array-back | Isomorphic , load-anywhere arrayify function
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kandi X-RAY | array-back Summary
Isomorphic, load-anywhere arrayify function
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QUESTION
so I'm basically trying to make a simple base game for a project. It's an array-backed grid in pygame which starts with a black background and a 10x10 array of green grid squares. what i'm trying to do is swap colors when a square is clicked, if the square was green, it turns to white and if the square was white, it turns to green. I am successfully able to change the color of a square once but I can't seem to swap it back. What am I doing wrong? Here's my code
...ANSWER
Answered 2020-Dec-17 at 17:44You accidently used the (==) compare equality operator instead of the (=) assignment operator:
grid[row][column]==0
QUESTION
I'm trying to do the following:
...ANSWER
Answered 2018-May-07 at 15:54Because it is clearly mentioned in the javadoc of keySet (emphasis mine)
[...] The set supports element removal, which removes the corresponding mapping from the map, via the Iterator.remove, Set.remove, removeAll, retainAll, and clear operations. It does not support the add or addAll operations.
Also, as mentioned in this comment not all collections have to implement this method.
More specifically - it (addAll of Collection) says,
UnsupportedOperationException - if the addAll operation is not supported by this collection
QUESTION
I am using MongoDB with Python. I used the following command to insert my documents:
...ANSWER
Answered 2017-Mar-27 at 01:21You are mixing up two concepts in mongo: sorting an array embedded in a document and sorting documents in a collection by some properties.
The post you link to refers to the former. You seem to be trying to permanently sort the documents in a collection.
Mongo stores documents in roughly the order they are written in. Like most databases it uses indexes to make retrieving data more efficient.
In the instance you describe, you would create an index on the age property of your document.
Use either the ensure_index or create_index functions:
QUESTION
I am studying algorithms on my own and turned to the Open Data Structures (C++ Edition) free ebook as a reference. In my attempt to master the topic, I am determined to finish all the challenges in the book. However, I am have a lot of trouble understanding how one particular unpopular algorithm could have O(1) for add() and remove.
One of the challenges is to create a random queue with these properties (exercises).
Exercise 2.2. Design and implement a RandomQueue. This is an implementation of the Queue interface in which the remove() operation removes an element that is chosen uniformly at random among all the elements currently in the queue. (Think of a RandomQueue as a bag in which we can add elements or reach in and blindly remove some random element.) The add(x) and remove() operations in a RandomQueue should run in constant time per operation.
The chapter deals with array-backed-list so the addition and removal of elements is rather trivial in that sense. However, the array sometimes has to be recreated for size. You are suppose to copy the old array to the new one. This essentially could be O(n). Also I believe that I need to utilize a circular array. So essentially I would have to shift indices within the array producing O(n-1) for time complexity.
I am very confused about how to calculate and measure these algorithms. The book does talk about O(m) but it is rather vague at times.
Theorem 2.2. An ArrayQueue implements the (FIFO) Queue interface. Ignoring the cost of calls to resize(), an ArrayQueue supports the operations add(x) and remove() in O(1) time per operation. Furthermore, beginning with an empty ArrayQueue, any sequence of m add(i, x) and remove(i) operations results in a total of O(m) time spent during all calls to resize().
Like how can you just ignore that, I am not making the connection on how you can just lop off that portion of the time complexity?
...ANSWER
Answered 2017-Jan-19 at 00:17The resize is usually scheduled to happen at exponentially distributed sizes. Say, when size is 8, get a size-16 storage, move the contents there and drop the old size-8 storage. When that fills up, get a size-32 storage, and so on. This way, the amortized complexity of inserting n elements is still O(n): in total, we moved the elements n + n/2 + n/4 + n/8 + ... times which is bounded by 2n and therefore linear.
The same can be done if resizing is needed when the structure shrinks: when a size-16 structure contains as little as 4 elements, get a size-8 storage, move the contents there and drop the old size-16 storage.
The powers of two can be substituted by powers of some other number, for example, 1.5. The tradeoff is how much memory is wasted versus how much time is spent moving the contents. Both are O(n), but we can raise the constant factor of one and lower the other.
Perhaps the book does not want to deal with amortized complexity just yet, and so avoids discussing the details of memory management.
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