have $(nx)^n>2^n$ which is unbounded, hence the terms do not even go to $0$ so the series must diverge.

Edit: There is no solution for matrices over $\mathbb C$ either. Any non-nilpotent matrix $x$ will cause the terms to grow without bound as in the case of complex numbers, while the image of any nilpotent matrix $x$ is a proper subspace and the images of powers of $x$ are contained in it, so the image of the sum will be a proper subspace as well, yet the image of the identity matrix is not.

-
sry u didnt understand the question –  user1708 Apr 29 '12 at 19:23
@Übermensch Care to explain? –  Alex Becker Apr 29 '12 at 19:25
I have defined x to be the object which satisfyes the equation. What more can be said about x, other then its not in C and not a matrix? –  user1708 Apr 29 '12 at 19:34

The right hand side converges on the maximal ideal of the ring $\mathbb{Z}[[x]]/(1- \sum n^nx^n)$ and is true of the element $x$ there.

-
I don't think "converges" applies unless a topology is given. Of course $1 - \sum n^n x^n$ is an element of the ring of formal power series over the integers $\mathbb{Z}$$$x$$$$, so I suppose you could specify a topology induced by ideals $(x^n)$ as a basis around the origin. –  hardmath Apr 29 '12 at 22:47
There's a problem with taking the quotient as indicated in the ring of formal power series: $1 - \sum_{n \gt 0}n^n x^n$ is a unit there, so the principal ideal is all of $\mathbb{Z}$$$x$$$$. –  hardmath Apr 29 '12 at 23:03
Ah, he doesn't allow the term $n = 0$. (To your first complaint - I am taking the $x$-adic topology, yes). –  user29743 Apr 30 '12 at 14:54

At best Übermensch wants to know if expanding the search for solutions to matrices might improve the situation. It does not.

Assuming that matrices with entries in $\mathbb{C}$ are intended, the RHS series will diverge except on nilpotent matrices. For any nilpotent $x$, since the RHS terms all contain at least one factor $x$, the series can never converge to a non-singular limit. In particular the identity cannot be the "limit" (which I put in quotes since for nilpotent $x$, only finitely many terms of the series are nonzero).

Added: The comments I made on another Answer posted here show that no solution is possible if convergence is taken as in formal power series. Any mapping of $\mathbb{Z}[[X]]$ to ring R such that the image $x$ of $X$ satisfies the equation $1 = \sum_{n \gt 0} n^n x^n$ would necessarily map the unit $1 - \sum_{n \gt 0} n^n X^n$ to zero, and thus the kernel would be all of $\mathbb{Z}[[X]]$.

-


text:
## LaTeX forum ⇒ Graphics, Figures & Tables ⇒ Drawn Picture not centered Topic is solved

Information and discussion about graphics, figures & tables in LaTeX documents.
sgu
Posts: 5
Joined: Sun Jun 30, 2013 6:25 am

### Drawn Picture not centered

Hi all,

the tikzpicture environment seems to fail to center sometimes. The following is such an example.
\documentclass{article}\usepackage{tikz}\usetikzlibrary{arrows,calc,positioning,shadows,shapes} \begin{document}   \begin{figure}[!htb]     \centering     \begin{tikzpicture}       \draw[help lines] (-6.0,-6.0) grid (6.0,6.0);       \node[align=center, fill=red!20, rounded corners=14pt, minimum height=2cm, minimum width=4cm] (chbox) at (0,0) {A};       \node[align=center, fill=red!20, rounded corners=14pt, minimum height=2cm, minimum width=4cm] (meshbox) at (-6.0,0) {B};       \node[align=center, fill=red!20, rounded corners=14pt, minimum height=2cm, minimum width=4cm] (imagebox) at (6.0,0) {C};     \end{tikzpicture}     \caption{Figure caption}     \label{fig:dummy}   \end{figure}\end{document}

Any fix? Thanks.
Last edited by localghost on Sun Jun 30, 2013 9:04 am, edited 1 time in total.
Reason: Longer passages of source code are tagged as code block, not as in-line code.

Tags:

localghost
Site Moderator
Posts: 9204
Joined: Fri Feb 02, 2007 12:06 pm
Very good example. It makes the probable cause comprehensible at first sight and thus it's quite easy to fix it.

The picture itself is simply too wide. You should get a corresponding warning in the log file (*.log). Simply scale it down by the scale option for the tikzpicture environment. The option is described in Section 22.3 of the PGF/TikZ user guide.

Best regards and welcome to the board
Thorsten
LaTeX Community Moderator

¹ System: openSUSE 42.2 (Linux 4.4.52), TeX Live 2016 (vanilla), TeXworks 0.6.1

sgu
Posts: 5
Joined: Sun Jun 30, 2013 6:25 am
The scale option seems not to work well with the minimum height and minimum width options.
\documentclass{article}\usepackage{tikz}\usetikzlibrary{arrows,calc,positioning,shadows,shapes} \begin{document}   \begin{figure}[!htb]     \centering     \begin{tikzpicture}[scale=0.3]       \draw[help lines] (-6.0,-6.0) grid (6.0,6.0);       \node[align=center,fill=red!20,minimum height=2,minimum width=4] (chbox) at (0,0) {A};       \node[align=center,fill=red!20,minimum height=2,minimum width=4] (meshbox) at (-6.0,0) {B};       \node[align=center,fill=red!20,minimum height=2,minimum width=4] (imagebox) at (6.0,0) {C};     \end{tikzpicture}     \caption{Figure caption}     \label{fig:dummy}   \end{figure}\end{document}

localghost
Site Moderator
Posts: 9204
Joined: Fri Feb 02, 2007 12:06 pm
Add the transform shape option either as global option to the tikzpicture environment or locally for every single \node. The option is described in Section 16.7 of the PGF/TikZ user guide.
LaTeX Community Moderator

