text:
I hit the hotel programs that matter to me and the only airline program I know that allows consolidation is Hawaiian (mentioned in that post over there <—) … but that's ok, we can still consolidate / leverage Bank Rewards points … which is where flexibility is king. Transfer to hotels, airlines, other people, buy Amazon.com gift certificates, or book travel directly and EARN more points … you name it you've got options. When it comes to frugal travel, last minute travel, family travel … flexibility is really what we want. If we can be flexible in our travel plans we can save a ton, and when we can't be flexible having points that can flex for us get us a LOT further.
To some degree we've already covered the big three (with key bullets below each one):
- American Express Membership Rewards
- Lots of points transfer partners
- Cannot transfer Membership Rewards points between accounts
- Can transfer points to hotel / airline partners in accounts of other people's names (so I can transfer from my MR account to my friend's Delta account)
- Chase Ultimate Rewards
- Not as many transfer partners as MR, but has good options and coverage across all three major airline alliances
- Can consolidate points between family members (nothing actually stopping transfers to others … there is a disclaimer however saying don't do it)
- Are not supposed to transfer points to hotel / airline partners in accounts of other people's names … but it works. Again, disclaimer text
- Citi ThankYou Points
- Most valuable "cash" point. All of the bank rewards points allow you to use them as cash … and for travel if you've got a Citi ThankYou Premier card you'll get 1.33 cents/point … that said, this is really about as good as it gets with ThankYou points.
- Hilton is the only transfer partner. One TYP gets you 1.5 Hilton Honors points. You'd do better converting some Hawaiian miles at a rate of 1:2 Hilton points.
- Can consolidate points between any other account (via phone call) but transferred points expire 90 days after transfer. When using points, the points that expire first are used first (which is nice here)
- U.S. Bank offers are tough to come by … and getting approved can be even harder, of course YMMV.
- Capital One doesn't have a real points program. Their "miles" and "points" are effectively cash value towards travel. Their cards are cash back cards that are marketed as "points and miles". All good and well. You'll earn 1-2% "cash back" on their cards. If you're the kind of person that keeps all your money in a savings account and never in the stock market or even a mutual fund then this is a good program for you. Its easy, and straightforward. You'll never fly in business or first class, and certainly not out of the country in either one of those.
Deal of the Day
For today's Deal of the Day TopCashback is donating $50 cash (to be deposited to your TopCashback.com account) for the best overall travel deal submitted. A great mileage run, mistake fare, partner promo, new card sign-up offer, or the special twist you figured out on a deal today. Add it to the comments section of this post (along with your First Name, Last Initial, and Airport Code) or this afternoon's Deal of the Day post by TODAY at 11:59pm ET to be eligible to win. That gives you two days to win!
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Woodworking Tools and Woodworking Machinery Article Archives
Wanna be lean? Just eliminate waste
Written by John English Monday, 17 November 2014 00:00
Waste has two meanings for a woodshop. It's the accumulated debris created by the manufacturing process (that is, the garbage), which includes sawdust, chips and cutoffs. But it's also a term used to describe the percentage of raw materials that don't get used in production — that is, the amount of money being spent on hardwoods, sheet stock, finishes and hardware that don't get used right away and sometimes even becomes the first kind of waste. For example, finishes with a shelf life often end up being discarded, as does dated hardware.
Bigger, better batteries are key to our future
Written by John English Sunday, 19 October 2014 00:00
As you read this, there are thousands of guys in North Dakota living in "man camps" that are eerily reminiscent of the 1930s. Back in 2000, Williston and Watford City were remote little hamlets on the windswept prairie, far removed from mainstream America. The biggest news in town was a stray steer on the highway or the librarian's new haircut. Now, thanks to something called fracking, these have become everything one imagines about boomtowns. They are at the center of the Bakken oilfield.
Cabinet Vision displays new 3-D software
Written by Jennifer Hicks Monday, 20 October 2014 00:00
Cabinet Vision introduced the Vortek Experience at IWF 2014, a software program that allows kitchen and bath manufacturers to display their products in an immersive 3-D visualization platform.
Steel City Tool Works trims product line
Written by Jennifer Hicks Monday, 20 October 2014 00:00
Leaders from Steel City Tool Works made several announcements about the machinery manufacturer's new direction at IWF 2014 in Atlanta, starting with the addition of Todd Damon, co-owner of Wood Werks Supply in Columbus, Ohio, as a managing partner.
DeWalt expands its storage solution lineup
Written by Jennifer Hicks Monday, 20 October 2014 00:00
DeWalt recently introduced a new rolling workshop along with several storage solutions to its existing lineup of related products to help woodworkers and other tradespeople stay organized while on or away from the job.
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2.3 Recursion
The idea of calling one function from another immediately suggests the possibility of a function calling itself. The function-call mechanism in Java supports this possibility, which is known as recursion. Recursion is a powerful general-purpose programming technique, and is the key to numerous critically important computational applications, ranging from combinatorial search and sorting methods methods that provide basic support for information processing (Chapter 4) to the Fast Fourier Transform for signal processing (Chapter 9).
Your first recursive program.The HelloWorld for recursion is to implement the factorial function, which is defined for positive integers N by the equation
N! = N × (N-1) × (N-2) × ... × 2 × 1N! is easy to compute with a for loop, but an even easier method in Factorial.java is to use the following recursive function:
You can persuade yourself that it produces the desired result by noting that factorial() returns 1 = 1! when N is 1 and that if it properly computes the value
public static int factorial(int N) { if (N == 1) return 1; return N * factorial(N-1); }
(N-1)! = (N-1) × (N-2) × ... × 2 × 1then it properly computes the value
N! = N × (N-1)! = N × (N-1) × (N-2) × ... × 2 × 1We can trace this computation in the same way that we trace any sequence of function calls.
factorial(5) factorial(4) factorial(3) factorial(2) factorial(1) return 1 return 2*1 = 2 return 3*2 = 6 return 4*6 = 24 return 5*24 = 120
Our factorial() implementation exhibits the two main components that are required for every recursive function.
- The base case returns a value without making any subsequent recursive calls. It does this for one or more special input values for which the function can be evaluated without recursion. For factorial(), the base case is N = 1.
- The reduction step is the central part of a recursive function. It relates the function at one (or more) inputs to the function evaluated at one (or more) other inputs. Furthermore, the sequence of parameter values must converge to the base case. For factorial(), the reduction step is N * factorial(N-1) and N decreases by one for each call, so the sequence of parameter values converges to the base case of N = 1.
Mathematical induction.Recursive programming is directly related to mathematical induction, a technique for proving facts about discrete functions. Proving that a statement involving an integer N is true for infinitely many values of N by mathematical induction involves two steps.
- The base case is to prove the statement true for some specific value or values of N (usually 0 or 1).
- The induction step is the central part of the proof. For example, we typically assume that a statement is true for all positive integers less than N, then use that fact to prove it true for N.
Euclid's algorithm.The greatest common divisor (gcd) of two positive integers is the largest integer that divides evenly into both of them. For example, the greatest common divisor of 102 and 68 is 34 since both 102 and 68 are multiples of 34, but no integer larger than 34 divides evenly into 102 and 68.
We can efficiently compute the gcd using the following property, which holds for positive integers p and q:
If p > q, the gcd of p and q is the same as the gcd of q and p % q.
The static method gcd() in Euclid.java is a compact recursive function whose reduction step is based on this property.
gcd(1440, 408) gcd(408, 216) gcd(216, 192) gcd(192, 24) gcd(24, 0) return 24 return 24 return 24 return 24 return 24
Towers of Hanoi.No discussion of recursion would be complete without the ancient towers of Hanoi problem. We have three poles and n discs that fit onto the poles. The discs differ in size and are initially arranged on one of the poles, in order from largest (disc n) at the bottom to smallest (disc 1) at the top. The task is to move the stack of discs to another pole, while obeying the following rules:
- Move only one disc at a time.
- Never place a disc on a smaller one.
To solve the problem, our goal is to issue a sequence of instructions for moving the discs. We assume that the poles are arranged in a row, and that each instruction to move a disc specifies its number and whether to move it left or right. If a disc is on the left pole, an instruction to move left means to wrap to the right pole; if a disc is on the right pole, an instruction to move right means to wrap to the left pole.
Exponential time.One legend says that the world will end when a certain group of monks solve the Towers of Hanoi problem in a temple with 64 golden discs on three diamond needles. We can estimate the amount of time until the end of the world (assuming that the legend is true). If we define the function T(n) to be the number of move directives issued by TowersOfHanoi.java to move n discs from one peg to another, then the recursive code implies that T(n) must satisfy the following equation:
T(n) = 2 T(n - 1) + 1 for n > 1, with T(1) = 1
Such an equation is known in discrete mathematics as a recurrence relation. We can often use them to derive a closed-form expression for the quantity of interest. For example, T(1) = 1, T(2) = 3, T(3) = 7, and T(4) = 15. In general, T(n) = 2^n - 1.
Knowing the value of T(n), we can estimate the amount of time required to perform all the moves. If the monks move discs at the rate of one per second, it would take more than one week for them to finish a 20-disc problem, more than 31 years to finish a 30-disc problem, and more than 348 centuries for them to finish a 40-disc problem (assuming that they do not make a mistake). The 64-disc problem would take more than 5.8 billion centuries.
Gray code.The playwright Samuel Beckett wrote a play called Quad that had the following property: Starting with an empty stage, characters enter and exit one at a time, but each subset of characters on the stage appears exactly once. The play had four characters and there are 2^4 = 16 different subsets of four elements; hence the title. How did Beckett generate the stage directions for this play? How would we do it for 5 actors, or more?
An n-bit Gray code is a list of the 2^n different n-bit binary numbers such that each entry in the list differs in precisely one bit from its predecessor. Gray codes directly apply to Beckett's problem because we can interpret each bit as denoting whether the integer corresponding to its bit position is in the subset. Changing the value of a bit from 0 to 1 corresponds to an integer entering the subset; changing a bit from 1 to 0 corresponds to and integer exiting the subset.
How do we generate a Gray code? A recursive plan that is very similar to the one that we used for the towers of Hanoi problem is effective. The n bit binary reflected Gray code is defined recursively as follows:
- the n-1 bit code, with 0 prepended to each word, followed by
- the n-1 bit code in reverse order, with 1 prepended to each word.
Recursive graphics.Simple recursive drawing schemes can lead to pictures that are remarkably intricate. For example, an H-tree of order n is defined as follows: The base case is null for n = 0. The reduction step is to draw, within the unit square three lines in the shape of the letter H four H-trees of order n-1, one connected to each tip of the H with the additional provisos that the H-trees of order n-1 are centered in the four quadrants of the square, halved in size. Program Htree.java takes a command-line argument n, and plots an order n H-tree using standard draw.
Brownian bridge.An H-tree is a simple example of a fractal: a geometric shape that can be divided into parts, each of which is (approximately) a reduced size copy of the original. The study of fractals plays an important and lasting role in artistic expression, economic analysis, and scientific discovery. Artists and scientists use them to build compact models of complex shapes that arise in nature and resist description using conventional geometry, such as clouds, plants, mountains, riverbeds, human skin, and many others. Economists also use fractals to model function graphs of economic indicators.
Program Brownian.java produces a function graph that approximates a simple example known as a Brownian bridge and closely related functions. You can think of this graph as a random walk that connects two points, from (x0, y0) to (x1, y1), controlled by a few parameters. The implementation is based on the midpoint displacement method, which is a recursive plan for drawing the plot within an interva [x0, x1]. The base case (when the size of the interval is smaller than a given tolerance) is to draw a straight line connecting the two endpoints. The reduction case is to divide the interval into two halves, proceeding as follows:
- Compute the midpoint (xm, ym) of the interval.
- Add to the y-coordinate of the midpoint a random value δ, chosen from the Gaussian distribution with mean 0 and given variance.
- Recur on the subintervals, dividing the variance by a given scaling factor s.
The shape of the curve is controlled by two parameters: the volatility (initial value of the variance) controls the distance the graph strays from the straight line connecting the points, and the Hurst exponent controls the smoothness of the curve. We denote the Hurst exponent by H and divide the variance by 2^(2H) at each recursive level. When H is 1/2 (divide by 2 at each level) the standard deviation is constant throughout the curve: in this situation, the curve is a Brownian bridge.
Pitfalls of recursion.With recursion, you can write compact and elegant programs that fail spectacularly at runtime.
- Missing base case.
The recursive function in
is supposed to compute Harmonic numbers, but is missing a base case:
public static double harmonic(int n) { return harmonic(n-1) + 1.0/n; }
- No guarantee of convergence.
Another common problem is to include within a recursive function a
recursive call to solve a subproblem that is not smaller.
The recursive function in
will go into an infinite recursive loop if it is invoked with an argument
N having any value other than 1:
public static double H(int N) { if (N == 1) return 1.0; return H(N) + 1.0/N; }
- Excessive space requirements.
Java needs to keep track of each recursive
call to implement the function abstraction as expected. If a function calls
itself recursively an excessive number of times before returning, the space
required by Java for this task may be prohibitive.
The recursive function in
correctly computes the nth harmonic number. However, we cannot use it
for large n
because the recursive depth is proportional to N, and this
creates a StackOverflowError for N = 5,000.
public static double harmonic(int n) { if (n == 0) return 0.0; return harmonic(n-1) + 1.0/n; }
- Excessive recomputation.
The temptation to write a simple recursive
program to solve a problem must always be tempered by the
understanding that a simple program might require exponential
time (unnecessarily), due to excessive recomputation.
For example, the Fibonacci sequence
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 ...is defined by the formula Fn = Fn-1 + Fn-2 for n ≥ 2 with F0 = 0 and F1 = 1. A novice programmer might implement this recursive function to compute numbers in the Fibonacci sequence, as in Fibonacci.java:
public static long F(int n) { if (n == 0) return 0; if (n == 1) return 1; return F(n-1) + F(n-2); }
Q. Are there situations when iteration is the only option available to address a problem?
A. No, any loop can be replaced by a recursive function, though the recursive version might require excessive memory.
Q. Are there situations when recursion is the only option available to address a problem?
A. No, any recursive function can be replaced by an iterative counterpart. In Section 4.3, we will see how compilers produce code for function calls by using a data structure called a stack.
Q. Which should I prefer, recursion or iteration?
A. Whichever leads to the simpler, more easily understood, or more efficient code.
Q. I get the concern about excessive space and excessive recomputation in recursive code. Anything else to be concerned about?
A. Be extremely wary of creating arrays in recursive code. The amount of space used can pile up very quickly, as can the amount of time required for memory management.
- What happens if you run factorial() with negative value of n? With a large value, say 35?
- Write a recursive program that computes the value of ln(N!).
Give the sequence of integers printed by a call to ex233(6):
public static void ex233(int n) { if (n <= 0) return; StdOut.println(n); ex233(n-2); ex233(n-3); StdOut.println(n); }
Give the value of ex234(6):
public static String ex234(int n) { if (n <= 0) return ""; return ex234(n-3) + n + ex234(n-2) + n; }
Criticize the folowing recursive function:
public static String ex235(int n) { String s = ex233(n-3) + n + ex235(n-2) + n; if (n <= 0) return ""; return s; }
- Given four positive integers a, b, c, and d, explain what value is computed by gcd(gcd(a, b), gcd(c, d)).
Explain in terms of integers and divisors the effect of the
following Euclid-like function.
public static boolean gcdlike(int p, int q) { if (q == 0) return (p == 1); return gcdlike(q, p % q); }
Solution. Returns whether p and q are relatively prime.
Consider the following recursive function.
public static int mystery(int a, int b) { if (b == 0) return 0; if (b % 2 == 0) return mystery(a+a, b/2); return mystery(a+a, b/2) + a; }
What are the values of mystery(2, 25) and mystery(3, 11)? Given positive integers a and b, describe what value mystery(a, b) computes. Answer the same question, but replace + with * and replace return 0 with return 1.
Solution. 50 and 33. It computes a*b. If you replace + with *, it computes a^b.
- Write a recursive program Ruler.java to plot the subdivisions of a ruler using StdDraw as in Program 1.2.1.
Solve the following recurrence relations, all with T(1) = 1. Assume N
is a power of two.
- Prove by induction that the minimum possible number of moves needed to solve the towers of Hanoi satisfies the same recurrence as the number of moves used by our recursive solution.
- Prove by induction that the recursive program given in the text makes exactly F(n) recursive calls to F(1) when computing F(N).
- Prove that the second argument to gcd() decreases by at least a factor of two for every second recursive call, then prove that gcd(p, q) uses at most log2 N recursive calls, where N is the larger of p and q.
Write a program AnimatedHtree.java that
animates the drawing of the H-tree.
Creative Exercises
- Binary representation. Write a program IntegerToBinary.java that takes a positive integer N (in decimal) from the command line and prints out its binary representation. Recall, in program 1.3.7, we used the method of subtracting out powers of 2. Instead, use the following simpler method: repeatedly divide 2 into N and read the remainders backwards. First, write a while loop to carry out this computation and print the bits in the wrong order. Then, use recursion to print the bits in the correct order.
