Explore Related Concepts


roster method set
Best Results From Wikipedia Yahoo Answers Youtube
From Wikipedia
In optimization, a problem is defined using an objective function to minimize or maximize, and a set of constraints
 g_1(x)\ge 0, \dots, g_k(x)\ge 0
that define the feasible region, that is, the set of all x to search for the optimal solution. Given a point x in the feasible region, a constraint
 g_i(x) \ge 0
is called active at x if g_i(x)=0 and inactive at x if g_i(x)>0. Equality constraints are always active. The active set at x is made up of those constraints g_i(x) that are active at the current point .
The active set is particularly important in optimization theory as it determines which constraints will influence the final result of optimization. For example, in solving the linear programming problem, the active set gives the hyperplanes that intersect at the solution point. In quadratic programming, as the solution is not mandatorily on one of the edges of the bounding polygon, an estimation of the active set gives us a subset of inequalities to watch while searching the solution, which reduces the complexity of the search.
In general an active set algorithm has the following structure:
 Find a feasible starting point
 repeat until "optimal enough"
 solve the equality problem defined by the active set (approximately)
 compute the Lagrange multipliers of the active set
 remove a subset of the constraints with negative Lagrange multipliers
 search for infeasible constraints
 end repeat
In optical fiber technology, the substitution method is a method of measuring the transmission loss of a fiber. It consists of:
 using a stable optical source, at the wavelength of interest, to drive a mode scrambler, the output of which overfills (drives) a 1 to 2 meter long reference fiber having physical and optical characteristics matching those of the fiber under test,
 measuring the power level at the output of the reference fiber,
 repeating the procedure, substituting the fiber under test for the reference fiber, and
 subtracting the power level obtained at the output of the fiber under test from the power level obtained at the output of the reference fiber, to get the transmission loss of the fiber under test.
The substitution method has certain shortcomings with regard to its accuracy, but its simplicity makes it a popular field test method. It is conservative, in that if it were used to measure the individual losses of several long fibers, and the long fibers were concatenated, the total loss obtained (excluding splice losses) would be expected to be lower than the sum of the individual fiber losses.
Some modern optical power meters have the capability to set to zero the reference level measured at the output of the reference fiber, so that the transmission loss of the fiber under test may be read out directly.
In computational complexity theory, the potential method is a method used to analyze the amortized time and space complexity of an algorithm. It can be thought of as a generalization of the accounting method and the debit method. It is useful in cases where it is hard to assign credit to specific elements of the structure.
The method
The potential function \phi is set to be a nonnegativevalued function from states of the data structure in question. If c_i represents the actual cost of the ith operation, which updates the data structure from state A_{i1} to state A_i, then the effective cost d_i of this operation is defined to be d_i = c_i + \phi(A_i)  \phi(A_{i1}) (i.e., the effective cost is the actual cost plus the difference in potential).
If \phi is chosen so that the starting state of the data structure has potential 0, then the sum of the effective costs d_1,\dots,d_n is greater than or equal to the sum of the actual costs c_1,\dots,c_n. So if an upper bound on the effective cost of each operation can be shown, this implies an upper bound on the total cost of n operations, which gives an upper bound on the amortized cost per operation.
Sample applications
Table expansion
Determining the amortized cost of table expansion can be solved using the potential method. Let the potential function be the MAX of (the number of occupied cells minus the number of vacant cells, or 0). A regular insert incurs O(1) cost and increases the potential by O(1); thus, the effective cost of a regular insert is O(1). An insert that causes reallocation incurs an actual cost of O(n) but decreases the potential by O(n) (i.e \phi(A_i)  \phi(A_{i1}) =  O(n) ), giving an effective cost of O(1) for this operation. Since each type of operation has O(1) effective cost, this implies an amortized cost of O(1) as well.
In the field of analysis of algorithms in computer science, the accounting method is a method of amortized analysis based on accounting. The accounting method often gives a more intuitive account of the amortized cost of an operation than either aggregate analysis or the potential method. Note, however, that this does not guarantee such analysis will be immediately obvious; often, choosing the correct parameters for the accounting method requires as much knowledge of the problem and the complexity bounds one is attempting to prove as the other two methods.
The accounting method is most naturally suited for proving a O(1) bound on time. The method as explained here is for proving such a bound.
The method
Preliminarily, we choose a set of elementary operations which will be used in the algorithm, and arbitrarily set their cost to 1. The fact that the costs of these operations may in reality differ presents no difficulty in principle. What is important, is that each elementary operation has a constant cost.
Each aggregate operation is assigned a "payment". The payment is intended to cover the cost of elementary operations needed to complete this particular operation, with some of the payment left over, placed in a pool to be used later.
The difficulty with problems that require amortized analysis is that, in general, some of the operations will require greater than constant cost. This means that no constant payment will be enough to cover the worst case cost of an operation, in and of itself. With proper selection of payment, however, this is no longer a difficulty; the expensive operations will only occur when there is sufficient payment in the pool to cover their costs.
Examples
A few examples will help to illustrate the use of the accounting method.
Table expansion
It is often necessary to create a table before it is known how much space is needed. One possible strategy is to double the size of the table when it is full. Here we will use the accounting method to show that the amortized cost of an insertion operation in such a table is O(1).
Before looking at the procedure in detail, we need some definitions. Let T be a table, E an element to insert, num(T) the number of elements in T, and size(T) the allocated size of T. We assume the existence of operations create_table(n), which creates an empty table of size n, for now assumed to be free, and elementary_insert(T,E), which inserts element E into a table T that already has space allocated, with a cost of 1.