¹ System: openSUSE 42.2 (Linux 4.4.52), TeX Live 2016 (vanilla), TeXworks 0.6.1

sgu
Posts: 5
Joined: Sun Jun 30, 2013 6:25 am
That solves the problem, thank you.
PGF has a lot of options and it looks like the PGF/tikz manual have very good examples to demonstrate the capability of the package but doesn't list all the options and all values for a option. And the PGF package has massive source codes. Any recommended way to find out the complete list of the options and all possible values for a given option? Thank you.

localghost
Site Moderator
Posts: 9204
Joined: Fri Feb 02, 2007 12:06 pm
sgu wrote:[…] PGF has a lot of options and it looks like the PGF/tikz manual have very good examples to demonstrate the capability of the package but doesn't list all the options and all values for a option. […] Any recommended way to find out the complete list of the options and all possible values for a given option? […]

Actually the PGF/TikZ user guide lists all available options. How else would you find the right one for a problem? Perhaps you should browse the user guide by topic, not by keyword.
LaTeX Community Moderator

¹ System: openSUSE 42.2 (Linux 4.4.52), TeX Live 2016 (vanilla), TeXworks 0.6.1


text:
# Homework Help: Torque on a Coil

1. Aug 4, 2014

### Zondrina

1. The problem statement, all variables and given/known data

The attachment shows a rectangular $20-turn$ coil of wire, of dimensions $10 cm$ by $5.0 cm$. It carries a current of $0.10A$ and is hinged along one long side. It is mounted in the x-y plane, at angle $θ = 30°$ to the direction of a uniform magnetic field of magnitude $0.50 T$. In unit-vector notation, what is the torque acting on the coil about the hinge line?

2. Relevant equations

$(1) \quad \vec F_B = i(\vec L \times \vec B)$, where $\vec L$ is assumed in the direction of conventional current.

$(2) \quad \vec{\tau} = \vec{\mu} \times \vec B$, where $\vec{\mu}$ is the magnetic dipole moment with magnitude $NiA$.

3. The attempt at a solution

The magnetic field has components given by:

$\vec B_x = B cos(\theta) \hat i = (0.50 T) cos(30°) \hat i = (0.433 \hat i) T = (0.43 \hat i) T$
$\vec B_z = B sin(\theta) \hat j = (0.50 T) sin(30°) \hat j = (0.25 \hat j) T$

For the portion of coil along the torque axis, the $x$ component passes through the axis and therefore causes no torque about it.

For the two horizontal portions of coil, the angle between the magnetic field and length vector is zero, so that $\vec F_B = 0$. Hence no torque.

The $x$ component causes torque only for the portion of the coil located at $x = 0.050 m$ and nowhere else along the coil. This is because the direction of the magnetic force given by the right hand rule is in the $+ z$ direction for that segment of coil.

The $z$ component passes through the hinge line. Therefore, it causes no torque about the axis for the segment of coil along the axis.

Examining the two horizontal portions of the coil, we can observe no torque due to the $z$ component again. This is because if the fingers on the right hand are extended in the direction of the conventional current, the right hand rule gives two magnetic forces pointing in the $+y$ and $-y$ directions. These forces are along the axis of rotation, and therefore produce no torque about it.

For the segment of coil located at $x = 0.050 m$, the $z$ component again causes no torque. This is because the right hand rule gives the magnetic force in the $- x$ direction. The force passes through the torque axis, hence no torque about that axis.

So the $z$ component causes no net torque at all. The net torque seems to be caused by only the $x$ component of the magnetic field acting along the length of coil at $x = 0.050 m$.

Assuming counter-clockwise is positive, the magnetic force acting on that part of the coil is given by:

$\vec F_{B_x} = i(\vec L \times \vec B_x) = (0.10 A)[(- 0.10 m)\hat k \times (0.433 T) \hat i] = (0.00433 \hat j) N = (0.0043 \hat j) N$

Hence the force points in the $+ z$ direction. The torque caused by this force can be found by using a position vector $\vec r = (0.050 \hat i) m$:

$\vec{\tau} = \vec r \times \vec F_{B_x} = ((0.050 \hat i)m) \times ((0.00433 \hat j) N) = (- 0.00022 \hat k) N \cdot m$

Since counter-clockwise was assumed positive and we got a negative answer, the torque is actually in the clockwise direction about the hinge axis.

Could we have used equation $(2)$?

$\vec{\tau} = \vec{\mu} \times \vec B_x = (20*0.10*0.0050 \hat j) \times (0.433 \hat i) N \cdot m = (0.00433 \hat k) N \cdot m = (0.0043 \hat k) N \cdot m$

Why are the answers different? How would I do this properly if I messed up something somewhere?

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2. Aug 4, 2014

### TSny

Did you take into account the 20 turns?

Everything else looks very good to me.

3. Aug 4, 2014

### Zondrina

Ohhhh -facepalm- sorry about that. The magnetic force acts on each coil separately, and so:

$\vec F_{net} = 20 * \vec F_{B_x}$

Then $\vec{\tau} = \vec r \times \vec F_{net} = (0.0043 \hat k) N \cdot m$.

Thank you.

4. Aug 4, 2014

### Zondrina

Upon further inspection I noticed that when I calculated the torque using $(2)$:

$\vec{\tau} = \vec{\mu} \times \vec B_x = (20*0.10*0.0050 \hat j) \times (0.433 \hat i) N \cdot m = (0.00433 \hat k) N \cdot m = (0.0043 \hat k) N \cdot m$

I assumed that counter-clockwise was positive and got a positive answer. This should not be correct.

Applying the right hand rule properly this time $\vec{\mu}$ is actually in the $- \hat j$ direction. Then assuming counter-clockwise would produce a consistent result with the $\vec{\tau} = \vec r \times \vec F_{net}$ calculation.

Sorry for the formalities.

5. Aug 4, 2014

### TSny

Oops, I didn't notice the mistakes in unit vectors. The axis of rotation is along the y axis. Your torque vector should therefore be parallel to the y-axis (either +j or -j).

μ is not in the -j direction. Notice how the axes are labeled.