- A4 paper. The width-to-height ratio of paper in the ISO format is the square root of 2 to 1. Format A0 has an area of 1 square meter. Format A1 is A0 cut with a vertical line into two equal halves, A2 is A1 cut with a horizontal line into in two halves, and so on. Write a program that takes a command-line argument n and uses StdDraw to show how to cut a sheet of A0 paper into 2^n pieces. Here's a nice illustration of A size formats.
- Permutations.
Write a program Permutations.java
that take a command-line argument n and prints out all n! permutations
of the n letters starting at a (assume that n is no greater
than 26).
A permutation of n elements is one of the n! possible
orderings of the elements.
As an example, when n = 3 you should get the following output.
Do not worry about the order in which you enumerate them.
bca cba cab acb bac abc
- Permutations of size k.
Modify Permutations.java to
PermutationsK.java so that
it takes two command-line arguments n and k, and prints out all
P(n, k) = n! / (n-k)! permutations that contain exactly k of the n
elements. Below is the desired output when k = 2 and n = 4.
You need not print them out in any particular order.
ab ac ad ba bc bd ca cb cd da db dc
- Combinations.
Write a program Combinations.java
that takes one command-line argument n and prints out all 2^n
combinations of any size. A combination is a subset of the
n elements, independent of order.
As an example, when n = 3 you should get the following output.
a ab abc ac b bc c
- Combinations of size k.
Modify Combinations.java to
CombinationsK.java so that
it takes two command-line arguments n and k, and prints out all
C(n, k) = n! / (k! * (n-k)!) combinations of size k.
For example, when n = 5 and k = 3 you should get the following output.
abc abd abe acd ace ade bcd bce bde cde
- Hamming distance.
The Hamming distance between two bit strings of length n is equal
to the number of bits in which the two strings differ.
Write a program that reads in an integer k and a bit string s
from the command line, and prints out all bit strings that
have Hamming distance at most k from s. For example if
k is 2 and s is 0000 then your program should print out
0011 0101 0110 1001 1010 1100
- Recursive squares.
Write a program to produce each of the following recursive patterns.
The ratio of the sizes of the squares is 2.2.
To draw a shaded square, draw a filled gray square, then an
unfilled black square.
RecursiveSquares.java gives a solution to part a.
- Pancake flipping. You have a stack of N pancakes of varying sizes on a griddle. Your goal is to re-arrange the stack in descending order so that the largest pancake is on the bottom and the smallest one is on top. You are only permitted to flip the top k pancakes, thereby reversing their order. Devise a scheme to arrange the pancakes in the proper order by using at most 2N - 3 flips. Hint: you can try out strategies here.
- Gray code.
Modify Beckett.java to print out the
Gray code (not just the sequence of bit positions that change).
Solution. GrayCode.java uses Java's string data type; GrayCodeArray.java uses a boolean array.
- Towers of Hanoi variant. Consider the following variant of the towers of Hanoi problem. There are 2N discs of increasing size stored on three poles. Initially all of the discs with odd size (1, 3, ..., 2N-1) are piled on the left pole from top to bottom in increasing order of size; all of the discs with even size (2, 4, ..., 2N) are piled on the right pole. Write a program to provide instructions for moving the odd discs to the right pole and the even discs to the left pole, obeying the same rules as for towers of Hanoi.
- Animated towers of Hanoi animation. Write a program AnimatedHanoi.java that uses StdDraw to animate a solution to the towers of Hanoi problem, moving the discs at a rate of approximately 1 per second.
- Sierpinski triangles.
Write a recursive program to draw the Sierpinski gasket.
As with Htree, use a command-line argument to control the
depth of recursion.
- Binomial distribution.
Estimate the number of recursive calls that would be used
by the code
public static double binomial(int N, int k) { if ((N == 0) || (k < 0)) return 1.0; return (binomial(N-1, k) + binomial(N-1, k-1)) / 2.0; }
- Collatz function.
Consider the following recursive function in
Collatz.java, which is related
to a famous unsolved problem in number theory, known as the
Collatz problem or the 3n + 1 problem.
public static void collatz(int n) { StdOut.print(n + " "); if (n == 1) return; if (n % 2 == 0) collatz(n / 2); else collatz(3*n + 1); }
For example, a call to collatz(7) prints the sequence
7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1as a consequence of 17 recursive calls. Write a program that takes a command-line argument N and returns the value of n < N for which the number of recursive calls for collatz(n) is maximized. Hint: use memoization. The unsolved problem is that no one knows whether the function terminates for all positive values of n (mathematical induction is no help because one of the recursive calls is for a larger value of the argument).
- Brownian island. Benoit Mandelbrot asked the famous question
How long is the coast of Britain?
Modify Brownian.java to get a program
that plots
Brownian islands,
whose coastlines resemble that of Great Britain. The modifications are simple:
first, change curve() to add a gaussian to the x coordinate as well
as to the y coordinate;
second, change main() to draw a curve from the point at the center of the
canvas back to itself. Experiment with various values of the arguments to
get your program to produce islands with a realistic look.
- Plasma clouds. Write a recursive program Program PlasmaCloud.java to draw plasma clouds, using the method suggested in the text. Remark: this technique can be used to produce synthetic terrain, by interpreting the color value as the altitude. Variants of this scheme are widely used in the entertainment industry to generate artificial background scenery for movies and the "Genesis effect" in Star Trek II and the "outline of the "Death Star" in Return of the Jedi.
- A strange function.
Consider McCarthy's 91 function:
public static int mcCarthy(int n) { if (n > 100) return n - 10; else return mcCarthy(mcCarthy(n+11)); }
Determine the value of mcCarthy(50) without using a computer. Give the number of recursive calls used by mcCarthy() to compute this result. Prove that the base case is reached for all positive integers n or give a value of n for which this function goes into a recursive loop.
- Recursive tree.
Write a program Tree.java
that takes a command-line arguemnt N and produces the following
recurisve patterns for N equal to 1, 2, 3, 4, and 8.
Web Exercises
- Does Euclid.java still work if the inputs can be negative? If not, fix it. Hint: Recall that % can return a negative integer if the first input is negative. When calling the function, take the absolute value of both inputs.
Write a recursive
program GoldenRatio.java that
takes an integer input N and computes
an approximation to the
golden ratio
using the following recursive formula:
f(N) = 1 if N = 0 = 1 + 1 / f(N-1) if N > 0
- Discover a connection between the golden ratio and Fibonacci numbers. Hint: consider the ratio of successive Fibonacci numbers: 2/1, 3/2, 8/5, 13/8, 21/13, 34/21, 55/34, 89/55, 144/89, ...
Consider the following recursive function. What is
mystery(1, 7)?
public static int mystery(int a, int b) { if (0 == b) return 0; else return a + mystery(a, b-1); }
Will the function in the previous exercise terminate for every pair of integers a and b between between 0 and 100? Give a high level description of what mystery(a, b) returns, given integers a and b between 0 and 100.
Answer: mystery(1, 7) = 1 + mystery(1, 6) = 1 + (1 + mystery(1, 5)) = ... 7 + mystery(1, 0) = 7.
Answer: Yes, the base case is b = 0. Successive recursive calls reduce b by 1, driving it toward the base case. The function mystery(a, b) returns a * b. Mathematically inclined students can prove this fact via induction using the identity ab = a + a(b-1).
Consider the following function. What does mystery(0, 8) do?
public static void mystery(int a, int b) { if (a != b) { int m = (a + b) / 2; mystery(a, m); StdOut.println(m); mystery(m, b); } }
Consider the following function. What does mystery(0, 8) do?
public static void mystery(int a, int b) { if (a != b) { int m = (a + b) / 2; mystery(a, m - 1); StdOut.println(m); mystery(m + 1, b); } }
- Repeat the previous exercise, but replace if (a != b) with if (a <= b).
What does mystery(0, 8) do?
public static int mystery(int a, int b) { if (a == b) StdOut.println(a); else { int m1 = (a + b ) / 2; int m2 = (a + b + 1) / 2; mystery(a, m1); mystery(m2, b); } }
What does the following function compute?
public static int f(int n) { if (n == 0) return 0; if (n == 1) return 1; if (n == 2) return 1; return 2*f(n-2) + f(n-3);
Write a program Fibonacci2.java
that takes a command-line argument N and prints out the first N
Fibonacci numbers using the following alternate definition:
F(n) = 1 if n = 1 or n = 2 = F((n+1)/2)2 + F((n-1)/2)2 if n is odd = F(n/2 + 1)2 - F(n/2 - 1)2 if n is even
What is the biggest Fibonacci number you can compute in under a minute using this definition? Compare this to Fibonacci.java.
Write a program
that takes a command-line argument N and prints out the first N
Fibonacci numbers using the
method proposed by Dijkstra:
F(0) = 0 F(1) = 1 F(2n-1) = F(n-1)^2 + F(n)^2 F(2n) = (2F(n-1) + F(n)) * F(n)
- Prove by mathematical induction that the alternate definitions of the Fibonacci function given in the previous two exercises are equivalent to the original definition.
- Write a program Pell.java that takes a command-line argument N and prints out the first N Pell numbers: p0 = 0, p1 = 1, and for n >= 2, pn = 2 pn-1 + pn-2. Print out the ratio of successive terms and compare to 1 + sqrt(2).
Consider the following function from program
public static void mystery(int n) { if (n == 0 || n == 1) return; mystery(n-2); StdOut.println(n); mystery(n-1); }
What does mystery(6) print out? Hint: first figure out what mystery(2), mystery(3), and so forth print out.
What would happen in the previous exercise if the base case
was replaced with the following statement?
if (n == 0) return;
Consider the following recursive functions.
public static int square(int n) { if (n == 0) return 0; return square(n-1) + 2*n - 1; } public static int cube(int n) { if (n == 0) return 0; return cube(n-1) + 3*(square(n)) - 3*n + 1; }
What is the value of square(5)? cube(5)? cube(123)?
Consider the following pair of mutually recursive functions.
What does g(g(2)) evaluate to?
public static int f(int n) { public static int g(int n) { if (n == 0) return 0; if (n == 0) return 0; return f(n-1) + g(n-1); return g(n-1) + f(n); } }
- Write program to verify that (for small values of n) the sum of the cubes of the first n Fibonacci numbers F(0)^3 + F(1)^3 + ... + F(n)^3 equals (F(3n+4) + (-1)^n * 6 * f(n-1)) / 10, where F(0) = 1, F(1) = 1, F(2) = 2, and so forth.
- Transformations by increment and unfolding.
Given two integers a ≤ b, write a program
that transforms a into b by a minimum
sequence of increment (add 1) and unfolding (multiply by 2) operations.
For example,
% java Sequence 5 23 23 = ((5 * 2 + 1) * 2 + 1) % java Sequence 11 113 113 = ((((11 + 1) + 1) + 1) * 2 * 2 * 2 + 1)
- Hadamard matrix.
Write a recursive program Hadamard.java that
takes a command-line argument n and plots an N-by-N Hadamard pattern where
N = 2n.
Do not use an array.
A 1-by-1 Hadamard pattern is a single black square.
In general a 2N-by-2N Hadamard pattern is obtained by aligning 4 copies
of the N-by-N pattern in the form of a 2-by-2 grid, and then
inverting the colors of all the squares in the lower right N-by-N copy.
The N-by-N Hadamard H(N) matrix is a boolean matrix with the remarkable
property that any two rows differ in exactly N/2 bits.
This property makes it useful for designing error-correcting codes.
Here are the first few Hadamard matrices.
- 8 queens problem.
In this exercise, you will solve the classic
8-queens problem: place 8
queens on an 8-by-8 chess board so that no two queens are in the same row,
column, or diagonal. There are 8! = 40,320 ways in which no two queens are
placed in the same row or column. Any permutation p[] of the integers
0 to 7 gives such a placement: put queen i in row i, column p[i].
Your program Queens.java should take
an integer command-line argument n and enumerate all solutions to the
n-queens problem by drawing the location of the queens in ASCII like
the two solutions below.
- if q[i] equals q[n]: two queens are placed in the same column
- if q[i] - q[n] equals n - i: two queens are on same major diagonal
- if q[n] - q[i] equals n - i: two queens are on same minor diagonal
- Another 8 queens solver. Program Queens2.java solves the 8 queens problem by implicitly enumeration all n! permutations (instead of the n^n placements). It is based on program Permutations.java.
- Euclid's algorithm and π. The probability that two numbers chosen from a large random set of numbers have no common factors (other than 1) is 6 / π2. Use this idea to estimate π. Robert Matthews use the same idea to estimate π by taken the set of numbers to be a function of the positions of stars in the sky.
- Towers of Hanoi variant II. (Knuth-Graham and Pathashnik) Solve the original Towers of Hanoi problem, but with the extra restriction that you are not allowed to directly transfer a disk from A to C. How many moves does it take to solve a problem with n disks? Hint: move n-1 smallest disks from A to C recursively (without any direct A to C moves), move disk n from A to B, move n-1 smallest disks from C to A (without any direct A to C moves), move disk N from B to C, and move n-1 smallest disks from A to C recursively (without any direct A to C moves).
- Towers of Hanoi variant III. Repeat the previous question but disallow both A to C and C to A moves. That is, each move must involve pole B.
- Towers of Hanoi with 4 pegs. Suppose that you have a fourth peg. What is the least number of moves needed to transfer a stack of 8 disks from the leftmost peg to the rightmost peg? Answer. Finding the shortest such solution in general has remained an open problem for over a hundred years and is known as Reve's puzzle.
Another tricky recursive function.
Consider the following recursive function.
What is f(0)?
public static int f(int x) { if (x > 1000) return x - 4; else return f(f(x+5)); }
- Checking if n is a Fibonacci number. Write a function to check if n is a Fibonacci number. Hint: a positive integer is a Fibonacci number if and only if either (5*n*n + 4) or (5*n*n - 4) is a perfect square.
- Random infix expression generator.
Run RandomExpression.java
with different command-line argument p between 0 and 1.
What do you observe?
public static String expr(double p) { double r = Math.random(); if (r <= 1*p) return "(" + expr(p) + " + " + expr(p) + ")"; if (r <= 2*p) return "(" + expr(p) + " * " + expr(p) + ")"; return "" + (int) (100 * Math.random()); }
- A tricky recurrence.
Define F(n) so that F(0) = 0 and F(n) = n - F(F(n-1)).
What is F(100000000)?
Solution: The answer is related to the Fibonacci sequence and the Zeckendorf representation of a number.
- von Neumann ordinal.
The von Neumann integer i is defined as follows: for i = 0, it is the
empty set; for i > 0, it is the set containing the von Neumann integers 0 to i-1.
0 = {} = {} 1 = {0} = {{}} 2 = {0, 1} = {{}, {{}}} 3 = {0, 1, 2} = {{}, {{}}, {{}, {{}}}}
- Subsequences of a string.
Write a program Subsequence.java
that takes a string command-line argument s and an integer
command-line argument k and prints out all subsequences
of s of length k.
% java Subsequence abcd 3 abc abd acd bcd
- Interleaving two strings.
Given two strings s and t of distinct characters,
print out all (M+N)! / (M! N!) interleavings, where M and N are the
number of characters in the two strings.
For example, if
s = "ab" t = "CD" abCD CabD aCbD CaDb aCDb CDab
- Binary GCD.
Write a program BinaryGCD.java
that finds the greatest common divisor of two positive integers
using the
binary gcd algorithm: gcd(p, q) =
- p if q = 0
- q if p = 0
- 2 * gcd(p/2, q/2) if p and q are even
- gcd(p/2, q) if p is even and q is odd
- gcd(p, q/2) if p is odd and q is even
- gcd((p-q)/2, q) if p and q are odd and p >= q
- gcd(p, (q-p)/2) if p and q are odd and p < q
- Integer partitions.
Write a program Partition.java
that takes a positive integer N as a command-line argument and
prints out all partitions of N.
A partition
of N is a way to write N as a sum of positive integers.
Two sums are considered the same if they only differ in the
order of their constituent summands.
Partitions arise in symmetric polynomials and group
representation theory in mathematics and physics.
% java Partition 4 % java Partition 6 4 6 3 1 5 1 2 2 4 2 2 1 1 4 1 1 1 1 1 1 3 3 3 2 1 3 1 1 1 2 2 2 2 2 1 1 2 1 1 1 1 1 1 1 1 1 1
- Johnson-Trotter permutations.
Write a program JohnsonTrotter.java
that takes an integer command-line argument n and prints all n! permutations
of the integer 0 through n-1 in such a way that consecutive permutations
differ in only one adjacent transposition (similar to way Gray code
iterates over combinations in such a way that consecutive combinations
differ in only one bit).
% java JohnsonTrotter 3 012 (2 1) 021 (1 0) 201 (2 1) 210 (0 1) 120 (1 2) 102 (0 1)
- Permutations in lexicographic order.
Write a program PermutationsLex.java
that take a command-line argument N and prints out all N! permutations
of the integer 0 through N-1 in lexicographic order.
% java PermutationsLex 3 012 021 102 120 201 210
- Derangements. A derangement is a permutation p[] of the integers from 0 to N-1 such that p[i] doesn't equal i for any i. For example there are 9 derangmenets when N = 4: 1032, 1230, 1302, 2031, 2301, 2310, 3012, 3201, 3210. Write a program to count the number of derangements of size N using the following recurrence: d[N] = (N-1) (d[N-1] + d[N-2]), where d[1] = 0, d[2] = 1. The first few terms are 0, 1, 2, 9, 44, 265, 1854, 14833, 133496, and 1334961.
- Tribonacci numbers. The tribonacci numbers are similar to the Fibonacci numbers, except that each term is the sum of the three previous terms in the sequence. The first few terms are 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81. Write a program to compute tribonacci numbers. What is the ratio successive terms? Answer. Root of x^3 - x^2 - x - 1, which is approximately 1.83929.
- Sum of first n Fibonacci numbers. Prove by induction that the sum of the first n Fibonacci numbers F(1) + F(2) + ... + F(N) is F(N+2) - 1.
- Combinational Gray Code. Print out all combination of k of n items in such a way that consecutive combinations differ in exactly one element, e.g., if k = 3 and n = 5, 123, 134, 234, 124, 145, 245, 345, 135, 235, 125. Hint: use the Gray code, but only print out those integers with exactly k 1's in their binary representation.
- Maze generation. Create a maze using divide-and-conquer: Begin with a rectangular region with no walls. Choose a random gridpoint in the rectangle and construct two perpendicular walls, dividing the square into 4 subregions. Choose 3 of the four regions at random and open a one cell hole at a random point in each of the 3. Recur until each subregion has width or height 1.
- Plasma clouds.
Program PlasmaCloud.java
takes a command-line argument N and produces a random
N-by-N plasma fractal using the midpoint displacement method.
Here's an 800-by-800 example. Here's a reference, including a simple 1D version. Note: some visual artifacts are noticeable parallel to the x and y axes. Doesn't have all of the statistical properties of 2D fractional Brownian motion.
- Fern fractal. Write a recursive program to draw a fern or tree, as in this fern fractal demo.
- Integer set partition. Use memoization to develop a program that solves the set partition problem for positive integer values. You may use an array whose size is the sum of the input values.
- Voting power. John F. Banzhaf III proposed a ranking system for each coalition in a block voting system. Suppose party i control w[i] votes. A strict majority of the votes is needed to accept or reject a proposal. The voting power of party i is the number of minority coalitions it can join and turn it into a winning majority coalition. Write a program VotingPower.java that takes in a list of coalition weights as command-line argument and prints out the voting power of each coalition. Hint: use Schedule.java as a starting point.
- Scheduling on two parallel machines. Program Schedule.java takes a command-line argument N, reads in N real number of standard input, and partitions them into two groups so that their difference is minimized.


