Adventure games, an established and popular genre among microcomputer enthusiasts, are generally divided into two categories: text adventures and graphics adventures. This dichotomy seems clear enough: a text adventure uses words alone, similar to books or radio plays, to create a picture in the mind. A graphic adventure, in contrast, draws these pictures for the eye. Each category claims its own adherents.
Text adventure aficionados assert that only verbal descriptions can provide a satisfactorily rich level of story telling, as they leave most of the visualization to imagination. For these people, a good adventure is like a good book-a reading experience.
Graphic adventure chauvinists point out correctly that text adventures employ a very static screen display. This is wasteful, they argue, in light of the capabilities of microcomputers. Graphics potential (and sound potential, for that matter), should not be left unexploited by the adventure program. Each faction has a well-taken point.
The fact is, however, that the division between text adventures and graphic adventures is not nearly as sharp as this, and some of the most interesting developments in microgames are taking place between the two poles. All graphics adventures employ text to one degree or another--either to augment the graphic display or as the basis for graphic augmentation. Let's examine these approaches more closely.
One method of constructing graphic adventures might be called the illustrated text adventure. This employs all the normal conventions of a text adventure, including text command input, though the text may be less descriptive and more to the point. In addition to this, the player is provided with an illustration of his or her current position via a high resolution picture. These are stored on disk and are called out as necessary by the main, text-oriented program.
The simile is close to the idea of a comic book. Each piece of the story has its own "frame" of picture and text. The point of view is that of the central character--the player sees the locations as if he were there.
Another approach results in the mapped adventure, wherein the player or players appear as symbols on a map. This map depicts the details of the location, indicating the type and placement of terrain, walls, objects, and enemies, among other things. Player movement is input through keyboard or joystick. As a character reaches the border of a screen map in any direction, either the screen begins to scroll in that direction, or a new location map is drawn. This creates a more omniscient perspective, with the player looking down on his character's movement from above.
Typically this type of adventure does not allow text input, but limits the player to a certain number of possible contingencies in a menu or command format. The comparison between text adventures and mapped adventures is close to that of a fill-in-the-blank vs. a multiple-choice test.
Text adventure enthusiasts feel the menu format is restrictive--one can only choose to move or stand still, flee or fight, take or drop, and so on. The mapped adventure began as an outgrowth of a certain fantasy role-playing game whose name we are not permitted to print. One obvious advantage of this format is the capability for multi-character play.
Until quite recently, the preference boiled down to a choice between the classic-style text adventure game and the fantasy role-playing adventure. However, hybrids are now being developed that now provide both types of enjoyment.
Truly hybrid adventure games of the future will offer the best aspects of all approaches to computer gaming. Textual description will provide detailed background and help set the mood. Maps will be available to indicate position (except in areas still unexplored). Computer graphics and sound will be called on where appropriate to animate action sequences. Arcade-style challenges will become part of the stories, calling for feats of coordination before the plot advances. And it won't be too long before the videodisc becomes a necessary peripheral for state-of-the-art adventuring.
This is still a ways off in the future. That is quite enough background, however, to examine the spectrum of graphic adventure software currently available for the Atari.
In the arcade game Asteroids, you use hand-eye coordination to zap as many asteroids as possible in the few minutes you are allotted. In Mission: Asteroid, your job is to destroy just one asteroid, and you have several hours in which to do it. The game allows you to input two-word commands, and displays a high resolution picture for every location, with a four-line text window at the bottom of the screen. Pressing return without entering text will kill the picture momentarily, giving you a chance to recall the last 24 lines of the text that have scrolled by. Tap return again and the picture will reappear.
The plot line of Mission: Asteroid is pretty easy to follow; it was designed as an introduction to a series of "Hi-Res Adventures" of much greater complexity. That is not to say that the game is easy to solve--it's not.
Briefly described, you are an astronaut, who is sent into deep space to destroy an asteroid headed toward earth. You have only a limited amount of time before the asteroid collides with earth and the game is over.
It is interesting to gauge the reactions of text adventure buffs to the over 100 pictures on the disk. Some enjoy them thoroughly, others use them as mnemonic devices, which eliminate the necessity to map the adventure on paper. Still others find the pictures a questionable, if not downright distracting, addition. The pictures are well-done--obviously a lot of time was spent executing them. They are pretty, but they do not achieve even the quality of a well-drawn comic book. This leaves me a little dissatisfied. I think it is fair to ask of any graphic adventure game: How well would the adventure stand on its own, if we were to delete all the pictures?
In the case of Mission: Asteroid, the plot line is kept intentionally simple, as it is intended to be an introduction to a series of graphic adventures, and it is an enjoyable program. As such, it would not be really entertaining without the pictures. The graphics serve to enhance the overall effect, but are unable to enrich the story significantly. In the intricate and involving stories of other adventures in the series, they stand a better chance of achieving this goal. I am certainly among those who feel illustrations can improve text adventures.
Apparently so is Scott Adams, who is in the process of re-releasing all twelve of his now-classic adventures as illustrated text adventures.
Ali Baba and the Forty Thieves attempts to move beyond the illustrated text adventure. It was designed for the Atari, in contrast to Mission: Asteroid, which is an Apple translation. The game makes use of some of the special features of the Atari, such as multi-channel sound. It is an example of a mapped adventure, wherein players are depicted on a multi-colored map. As the players move to new locations, new areas of the map are displayed.
When you first sit down with Ali Baba, you may not get up for several hours. The game shows strong potential when you see it for the first time. It allows for a total of 17 characters to participate simultaneously in a search to rescue a kidnapped princess.
This adventure leans heavily in the direction of a fantasy role-playing game, assigning weights to the attributes of each character.
From my experience, this makes for a much more lively game when friends are sitting in. Instead of everyone in the room discussing what the sole character in a text adventure should do, everyone can be his own character, interacting within the adventure as well as with each other. The discussion takes on new depth as individuals decide what course to take for themselves.
In Ali Baba you can choose to become one of the set of humans, elves, halflings or dwarves that are profiled in the extensive documentation that accompanies the game. Each of these sets of beings is represented by a different symbol on the screen. Using the keyboard or joysticks to input movement, players can explore the caverns, palaces, passages, and treasure rooms of the game--and happen upon the dangerous inhabitants therein.
The game is immediately addictive, and is really tough. The first thing the novice should do is turn the "monster recurrence" level down to zero. This will keep him from becoming utterly, bogged down in fending off attackers.
You'll spend hours wandering around, acquiring treasures, buying weapons and armor, fighting enemies and thieves, and searching for the princess. Characters can be reincarnated if they are snuffed out, and reinforcements can be called in when the going gets really heavy.
As in other role-playing games, each character has unique attributes. Elves are fast and hard to hit; dwarves are slow and clumsy but when they connect they pack a heavy blow. This richness of character is the strong point of the game.
The adventure has some other interesting features. It can be configured to play through nested menus, controlled completely by one or more joysticks, eliminating keyboard entry completely. Characters can be retired as well as reincarnated, and in fact Ali himself can be retired if no one wishes to take his role. Characters other than players wander through the game, and will sometimes attack each other, allowing players to make a fast getaway. The map graphics, Arabic-style typeface, music, and sound effects are quite good.
The problem with Ali Baba is the way it ends. After much exploration and fighting, you will discover the princess, and fight valiantly to return her to the king. When you manage to do so, the king thanks you, gives you some gold, and invites you to go back to amass more treasure. And that's it. It's a rather sudden let down, and it leaves players without the feeling they have mastered a difficult puzzle.
The documentation states that All Baba, himself a relatively weak, slow, and unskilled character in the adventure, can reach and rescue the princess unaided, and without once raising his sword to fight. As it still seems impossible to me after several hours of trying, this poses the challenge of a real puzzle. Realizing how much more fun it is to play this type of game with other people. I think it's a shame that the challenge evaporates so quickly in the multiplayer game.
Still, fans of role-playing games may really enjoy Ali Baba. It employs some nice concepts in its execution. For those protracted sessions, multiple games can be saved to disk.
Purist adventurers who feel arcade games are beneath them may wince at the inclusion of Action Quest in an article on adventure games. It is, in fact, a rather radical departure from the format of traditional adventure games. Action Quest, despite the uninspired title, deftly exhibits some of the qualities of the mapped adventure: it draws a map of current location, provides a running status report, and allows character movement to be input by joystick. Pull on the stick, and your character scrolls smoothly across the screen. Head for a portal, and a map of your new room location appears on the screen. But the game transcends the typical adventure from here on in.
Action Quest is a one player arcade-style game within an adventure format--and it is addictive, well-paced, and fun.
The game is divided into five levels of six rooms each. If you complete all the required actions of each room, including gathering treasures, completing obstacle courses and traversing mazes, you may advance to the next, more advanced level of play.
You encounter numerous monsters, dodge bullets, run through rooms with walls closing in on two sides, and happen upon mystifying puzzles that must be solved. You carry a gun which you aim with the stick (this takes time to master), and while without it you wouldn't last too long, I still would not call this a "shoot'em-up" type game. You have ten lives, which is not as generous as it sounds, considering the rate at which you expend them.
Each room has a name, which gives a clue concerning what you need to do to get through it. The only problem is that by the time you have read the name for the first time, you may already have been skewered by some bizarre creature.
You play against the clock as well as attempting to get through as many levels of the game as you can. This ensures you may still enjoy the game even after you have completed the adventure. I do not wish to instill the idea, however, that you will get all the way through Action Quest in any short amount of time. Mastery requires the acquisition of some formidable skills. At the end of the game your score is tallied, and you are assigned a rating on the basis of time used, lives expended, and treasures amassed.
Perhaps it is a stretch to label Action Quest as an adventure game, but it is an exciting move in the right direction. The game is strong in one of the fundamental aspects of computer gaming: building toward a goal.
Though the sound is somewhat unsophisticated, the graphic animation is well-executed. Your character is an undulating ghost, and as remaining in one room for too long can be fatal, it begins to fade slowly when you enter a room. If the ghost disappears completely, it costs you a life. Sometimes you must shuttle between rooms quickly to avoid "suffocating" in this manner.
The author of Action Quest has indicated that a sequel with greater challenges, as well as more sophisticated graphics is in the works. I, for one, am looking forward to it.


