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# 224 28.2 Simultaneity And Time Dilation

### Summary

• Describe simultaneity.
• Describe time dilation.
• Calculate $\boldsymbol{\gamma}$.
• Compare proper time and the observer’s measured time.

Do time intervals depend on who observes them? Intuitively, we expect the time for a process, such as the elapsed time for a foot race, to be the same for all observers. Our experience has been that disagreements over elapsed time have to do with the accuracy of measuring time. When we carefully consider just how time is measured, however, we will find that elapsed time depends on the relative motion of an observer with respect to the process being measured.

# Simultaneity

Consider how we measure elapsed time. If we use a stopwatch, for example, how do we know when to start and stop the watch? One method is to use the arrival of light from the event, such as observing a light turning green to start a drag race. The timing will be more accurate if some sort of electronic detection is used, avoiding human reaction times and other complications.

Now suppose we use this method to measure the time interval between two flashes of light produced by flash lamps. (See Figure 2.) Two flash lamps with observer A midway between them are on a rail car that moves to the right relative to observer B. Observer B arranges for the light flashes to be emitted just as A passes B, so that both A and B are equidistant from the lamps when the light is emitted. Observer B measures the time interval between the arrival of the light flashes. According to postulate 2, the speed of light is not affected by the motion of the lamps relative to B. Therefore, light travels equal distances to him at equal speeds. Thus observer B measures the flashes to be simultaneous.

Now consider what observer B sees happen to observer A. Observer B perceives light from the right reaching observer A before light from the left, because she has moved towards that flash lamp, lessening the distance the light must travel and reducing the time it takes to get to her. Light travels at speed $\boldsymbol{c}$ relative to both observers, but observer B remains equidistant between the points where the flashes were emitted, while A gets closer to the emission point on the right. From observer B’s point of view, then, there is a time interval between the arrival of the flashes to observer A. From observer B’s point of view, then, there is a time interval between the arrival of the flashes to observer A. In observer A’s frame of reference, the flashes occur at different times. Observer B measures the flashes to arrive simultaneously relative to him but not relative to A.

Now consider what observer A sees happening. She sees the light from the right arriving before light from the left. Since both lamps are the same distance from her in her reference frame, from her perspective, the right flash occurred before the left flash. Here a relative velocity between observers affects whether two events are observed to be simultaneous. Simultaneity is not absolute

This illustrates the power of clear thinking. We might have guessed incorrectly that if light is emitted simultaneously, then two observers halfway between the sources would see the flashes simultaneously. But careful analysis shows this not to be the case. Einstein was brilliant at this type of thought experiment (in German, “Gedankenexperiment”). He very carefully considered how an observation is made and disregarded what might seem obvious. The validity of thought experiments, of course, is determined by actual observation. The genius of Einstein is evidenced by the fact that experiments have repeatedly confirmed his theory of relativity.

In summary: Two events are defined to be simultaneous if an observer measures them as occurring at the same time (such as by receiving light from the events). Two events are not necessarily simultaneous to all observers.

# Time Dilation

The consideration of the measurement of elapsed time and simultaneity leads to an important relativistic effect.

### Time dilation

Time dilation is the phenomenon of time passing slower for an observer who is moving relative to another observer.

Suppose, for example, an astronaut measures the time it takes for light to cross her ship, bounce off a mirror, and return. (See Figure 3.) How does the elapsed time the astronaut measures compare with the elapsed time measured for the same event by a person on the Earth? Asking this question (another thought experiment) produces a profound result. We find that the elapsed time for a process depends on who is measuring it. In this case, the time measured by the astronaut is smaller than the time measured by the Earth-bound observer. The passage of time is different for the observers because the distance the light travels in the astronaut’s frame is smaller than in the Earth-bound frame. Light travels at the same speed in each frame, and so it will take longer to travel the greater distance in the Earth-bound frame.

To quantitatively verify that time depends on the observer, consider the paths followed by light as seen by each observer. (See Figure 3(c).) The astronaut sees the light travel straight across and back for a total distance of $\boldsymbol{2D}$, twice the width of her ship. The Earth-bound observer sees the light travel a total distance $\boldsymbol{2s}$. Since the ship is moving at speed $\boldsymbol{v}$ to the right relative to the Earth, light moving to the right hits the mirror in this frame. Light travels at a speed $\boldsymbol{c}$ in both frames, and because time is the distance divided by speed, the time measured by the astronaut is

$\boldsymbol{\Delta t_0 =}$$\boldsymbol{\frac{2D}{c}}.$

This time has a separate name to distinguish it from the time measured by the Earth-bound observer.

### Proper Time

Proper time $\boldsymbol{\Delta t_0}$ is the time measured by an observer at rest relative to the event being observed.

In the case of the astronaut observe the reflecting light, the astronaut measures proper time. The time measured by the Earth-bound observer is

$\boldsymbol{\Delta t =}$$\boldsymbol{\frac{2s}{c}}.$

To find the relationship between $\boldsymbol{\Delta t_0}$ and $\boldsymbol{\Delta t}$, consider the triangles formed by $\boldsymbol{D}$ and $\boldsymbol{s}$. (See Figure 3(c).) The third side of these similar triangles is $\boldsymbol{L}$, the distance the astronaut moves as the light goes across her ship. In the frame of the Earth-bound observer,

$\boldsymbol{L =}$$\boldsymbol{\frac{v \Delta t}{2}}.$

Using the Pythagorean Theorem, the distance $\boldsymbol{s}$ is found to be

$\boldsymbol{s=}$$\boldsymbol{\sqrt{D^2 + {(\frac{v \Delta t}{2}})^2}}.$

Substituting $\boldsymbol{s}$ into the expression for the time interval $\boldsymbol{\Delta t}$ gives

$\boldsymbol{\Delta t =}$$\boldsymbol{\frac{2s}{c}}$$\boldsymbol{=}$$\boldsymbol{\frac{2 \sqrt{D^2 + (\frac{v \Delta t}{2})^2}}{c}}.$

We square this equation, which yields

$\boldsymbol{(\Delta t)^2 =}$$\boldsymbol{\frac{4(D^2 + \frac{v^2 (\Delta t)^2}{4})}{c^2}}$$\boldsymbol{=}$$\boldsymbol{\frac{4D^2}{c^2}}$$\boldsymbol{+}$$\boldsymbol{\frac{v^2}{c^2}}$$\boldsymbol{(\Delta t)^2}.$

Note that if we square the first expression we had for $\boldsymbol{\Delta t_0}$, we get $\boldsymbol{(\Delta t_0)^2 = \frac{4D^2}{c2}}$. This term appears in the preceding equation, giving us a means to relate the two time intervals. Thus,

$\boldsymbol{(\Delta t)^2 = (\Delta t_0)^2 +}$$\boldsymbol{\frac{v^2}{c^2}}$$\boldsymbol{(\Delta t)^2}$.

Gathering terms, we solve for $\boldsymbol{\Delta t}$:

$\boldsymbol{(\Delta t)^2 (1 -}$$\boldsymbol{\frac{v^2}{c^2}}$$\boldsymbol{)= (\Delta t_0)^2}.$

Thus,

$\boldsymbol{(\Delta t)^2 =}$$\boldsymbol{\frac{(\Delta t_0)^2}{1 - \frac{v^2}{c^2}}}.$

Taking the square root yields an important relationship between elapsed times:

$\boldsymbol{\Delta t =}$$\boldsymbol{\frac{\Delta t_0}{1 - \frac{v^2}{c^2}}}$$\boldsymbol{= \gamma \Delta t_0},$

where

$\boldsymbol{\gamma =}$$\boldsymbol{\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}}.$

This equation for $\boldsymbol{\Delta t}$ is truly remarkable. First, as contended, elapsed time is not the same for different observers moving relative to one another, even though both are in inertial frames. Proper time $\boldsymbol{\Delta t_0}$ measured by an observer, like the astronaut moving with the apparatus, is smaller than time measured by other observers. Since those other observers measure a longer time $\boldsymbol{\Delta t}$, the effect is called time dilation. The Earth-bound observer sees time dilate (get longer) for a system moving relative to the Earth. Alternatively, according to the Earth-bound observer, time slows in the moving frame, since less time passes there. All clocks moving relative to an observer, including biological clocks such as aging, are observed to run slow compared with a clock stationary relative to the observer.

Note that if the relative velocity is much less than the speed of light ($\boldsymbol{v< The equation [latex]\boldsymbol{\Delta t = \gamma \Delta t_0}$ also implies that relative velocity cannot exceed the speed of light. As $\boldsymbol{v}$ approaches $\boldsymbol{c}$, $\boldsymbol{\Delta t}$ approaches infinity. This would imply that time in the astronaut’s frame stops at the speed of light. If $\boldsymbol{v}$ exceeded $\boldsymbol{c}$, then we would be taking the square root of a negative number, producing an imaginary value for $\boldsymbol{\Delta t}$.