text:
Letters with Pooh | iPad App Review
October 18, 2012 by admin - Jennifer
Filed under Reviews
Company: Disney Learning
Item Reviewed: Letters with Pooh
Letters with Pooh is a Disney Learning interactive app filled with delightful activities designed to help young children identify and write letters of the alphabet. Children will join Winnie the Pooh on his quest for honey and help him reach a honeybee's hive by tracing letters and collecting letter balloons. Finally, they will assist Pooh's friends in supplemental activities that reinforce the letters they traced and collected during Pooh's honey-hunting adventure.
- Practice tracing and writing upper- and lowercase letters to help Pooh find honey!
- Record all the letters of the alphabet in your own voice!
- Play fun, character-driven activities that reinforce letter identification!
- Create your own Winnie the Pooh scenes with interactive stickers!
- Guided skill instruction for up to four players
- Comprehensive parents' section with content leveling and progress monitoring
- Whimsical Winnie the Pooh artwork!
- Professional narration, music and charming sound effects!
My Thoughts
My daughter is actively trying to learn her letters and seems to do so much better when that learning time is fun while being educational. That is why I like Letters with Pooh. She gets to learn her letters as she plays with the beloved Winnie the Pooh characters and it has helped her so much. She can now write about half of her letters after tracing them on the app and she is so proud of herself when she does it. I like this app because it is portable, she can keep learning while out and about with me and she doesn't have to carry all the papers, crayons and pencils she use to before these types of apps have come out for the iPad. A definite favorite with us!