Note: Adapted from "Chez Panisse Desserts" by Lindsey Remolif Shere. Do not use a black tart pan or the crust may burn. The tart shell and the pastry cream can be made in advance. All you really need to do the day of serving is bake the shell, prepare the nuts and caramel, section the oranges and put the tart together. If possible, assemble the tart just before serving. You can substitute navel oranges. This recipe uses half the pastry cream. Set the rest aside for another use.
Step 1Mix the flour, sugar, salt and lemon peel. Cut the butter into half-inch slices and work it into the flour mixture with your hands or a pastry blender until the butter is mostly cornmeal-size pieces and the mixture begins to hold together; the softer your butter is, the faster this will happen.
Step 2Combine the water and the vanilla and work it into the flour-butter mixture just until the pastry is blended and will hold together if you press it. Gather it into a ball and wrap it in plastic. Let it rest for 30 minutes so the flour will absorb the moisture more completely.
Step 3Press the pastry into a 9-inch tart pan, making sure that you have a layer of even thickness over the bottom and the sides. If the thickness is uneven, some parts will bake too much before other parts are cooked. Set the shell in the freezer for 30 minutes or overnight, wrapped in foil.
Step 4Bake in a 375-degree oven for about 25 minutes, or until the shell is light golden brown and baked all the way through. Cool before filling.
Step 1Scald the milk -- that is, heat it to just under boiling. Mix the flour and sugar in a heavy non-corroding saucepan. Beat the egg yolks until thick and light colored, about 2 to 3 minutes in a mixer. Thoroughly whisk the hot milk into the mixed flour and sugar and cook over medium heat, stirring constantly, until the mixture has boiled for 1 minute or 2.
Step 2Whisk a little of the mixture into the egg yolks to warm them and stir back into the flour mixture. Mix well, being sure to incorporate all the flour mixture from the sides of the pan, and cook over medium heat, stirring constantly, until the pastry cream begins to hold a slight shape (170 degrees). It is important to cook the pastry cream thoroughly after adding the egg yolks, but don't let it boil. Cook until slightly thickened, about 10 minutes.
Step 3Remove from the heat and stir in the butter, then put through a medium-fine strainer. Cool and whisk occasionally to keep a crust from forming, or cover with plastic and refrigerate until cool, then whisk to smooth it out. Stir in vanilla to taste. Don't over-beat after cooking or the cream will thin out -- a useful thing to remember, because you can thin it, if it is too thick, by beating it. Beat it when cold, just before using, to smooth and thin slightly -- it should look shiny.
Step 1Shell the pistachios and toast them in a 350-degree oven for about 5 minutes, just long enough to bring out their flavor but not so long that they lose their green color. Cool them slightly, rub off their skins, chop them into one-sixteenth-inch to one-eighth-inch pieces and set them aside for the garnish.
Step 2Section the blood oranges into a small bowl, removing peel and membranes. Cover with their juice, cover tight and refrigerate if you are not making the tart immediately.
Step 3Put the sugar, butter and water into a small, light-colored saucepan and cook over medium heat until it becomes a light golden caramel, about 4 minutes. Pour it quickly into the bottom of the tart shell and swirl it over the bottom. It probably won't cover the bottom completely, but try to get an even covering by pouring the caramel in circles. This caramel makes a nice, crunchy texture contrast in the tart.
Step 4Stir a little Armagnac, Cognac or brandy into the pastry cream. Be careful not to beat it too much or it will thin out. Spread the pastry cream in a layer over the caramel in the bottom of the shell.
Step 5Dry the orange sections on a towel or paper towel, then arrange them over the pastry cream, overlapping the sections slightly. Begin from the outside of the tart and turn the sections in opposite directions in each successive ring. If the center is too small to make a final circle, fill it with small sections, perhaps in the form of a rose or an open flower. Sprinkle pistachios around the outside edge or around the center row of sections, or wherever you think they will look best.
310 calories; 5 grams protein; 34 grams carbohydrates; 2 grams fiber; 18 grams fat; 10 grams saturated fat; 121 mg. cholesterol; 93 mg. sodium.
There are very few things as flat-out gorgeous in market stalls now as blood oranges, those glorious citrus orbs that darken over the course of spring until the exterior can be almost as deep red as the fruit inside. But they won't be here for long, as this dark color also heralds the approaching end of the season: A folksy caveat marks Easter as a the signal of the end, the blood of Christ a reminder that the blood oranges will soon be gone from the trees.
Moros, the type most widely grown in California and likely the only ones you'll find this late in the season (which started in January), have only a few weeks left. But now's the perfect time to get them. Earlier in the season, before they've had time to ripen thoroughly, they're more tart and tangy. You have to search out the darker ones, rolling them around in their bins and crates like enormous avant-garde marbles. No need for that now, though; they're almost uniformly at their peak.
"I don't bring the oranges until they're really dark on the outside," says Armando Garcia of Garcia Organic Farm in Fallbrook at the Santa Monica farmers market Saturday. He says they're sweetest now, when the colors from the inside have bled out until the skin gets almost purple. But, as with all fruit, it's a delicate balance. You don't want to wait too long: The fruit begins to dry out after it reaches its peak, and the flavor can lose its bright citrus notes and become murky.
This is a terrific season for blood oranges, as it's the less productive year of what fruit producers call an "alternate bearing" crop. Last year's yield was huge, so this year the crop is much smaller, Bob Polito of Polito Family Farms in Valley Center explains. But he also says that a smaller crop can often produce more desirable fruit: "This year the quality seems to be better; they're darker, and when they have that color, they're the best."
So grab them now, as they're easy to find and pretty inexpensive: around $2 per pound at farmers markets and some grocery stores.
You can eat them out of hand as you would any orange, or peel and section them, or cut them into glorious segments like jeweled wedges. They're fabulous tossed into a fruit salad with the strawberries that are also at their peak. Or, with a handful of arugula and a crumble of goat cheese, they're a delightful first course, particularly with a vinaigrette made with the garnet-colored juice.
Juice them to make a fresh, tangy sorbet or use them for a stunning blood-red martini, a curl of the rind like a tongue of flame inside.
There aren't too many chefs who get as excited about blood oranges as Lindsey Remolif Shere, longtime pastry chef, now retired, at Berkeley's Chez Panisse. She loves them in sorbets and compotes, in tarts and cold citrus soups. "To me, blood oranges have a flavor of raspberries that's sometimes quite strong; their flavor is richer and more complex than plain navels or Valencias," she writes in a recent e-mail. She should know: Not content simply to write a dessert cookbook, she and the chefs at Chez Panisse penned one about fruit as well.
Shere's blood orange tart, from "Chez Panisse Desserts," is gorgeous, with the blood orange segments arranged like a wide-open rose on the top. The flavor of the blood oranges plays beautifully with a hidden, thin layer of caramel that gives unexpected crunch. A pillowy pastry cream and buttery short crust shell pull it all together.
But if you want something more market-sudden, try the compote from "Chez Panisse Fruit." A bowl of strawberries and crimson citrus segments, with a thatch of candied peel like a little nest on top, it's a simple tribute to fruit at the peak of its season. It looks stunning, like a jewelry box filled with garnets, and the flavors are remarkable. And it's so easy, especially if you candy the peel ahead, that it comes together in a flash.
Which is a good thing, considering how little time there is left in the season. You can't stockpile blood oranges in the fridge the way you can Meyer lemons -- they dry out too quickly. So do what many chefs do and buy them now, juice them and freeze the juice for use year-round. Frozen in ice cube trays, it'll last for months.


Thick-It, a provider of dysphagia nutrition solutions, has changed the name of its concentrated formula to Thick-It Concentrated.
The change was fueled by Thick-It's desire to differentiate Thick-It Original from Thick-It Concentrated (previously known as Thick-It 2).
The concentrated formula offers 20 percent more servings in a 36-ounce canister compared to the same canister size of Thick-It Original. This allows consumers to use less product to thicken food or beverage to reach the appropriate consistency. Thick-It Original requires 3 ½ tsp. to thicken 4 ounces of water, achieving nectar consistency, whereas Thick-It Concentrated requires 3 tsp. to thicken 4 ounces of water.
Thick-It is passionately dedicated to and solely focused on creating a suite of innovative, science-based, nutritious and appealing food and beverage options to improve the lives of people with swallowing disorders.
Consumers can expect to see new packaging for Thick-It Concentrated this August. The product formula, high quality, UPCs and manufacturing codes as well as the ordering process, will stay the same.
To find Thick-It products near you, click here. For more information, follow the brand on Facebook at Facebook.com/ThickIt, on Twitter at @ThickIt_PFI and on LinkedIn at Thick-It Nutrition Innovation for Dysphagia.
Thick-It is a registered trademark of Kent Precision Foods Group, Inc., with the goal of continually researching, innovating and educating on dysphagia nutrition products. Thick-It offers xanthan- and starch-based food and beverage thickeners, including Thick-It Original Thickener, the number one selling retail brand of food and beverage thickeners, introduced in 1987; ready-to-drink, pre-thickened beverages available in water, juice, coffee and tea varieties; and ready-to-eat pureed foods made from real food ingredients.


