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Linear probing

Linear probing
ECE 250 Algorithms and Data Structures Linear Douglas Wilhelm Harder, M.Math. LEL Department of Electrical and Computer Engineering University of Waterloo probing Waterloo, Ontario, Canada ece.uwaterloo.ca dwharderalumni.uwaterloo.ca © 20062013 by Douglas Wilhelm Harder. Some rights reserved.Linear probing 2 Outline Our first scheme for open addressing: – Linear probing—keep looking ahead one cell at a time – Examples and implementations – Primary clustering – Is it working looking ahead every k entriesLinear probing 3 Linear Probing The easiest method to probe the bins of the hash table is to search forward linearly Assume we are inserting into bin k: – If bin k is empty, we occupy it – Otherwise, check bin k + 1, k + 2, and so on, until an empty bin is found • If we reach the end of the array, we start at the front (bin 0)Linear probing 4 Linear Probing Consider a hash table with M = 16 bins Given a 3digit hexadecimal number: – The leastsignificant digit is the primary hash function (bin) – Example: for 6B72A , the initial bin is A and the jump size is 3 16 Linear probing 5 Insertion Insert these numbers into this initially empty hash table: 19A, 207, 3AD, 488, 5BA, 680, 74C, 826, 946, ACD, B32, C8B, DBE, E9C 0 1 2 3 4 5 6 7 8 9 A B C D E FLinear probing 6 Example Start with the first four values: 19A, 207, 3AD, 488 0 1 2 3 4 5 6 7 8 9 A B C D E FLinear probing 7 Example Start with the first four values: 19A, 207, 3AD, 488 0 1 2 3 4 5 6 7 8 9 A B C D E F 207 488 19A 3ADLinear probing 8 Example Next we must insert 5BA 0 1 2 3 4 5 6 7 8 9 A B C D E F 207 488 19A 3ADLinear probing 9 Example Next we must insert 5BA – Bin A is occupied – We search forward for the next empty bin 0 1 2 3 4 5 6 7 8 9 A B C D E F 207 488 19A 5BA 3ADLinear probing 10 Example Next we are adding 680, 74C, 826 0 1 2 3 4 5 6 7 8 9 A B C D E F 207 488 19A 5BA 3ADLinear probing 11 Example Next we are adding 680, 74C, 826 – All the bins are empty—simply insert them 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 826 207 488 19A 5BA 74C 3ADLinear probing 12 Example Next, we must insert 946 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 826 207 488 19A 5BA 74C 3ADLinear probing 13 Example Next, we must insert 946 – Bin 6 is occupied – The next empty bin is 9 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 826 207 488 946 19A 5BA 74C 3ADLinear probing 14 Example Next, we must insert ACD 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 826 207 488 946 19A 5BA 74C 3ADLinear probing 15 Example Next, we must insert ACD – Bin D is occupied – The next empty bin is E 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 826 207 488 946 19A 5BA 74C 3AD ACDLinear probing 16 Example Next, we insert B32 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 826 207 488 946 19A 5BA 74C 3AD ACDLinear probing 17 Example Next, we insert B32 – Bin 2 is unoccupied 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 826 207 488 946 19A 5BA 74C 3AD ACDLinear probing 18 Example Next, we insert C8B 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 826 207 488 946 19A 5BA 74C 3AD ACDLinear probing 19 Example Next, we insert C8B – Bin B is occupied – The next empty bin is F 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 20 Example Next, we insert D59 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 21 Example Next, we insert D59 – Bin 9 is occupied – The next empty bin is 1 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 22 Example Finally, insert E9C 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 23 Example Finally, insert E9C – Bin C is occupied – The next empty bin is 3 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E9C 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 24 Example Having completed these insertions: – The load factor is l = 14/16 = 0.875 – The average number of probes is 38/14 ≈ 2.71 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 25 Resizing the array To double the capacity of the array, each value must be rehashed – 680, B32, ACD, 5BA, 826, 207, 488, D59 may be immediately placed • We use the leastsignificant five bits for the initial bin 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 ACD B32 D59 5BALinear probing 26 Resizing the array To double the capacity of the array, each value must be rehashed – 19A resulted in a collision 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 ACD B32 D59 5BA 19A Linear probing 27 Resizing the array To double the capacity of the array, each value must be rehashed – 946 resulted in a collision 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 ACD B32 D59 5BA 19A Linear probing 28 Resizing the array To double the capacity of the array, each value must be rehashed – 74C fits into its bin 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 74C ACD 946 B32 D59 5BA 19A Linear probing 29 Resizing