Applications Of The Pigeonhole Principle Mathematics Essay

Applications Of The Pigeonhole Principle Mathematics Essay We begin our discussion with a common daily embarrassing moment. Suppose that in ones dresser drawer, he has socks of three different colours (all placed in messy order). Having to get up early in the morning while it is still dark, how does he ensure that he gets a matching pair of same coloured socks in the most convenient way without disturbing his partner? While, the answer is simple! He just has to take 4 socks from the drawer! The answer behind this is of course, the Pigeonhole Principle which we will be exploring in this Maths Project. What is the Pigeonhole Principle then? Let me give you an example to illustrate this principle. For instance, there are 3 pigeonholes around. A pigeon is delivering 4 mails and has to place all its mails into the available pigeonholes. With only 3 pigeonholes around, there bound to be 1 pigeonhole with at least 2 mails!. Thus, the general rule states when there are k pigeonholes and there are k+1 mails, then they will be 1 pigeonhole with at least 2 mails. A more advanced version of the principle will be the following: If mn + 1 pigeons are placed in n pigeonholes, then there will be at least one pigeonhole with m + 1 or more pigeons in it. The Pigeonhole Principle sounds trivial but its uses are deceiving astonishing! Thus, in our project, we aim to learn and explore more about the Pigeonhole Principle and illustrate its numerous interesting applications in our daily life. We begin with the following simple example: 2. Pigeonhole Principle and the Birthday problem We have always heard of people saying that in a large group of people, it is not difficult to find two persons with their birthday on the same month. For instance, 13 people are involved in a survey to determine the month of their birthday. As we all know, there are 12 months in a year, thus, even if the first 12 people have their birthday from the month of January to the month of December, the 13th person has to have his birthday in any of the month of January to December as well. Thus, we are right to say that there are at least 2 people who have their birthday falling in the same month. In fact, we can view the problem as there are 12 pigeonholes (months of the year) with 13 pigeons (the 13 persons). Of course, by the Pigeonhole Principle, there will be at least one pigeonhole with 2 or more pigeons! Heres another example of the application of Pigeonhole Principle with peoples relationship: 3. Pigeonhole Principle and problems on relations Assume that the relation `to be acquainted with is symmetric: if Peter is acquainted with Paul, then Paul is acquainted with Peter. Suppose that there are 50 people in the room. Some of them are acquainted with each other, while some not. Then we can show that there are two persons in the room who have equal numbers of acquaintances. Lets assume that there is one person in the room that has no acquaintance at all, then the others in the room will have either 1, 2, 3, 4, à ¢Ã¢â€š ¬Ã‚ ¦, 48 acquaintance, or do not have acquaintance at all. Therefore we have 49 pigeonholes numbered 0, 1, 2, 3, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦, 48 and we have to distribute between them 50 pigeons. So, there are at least two persons that have the same number of acquaintance with the others. Next, if everyone in the room has at least one acquaintance, we will still have 49 pigeonholes numbered 1, 2, 3, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦, 48, 49 and we have to be distribute between them 50 pigeons! Also, we can apply the Pigeonhole Principle in the proving of numerical properties. The following are two of such examples: 4. Pigeonhole Principle and divisibility Consider the following random list of 12 numbers say, 2, 4, 6, 8, 11, 15, 23, 34, 55, 67, 78 and 83. Is it possible to choose two of them such that their difference is divisible by 11? Can we provide an answer to the problem by applying the Pigeonhole Principle? There are 11 possible remainders when a number is divided by 11: 0, 1, 2, 3, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦.., 9, 10. But we have 12 numbers. If we take the remainders for pigeonholes and the numbers for pigeons, then by the Pigeon-Hole Principle, there are at least two pigeons sharing the same hole, ie two numbers with the same remainder. The difference of these two numbers is thus divisible by 11! In fact, in our example, there are several answers as the two numbers whose difference is divisible by 11 could be 4 15; 34 67 or 6 83. 