6. Aug 4, 2014

### Zondrina

Ah I was curious about that. I switched my vectors around and got the right answer though.

From now on I'll probably re-define the axes for simplicity.

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text:
Issue No. 03 - March (1972 vol. 21)
ISSN: 0018-9340
pp: 260-268
Bruce J. Hansen , Raytheon Data Systems, Norwood, Mass. 02067.
Jack Sklansky , School of Engineering, University of California, Irvine, Calif. 92664.
Robert L. Chazin , Department of Mathematics, University of California, Irvine, Calif. 92664.
ABSTRACT
The minimum-perimeter polygon of a silhouette has been shown to be a means for recognizing convex silhouettes and for smoothing the effects of digitization in silhouettes. We describe a new method of computing the minimum-perimeter polygon (MPP) of any digitized silhouette satisfying certain constraints of connectedness and smoothness, and establish the underlying theory. Such a digitized silhouette is called a regular complex,'' in accordance with the usage in piecewise linear topology. The method makes use of the concept of a stretched string constrained to lie in the cellular boundary of the digitized silhouette. We show that, by properly marking the virtual as well as the real vertices of an MPP, the MPP can serve as a precise representation of any regular complex, and that this representation is often an economical one.
INDEX TERMS
CITATION
Bruce J. Hansen, Jack Sklansky, Robert L. Chazin, "Minimum-Perimeter Polygons of Digitized Silhouettes", IEEE Transactions on Computers, vol. 21, no. , pp. 260-268, March 1972, doi:10.1109/TC.1972.5008948


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# Another seemingly simple integral

• February 25th 2009, 01:02 PM
fattydq
Another seemingly simple integral
The definite integral from -1/2 to 1/2 of 9/the square root of 1-x^2 dx

I don't even know where to begin in this one. Finding the antiderivative of that seems impossible...
• February 25th 2009, 01:05 PM
Krizalid
$\int_{-1/2}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}=2\int_{0}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}.$

Look at a table of integrals, that primitive is the arcsin function.
• February 25th 2009, 01:10 PM
fattydq
Quote:

Originally Posted by Krizalid
$\int_{-1/2}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}=2\int_{0}^{1/2}{\frac{9}{\sqrt{1-x^{2}}}\,dx}.$

Look at a table of integrals, that primitive is the arcsin function.

I know that, but what did you just do knowing that? Where did the 2 come from and why did the -1/2 just disappear? And I know that 1/the sq root of 1-x^2 but it's a 9 not a 1, so where does this come into play?
• February 25th 2009, 01:22 PM
Krizalid
$\int_{-a}^af=2\int_0^af$ provided that $f$ is an even function and $a\in\mathbb R.$
• February 25th 2009, 01:26 PM
Abu-Khalil
Quote:

Originally Posted by fattydq
I know that, but what did you just do knowing that? Where did the 2 come from and why did the -1/2 just disappear? And I know that 1/the sq root of 1-x^2 but it's a 9 not a 1, so where does this come into play?

$\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=
\int_{-\frac{1}{2}}^{0}\frac{9}{\sqrt{1-x^2}}dx+\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx$
,
but $\frac{9}{\sqrt{1-(-x)^2}}=\frac{9}{1-x^2}\Rightarrow \int_{-\frac{1}{2}}^{0}\frac{9}{\sqrt{1-x^2}}dx=\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx$.
Therefore $\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=
2\int_{0}^{\frac{1}{2}}\frac{9}{\sqrt{1-x^2}}dx=18\int_{0}^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^2}}$
.

PD: Late >.<
• February 25th 2009, 01:44 PM
fattydq
Quote:

Originally Posted by Krizalid
$\int_{-a}^af=2\int_0^af$ provided that $f$ is an even function and $a\in\mathbb R.$

I love calculus, because there's rule's for everything that I'm just supposed to know magically, intuitively. I swear my professor never mentioned anything like that.


text:
# How to prove Shortest Common Superstring is NP-Hard

After some research and many youtube videos I have learnt that to prove a problem is NP-Hard; you would need to reduce that problem to known NP-Hard problems such as Subset Sum Problem, Halting Problem, Satisfiability Problem or Traveling Salesman Problem. Now my problem is Shortest Common Superstring

• Input: A finite set R = {r1, r2, ..., rm} of binary strings (sequences of 0 and 1); positive integer k.
• Question: Is there a binary string w of length at most k such that every string in R is a substring of w, i.e. for each r in R, w can be decomposed as w = w0rw1 where w0, w1 are (possibly empty) binary strings?

In this link it states that the problem is NP-Hard(https://www.geeksforgeeks.org/shortest-superstring-problem/), besides I am using the greedy algorithm stated there to solve the problem.

So I need help with choosing which NP-Hard Algorithm to reduce to and a way to reduce it. I do not expect a full solution(even though it would be welcome), just guidance would be enough. I honestly do not have much clue on how to go about it.

• The reduction actually goes in the other direction: You pick a problem that you already know to be NP-hard, and reduce it to the problem you want to prove NP-hard. This is a common mistake. – j_random_hacker Dec 15 '19 at 2:31
• (This is a common mistake - getting the reduction direction wrong, that is.) – greybeard Dec 15 '19 at 3:39

Theorem 1 states that Hamilton path problem can be reduced to SCS even if all strings have fixed length $$H \ge 3$$.


text:
# Overte identitu 1-sinx cosx = cosx 1 + sinx

My third problem solving.

[(1 + sin x)(1 + sin x) + (cos x)(cos x)]/[(cos x)(1 + sin x)] = 2 sec x (1 + 2 sin x + sin^2 x + cos^2 x)/[(cos x)(1 + sin x)] = 2 sec x. where: sin^2 x + cos^2 x = 1 Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. 1- (sin (r) - cos (a 2 cos (r Determine the identity to ︵) tan (x) o cos (x) o sec (a) o sin(x) none of these Get more help from Chegg Solve it with our pre-calculus problem solver and calculator 1 Sinx Cosx. How do you verify this identity: (cosx)/(1 sinx) (1 sinx iit 1981 find the solution of cosx = 1 youtube show that tanx secx 1/tanx is equivalent to what integration x sinx/1 cosx? quora solve equation 15 Oct 2016 By multiplying both numerator and denominator by 1+sinx and using the difference of squares the result follows quickly.

