## where on earth – Houses on a hill by a beach: what city or neighborhood is this?

What city or neighborhood is in the following image, which depicts houses on a hill by a beach?

The image is cropped from a Playboy magazine photoshoot of model Katie Vernola. The photoset is entitled “Pretty Chic” and was released on April 24, 2012. I have cropped out the model to show only the neighborhood in the background. Because Playboy is an American magazine, I suspect the location is in the United States (though that is not certain because Playboy sometimes uses international locations).

I have tried reverse-image-searching, but Google Images gives no useful results, and TinEye just returns the photoset I’ve already described above.

I also tried asking at http://www.where-is-this.com, but the mods there removed my picture because they thought it wasn’t interesting enough. 🙁

## NGClouds.com – 1GB ESXI VPS for \$3.50/mo in Amsterdam, Instanbul or Kansas City

Kevin from NGClouds.com sent in their first LowEndBox exclusive offer recently and we’re excited to be able to share it with our users. NGClouds.com utilizes VMWare as their virtualization engine, making this somewhat unique in today’s market place given that so many providers have opted for SolusVM/KVM as of late. The 1GB and 2GB packages are available in three locations overall, Amsterdam, Istanbul and Kansas City.

Their WHOIS is public and you can find their ToS/Legal Docs on their website. They preferred method of payment is Credit/Debit Card.

Given that this is the first time NGClouds.com is being featured on LowEndBox we recommend paying by month only. Please share your feedback in the comments below.

The exclusive VMWare offers:

 VMWare 1 GB 1 GB RAM 1x vCPU 20GB SSD 1TB transfer 1Gbps uplink 1x IPv4 VMware ESXI Coupon: LEBSpecial \$3.50/month (ORDER) VMWare 2 GB 2 GB RAM 2x vCPU 25GB SSD 2TB transfer 1Gbps uplink 1x IPv4 VMware ESXI Coupon: LEBSpecial \$7/month (ORDER)

Network and hardware information after the break!

Be sure to leave your comments if you are a customer or have questions!

NETWORK INFO:
Datacenter Name – Amsterdam (Netherlands)
Test IPv4: 185.229.243.131
Test file: https://lg.nl.ngclouds.com/100MB.test
Looking Glass: https://lg.nl.ngclouds.com

Datacenter Name – Istanbul (Turkey)
Test IPv4: 185.219.135.4
Test file: https://lg.tr.ngclouds.com/100MB.test
Looking Glass: https://lg.tr.ngclouds.com

Datacenter Name – Kansas City (USA)
Test IPv4: 185.202.172.4
Test file: https://lg.us.ngclouds.com/100MB.test
Looking Glass: https://lg.us.ngclouds.com

Hardware Node:

### Jon Biloh

I’m Jon Biloh and I own LowEndBox and LowEndTalk. I’ve spent my nearly 20 year career in IT building companies and now I’m excited to focus on building and enhancing the community at LowEndBox and LowEndTalk.

## This is a considerably hard analytical problem

The question you linked, although already quite hard, is easier because it can be computed as the probability of a dice being in one set, and not in the other. In your question, this is done in a one-to-one basis, as you mentioned, and this is a considerably hard problem.

## Doesn’t matter – do it the easy way

Unless you need the analytical expression for something, I recommend going with Monte Carlo:
You basically run a random simulation a bunch of times, and then see the distribution of that. This code for MATLAB does it – I am sure someone can translate it to Python or something that doesn’t need to be paid, but I’m more familiar with MATLAB haha

``````N_it = 10^5;

N_action = 6;
N_danger = 5;

bins = 1:6;

for i = 1:N_it

Action_Dice = randi((1 6), N_action, 1);
Danger_Dice = randi((1 6), N_danger, 1);

Action_Dice_Count = hist(Action_Dice, bins);
Danger_Dice_Count = hist(Danger_Dice, bins);

Result_Count = Action_Dice_Count - Danger_Dice_Count;

Result = find(Result_Count > 0, 1, 'last');

if(Result)
Result_Save(i) = Result;
else
Result_Save(i) = 0;
end
end
``````

So, what this code does is basically: roll a set number of action dice and danger dice, count how many of each dice you got, subtract the number of rolled danger dice from the number of rolled action dice, then find the last value which is non-zero (i.e., the highest Action Dice remaining) and finally, if no such action dice exists, it sets the result to 0.

As an example, for 6 Action Dice and 5 Danger Dice, this is what it looks like:

Why does it look like this? In the example, I used 6 action dice and 5 danger dice. The probability that there is no remaining action dice is, evidently, zero, since we will always have one remaining dice. But not only that, there is a high probability on 6, why? Because every time there is a 6 remaining, that 6 will be chosen. This is actually an easier analytical problem. Let $$X$$ denote the number of sixes appearing in the Action Dice pool, and $$Y$$ be the number of sixes appearing in the Danger Dice pool. We are interested in $$P(X > Y)$$. But what is nice in this case is that we know the exact distribution of both $$X$$ and $$Y$$, which are binomial distributions with the number of trials being equal to the number of dice in the pool, and probability in each trial equal to 1/6.

We can then compute $$P(X – Y)$$, which can be done by the convolution of $$P(X)$$ and $$P(-Y)$$, and finally compute $$P(X – Y > 0)$$. I won’t bother with all this computation here, but if you do it, you will find out that the probability is 0.383367984110654. The probability found by code was 0.386, so, close enough, the code seems to make sense.

