Jumble | An Android service implementation of the Mumble protocol

For every Ask message sent to an actor, akka creates a proxy ActorRef whose sole responsibility is to process one single message. This temp "actor" is initialized with a promise, which it needs to complete on message processing.

The source code of it is found here

but the main details are

The simple way to do this is simply to take one point over N in each vectors given to quiver.

For a given np.array, you can do this using the following syntax: a[::N]. If you have multiple dimensions, repeat this in each dimension (you can give different slip for each dimension): a[::N1, ::N2].

In your case:

Once you've split the words, you can use sample() to rescramble the letters, and then paste0() with collapse="", to concatenate back into a 'word'

The startup object has to be Sub Main for the Startup event to work, not MainForm (or any form at all).
(Thanks to @Hans Passant for the tip.)

If you want to change which one is the main form later on you have to do the following:

1. Close your project in Visual Studio.
2. Open your project's folder in explorer.
3. In it, open folder My Project.
4. Open Application.myapp with any text editor.
5. Change the form between the tags to your (new) main forms name.
6. Save and close.
7. Open ApplicationDesigner.vb with any text editor.
8. Find the following line and change YourMainFormsName to your (new) main forms name:

As I understand it you are trying to identify all possible matches for the jumbled string in your list. You could sort the letters in the jumbled word and match the resulting list against sorted lists of the words in your data file.

I see a lot of repetition in your code, so the first thing I'll think of would be: "can I use a loop to simply this code?" and the answer is yes!

Since your code repeatedly used the six subjectTypes and the keySubject depends on the subject type, creating a list of the six types then use next() with a generator expression should simplify the over abundance of if's (If there weren't any correlations, a dictionary would work instead). Also, instead of array_data, you can use an if-elif-else clause to prevent an extra block level.

You are defining X as an array (you are using brackets). That's why you are obtaining an error. Change the way you define X and it should work:

Since you want to put the matchups back into the same df, the teams need to go into separate columns (otherwise it will throw an error about uneven row counts).

Instead of multiple splits, use a single str.extract with the following patterns:

• .*[AM|PM] -- match everything up to AM or PM (but don't capture it)
• ([0-9]+[a-zA-Z ]+) -- capture 1+ numbers and 1+ letters/spaces (away team)
• ([0-9]+[a-zA-Z ]+) -- capture 1+ numbers and 1+ letters/spaces (home team)

To prove that it's correct, you have to find some sort of invariant. Something that's true during every pass of the loop.

Looking at it, after the very first pass of the inner loop, the largest element of the list will actually be in the first position.

Now in the second pass of the inner loop, i = 1, and the very first comparison is between i = 1 and j = 0. So, the largest element was in position 0, and after this comparison, it will be swapped to position 1.

In general, then it's not hard to see that after each step of the outer loop, the largest element will have moved one to the right. So after the full steps, we know at least the largest element will be in the correct position.

What about all the rest? Let's say the second-largest element sits at position i of the current loop. We know that the largest element sits at position i-1 as per the previous discussion. Counter j starts at 0. So now we're looking for the first A[j] such that it's A[j] > A[i]. Well, the A[i] is the second largest element, so the first time that happens is when j = i-1, at the first largest element. Thus, they're adjacent and get swapped, and are now in the "right" order. Now A[i] again points to the largest element, and hence for the rest of the inner loop no more swaps are performed.

So we can say: Once the outer loop index has moved past the location of the second largest element, the second and first largest elements will be in the right order. They will now slide up together, in every iteration of the outer loop, so we know that at the end of the algorithm both the first and second-largest elements will be in the right position.

What about the third-largest element? Well, we can use the same logic again: Once the outer loop counter i is at the position of the third-largest element, it'll be swapped such that it'll be just below the second largest element (if we have found that one already!) or otherwise just below the first largest element.

Ah. And here we now have our invariant: After k iterations of the outer loop, the k-length sequence of elements, ending at position k-1, will be in sorted order:

After the 1st iteration, the 1-length sequence, at position 0, will be in the correct order. That's trivial.

After the 2nd iteration, we know the largest element is at position 1, so obviously the sequence A[0], A[1] is in the correct order.

Now let's assume we're at step k, so all the elements up to position k-1 will be in order. Now i = k and we iterate over j. What this does is basically find the position at which the new element needs to be slotted into the existing sorted sequence so that it'll be properly sorted. Once that happens, the rest of the elements "bubble one up" until now the largest element sits at position i = k and no further swaps happen.

Thus finally at the end of step N, all the elements up to position N-1 are in the correct order, QED.