text:
H.E Dr Thani bin Ahmed Al Zeyoudi
Minister of Climate Change and Environment
His Excellency Dr Thani bin Ahmed Al Zeyoudi was appointed the Minister of Climate Change and Environment for the United Arab Emirates in February 2016. In this role, he oversees the Ministry's mission to spearhead the UAE's drive to mitigate and adapt to the impact of climate change and protect the country's unique ecosystems through developing and implementing effective measures, policies and initiatives. In addition to his ministerial responsibilities. Previously, he served as Permanent Representative of the UAE to IRENA and Director of the Department of Energy and Climate Change at the Ministry of Foreign Affairs and International Cooperation. Earlier, as Project Engineer at Masdar, His Excellency Dr Al Zeyoudi worked to advance renewable energy technologies and solutions. During this time, he also played an instrumental role in the UAE's successful 2009 bid to host IRENA, the first international organization dedicated to renewable energy. He started his career as Reservoir Engineer at the Abu Dhabi Marine Operating Company (ADMA-OPCO). He heads the UAE Council for Climate Change and Environment, and National Committee of Biosecurity. He is also a member of the Board of Directors of the Environment Agency – Abu Dhabi and the Trustees of the Khalifa International Award for Date Palm and Agricultural Innovation, the jury and Selection Committees of the Zayed Sustainability Prize, and the Board of Directors of the Global Green Growth Institute. In addition, His Excellency Dr Al Zeyoudi is Honorable Advisor at the China Biodiversity Conservation and Green Development Foundation. His Excellency Dr Thani bin Ahmed Al Zeyoudi holds a PhD in Project and Program Management from SKEMA Business School in France, an MBA from the New York Institute of Technology in the US, an MSc in Project Management from the British University in Dubai and a bachelor's degree in Petroleum Engineering from the University of Tulsa in the US. In 2015, His Excellency Dr Al Zeyoudi received the first Gulf Cooperation Council Excellence Award in recognition of his pioneering efforts in renewable energy.

redpajama_set_name: RedPajamaCommonCrawl


text:
Astronomia
 3152 Jones – asteroide della fascia principale

Geografia
 Jones  – municipalità delle Filippine nella Provincia di Isabela
 Jones  – comune degli Stati Uniti d'America in Oklahoma
 Contea di Jones  – contea degli Stati Uniti d'America nello stato della Carolina del Nord
 Contea di Jones  – contea degli Stati Uniti d'America nello stato del Dakota del Sud
 Contea di Jones  – contea degli Stati Uniti d'America nello stato della Georgia
 Contea di Jones  – contea degli Stati Uniti d'America nello stato dello Iowa
 Contea di Jones  – contea degli Stati Uniti d'America nello stato del Mississippi
 Contea di Jones  – contea degli Stati Uniti d'America nello stato del Texas
 Jones  – piattaforma glaciale antartica

Persone

Personaggi immaginari
 Jones  – personaggio de La fattoria degli animali
 Jones  – personaggio del manga e anime Le bizzarre avventure di JoJo
 Freddy Jones  – personaggio di Scooby-Doo
 Larvell Jones  – personaggio della serie cinematografica di Scuola di polizia, dell'omonima serie animata e dell'omonimo spin off televisivo
 Professor Jones, personaggio della serie animata Freakazoid
 Indiana Jones  – personaggio di quattro film, una serie televisiva e diversi videogiochi
 Mr. Jones  – personaggio Disney, vicino di casa e nemico di Paperino
 Secco Jones  – personaggio de I Simpson
 Bridget Jones – personaggio letterario e cinematografo creato da Helen Fielding
 Davy Jones – personaggio leggendario e letterario
 Davy Jones – personaggio della serie cinematografica di Pirati dei Caraibi

Pagine correlate
Mr. Jones
Michael Jones
Mike Jones
Owen Jones

redpajama_set_name: RedPajamaWikipedia


Over Christmas my sister and her husband made the journey from southern Saskatchewan up to boomtown Alberta. We spoke about the changes taking place in some of the small towns, not all for the better. Even places like North Dakota are noticing what happens with a boom. The problem with finding accomodation and when you find it - the unexpected high cost. Young men making one hundred grand a year and more up in Ft. McMurray.
And for some, it sounds pretty great. For others its a nightmare.
At age 10 our family lived in Beverly - which was a coal mining community and one of the faster growing towns - that was in 1957 - and I recall going from an outdoor biffy to indoor plumbing, from wooden walks to paved walks and paved roads. Lots of dust and lots of noise. Things were changing. Progess had arrived at our town. But as Singer-Songwriter tells us - "I can see the sun setting fast but as they say, nothing good ever lasts..." and certainly things would never be the same as time went by.
For a time some of the small towns in Alberta seem to be reaching the bitter end, such as the town of Ryley - which is now becoming the place for Edmonton to dump its garbage. Many other small towns and villages are noticing an influx of both good guys and bad. Those who want to get rich by work and those who try to get rich by home invasions, threats, drugs and the list goes on.