text:
The T3 Morning Market Call: Use Reversal Strategy to Catch a Bounce
A reversal can sometimes lead to a day-and-a-half move or more, depending on each individual set-up.
World markets are mixed this morning as investors try to figure out what's next. The signal to pare down some risk on a short time-frame came when the
(INDEXSP:.INX) couldn't hold 1698 on August 6. The market did have a gap and go to the downside Thursday, and then hit a natural target at the 50-day MA on Friday.
Today, futures are hovering around flat line as oscillators are in the type of oversold territory that we've seen only twice in 2013. The oversold conditions make it very hard to sit short, but playing a bounce can also be very tricky. If you are going to look to catch a bounce, I would use a "Red Dog Reversal" type strategy, which is a method for buying a stock vs. a level with a plan. A reversal can sometimes lead to a day-and-a-half move or more, depending on each individual set-up.
SPDR S&P 500 ETF Trust
(NYSEARCA:SPY) has a pivot low from Friday at $165.50 and then your next point of reference under this is $164.63. The next major spot is the gap from July 8 that has support at $163.50 and that also lines up to the 100-day MA at $163.20. If a bounce starts, you have resistance at $166.63 and then the gap at $167.43.
Industrials ETF
(NYSEARCA:XLI) showed relative strength after last week's sell-off in the market, as it's still hovering around the 21-day MA and well above the 50-day MA. This group hasn't got much volatility as it's been on a steady uptrend since November 2012. It's trying to build a base at $44.77 area. Below that, $44.24 is a bigger support zone.
Russell 2000 ETF
(NYSEARCA:IWM) also saw a big gap down on Thursday but is still holding above its 50-day MA. However, the IWM doesn't look like it has found its footing yet. Another push lower into the 50-day MA, which stands at $101.16, could give us a new pivot to trade against as this is a big support area. This sector could be an avoid until then.
Retailers ETF
(NYSEARCA:XRT) was hovering near highs and showing relative strength during the months of July/August until last week it dropped below the 50-day following a big gap down. It's struggling to get back above this key moving average. The longer it stays below the 50-day at $53.89, the higher the probability it could see lower prices.
Financials ETF
(NYSEARCA:XLF) got hit big and broke below its 50-day MA. It's struggling around this key moving average. If it can't find a friend at this $20 level, it could retest the 100-day MA at around $19.51.
Homebuilders ETF
(NYSEARCA:XHB) traded inversely with the market last week as it saw a strong bounce on Thursday to get back above the 200-day moving average. It had nice follow-through on Friday but also finished on lows. This was a good example of a Red Dog Reversal day-and-a-half strategy. It's currently sitting at the 200-day MA at $28.87. If dip buyers don't step in at this area, the chances are it could be a short-lived bounce that could lead to lower prices.
Utilities ETF
(NYSEARCA:XLU) is a defensive, dividend-oriented sector that suffers in a rising interest rate environment. The XLU has been leading the market down since May and experienced another sharp sell-off last week, which brought it back to retest the 200-day MA at $37.56. It has bigger support at $35.80 from June 21's pivot low. Overall, this is among the weakest sectors that need time to repair.
Let's look at what's going on in high beta tech.
(NASDAQ:AAPL) digested the huge move on Tuesday's Icahn tweet, as it continued to stay above $493.40 level. It was impressive to see this stock holding up well during the big pullback of the market. We now have Friday's low of $498.86 as the new pivot to trade against. If it continues to hold above this pivot low, we could see a quick push through last week's pivot high of $504.25. The stock has been a great technical long since it reported earnings. I believe it could see the $535 area.
(NASDAQ:NFLX) continued to find buyers at the 21-day MA. The stock retested this key short-term moving average at $251ish on Thursday, then buyers stepped in at this level early Friday to lift the stock up. The stock closed well off of lows. It could see some continuation above $260.50.
(NYSE:LNKD) entered its earnings gap after gapping down on Thursday's open then had some downside follow-through on Friday to retest the 21-day, where it quickly met some buyers. However, it couldn't find momentum above $230 as it hovered in front of this resistance area for the whole session on Friday. Use this level as the new action area. This stock needs time.
(NASDAQ:GOOG) showed relative weakness as it broke below the 100-day MA on Thursday and struggled to reclaim this key moving average on Friday. Use Friday's low of $856.54 as the new point of reference. Overall, the action has been bearish and a bit of an avoid since its outside day in the $905 area
(NASDAQ:TSLA) broke above the downtrend resistance that was in control since the earnings peak of $158.88 on a gap up on Friday. However, it has some resistance at $144 area. It held above the opening's gap at $141 well showing some commitment. The longer it holds above Friday's low of $141, the higher the probability we could see some upside momentum above $144.
(NASDAQ:FB) also showed some relative strength on Friday, as it saw a big gap up and held above it. The longer it stays above $36.90, the higher the probability we could see higher prices in the coming sessions. A break and close above $37.50 could bring in some buyers.
Metals have performed well since making lows on June 28, then gave us another nice set up on August 14.
Gold (NYSEARCA:GLD) had a nice igniting move on Thursday and digested very well on Friday even though it ran into some resistance from the 100-day MA at $133.46. A break and close above this level could open the door for some upside follow-through. Next resistance is standing at $136.75.
Silver (NYSEARCA:SLV) showed relative strength as it had a bigger move and is already well above the 100-day MA at $21.55. It held in well during Friday's session, showing some commitment. The longer it stays above the 100-day MA, the higher the probability we could see some continuation above $22.59. Next real resistance doesn't come in until $23.70. Above that we have the 200-day MA at $25.76ish.
Bonds continue to drift lower, pushing the Inverse
ProShares UltraShort 20+ Year Treasury ETF
(NYSEARCA:TBT) higher. This could be foreshadowing market expectations that the Fed will taper QE in September, but overall we have been getting mixed signals about potential tapering.
My game plan today will be to look for some type of bounce, but I will keep it pretty light. I hope everyone is enjoying their summer.
Scott J. Redler is long SPY puts, IBM puts.