5. Pigeonhole Principle and numerical property We can also apply the Pigeonhole Principle in determining useful numerical properties. Consider a sequence of any 7 distinct real numbers. Is it possible to select two of them say x and y, which satisfy the inequality that 0 The problem sounds difficult as we may need to consider more advanced calculus and trigonometrical methods in the determination of the result. Well, to answer the above problem, one will be surprised to know that we just need a simple trigonometrical identity and apply the Pigeonhole Principle! Before proceeding to answer the problem, we first note that given any real number x, we can always find a real number a where n1 = tan a1, n2 = tan a2, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦.., n7 = tan a7 Now, if we were to divide the interval (-p, p) into 6 equal intervals, we obtain the following sub-intervals: ( -p, -p ), [ -p, -p ), [ -p, 0 ) , [ 0, p ), [p,p ) and [p, p ). For the 7 distinct numbers a1, a2, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦.a7, by the Pigeonhole Principle, there should be two values say, ai and aj such that aI > aj and ai aj are in the same interval! For these two values ai and aj, we should have 0 We may recall an important trigonometrical identity: B ) = . Thus, if ni = ai and nj = aj , then = = tan ( ai aj ) As 0 0 and so, 0 which is the result we are seeking! We may also apply the Pigeonhole Principle in the proving of useful daily geometrical results.. The following examples illustrate such usages: 6. Pigeonhole Principle and Geometry a. Dartboard applications Another common type of problem requiring the pigeonhole principle to solve are those which involve the dartboard. In such questions, a given number of darts are thrown onto a dartboard, the general shape and size of which are known. Possible maximum distance between two certain darts is then to be determined. As with most questions involving the pigeonhole principle, the hardest part is to identify the pigeons and pigeonholes. Example 1: Seven darts are thrown onto a circular dartboard of radius 10 units. Can we show that there will always be two darts which are at most 10 units apart? To prove that the final statement is always true, we first divide the circle into six equal sectors as shown; Allowing each sector to be a pigeonhole and each dart to be a pigeon, we have seven pigeons to go into six pigeonholes. By pigeonhole principle, there is at least one sector containing a minimum of two darts. Since the greatest distance between two points lying in a sector is 10 units, the statement is proven to be true in any case. In fact, it is also possible to prove the scenario with only six darts. In such a case, the circle is this time divided into five sectors and all else follows. However, take note that this is not always true anymore with only five darts or less. Example 2: Nineteen darts are thrown onto a dartboard which is shaped as a regular hexagon with side length of 1 unit. Can we prove that there are two darts within units of each other ? Again, we identify our pigeonholes by dividing the hexagon into six equilateral triangles as shown below. With the six triangles as our pigeonholes and the 19 darts as pigeons, we find that there must be at least one triangle with a minimum of 4 darts in it. Now, considering the best case scenario, we will have to try an equilateral triangle of side 1 unit with 4 points inside. If we try to put the points as far apart from each other as possible, we will end up assigning each of the first three points to the vertices of the triangle. The last point will then be at the exact centre of the triangle. As we know that the distance from the centre of the triangle to each vertex is two-third of the altitude of this triangle, that is, units, we can see that it is definitely possible to find two darts which are units apart within the equilateral triangle! b. Encompassing problems Consider the following problem: 51 points are placed, in a random way, into a square of side 1 unit. Can we prove that 3 of these points can be covered by a circle of radius units ? To prove the result, we may divide the square into 25 equal smaller squares of side units each. Then by the Pigeonhole Principle, at least one of these small squares (so call pigeonholes) should contain at least 3 points (ie the pigeons). Otherwise, each of the small squares will contain 2 or less points which will then mean that the total number of points will be less than 50 , which is a contradiction to the fact that we have 51 points in the first case ! Now the circle circumvented around the particular square with the three points inside should have radius = = = It will be worthwhile to note the above technique can be useful in analyzing accuracy of weapons in shooting practices and tests. Next, we will like to proceed to a more creative aspect of the application of Pigeonhole Principle by showing how it can be used to design interesting games: 7. Application of pigeonhole principle in card games We like to introduce