text:
An FPT haplotyping algorithm on pedigrees with a small number of sites
Duong D Doan1 &
Patricia A Evans1
Genetic disease studies investigate relationships between changes in chromosomes and genetic diseases. Single haplotypes provide useful information for these studies but extracting single haplotypes directly by biochemical methods is expensive. A computational method to infer haplotypes from genotype data is therefore important. We investigate the problem of computing the minimum number of recombination events for general pedigrees with a small number of sites for all members.
We show that this NP-hard problem can be parametrically reduced to the Bipartization by Edge Removal problem with additional parity constraints. We solve this problem with an exact algorithm that runs in time, where n is the number of members, m is the number of sites, and k is the number of recombination events.
This algorithm infers haplotypes for a small number of sites, which can be useful for genetic disease studies to track down how changes in haplotypes such as recombinations relate to genetic disease.
Human genomes contain two copies of each chromosome. Research shows that single chromosomes, called haplotypes, are useful to study complex genetic diseases. While genomic data, called genotypes, are abundant and easy to collect, haplotypes are rare and much more difficult to obtain by a biochemical method. Therefore, computationally inferring haplotypes from genotype data, called haplotyping, is necessary. Genotypes can be obtained from a population group where relationships between members are unknown or from a family pedigree with known relationships between members. We only consider pedigree data.
In the absence of recombination events, haplotypes of members in a pedigree follow the Mendelian law of inheritance, where the two haplotypes of a child are transferred from its parents, one haplotype from its father and the other from its mother. Various haplotyping algorithms exist for non-recombinant pedigree data [1, 2], especially a linear algorithm for tree pedigrees [1] and a near-linear algorithm for general pedigrees [2]. Haplotype inference is complicated by recombination events and the complex structures of the data. In recombination events, complementary parts of both of a parent's haplotypes can be inherited as a single combined haplotype of a child. Structures of the pedigree can be complex, where there are multiple inheritance paths between some family members.
When recombination events are allowed, the problem of inferring haplotypes for pedigrees with the minimum number of recombination events is NP-hard, even for general pedigrees with only two sites or tree pedigrees with multiple sites [3]. For reconstructing haplotype configurations for pedigree data, Qian and Beckmann [4] proposed a rule-based algorithm with a time complexity O(2 dn2m3), for n members, m sites, and family size ≤ d. The main principle of their algorithm is that the best haplotype configuration for pedigree data is the one that minimizes the number of recombination events (the MRHC problem). Li and Jiang [5] proposed an integer linear programming (ILP) formulation for the MRHC problem. When the number of recombination events is strictly smaller than a positive number k, an O(mn · logk+1n) time probabilistic algorithm is given on tree pedigrees [6]. Doan and Evans [7] presented an O(2 k · n2) time fixed-parameter algorithm for general pedigrees where each member has two sites, a special case of the problem that is still NP-complete.
We study the haplotype inference for general pedigrees with recombination events when the number of recombination events k and the number of sites m in an input pedigree are small. We also assume that there are no data missing and no data errors. We prove that our problem can be reduced to the problem of finding the line index of a signed graph[8] with additional parity constraints. We further show that finding the line index of a signed graph can also be reduced to the Graph Bipartization by Edge Removal (GBER) problem with parity constraints. The GBER problem is fixed-parameter tractable, but the existing solution [9] cannot satisfy the additional parity constraints. We present an algorithm that solves the problem while still satisfying the additional constraints, and thus show that the Recombinant Haplotype Configuration problem can be solved by a fixed-parameter algorithm with a running time of , for n members, m sites, and k recombination events. This result extends our prior work for pedigrees with two sites to an arbitrary small number of sites.
A member is an individual. A set of members is called a family if it includes only two parents and their children; it is a parent-offspring trio (hereafter a trio) if only two parents and one child are considered. A set of families connected through known family relationships is a pedigree.
In diploid organisms, a cell contains two copies of each chromosome. The description data of the two copies are called a genotype while those of a single copy are called a haplotype. A specific location in a chromosome is called a sit e and its state is called an allele. There are two main types of sites, microsatellites and single nucleotide polymorphisms. A microsatellite site has several different states while a single nucleotide polymorphism (SNP) site has exactly two possible states, denoted by 0 and 1. Only SNPs with two possible states are considered in this paper, as in other works on haplotype inference.
If the states at a specific site in two haplotypes are the same, then this site is a homozygous site (0-0 or 1-1); if they differ, it is heterozygous (0-1 or 1-0). Two haplotypes combine together to form one genotype. Each member u has two haplotypes, denoted by h 1 u and h 2 u , which are vectors of 0 and 1's of length m, where m is the number of sites. The genotype of u, g u , is a vector of 0's, 1's and 2's of length m, where g u [i] = 0 means h 1 u [i] = 0 = h 2 u [i], g u [i] = 1 means h 1 u [i] = 1 = h 2 u [i], and where g u [i] = 2 means {h 1 u [i]; h 2 u [i]} = {0, 1}. We say h 1 u and h 2 u are consistent with g u . The complement haplotype of a haplotype h at a heterozygous site is denoted by , where so, and .
When there is no recombination event in a pedigree, a child member receives one entire haplotype from its father and another entire haplotype from its mother. Figure 1a shows member c receiving the entire left haplotype of parental member u and the entire left haplotype of parental member v. However, during the meiosis process, haplotypes of a parent sometimes shuffle due to the crossover of chromosomes and one of the shuffled copies is transferred to the child. This phenomenon is called a recombination and the result is called a recombinant. Figure 1b shows a recombination event between site 1 and site 2 of member u. As the result, member c receives a combined haplotype from site 1 of the left haplotype, and from sites 2 and 3 of the right haplotype of member u.
Non-recombination vs. recombination, showing haplotypes of members.
The problem in this paper is to find the haplotypes h 1 u and h 2 u for all members u that minimize the number of recombination events, given their genotypes g u . A set of haplotypes found for all members is called a haplotype configuration. When g u [i] = 0 or 1, then h 1 u [i] and h 2 u [i] are known, but if g u [i] = 2, we may not yet know the value of h 1 u [i] and h 2 u [i], in which case we give them the value "?", and say that the site is unresolved. Our problem is defined as follows.
RHC opt : Given the genotypes of a general pedigree P containing n members, where each member has m sites (m is small), find a haplotype configuration that minimizes the number of recombination events.
This optimization problem, called Recombination Haplotype Configuration (RHC opt ) which is identical to MRHC, was proven NP-hard [3]. We investigate the corresponding decision version of RHC opt .
RHC k : Given positive integers k and the genotypes of a general pedigree P containing n members, where each member has m sites (m is small), is there a haplotype configuration with at most k recombination events explaining P ?
In this paper, we use u, v and c to represent members, from 1 to n; and i and j to represent sites, from 1 to m.
Setting Up Graphs
Given a general pedigree with n members, where each member has m sites, we set up a pedigree graph G = (V, E) and parity-constraint sets S pc to compute the minimum number of recombination events in the pedigree. A recombination event can only be detected if there is at least one heterozygous site on each side of a recombination breakpoint, e.g. we cannot detect if a recombination event happens between homozygous sites 1 and 3 of member u in Figure 2a because the states at the two haplotypes for each homozygous site are the same. The graph captures constraints between pairs of closest heterozygous sites and pairs of closest homozygous sites, which will enable the detection of possible recombination events in pedigrees. A vertex in the pedigree graph represents a pair of homozygous sites or a pair of heterozygous sites, and is colored to represent the relationship between the haplotypes of the sites.
Pedigree graph created from pedigree structure and genotype data.
Pedigree Graph
Create grey vertices
Let i be a heterozygous site in a member u (i = 1, ..., m - 1). Let j > i be the closest heterozygous site to i in u. We create a vertex u ij from site i and site j and label this vertex grey. A grey vertex is an unresolved vertex and will later be resolved green if h1 u [i] = h 1 u [j] = 0 or h 1 u [i] = h 1 u [j] = 1. It is resolved red otherwise. The resolution of a grey vertex depends on its adjacent vertices. Figure 2b shows a grey vertex u45 created from sites 4 and 5 of u in Figure 2a.
Create red and green vertices
Let i be a homozygous site in a member u (i = 1,..., m - 1). Let j > i be the closest homozygous site to i in u. We create a vertex u ij from site i and site j, and label this vertex red if g u [i] ≠ = g u [j] and green if g u [i] = g u [j]. A red or green vertex is a resolved vertex. Figure 2 shows a red vertex u12 created from sites 1 and 2, and a green vertex u23 from sites 2 and 3.
Insert positive edges
We insert positive edges between a parent member u and its direct child member v. For each vertex u ij in u, if there is a vertex v ij in v we insert a positive edge between u ij and v ij . If there is no vertex v ij in v and i and j are both homozygous sites or both heterozygous sites in v, we create a vertex v ij in v and label this vertex properly, inserting a positive edge between u ij and v ij . We call v ij a supplementary vertex as it is created by the need of member u.
Similarly, for each vertex v ij in v, if there is no vertex u ij in u, and i and j are both homozygous sites or both heterozygous sites in u, we create a supplementary vertex u ij in u and label this vertex properly, inserting a positive edge between u ij and v ij . Figure 2b shows four positive edges linking u12 and c12 that is created from heterozygous sites 1 and 2 of member c, u23 and c23, v12 and c12, v23 and c23.
A positive edge between vertices u ij and v ij means vertex u ij and v ij should be resolved with the same color (both red or both green) unless a recombination event occurs in u. The reason for this is that if there is no recombination event in u, then v receives one full haplotype from u and another full haplotype from another parent. Therefore, the label of u ij and the label of v ij should be the same if there is no recombination event; otherwise, there is a recombination event in u. If u ij is a resolved vertex forming from two homozygous sites i and j and there is a positive edge between u ij and a grey vertex v ij , we color v ij the same as the color of u ij , since a recombination event at u ij is not detectable and does not affect the color of v ij .
Insert negative edges
We insert negative edges between two parents u and v of a common child c. If u ij is a vertex in u but there is not a vertex c ij in c (sites i and j are one homozygous and one heterozygous in c), two situations happen. If there is a vertex v ij in v, we insert a negative edge between u ij and v ij . Otherwise, if there is no vertex v ij in v and i and j are both homozygous sites or both heterozygous sites, we create a supplementary vertex v ij in v and label it properly. We insert a negative edge between u ij and v ij . Similarly, if v ij is a vertex in v but there is not a vertex c ij in c, there are two situations. If there is no vertex u ij in u, and i and j are both homozygous or both heterozygous, we create a supplementary vertex u ij in u, and insert a negative edge between u ij and v ij . Figure 2b shows a negative edge linking u45 and v45.
A negative edge between u ij and v ij means vertices u ij and v ij should be resolved with different colors unless a recombination event occurs in one parent of c. This phenomenon can be explained as follows. If there is no recombination event and u ij and v ij have the same label (both red or both green), then sites i and j of c must be both homozygous or both heterozygous based on the Mendelian law of inheritance. Because sites i and j of c are one homozygous and one heterozygous, one recombination occurs if u ij and v ij have the same label when resolved, but no recombination event occurs if they are resolved differently.
Create additional vertices
Consider a grey vertex u ij in u (i < j). It is possible that u ij has no incident edge but there is one recombination event occurring between site i and j. In this case none of the other two members in the trio has a vertex created for site i and j. We delete vertex u ij and create an additional vertex to capture the recombination event. Let j' be the closest heterozygous site from j in u (j < j'), where i and j' are both heterozygous sites or both homozygous sites in at least one member among the other two members, say v. If there is no vertex u ij' in u, we create an additional grey vertex uij'in u and create a supplementary vertex c ij' from sites i and j' in c if it does not exist. We color c ij' properly and insert a corresponding edge (positive or negative) between u ij' and v ij' depending on the relationship between u and v. Figure 2c shows an additional vertex u14 created represented by a dashed edge between sites 1 and 4. A negative edge is inserted between u14 and v14.
Pedigree graph G = (V, E) created as described above is an undirected graph. Each vertex y ∈ V has three possible labels, red, green, and grey. Each edge e(y, z) ∈ E is either a positive edge, e ∈ E pos , or a negative edge, e ∈ E neg , with E = E pos ∪ E neg . Graph G, set up this way, is a signed graph [8]. Let N(y) be the set of adjacent vertices of y. Let w(e) be the weight of edge e. If e is a positive edge, w(e) = +1. If e is a negative edge, w(e) = -1.
Observation 1. There are at most O(n · m2) vertices and O(n · m2) edges in the pedigree graph. Each member has m sites. The total number of vertices created from pairs of sites for each member is O(m2). The whole pedigree graph with n members has O(n · m2) vertices. A vertex has at most two positive edges linking it to two vertices in its parents. Therefore, the number of positive edges is linear in the number of vertices. The number of negative edges is also linear to the number of vertices. Thus the number of edges in the pedigree graph is O(n · m2).
Parity-Constraint Sets
When a supplementary grey vertex u ij is created in u by the need of an adjacent member, there must be more than one grey vertex already created from site i to site j in u. It is important to ensure that these grey vertices and u ij when resolved will not result in an odd number of red vertices. Recall that a grey vertex is resolved red if h 1 u [i] ≠ h 1 u [j]. In other words, the value of h 1 u flips from 0 to 1 and vice versa for a red vertex u ij . Therefore there is a parity conflict if the number of red vertices from site i to site j including u ij is odd.
In Figure 3a, there are five grey vertices created for member u where vertices u12, u23, u34 and u45 are created from closest heterozygous sites, and a supplementary vertex u15 is created for a member adjacent to u. Figure 3b shows an invalid solution with three resolved red vertices u23, u34 and u15 in member u. A valid solution with an even number of red vertices is shown in Figure 3c.
Parity conflict between vertices within each member.
We create parity-constraint sets S pc to capture parity constraints between each supplementary vertex and other vertices within each member. Let u ij be a supplementary vertex and u ip , ..., u qj be grey vertices from site i to site j. These vertices form a parity-constraint set, and its total number of red vertices must be even. There are O(m2) parity-constraint sets in each member and O(nm2) parity-constraint sets for the whole pedigree graph. A valid solution for RHC k must ensure that the number of red vertices in each parity-constraint set is even.
Signed Graph