text:
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text:
Monday January 27, 2014 – Monday January 27, 2014
Take a journey with your 3- to 5-year old; learning ways to gain strength, flexibility and exercise through yoga. This 3-week series is a gentle way to introduce your child to a lifelong habit of staying healthy and active! Registration includes one child and one adult.
January 27, 5:30pm-6:15pm
February 3, 5:30pm-6:15pm
February 10, 5:30pm-6:15pm


text:
Submit News
UC Researchers Produce First Quantitative Estimate of Fukushima Radiation Leak
Tiffany Kaiser
August 16, 2011 11:47 AM
comment(s) - last by
Aug 20
Recipient E-mail
Sender E-mail
(Source: ibtimes.com)
UC San Diego researchers concluded that 400 billion neutrons were released per square meter surface of the cooling pools at Fukushima Daiichi
9.0-magnitude earthquake
that shook Japan and crippled the reactor at Fukushima Daiichi nuclear plant back in March caused quite a bit of havoc with the release of
radioactive water
contamination of crops
and of course, the thousands of lives lost. At the same time, news networks like
sensationalized reports, causing unnecessary nuclear-related fear. U.S. senators even demanded the Nuclear Regulatory Commission to
repeat an expensive inspection
of U.S. nuclear plants.
In an attempt to clear some confusion and understand exactly how much radiation actually leaked from the damaged nuclear reactor at the Fukushima Daiichi nuclear plant in Japan on March 11, atmospheric chemists at the University of California, San Diego, have produced the first quantitative estimate of how much radiation actually leaked from the reactor.
Mark Thiemens, study leader and Dean of the Division of Physical Sciences at UC San Diego, along with post-doctoral researcher Antra Priyadarshi, and a team of researchers, observed the amount of radioactive sulfur in the air soon after the earthquake in Japan and was able to report a quantitative measurement of the amount of radiation leaked.
When fuel rods melt, products like neutrons leak from the fuel rods. Seawater is used to cool the hot reactors, and absorbs the leaked neutrons. These neutrons "collide" with chloride ions in the seawater, which results in the loss of a proton out of the nucleus of a chloride atom and turns the atom into a radioactive form of sulfur. Most of this vaporizes into steam when the saltwater comes into contact with the hot reactors, and to avoid explosions due to the collection of hydrogen, operators vent the steam into the atmosphere. Once in the air, the sulfur reacts with oxygen to create sulfur dioxide gas and eventually sulfate particles.
On the other side of the Pacific Ocean in La Jolla, California on March 28, 2011, Thiemens and his team noticed an "unprecedented spike" in radioactive sulfur in the air. They used a model, which was based on the NOAA's observations of atmospheric conditions, to determine the path the air took to get to California over the previous 10 days, and found that it had come from Fukushima Daiichi.
The next step was to calculate
how much radiation had leaked
from the reactor based on the path over the Pacific Ocean. They took into account that some sulfate particles had fallen into the ocean or decayed along the way, and concluded that 400 billion neutrons were released per square meter surface of the cooling pools. They predicted that this occurred between March 13, 2011 and March 20, 2011. March 13 was when operators began flooding the reactor with seawater.
"You know how much seawater they used, how far neutrons will penetrate into the seawater and the size of the chloride ion," said Priyadarshi. "From that, you can calculate how many neutrons must have reacted with chlorine to make radioactive sulfur."
To achieve the levels observed in California, the team said the concentrations a kilometer above the ocean close to Fukushima must have been 365 times above normal levels. Over the four days that the team took measurements, which ended March 28, Thiemens measured 1501 atoms of radioactive sulfur in sulfate particles per cubic meter of air. They mentioned that this was the highest they had seen in two years of observations and recordings.
According to the researchers, the radioactive sulfur observed was produced by partially melted nuclear fuel in the storage ponds or reactors. While cosmic rays can produce radioactive sulfur, the team noted that these rays rarely mix into the layer of air right above the ocean.
Despite the high levels of radioactive sulfur recorded in California, Thiemens and his team said these levels were
not dangerous to human health
"Although the spike that we measured was very high compared to background levels of radioactive sulfur, the absolute amount of radiation that reached California was small," said Thiemens. "The levels we recorded aren't a concern for human health. In fact, it took sensitive instruments, measuring radioactive decay for hours after lengthy collection of the particles, to precisely measure the amount of radiation."
This study
was published in
Proceedings of the National Academy of Sciences
remember me
This article is over a month old, voting and posting comments is disabled
RE: If radiation is harmless, LETS ALL EAT SOME PLUTONIUM!
Here's some more truth speak for you...
1. We KNOW radiation exists naturally. In all of us. Mostly in the form of Potassium-40.
2. We KNOW that UV radiation is necessary for the production of Vitamin D. If you could eliminate it completely you'd be like any number of fat kids who don't go outside anymore and have a Vitamin D deficiency.
3. Okay I don't really have an argument against this one.
As for the types of cancer increase from various sources, from what I understand there's a bit too many variables to make a simple chart for that. If you are truly concerned, stock up on iodine tablets. Your body will readily absorb iodine (radioactive or not) up to a certain point. If you're fully stocked on non-radioactive iodine your body will excrete the rest.
Many other radioactive materials aren't readily used in bodily processes so you only have to work about the radioactive particles themselves. Here what matters is the energy of the particle.
If I recall correctly (and I could easily be wrong lol) the heavier the particle the less energy it typically has. In most cases heavy alpha particles won't have enough energy to pierce the skin. Beta and gamma radiation have the potential to be more harmful as they can penetrate deeper.
------------------------------------------------- ----------
Perhaps more useful information for you (and all of us), would have been if the study provided the data in the form of Sieverts.
Straight from Wiki:
The sievert (symbol: Sv) is the International System of Units (SI) SI derived unit of dose equivalent radiation. It attempts to quantitatively evaluate the biological effects of ionizing radiation as opposed to the physical aspects, whichgray. are characterised by the absorbed dose, measured in
...so that would have been far more useful in determining if this amount of radiation was actually harmful. It also would have been far less sensational.
"Then they pop up and say 'Hello, surprise! Give us your money or we will shut you down!' Screw them. Seriously, screw them. You can quote me on that." -- Newegg Chief Legal Officer Lee Cheng referencing patent trolls
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Davor Zorc in System Utilities / File & Disk Management
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Hey there, time traveller!
This article was published 8/7/2014 (719 days ago), so information in it may no longer be current.
The year is 1999, and Anatolia is a Bosnian immigrant living in Edmonton, obliged to address her classmates in an English as a Second Language course on some benign topic chosen by her teacher as an exercise in public speaking. So she talks about her job at Superstore and the friends she has made. But over the course of an hour, Anatolia reveals aspects of a tragic past experienced in the madness of the Bosnian war from 1992 to '95.
Edmonton writer-director Kenneth Brown has fashioned a beguiling solo show that starts as a sweetly inspirational story of the immigrant experience and morphs into a dramatic wallop as Anatolia lets her guard down and relates an all-too-credible story of encountering both heroism and pure evil. In this capacity, actress Candice Fiorentino really delivers the goods, delineating a character arc that takes her from a cute comic figure to a tragic heroine in breathtakingly subtle increments.
— Randall King