the application of pigeonhole principle in two exciting card tricks: a. Combinatorial Card Trick : Heres the trick: A magician asks an unsuspecting observer to randomly choose five cards from a standard deck of playing cards. The participant does not show these cards to the magician, but does show them to the magicians accomplice. The accomplice looks at the five cards, chooses four of them, and shows these four to the magician in a certain ordered manner. The magician immediately identifies the fifth hidden card. How does the trick work? The following is an explanation of our working strategy: (1) First of all, notice that in any hand of five cards there must be two cards of the same suit (an application of Pigeonhole Principle). The first card that the accomplice shows to the magician is one of these two cards. The other card of the same suit is never shown it is the mystery card, the card which the magician must discover. Thus, the accomplice can easily communicate the suit of the hidden card: the hidden card has same suit as the first card shown to the magician. Specifying the rank of the mystery card (ie its value) is a little trickier but can be accomplished with a little circular counting manner which we will explained below Number the cards in a suit circularly from 1(ace) to 11 (jack), 12 (queen) and 13 (king) so that 1 follows 13 i.e. the list is ordered in a clockwise direction. Now, given any two cards A and B, define distance (A,B) as the clockwise distance from A to B. It is easy to see that for any two cards A and B either distance(A,B) or distance(B,A) must always be less than or equal to 6. Again as an application of the Pigeonhole Principle, we note that if they are both 7 or more, then there will be at least 2 x 7 = 14 cards in a standard suit of cards!! Example Cards: 3 and Jack (11) distance(Jack, 3) = 5; distance (3, Jack) = 8 Cards: Ace(1) and 7 distance (Ace, 7) = 6; distance (7,Ace) = 7 (2) Our working strategy thus proceeds as follows.: From those two cards of the same suit, A and B, the accomplice shows the magician card A such that distance(A, B) is 6 or less. For example, given the choice between the three of clubs and the Jack of clubs, the accomplice reveals the Jack (since distance (Jack ,3) = 5 and distance(3, Jack)= 8). The three of clubs remains hidden. If the two same-suit cards are the five of hearts and the six of hearts, the accomplice chooses the five (since distance (5,6) = 1 but distance (6,5) = 12) leaving the six of hearts as the mystery card. (3) Finally, the accomplice arranges the last three cards to encode a number from 1 to 6 the distance from the value of first card to that of the hidden card. A quick calculation allows the magician to discover the value of the mystery card. Notice that although the magician must decode only one of 6 possibilities, it should not present a problem, even to the slowest of magicians. To facilitate the explanation for the last step involved, we may assign each card a number from 1 to 52 for ranking purpose. For example, the ace of spade can be numbered 1 (the highest ranking card), ace of heart numbered 2, ace of club numbered 3, ace of diamond numbered 4, king of spade numbered 5, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦.., queen of spade numbered 9, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦., jack of spade numbered 13, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦., 10 of spade numbered 17, à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦. , à ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦Ãƒ ¢Ã¢â€š ¬Ã‚ ¦. , 2 of diamond numbered 52 (the lowest ranking card). We will now proceed to explain the last step using the following example: Example: Suppose the five cards chosen are the following: 3 of Hearts (numbered 46) 5 of Spades (numbered 37) 6 of Clubs (numbered 35) 7 of Hearts (numbered 30) 2 of Diamonds (numbered 52) The accomplice notices that the 3 and the 7 have the same suit hearts. Since the distance( 3 ,7) = 4 and distance(7, 3) = 9, the accomplice chooses the 3 as the first card to show the magician, leaving the 7 of hearts as the hidden card. The magician now knows that the suit of the mystery card is hearts. The accomplices next task is thus to let the magician know that he must add the value 4 to the number 3 to obtain the final value of 7 for the hidden card! How can he achieve this? Basically, he can arrange the other three cards in 3! = 6 ways. Based on the numbering method explained earlier, the 3 remaining cards can be ranked 1st, 2nd and 3rd . In our example, the 6 of Clubs will be ranked 1, the 5 of Spades will be ranked 2 and the 2 of Diamonds will be ranked 3. The accomplice may agree with the magician earlier that the arrangement of these 3 cards represent specific numbers as shown below: Order in which 3 