There is considerable experimental evidence that the equation $\boldsymbol{\Delta t = \gamma \Delta t_0}$ is correct. One example is found in cosmic ray particles that continuously rain down on the Earth from deep space. Some collisions of these particles with nuclei in the upper atmosphere result in short-lived particles called muons. The half-life (amount of time for half of a material to decay) of a muon is $\boldsymbol{1.52 \;\mu \textbf{s}}$ when it is at rest relative to the observer who measures the half-life. This is the proper time $\boldsymbol{\Delta t_0}$. Muons produced by cosmic ray particles have a range of velocities, with some moving near the speed of light. It has been found that the muon’s half-life as measured by an Earth-bound observer ($\boldsymbol{\Delta t}$) varies with velocity exactly as predicted by the equation $\boldsymbol{\Delta t = \gamma \Delta t_0}$. The faster the muon moves, the longer it lives. We on the Earth see the muon’s half-life time dilated—as viewed from our frame, the muon decays more slowly than it does when at rest relative to us.

### Example 1: Calculating Δt for a Relativistic Event: How Long Does a Speedy Muon Live?

Suppose a cosmic ray colliding with a nucleus in the Earth’s upper atmosphere produces a muon that has a velocity $\boldsymbol{v = 0.950 c}$. The muon then travels at constant velocity and lives $\boldsymbol{1.52 \;\mu \textbf{s}}$ as measured in the muon’s frame of reference. (You can imagine this as the muon’s internal clock.) How long does the muon live as measured by an Earth-bound observer? (See Figure 4.)

Strategy

A clock moving with the system being measured observes the proper time, so the time we are given is $\boldsymbol{\Delta t_0 = 1.52 \;\mu \textbf{s}}$. The Earth-bound observer measures $\boldsymbol{\Delta t}$ as given by the equation $\boldsymbol{\Delta t = \gamma \Delta t_0}$. Since we know the velocity, the calculation is straightforward.

Solution

1) Identify the knowns. $\boldsymbol{v=0.950c}$, $\boldsymbol{\Delta t_0 = 1.52 \;\mu \textbf{s}}$

2) Identify the unknown. $\boldsymbol{\Delta t}$

3) Choose the appropriate equation.

Use,

$\boldsymbol{\Delta t = \gamma \Delta t_0},$

where

$\boldsymbol{\gamma =}$$\boldsymbol{\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}}.$

4) Plug the knowns into the equation.

First find $\boldsymbol{\gamma}$.

$\begin{array}{r @{{}={}}l} \boldsymbol{\gamma} & \boldsymbol{\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}} \\[1em] & \boldsymbol{\frac{1}{\sqrt{1 - \frac{(0.950c)^2}{c^2}}}} \\[1em] & \boldsymbol{\frac{1}{\sqrt{1 - (0.950)^2}}} \\[1em] & \boldsymbol{3.20}. \end{array}$

Use the calculated value of $\boldsymbol{\gamma}$ to determine $\boldsymbol{\Delta t}$.

$\begin{array}{r @{{}={}}l} \boldsymbol{\Delta t} & \boldsymbol{\gamma \Delta t_0} \\[1em] & \boldsymbol{(3.20)(1.52 \;\mu \textbf{s})} \\[1em] & \boldsymbol{4.87 \;\mu\textbf{s}} \end{array}$

Discussion

One implication of this example is that since $\boldsymbol{\gamma = 3.20}$ at $\boldsymbol{95.0 \%}$ of the speed of light ($\boldsymbol{v = 0.950c}$), the relativistic effects are significant. The two time intervals differ by this factor of 3.20, where classically they would be the same. Something moving at $\boldsymbol{0.950c}$ is said to be highly relativistic.

Another implication of the preceding example is that everything an astronaut does when moving at $\boldsymbol{95.0 \%}$ of the speed of light relative to the Earth takes 3.20 times longer when observed from the Earth. Does the astronaut sense this? Only if she looks outside her spaceship. All methods of measuring time in her frame will be affected by the same factor of 3.20. This includes her wristwatch, heart rate, cell metabolism rate, nerve impulse rate, and so on. She will have no way of telling, since all of her clocks will agree with one another because their relative velocities are zero. Motion is relative, not absolute. But what if she does look out the window?

### Real-World Connections

It may seem that special relativity has little effect on your life, but it is probably more important than you realize. One of the most common effects is through the Global Positioning System (GPS). Emergency vehicles, package delivery services, electronic maps, and communications devices are just a few of the common uses of GPS, and the GPS system could not work without taking into account relativistic effects. GPS satellites rely on precise time measurements to communicate. The signals travel at relativistic speeds. Without corrections for time dilation, the satellites could not communicate, and the GPS system would fail within minutes.

An intriguing consequence of time dilation is that a space traveler moving at a high velocity relative to the Earth would age less than her Earth-bound twin. Imagine the astronaut moving at such a velocity that $\boldsymbol{\gamma =30.0}$, as in Figure 5. A trip that takes 2.00 years in her frame would take 60.0 years in her Earth-bound twin’s frame. Suppose the astronaut traveled 1.00 year to another star system. She briefly explored the area, and then traveled 1.00 year back. If the astronaut was 40 years old when she left, she would be 42 upon her return. Everything on the Earth, however, would have aged 60.0 years. Her twin, if still alive, would be 100 years old.

The situation would seem different to the astronaut. Because motion is relative, the spaceship would seem to be stationary and the Earth would appear to move. (This is the sensation you have when flying in a jet.) If the astronaut looks out the window of the spaceship, she will see time slow down on the Earth by a factor of $\boldsymbol{\gamma =30.0}$. To her, the Earth-bound sister will have aged only 2/30 (1/15) of a year, while she aged 2.00 years. The two sisters cannot both be correct.

As with all paradoxes, the premise is faulty and leads to contradictory conclusions. In fact, the astronaut’s motion is significantly different from that of the Earth-bound twin. The astronaut accelerates to a high velocity and then decelerates to view the star system. To return to the Earth, she again accelerates and decelerates. The Earth-bound twin does not experience these accelerations. So the situation is not symmetric, and it is not correct to claim that the astronaut will observe the same effects as her Earth-bound twin. If you use special relativity to examine the twin paradox, you must keep in mind that the theory is expressly based on inertial frames, which by definition are not accelerated or rotating. Einstein developed general relativity to deal with accelerated frames and with gravity, a prime source of acceleration. You can also use general relativity to address the twin paradox and, according to general relativity, the astronaut will age less. Some important conceptual aspects of general relativity are discussed in Chapter 34.2 General Relativity and Quantum Gravity of this course.

In 1971, American physicists Joseph Hafele and Richard Keating verified time dilation at low relative velocities by flying extremely accurate atomic clocks around the Earth on commercial aircraft. They measured elapsed time to an accuracy of a few nanoseconds and compared it with the time measured by clocks left behind. Hafele and Keating’s results were within experimental uncertainties of the predictions of relativity. Both special and general relativity had to be taken into account, since gravity and accelerations were involved as well as relative motion.

1: What is $\boldsymbol{\gamma}$ if $\boldsymbol{v = 0.650c}$?

2: A particle travels at $\boldsymbol{1.90 \times 10^8 \;\textbf{m/s}}$ and lives $\boldsymbol{2.10 \times 10^{-8} \;\textbf{s}}$ when at rest relative to an observer. How long does the particle live as viewed in the laboratory?

# Section Summary

• Two events are defined to be simultaneous if an observer measures them as occurring at the same time. They are not necessarily simultaneous to all observers—simultaneity is not absolute.
• Time dilation is the phenomenon of time passing slower for an observer who is moving relative to another observer.
• Observers moving at a relative velocity $\boldsymbol{v}$ do not measure the same elapsed time for an event. Proper time $\boldsymbol{\Delta t_0}$ is the time measured by an observer at rest relative to the event being observed. Proper time is related to the time $\boldsymbol{\Delta t}$ measured by an Earth-bound observer by the equation
$\boldsymbol{\Delta t =}$$\boldsymbol{\frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}}$$\boldsymbol{= \gamma \Delta t_0},$

where

$\boldsymbol{\gamma =}$$\boldsymbol{\frac{1}{\sqrt{1- \frac{v^2}{c^2}}}}.$
• The equation relating proper time and time measured by an Earth-bound observer implies that relative velocity cannot exceed the speed of light.
• The twin paradox asks why a twin traveling at a relativistic speed away and then back towards the Earth ages less than the Earth-bound twin. The premise to the paradox is faulty because the traveling twin is accelerating. Special relativity does not apply to accelerating frames of reference.
• Time dilation is usually negligible at low relative velocities, but it does occur, and it has been verified by experiment.