Statement Rule 8 (1 - sin²x) tan x cosa sino tan o Select Rule cot seco CSC Validate л (0) … 07.12.2007 Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. 06.07.2012 20.10.2014 cosx(1＋sinx)の微分の仕方を教えてください。 数学 y=cosx/1-sinx (0≦x≦2π)の関数の極値を求めよ の解答を教えて下さい。 20.03.2010 Solution for Prove the identity. 1+ cosx + sinx sinx =2 cscx 1+ cosx 03.08.2013 1/(1+sinx+cosx)の不定積分を手早く求める方法です。 plzz answer fast prove the identity sinx- tanx / sinx * tanx = cosx -1 / sinx - Math - Introduction to Trigonometry 19.01.2011 Similar Questions. trigonometry.

## Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.

Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor. Feb 04, 2010 · Quite simple. We use (a - b)(a + b) = a^2 - b^2.


text:
# How can heavier elements be made from hydrogen atoms?

In a JRE podcast from a few years ago, talking about the idea of a universal constructor, Sam Harris said that we could go to a near-vacuum area of deep space, collect nothing but hydrogen atoms, build a fusion reactor, create heavier elements, and based on those elements, create a universal constructor and then build anything.

I understand that standard hydrogen atoms are made up of a proton and an electron (no neutron), while other heavier elements require neutrons.

So is it possible to create heavier elements from nothing but hydrogen? How?

• Stars do it all the time. – badjohn May 16 '20 at 7:01
• Did he estimate how long it would take to 'collect' enough stuff to build a reactor? Did he elaborate how to build a fusion reactor from hydrogen without forming a star? – user257090 May 16 '20 at 7:24

Saying "We" could do it is probably a bit ambitious, but stars collect hydrogen from space in gravitational collapse, and form into natural fusion reactors. The first step in stars like the Sun is the pp-n chain (diagrams from Structures of the Sky which contains more details), which creates helium from hydrogen

Variants of this process dominate in larger stars. In stars of mass greater than about 1.3 solar masses, the CNO cycle dominates. This is a catalytic process, in which protons are added to an atomic nucleus, which then decays releasing Helium.

Subsequently, Helium burning initiates the alpha ladder

The alpha ladder adds further helium nuclei to create a range of heavier elements, known as alpha elements whose most abundant isotopes are integer multiples of four, the mass of the alpha particle or helium nucleus. The alpha ladder starts with carbon-12 and produces oxygen-16, neon-20, magnesium-24, silicon-28, sulphur-32, argon-36, calcium-40, titanium-44, chromium-48 terminating at iron-52 since the further fusion of alpha particles consumes energy. This is the reason for the abundance of iron in the universe. Interactions above oxygen contribute little to the energy produced by stars.

These are the main processes in stars. Other elements can be made in less common processes, and in supernovae.

The first step in the process is to create helium. Here's how the Sun and many other stars do it: Proton-proton chain reaction.

• Ok, so one of the protons is converted into a neutron through beta plus decay while releasing a positron and an electron neutrino – user264571 May 16 '20 at 7:17
• Yes, you need to stop thinking of protons and neutrons as immutable little balls. – badjohn May 16 '20 at 7:20

It's thanks to the weak nuclear force. Protons can turn into neutrons and vice-versa through weak force interactions.

The neutron is more massive than the proton, so free neutrons decay into protons (accompanied by an electron and an anti-neutrino). However, neutrons can be stable to beta decay when trapped in nuclei with other nucleons and this offers the possibility of protons turning into neutrons via the inverse process - electron capture or the emission of a positron.

The latter is the dominant process in stars, but the former might be easier if one were to do it artificially. It would require compressing the hydrogen to densities of around $$10^{11}$$ kg/m$$^3$$, but would have no temperature constraints. At these densities, the electrons would be degenerate and the Fermi energy would be high enough to initiate electron capture with the protons. The free neutrons would then react with protons to produce deuterons. The positron emission route has to proceed via the intermediary of an unstable di-proton. This requires very high temperatures to form and along with the improbability of a beta decay occurring whilst the di-proton exists, is why hydrogen burning is slow in the Sun.


text:
# Solving just a Trigonometric Equation!

Geometry Level 5

$\large{20\sin(x) - 21\cos(x) = 81y^2 -18y + 30}$

How many ordered pairs of real $$(x,y)$$ exists satisfying the above equation subjected to the constraint that $$-2015 < x < 2015$$.

×

Problem Loading...

Note Loading...

Set Loading...


text:
NIPS 2018
Sun Dec 2nd through Sat the 8th, 2018 at Palais des Congrès de Montréal
Paper ID: 1652 Contextual Combinatorial Multi-armed Bandits with Volatile Arms and Submodular Reward

### Reviewer 1

The paper introduces a novel algorithm with regret guarantees for stochastic contextual combinatorial multi-armed bandits for volatile arms and submodular rewards. The dependence on dimension of the space of their regret bound is not great and in general constants seem large. A discussion of these quantities is warranted here. The main drawback is that there is no comparison to other bounds. It seems that they can instantiate their algorithm to simpler settings that have been analyzed by previous work. How does their bound compare to the regret of previous algorithms in these simpler setting? Do they have any guess at the lower bound? Splitting the context space by hypercubes seems expensive. Is the algorithm computationally efficient? Please add a discussion on this. As for the experiments, it would have been nice to have seen results for more than one dataset. I understand their reasoning for their definition of regret, but is it necessary in their analysis to include the factor (1-1/e)?

