## international travel – Is 40 minutes stop time between flights at Munich Airport on two Lufthansa flights sufficient?

40 minutes would be very tight for this connection. Munich is your first port of entry into the Schengen area, so you will have to go through immigration control in Munich. Depending on your nationality and the time of day, this may take a little longer.

If Lufthansa is willing to sell this to you in one ticket, it's a legal connection and you have a decent chance of doing so. But there is no room for error: if something goes wrong, is delayed or just a little longer than usual, you will miss it. If you miss it, LH will book the next available flight for free, but you may miss your cruise, so I don't recommend it.

There is NO WAY you can do with two separate tickets.

## plotting – How to plot the solution of first-order time-dependent differential equations with another parameter not with time

I am trying to trace the solution given in the code with regard to "g0", not with time "t". I don't know what to do to get the plot between the solution and the value "g0" which vary from 0 to 2. The initial conditions are also included in the text. The only thing I want is to plot between the solution and "g" for any arbitrary value of "t". If anyone can solve this problem, it's welcome.

w1 = 1;
gma1 = 0.005;
n1 = 1;
gma2 = 0.005;
G1 = 0.005;
delc = 1;
k1 = .1;
k2 = 0.1;
a1 = 0.07;
a2 = 0.58;
k0 = 0.1;
Q1 = 1.268;
del0 = 1;
N1 = 1;
ome = 1;
M1 = del0*(1 - Cos(ome*t));
s = ParametricNDSolveValue({V11'(t) - V21(t)*w1 - V12(t)*w1 == 0,
V12'(t) - V22(t)*w1 + w1*V11(t) + gma1*V12(t) -
Sqrt(2)*G1*a1*V13(t) - Sqrt(2)*G1*a2*V14(t) == 0,
V13'(t) - V23(t)*w1 + k1*V13(t) +
Sqrt(2)*G1*a2*V11(t) - (-G1*Q1 + delc)*V14(t) == 0,
V14'(t) - V24(t)*w1 + k1*V14(t) -
Sqrt(2)*G1*a1*V11(t) - (G1*Q1 - delc)*V13(t) == 0,
V21'(t) + V11(t)*w1 + gma1*V21(t) - Sqrt(2)*G1*a1*V31(t) -
Sqrt(2)*G1*a2*V41(t) - w1*V22(t) == 0,
V22'(t) + V12(t)*w1 + gma1*V22(t) - Sqrt(2)*G1*a1*V32(t) -
Sqrt(2)*G1*a2*V42(t) + w1*V21(t) + gma1*V22(t) -
Sqrt(2)*G1*a1*V23(t) - Sqrt(2)*G1*a2*V24(t) - gma2*(2*n1 + 1) ==
0, V23'(t) + w1*V13(t) + gma1*V23(t) - Sqrt(2)*G1*a1*V33(t) -
Sqrt(2)*G1*a2*V43(t) + k1*V23(t) +
Sqrt(2)*G1*a2*V21(t) - (-G1*Q1 + delc)*V24(t) == 0,
V24'(t) + V14(t)*w1 + gma1*V24(t) - Sqrt(2)*G1*a1*V34(t) -
Sqrt(2)*G1*a2*V44(t) + k1*V24(t) -
Sqrt(2)*G1*a1*V21(t) - (G1*Q1 - delc)*V23(t) == 0,
V31'(t) + k1*V31(t) +
Sqrt(2)*G1*a2*V11(t) - (-G1*Q1 + delc)*V41(t) - w1*V32(t) == 0,
V32'(t) + k1*V32(t) +
Sqrt(2)*G1*a2*V12(t) - (-G1*Q1 + delc)*V42(t) + w1*V31(t) -
Sqrt(2)*G1*a2*V34(t) - Sqrt(2)*G1*a1*V33(t) + gma1*V32(t) == 0,
V33'(t) + k1*V33(t) +
Sqrt(2)*G1*a2*V13(t) - (-G1*Q1 + delc)*V43(t) + k1*V33(t) +
Sqrt(2)*G1*a2*V31(t) - (-G1*Q1 + delc)*V34(t) - k0 == 0,
V34'(t) + k1*V34(t) +
Sqrt(2)*G1*a2*V14(t) - (-G1*Q1 + delc)*V44(t) + k1*V34(t) -
Sqrt(2)*G1*a1*V31(t) - (G1*Q1 - delc)*V33(t) == 0,
V41'(t) + k1*V41(t) -
Sqrt(2)*G1*a1*V11(t) - (G1*Q1 - delc)*V31(t) - w1*V42(t) == 0,
V42'(t) + k1*V42(t) +
Sqrt(2)*G1*a1*V12(t) - (G1*Q1 - delc)*V32(t) + w1*V41(t) -
Sqrt(2)*G1*a2*V44(t) - Sqrt(2)*G1*a1*V43(t) + gma1*V42(t) == 0,
V43'(t) + k1*V43(t) -
Sqrt(2)*G1*a1*V13(t) - (G1*Q1 - delc)*V33(t) + k1*V43(t) +
Sqrt(2)*G1*a2*V41(t) - (-G1*Q1 + delc)*V44(t) == 0,
V44'(t) + k1*V44(t) -
Sqrt(2)*G1*a1*V14(t) - (G1*Q1 - delc)*V34(t) + k1*V44(t) -
Sqrt(2)*G1*a1*V41(t) - (G1*Q1 - delc)*V43(t) - k0 == 0,
V11(0) == 1, V12(0) == 1, V13(0) == 0, V14(0) == 0, V21(0) == 0,
V22(0) == 1, V23(0) == 0, V24(0) == 0, V31(0) == 0, V32(0) == 0,
V33(0) == 0, V34(0) == 0, V41(0) == 0, V42(0) == 0, V43(0) == 0,
V44(0) == 0}, {V11, V12, V13, V14, V21, V22, V23, V24, V31, V32,
V33, V34, V41, V42, V43, V44},{t, 0, 100},g0);
P2 = Plot({Evaluate(1/2*(V11(t) + V22(t) - 2*V12(t))^(-1) /. s)}, {t,
0, 60}, PlotRange -> {0, 1}, Frame -> True,
FrameLabel -> {Style("Time", Bold, 20),
Style(" !(*SubscriptBox((S), (q)))", Bold, 20)},
FrameTicksStyle -> Directive(FontSize -> 20),
PlotStyle -> {Thickness(0.0005), Thickness(0.008)})