### Conceptual Questions

1: Does motion affect the rate of a clock as measured by an observer moving with it? Does motion affect how an observer moving relative to a clock measures its rate?

2: To whom does the elapsed time for a process seem to be longer, an observer moving relative to the process or an observer moving with the process? Which observer measures proper time?

3: How could you travel far into the future without aging significantly? Could this method also allow you to travel into the past?

### Problems & Exercises

1: (a) What is $\boldsymbol{\gamma}$ if $\boldsymbol{v = 0.250c}$? (b) If $\boldsymbol{v = 0.500c}$?

2: (a) What is $\boldsymbol{\gamma}$ if $\boldsymbol{v = 0.100c}$? (b) If $\boldsymbol{v = 0.900c}$?

3: Particles called $\boldsymbol{\pi}$-mesons are produced by accelerator beams. If these particles travel at $\boldsymbol{2.70 \times 10^8 \;\textbf{m/s}}$ and live $\boldsymbol{2.60 \times 10^{-8} \;\textbf{s}}$ when at rest relative to an observer, how long do they live as viewed in the laboratory?

4: Suppose a particle called a kaon is created by cosmic radiation striking the atmosphere. It moves by you at $\boldsymbol{0.980c}$, and it lives $\boldsymbol{1.24 \times 10^{-8} \;\textbf{s}}$ when at rest relative to an observer. How long does it live as you observe it?

5: A neutral $\boldsymbol{\pi}$-meson is a particle that can be created by accelerator beams. If one such particle lives $\boldsymbol{1.40 \times 10^{-16} \;\textbf{s}}$ as measured in the laboratory, and $\boldsymbol{0.840 \times 10^{-16} \;\textbf{s}}$ when at rest relative to an observer, what is its velocity relative to the laboratory?

6: A neutron lives 900 s when at rest relative to an observer. How fast is the neutron moving relative to an observer who measures its life span to be 2065 s?

7: If relativistic effects are to be less than 1%, then $\boldsymbol{\gamma}$ must be less than 1.01. At what relative velocity is $\boldsymbol{\gamma =1.01}$?

8: If relativistic effects are to be less than 3%, then $\boldsymbol{\gamma}$ must be less than 1.03. At what relative velocity is $\boldsymbol{\gamma = 1.03}$?

9: (a) At what relative velocity is $\boldsymbol{\gamma = 1.50}$? (b) At what relative velocity is $\boldsymbol{\gamma = 100}$?

10: (a) At what relative velocity is $\boldsymbol{\gamma = 2.00}$? (b) At what relative velocity is $\boldsymbol{\gamma =10.0}$?

11: Unreasonable Results

(a) Find the value of $\boldsymbol{\gamma}$ for the following situation. An Earth-bound observer measures 23.9 h to have passed while signals from a high-velocity space probe indicate that 24.0 h have passed on board. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

## Glossary

time dilation
the phenomenon of time passing slower to an observer who is moving relative to another observer
proper time
$\boldsymbol{\Delta t_0}$. the time measured by an observer at rest relative to the event being observed: $\boldsymbol{\Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma \Delta t_0}$, where $\boldsymbol{\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}}$
this asks why a twin traveling at a relativistic speed away and then back towards the Earth ages less than the Earth-bound twin. The premise to the paradox is faulty because the traveling twin is accelerating, and special relativity does not apply to accelerating frames of reference

### Solutions

1: $\boldsymbol{\gamma =}$$\boldsymbol{\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}}$$\boldsymbol{=}$$\boldsymbol{\frac{1}{1 - \sqrt{\frac{(0.650c)^2}{c^2}}}}$$\boldsymbol{= 1.32}$

2: $\boldsymbol{\Delta t =}$$\boldsymbol{\frac{{\Delta _t}^0}{\sqrt{1 - \frac{v^2}{c^2}}}}$$\boldsymbol{=}$$\boldsymbol{\frac{2.10 \times 10^{-8} \;\textbf{s}}{\sqrt{1 - \frac{(1.90 \times 10^8 \;\textbf{m/s})^2}{(3.00 \times 10^8 \;\textbf{m/s})^2}}}}$$\boldsymbol{= 2.71 \times 10^{-8} \;\textbf{s}}$

Problems & Exercises

1: (a) 1.0328

(b) 1.15

3: $\boldsymbol{5.96 \times 10^{-8} \;\textbf{s}}$

5: $\boldsymbol{0.800c}$

7: $\boldsymbol{0.140c}$

9: (a) $\boldsymbol{0.745c}$

(b) $\boldsymbol{0.99995 c}$ (to five digits to show effect)

11: (a) 0.996

(b) $\boldsymbol{\gamma}$ cannot be less than 1.

(c) Assumption that time is longer in moving ship is unreasonable.


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# Nonlocal hydrodynamic transport and collective excitations in Dirac fluids

@article{Kiselev2020NonlocalHT,
title={Nonlocal hydrodynamic transport and collective excitations in Dirac fluids},
author={Egor I. Kiselev and Joerg Schmalian},
journal={Physical Review B},
year={2020}
}
• Published 1 July 2020
• Physics
• Physical Review B
We study the response of a Dirac fluid to electric fields and thermal gradients at finite wave-numbers and frequencies in the hydrodynamic regime. We find that non-local transport in the hydrodynamic regime is governed by infinite set of kinetic modes that describe non-collinear scattering events in different angular harmonic channels. The scattering rates of these modes $\tau_{m}^{-1}$ increase as $\|m\|$, where $m$ labels the angular harmonics. In an earlier publication, we pointed out that…
5 Citations

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Question

# $$\Box ABCD$$ is a parallelogram. The ratio of $$\angle A$$ and $$\angle B$$ of this parallelogram is $$5 : 4$$. Find the measure of $$\angle B$$.

Solution

## Here, $$ABCD$$ is a parallelogram.Let the measure of $$\angle A$$ and $$\angle B$$ be $$5x$$ and $$4x$$.We know that in parallelogram sum of measure of adjacent angles is supplementary.$$\therefore$$  $$\angle A+\angle B=180^o$$$$\Rightarrow$$  $$5x+4x=180^o$$$$\Rightarrow$$  $$9x=180^o.$$$$\Rightarrow$$  $$x=20^o.$$Measure of $$\angle B=4x=4\times 20^o=80^o.$$ Maths

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Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 7 Quiz Answer Key.

Use the table to find each ratio. _____________

Question 1.
white socks to brown socks Color _____________
Write the ratio of white socks to brown socks in three different ways

part to part
8 : 5 , $$\frac{8}{5}$$, 8 white socks to 5 brown socks
8 : 5 = Final solution

Question 2.
blue socks to non blue socks _____________
Solution to this example is given below
Non blue socks = 8 + 6 + 5 = 19
Write the ratio of blue socks to non blue socks in three different ways

part to part
4 : 19, $$\frac{4}{9}$$, 4 blue socks to 19 non blue socks
4 : 19 = Final Solution

Question 3.
black socks to all of the socks _____________
Solution to this example is given below
All of the socks = 8 + 6 + 4 + 5 = 23
Write the ratio of black socks to alt of the socks in three different ways

part to part
6 : 23, $$\frac{6}{23}$$ 6 black socks to 23 all of the socks
6 : 23 = Final solution

Question 4.
Find two ratios equivalent to the ratio in Exercise 1.
There are 8 white socks and 5 brown socks so the ratio of white socks to brown socks is 8 : 5 which can be written
as $$\frac{8}{5}$$ 2 equivalent ratios of this are $$\frac{8 \times 2}{5 \times 2}=\frac{16}{10}$$ and $$\frac{8 \times 3}{5 \times 3}=\frac{24}{15}$$

7.2 Rates

Find each rate.