### Reviewer 2

This paper studies the stochastic contextual combinatorial multi-armed bandit with volatile arms (i.e. arms can become available or unavailable in each round) and submodular rewards (i.e the reward observed for the selected arms is not a sum of their individual rewards, but a diminishing function of their returns, dependent on the relationship between selected arms). The authors very clearly introduce the reader to the problem and to the key relevant topics (the contextual, combinatorial and volatile bandits, and submodular functions). They successfully explain their main contribution (a contextual combinatorial multi-armed bandit algorithm compatible with submodular reward functions and volatile arms), by explicitly stating the novelty of their work: "[17] considers a special submodular reward function that depends on a linear combination of individual rewards and assumes a constant set of arms which is very different from the volatile arms in our work". As such, the significance of their work is important, since they consider (a) general submodular reward functions (as in equation (1)), and (2) allow for arms to dynamically become available or unavailable, all within the contextual-combinatorial bandit setting. Their proposed algorithm is based on ideas from the contextual bandit with similarity information literature. They divide the arms into different groups based on their context, learn the expected quality/reward for each group of arms, and assume that arms with similar context have similar properties (as in Lipschitz bandits via the Holder condition). The authors propose a heuristic to explore under-explored groups (hypercubes in space), in combination with the greedy algorithm (1) to solve the submodular function maximization. The key parameters of their algorithm are the hypercube partition sizes $h_T$ and the control policy $K(t)$, which they specify in connection with their theoretical regret bound. Some insights into the tightness of the bound, and its connection to the stochastic pattern of the arm arrival are provided. The authors have provided in their rebuttal further comments on the impact of context space dimension in the regret bound, which would be interesting to add to the final manuscript. However, it is unclear how realistic the budget assumption in corollary 1 is. The authors argue in their rebuttal that it is realistic in some practical applications (such as crowdsourcing). However, it is likely that in other applications, having a smaller budget than the available arms at every time instant would not apply. Thus, it would be interesting if the authors include such discussion on their final manuscript. Nevertheless, further analysis (both theoretical and empirical) of realistic arm arrival patterns would increase the impact of their work. The evaluation section focuses on a real problem of crowdsourcing, which is specifically re-casted to fit their problem formulation. They conclude that (a) contextual-combinatorial bandits perform better than non-contextual ones, (2) not considering submodularity incurs an obvious reward loss (because the underlying reward is indeed submodular); and (3) increasing the budget reduces regret (no learning required). These results are not surprising and in some sense expected, since the problem set up is specifically targeted to the designed framework. On the one hand, the authors have agreed to extend the evaluation of their algorithm to longer runs (as only 200 time slots are shown in Figure 3). By doing so, they can empirically evaluate the presented regret bound (as it is not clear from the provided figures whether it is sublinear, or how it compares to the theoretical bound). On the other, I would still encourage the authors to consider simulated scenarios where they can empirically evaluate the impact of stochastic pattern of the arm arrivals. Finally, their algorithm is based on static partitions of the space (hypercubes $h_t$). I thank the authors for the insights provided in their rebuttal regarding (a) the incurred algorithm complexity and (b) extending the partition to be non-homogeneous and/or dynamic. As such, I encourage the authors to add them to the final manuscript.

### Reviewer 3

This paper studies the contextual combinatorial multi-armed bandit (MAB) problem, where the goal is to select in each round the subset of arms that maximize the expected cumulative reward. The main contribution of the paper is the formulation and analysis of the CC-MAB algorithm, which addresses this contextual combinatorial MAB problem for the setting where the reward function is submodular and the set of arms can change between rounds. Although the algorithm builds on previously well-studied MAB frameworks, it has a relatively significant contribution by bringing together the combinatorial MAB setting, with the use of contexts, and with the challenging setting of "volatile arms". The paper is well written and the studied problem is interesting and quite well motivated for real-world applications. On the other hand, the experiments seem a bit lacking. I wonder why there was no empirical comparison to combinatorial MAB algorithms (for instance, from references [5,7])? Also, none of the considered benchmarks considers the submodularity of the reward function, thus, as it is pointed out in the paper, they incur obvious reward loss. More details about the comparison of CC-MAB behavior compared to submodular MAB algorithms (for instance, the one proposed in [Chen, Krause, Karbasi, NIPS 2017]) would be appreciated. Minor comments/typos: - line 172: performs *the* following steps? - line 221: budge -> budget ======================= Post author response edit: I have read the rebuttal and thank the authors for addressing some of my concerns.


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## c1c9m9h1 Group Title (2x^2+xy+y^2)dx + 2x^2 dy=0 How Do I Solve This 1st Order DifEQ? Help Please!! one year ago one year ago

• This Question is Open
1. matricked Group Title

yup this 1st Order DifEQ

2. matricked Group Title

you can use y=vx (for homogeneous equations) to solve it

3. c1c9m9h1 Group Title

I don't understand how you determine if its homogeneous, separable...but I understand if its an exact equation...

4. matricked Group Title

then dy =xdv +vdx

5. c1c9m9h1 Group Title

I have 6 problems that i have to do and i have the answers but i just dont understand how to do the work to get the answer unless its an exact equation..

6. matricked Group Title

see the sum of the powers of x and y is uniform (here its 2 ) throughout hence homogeneous

7. c1c9m9h1 Group Title

Great! I'm Learning Something Finally!

8. matricked Group Title

welcome

9. matricked Group Title

after using y=vx and dy =xdv +vdx the resultant will become a seperable one

10. c1c9m9h1 Group Title

Is there a certain formula that is out there where i can solve these problems if they are separable or homogenous or any other thing out there?? I really want to learn this stuff

11. abb0t Group Title

You cannot solve this using separation of variables.

12. c1c9m9h1 Group Title

Yes your right, I understand that its not an exact equation..

13. c1c9m9h1 Group Title

I have 6 problems that are due in 45 mins... I need help!!!