remaining cards are shown Number represented by the arrangement 1, 2, 3 1 1, 3, 2 2 2, 1, 3 3 2, 3, 1 4 3, 1, 2 5 3, 2, 1 6 Thus in our example, the accomplice should display the cards in the following manner: firstly, the 5 of Spades, then the 2 of Diamonds and lastly, the 6 of Clubs ! b. Permutation Card Trick: Heres the trick: A magician asks an unsuspecting observer to randomly arrange 10 cards which are labelled 1 to 10 in a hidden face down manner. The participant does not show the arrangement of these cards to the magician, but does show them to the magicians accomplice. The accomplice looks at the ten cards and flips over six of the cards in a certain ordered manner to reveal their values to the magician. The magician immediately identifies the values of the four remaining unknown cards. How does the trick work? We first note that by applying the Pigeonhole Principle, we can show that in any permutation of 10 distinct numbers there exists an increasing subsequence of at least 4 numbers or a decreasing subsequence of at least 4 numbers. (refer next section of our discussion). These are the numbers that remain hidden in our trick. The magician will know that the sequence is increasing if the accomplice flips over the other six cards from the left to right and it is decreasing if the other six cards are flipped over from the right to the left. We will now proceed to explain the trick behind the game: The trick behind the game: Given any sequence of mn+1 real numbers, some subsequence of (m+1) numbers is increasing or some subsequence of (n+1) numbers is decreasing. We shall prove the result by Contradiction method. Assume that the result is false. For each number x in the sequence, we have the ordered pair (i,  j), where i is the length of the longest increasing subsequence beginning with x, and j is the length of the longest decreasing subsequence ending with x. Then, since the result is false, 1  £ i  £ m and 1  £ j  £ n. Thus we have mn+1 ordered pairs, of which at most mn are distinct. Hence by the Pigeonhole Principle, two members of the sequence, say a and b, are associated with the same ordered pair (s,  t). Without loss of generality, we may assume that a precedes b in the sequence. If a Thus, in our trick, we should have an increasing subsequence of at least (3+1) numbers or a decreasing subsequence of at least ( 3+ 1) numbers in a permutation of (33+ 1) distinct numbers! Here is an example of how the trick can be performed: Example Suppose the participant arranges the 10 cards in the following manner (value faced down from left to right): 3, 5, 8, 10, 1, 7, 4, 2, 6, 9. Upon careful inspection, the accomplice notices that an increasing subsequence can be 3, 5, 8, 10 while a decreasing subsequence can be 10, 7, 4, 2. If he decides to use the increasing subsequence, he should leave the first four cards untouched and flips the other six cards over in a leftward manner as shown: 1 7 4 2 6 9 Direction of flip The magician on realising that the four missing numbers are 3, 5, 8 and 10 and the leftward direction of flip, will thus proclaim the 4 hidden numbers to be 3, 5, 8, and 10 respectively! If the accomplice decides to use the decreasing subsequence, he should leave the cards bearing the numbers 10, 7, 4, 2 untouched and flips the other six cards over in a rightward manner as shown: 3 5 8 1 6 9 Direction of flip The magician on realising that the four missing numbers are 2, 4, 7 and 10 and the rightward direction of flip, will thus proclaim the 4 hidden numbers (from left to right) to be 10, 7, 4, 2 respectively! 8. Conclusion Although the Pigeonhole Principle seems simple and trivial, it is extremely useful in helping one to formulate and facilitate calculation and proving steps for numerous important Mathematical results and applications. We have included just a substantial amount of its applications in our project discussion. More importantly, we will like to show that a simple Mathematical concept like the Pigeonhole Principle does have numerous interesting and beneficial application in our daily life! ~ ~ ~ ~ ~ End of Report ~ ~ ~ ~ ~

Contrasting Old Mother Savage and The Tell-Tale Heart Essay -- compari