Question 5.
Earl runs 75 meters in 30 seconds. How many meters does Earl run per second? _____________
Find the unit rate. How many meters Earl ran in a second

Divide to find the unit rate
$$\frac{75 \text { meters }}{30 \text { seconds }}=\frac{2.5 \text { meters }}{1 \text { second }}$$

Earl ran 2.5 meters per second
2.5 = Final solution

Question 6.
The cost of 3 scarves is $26.25. What is the unit price? _____________ Answer: Find the unit rate. What is the unit price? Divide to find the unit rate $$\frac{26.25 \text { dollars }}{3 \text { scarves }}=\frac{8.75 \text { dollars }}{1 \text { scarf }}$$ The cost of 1 scarf is 8.75 dollars 8.75 = Final solution 7.3 Using Ratios and Rates to Solve Problems Question 7. Danny charges$35 for 3 hours of swimming lessons. Martin charges $24 for 2 hours of swimming lessons. Who offers a better deal? Answer: Evaluate the cost of 1 hour of swimming. Danny charges$35 for 3 hours so $$\frac{\ 35}{3}$$ = $11.6 per hour while Martin charges$24 for 2 hours so $$\frac{\ 24}{2}$$ = $12 per hour. Since 12 > 11.6, Danny’s charges less so offers a better deal. Danny offers a better deal. Question 8. There are 32 female performers in a dance recital. The ratio of men to women is 3:8. How many men are in the dance recital? Answer: Multiply the numerator and denominator by the $$\frac{4}{4}$$. $$\frac{3}{8}=\frac{3}{8} \cdot \frac{4}{4}=\frac{12}{32}$$ There are 12 men. Essential Question Question 9. How can you use ratios and rates to solve problems? Answer: Ratios and rates are used to solve problems as they numerically define the relationship between the 2 quantities under consideration. Texas Go Math Grade 6 Module 7 Mixed Review Texas Test Prep Answer Key Selected Response Question 1. Which ratio is not equivalent to the other three? (A) $$\frac{2}{3}$$ (B) $$\frac{6}{9}$$ (C) $$\frac{12}{15}$$ (D) $$\frac{18}{27}$$ Answer: (C) $$\frac{12}{15}$$ Explaination: It can be seen $$\frac{2}{3}=\frac{2 \times 3}{3 \times 3}=\frac{6}{9}$$ and $$\frac{2}{3}=\frac{2 \times 9}{3 \times 9}=\frac{18}{27}$$, so option C. Question 2. A lifeguard received 15 hours of first aid training and 10 hours of cardiopulmonary resuscitation (CPR) training. What is the ratio of hours of CPR training to hours of first aid training? (A) 15:10 (B) 15:25 (C) 10:15 (D) 25:15 Answer: (C) 10:15 Explaination: The ratio of hours of CPR training to hours of first aid training is 10 : 15, therefore Option C. Question 3. Jerry bought 4 DVDs for$25.20. What was the unit rate?
(A) $3.15 (B)$4.20
(C) $6.30 (D)$8.40
(C) $6.30 Explaination: The divisor has one decimal place, so multiply both the dividend and the divisor by 10 so that the divisor is a whole number 4 × 10 = 40 25.20 × 10 = 252 Divide: Question 4. There are 1,920 fence posts used in a 12-kilometer stretch offence. How many fence posts are used in 1 kilometer offence? (A) 150 (B) 160 (C) 155 (D) 180 Answer: (B) 160 Explaination: Solution to this example is given below 160 fence posts are used in 1 kilometer of fence. Question 5. Sheila can ride her bicycle 6,000 meters in 15 minutes. How far can she ride her bicycle in 2 minutes? (A) 400 meters (B) 600 meters (C) 800 meters (D) 1 ,000 meters Answer: (C) 800 meters Explaination: Solution to this example is given below So in 1 minute can ride 400 meters Therefore in 2 minutes, she can travel a distance of 800 meters. Question 6. Lennon has a checking account. He withdrew$130 from an ATM Tuesday. Wednesday he deposited $240. Friday he wrote a check for$56. What was the total change in Lennon’s account?
(A) – $74 (B)$54
(C) $166 (D)$184
(D) $184 Explaination: First, he withdrew$130. So the change in the account is – $130. After that, he deposited$240. So the change in the account is
– $130 + 240 =$110
And finally, he wrote a check for $56. So the total change in the account is$240 + (- 56) = \$ 184

Question 7.
Cheyenne is making a recipe that uses 5 cups of beans and 2 cups of carrots. Which combination below uses the same ratio of beans to carrots?
(A) 10 cups of beans and 3 cups of carrots
(B) 10 cups of beans and 4 cups of carrots
(C) 12 cups of beans and 4 cups of carrots
(D) 12 cups of beans and 5 cups of carrots
(B) 10 cups of beans and 4 cups of carrots

Explaination:
The combination given in option B is according to the ratio of the recipe. Both items are increased by the same constant 2, that is are doubted.

Question 8.
$$\frac{5}{8}$$ of the 64 musicians in a music contest are guitarists. Some of the guitarists play jazz solos, and the rest play classical solos. The ratio of the number of guitarists playing jazz solos to the total number of guitarists in the contest is 1:4. How many guitarists play classical solos in the contest?
(A) 10
(B) 20
(C) 30
(D) 40
Multiply the numerator and denominator by $$\frac{8}{8}$$:
$$\frac{5}{8}=\frac{5}{8} \cdot \frac{8}{8}=\frac{40}{64}$$
There are 40 guitarists.
Multiply the numerator and denominator by $$\frac{10}{10}$$:
$$\frac{1}{4}=\frac{1}{4} \cdot \frac{10}{10}=\frac{10}{40}$$
10 guitarists play classical solos.

Gridded Response

Question 9.
Mikaela is competing in a race in which she both runs and rides a bicycle. She runs 5 kilometers in 0.5 hour and rides her bicycle 20 kilometers in 0.8 hour. At this rate, how many kilometers can Mikaela ride her bicycle in one hour?

To find the solution, multiply this unit rate by $$\frac{0.62}{0.62}$$.
$$\frac{25}{1}=\frac{25}{1} \cdot \frac{0.62}{0.62}=\frac{15.5}{0.62}$$


text:
# crsfml

Bindings to SFML multimedia/game library
2.5.2 Latest release released
oprypin/crsfml
267 10 1
Oleh Prypin

## Introduction

CrSFML is a library that allows SFML to be used with the Crystal programming language. SFML is a library written in C++, so CrSFML also needs to ship thin C bindings to SFML.

To quote the official site of SFML,

SFML provides a simple interface to the various components of your PC, to ease the development of games and multimedia applications.

Indeed, SFML is most often used to make video games. It provides features such as hardware-accelerated 2D graphics, handling keyboard, mouse and gamepad input, vector and matrix manipulation, managing windows (can also be used as a base for OpenGL drawing), working with multiple image formats, audio playback and recording, basic networking... Check out some demos of CrSFML to see what it can do.

CrSFML consists almost entirely of automatically generated code, based on SFML's header files. Read more about generated code and API differences between SFML and CrSFML.

## Installation

First, install SFML.

Then create a shard.yml file in your project's folder (or add to it) with the following contents:

name: awesome-game
version: 0.1.0

dependencies:
crsfml:
github: oprypin/crsfml
version: ~> 2.5.2

Resolve dependencies with Shards:

shards install

During installation this will invoke make to build the C++ wrappers as object files. If that fails, see Custom SFML location.

Try running an example:

cp lib/crsfml/examples/snakes.cr .
crystal snakes.cr

Now you're ready for the tutorials!

##### Windows

Crystal does not officially support Windows, but CrSFML supports it and is perfectly usable already. See a video detailing the full setup.

### Prerequisites

The C++ wrappers require a C++ compiler (C++03 will do).

#### Install SFML

SFML must be installed, with the version that matches SFML_VERSION in src/version.cr (usually latest). If it doesn't, no need to look for an older release of CrSFML, just re-generate the sources for your version. SFML versions 2.3.x through 2.5.x are supported by CrSFML.

There are detailed official instructions on how to install SFML manually, but on many systems there are easier ways.

If SFML is not installed in a global/default location, see Custom SFML location.

##### Linux

Many Linux-based systems provide SFML through their package manager. Make sure to install the -dev packages if there is such a separation in your Linux distribution of choice.

Note that most often the packages provided by Linux distributions are outdated. If you're installing an older version of SFML (not recommended), make sure that it's still supported by CrSFML. You will need to re-generate the sources.

##### Mac

The easiest way to install SFML on macOS is through the Homebrew package manager:

brew update
brew install sfml

It can also be installed by copying binaries, as described in official instructions.

##### Windows

Downloading the official binaries ("Visual C++ 15 (2017) - 64-bit") will do. Check out the video on how to set things up on Windows.

##### From source

cmake .
cmake --build .
sudo make install  # optional!

In some cases the dependencies are bundled with SFML, otherwise see the official build instructions.

### Custom SFML location

If SFML's headers and libraries are not in a path where the compiler would look by default (and the defaults usually work only on Linux), additional steps are needed.

First, before building the extensions (make) or generating sources, you need to configure the include path:

export SFML_INCLUDE_DIR=/full/path/to/sfml/include

Windows equivalent:

set INCLUDE=%INCLUDE%;C:\path\to\sfml\include

Setting these variables beforehand can also fix shards install.