14. JamesJ Group Title

Let's start from the beginning (2x^2+xy+y^2)dx + 2x^2 dy=0 implies that dy/dx = -(2x^2+xy+y^2)/(2x^2) and hence dy/dx = -1 - (1/2)(y/x) - (y/x)^2

15. JamesJ Group Title

This is what is called a homogeneous equation. It is one of the meanings of the word homogeneous for differential equations. The substitution 'trick' for such equations is to write v = y/x and hence y = vx which implies y' = v + xv' Therefore we can rewrite the equation above as v + xv' = - (1 + (1/2)v + v^2) This now IS a separable equation which you can solve using standard techniques to find the function v(x). Once you have done that, substitute back v(x) = y(x)/x to solve for y(x)

16. JamesJ Group Title

***I dropped a half in front of the (y/x)^2 term, so v + xv' = -(1/2) ( 2 + v + v^2 )

17. JamesJ Group Title

Make sense?

18. c1c9m9h1 Group Title

I keep getting different answers though when i do that... i have the solutions just can get there while showing my work

19. c1c9m9h1 Group Title

Anyone who can help solve these problems will be a life saver!

20. JamesJ Group Title

Well, v + xv' = -(1/2) ( 2 + v + v^2 ) implies -2x v' = v^2 + 3v + 2 = (v+2)(v+1) hence separating $-2\int \left( \frac{A}{v+2} + \frac{B}{v+1} \right) dv = \int \frac{dx}{x}$ You take it from there.


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## Psalm 23: English to Maori

The Lord is my shepherd; I shall not want. 2 He maketh me to lie down in green pastures; he leadeth me beside the still waters. 3 He restoreth my soul: he leadeth me in the paths of righteousness for his name’s sake. 4 Yea, though I walk through the valley of the shadow of […]

## Taming a Beast called UNIX

The year was 1988. Voices hushed in trepidation, the apprentices gathered in a dingy corner of the Engineering building. In the undergraduate computer lab we assembled to offer sacrifices to the mainframe, hunched over the fading amber or green glow of hesitant dumb terminals, all connected to the mighty UNIX (VAX/VMS) mainframe. Poor first year […]

## Fermat’s Last Theorem

Fermat’s Last Theorem states that:

There are no positive integers $x, y,$ and $z$ with $x^n + y^n = z^n$ if $n$ is an integer greater than 2.

For $n = 2$ there are many solutions:

( 3, 4, 5) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17)
( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85)
(16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65)
(36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97) … the Pythagorean triples.

In the margin of his copy of the Arithmetica of Diophantus the French jurist Fermat wrote circa 1637 that for greater $n$ no such triples can be found; he added that he had a marvelous proof for this, which however the margin was too small to contain:

Cubum autem in duos cubos, aut quadrato-quadratum in duos quadrato-quadratos, et generaliter nullam in infinitum ultra quadratum potestatum in duos ejusdem nominis fas est dividere; cujus rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caparet.

Some of the greatest mathematical minds in history have applied their genius to this problem, none succeeded. Every other result which Fermat had announced in like manner had long ago been dealt with; only this one, the last, remained. Finally, 358 years after Fermat wrote his theorem, Andrew Wiles saw an application of elliptic curves and modular forms (20th century mathematical constructs, not known to earlier mathematicians) that could contribute to the proof.

Theorem:

Let $n, a, b, c, \in \mathbb{Z}$ with $n > 2$.
If $a^n + b^n = c^n$ then $abc = 0.$

Proof:

The proof follows a program formulated around 1985 by Frey and Serre[F,S]. By classical results of Fermat, Euler, Dirichlet, Legendre, and Lamé, we may assume that $n = p$, an odd prime ≥ 11. Suppose $a, b, c \in \mathbb{Z} , abc \neq 0,$ and $a^p + b^p = c^p$. Without loss of generality we may assume that $2|a$ and $b \equiv 1 \bmod 4$. Frey [F] observed that the elliptic curve $E : y^2 = x(x -a^p )(x + b^p)$ has the following ‘remarkable’ properties:
(1) $E$ is semistable with conductor $N_E = \prod_{{l}|abc} {l}$ ; and
(2) $\overline{\rho}_{E,p}$ is unramified outside $2p$ and is flat at $p$ .
By the modularity theorem of Wiles and Taylor-Wiles [W,T-W], there is an eigenform $f \in S_2 (\Gamma_0(N_E))$ such that $\rho_{f,p} = \rho_{E,p}.$ A theorem of Mazur implies that $E,p$ is irreducible, so Ribet’s theorem [R] produces a Hecke eigenform $g \in S_2 (\Gamma_0(2))$ such that $\rho_{g,p} \equiv \rho_{f,p} \bmod {\tt p}$ for some ${\tt p}|p.$ But $X_0(2)$ has genus zero, so $S_2 (\Gamma_0(2)) = 0$. This is a contradiction and Fermat’s Last Theorem follows. Q.E.D.

References: Continue reading “Fermat’s Last Theorem”

## RE: Darwin & The Deity

Faith and Theology: Ten propositions on Darwin and the deity Should it be of concern to Christians that Darwin was never more than a nominal believer? – well, are you concerned whether your surgeon, mechanic, or hair stylist goes to church? Of course not. Your only concern is that she wields a scalpel, wrench, or […]

## Unix bash snippet: global replace text in multiple files

Unlike this overkill solution in python, bash is the best tool for this sort of task. The following script searches all files the current directory and subdirectories, and replaces all instances of text “abc” with text “xyz”. grep and sed are your friends…

## USA is going crazy.

Rick Santelli vented a forceful and furious rant against Obama’s multi-trillion stimulus package — many people suspect it will only cause further distortions in a deeply troubled economy, and the only realistic solution is to let people suffer the consequences of vast overspending. Anything else is just an attempt to defy the law of gravity. Personally I don’t know — I hope America can reinvent itself, but there’s a real sense of panic emerging from the States right now.

Even if it was some kind of staged PR exercise, what Santelli actually said is entirely correct. I’ve been saving doggedly all these years, watching everyone jump on board the property gravy train, and there’s NO WAY that I would willingly subsidize them. As Peter Schiff said, the President should be defending the dollar, not embarking on mad inflationary schemes. Obama’s plan is fundamentally unjust.