Contrasting Old Mother Savage and The Tell-Tale Heart Writers may use different techniques to get the same effect out of the audience. In the short story, "Old Mother Savage" by Guy Du Maupassant, a tragic story of a woman who losses everything is told. The story is scary in that it has an ending that one would not expect. Also, it can be looked at as a sad story because the mother seems to be sad throughout the entire story. At the end the only thing that she has to be satisfied about is that her murdering four young men can make other women feel how she felt when she found out about the death of her son. This story can be compared to Edgar Allen Poe's "The Tell-Tale Heart", when you talk about the strategies that both authors use to make the audience frightened. They both describe scenes in full detail to give the effect of disgust. However, Du Maupassant, makes the audience feel sorry for the mother in this story turning it into a tragedy instead of horror. The story starts out with two men walking through a forest. One of the men recognizes an abandoned house. The house is described as "...a skeleton still standing, yet ruined and sinister" (Du Maupassant, 1). The speaker asks the man he is walking with what happened to the people who lived in it. The other starts explaining that the father was killed and that during the war, the son was sent to fight leaving the mother by herself. It was said that no one bothered her since everyone in the town thought she had money. It was said that she hardly ever laughed, but that was normal for women of that time: "The women suffer with sad and restricted souls, their life being solemn and hard" (Du Maupassant, 2). With this thought in mind it seems as if the peopl... ...t the woman as being a hero. She is what we consider a "good guy" not because she has killed innocent people, but because she has taken charge of a situation, which is out of the ordinary for women to do. This is a far contrast from Poes' ending. In his story the speaker confesses to killing the old man because the mans' heart, which at that point the reader knows is the speakers conscious is annoying him. At the end of his story the audience is glad that the speaker is caught. Both "Old Mother Savage", by Guy Du Maupassant and "The Tell-Tale Heart", by Edgar Allen Poe, offer a look into the other side of tragedies. In both we get to see the reasoning behind the killings of innocent people. The difference between the two is in one case the audience is left feeling sad for the killer, while in the other we are glad that justice is served.

Impact of gamma rays on the germination

To find out LD50 dose for the seed and urther investigate the influence on germination and seedling parameters. The experimental results revealed that the percentage of germination had decreased after irradiation and the effect become stronger with increase of gamma dose. Parameters such as germination percentage, speed ot germination, mean daily germination, peak value and germination value had significantly decreased with increased irradiation doses. Similarly seedling parameters viz. , Root length, Shoot length, Vigour index and Root/Shoot length ratio expressed higher reduction at higher doses as compared to non irradiated control.The study clearly indicated ncrease in the deleterious effects of gamma irradiation at regular intervals, with attainment of LD50 at a dose of 1. 50 kGy. Keywords: Gamma irradiation, Groundnut, Seed germination, Groundnut (Arachis hypogaea L. ) is popularly known as peanut. It is one of the world ‘s most popular oil seed crops, cultivated in more than 100 countries of six continents. It is the single largest source of edible oils in India and constitutes roughly about 50 percent of the total oilseeds production.Among the major Groundnut growing states there has been consistent increase in area under cultivation in Andhra Pradesh. The groundnut seed mainly comprised of protein, fat, carbohydrate which make it sensitive to radiation induced stress. Among the environmental stresses, the radiation is the most important factor, which limits production of groundnut. This would result in drastic reduction in crop yield and magnitude of reduction would depend on groundnut varieties. Not only the yield of Groundnut but also the quality of products decreases under radiation stress.The seed stage is a convenient phase in the plant's life cycle for use in radiological studies to determine relative radio sensitivity of species and the effects of various actors on radio sensitivity. Earlier experiments in this field have indicated that io nizing radiation could cause permanent genetical effects, lethal or beneficial mutations, morphological modifications and other effects in plants. Several factors may be involved in the inhibition of germination and the growth of the plants from seeds following their exposure to high irradiation doses.A number of radiobiological parameters are commonly used in early assessment of effectiveness of radiation. Methods based on physiological changes such as inhibition of seed germination and hoot and root elongation have been reported for detection of irradiated legumes. Therefore, in present study the response of groundnut seed (cv. Narayan') to gamma radiation stress on germination and seedling parameters of groundnut was investigated compared to non irradiated seed. ASIAN J. EXP. BIOL. SCI. VOL 4 (1) 2013 61 Impact of Gama Rays on the Seed Germination and Seedling Parameters of Groundnut (Arachis Hypogaea L. †¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦Ã¢â‚¬ ¦ M. Aparna et al. MATERIALS AND METHODS T he material for the present study comprised of seed of groundnut variety Narayan'. 