the array To double the capacity of the array, each value must be rehashed – 3AD resulted in a collision 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 74C ACD 3AD 946 B32 D59 5BA 19A Linear probing 30 Resizing the array To double the capacity of the array, each value must be rehashed – Both E9C and C8B fit without a collision – The load factor is l = 14/32 = 0.4375 – The average number of probes is 18/14 ≈ 1.29 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 C8B 74C ACD 3AD 946 B32 D59 5BA 19A E9CLinear probing 31 Marking bins occupied How can we mark a bin as occupied Pointers nullptr Positive integers 1 Floatingpoint numbers NaN Create a privately stored static object that does not Objects compare to any other instances of that class Suppose we’re storing arbitrary integers – Should we store –1938275734 in the hopes that it will never be inserted into the hash table – In general, magic numbers are bad—they lead to spurious errors A better solution: th – Create a bit vector where the k entry is marked th true if the k entry of the hash table is occupiedLinear probing 32 Searching Testing for membership is similar to insertions: Start at the appropriate bin, and searching forward until 1. The item is found, 2. An empty bin is found, or 3. We have traversed the entire array 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C 3AD ACD C8B The third case will only occur if the hash table is full (load factor of 1)Linear probing 33 Searching Searching for C8B 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 34 Searching Searching for C8B – Examine bins B, C, D, E, F – The value is found in Bin F 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 35 Searching Searching for 23E 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 36 Searching Searching for 23E – Search bins E, F, 0, 1, 2, 3, 4 – The last bin is empty; therefore, 23E is not in the table 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 × 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 37 Erasing We cannot simply remove elements from the hash table 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 38 Erasing We cannot simply remove elements from the hash table – For example, consider erasing 3AD 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C 3AD ACD C8BLinear probing 39 Erasing We cannot simply remove elements from the hash table – For example, consider erasing 3AD – If we just erase it, it is now an empty bin • By our algorithm, we cannot find ACD, C8B and D59 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C ACD C8BLinear probing 40 Erasing Instead, we must attempt to fill the empty bin 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C ACD C8BLinear probing 41 Erasing Instead, we must attempt to fill the empty bin – We can move ACD into the location 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C ACDACD C8BLinear probing 42 Erasing Now we have another bin to fill 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C ACD C8BLinear probing 43 Erasing Now we have another bin to fill – We can move ACD into the location 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C ACD C8B C8BLinear probing 44 Erasing Now we must attempt to fill the bin at F – We cannot move 680 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C ACD C8BLinear probing 45 Erasing Now we must attempt to fill the bin at F – We cannot move 680 – We can, however, move D59 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 D59 B32 E93 826 207 488 946 19A 5BA 74C ACD C8B D59Linear probing 46 Erasing At this point, we cannot move B32 or E93 and the next bin is empty – We are finished 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 E93 826 207 488 946 19A 5BA 74C ACD C8B D59Linear probing 47 Erasing Suppose we delete 207 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 E93 826 207 488 946 19A 5BA 74C ACD C8B D59Linear probing 48 Erasing Suppose we delete 207 – Cannot move 488 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 E93 826 488 946 19A 5BA 74C ACD C8B D59Linear probing 49 Erasing Suppose we delete 207 – We could move 946 into Bin 7 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 E93 826 946 488 946 19A 5BA 74C ACD C8B D59Linear probing 50 Erasing Suppose we delete 207 – We cannot move either the next five entries 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 E93 826 946 488 19A 5BA 74C ACD C8B D59Linear probing 51 Erasing Suppose we delete 207 – We cannot move either the next five entries 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 E93 826 946 488 D59 19A 5BA 74C ACD C8B D59Linear probing 52 Erasing Suppose we delete 207 – We cannot fill this bin with 680, and the next bin is empty – We are finished 0 1 2 3 4 5 6 7 8 9 A B C D E F 680 B32 E93 826 946 488 D59 19A 5BA 74C ACD C8BLinear probing 53 Erasing In general, assume: – The currently removed object has created a hole at index hole – The object we are checking