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99 Blessings – An Invitation to Life
I am fascinated with a monk called Brother David Steindl-Rast! After all my research on gratitude I have come to my own personal conclusion that he is likely the world leader on the subject of gratitude. You can read more, and an interview with him here .
This month he has just released his new book 99 Blessings – An Invitation to Life. There are 99 little prayers directed toward 'Source of all Blessings' and it is such a yummy little book. Beautiful to look at, beautiful to touch and beautiful, as well as life-giving, to read.
A little sample for you:
Blessing 27
'you bless us with us with warning voices,
-poets, prophets, thinkers who dare to speak against the current
of accepted norms, to question, at risk of their good standing, their careers,
even their lives.
May I have ears to hear these lonely voices of Common Sense,
so often nearly drowned out by the din of public opinion, and to heed them,
take their message to heart,
weigh it, and let it change my life.
I hope you get to read this lovely book – 99 Blessings.


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The human compulsion to create a visual record of our
experiences and emotions can be traced back to earliest man and his crude cave
paintings. Evolution has led us from the Dark Age to the Digital Age, yet our
need to create is unabated. Our tools for creation have also evolved, and today
we will be looking at the digital equivalent of pencil and paper—paint and
canvas—the Intuos3 tablet from Wacom.
The Intuos3 is the latest in a long line of high quality
tablets from Wacom. It sports a new
modern industrial design that includes a gray back-painted transparent surface,
matte-finish active area, and built-in palm rest with tapered sides to make it
not only elegant, but comfortable as well.
The Intuos series has many advantages over the Graphire series such as
larger sizes, two times the levels of pressure sensitivity, tilt features, and much
more. The Intuos3 also makes some
improvements to the former leader, Intuos2, with an increase in coordinate
resolution, the addition of touch strips and express keys, and an updated
software bundle.
The Intuos3 that we are reviewing today is the middle of the
road 6x8 version which is flanked by the 4x5 and 9x12 versions. Each has the same features with the exceptions
that the tablet and active area are larger, and on the 4x5 version there is
only one set of express keys and a touch strip.
Looking at the specifications, it is clear that
developers at Wacom do not joke around when it comes to packing in the
features. The tablet, along with its new design,
includes express keys and an innovative touch strip which Wacom is
their new non-dominant-hand capabilities—fancy words—with a lot of
function. This allows you to draw freely with one hand while using the
ExpressKeys and Touch Strip to control the application with the other
Continuing with Wacom tradition, neither the pen or mouse need
batteries or cords—allowing for a hassle-free work environment. Every inch of the tablet has been designed
with comfort and usability in mind. The
tablet, pen, and mouse are all ambidextrous—allowing both right and left handed
users to take full advantage of their features.
The Intuos3 grip pen has a larger grip area for comfort and comes with a
conical two-way pen stand to keep the pen from sustaining any undue
damage. The mouse that comes with the
Intuos3 is a 5-button mouse that doesn't rely on a ball or optics.
Additionally, the tablet comes with a very nice bundle of
software which contains Adobe Photoshop Elements 3, nik Color Efex Pro 2 IE by
nik multimedia, Painter Essentials 2 by Corel, and Wacom Brushes 2.


