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External Sorting

External Sorting
ECE 250 Algorithms and Data Structures External Sorting Douglas Wilhelm Harder, M.Math. LEL Department of Electrical and Computer Engineering University of Waterloo Waterloo, Ontario, Canada ece.uwaterloo.ca dwharderalumni.uwaterloo.ca © 20062013 by Douglas Wilhelm Harder. Some rights reserved.External Sorting 2 Outline In this topic we will look at external sorting: – Suppose we are sorting entries stored in secondary memory – What if we can’t load everything into main memory – Can we sort sections of the unsorted dataExternal Sorting 3 Situation 8.10 Suppose we are sorting n entries stored in secondary memory (a hard drive or tape drive) than cannot be loaded into main memory – These secondary storages are block addressable – Assume each block is 4 KiB and there are b entries per block – Note, 4 KiB blocks are a common division of both hard drives and of main memory pagesExternal Sorting 4 Naming Convention 8.10.1 Convention: – Lowercase variables will sort the number of entries being sorted • Entries being sorted, entries per block, etc. – Uppercase variables will store the size of larger structuresExternal Sorting 5 Strategy 8.10.2 First, assume we can load M blocks at a time into main memory – Divide the data we are sorting into N sections of M blocks – Each section has m = Mb entries and n = NMb – Here, we are loading the first M = 10 blocks into main memoryExternal Sorting 6 Strategy 8.10.2 On these m = Mb entries, apply an inplace sorting algorithm such as quicksort or merge sort – The run time is Q( m ln(m) )External Sorting 7 Strategy 8.10.2 Write these M sorted blocks back into secondary memoryExternal Sorting 8 Strategy 8.10.2 Repeat this process for all N blocks – The run time is Q( Nm ln(m) ) = Q( n ln(m) ) External Sorting 9 Strategy 8.10.2 Next, load the first block of each of the N sections into corresponding input buffers allocated in main memory – If we had 2 GiB of main memory, we could load 512 such input buffers – We would need to sort more than 1 EiB before this was not possible 31 2GiB 2 19 – The math: 2GiB 2GiB 2GiB 21EiB 12 4KiB 2External Sorting 10 Strategy 8.10.2 Now, perform an Nway merge of these N sorted blocks into a single blocksized output bufferExternal Sorting 11 Strategy 8.10.2 When this output buffer is full, we will write it into secondary memory and start again with an empty buffer – Because we are using merging, we cannot do this inplace on secondary memory—we will require Q(n) additional secondary memoryExternal Sorting 12 Strategy 8.10.2 Whenever one of the N input buffers is entirely merged, we load the next block from the corresponding section into the input bufferExternal Sorting 13 Strategy 8.10.2 We continue this process of storing the merged block into secondary memory and reading subsequent blocks into main memoryExternal Sorting 14 Strategy 8.10.2 We continue this process of storing the merged block into secondary memory and reading subsequent blocks into main memory – At some point we will be merging the last blocks of each of the sectionsExternal Sorting 15 Strategy 8.10.2 When the last block is written out, the collection of MN blocks now contain a sorted list of the initial n entriesExternal Sorting 16 Runtime Analysis 8.10.3 With merge sort, we saw that merging was an Q(n + n ) operation 1 2 when merging lists of size n and n 1 2 – From slide 8, this suggests the run time is Q( n ln(m) ) – If m is sufficiently small, this suggests the runtime is Q(n) External Sorting 17 Runtime Analysis 8.10.3 This, however, is a mathematical fallacy… – A binary merge can be performed in linear time, an Nway merge cannot – The best we can hope for is using a binary minheap • Each will require O(ln(N)) time for a total of Q( n ln(N) ) Thus, the total run time is Q( n ln(m) + n ln(N) ) = Q( n ln(mN) ) = Q( n ln(n) ) Recall that n = mNExternal Sorting 18 Runtime Analysis 8.10.3 To be complete, we should also consider the run time associated with loading and saving data to the disk: – Each item of data is copied from or to the external disk four times – We will assume the memory size of each item is fixed and reasonable – Thus, the run time of these operations is Q(n) – If we are considering the size of the objects to be variable, say k bytes, we would have to revise our run time to Q( n ln(n) + nk )External Sorting 19 Additional Remarks 8.10.4 With 4 KiB blocks and 2 GiB of available main memory, we determined we could sort up to 1 EiB of data What if our available memory was smaller, say 1 MiB 20 1MiB 2 8 1MiB1MiB1MiB 2 256MiB 12 4KiB 2 Solution – Divide the data into 256 MiB blocks – Sort each of these individually use external sorting – Now merge these using the same strategy • We are restricted by the number of blocks we can merge – This process can be repeated arbitrarily often—doing so K times allows K K us to sort 20   1MiB 2 8K 1MiB1MiB 2 MiB   12 4KiB 2  External Sorting 20 Additional Remarks 8.10.4 We have focused on one implementation of using blocksized buffers—optimizations, however, are always possible: – Performing the initial sorting (quick sort) in parallel – Storing at least a portion of the data in solidstate drives – Using larger input and output buffers (useful if the data is contiguous on the hard drive)External Sorting 21 Summary This topic covered external sorting – Divide the n entries being sorted into N sections that fit in main memory – Sort each of the N section and write to secondary memory – Load blocks from each section as necessary into main memory – Merge the sections using an Nway merge – Write the merged blocks to secondary memory – The run time is Q(n ln(n))External Sorting 22 References nd Donald E. Knuth, The Art of Computer Programming, Volume 3: Sorting and Searching, 2 Ed., Addison Wesley, 1998, §5.4, pp.248379. Wikipedia, https://en.wikipedia.org/wiki/Externalsorting Special thanks to Prof. Ran Ginosar who made some observations and suggestions and who led me to finding an error in my calculations These slides are provided for the ECE 250 Algorithms and Data Structures course. The material in it reflects Douglas W. Harder’s best judgment in light of the information available to him at the time of preparation. Any reliance on these course slides by any party for any other purpose are the responsibility of such parties. Douglas W. Harder accepts no responsibility for damages, if any, suffered by any party as a result of decisions made or actions based on these course slides for any other purpose than that for which it was intended.
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