is located at the position index and has a hash value of hashLinear probing 54 Erasing The first possibility is that hole index – In this case, the hash value of the object at index must either • equal to or less than the hole or • it must be greater than the index of the potential candidate – Remember: if we are checking the object at location index, this means that all entries between hole and index are both occupied and could not have been copied into the holeLinear probing 55 Erasing The other possibility is we wrapped around the end of the array, that is, hole index – In this case, the hash value of the object at index must be both • greater than the index of the potential candidate and • it must be less than or equal to the hole In either case, if the move is successful, the Now becomes the new hole to be filledLinear probing 56 Black Board Example Using the last digit as our hash function—insert these nine numbers into a hash table of size M = 10 31, 15, 79, 55, 42, 99, 60, 80, 23 Then, remove 79, 31, 42, and 60, in that orderLinear probing 57 Primary Clustering We have already observed the following phenomenon: – With more insertions, the contiguous regions (or clusters) get larger 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 C8B 74C ACD 3AD 946 B32 D59 5BA 19A E9C This results in longer search timesLinear probing 58 Primary Clustering We currently have three clusters of length four 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 C8B 74C ACD 3AD 946 B32 D59 5BA 19A E9CLinear probing 59 Primary Clustering There is a 5/32 ≈ 16 chance that an insertion will fill Bin A 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 C8B 74C ACD 3AD 946 B32 D59 5BA 19A E9CLinear probing 60 Primary Clustering There is a 5/32 ≈ 16 chance that an insertion will fill Bin A – This causes two clusters to coalesce into one larger cluster of length 9 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 747 C8B 74C ACD 3AD 946 B32 D59 5BA 19A E9CLinear probing 61 Primary Clustering There is now a 11/32 ≈ 34 chance that the next insertion will increase the length of this cluster 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 747 C8B 74C ACD 3AD 946 B32 D59 5BA 19A E9CLinear probing 62 Primary Clustering As the cluster length increases, the probability of further increasing the length increases 0 1 2 3 4 5 6 7 8 9 A B C D E F 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 680 826 207 488 946 747 C8B 74C ACD 3AD 946 B32 D59 5BA 19A E9C In general: – Suppose that a cluster is of length ℓ – An insertion either into any bin occupied by the chain or into the locations immediately before or after it will increase the length of the chain   2 – This gives a probability of MLinear probing 63 Runtime analysis The length of these chains will affect the number of probes required to perform insertions, accesses, or removals It is possible to estimate the average number of probes for a successful search, where l is the load factor: 1   1 1    2 1  l   For example: if l = 0.5, we require 1.5 probes on average nd Reference: Knuth, The Art of Computer Programming, Vol. 3, 2 Ed., Addison Wesley, 1998, p.528.Linear probing 64 Runtime analysis The number of probes for an unsuccessful search or for an insertion is higher:   1 1   1  2 2   1  l   2 For 0 ≤ l ≤ 1, then (1 – l) ≤ 1 – l, and therefore the reciprocal will be larger – Again, if l = 0.5 then we require 2.5 probes on average nd Reference: Knuth, The Art of Computer Programming, Vol. 3, 2 Ed., Addison Wesley, 1998, p.528.Linear probing 65 Runtime analysis The following plot shows how the number of required probes increasesLinear probing 66 Runtime analysis Our goal was to keep all operations Q(1) Unfortunate, as l grows, so does the run time One solution is to keep the load factor under a given bound If we choose l = 2/3, then the number of probes for either a successful or unsuccessful search is 2 and 5, respectivelyLinear probing 67 Runtime analysis Therefore, we have three choices: – Choose M large enough so that we will not pass this load factor • This could waste memory – Double the number of bins if the chosen load factor is reached • Not available if dynamic memory allocation is not available – Choose a different strategy from linear probing • Two possibilities are quadratic probing and double hashingLinear probing 68 Summary This topic introduced linear problem – Continue looking forward until an empty cell is found – Searching follows the same rule – Removing an object is more difficult – Primary clustering is an issue – Keep the load factor l ≤ 2/3Linear probing 69 References Wikipedia, http://en.wikipedia.org/wiki/Hashfunction 1 Cormen, Leiserson, and Rivest, Introduction to Algorithms, McGraw Hill, 1990. rd 2 Weiss, Data Structures and Algorithm Analysis in C++, 3 Ed., Addison Wesley. 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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