A graph G = (V, E) is a signed graph if it has both positive and negative edges (E = E pos ∪ E neg ) [8], where w(e pos ) = 1 and w(e neg ) = - 1. Let (V1, V2) be a partition of V , and E* be the set of edges between V1 and V2. The line index of the cut (V1, V2) is defined as:
The line index of graph G is defined as:
The decision version of the line index of graph G is defined as follows.
LineIndex k : Given a signed graph G and a positive integer k, is there a line index of G at most k? Given a pedigree graph G = (V, E), the RHC k problem can be solved by determining if we can label every grey vertex in G either red or green such that if we partition the set of vertices V into (V red , V green ) and let E* be the set of edges between V red and V green then
and this partition (V red , V green ) must satisfy parity-constraint sets S pc .
Given a pedigree graph, any two adjacent members linked by a positive edge should be in the same set of the partition, and any two adjacent members linked by a negative edge should be in different sets. Any edge whose constraint is not satisfied represents a recombination event between the two adjacent members, or, in the case of a negative edge having endpoints in the same partition, between one parent and the child. Equation 3 thus counts the number of recombination events in the whole pedigree and ensures that it is at most k.
Clearly, the RHC k problem can be reduced to the LineIndex k problem with additional parity-constraint sets S pc on its vertices. We will show that the LineIndex k problem can be reduced to the GBER problem, a classic NP-complete problem that is fixed-parameter tractable. The RHC k can therefore be solved through the GBER problem with additional parity-constraint sets S pc .
Theorem 1 A pedigree has at most k recombination events if and only if its corresponding signed graph has the line index of size at most k.
Proof 1 We will show that one recombination event in the pedigree corresponds to exactly one negative edge within each set of the partition of vertices or one positive edge between the sets of the partition of vertices in the signed graph.
⇒ Consider a recombination event in member u. To detect this recombination event there must be at least one heterozygous site on each side of the recombination breakpoint. Let i and j be the two closest heterozygous sites on the two sides of the recombination breakpoint. There are three possible types of vertices associated with this recombination event: a grey vertex u ij , an additional vertex u ij' , and supplementary vertices u pq (p ≤ i, j ≤ q).
If vertex u ij has an incident positive edge to a vertex c ij , the color u ij should be different from the color of c ij because of the recombination event and the positive edge between them would cross between sets of the partition. On the other hand, if u ij has an incident negative edge to a vertex v ij , the color u ij and v ij should be the same because of the recombination event and the negative edge between them would be within the same set of vertices. In both cases the line index increases by one. An additional vertex u ij' replaces u ij when u ij has no incident edge. The resolution of an additional vertex u ij' is similar to that of u ij . Consider a supplementary vertex u pq constrained by a parity-constraint set S pc where u pq has an incident positive edge to a vertex c pq . The color u pq is determined by the swap of values in h 1 u by red vertices and recombination events from p to q, including the recombination from i to j. If no more recombinations happen, u pq and c pq must have the same color and the line index of the signed graph is the same. If u pq and c pq have different colors, there must be another recombination from sites p to q and the line index increases by one. A similar explanation follows for u pq with an incident negative edge.
⇐ A negative edge links two vertices of two parents in a trio, and the two vertices are supposed to have different colors based on the Mendelian law of inheritance. Similarly, a positive edge links two vertices of a parent and a child and the two vertices are supposed to have the same color. Therefore, if a negative edge linking two vertices with the same color or a positive edge linking two vertices with different colors, one recombination event must happen.
Fixed-Parameter Algorithm
A NP-hard problem cannot be solved by a polynomial time algorithm unless P = NP. However, if we can restrict some parameters of the problem to small values, the running time of an algorithm for the problem can potentially be greatly reduced [10]. In this case, the problem is a parameterized problem and an algorithm that can solve the parameterized problem efficiently is a fixed-parameter algorithm, defined as follows [10].
Definition 1 A parameterized problem is a language L ⊆ Σ* × Σ*, where Σ is a finite alphabet and Σ* is the set of all strings over that alphabet. The second component is called the parameter of the problem.
Practically, the parameter is a nonnegative integer or a set of nonnegative integers and therefore L ⊆ Σ* × ℕ. For (x, k) ∈ L, the size of the input is n = |(x, k)|, and the parameter is k.
Definition 2 A parameterized problem L is fixed-parameter tractable (in class FPT) if it can be determined in f(k)· nO(1)time whether or not (x, k) ∈ L, where n is the size of the input and f is a computable function only depending on k.
Transforming to Bipartization by Edge Removal Problem
We review an important property of a signed graph given by [8].
Theorem 2 Let G be a signed graph. If we replace each edge with weight w(e) > 0 by two consecutive edges with weight -w(e) to get a graph G' then l(G) = l(G').
Proof 2 Suppose (V1, V2) is a cut of G such that l(V 1 , V 2 ) = l(G). We replace each positive edge e(u, v) by two consecutive negative edges e(u, y) and e(y, v), where w(e(u, y)) = w(e(y, v)) = - w(e(u, v)) and y is a new vertex adjacent only to u and v. If u and v belong to the same set of vertices in the partition we put y into the other set. If u and v belong to different sets, we can arbitrarily put y into the same set as either u or v. In all of the cases above we find the corresponding cut of G', such that . Therefore l(G') ≥ l(G).
Conversely, if and y is a new vertex, then at least one edge incident to y is in the cut. We can find a corresponding cut of G, (V1, V2) such that . Therefore l(G') ≥ l(G). Taken together, we get l(G') = l(G).
The pedigree graph is transformed into a new graph by replacing every positive edge by two consecutive negative edges and adding new intermediate vertices (dum vertices). We obtain a new weighted graph G' with all negative edges. This transformation does not affect the parity-constraint sets S pc . The graph G' still has only O(n · m2) vertices and O(n · m2) edges. Equation 3 becomes
This equation is to ensure that the total number of edges within V1 and edges within V2 is at most k. Removing these edges will make the graph bipartite.
To make the GBER algorithm [9] works on our partially colored graph, we merge all red vertices into one red vertex and all green vertices into one green vertex. We relabel the merged red vertex and the merged green vertex into two grey vertices, and insert k + 1 negative edges between them. This transformation does not affect the parity-constraint set S pc . We further transform our negative graph into a new graph with all positive edges by multiplying the weight of every edge by -1. Our problem becomes the GBER problem [9] with additional parity-constraint set S pc . The k-Bipartization by Edge Removal problem is defined as follows.
Definition 3 Given a graph G = (V, E) and a positive integer k, is there a set C ⊆ E with |C| ≤ k whose removal produces a bipartite graph?
GBER is a classical NP-hard problem [11] and is in FPT [9].
FPT Algorithm for Bipartization by Edge Removal
There are many techniques to solve an FPT problem such as kernelization, depth-bounded search trees, dynamic programming, crown reduction, greedy localization, and iterative compression. The iterative compression technique is used by Guo et al. [9] to solve the GBER problem with a running time of O(2 k · |E|2), where |E| is the number of edge in the graph and k is the number of edges to be deleted to make the graph bipartite. However, this algorithm does not enforce our parity constraints that require the number of red vertices in each set to be even. We thus need to modify this algorithm [9] to solve the RHC k problem while respecting the additional parity-constraint sets S pc .
Given a graph G = (V, E) where E = {e1, ..., e m }, let G i be a graph induced by edges {e1, ..., e i } of G (1 ≤ i ≤ m). If i = 1, the optimal edge bipartization set of G1 is empty. If i > 1, let X be an optimal edge bipartization set of G i = G[e1, ..., e i ] and |X| = k'. Consider graph Gi+1= G[e1, ..., ei+1]. If X is not an optimal edge bipartization set for Gi+1then X' = X ∪ {ei+1} is clearly an edge bipartization set for Gi+1. From the edge bipartization set X' of size k' + 1, we find an edge bipartization set of size at most k' or show that no such edge bipartization set of size at most k' exists. The algorithm assumes that an edge bipartization Y which is smaller than X' must be disjoint from X', Y ∩ X' = ∅. This assumption can be made without loss of generality by a simple graph transformation, replacing each edge in X' by three consecutive edges and choosing the middle edge to be in the new X'. This graph transformation preserves the parities of lengths of all cycles and does not affect the parity constraint sets S pc . Therefore the transformed graph has an edge bipartization set of size k' if and only if the original graph has an edge bipartization set of size k'. Let mapping Φ: V (X') → {A, B} be a valid partition of V (X') if for each {y, z} ∈ X, we have Φ(y) ≠ Φ(z). Let AΦ be Φ-1(A) and BΦ be Φ-1(B). We enumerate all 2 k' valid partitions Φ of V (X'). For each valid partition Φ we find a minimum edge cut Y in G\X' between AΦ and BΦ. In other words, we use X' to partially color G and from the partially colored graph we compute a smaller bipartization set Y. This compression step is the core of the algorithm.
Theorem 3[9]Consider a graph G = (V, E) and a minimal edge bipartization set X' for G. For a set of edges Y ⊆ E with X' ∩ Y = ∅, the following are equivalent:
(1) Y is an edge bipartization set for G.
(2) There is a valid partition Φ for V (X') such that Y is an edge cut in Gn\X' between AΦ= Φ-1(A) and BΦ= Φ-1(B).
Consider a graph G in Figure 4a where ⊕ denotes a red vertex, ∅ a green vertex, and O a grey vertex. A minimal edge bipartization set X' of size 4 illustrated by dashed lines is given in Figure 4b. We compute a mincut Y for G\X' as in Figure 4c. Set Y is the edge bipartization set of size 3 for G in Figure 4d.
Compression step.
It remains to find a minimum edge cut Y between AΦ and BΦ that satisfies
|Y| ≥ k' and
graph G i with set Y satisfies parity-constraint sets S pc .
(s-t) Mincuts with parity constraints
A minimum edge cut Y between AΦ and BΦ can be computed in O(k' · |E|) time by the Edmonds-Karp algorithm [12] by finding at most k' augmenting paths; each path takes O(|E|) time to find. If no min edge cut Y of size k' is found, we skip the current partition Φ and check a new valid partition. If a min edge cut Y of size k' is found, we need to check if G i bipartized by Y satisfies the parity-constraint sets S pc . Note that there can be many mincuts Y of size k' between AΦ and BΦ, and it is possible that the current mincut Y found does not make G i satisfy S pc while another mincut Y of size k' makes G i satisfy S pc . However, enumerating all mincuts in a graph is expensive. Consider a simple directed graph with n disjoint paths of length 2 from a source s to a sink t, where the weight of each edge is 1. Each (s-t) mincut has weight n and we have up to 2 n (s-t) mincuts. If a graph is an undirected graph, we replace each undirected edge by two directed edges with opposite directions and the number of (s-t) mincuts is still 2 n . Therefore enumerating all (s-t) mincuts in a graph in polynomial time, or in FPT, is impossible.
We do not enumerate all mincuts. Instead, we examine the structure of all mincuts in a graph by an algorithm in [13]. Given a graph G = (V, E) including a source s and a sink t, where each directed edge (i, j) ∈ E has a capacity c ij , an (s-t) cut (S, S') is a cut where S' = V - S, s ∈ S and t ∈ S'. If a graph is not directed, we replace every undirected edge by two oppositely directed edges. If a graph has multiple sources and sinks, we can transform the graph into a new graph with only a single source and a single sink by inserting edges of ∞ weights from a super source s to all sources, and from all sinks into a super sink t. Flows and mincuts in the new and old graphs correspond [12].
An (s-t) mincut is an (s-t) cut where the total capacity of all the edges between S and S' is minimum. We will call an (s-t) mincut a mincut hereafter. Ford and Fulkerson [12] show that the value of a minimum cut between s and t is equal the value of the maximum flow from s to t. Consider a binary relation R on V , a subset of vertices V' ⊆ V is a closure for R if and only if for any two vertices i and j in V with iRj and i ∈ 2 V' we also have j ∈ V'. Given a relation iRj, we say that i is the predecessor of j and j is a successor and i. Picard and Queyranne [13] present the relationship between mincuts and closures as follows.
Theorem 4[13].
Let f be a maximum flow in G. Define a relation R on the set of vertices V as follows:
iRj iff (i, j) ∈ E and f ij < c ij , or (j, i) ∈ E and f ji > 0.
Then a cut (S, S') separating s from t is a minimum cut if an only if S is a closure for R containing s and not t.
Suppose we find a maximum flow in a graph by the Edmonds-Karp algorithm [12]. Clearly, the residual graph G r = (V, E r ) of G is defined by relation R where edge (i, j) ∈ E r iff iRj. We find strongly connected components in G r and shrink each of them into a single vertex. Finding strongly connected components of a directed graph G r can be done in O(V + E) time using two depth first searches, one search on G r and the other search on the transpose graph of G r [12].
Let V' be the reduced vertex set of V , we define a relation on V' by iff iRj for some , , and . We eliminate component S containing source s and its successor components, and eliminate component T containing sink t and its predecessor components. Combining S and all successor components with any closure induced from the remaining components will produce a mincut. When the number of sites m is small, we can check if a member can satisfy its parity-constraint sets by a backtracking search on at most O(m2) components. Since the parity constraints involve vertices for an individual member, these searches can be done independently. Therefore we need to examine if a valid partition Φ satisfies S pc on at most cuts for the whole pedigree.
Theorem 5 The RHC k problem is solvable in time.
Proof 3 Setting up the pedigree graph G = (V, E) takes O(|V|) time, where |V| = |E| = O(nm2). Generating parity-constraint sets S pc takes O(nm3 ). Transforming the pedigree graph into a graph with all negative edges takes O(|E|) time. The GBER problem can be solved by trying at most 2 k valid partitions Φ . For each partition, we can find the first mincut in O(k ·|E|) time by finding at most k augmenting paths using Edmonds-Karp algorithm. We can find strongly connected components in O(|E|) time. We do backtracking in at most cuts for each member to check if one can satisfy S pc ; each check takes O(|E|) time. Therefore, checking each partition takes . The overall time complexity of the algorithm is .
We have shown that given a general pedigree with n members, m sites, and k recombination events, where m and k are small, the haplotype inference can be done in time.
While not yet implemented, this algorithm should be implemented fairly easily. We only need to create a pedigree graph from input data according to the given construction and then transform the graph into the graph bipartization by edge removal with additional pedigree constraints, which can be tackled by making the appropriate modifications to an existing software package [14]. Future work will investigate the performance of the algorithm with simulated and real data.
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This research was funded by the Natural Sciences and Engineering Research Council of Canada through Discovery Grant 204923 to P.A. Evans.
Faculty of Computer Science, University of New Brunswick, Fredericton, New Brunswick, Canada
Duong D Doan & Patricia A Evans
Duong D Doan
Patricia A Evans
Correspondence to Patricia A Evans.
DDD designed the algorithm and drafted the manuscript. PAE supervised the research, assisted in crafting the algorithm and polished the manuscript. Both authors read and approved the final manuscript.
Doan, D.D., Evans, P.A. An FPT haplotyping algorithm on pedigrees with a small number of sites. Algorithms Mol Biol 6, 8 (2011). https://doi.org/10.1186/1748-7188-6-8
Received: 15 August 2010
Recombination Event
Positive Edge
Negative Edge
Heterozygous Site