I Bought an Expensive House. My Bad, Not Yours

The only people affected by plummeting real estate prices are the ones who bought a house that cost more than they could afford, hoping for a spike in value so they could sell at a profit or take out a new loan based on an increased value. Their home wasn’t just a place to live; it was an investment they thought they could liquefy at will. If we’re saving these poor souls from the 26.7% drop in their investment, we should give twice as much aid to everyone who has lost approximately 50% in the stock market since its peak.

Transcript over the jump.. Continue reading “USA is going crazy.”

## RE: THIS IS LIFE, Get Creative.

6 Truths to Wake Up to Right Now | steve-olson.com 1. This is Life Quite possibly the most obvious but most eye-opening advice I can ever offer people is the statement: this is life. We may have more than one shot, but nobody knows that for certain so we have to make the most of […]


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11.052002

# Find the solutions to the equation -5/2x+3/4x=-34

Jaye Mathematics 2 3

Find the solutions to the equation -5/2x+3/4x=-34

Posted by Jaye | Posted at May 11, 2002 | Categories: Mathematics

$-frac{5}{2}x+frac{3}{4}x=-frac{3}{4}\-frac{5 imes2}{2 imes2}x + frac{3}{4}x=-frac{3}{4}\frac{-10}{4}x+frac{3}{4}x=-frac{3}{4}\frac{-10+3}{4}x=-frac{3}{4}\-frac{7}{4}x=-frac{3}{4}\-frac{7}{4}x imes -4 =-frac{3}{4} imes -4\7x = 3\oxed{f{x =frac{3}{7}}}$


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Point A is at (-7 ,4 ) and point B is at (2 ,-1 ). Point A is rotated pi  clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?

Feb 22, 2018

Due to rotation of A distance reduced between A and B by 4.4646

Explanation:

A((-7),(4)) -> A’((7),(-4))

$\vec{A B} = \sqrt{{\left(- 7 - 2\right)}^{2} + {\left(4 + 1\right)}^{2}} = 10.2956$

vec (A’B) = sqrt((7-2)^2 + (-4+1)^2) = 5.831

Distance change

vec(AB) - vec (A’B) = 10.2956 - 5.831 = 4.4646


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# 2014-12 Rational ratios in a triangle

Determine all triangles ABC such that all of $$\frac{AB}{BC}, \frac{BC}{CA}, \frac{CA}{AB}, \frac{\angle A}{\angle B}, \frac{\angle B}{\angle C}, \frac{\angle C}{\angle A}$$ are rational.

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# 2012-18 Diagonal

Let $$r_1,r_2,r_3,\ldots$$ be a sequence of all rational numbers in $$(0,1)$$ except finitely many numbers. Let $$r_j=0.a_{j,1}a_{j,2}a_{j,3}\cdots$$ be a decimal representation of $$r_j$$. (For instance, if $$r_1=\frac{1}{3}=0.333333\cdots$$, then $$a_{1,k}=3$$ for any $$k$$.)

Prove that the number $$0.a_{1,1}a_{2,2}a_{3,3}a_{4,4}\cdots$$ given by the main diagonal cannot be a rational number.

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# 2010-5 Dependence over Q

Find the set of all rational numbers x such that 1, cos πx, and sin πx are linearly dependent over rational field Q.

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# 2009-9 min or max

Suppose that * is an associative and commutative binary operation on the set of rational numbers such that

1. 0*0=0
2. (a+c)*(b+c)=(a*b)+c for all rational numbers a,b,c.

Prove that either

1. a*b=max(a,b) for all rational numbers a,b, or
2. a*b=min(a,b) for all rational number a,b.
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# 2009-7 A rational problem

Let n>1 be an integer and let x>1 be a real number. Prove that if
$$\sqrt[n]{x+\sqrt{x^2-1}}+\sqrt[n]{x-\sqrt{x^2-1}}$$
is a rational number, then x is rational.

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Let a, b, c, d be positive rational numbers. Prove that if $$\sqrt a+\sqrt b+\sqrt c+\sqrt d$$ is rational, then each of $$\sqrt a$$, $$\sqrt b$$, $$\sqrt c$$, and $$\sqrt d$$ is rational.


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# Higher dimension operator in free Dirac Lagrangian

When discussing higher dimensional operators in a theory with fermions, why do I never see anyone ever talk about the dimension five operator $\partial_\mu\bar\psi\partial^\mu\psi$?

How does the Fermion field behave when such a strange operator is its Lagrangian?

$$\mathcal{L}=\bar\psi(i\partial_\mu\gamma^\mu-m)\psi+\frac{1}{\Lambda}\partial_\mu\bar\psi\partial^\mu\psi$$

Since it is still quadratic in $\psi$, I expect this to be fairly easy to analyze. What happens when such a term is added?

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I'm not an effective field theory guy myself, but you can pass the derivative to the other field and get $\overline{\psi} \partial^2 \psi.$ But then, using on-shellness (which for some reason is accepted in that field), $\psi$ obeys the Klein-Gordon-equation, so $\partial^2 \psi = \pm m^2 \psi$. So up to off-shell stuff going on, you've written down a mass term for the fermion. – Vibert Dec 7 '12 at 5:34
The equation of motion for $\psi$ would be $(\Lambda^{-1}\Box - i\gamma^\mu\partial_\mu + m)\psi=0$. I believe this violates the positivity postulates of quantum mechanics. See Weinberg, Quantum Theory of Fields, Vol I, Ch 10.7. The current $\overline{\psi}\gamma^\mu\psi$ isn't conserved either. – MarkWayne Dec 7 '12 at 6:34

This particular extra term may be removed by a field redefinition $$\psi\to \psi' = \psi - K \cdot \gamma^\mu \partial_\mu \psi$$ for an appropriate value of $K\sim 1/\Lambda$, up to terms that are even higher dimension operators. This also modifies the mass. This field redefinition is an explicit off-shell way to realize Vibert's comment that one is just modifying the mass in which he assumed the equations of motion.