Note that this same strategy of analytically computing the other probabilities will not help for other values than 6, since, not only you need that, for example, the number of fives in the Action pool should be larger than the number in the Danger pool, you also need that the number of sixes in the action pool is smaller than or equal to the number in the danger pool. For 5 it is already a bunch of complicated conditional probabilities, and it will get harder and harder for lower numbers, since each time you need to consider the numbers above it.

## Validating the Code

I can only think about one trivial case, so we can check whether the code makes sense for that case at least. Consider only 1 Action Dice and 1 Danger Dice. In this case, the probability that they are equal is 1/6, and in that case the result is 0 (i.e., no action dice remaining). Otherwise, we have an equal probability for any dice, which is 5/36.

The values resulting from the code are, in fact, approximately 1/6 for 0 and 5/36 for the remaining.

## safety – In Germany, Is/are there any city or a special region in a city which is/are dangerous for non-white people to go to?

In Germany, Is/are there any city or a special region in a city which is/are dangerous for non-white people to go to?

I mean, is/are there any city or an area in a city in which predominantly white supremacists live and they are antagonistic to non-white people and are/is dangerous as a result?

## seo – Single home landing page versus city specific CTA

I have several city specific pages, more like extended suburbs specific pages, each with their own CTA. My home page uses a similar CTA as these other cities, but is more generic about location.

My city specific pages rank well for the exact keywords, but I’m wondering if I’m diluting everything by having similar but not identical content on these pages?

While I intentionally try to write different content, including content that is specific to the city, the theme of each page is substantially the same.

Would I be better off using canonical references on my city pages pointing them to my home page?

## optimization – Visit given city before a given cumulative distance in the traveling salesman problem

I would like to add an additional constraint to the traveling salesman problem: that a given city is visited within a given distance (say `100`) from start. Is there a way to do this? My question is related to this unanswered CS question.

I have a mixed integer program using the R package `CVXR` that find the shortest route without subroutes (see below). The city order is represent in the vector `node_order`. The strategy I’ve pursued so far is:

1. Re-organize `node_order` so that the index is the order and the value is the city id
2. Look up the associated distances in `distances`
3. Compute a vector with the cumulative sum of these distances.
4. Add the constraint that city `i` must occur before the first index in (3) exceeding the distance constraint for that city.

The issues I’ve encountered with this approach is that I have not found a way to include finding-index-by-value into the optimization in `CVXR`. This is needed in both step (1) and (4) above. Maybe this is possible after all, or there is another approach? I am willing to use other packages than `CVXR` and other software than `R`.

``````library(CVXR)

# Make distances
N = 10
distances = matrix(1:(N*N), ncol = N)

# Flag 1 iff we travel that path. 0 otherwise
do_transition = Variable(N, N, boolean = TRUE)

# Minimize the total duration of the traveled paths.
objective = Minimize(sum(do_transition * distances))

# Only go one tour. Order is 1:(N-1)
node_order = Variable(N-1, integer = TRUE)
ii = t(node_order %*% matrix(1, ncol = N - 1))  # repeat as N rows
jj = node_order %*% matrix(1, ncol = N - 1)  # repeat as N cols

# Constraints
constraints = list(
do_transition * diag(N) == 0,  # Disallow transitions to self (diagonal elements)
sum_entries(do_transition, 1) == rep(1, N),  # Exactly one entrance to each node
sum_entries(do_transition, 2) == rep(1, N),  # Exactly one exit from each node
(jj - ii) + N * do_transition(2:N, 2:N) <= N - 1,  # One tour constraint (no subtours)
node_order >= 1,   # This interval represents order as ranks (1 to N-1)
node_order <= N-1
)

# Find optimum
solution = solve(Problem(objective, constraints))
``````

A bit of code pertaining to my current (unsuccessful) attempts:

``````# Get tour order
#tour = order(c(NA, result$$getValue(node_order))) # R solution tour = rep(NA, N-1) tour(result$$getValue(node_order)) = 2:N

# Get tour distances
distances_optim = diag(distances(tour, tour(2:N)))

# Tour cumulative distances
distances_cumul = cumsum_axis(distances_optim)
``````

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## terminology – Is the term “service city” correct for the main city served by an airport placed outside of that city?

Some airports of big cities are, in fact, technically speaking in a “very near municipality” which is not the main city itself.

For example the airport BCN serving “Barcelona” is, in fact, placed in the city “El Prat del Llobregat” which is 15 Km south from Barcelona.

Or for example the airport CDG serving “Paris” is, in fact, at the town “Roissy” 25 Km northeast from Paris.

I “think” (but can’t confirm) I have heard the term “Service City” to refer to the “main city” the airport is conceptually associated to. For example to say “CDG’s service city is Paris”.

I’m trying to confirm this terminology with no luck.

Can anyone confirm, if this term is correct and provide any public source defining the term? If the term is not correct, what would be the correct one?

## algorithms – Three City Scheduling

I came across the following interview question

There are 2N people a company is planning to interview. The cost of
flying the i-th person to city A is costs[i][0], and the cost of
flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that
exactly N people arrive in each city.

The solution to this involves greedy approach, where we sort the array based on the “profit” parameter. profit of choosing city A for a candidate `i` is defined as `costs[i][1] - costs[i][0]` and choose the top half elements from the sorted array to go to A and rest to B.

What if this question is modified to 3 cities and you have find optimal partition of n/3 chunks? Will greedy algorithm still work?