Then, whenever using CrSFML, you need to configure the path to SFML libraries so the linker can find them. To apply these for the current shell session, run:

export LIBRARY_PATH=/full/path/to/sfml/lib     # Used during linking
export LD_LIBRARY_PATH=/full/path/to/sfml/lib  # Used when running an executable

Windows equivalent:

set LIB=%LIB%;c:\path\to\sfml\lib
set PATH=%PATH%;c:\path\to\sfml\bin

CrSFML's top-level scripts also need the include path to work. E.g. crystal generate.cr -- /full/path/to/sfml/include.

### Generating sources

CrSFML's sources come almost entirely from a generator program. They are based on a particular version of SFML. But as sources for the latest version are already bundled, usually you don't need to do this. More details.

As this is out of scope for Shards, let's download the repository separately (then use CrSFML without Shards).

git clone https://github.com/oprypin/crsfml
cd crsfml

Then we can generate the sources, either directly with crystal generate.cr or as part of the build process:

touch generate.cr
make

If run successfully, this generates all the source files and also compiles the C++ wrapper.

### CrSFML without Shards

It's also possible to use CrSFML outside of Shards, as with any library. One option is to directly create a symbolic link to CrSFML in your project's lib folder.

mkdir lib
ln -s /full/path/to/crsfml lib/crsfml

Another option is to export CRYSTAL_PATH=/full/path/to a directory that contains the crsfml directory.

## CrSFML as bindings to SFML

### API differences between SFML and CrSFML

The API of CrSFML (a library for Crystal) attempts to be similar to SFML (a C++ library), but some general changes are present:

• Methods are renamed to snake_case.
• Getter, setter methods are changed:
• x.getSomeProperty() becomes x.some_property.
• x.isSomeProperty(), x.hasSomeProperty() become x.some_property?.
• x.setSomeProperty(v) becomes x.some_property = v.
• Structs in Crystal are always passed by copy, so modifying them can be problematic. For example, my_struct.x = 7 is fine but array_of_structs[2].x = 5 will not work. To work around this, copy the whole struct, modify it, then write it back. Better yet, avoid the need to modify structs (work with them like with immutable objects).
• Member functions, such as loadFromFile, that are used for initialization, each have a corresponding shorthand class method (from_file) that raises SF::InitError on failure.
• SFML sometimes uses enum values as bitmasks. You can combine them using the | operator.
• enum members are exposed at class level, so instead of SF::Keyboard::Code::Slash you can use SF::Keyboard::Slash.
• SFML sometimes requires that an instance must remain alive as long as it is attached to the object. For example, a textured shape will cause errors if the texture object is destroyed. CrSFML prevents this problem by keeping a reference to the object.
• The Event union and EventType enum are represented as a class hierarchy. Instead of ev.type == SF::Event::Resized use ev.is_a?(SF::Event::Resized); instead of ev.size.width use ev.width.
• Instead of subclassing Drawable, include the Drawable module with an abstract draw method.
• Most of the API documentation is taken directly from SFML, so don't be surprised if it talks in C++ terms.

### The C++ wrapper

The interface of the C++ → C wrapper (which Crystal ultimately binds to) consists entirely of simple functions that accept only native types (such as float, uint32_t, char*) and untyped pointers (void*). The untyped pointers are never exposed to the user, only to other auto-generated parts of the code. The function names consist of the original SFML class name, the function name itself, and a base62 hash of the parameter types. Return types are never used; instead, the output is done into a pointer (which is usually the last argument of the function), but, as usual, the memory allocation is the caller's job. The first argument of each function is a pointer to the receiver object (if applicable).

Abstract classes are implemented by exposing a collection of global callback variables, which must be set by the user if they want to use the corresponding class. The callback's first argument is the object, and some arguments are pointers that need to be assigned to inside the callback implementation (because return values are not used).

Compilation of the C++ extensions is based only on SFML's header files, these are made into object files, and all the linking is deferred to the final linker invocation done by Crystal.

#### Why not CSFML?

CSFML is a great library that allows SFML to be used with C. It goes to great lengths to be human-friendly and does a good job of converting C++ idioms to C idioms. In the past CrSFML used to be based on it, but after a while it became apparent that the advantages of CSFML's nice interface are also disadvantages when constructing (especially auto-generated) bindings that attempt to look as close to the real SFML as possible.

Many details about functions' signatures are lost, as well as function overloads. Names of data types had to be simplified (not namespaced). And many other such small things that bring the frustration of having to reconstruct the details of the original SFML interface based on the simplified CSFML interface.

There are many aspects that prevent an efficient implementation from the standpoint of bindings, most importantly, CSFML takes memory allocation into its own hands, so any object creation in CrSFML involved allocation of two objects on the heap by two different libraries, and every interaction with it had to go through at least two pointers. Structs in CSFML are actually completely separate data types and they have to be constantly be converted between a "SFML-struct" and a "CSFML-struct".

Instead of that, the C++ → C wrapper passes the bare SFML data types directly through untyped pointers, and relies on the higher-level binding to deal safely with them. In case of structs the data layout is mirrored, in case of classes the pointers remain completely opaque.

Not to forget that the wrapper is made automatically, so it can be quickly updated to any SFML release and prevents human error that could happen when implementing CSFML.

## Credits

CrSFML was made by Oleh Prypin.

This library uses and is based on SFML.

Thanks to Alan Willms for translating tutorials to Crystal.

crsfml:
github: oprypin/crsfml
version: ~> 2.5.2
Crystal >= 0.25.1

### Releases

Show all 20 releases
Last synced .


text:
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

JH

# (a) What is the difference between a sequence and a series?(b) What is a convergent series? What is a divergent series?

## (a) A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers.(b) A series is convergent if the sequence of partial sums is a convergent sequence. A series is divergent if it is not convergent.

Sequences

Series

### Discussion

You must be signed in to discuss.
##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp

### Video Transcript

for Part A. We'd liketo list the difference between a sequence and a Siri's. Well, the sequence usually denoted something like this, braces a N. Then you have your starting point, your ending point, or you could denoted by listing its values in order. It's just the order list of numbers, however. A Siri's Siri's will arise from a sequence. So from the sequence, we can go to a new sequence which will call us in and goes from one to infinity. And let's define the ends firm of this new sequence. SN to just be the sum of the first and values of the original sequence. So, for example, as one is a one as to is a one plus a two and so on that here, the Siri's will be the limit as n goes to infinity of s end. So that's the difference is a sequence is this is a list of numbers in order. A Siri's is a limit of a sequence. So for part B was a convergent Siri's well, a Siri's will converge. So let's write it this way. Brackets or braces. Excuse me. SN from one to infinity converges if till limit as and goes to infinity. SN is a real number. So again by this sn, we still mean what we met in part made, that s and is the sum of the first and values. So in other words, this is saying so. This is equivalent to saying the limit and goes to infinity of a one plus dot, dot, dot plus a M. Israel is a real number, and as you could see, is you and go to infinity. This finite sum here in the apprentices well, get closer and closer to the infinite sum. And that's the limit of us. And so this is what it means to converge means a CZ. You take more, more sums. The women exist. Otherwise, the limit of SN is not a real number. So let's just right. It does not exist. Or you could just say its equal two equals infinity or minus infinity and any of these cases. So the limit is not a real number. Then we say diversions, and that's your final answer.

JH

#### Topics

Sequences

Series

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp


text:
# Tag Info

36

Edit: Steven's answer seems to be in the right direction, so here's my updated answer: As such, it seems the best fit for the final answer would be

31

I notice straight away that This means that even without knowing the cipher, I was able to work out a lot of these right off the bat, and from that I was able to work out that the cipher is I won't write out the sums for every single item (at least, not right now), but here are the deciphered items: Movies: The odd one out is Influential People: The ...

30

Ok I think I have an overall answer that I'm happy with all the steps of: Previous answer/working for posterity: Ok I think I know how to solve it, and I want to get in early...

27

Is it the 3rd owl in the 6th row? The pattern I have seen is that So this is a copycat owl who needs to be sho-who-ed away before he learns the ways of this flock. To clarify:

26

The word that doesn't belong is The crossword looks like this: How I solved it:

25

Due to the occurrence of $0.4\overline{8}$ rather than $\frac{22}{45}$ I think it could be Because An alternative, quirky one that I kind of like, but definitely not the intended one: Because Or how about Because I'll squeeze another in here... Because

23

These questions are why many people ridicule IQ tests. I'm guessing the thinking behind number 1 is that Sydney is the only one that's not a national capital. But you would make a case for any of these. My first thought on seeing the question was that it was Buenos Aires, as it's the only one in the Western Hemisphere. Moscow is the only one in a formerly ...