Mark Wayne is also right that (despite the equivalence, up to even higher-order terms), the propagator violates positivity of quantum mechanics (we really mean the positive norm of states with particle excitations which is linked to positive probabilities, a "must": excitations with negative norm are known as "[bad] ghosts"). If one writes down the full propagator for this (free) theory, one gets some additional poles which have a coefficient of the wrong sign. However, these pathologies occur at $p^2\sim \Lambda^2$ where we expect the theory to break down, anyway: the OP wrote the extra term as a correction in effective field theory that is meant to be used at energies $p^2\ll \Lambda^2$. This pathology with non-positivity may be fixed by new physics near $\Lambda$ and because of the equivalence, indeed, we can see that we may adjust the even higher-order terms (which may also arise from integrating out other fields and interactions) so that the full theory is exactly equivalent to the ordinary massive Dirac fermion and therefore has no ghosts.

That means that as long as we use this as an effective field theory, knowing that it may have further higher-order modification and new physics in general that kicks in near $p^2\sim \Lambda^2$, it is healthy as an effective field theory. Again, if we wanted to use it as a full theory even at energies of order $\Lambda$, it would be inconsistent.

In general, higher-order "free" terms like that – similarly $\square\phi\cdot \square\phi$ for bosons etc. – have been proposed as ways to make the propagators softer in the UV, like $1/p^4$ instead of $1/p^2$, which would improve the convergence of Feynman diagrams. The price is that if such a theory is to be taken seriously near $\Lambda$, the scale determining the size of the higher-dimensional operators, then this theory leads to new negative-norm excitations which is an inconsistency. So the addition of these new terms makes the theory break down "earlier", despite the intriguing softening of the UV problems.

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@QuantumDot The conclusion of this answer is correct. However, the problem with higher order derivatives in free actions is the unboundedness of the Hamiltonian. The problem with negative norm states only shows up when one does not quantize the theory properly (and even in this case the vacuum is not normalizable). This is a very spread mistake which does not have any consequence because in both cases the theory needs a completion at $\Lambda$. – Diego Mazón Dec 7 '12 at 8:16
Dear @drake, if you allow, I would disagree. In some cases, you may say that to "quantize the theory properly" means to avoid the negative norms at any cost, and convert the minus signs to unboundedness of energy. You would have this freedom if the higher-derivative term were actually the leading one. But in this case, the norms are already determined by the ordinary kinetic and mass terms and we're just adding corrections. So we're not able to change the overall sign of the norm of the excitations. – Luboš Motl Dec 7 '12 at 8:22
Instead, we see new poles and we may compute the relative sign of the poles - the coefficients of the new ones and the old ones - and regardless of conventions, the relative sign is minus. So one of the poles of the propagator inevitably corresponds to wrong-sign-of-norm excitations regardless of any conventions. Quite generally, I think it's right to treat Lagrangians with the wrong sign of the kinetic terms to describe negative-norm (yes, very pathological) excitations and it is bad practice to pretend that the wrong sign of kinetic terms is "just" about the indefinite energy. – Luboš Motl Dec 7 '12 at 8:24
In the usual kinetic+mass terms, it's simply appropriate to read the coefficients in the standardized way: the coefficient of the kinetic term decides about the norm of the excitations of the field; the mass term describes the mass (which may be tachyonic in some cases, for unusual signs). The field redefinitions needed to claim that the wrong sign of the kinetic term still allow all the excitations to be positive norm aren't really allowed, so a theory with physical excitations with a wrong sign of the kinetic terms is already inconsistent as a physical theory. Denying it is wrong. – Luboš Motl Dec 7 '12 at 8:27
Dear LubosMotl, if one quantizes the theory correctly (annihilation operators go with $e^{-i(-|E|)t}$, with $-|E|$ the energy of the mode), one obtains a theory with positive definite states and unbounded from bellow Hamiltonian. The $i\epsilon$ terms of the new poles have the correct sign even though the $1/(p^2-m^2)$ has the wrong sign (because the energy of the mode is negative). If you look at the classical theory where everything is clear, the Hamiltonian is also unbounded from bellow. So if one quantizes the theory in the way you like, one obtains a wrong classical limit. Kind regards. – Diego Mazón Dec 7 '12 at 9:28


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# Cantor theorem on orders

It is "a well-known theorem of Cantor", said Sierpinski (circa 1920), that every countable total order can be imbedded in the rationals, and he proceeds to demonstrate that, assuming the continuum hypothesis, it is possible to construct a similar "universal order" of cardinal $\aleph_1$. I have two questions : 1) Where did Cantor prove his theorem 2) Can CH be weakened in the previous result ? For reference, Sierpinski construct an ordering on binary sequences indexed by countable ordinals (up to some ordinal $<\omega_1$), having the desired property than any countable Dedekind cut is separated by some element ; this is actually the same order than the one on surreal numbers born before day $\omega_1$ (as easily shown by the Gondor construction)

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Concerning question 2:

a. It is consistent that $2^{\aleph_0}= \aleph_2$ and there is a universal order of size $\aleph_1$.

b. It is consistent that $2^{\aleph_0}= \aleph_2$ and there is no universal order of size $\aleph_1$.

See Kojman+Shelah, JSL. preprint

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In his 1895/97 papers "Beiträge zur Begründung der transfiniten Mengenlehre", Cantor proved that any two countable dense orders without endpoints are isomorphic, according to this article by Charles Silver. The method used by Cantor also shows that every countable linear order embeds into the rationals. I do not have Cantor's paper here, so I am not sure if Cantor stated this result. –  Goldstern Jul 11 '12 at 22:50

Sorry, almost all pertinent answers (even with the exact reference of the Sierpinski article) were given in the article Universal order type ; so the only question I have left is "where did Cantor state it ?"

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