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  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
  <option value="6">6</option>
  <option value="7">7</option>
  <option value="8">8</option>
  <option value="9">9</option>
</select>

I just can't figure out why this happens or how I can resolve this. Do you guys have any ideas?

A: There shouldn't be anything wrong with this. I tested in Chrome and Firefox. Maybe it is because of <option value="0"></option> because it is empty. but that should work.

<select id="cardexpiremonth" style="width: 100px;">

    <option value="1">1</option>
    <option value="2">2</option>
    <option value="3">3</option>
    <option value="4">4</option>
    <option value="5">5</option>
    <option value="6">6</option>
    <option value="7">7</option>
    <option value="8">8</option>
    <option value="9">9</option>
    <option value="10">10</option>
    <option value="11">11</option>
    <option value="12">12</option>
</select>

redpajama_set_name: RedPajamaStackExchange


So where is everyone getting their batteries these days?
Re: So where is everyone getting their batteries these days?
I picked up some Solarforce RCR123s and 2 Solarforce 18650s from ebay. Also some Ultrafire 18650s from ebay.
Pretty much anything that says AW on it. I'm also liking the 26650 cell size!
I buy my AA batteries from Dollar General (20 alkalines for $5), CVS pharmacy, various supply houses (for industrial batteries).
academy sports and outdoors is also a good place for 123A size.
Ever since I discovered that Dell laptop batteries are made up of my favorite Li-ion, the 18650, I haven't bought a single one. Disassembly is a snap, batteries are usually perfectly fine (seems the electronics die long before the batteries themselves), and in the event I overdraw one, I'll toss it in the recycle bin at Best Buy.
the disposal of batteries is big issue , lead is very harmful.
Seems batteries can not be delivered sometimes.


text:
"The Hand of the Lord"
Job 12:9 "Which of all these does not know that the hand of the Lord has done this?"
Some years ago the most exquisite diamond in the history of the world was found in Africa. Soon it was presented to the king of England to adorn the crown of state. The monarch sent it to Amsterdam to be cut by an expert stonecutter. The latter took the precious gem, cut a notch in it, and struck a hard blow with his instrument. Now the superb jewel lay in his hand split in two pieces. What recklessness! What waste! What carelessness!
Not at all! For weeks that blow had been planned. Models and drawings were made of the stone. Its quality, its defects, its cleavage lines had been studied with care. The blow was but the climax of the lapidary's skill. By it he brought the stone to its most perfect radiance and splendor. From the two halves came the two magnificent gems that the skilled eye of the cutter saw hidden in the stone.
Redeemed through Christ, we are God's precious possession. Yet sometimes He lets a stinging blow fall upon our lives. We cry out in agony: it all seems an appalling error. But no, for to God we are the most precious jewel in the world. Some day we shall emblazon the diadem of our King. By His grace, the blow that falls only fits us better for the splendor of the crown of His glory.
PRAYER: Lord Jesus Christ, discipline us in love and truly fit us for the crown of life. Amen.
Today's Bible Readings: Job 3-4 Matthew 8:1-17


Prepare a planned announcement before you make staff cutbacks.
1 How Does Project Management Reduce Customer Complaints?
Customer dissatisfaction may happen for a variety of reasons, some of which are out of your control. Although customer complaints may happen in any small business, steps can be taken to reduce them. Employees in any industry may be proactive at reducing complaints, which increases customer satisfaction.
To prevent and reduce customer complaints, it helps to know what potential problems the customer may have. If you know the root cause of the problem, you can take steps to fix it before the customer complains. Employees should play the part of a customer to test out the service of the organization. Doing so allows employees to see how customers view the organization. This also offers the ability to identify any potential issues that may lead to complaints.
Certain customer complaints may be due to high expectations or expecting more than your organization can offer. By letting customers know what to expect, employees can reduce the number of complaints. To do this effectively, employees must be able to meet the expectations that are set. This may be accomplished by underpromising and overdelivering, which means setting slightly lower expectations so that you can easily exceed them.
Proving the customers with a way to offer feedback about your organization is an effective way of reducing complaints. This may be done in a variety of ways, depending on the organization. Employees may directly ask the customer if all needs were met or if they were dissatisfied in any way. A feedback form may be given to customers or placed in a high-traffic area. Receipts can also include a URL or phone number where customers can offer feedback. When employees receive customer feedback, acting on that feedback may reduce or prevent complaints.
Customer complaints are an inevitable part of business, regardless of the steps you take to reduce them. When you get a complaint, it's important to handle it properly so that you retain the customer. Listen carefully to the complaint and repeat it back to the customer to make sure you understand the problem correctly. Apologize for the problem without making excuses for it and offer a resolution. You should also thank the customer for bringing the problem to your attention.
Lisse, Jamie. "How Employees Can Reduce Customer Complaints." Small Business - Chron.com, http://smallbusiness.chron.com/employees-can-reduce-customer-complaints-15841.html. Accessed 21 April 2019.
Why Do Businesses Need Customer Complaints?