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Sept. 2, 2000
Washington head coach Rick Neuheisel
General Remarks: " Certainly excited to be 1-0. Excited about a lot of good things that I saw on the field. I guess that would be the ultimate in a wake-up call, the 83 yard run to start the game. It certainly got our attention. I was proud of the fact that no one panicked on the sideline. They caught us with a new formation and we didn't get the safety down to help us out. But we fixed that and I thought that minus that we played pretty good. They're a skilled team. They threw the ball well. I thought that the quarterback Welsh was very talented passer. He kept the ball in front of him and he had talented receivers. Every time the ball was close to them they made pretty good catches. Therefore they were able to keep the ball moving and got over 400 yards total offense. I think at the end of the game they were doing things to try to work on their offense. On our side of side of the ball I thought we played well at times and sporadic at times. Some things became glaring and we'll just have to address those things tomorrow when we watch the film. I know you're going to ask what those things are, but I'm not going to tell you because we've got another game to play and I don't want to give them a head start. But we've got some things to address and I'm anxious to get to the tapes. But I know it certainly beats where we were a year ago. It's always the case that anybody can beat anybody in this game so we don't take the victory lightly."
"The first game is a difficult game because we don't know how tired we are. I thought that we were fine. I thought that the 83 yard run, certainly got out attention and got us into the game."
"I was proud of the way the kids caught the ball. We need to do some things to get open."
On John Anderson: "I had hoped that the preseason inconsistency would settle itself out. I saw some improvement later in the week, but basically. But when you're talking about the kicking, you're talking about a trio of people. You're talking about a snapper, holder and a kicker. That trio didn't perform fully effectively. It got better in the second half. We missed two field goals in the first half and they raised some concern. Thankfully he didn't let that bother him mentally. John is a strong mentally. I don't worry about that, but we do need to work on that."
On Marcus Tuiasosopo after the first series: "Marcus is going to be the focal part of our offense whether things are going well or not. I thought he made some real good plays. I know that he would like two or three of his throws back. It's just nice to be improving with a victory rather than vise versa."
On the rushing game: "At times I thought we were great. We tackled well, we stripped the ball well. We caused some fumbles. I thought we did some really good things. But I want to see if we can get some more pressure without having to do a full out blitz. I'll have to wait and look at the video to see if we are improving there."
On Coach Moses: "Coach Moses has a good feel for that and I trust his instincts. And I trust his instincts as to how we play those tailbacks. The way to keep people happy is to win games. I trust him and if I feel like I've got an instinct as to who should go in the game I'll go and tell him."
On the offensive linemen changes: "In the early portion of your season you make some changes until you settle in a realize who your best five and most consistent are. It takes a while and you don't really learn that in practice, it's difficult. But after time you determine who is doing the best job."
On Ben Mahdavi: "Well I gave him a scholarship didn't I? I thought he was a pretty good player. He's got that high revving engine that causes him to be where he needs to be most of the time. The blocked punt was certainly a big play."
On Hakim's interception: "I'm still a little bewildered as to why it wasn't a fumble. But it was great. The defense did a nice job in terms of turnovers. Offensively we did a nice job holding on to the ball, but I'm a little disappointed we had an interception so late in the game because we had a wide-open guy. But I think we'll learn from that."
On Anderson's kicks: "Part of our plan is to do kangaroo kicks, which is short. Some of them were done without being called. Again John has got to bear down and become more consistent as a kicker."
On the cornerbacks: "I think they played pretty well. You ask yourself what you want to try to do defensively. Do you want to let them nickel and dime you and try to get to the passer and make plays. And just don't get beat by the big play. I know Chris Massey is sick about the play that got over the top of him, but he'll learn from that. Massey, Derrick Johnson and Rock Alexander are all true freshman. We're getting better. We've got a great group of athletes that are going to be great. I've got a great corner coach and I know he'll do well."
On Derrick Johnson's touchdown: "Keith calls all the plays. I thought it was a great call at a great time. It was a nice play at a nice time."
On the special teams: "They did some really good things and then we need to work on some things. We need to work on our kick off coverage. Didn't get many opportunities on punt return."
Washington Player Quotes
(On the team's reaction after Idaho's first touchdown): "I was scared after they got that first touchdown because we knew they were a good team. I think after that happened though everyone relaxed. We had a lot of young guys out there so there were a lot of first-game jitters, especially for myself because that was my first game at linebacker. We settled down after that though and played well."
(On The Blocked Punt): "It was just a regular punt return with an overload to the right. The upback told the center to go right so I thought to myself, `there is no way that I am going to get blocked.' I knew before the snap that I was going to get through."
(On his first touchdown as a Husky): "I was overwhelmed but just happy that I could contribute. We came out on that play with a little trick under our sleeve. We wanted to catch them off guard by waiting to go out on the field until the play clock went down and it worked out because we caught them on their heels a little."
PAUL ARNOLD (On playing a bigger role in the offense this season): "Everyone has been saying that I'm the main guy so I'm going to be fine but when I got thrown in there I was really nervous and it showed early when I made a lot of mistakes but I was able to rally. Coach, instead of saying 'you suck', was really supportive and said 'it's over with, pick it up in the second half''. That made me feel better and I came out and played a lot better in the second half."
(On playing with little practice time): "I definitely took my lumps and that's what happens to you in this game. I just had to go out there and hack it out. I didn't play that well and when you don't practice then you figure that it's all going to catch up to you and it did. But just going out there and fighting it out was a victory for me."
(On how this game prepares the Huskies for Miami and receiver Santana Moss): "It's great that we took our lumps this week and we have a lot to work on in the next week of practice. I look forward to the challenge of facing Moss. I have only one direction to go and that's up. I'm happy I competed well and it was a tough game but Idaho was by no means a pushover."
(On how the receivers performed): "They came out and did a great job. Todd Elstrom made some great catches. Wondame Davis made some great adjustments off the safety pressure and Wilbur Hooks did a great job. The couple plays we didn't make, that was my fault. We had guys wide open and I didn't hit them I lofted the ball a couple times. I'm happy for those guys. They were a question mark but they stuck together and worked hard and I think it showed today."
(On running the option): "Coach Gilbertson called some great plays, he's got a lot of them . We don't worry about not running certain plays to keep Miami in the dark. We just go out and play and the option just gives Miami one more thing to worry about. The offensive line did a great job, especially on Derrick Johnson's touchdown. I flipped the ball to Derrick and looked back after I got hit and the offensive line just flowed. It was a great time to run that play too, being right after a turnover."
(On His Touchdown): "It was just a simple post route and I was fortunate to get inside of the man and get good position and Marques was threw a ball up there that I could get good position on and I brought it down."
(On Almost Scoring A Touchdown): " That was the longest run towards the ball I have ever had. All I saw was the ball and the whole field, I was going so fast that it was tough to pick up, but I blew it and kicked it out of bounds."
Idaho Head Coach Tom Cable
GENERAL REMARKS: "I really don't think if you turn the ball over five times you can expect to win those games. I believe we probably gave them 20 points by simply turning the football over. That's hard to swallow right now because we feel like we put a lot of time into ball security and our kicking game. We've got work to do, we've got to back and coach better and take a little bit more responsibility with ball security with the players. I felt like we went out there and played, we moved the ball, they forced us around on defense. But, we did give them 20 points."
ON A "MORAL VICTORY": "I think we're gonna be a good football team when we get a couple of things worked out as coaches and players. The moral victory thing is not a big deal to us, we came here to win a game, to play a good game. We got beat today and turned over five times."
ON THE OPENING TOUCHDOWN PLAY: "We thought we would come out and try to set the tempo with them. We felt like if we didn't make a statement by trying to be physical right in the beginning of the game, they would see that. Our plan was to come out and see if we could knock them off the ball early. It was really a great thing, we blocked well and Willie (Alderson) made a great cut and out-ran everybody."
ON HALFTIME LOCKER ROOM MOOD: "We really felt like we had already given them 13 points in the first half by turning the ball over. So we were disappointed from that standpoint, but we also felt like we were starting to push them off the ball offensively. We really weren't settled in, but the flow and the direction of the ball game was what we wanted."
ON UW DEFENSIVE STANDOUTS: "The safety (Hakim Akbar) we knew he was a good player coming in. I saw him make some plays in space when we thought we had a chance and he would show up and make a play. Line of scrimmage, I felt like we were fine offensively."
ON IDAHO QB JOHN WELSH: "I think for his first time out, he missed a couple of hot plays. It was his first time in the system and we'll get it straight. He picked a couple up, but he also missed a couple. He stood there and he threw some balls a couple of times in traffic where he made some huge throws. He's a good quarterback and he's really going to do something for us."
ON THE OPPONENT MATCH-UP: "Obviously we were concerned (about the match-up) because I knew since day one that the defensive line was something to look at. We got beat on the option a couple of times today and we did not play our option responsibility. I'm still concerned about the depth issue, some young guys played today. I think we belonged in this stadium today."
ON STATUS OF TEAM: "One thing I was really proud of is that I think we were in really great shape. I didn't see us get obliterated there in the second half when we didn't have any juice left. We were still pushing on the offensive side of the ball and I am proud of that."
ON FIRST GAME AS HEAD COACH: "Coming out of the tunnel was really emotional for me. This has been my dream to be the head coach at Idaho. It was tough to get out there, get settled and get going. I'll have to go back and re-think things and we'll be better next week."
Idaho Players Quotes
(On Idaho Offense): "We could have run our offense better even with the five turnovers if it weren't for the little things which we will work on starting now.
(On first play of the game): "I just handed off the ball to Willie (Alderson) and let him do his thing. It is something we have practiced so I knew it would work."
(On the first half): "We stopped Washington good on defense. I thought we were in it until the second half. I became much more comfortable as the first half went on. I had a lot of time to pass. I leave this game knowing we competed hard. We are on a mission for the rest of the season to win."
(On young Idaho team): "We learned a lot as a team today. A lot of the young guys did a good job of not losing control. We should we could play tough offensively and defensively. I am excited for the next few games.
(On the experience of playing at Husky Stadium): "I love the atmosphere here. We were not intimidated by the crowd or the stadium itself."
(On UW Quarterback Marques Tuiasosopo): "He is a good athlete. I think we fought hard against him and we are not going to hang our heads about how we played against him."
(On #14 ranking of UW): "I can't really tell if they deserve this rank, but they have good athletes and big guys. Obviously more people think they deserve to be #14 than me."
(On the first play of the game): "We executed it perfectly. Our offensive line made the blocks we needed. This play is something we have practiced and I just envisioned it how we have practiced it."
(On Idaho turnovers): "We can't have five turnovers and win over any team. We need to work on our ball security. Washington adjusted well to our running game and that hurt us also."
(How prepared for Washington): "We watched a lot of film to prepare for their throwing ability. We prepared for the speed of their offensive line also."
(On Idaho turnovers): "On defense when we make stops, we have to keep making them. It is something we can't give up on. Football is not just defense and offense though either. We win and lose as a group. I guess we can't pin this loss on one certain area."