22

For question 1, the odd city out is For question 2, the answer is For question 3, the odd metal out is For question 4, the answer is

20

Perhaps: Another possibility:

18

The cone is the odd one out because it is the only one that isn't smooth and the only one that's of infinite extent (at least if you're a mathematician). The torus is the odd one out because it's the only one that isn't simply connected. The sphere is the odd one out because it's the only one on which euclidean symmetries act transitively (i.e., all its ...

18

I can sympathize. This is from The Guardian, 2015: From left to right, starting with image 2, we have: no outline not a square green small leaving image 1 because it is the only image that is not an odd one out! Which I think qualifies it as an odd one out. So the answer given might not be unique, which makes some of the puzzles unsolvable.

14

14

If you're positive, I'm so super attractive- My flow is current. I am high spy, Or a big help on the road. I can be for Sky. Typical package, You can do me in a ring, But I'm just a square Rock, or metallic? Holder of keys, or your vows, Or maybe a sound? I think the one that doesn't fit is

14

This sounds like Namely... My second older brother was a dangerous man, a pig who gorged himself on all our land. He hurt our mother’s feelings (called her a cow), so she hurt him back from belly to brow. My oldest sister was wise, her stare — penetrating. It pinned him to the ground. He tried to say sorry but died, while my oldest brother ran on foot ...

14

Obviously wrong answer ... but it amuses me and it might amuse OP too. Perhaps the book that's in the wrong place is because

12

Like Votbear, the first thing I noticed is Since my first guess is apparently wrong, my second guess would be

11

I would say

11

The Situation that you are talking about is a lot more common than you know. And it's not only limited to odd-one-out puzzles. There are riddles which are basically "guess what I am thinking" puzzles. Odd-one-out puzzles usually land in this basket. There is nothing wrong with your thinking or your reasoning. The fault, almost always, lies with the question....

11

The odd one out is First I’ll need to mention that the puzzles, in order, are The solution: Rough explanations of the others: 1) 2) 4) 5)

11

There may be more than one connection, but one odd-one-out is: If the rule is the common words form a

9

The topologically equivalent explanation makes sense to me, since all the objects except the torus can be transformed into eachother. On the other hand, as mentioned, there can be many different explanations. The cone is the only one with a flat surface. The ellipsoid is the only one that can't be described using only circles. But if you're a topologist, ...

9

8

This might help you visualize no. 2 a bit better: Although you could probably find different patterns, to me this seems like the simplest way to describe the progression. I believe such sequences are fairly common in IQ tests. For example, if you had to continue the sequence: 3, 1, 0, 2, -3, 3, -6, 4, ?, ? a simple approach would be to There may easily ...

8

because

8

Partial Answer: I. II. III. V.

8

Is an odd one out because The others:

8

Depending on interpretation, any of them is the odd one out: A: B: C: D: E:

8


text:
# A Framework for Chosen IV Statistical Analysis of Stream Ciphers

@inproceedings{Englund2007AFF,
title={A Framework for Chosen IV Statistical Analysis of Stream Ciphers},
author={H{\aa}kan Englund and Thomas Johansson and Meltem S{\"o}nmez Turan},
booktitle={INDOCRYPT},
year={2007}
}
• Published in INDOCRYPT 9 December 2007
• Computer Science
Saarinen recently proposed a chosen IV statistical attack, called the d-monomial test, and used it to find weaknesses in several proposed stream ciphers. In this paper we generalize this idea and propose a framework for chosen IV statistical attacks using a polynomial description. We propose a few new statistical attacks, apply them on some existing stream cipher proposals, and give some conclusions regarding the strength of their IV initialization. In particular, we experimentally detected… Expand
107 Citations
Chosen IV Statistical Analysis for Key Recovery Attacks on Stream Ciphers
• Computer Science
• AFRICACRYPT
• 2008
Evidence is given that the present analysis is not applicable on Grain-128 or Trivium with full IV initialization, and it is experimentally demonstrated how to deduce a few key bits. Expand
A New Chosen IV Statistical Attack on Grain-128a Cipher
• Computer Science
• 2017 International Conference on Cyber-Enabled Distributed Computing and Knowledge Discovery (CyberC)
• 2017
A new chosen IV statistical attack is proposed that enables us to distinguish between random sequences and keystreams of Grain-128a up to 169 initial rounds with the 2^26 computational complexity and can reveal the weaknesses that are not possible to be found by d-monomial tests. Expand
High order differential attacks on stream ciphers
• Mathematics, Computer Science
• Cryptography and Communications
• 2012
This work reviews the various techniques of differential cryptanalysis and translates them into the terminology of high order derivatives introduced by Lai, and naturally suggests generalizations and refinements such as conditional differential crypt analysis. Expand
A new chosen IV statistical distinguishing framework to attack symmetric ciphers, and its application to ACORN-v3 and Grain-128a
• Mathematics, Computer Science
• IACR Cryptol. ePrint Arch.
• 2017
A new attack framework based upon cube testers and d-monomial test that is applicable to all symmetric ciphers and hash functions and can reveal weaknesses that are not possible to find by other statistical tests is proposed. Expand
A Distinguish attack on Rabbit Stream Cipher Based on Multiple Cube Tester
This paper shows for a mature cube the authors could easily identify weak subcubes which increase the probability of distinguishing for an unknown secret key, and represents with 2 complexity, using one iteration of next state function the keystream is completely distinguishable from random. Expand
Design and Implementation of a Statistical Testing Framework for a Lightweight Stream Cipher
• Computer Science
• 2020
UWS is used to illustrate the effect of the LFSR choice on possibly distinguishing attacks on the SG and confirm that the proposed UWS scheme is a viable measure of the cryptographic strength of a stream cipher. Expand
Improved Greedy Nonrandomness Detectors for Stream Ciphers
• Computer Science
• ICISSP
• 2017
An improved algorithm is constructed to determine the subset of key and IV-bits used in the maximum degree monomial test, which is generic, and can be applied to any stream cipher. Expand
New Distinguishers Based on Random Mappings against Stream Ciphers
• Mathematics, Computer Science
• SETA
• 2008
A new framework of randomness testing based on random mappings is proposed and three new distinguishers of coverage test, ρ-test and DP-coverage test are presented and applied on Phase III Candidates of eSTREAM project. Expand
A Chosen IV Related Key Attack on Grain-128a
• Computer Science
• ACISP
• 2013
This paper presents a Key recovery attack on Grain-128a, in a chosen IV related Key setting, and shows that using around γ·232 (γ is a experimentally determined constant and it is sufficient to estimate it as 28) related Keys andγ·264 chosen IVs, it is possible to obtain 32·γ simple nonlinear equations and solve them to recover the Secret Key in Grain- 128a. Expand
NOCAS : A Nonlinear Cellular Automata Based Stream Cipher
• Computer Science, Mathematics
• Automata
• 2011
The proposed cipher is shown to be resistant against known existing attacks and designed to produce $\mathrm{2^{128}}$ random keystream bits and initialization phase is made faster 4 times than that of Grain-128. Expand

#### References

SHOWING 1-10 OF 19 REFERENCES
Chosen-IV Statistical Attacks on eSTREAM Stream Ciphers
d-Monomial tests are statistical randomness tests based on Algebraic Normal Form representation of a Boolean function, and were first introduced by Filiol in 2002. We show that there are strongExpand
Decim v2
In this paper, we present Decim, a stream cipher hardwareoriented selected for the phase 3 of the ECRYPT stream cipher project eSTREAM. As required by the initial call for hardware-oriented streamExpand
A New Statistical Testing for Symmetric Ciphers and Hash Functions
A new statistical testing of symmetric ciphers and hash functions which allow us to detect biases in a few of these systems is presented, showing that AES, DES, Snow, and Lili-128 fail the tests wholly or partly and thus present strong biases. Expand
Resynchronization Weaknesses in Synchronous Stream Ciphers
• Computer Science
• EUROCRYPT
• 1993
A powerful general attack on nonlinearly filtered linear (over Z2) systems is presented and this attack is further refined to efficiently cryptanalyze a linear system with a multiplexer as output function. Expand
Extending the Resynchronization Attack
• Computer Science
• Selected Areas in Cryptography
• 2004
This paper extends attacks on ciphers using the standard attack with cryptanalytic techniques such as algebraic attacks and linear cryptanalysis, and shows that using linear resync mechanisms should be avoided. Expand
Weaknesses in the Key Scheduling Algorithm of RC4
• Computer Science
• Selected Areas in Cryptography
• 2001
It is shown that RC4 is completely insecure in a common mode of operation which is used in the widely deployed Wired Equivalent Privacy protocol (WEP, which is part of the 802.11 standard), in which a fixed secret key is concatenated with known IV modifiers in order to encrypt different messages. Expand
A Stream Cipher Proposal: Grain-128
• Computer Science
• 2006 IEEE International Symposium on Information Theory
• 2006
A new stream cipher, Grain-128, is proposed. The design is very small in hardware and it targets environments with very limited resources in gate count, power consumption, and chip area. Grain-128Expand
On the Resynchronization Attack
• Mathematics, Computer Science
• FSE
• 2003
The resynchronization attack on stream ciphers with a linear next-state function and a nonlinear output function is further investigated and an efficient branching algorithm for reconstructing this function along with the secret key is proposed and analyzed. Expand
Trivium Specifications ?
This document specifies Trivium, a hardware oriented synchronous stream cipher which aims to provide a flexible trade-off between speed and area. The description of the cipher is followed by someExpand
A Statistical Test Suite for Random and Pseudorandom Number Generators for Cryptographic Applications
• Mathematics
• 2000
Abstract : This paper discusses some aspects of selecting and testing random and pseudorandom number generators. The outputs of such generators may he used in many cryptographic applications, such asExpand