Payment card terminals - getting money from all your clients!
A familiar situation: the client goes away because they don't have cash and cards are not accepted. With a card terminal, your business can receive money from all its clients. Both those with a wallet full of the 50s, and those who beep their phone or a wristband.
Experience speaks louder than words, so just try it out.
Use all card terminals for free for the first 6 months.
Commission for each transaction of 0.85% + 0.01 EUR.
No expenses to buy the terminal - the cost of use covers the terminal, maintenance, replacement, and consultation.
Additional receipt paper for free.
A retailers' portal with a review of customer transactions is free to use.
Get to know the 5 types of terminal and choose the best one for you.
Citadele payment card terminals are a good investment even for the most demanding retailers. Prove it? Easy!
Stores and service providers - card acceptance, integration with till systems, refund money.
Hotels and car rental - make a cash reservation, pre-authorisation or make a transaction without needing the physical card.
Restaurants - bring the terminal to the table, allow the payment of tips.
Businesses which sell off-site - mobile terminals with wireless internet.
Anyone who serves overseas guests - earn money by allowing clients to pay in their own currency, receiving a portion of the currency conversion commission fee yourself.
Those who have information and can use it are successful! That's why our clients can use the Retailers' Portal and review all their card transactions for free.
It's a convenient, simple and secure way to check up on your business' payment card transactions.
Access information on card transaction figures and bank commissions within the past 2 years.
Receive 8 kinds of daily and monthly reports for each terminal or the whole business. You will see information on cash payout services, fraud, and reclamation instances, as well the sale of electronic pre-payment codes.
Use our specialists to help with any terminal servicing issues and other questions.
Service provided by the bank's business partner SIA Worldline Latvia.
It's easy - we'll come to you for a consultation.
Sign a contract and open a Citadele account if you are not already a client.
After signing a contract, your new card terminal will be installed within 3 days in Riga and 5 days outside of Riga. Money will come flooding in!
What if a client doesn't pay by card?
Accept cash. Pay it into your account with a deposit card via an ATM.
Conveniently deposit and withdraw cash from ATMs. You can also get a card with just a cash depositing function - a cash deposit card*.
Control employees' card transactions. Set a daily limit for each card on purchases and cash withdrawal from ATMs and customer service centers.
If necessary, use to pay for goods and services.
Fee: 1.25 Euro per month. *The cash deposit card is only for depositing cash in Citadele ATMs and receiving a receipt for the transaction. A cash deposit card cannot be used for the following: transactions at POS and online, ATM withdrawals or balance checks.


Wittenberg junior setter Karen Wildemann (Columbus, IN/North) was the recipient of the Elite 90 award for the 2017 NCAA Division III Women's Volleyball Championship.
Wildemann, a Biology and Finance major with a 4.0 GPA, was presented the award during the 2017 NCAA Div. III Women's Volleyball Championship Banquet on Wednesday evening in Grand Rapids, Michigan.


text:
Q: HTML 5 video tag: is it possible to set src to a network resource? For an internal website, I'm trying to show an inline video of an avi file on the network.
<video class="VideoTagLink" src="file://\\network\path\file.avi" controls="controls"></video>

In IE9 I get a red X and Chrome only shows a play button -- is this a supported scenario in HTML 5?
edit:
I have changed my html to this:
<video class="VideoTagLink" controls="controls">
<source src="http://localhost:99/Handlers/GetVideo?path=\\network\path\file.avi" type="video/x-msvideo">
</video>

and my handler is just:
public ActionResult GetVideo(string path)
    {
        return base.File(path, "video/x-msvideo");
    }

So I believe i'm sending the right content type. I verified in IIS that *.avi is mapped to video/x-msvideo. If I navigate to that src URL directly in either browser, I get a download prompt for the video file, so I assume that works fine.

A: See HTML5 Video Error - Internet Explorer 9. This happens because IE9 cannot determine the content type, and your file server is not sending a Content-Type header.
If you serve the file via HTTP, you can configure your web server to specify the correct content type.


RANTAI KMC X11 EPT SILVER | .:: pitstop-bikes ::.
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This comfortable, foam-padded Assure transfer belt has both vertical and horizontal handgrips that offer seven holding positions. This safety belt can be used as a transfer device or as a walker belt. A fixed-lock buckle allows the tension to be adjusted without unfastening. The lining is made of an anti-slip fabric to reduce upward movement. Transfer belt is machine washable. Size by waist circumference.


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We have experience in promoting products, events, venues and restaurants. Every project is a new experience and will require a unique approach. Tell us about yours and we will contact you on how we can go about it.
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text:
Tag Archives: Laura Tipper
Double-edged War Puns
Belgrade Theatre, Coventry, Saturday 8th December, 2018
It's become quite a tradition at the Belgrade that while the panto is on in the main house, the B2 studio hosts an alternative, something for the grown-ups. This year, writer Nick Walker chooses the centenary commemorations of the end of the First World War and of the start of the women's suffrage movement as the basis for this pun-riddled romp.
As ever, the script is jam-packed with groanworthy gags, delivered with the rapidity and subtlety of a machine gun, as it tells the story of four men enlisted to go to the Front to rescue a troupe of actresses. The cast is entirely female – the reason for which becomes apparent by the end.
Laura Tipper sings sweetly as Bell, and harumphs horribly as Sidebottom, complete with period moustache. Aimee Powell is dashing as Ashwell, dapper in black tie and tails. Kimisha Lewis shows her versatility as Flowers, a German, and a balletic Red Baron. Miriam Grace Edwards is magical as stage magician Mickey… The ladies have several roles each and are well-matched for talent and likeability.
Walker's clever script has a repeating plot device, taking us back time and again to a music hall, interspersed with scenes of action and espionage reminiscent of a John Buchan. Director Katy Stephens, a veteran of several of these shows, paces the delivery to perfection. There is a silent-movie type sequence involving a bomb in a French restaurant that is superb, and a break from the otherwise relentless barrage of bad jokes. ("Is it snails?" "No, this is a fast food restaurant.")
It's not all daftness and running around. Walker, recognising the solemnity of the occasion, provides a sucker punch ending. We've all seen how Blackadder turned out; here the impact is equally if not more powerful as it is revealed that the characters are all based on real women, and there really was a mission to rescue the actresses. The final moments commemorate the contributions of women to the war effort and the sacrifices they made, something that many of the events we have seen over the past four years have overlooked.
Delightfully corny, rib-ticklingly daft, and ultimately sobering, this is a solid hour of entertainment with a powerful message.
Leave a comment	| tags: Aimee Powell, Belgrade Theatre, Coventry, Katy Stephens, Kimisha Lewis, Laura Tipper, Miriam Grace Edwards, Nick Walker, Over The Top, review	| posted in Review, Theatre Review
All Puns Blazing
THE GOOD, THE BAD AND THE UGLY SISTERS
B2, Belgrade Theatre, Coventry, Friday 8th December, 2017
One Christmas tradition that doesn't get me bah-humbugging all the way home, is the Belgrade Theatre's annual alternative production to the (excellent) pantomime in the main house. The B2 studio becomes home to a show for the grown-ups, in a genre- as well as gender-bending cavalcade of bad jokes. This year, riffing on Cinderella, writer-director Nick Walker gives us a Western with a cast of four women, playing cowboys. There is a plot, a chase to beat the bad guy to some buried treasure, and along the way we encounter a range of tropes (the saloon, the train, the Native American guide) as well as a host of larger-than-life characters performed by this versatile and industrious quartet.
Doc (the mighty Katy Stephens) is our protagonist and narrator. Such is her wry charm, we let her get away with the worst puns imaginable without rising up and lynching her. She is supported by the Magnificent Three: Miriam Edwards, Laura Tipper and Aimee Powell, in this relentless barrage of fun. Some of the jokes are as old as the hills and the corn is as high as an elephant's eye, but there is plenty of invention in the pun-fire, lots of new material to make groan-ups of us all.
Walker evidently spends the rest of the year writing jokes for Christmas crackers, and is possessed of a particular kind of genius. For example, the treasure could be silver, could be gold – it could be either ore.
I can hear you groaning from here. This type of thing is perhaps an acquired taste. It is certainly right up my alley.
Performing with indefatigable brio, the cast pull out all the stops to keep the laughs coming, and the knowing looks add to the fun. We are not expected to take a second of it seriously – but the cast certainly do, playing with commitment and skill – the comic timing is superb; and the production values are certainly no joke. The Belgrade's in-house production services dress the show in quality costumes. I love the tumbleweeds that punctuate the script's worst excesses and the horses are hot to trot. A simple but effective set with a sunset backcloth serves for all locations, allowing the performers to do most of the work, while the sound effects (Rob Clews) and the lighting (Chris Munn) evoke the genre while augmenting the humour.
It's an hour of fantastic fun and it makes me think we don't see many Westerns on the stage. Yes, there are musicals and opera set in the Wild West but no 'straight' plays? It's a gap in the market perhaps I can head off at the pass…
The Magnificent Four: Laura Tipper, Aimee Powell, Katy Stephens and Miriam Edwards (Photo: Robert Day)
Leave a comment	| tags: Aimee Powell, B2 Studio, Belgrade Theatre Coventry, Chris Munn, Katy Stephens, Laura Tipper, Miriam Edwards, Nick Walker, review, Rob Clews, The Good the Bad and the Ugly Sisters	| posted in Theatre Review


text:
Q: Custom Sorting an array in java using Two values , x and y I am trying to sort the array based on x values in case x value is same , sort it according to y. After sorting print the values I get an exception, I am calling compare method by overriding compare method of Comparator Interface.
class coder
{
    int x,y,index;
}
public class CoderRating  implements Comparator<coder>{
    public int compare(coder A,coder B)
    {
        if(A.x==B.x)
        {
            if(A.y<B.y)
            return -1;
            else if(A.y>B.y)
            return 1;
            else
            return 0;
        }

        else if(A.x<B.x)
        return -1;

        else
        return 1;
    }

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        coder []c=new coder[n];

        for(int i=0;i<n;i++)
        {
            c[i]=new coder();
            c[i].x=sc.nextInt();
            c[i].y=sc.nextInt();
            c[i].index=i;
        }
        Arrays.sort(c);
        for(int i=0;i<n;i++)
        {
            System.out.println(c[i].x + "  "+c[i].y );
        }
    }

}

A: You're calling the wrong overload of sort, it should be:
Arrays.sort(c, new CoderRating());

Currently, you're calling this sort method instead of this, specifying your comparator hence the issue.