## post-processing – Do I lose information every time I add an adjustment layer to an 8-bit image in Photoshop?

There may be less color in the result, yes. In a pixel, each color channel can only take one of the 256 values ​​(2⁸). Each adjustment transforms these 256 values ​​into 256 other values. But there is a constraint: for any pair of two input values, the higher of the two must remain the higher (otherwise you get an effect called solarization). If you use curves, it means that your curve cannot go down from left to right.

Now consider a value somewhere in the middle, for example 128. Your adjustment moves it to 140. The 128 values ​​less than 128 are mapped to 128 new positions between 0 and 140 (of course, this means that some values ​​less than 140 don't will not be And the remaining 128 values ​​(128 and higher) are now mapped to 116 values, so you started with 256 values, and you now have 128 + 116 = 244 values.

This can lead to what is called a hair-shaped histogram:

The above histogram is a black-white linear gradient, in which, using curves, I moved 128 to 140. The gaps to the left are the 12 values ​​that are left blank (128 to 140 ) while the teeth on the right are result values ​​which correspond to two input values ​​(12 of them, 128 in 116).

Each of these adjustments eats a little color. It is not too visible at first because you have three color channels and the missing values ​​do not apply to all three channels at the same time (sometimes you can even get more colors …), but in flat areas with uniform colors, you will get visible bands.

## algorithms – Why is the execution time with a loop of this structure considered O (log n)

I have used the search function and a fair amount of Google searches, but I could not get a direct answer on how a loop from the form below is translated to an appropriate summation where the function derived from the summation is: $$O ( log n)$$.