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Almost all of the attractions in Nashville, Tennessee, offer fun for the entire family. It is a favored travel destination for families. From history to shopping and cultural experiences, there isn't anything that Nashville doesn't have. If you are thinking about vacation travel this summer, then here are three things to do that you won't want to miss.
Shopping in Nashville
It's a given that if you travel to Nashville, that you have to stop and shop. There is the trendy Green Hills Avenue area and lots of unique shops in the downtown area. According to Simon Mall, Opry Mills is Tennessee's "larget value retail, shopping, dining and entertainment destination" in the state. Take a look at the website for Opry Mills to find out more about the mall and its attractions.
Flamingos at the Nashville Zoo
Flamingos like these are just the start of the animals that you will see at the Nashville Zoo. There are African elephants, salt water and freshwater fish, cougars and the Short-tailed Leaf Nosed Bat that call the Nashville Zoo home. Travel to Nashville may not be complete with a zoo trip. Find out about admission and hours at the website for the Nashville Zoo.
Dulcimers and more at the Tennessee State Museum
The Tennessee State Museum has no admission fee. Exhibits include items that belonged to Presidents Andrew Jackson and Andrew Johnson. There are also items that are associated with a free black family that lived in the Knoxville area before 1860. Other exhibits include items from famous Tennesseans Alvin York and David Crockett. Information on hours and days of operation can be found online.


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I wore Illamasqua Lament as a base, and used a dotting tool for polka dots. The green dots are Illamasqua Milf and the purple ones are Illamasqua Jo'Mina. All three colors were just lovely to use. They each dried very quickly, so I had no problems with drag when I applied top coat.
I think that polka dots are just perfect for a cheery manicure. They are always cute, and they aren't hard to do at all! That is my favorite type of manicure :D
Let me know what you think of this mani, and come back tomorrow to see my take on a Valentine's Day look!
- Aly


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Just worrying that your family will be the next tragic news headline being discussed around the water cooler in Island Park isn't going to protect them. The easiest way to take control over your family's safety is to get an ADT Monitored Home Security System installed as soon as possible. Each day you wait to get your ADT monitored system in Idaho only increases the likelihood of your loved ones turning into victims of the predators who would like to do them harm.
ADT stands behind their home alarm monitoring in Idaho with the confidence that comes from over 100 years of securing households nationwide. That's one of the reasons that ADT is able to provide you with a Theft Protection Guarantee when purchasing your Home Security System. But if your Island Park house is successfully burglarized while your alarm is armed, ADT will pay up to $500 of the insurance deductible.
ADT's security experts can act right away in sending the closest Island Park law enforcement officers. Simultaneously, an ADT security professional will contact you by various means, such as a direct voice intercom connection, to assess the situation and take the most appropriate measures in making sure that you and your family in Idaho remain safe. And because ADT Monitored Home Security Systems in Idaho can connect with ADT's monitoring services using a dedicated cellular connection, you don't need to worry about devious home invaders who often make attempts to bypass alarm systems by severing your phone lines. No other home security company in Island Park can offer the level of protection possible by having an ADT Monitored Home Security System.
Drive through any Island Park neighborhood, and you are likely to see "Protected By" signs from a number of home security companies, cautioning possible burglars and home invaders that they're boosting their odds of getting caught. But the houses that most criminals tend to avoid are those with signs that announce they are protected by an ADT Monitored Home Security System. But what is it about an ADT monitored alarm that strikes fear into the hearts of crooks in Idaho? An ADT Monitored Home Security System offers home protection technology that is simply unmatched in ID.
Get an ADT monitored security system in Island Park ID and help protect your family today.


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GCC power grid headquarters dedicated in Dammam
April 2, 2014
Under the patronage of Eastern Province Governor Prince Saud bin Nayef bin Abdulaziz, the new headquarters of the Gulf Cooperation Council (GCC) Power Grid was dedicated in Dammam today. The inauguration was carried out by Minister of Water and Electricity Abdullah Al-Hussein, other GCC ministers in charge of electricity, and GCC Secretary General Dr. Abdullatif Al-Zayani.
Addressing the gathering, Undersecretary of the Kuwaiti Ministry of Electricity and Water Ahmed Al-Jassar expressed thanks to the Custodian of the Two Holy Mosques King Abdullah bin Abdulaziz for providing the land for the project.
During his remarks, Al-Zayani said the initiative is more evidence that GCC leaders are committed to furthering cooperative and developmental projects for the benefit of their peoples.


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Republicans and Democrats are all out trying to gain support before the big day. Last week, Democratic Candidate, Mike Weaver held a public discussion with former US Senator Max Cleland about the war in Iraq. Now Laura Bush is lending her support to help re-elect Republican Congressman Ron Lewis.
The First Lady spoke for about 30-minutes and she was joined by several other candidates for local races. Mrs. Bush, thanked Lewis for his support of the President's tax cuts and his support of the military. Lewis's opponent, Mike Weaver held a similar event last week, with Max Cleland.
With the race for the Second Congressional District gaining national attention, both parties are hosting high level party members to gain votes before next Tuesday.
In 1994 Lewis won the critical Second Congressional seat, helping to push the Republican party into the majority in Washington. This year Democrats view the house seat as a pivitol spot to gain if they hope to take back the majority on Capitol Hill.
The race between Lewis and Weaver is a heated one because it is widely believed this is the first real competition for the incumbent Congressman.


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Get a QuoteRetrieve a Saved Quote
Find an AgentMore search options
Whole Life Insurance will provide protection from the day you purchase the policy until you die. In addition, premiums will remain the same throughout the life of the policy.
A Whole Life Insurance policy may let you
- Earn annual dividends1, which will be paid in cash, left to accumulate interest, or used to reduce premiums or purchase additional coverage.
- Borrow the cash value, or receive dividends as loans (in some cases).
- Appoint a charity as a beneficiary of your policy.
Additionally, your survivors can use the proceeds to
- Pay off mortgages and other outstanding debts.
- Avoid having to sell assets or incur debt to cover estate taxes and other expenses.
Other policy options
There are several options that can be added on as well:
- A waiver of premium. This allows the premium to be waived after six months of total disability.
- An accidental death benefit rider. This increases the death benefit in case of accidental death.
- A payor benefit. This provides continuous premium payments for a specified time if the payor of a child's life insurance policy dies or becomes totally disabled.
- A guarantee of insurability. This allows for the purchase of additional life insurance at certain intervals without proof of good health.
- A paid-up additional insurance rider. This allows for the purchase of additional life insurance, with proof of good health, while premium is still being paid. These additions can increase your policy's surrender value and death benefit and can also earn dividends.1
1 Dividends are not guaranteed.
The product information contained on this website is informational only and not a statement of contract. All coverage options are subject to the provisions of the policy purchased and details of the policyholder's situation.
Shelter Life Insurance Company, 1817 West Broadway, Columbia, MO 65218
Policy form #L-001.12 in AR, OK, TN; L-001.13 in CO, IA, KS, KY, MS, MO, NV; L-001.23 in IL; L-001.15 in IN; L-001.16 in LA; L-001.22 in NE. Not available in all states.


text:
Programs developed by googale
googale Toolbar is an useful toolbar that helps you to receive the freshest content directly in your browser ( Google Chrome, Internet Explorer, Mozila Firefox) . In the same time you can receive informations about the weather and forecast. Also you can look over the internet for thousands of application for your browser.


text:
A. Bertha "Bert" Smith, 96, died Sunday, February 10, 2013 at her residence in Liberal.
She was born April 16, 1916 in El Paso , Texas the daughter of Alexander David and Adela McLean.
She lived in El Paso , TX , Sun Lakes , AZ and Austin , TX before moving to Liberal to be with her daughter.
On Oct. 4, 1947, she married Eugene P. Smith in El Paso, TX. He died in May 19, 1953.
Survivors include:
Son - David Smith and wife Jan of Austin, TX;
Daughter - Shirley Scott and husband Don of Liberal, KS;
3 Grandchildren - Martha Scott, Wichita, KS; Michael Scott, Liberal, KS;
Alex Smith, Los Angeles, CA.
2 Great-Grandchildren - Maxwell Smith and Holly Smith
Step-son – Gerald Smith of Anthony , TX
Step-daughter - Mary Gail Abbott of Albuquerque , NM
She was preceded in death by a sister Nettie Harms, 2 brothers
William and Alex McLean, a granddaughter, Kristen Smith and a step-son, Glenn Smith
Memorial Service will be held at 2:00 pm Thursday, Feb. 14th at St. Andrews Episcopal Church with Fr. Charles Schneider presiding. Cremation has taken place.
Memorial Contributions may be sent to St. Andrews Episcopal Church or the Arthritis Foundation in care of Brenneman Funeral Home 1212 W. 2nd Liberal , Kansas 67901
Condolences may be sent to the family at www.brennemanfuneralhome.com
©2016 Seward County Broadcasting Company, Inc.
1410 N. Western Ave. Liberal, KS 67901


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Our consumerist culture appears to be driving an unsustainable level of consumption of the world's resources. Unless we believe in miraculous development of new technology, we must fundamentally change our behaviour to consume - and pollute - substantially less. But how do we change? In simply existing, we consume. It seems the issue is one of developing a new mindfulness to the world around us – one which is not only contemporary but one which in its essence embraces the world as something which, at least in part out of our own mortality, is 'constantly becoming'. Traces of Possibility is a journey in search of signs and space for such a sensibility.
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