What exactly defines the size of bucket while sorting? Like in counting the size
will be from 0 to max and in radix the bucket size is 0-9.
The amount of "buckets" in Bucket Sort corresponds to and depends on the range of possible values that are to be sorted. For example, if I am considering the following values:
1, 4, 10000, 5, 12
The range of the values I am sorting is 9999.
range = (highest value) - (lowest value) = 10000 - 1 = 9999 buckets
This means that I will need 9999 buckets to sort these values as efficiently as possible. This is because Bucket Sort is NOT a comparison sort. This means that values are not compared to each other to determine their order. They are simply placed in buckets that represent their values. Bucket Sort is hailed to be O(n) because of this, but it actually tends to be much less efficient due to the amount of buckets needed to be created.
With the previous example we only were sorting 5 values, but needed to create upwards of 10,000 buckets. This is grossly inefficient when bucket amounts are within the time complexities of O(N^2), which makes buckets sort just as time costly (if not, worse) than the traditional comparison sorting algorithms. However, if the conditions are right, then Bucket Sort may be a good choice.
Related
I have a dataset comprised of n unsorted tuples representing numbers (let's say specific color codes) and their frequency (number of times of appearance).
I want to find an exact median of the numbers with worst case complexity of O(n).
For example:
dataset: {(5000, 8000), (25000, 4000), (9, 9000)} median: 5000
dataset: {(7000, 4), (23000, 400), (3000, 9000), (2500, 12000), (19000, 350), (500, 9000)....} median: ?
Failed attempts so far:
"Decompress" the list (so that it looks like this: {7000, 7000, 7000, 7000, 23000, 23000...}) and then sort it. Problem is - it takes Ω(nlogn), and probably more since the frequencies can be very large and doesn't have any upper bound.
Try to use QuickSelect over the data. To ensure O(n) time complexity we must guaranty good pivot selection. To do so I thought about Median of medians (supposedly O(n)) with the data - but I couldn't figure out how to do that without decompressing, thus making it potentially more than O(n).
Is there a way to manipulate the tuple list so that it wouldn't be decompressed and still use median of medians or another way to find the median?
End note: I don't want to assume anything about the dataset - amount of tuples, confined range of numbers/frequencies, etc).
Use quickselect on the values, and only pay attention to the frequencies in determining which half to keep.
Your ideal pivot is one that splits the list values in half. Because that will cut the work of the next pass in half. Where that split happens to in the whole dataset doesn't particularly. Because your goal is to get it down to the one value you want, and then you're done.
This means that for median of medians you can ignore frequencies entirely in selecting a pivot. Then pay attention to frequencies when deciding which side of the pivot to keep. And ignore frequencies again while selecting the next pivot.
I am trying to learn the bucket sort algorithm for integers and am quite confused about the number of buckets part.
I have looked at various sources on the internet and I have seen that we should calculate the number of buckets as follows:
Range = Max-Min
Number of buckets = Range/Length of the array.
But my confusion is that if we follow the above formula, then for an array [9,1] we have range=8 and no of buckets is 8/2=4, which seems to be unnecessary. On the other hand, if we have [9,8,7,6,5,4,3,2,1], we get range=8 and no of buckets is 8/9=1, which will force every number in a single bucket and since it is in reverse order, it'll lead to worst case. So is there any right way to calculate the no of buckets? or am I understanding it wrong?
When the distribution of keys is sparse in bucket sort, there may be a lot of empty buckets.
How could we retrieve the sorted list (i.e., achieve the concatenation operation) efficiently?
We want to implement an bucket based priority queue, but the search for the first non-empty bucket may take lots of time. So we wonder a smarter way to do so.
For example, if we got a list with millions of 10, 1000, 50000, 100000, 6400000, 10000000 and so on, how could we retrieve the sorted list by using bucket sort?
Another tougher example would be, 1, 100, 101, ..., 999, 1000, 100000, 100001, ... 999999, 1000000, 100000000, 100000001, ..., 199999999.
There could be even harder cases that the distributions within some segments are dense, but there might be huge gaps between segments.
Your application must be special. If buckets are sparse, one might expect that you would only have one or two items per bucket on average. If so, then the bucket sort isn't doing you any good -- just put the items in a heap.
If buckets are not really that sparse, i.e., if the number of buckets is <= a few times the number of items, then the bucket sort suffices -- iterate through the buckets in order and the cost will be O(N) in the number of items.