100 seeds were taken in 0. 1 mm thick polythene bags of 1 5 X 22cm dimension and ealed. The bags were exposed to gamma irradiation wit n doses ot O 1. 30, 1. 50, 1. 70, 1. 90, 2. 10 and 2. 30 kGy. Samples were irradiated in continuous gamma sterilization plant (GC 5000, designed by Board of Radiation and Isotope Technology, Mumbai) with 444 TBq (12000Ci) and Cobalt60 source with a specific activity of 3. 01 kGy/hour at Quality Control Laboratory, Acharya N.G. Ranga Agricultural University, Rajendranagar, Hyderabad- 500 030 and were compared with the observations made on untreated control. The material for irradiation was placed in an irradiation chamber located in vertical drawer inside the Lead flask. Radiation field was provided by a set of stationary Cobalt60 source placed in a cylindrical cage. The source was doubly encapsulated in corrosion resistant stainless steel pencils and was tested in acco rdance with international standards. Two access holes of 8 mm diameter were provided of service sleeves for gasses, thermocouple etc. Mechanism for rotating/stirring samples during irradiation is also incorporated. The quantity of absorbed dose (kGy) can be defined as the amount of energy absorbed per unit mass of the matter at the point of interest. The experiment was carried out as per Completely Randomized Design CRD). The irradiated seed along with nonirradiated control were sown in petridishes in the laboratory. Data on germination and seedling parameters were recorded seven days after sowing under ambient condition. Germination percentage was calculated using the formula as per ISTA [10].Speed of germination of the given sample was calculated according to the formula given by Maguire [16]. Similarly, other germination parameters viz. , Mean daily germination, Peak value (Edwards [9]) and Germination value (Czebator [7]) were calculated. Seedling parameters like Shoot and root length were measured using ten seedlings ollected at random from each sample on 7th day from the seeds subjected to germination test. The shoot and root length were measured in centimeters (cm) using a scale and root/shoot length ratio was calculated using the estimates of seedling length.

Ergative Verbs and Processes in English Grammar

In grammar and morphology, ergative  is a verb that can be used in a construction in which the same noun phrase can serve as a subject when the verb is intransitive, and as a direct object when the verb is transitive. In general, ergative verbs tend to communicate a change of state, position, or movement. In an ergative language (such as Basque or Georgian, but not English), ergative is the grammatical case that identifies the noun phrase as the subject of a transitive verb. R.L. Trask draws this broad distinction between ergative languages and nominative languages (which include English): Roughly, ergative languages focus their articulation on the agency of the utterance, while nominative languages focus on the subject of the sentence (Language and Linguistics: The Key Concepts, 2007). Etymology:  From the Greek, working Observation on ​the Modern American Usage   In the mid-20th century, grammarians devised the term ergative to describe a verb that can be used (1) in the active voice with a normal subject (actor) and object (the thing acted on) [I broke the window]; (2) in the passive voice, with the recipient of the verbs action as the subject of the sentence (and most often the actors becoming the object of a by-phrase) [the window was broken by me]; or (3) in what one textbook called the third way, active in form but passive in sense [the window broke]. Ergative verbs show remarkable versatility. For example, you might say that he is running the machine or the machine is running, she spun the top or the top spun, the crew decided to split the rail or the rail split at that point.(Bryan Garner, Garners Modern American Usage. Oxford University Press, 2009) Downing and Locke on Ergative Pairs When the Affected object of a transitive clause (e.g. the bell) is the same as the Affected subject of an intransitive clause, we have an ergative alternation or ergative pair, as in I rang the bell (transitive) and the bell rang (intransitive). . . . English marks both the subject of an intransitive clause and that of an intransitive clause as nominative, and the object of the transitive as accusative. We can see this in the two meanings of leave: he left (went away, intrans.), he left them (abandon trans.). . . .Ergative pairs account for many of the most commonly used verbs in English, some of which are listed below, with examples: burn Ive burned the toast. The toast has burned.break The wind broke the branches. The branches broke.burst She burst the balloon. The balloon burst.close He closed his eyes. His eyes closed.cook Im cooking the rice. The rice is cooking.fade The sun has faded the carpet. The carpet has faded.freeze The low temperature has frozen the milk. The milk has frozen.melt The heat has melted the ice. The ice has melted.run Tim is running the bathwater. The bathwater is running.stretch I stretched the elastic. The elastic stretched.tighten He tightened the rope. The rope tightened.wave Someone waved a flag. A flag waved. Within this alteration — described here as an ergative pair — there is a set of basically intransitive volitional activities (walk, jump, march) in which the second participant is involved either willingly or unwillingly. The control exerted by the Agent predominates in the causative-transitive: He walked the dogs in the park. The dogs walked.He jumped the horse over the fence. The horse jumped over a fence.The sergeant marched the soldiers. The soldiers marched. It is also possible to have an additional agent and an additional causative verb in the transitive clauses of ergative pairs; for example, The child got his sister to ring the bell, Mary made Peter boil the water.(Angela Downing and Philip Locke, English Grammar: A University Course. Routledge, 2006) The Difference Between Transitive Processes and Ergative Processes What distinguishes a transitive from an ergative process? Characteristic of transitive processes (e.g., chase, hit, kill) is that they are Actor-centered: their most central participant is the Actor, and the Actor-Process complex is grammatically more nuclear and relatively more independent ([Kristin] Davidse 1992b: 100). The basic Actor-Process complex can be extended only to include a Goal, as in The lion is chasing the tourist. Ergative processes such as break, open and roll, in contrast, are Medium-centered, with the Medium as most nuclear participant (Davidse 1992b: 110) (e.g., The glass broke). The basic Medium-Process constellation can only be opened up to include an Instigator, as in The cat broke the glass. While the transitive Goal is a totally inert Affected, the ergative Medium co-participates in the process (Davidse 1992b: 118). In ergative one-participant constructions such as The glass broke, this active coparticipation of the Medium in the process is foregrounded and the Medium is presented as semi- or quasi-autonomous (Davidse 1998b).(Liesbet Heyvaert, A Cognitive-Functional Approach to Nominalization in English. Mouton de Gruyter, 2003) Ergative Languages and Nominative Languages An ergative language is one in which the subject of an intransitive verb (e.g., Elmo in Elmo runs home) is treated in grammatical terms (word order, morphological marking) similarly to the patient of a transitive verb (e.g., Bert in Elmo hits Bert) and differently from the agent of a transitive verb (Elmo in Elmo hits Bert). Ergative languages contrast with nominative languages such as English; in English, both the subject of the intransitive verb (Elmo runs home) and the agent of a transitive verb (Elmo hits Bert) are placed before the verb, whereas the patient of a transitive verb is placed after the verb (Elmo hits Bert).(Susan Goldin-Meadow, Language Acquisition Theories. Language, Memory, and Cognition in Infancy and Early Childhood, ed. by Janette B. Benson and Marshall M. Haith. Academic Press, 2009) Example Sentences In English, for example, the grammar in the two sentences Helen opened the door and The door opened is quite different, though the agency of the event might be thought of as being the same. A language with an ergative case would articulate these relationships very differently. Examples of ergative languages include Basque, Inuit, Kurdish, Tagalog, Tibetan and many native Australian languages like Dyirbal.(Robert Lawrence Trask and Peter Stockwell, Language and Linguistics: The Key Concepts, 2nd ed. Routledge, 2007) From Diversity and Stability and Language [E]rgativity is a recessive feature (Nichols 1993), that is, a feature which is almost always lost by at least some daughter languages in a family and is not readily borrowed in contact situations. Thus, though not always inherited, when found in a language it is more likely to have been inherited than borrowed. Therefore, ergativity can be an important component of the grammatical signature of a language family: not every daughter language has it, but its mere presence in several or most languages of the family helps characterize the family and identify languages belonging to the family.(Johanna Nichols, Diversity and Stability in Language. The Handbook of Historical Linguistics, ed. by Brian D. Joseph and Richard D. Janda. Blackwell, 2003) Pronunciation: ER-ge-tiv