The following pseudocode illustrates the table insertion procedure: function table_insert(T,E) if num(T) = size(T) U := create_table(2 × size(T)) for each F in T elementary_insert(U,F) T := U elementary_insert(T,E)
Without amortized analysis, the best bound we can show for n insert operations is O(n^{2}) — this is due to the loop at line 4 that performs num(T) elementary insertions.
For analysis using the accounting method, we assign a payment of 3 to each table insertion. Although the reason for this is not clear now, it will become clear during the course of the analysis.
Assume that initially the table is empty with size(T) = m. The first m insertions therefore do not require reallocation and only have cost 1 (for the elementary insert). Therefore, when num(T) = m, the pool has (3  1)×m = 2m.
Inserting element m + 1 requires reallocation of the table. Creating the new table on line 3 is free (for now). The loop on line 4 requires m elementary insertions, for a cost of m. Including the insertion on the last line, the total cost for this operation is m + 1. After this operation, the pool therefore has 2m + 3  (m + 1) = m + 2.
Next, we add another m  1 elements to the table. At this point the pool has m + 2 + 2×(m  1) = 3m. Inserting an additional element (that is, element 2m + 1) can be seen to have cost 2m + 1 and a payment of 3. After this operation, the pool has 3m + 3  (2m + 1) = m + 2. Note that this is the same amount as after inserting element m + 1. In fact, we can show that this will be the case for any number of reallocations.
It can now be made clear why the payment for an insertion is 3. 1 goes to inserting the element the first time it is added to the table, 1 goes to moving it the next time the table is expanded, and 1 goes to moving one of the elements that was already in the table the next time the table is expanded.
We initially assumed that creating a table was free. In reality, creating a table of size n may be as expensive as O(n). Let us say that the cost of creating a table of size n is n. Does this new cost present a difficulty? Not really; it turns out we use the same method to show the amortized O(1) bounds. All we have to do is change the payment.
When a new table is created, there is an old table with m entries. The new table will be of size 2m. As long as the entries currently in the table have added enough to the pool to pay for creating the new table, we will be all right.
We cannot expect the first \frac{m}{2} entries to help pay for the new table. Those entries already paid for the current table. We must then rely on the last \frac{m}{2} entries to pay the cost 2m. This means we must add \frac{2m}{m/2} = 4 to the payment for each entry, for a total payment of 3 + 4 = 7.
From Yahoo Answers
Answers:In roster form, the elements are listed out as such. In rule method, the rule for building the set is specified 1) roster form : {1,2,3,4} rule method { x : x belongs to natural numbers and 3 < x < 5 } 2) roster form : { 1,2,3,4,6,9,12,18 } rule method {x : x/6 belongs to N and 5 < x <= 18}
Answers:Start with different sets of numbers. The natural numbers are the set of positive whole numbers. The symbol for the set of natural numbers is . The integers are the set of whole numbers (so that's the natural numbers and the negative whole numbers). We write . The rational numbers are the set of numbers that can be written as a fraction (NOTE: this includes the integers as all numbers can be written as being over 1.) So these numbers are the integers and all the fractions. We write . The real numbers are all the numbers. So this is the rational numbers, but also all the numbers in between. The real numbers consist of the whole number line. We write . The above sets of numbers are special sets, so they have their own symbols ( , , , ) but you can write any collection of numbers as a set. We always write sets in curly brackets, { }. We could just as easily write the set of natural numbers: = {1, 2, 3, ...} where the dots show that it goes up in the same manner (in this case ones) infinitely. But for your example above, numbers 3 through to 10, we would write {3, 4, 5, 6, 7, 8, 9, 10} or {3, 4, 5, ..., 10}. The numbers between 1 and 3 would be {0, 1, 2} assuming that we aren't including the 1 and the 3. Now, the real numbers without 0 would be \{0} where the \ sign means "without". So it means the real numbers without the set {0}. Other things you may come across are union, and intersection, you use the union sign to "add" sets together. So if you wanted the natural numbers and 5, for example, you could write {5}. or {1, 2, 3, ...} {5} the intersection of 2 sets is the stuff in both sets so if you had the following 2 sets: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} then {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} = {2, 4, 6, 8, 10}. now we can have a go at negative even numbers. We know that an even number is divisible by 2. So any even number can be written 2n where n is a whole number, or an integer in fact. First we find the set of even numbers {2n : n } where the means that n is an element of , the integers. Then we can find the set of positive even numbers: {2n : n } so the set of negative even numbers is {2n : n } Inequalities: < means less than, for example: 3 < 5 > means greater than, for example: 5 > 3 means less than or equal to. So 3 5, but 5 5 as well. means greater than or equal to. So 5 3, but 5 5 as well. When using equalities we can solve equations, for example: 2x + 4 = 10 then 2x = 6 x = 3 we can solve inequalities in the same way: 2x + 4 < 10 2x < 6 x < 3 the only difference being that if you divide by a negative number then you have to swap the inequality round: 2x + 5 < 3 2x < 2 x > 1 this actaully makes sense though, because when you get to 2x < 2 then instead of dividing by 2, you can add "2x + 2" to each side: 2 < 2x then divide by 2: 1 < x so x > 1. The symbol means "infinity" but I can't think what to tell you about that at the moment! If you haven't seen this stuff before, then it's a lot to take in and learn. I'm sorry that your teacher is so unhelpful, but maybe you can find a nicer teacher and explain your problem to them?
Answers:In roster form, you actually list all the elements. A = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z} is in roster form. The rule for this set is that A is the letters of the alphabet.
Answers:a collection of a distinct object considered as whole. describing a set. 3 methods. a. by word b. by listing or roster method c. by setbuilder rotation or rule defining method. hope those describing methods help. i was looking for other methods. need more learning. haha!
From Youtube