text:
771 Pages

Structure of sunEdit

 Name depth density Temperature remark Core 0~ 0.25Rs 150,000 kg/m³ (150 times the density of water on Earth) 13,600,000 kelvin or 0.2Rs Radiative zone 0.25~0.7 The density drops a hundredfold (from 20 g/cm³ to only 0.2 g/cm³) from the bottom to the top of the radiative zone.[1] from 7,000,000 °C to about 2,000,000 °C Convective zone 0.7~ 1.0 0.2 g/m³ (about 1/10,000th the density of air at sea level).[2] the temperature has dropped to 5,700° K Photosphere tens to hundreds of kilometers thick ~1023 m−3 (this is about 1% of the particle density of Earth's atmosphere at sea level).[1] 6,000 K,

Core of SunEdit

Main article: Solar core

The core of the Sun is considered to extend from the center to about 0.2 solar radii. It has a density of up to 150,000 kg/m³ (150 times the density of water on Earth) and a temperature of close to 13,600,000 kelvin (by contrast, the surface of the Sun is around 5,800 kelvin). Recent analysis of SOHO mission data favors a faster rotation rate in the core than in the rest of the radiative zone.[3] Through most of the Sun's life, energy is produced by nuclear fusion through a series of steps called the p–p (proton–proton) chain; this process converts hydrogen into helium. The core is the only location in the Sun that produces an appreciable amount of heat via fusion: the rest of the star is heated by energy that is transferred outward from the core. All of the energy produced by fusion in the core must travel through many successive layers to the solar photosphere before it escapes into space as sunlight or kinetic energy of particles.

About 3.4×1038 protons (hydrogen nuclei) are converted into helium nuclei every second (out of ~8.9×1056 total amount of free protons in the Sun), releasing energy at the matter–energy conversion rate of 4.26 million tonnes per second, 383 yottawatts (3.83×1026 W) or 9.15×1010 megatons of TNT per second. This actually corresponds to a surprisingly low rate of energy production in the Sun's core—about 0.3 W/m³ (watts per cubic meter). This is less power than generated by a candle. Power density is about 6 µW/kg of matter. For comparison, the human body produces heat at approximately the rate 1.2 W/kg, roughly a million times greater per unit mass. The use of plasma with similar parameters for energy production on Earth would be completely impractical—even a modest 1 GW fusion power plant would require about 170 billion tonnes of plasma occupying almost one cubic mile. Hence, terrestrial fusion reactors utilize far higher plasma temperatures than those in Sun's interior.

The rate of nuclear fusion depends strongly on density and temperature, so the fusion rate in the core is in a self-correcting equilibrium: a slightly higher rate of fusion would cause the core to heat up more and expand slightly against the weight of the outer layers, reducing the fusion rate and correcting the perturbation; and a slightly lower rate would cause the core to cool and shrink slightly, increasing the fusion rate and again reverting it to its present level.

The high-energy photons (gamma rays) released in fusion reactions are absorbed in only few millimetres of solar plasma and then re-emitted again in random direction (and at slightly lower energy)—so it takes a long time for radiation to reach the Sun's surface. Estimates of the "photon travel time" range between 10,000 and 170,000 years.[4]

After a final trip through the convective outer layer to the transparent "surface" of the photosphere, the photons escape as visible light. Each gamma ray in the Sun's core is converted into several million visible light photons before escaping into space. Neutrinos are also released by the fusion reactions in the core, but unlike photons they rarely interact with matter, so almost all are able to escape the Sun immediately. For many years measurements of the number of neutrinos produced in the Sun were lower than theories predicted by a factor of 3. This discrepancy was recently resolved through the discovery of the effects of neutrino oscillation: the Sun in fact emits the number of neutrinos predicted by the theory, but neutrino detectors were missing 2/3 of them because the neutrinos had changed flavor.

From about 0.25 to about 0.7 solar radii, solar material is hot and dense enough that thermal radiation is sufficient to transfer the intense heat of the core outward.[1] In this zone there is no thermal convection; while the material grows cooler as altitude increases (from 7,000,000 °C to about 2,000,000 °C) this temperature gradient is less than the value of adiabatic lapse rate and hence cannot drive convection.[2] Heat is transferred by radiationions of hydrogen and helium emit photons, which travel only a brief distance before being reabsorbed by other ions.[1] The density drops a hundredfold (from 20 g/cm³ to only 0.2 g/cm³) from the bottom to the top of the radiative zone.[1]

Between the radiative zone and the convection zone is a transition layer called the tachocline. This is a region where the sharp regime change between the uniform rotation of the radiative zone and the differential rotation of the convection zone results in a large shear—a condition where successive horizontal layers slide past one another.[5] The fluid motions found in the convection zone above, slowly disappear from the top of this layer to its bottom, matching the calm characteristics of the radiative zone on the bottom. Presently, it is hypothesized (see Solar dynamo), that a magnetic dynamo within this layer generates the Sun's magnetic field.[2]

Convective zoneEdit

In the Sun's outer layer, from its surface down to approximately 200,000 km (or 70% of the solar radius), the solar plasma is not dense enough or hot enough to transfer the heat energy of the interior outward via radiation (in other words it is opaque enough). As a result, thermal convection occurs as thermal columns carry hot material to the surface (photosphere) of the Sun. Once the material cools off at the surface, it plunges back downward to the base of the convection zone, to receive more heat from the top of the radiative zone. At the visible surface of the Sun, the temperature has dropped to 5,700° K and the density to only 0.2 g/m³ (about 1/10,000th the density of air at sea level).[2]

The thermal columns in the convection zone form an imprint on the surface of the Sun, in the form of the solar granulation and supergranulation. The turbulent convection of this outer part of the solar interior gives rise to a "small-scale" dynamo that produces magnetic north and south poles all over the surface of the Sun.[2] The Sun's thermal columns are Bénard cells and therefore tend to be hexagonal prisms.[6]

PhotosphereEdit

The visible surface of the Sun, the photosphere, is the layer below which the Sun becomes opaque to visible light.[7] Above the photosphere visible sunlight is free to propagate into space, and its energy escapes the Sun entirely. The change in opacity is due to the decreasing amount of H ions, which absorb visible light easily.[7] Conversely, the visible light we see is produced as electrons react with hydrogen atoms to produce H ions.[8][9] The photosphere is actually tens to hundreds of kilometers thick, being slightly less opaque than air on Earth. Because the upper part of the photosphere is cooler than the lower part, an image of the Sun appears brighter in the center than on the edge or limb of the solar disk, in a phenomenon known as limb darkening.[7] Sunlight has approximately a black-body spectrum that indicates its temperature is about 6,000 K, interspersed with atomic absorption lines from the tenuous layers above the photosphere. The photosphere has a particle density of ~1023 m−3 (this is about 1% of the particle density of Earth's atmosphere at sea level).[1]

During early studies of the optical spectrum of the photosphere, some absorption lines were found that did not correspond to any chemical elements then known on Earth. In 1868, Norman Lockyer hypothesized that these absorption lines were because of a new element which he dubbed "helium", after the Greek Sun god Helios. It was not until 25 years later that helium was isolated on Earth.[10]

Inverse biquadrate force and Photogravitational massEdit

In the inverse biquadrate scheme, Gravitational mass of the Sun is given as follows,

$M_s = (d_{ms})^5 (\omega_{ms})^2 / G''$

where G" =9.431 $10^8 m^5/kg/sec^2$

Gravitational density of moon is $228.4 ton/m^3$ Gravitational density of sun is $6.78*228.4 ton/m^3$