Example of a for loop:

int j = 0;
for (int i = 1; i <= n; i *= 2) {
j = j + n * 2;
}

So I understand that in the loop, we have 3 operations (multiplication, addition, then assignment).

I understand that the clue $$i$$ who goes from $$(1, 2, 4, 8, 16, 32, 64, 128, 256)$$... ($$i$$ is only equal to powers of 2, up to $$n$$).

So basically the range of $$i$$ seems to come from $$(2 ^ m, n)$$ and $$m in (0, log_2n)$$ right?

Also, $$i ^ m so the loop runs $$log_2n - 0 + 1 = log_2n + 1$$ times no? How are we going to express this in summation notation?

Is this it (just guessing, I don't know if this is true):

$$sum limits_ {i = 0} ^ {log_2n + 1} 3$$ since we have 3 operations? If that is the answer, why do we keep the upper limit of the sum $$log_2n$$ instead of $$n$$? Also, I know that we generally use $$log n$$ but just put in base 2 to help clear things up in my own head.

Could someone show how the summation of the for loop is correctly written?

## Nikon – My camera's exposure is like broken every time I use the flash?

regardless of the setting of my camera every time I use the flash, the exposure becomes super high. I am trying to change my ISO and my aperture in but it does not help. I have a Nikon D3400, could this be a sensor problem? It was working just a few days ago and I took it back to find it completely messed up. As long as I don't use the flash, it works perfectly fine what could be wrong?

## transfer time to Amsterdam airport

I have a flight from Istanbul to Gdansk with transfer to Amsterdam but at Schiphol Airport I only have one hour. Will I be enough to get on another plane? will they also check my passport at the airport? Both flights are with KLM btw.

## python – Complete time series (days) in the data of a DataFrame grouped by group by

I have the following Dataframe, with the name Data:

I applied df.groupby (& # 39; site & # 39;) to classify the data according to this functionality.
After having classified it, I want to complete, for all records, the column "date" day after day.

grouped = Datos.groupby('site')

The procedure that I think I should follow will be:
1. Generate a complete sequence between the start date and the end date. (Stage completed).

for site in grouped:
dates = ('2018-01-01', '2020-01-17')
#dates = (grouped.date)
startDate = datetime.datetime.strptime( dates(0), "%Y-%m-%d")
endDate   = datetime.datetime.strptime( dates(-1),"%Y-%m-%d")
days = (endDate - startDate).days  # how many days between?
allDates = {datetime.datetime.strftime(startDate+datetime.timedelta(days=k),
"%Y-%m-%d"):0 for k in range(days+1)}
1. Compare this sequence with the & # 39; date & # 39; column from my group. (& # 39; Site & # 39;) and add those that are not present do not correspond to the dates of & # 39; date & # 39 ;.
2. Write a function or loop that allows you to update the & # 39; date & # 39; column with the new dates and also fill in the missing values ​​with 0. For this I will use:
(grouped.apply(agregar_dias))

Until now, I have only managed to complete step 1, so I ask for your help and knowledge to complete steps 2 and 3.
I would very much appreciate your always important help.
Greetings

## c # – Ambient time context to help unit tests

The following test shows the functionality of Time class that is supposed to be used instead of DateTime.Now.

(TestClass)
public class Time_Should
{
(TestMethod)
{
Assert.IsTrue(Time.Now > new DateTime(2020, 1, 1));
using (new Time(2010, 1, 1))
{
Assert.AreEqual(new DateTime(2010, 1, 1), Time.Now);
using (new Time(2011, 1, 1))
Assert.AreEqual(new DateTime(2011, 1, 1), Time.Now);

Assert.AreEqual(new DateTime(2010, 1, 1), Time.Now);
}
Assert.IsTrue(Time.Now > new DateTime(2020, 1, 1));
}
}

The library code:

public class Time : IDisposable
{
static AsyncLocal

## Is a sufficient 2 hour transit time to Munich from Canada

The title says it all. Is 1h 55m enough for a stopover in Munich for a flight Montreal> Munich> Athens?