A: You are calling the Arrays.sort on coder object and not passing any comparator object so Arrays.sort is looking for comparable in Coder class.
2 solutions to your problem:

*

*change: Pass the comparator which you want while sorting coder array.
Change Arrays.sort(c) to  Arrays.sort(c,new CoderRating());

*implement comparable in coder. Override the compareTo method and move the code of compare method inside CoderRating  to the compareTo method in Coder.
PS: For your code cleanup :
Change:
       `if(A.y<B.y)
            return -1;
        else if(A.y>B.y)
            return 1;
        else
            return 0;`

can be changed to one liner:
return Integer.compare(A.y, B.y);
Output:
3
1 2
2 3
2 1
1 2
2 1
2 3

A: You are not calling the sorting algorithm with the Comparator you have implemented:
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
coder []c=new coder[10];

for(int i=0;i<n;i++)
{
    c[i]=new coder();
    c[i].x=sc.nextInt();
    c[i].y=sc.nextInt();
    c[i].index=i;
}
Arrays.sort(c, new CoderRating()); //this is how it should be called
for(int i=0;i<n;i++)
{
    System.out.println(c[i].x + "  "+c[i].y );
}

output:


text:
12/26/19 El Pollo Loco Improves Access to Better-For-You Food with New Pollo Fit Bowls
12/16/19 El Pollo Loco Restaurants are the New Canvases for Los Angeles Murals
11/01/19 El Pollo Loco Brings the Holidays to Life with Three Classic Holiday Offerings Worthy of Becoming a New Tradition
10/03/19 El Pollo Loco to Donate a Taco for Every Taco Sold on National Taco Day
09/16/19 El Pollo Loco Revives the Latino Mural Legacy for Hispanic Heritage Month
09/04/19 El Pollo Loco Leads Company-Wide Effort to Donate Surplus Food
08/29/19 El Pollo Loco Fires Up $5 Combos Menu with New Offerings
07/09/19 El Pollo Loco Introduces a New Way to Nacho with the $5 Crunchy Nacho Overstuffed Quesadilla
04/02/19 El Pollo Loco Strengthens Hometown with Day of Community Service to Honor Cesar Chavez Day
03/26/19 El Pollo Loco Names Miguel Lozano as New Chief Operating Officer
03/04/19 El Pollo Loco Brings the Taste of Tradition with New, Hand-Rolled Enchiladas
03/01/19 El Pollo Loco Announces the Grand Opening of New Restaurant in Concord, CA
02/18/19 El Pollo Loco Announces the Grand Opening of New Restaurant in San Luis Obispo, CA
01/02/19 El Pollo Loco Debuts Crave-Worthy Fire-Grilled Chicken Nachos to Kick-off the New Year


text:
Sports Nutrition 2016: Year in Review
Darren M
*The sports nutrition industry made some important advances in both sales and technology in2016, but the market still faces regulatory and legal issues. Market research indicates that the sports nutrition sector can be expected to grow from its 2015 appraisal of $28 billion to over $50 billion by2022 (P&S Market Research). Analysts predict an 8.4% compounded annual growth in that time frame.Euromonitor International (EI) reports that the US was the largest sports nutrition market, at 12%growth. EI reported that the industry's expanding consumer demographic is the primary driver, with 25to 34-year-olds as the main consumer demographic. The report noted middle-aged adults today are more active and are turning to the sports nutrition industry to enhance their nutritional fitness regimens. Sales Demographic Trends At the Expo West convention in Anaheim, California this year, David Sandler of Strengthpro.com said consumers across many sports nutrition subgroups want advanced products that work and that are scientifically proven safe. Branded ingredients are gaining in popularity. Sandler told interviewers that consumers want new technology in product delivery. Innovative products feature sophisticated flavor technologies and methods of delivery, and "they taste better and are easier to use."One such new product application is Regimen Whey with GPAC Technology as 2016's winner of Supply Side West Editor's Choice Award. Regimen Whey features a convenient protein pouch which also helps with dosage control.
New Technologies & New Ingredients Much innovation is also evident in modern ingredients. Two exciting new ingredients for 2016 include cucumber extract and whey flavor technology. Glanbia Nutritionals introduced EasyFlav Whey, which makes traditional carbohydrates obsolete and delivers protein and flavor for a promising new whey product with more protein and 30% less carbs. This new technology is made possible by vegetable-protein-neutralizing techniques from EasyFlav and has been adapted to help food manufacturers incorporate high concentrations of whey while still delivering a consistent and enjoyable flavor. XSTOSolutions and Naturalea SA introduced Cuvitus, an exceptional new cucumber extract, to the North American market. Known in Europe as Crownhealth Gherkin, it is a whole fruit extract derived from Cucumis Sativa. It was designed to give relief from muscle and joint pain after exercise. It was developed with the help of university research partners, is created through a proprietary extraction procedure, and helps to regulate temporary inflammation resulting from the rigors of exercise. Other emerging ingredients include shilajit, a mineral found in the Himalayan Mountains useful as a testosterone and nitric oxide booster for longer periods of exercise and better performance. Plant protein continued to gain momentum in the market, while insect protein also continued to gain acceptance in many sports products. This unusual protein is derived primarily from crickets, which are considered a plentiful and sustainable source that can be made into an easy to use and good tasting powder.
The Business of Sports Nutrition Product technology developments are always an objective of formulators and retailers, but there is much activity on the supply side helping to shape the brands that resonate most with consumers.Companies continue to form partnerships to introduce new products to the market. At the "Business of Sports Nutrition" workshop in October, Europa CEO, Eric Hillman explained that two merging manufacturers could form the biggest sports nutrition distributor in the US, with 45 years of experience and 16,500 customers, and $400 million in revenues between them. Another significant acquisition was made by American Securities of the Milk Specialties company last summer.
Competitor lawsuits and federal legal cases are an ongoing problem for the industry. Among brands taking competitors to court in 2016, were Capstone Nutrition's claim against MusclePharm for breaches of contract and failure to pay for $22 million worth of products. MusclePharm has been restructuring its business model in response. Its management settled with Capstone for $11 million and the rights to purchase 1.2 million in stocks in November. These are just a sample of the year's litigation in sports nutrition. Of course, federal actions have also been a significant stumbling block for many brands and manufacturers for as long as the sports nutrition industry has existed. Federal agencies have taken action against anabolic steroid products with the Anabolic Steroid Control Act of 2004. They took aim at prohormones by drafting the Designer Anabolic Steroid Control Act of 2014 into law.
2016 witnessed sweeping federal action against companies selling adulterated products illegally containing pharmaceuticals or anabolics. The executives of those companies were also targeted. In July of this year, the DOJ and FDA indicted two sports nutrition executives for conspiracy to sell steroids illegally.The charges carry potentially substantial fines and punishments. The FDA was unsuccessful in its attempts to forward claims against a dietary supplement contract manufacturing firm tied to alleged illegal activities involving a DMAA based sports supplement that was linked a 2013 liver disease outbreak.
Growing Pains Internationally Naturally, as sports nutrition makes gains around the world, including a Chinese market that is champing at the bit for access to foreign brands, regulatory compliance may become more of a challenge. China presented new sports supplement safety standards to address a problem with unauthorized selling in the region- especially via online sources dealing in counterfeit products.This year in review is really just a sample of the many significant changes, advances, and challenges experienced by the sports nutrition industry this 2016.
While the advances have been momentous, and the setbacks have been very disruptive, it's fair to say that any industry involved in the advancement of sports nutrition and health, outside the purview of the medical community, is bound to face significant stumbling blocks. But the advances, not to exclude the sharply piqued interest among consumers to achieve superior sports and nutritional wellness, are sufficiently promising- and the reality of the benefits are so powerful- that the industry can only expect to grow by leaps and bounds. ~
*Ghostwritten via WriterAccess.com Sources: http://www.euromonitor.com/sports-nutritionhttps://jissn.biomedcentral.comhttp://www.naturalproductsinsider.com/blogs/supplement-perspectives/2016/12/dietary-supplements-2016-year-in-review.aspx
Dover, Delaware, United States •
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greenrabbitsland: What's your guilty pleasure?
Już mi ślinka leci , a tak daleko do tych lodów.
Oh yes, the Danes know how to make yummie ice cream! no diet version there he he! Love the accordtion rosettes and the happy feel to your page. Thanks ever so much for sharing this delightful summer memory with us at OUAS!
It's such a lovely and summery lay-out. I really love your beautiful design.
absolutely wonderful! Thank you so much for playing with us over OUAS!
Such a lovely page! Thank you so much for playing along with us at OUAS!
Such a beautiful Summery page and I love your use of vibrant colours!! Thanks so much for sharing with us at OUAS!
I love ice cream, too :) What a fun and beautiful layout! So glad you joined us at OUAS!
Fun summer page with happy colors!
Thanks for playing with us at OUAS this month!


ORONO – University of Maine alumnus David Demsey will return to his alma mater to play the saxophone with the Bangor Symphony Orchestra on Saturday, Feb. 1 and host a master class with six UMaine School of Performing Arts saxo phone students on Jan. 31. During his visit, Demsey will also work with the UMaine Jazz Ensemble.
Demsey is professor of music and coordinator of jazz studies at William Paterson University in Wayne, N.J. After graduating with a bachelor's degree in music education from UMaine in 1977, he received a master's in music in saxophone from the Juilliard School and earned a doctorate at the Eastman School of Music.
Demsey's master class is being offered from 3 to 5 p.m. Friday, Jan. 31 in room 100 of the Class of 1944 Hall.
Ethan Manning of Vernon, Vt.
Michael Milazzo of Williamstown, Mass.