If you have many items per non-empty bucket AND many buckets per item, then you probably want to explain your use case, but when I've seen this in the past it has been reasonable to insert each bucket into a heap when it becomes non-empty.
The simple answer to your question is "Not without an additional data structure to keep track of which buckets have items."
There are multiple ways to do a bucket sort. The "best" depends a lot on the range of keys, the number of items, and the number of unique items. If your range is 0 to 1,000,000 and you know that you'll have, say, 50% unique, then a single array of 1,000,000 buckets is easy to work with, you don't waste too much space, and you don't waste a lot of time skipping over empty buckets.
But if you're talking a range of hundreds of millions that is very sparsely populated, you end up wasting a lot of memory and considerable time skipping over empty buckets. In extreme cases, you can't even allocate an array large enough to cover the entire range.
Another common way to implement a bucket sort is with a dictionary of hash map. The idea is:
initialize empty hash map
for each item in list
if key already in hash map
add item to that bucket
else
create new bucket in hash map
Of course, you then have to sort the buckets by key once you're done populating, but sorting a few thousand (if that) buckets takes trivial time. And you don't end up wasting gigabytes of memory on empty buckets.
When I built a bucket-based priority queue, I used the dictionary approach. I maintained a dictionary keyed by index, and added each item to the proper bucket. I also maintained a simple binary heap of the buckets. So adding an item to the heap became:
if item.key exists in dictionary
dictionary[item.key].add(item) // adds item to bucket
else
{
dictionary.add(item.key, item) // creates a new bucket
heap.push(dictionary[item.key]) // pushes the bucket onto the heap
}
And removing an item from the heap becomes:
bucket = heap.peek()
item = bucket.getFirst()
if (bucket.count() == 0)
{
// bucket is empty. Remove from heap and from dictionary
heap.pop()
dictionary.remove(item.key)
}
return item
This performs quite well. Because my keys were sparse and the buckets heavily filled, it was rare that the heap itself got any activity. Most of the activity involved adding things to and removing things from buckets that were already in the heap. The only time the heap got exercise was when a bucket was emptied, or when I added a new bucket. So on average, both insert and remove were very close to O(1).
This worked well for me because my range of keys was very large (10-character alphanumeric), the number of individual items in the hundreds of millions, or billions, but the number of unique keys used at any time was in the thousands. There is some slight overhead is the dictionary indirection, but that's more than offset by the savings of working with a heap of a few thousand, rather than hundreds of millions, of items.
I'm doing an algorithms course at uni, and I read the following sentence on Introduction to Algorithms 3ED, p200:
...bucket sort is fast because it assumes something about the input. Whereas counting sort assumes that the input consists of integers in a small range, bucket sort assumes that the input is generated by a random process that distributes elements uniformly and independently over the interval [0,1)
Why is it that the input has to be in [0,1)? Why can't any uniformly distributed sequence be sorted using bucket sort?
I imagine that the interval [0, 1) is used in order to obtain a theoretical result. Notice, however that any interval can easily be converted to the given interval so there is no loss of generality. That is, in practice any uniformly distributed sequence can be sorted using bucket sort.
The text given in your question only points out the conditions about the input for counting sort and bucket sort. For counting sort the assumption about the input is that the range of integers in the list for sorting is very small. Where as the statement about bucket sort make another assumption about the input values. Here the range can be arbitrarily large but the distribution of the numbers in this range should be uniform. The value [0,1) in the statement about bucket sort do not mean the range in which bucket sort is effective. It simply tells you the nature of the input values.
I'm trying to improve my bucket sort for large number over 10000.I'm not quite sure, why my code isn't performing well on large numbers.
My Bucket Sort algorithm for array of size n:
Create array of linked list of size n
Calculate range for numbers
Calculate interval for each bucket
Calculate index for bucket, where to put particular number
(Problem: I calculated index by constantly subtracting interval from number and increment counter,every time i subtract interval.Counter is the index)
I believe this particular way of finding index takes very long for large numbers.
How can i improve finding index of buckets?
P.S. i heard there's way to preprocess array and find min and max number of array. Then calculate index by subtracting particular number from min. index=number-min. I didn't quite get the idea of calculating index.
Questions:
1. Is this efficient way to find index?
2. How do i handle cases when i have array of size 4, and numbers 31,34,51,56? 31 goes to bucket 0,34 goes to bucket 3, how about 51 and 56?
3. Is there any other way to calculate index?
You can find your index faster through Division. Index = value / interval. If the first interval starts at 'min' instead of 0, then use (value-min) as the numerator.