Above values are photo-gravitational mass which is given as follows,

$m_{pg} = m_i T_s /T_0$

And the inertial density of the sun is determined with structure of the sun, Example is given for Structure of the Earth

Simple calculation gives a temperature of radiative zone about 23067K with gravitational mass  density of 1548$ton/m^3$

 Name Depth Density Temperature density(pg) Temperature(D/Dpg) Remark Core 0~ 0.25Rs 150 ton/m³ 13,600,000 kelvin 1,148,563.5 2,281,812.8°K 0.2Rs Radiative zone 0.25~0.7 20 g/cm³ 4,500,000 °C 1,548 23,067°K 0.2 g/cm³ Convective zone 0.7~ 1.0 0.2 g/cm³ 5,700° K 3.826 5,700°K Photosphere 50 kilometer ~1023 m−3 6,000 K, - 6,000°K 10s~100s


text:
시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
1 초 128 MB 10 2 2 20.000%

## 문제

Let p1, p2, . . . , pn be n points on the plane. We have m rules of form pi rel pj , each inform us that the relation rel holds among the locations of points pi and pj on the plane. For example, “pi NE pj” indicates that point pj is located NorthEast of point pi. There are eight different relations {N, E, S, W, NE, NW, SE, SW}, corresponding to the eight directions on the plane. Let (xi, yi) and (xj , yj ) be the coordinates of pi, and pj respectively. Then pi rel pj exactly means one of the following, depending on the value of rel:

1. N stands for North. This means that xj = xi and yj > yi
2. E stands for East. This means that xj > xand yj = yi
3. S stands for South. This means that xj = xi and yj < yi
4. W stands for West. This means that xj < xi and yj = yi
5. NE stands for NorthEast. This means that xj > xi and yj > yi
6. NW stands for NorthWest. This means that xj < xi and yj > yi
7. SE stands for SouthEast. This means that xj > xi and yj < yi and
8. SW stands for SouthWest. This means that xj < xi and yj < yi

The problem is to determine whether it possible to locate p1, p2, . . . , pn on the plane so that all given rules are satisfied.

## 입력

The first line of the input contains a single integer t (1 ≤ t ≤ 20) which is the number of test cases in the input. The first line of each test case contains two integers n (2 ≤ n ≤ 500) which is the number of points and m (1 ≤ m ≤ 104) which is the number of rules. In each of the following m lines, there is one rule of the form i rel j which means that pi has relation rel with pj .

## 출력

The output contains one line per each test case containing one of the words ‘POSSIBLE’ or ‘IMPOSSIBLE’ indicating if the set of points in the test case can be located on the plane according to the given rules.

## 예제 입력

2
3 2
1 N 2
2 N 1
6 6
1 E 2
1 E 3
2 N 4
3 NW 5
4 SW 6
6 NE 5

## 예제 출력

IMPOSSIBLE
POSSIBLE


text:
# Lower bound for determinant and permanent

In light of the recent chasm at depth-3 result (which among other things yields a $2^{\sqrt{n}\log{n}}$ depth-3 arithmetic circuit for the $n \times n$ determinant over $\mathbb{C}$), I have the following questions : Grigoriev and Karpinski proved an $2^{\Omega{(n)}}$ lower bound for any depth-3 arithmetic circuit computing the Determinant of $n \times n$ matrices over finite fields (which I guess, also holds for the Permanent). Ryser's formula for computing the Permanent gives a depth-3 arithmetic circuit of size $O(n^2 2^n) = 2^{O(n)}$. This shows that the result is essentially tight for depth-3 circuits for the Permanent over finite fields. I have two questions:

1) Is there a depth-3 formula for the determinant analogous to Ryser's formula for the Permanent?

2) Does a lower bound on the size of arithmetic circuits computing the Determinant polynomial \textit{always} yield a lower bound for the Permanent polynomial?(Over $\mathbb{F}_2$ they are the same polynomials).

Though my question currenly is regarding these polynomials over finite fields, I would also like to know the status of these questions over arbitrary fields.

• That's interesting.... recently (eccc.hpi-web.de/report/2013/026) a $2^{O(n^{1/2}\log n})$ upper bound has been proved over the complex numbers. So there is somehow a huge difference in characteristic zero and finite fields... Feb 15, 2013 at 0:05
• I should have mentioned the new result. I was reading the paper and I wanted to know what can be inferred from known results for the finite field case. Will update the question to include the paper. Feb 15, 2013 at 4:01
• Are there similar/any lower bound known for determinant/permanent in case of depth 3 circuits over fields of characteristic zero? Jun 5, 2014 at 15:31
• Over characteristic zero, AFAIK, the best lower bound is $\Omega(n^2)$ for the elementary symmetric function(and also the determinant polynomial) due to Shpilka and Wigderson. Check cs.technion.ac.il/~shpilka/publications/… Jun 8, 2014 at 8:33

Permanent is complete for VNP under p-projections over any field not of characteristic 2. This provides a positive answer to your second question. If this reduction were linear, it would give a positive answer to your first question, but I believe that remains open.

In more detail: there is some polynomial $q(n)$ such that $det_n(X)$ is a projection of $perm_{q(n)}(Y)$, i.e. there is a certain substitution sending each variable $y_{ij}$ either to a variable $x_{k\ell}$ or a constant such that after this substitution the $q(n) \times q(n)$ permanent is computing the $n \times n$ determinant.

1) Thus Ryser's formula yields a depth 3 formula (depth does not increase under projections because the substitutions can be done on the input gates) of size $2^{O(q(n))}$ for determinant. UPDATE: As @Ramprasad points out in the comments, this only gives something nontrivial if $q(n) = o(n \log n)$, since there is a trivial depth 2 formula of size $n \cdot n! = 2^{O(n \log n)}$ for det. I am with Ramprasad in that the best I know is the reduction through ABPs, which yields $q(n)=O(n^3)$.

2) If the $m \times m$ permanent can be computed - again, over some field of characteristic not 2 - by a circuit of size $s(m)$, then $n \times n$ determinant can be computed by a circuit of size $s(q(n))$. So a lower bound of $b(n)$ on circuit-size for $det_n$ yields a lower bound of $b(q^{-1}(n))$ on circuit-size for the permanent (that's $q$ inverse, not $1/q(n)$). The above-mentioned $q(n)=O(n^3)$ yields a $b(n^{1/3})$ perm lower bound from a $b(n)$ det lower bound.

• Just want to point out that the determinant being a projection of a polynomially larger permanent doesn't quite yield much. The determinant of course has a trivial $n!$ sized circuit. So even showing that the $n\times n$ determinant is a projection of a $n^2\times n^2$ permanent does not yield anything non-trivial via Ryser's formula. I guess, for your proof strategy, one needs to show that the $q(n) = O(n)$, but I don't see how to get this from the usual reduction. AFAIK, no depth-3 circuit asymptotically smaller than $n!$ is known for the determinant over finite fields. Feb 15, 2013 at 5:58
• @Ramprasad: There is a projection of $DET_n$ to $PERM_{O(n)}$ in the general case over arbitrary fields right? So implementing this reduction in depth-3 is the hurdle - is that what you mean? Feb 15, 2013 at 11:18
• @Nikhil: Is there such a projection?! If that were true, then of course we would immediately have a $2^{O(n)}$ depth-3 circuit for the determinant by just using Ryser's formula (which wasn't known prior to the chasm-at-depth-3 result). The only reduction I know is to take the ABP for the determinant (whcih is $O(n^3)$-sized) and write that as a projection of a $O(n^3)$ sized permanent. I would be very surprised of a reduction to $O(n)$-sized permanent holds. Feb 15, 2013 at 11:30
• I'm fairly sure that is a typo/error in the article (but I shall check with Manindra though). Avi Wigderson's talk (PPT) during Valiant's 60th birthday celebrations is one of the places where it was stated that improving $n!$ for the depth-3 complexity of determinant was unknown. Depth-3 circuits over finite fields is a curious example where the best upper bound for the permanent is smaller than the determinant! Feb 15, 2013 at 14:49
• Feb 15, 2013 at 15:26

It is very possible that the determinant is, in a way, harder than the permanent. They are both polynomials, the Waring Rank(sums of n powers of linear forms) of the permanent is roughly 4^n, Chow Rank(sums of products of linear forms) is roughly 2^n. Clearly, Waring Rank \leq 2^{n-1} Chow Rank. For the determinant, those numbers are just lower bounds. On the other hand, I proved a while ago that the Waring rank of the determinant is upper bounded by (n+1)! and this might be close to the truth.