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I recently heard about ternary search in which we divide an array into 3 parts and compare. Here there will be two comparisons but it reduces the array to n/3. Why don't people use this much?
Actually, people do use k-ary trees for arbitrary k.
This is, however, a tradeoff.
To find an element in a k-ary tree, you need around k*ln(N)/ln(k) operations (remember the change-of-base formula). The larger your k is, the more overall operations you need.
The logical extension of what you are saying is "why don't people use an N-ary tree for N data elements?". Which, of course, would be an array.
A ternary search will still give you the same asymptotic complexity O(log N) search time, and adds complexity to the implementation.
The same argument can be said for why you would not want a quad search or any other higher order.
Searching 1 billion (a US billion - 1,000,000,000) sorted items would take an average of about 15 compares with binary search and about 9 compares with a ternary search - not a huge advantage. And note that each 'ternary compare' might involve 2 actual comparisons.
Wow. The top voted answers miss the boat on this one, I think.
Your CPU doesn't support ternary logic as a single operation; it breaks ternary logic into several steps of binary logic. The most optimal code for the CPU is binary logic. If chips were common that supported ternary logic as a single operation, you'd be right.
B-Trees can have multiple branches at each node; a order-3 B-tree is ternary logic. Each step down the tree will take two comparisons instead of one, and this will probably cause it to be slower in CPU time.
B-Trees, however, are pretty common. If you assume that every node in the tree will be stored somewhere separately on disk, you're going to spend most of your time reading from disk... and the CPU won't be a bottleneck, but the disk will be. So you take a B-tree with 100,000 children per node, or whatever else will barely fit into one block of memory. B-trees with that kind of branching factor would rarely be more than three nodes high, and you'd only have three disk reads - three stops at a bottleneck - to search an enormous, enormous dataset.
Reviewing:
Ternary trees aren't supported by hardware, so they run less quickly.
B-tress with orders much, much, much higher than 3 are common for disk-optimization of large datasets; once you've gone past 2, go higher than 3.
The only way a ternary search can be faster than a binary search is if a 3-way partition determination can be done for less than about 1.55 times the cost of a 2-way comparison. If the items are stored in a sorted array, the 3-way determination will on average be 1.66 times as expensive as a 2-way determination. If information is stored in a tree, however, the cost to fetch information is high relative to the cost of actually comparing, and cache locality means the cost of randomly fetching a pair of related data is not much worse than the cost of fetching a single datum, a ternary or n-way tree may improve efficiency greatly.
What makes you think Ternary search should be faster?
Average number of comparisons:
in ternary search = ((1/3)*1 + (2/3)*2) * ln(n)/ln(3) ~ 1.517*ln(n)
in binary search = 1 * ln(n)/ln(2) ~ 1.443*ln(n).
Worst number of comparisons:
in ternary search = 2 * ln(n)/ln(3) ~ 1.820*ln(n)
in binary search = 1 * ln(n)/ln(2) ~ 1.443*ln(n).
So it looks like ternary search is worse.
Also, note that this sequence generalizes to linear search if we go on
Binary search
Ternary search
...
...
n-ary search ≡ linear search
So, in an n-ary search, we will have "one only COMPARE" which might take upto n actual comparisons.
"Terinary" (ternary?) search is more efficient in the best case, which would involve searching for the first element (or perhaps the last, depending on which comparison you do first). For elements farther from the end you're checking first, while two comparisons would narrow the array by 2/3 each time, the same two comparisons with binary search would narrow the search space by 3/4.
Add to that, binary search is simpler. You just compare and get one half or the other, rather than compare, if less than get the first third, else compare, if less than get the second third, else get the last third.
Ternary search can be effectively used on parallel architectures - FPGAs and ASICs. For example if internal FPGA memory required for search is less than half of the FPGA resource, you can make a duplicate memory block. This would allow to simultaneously access two different memory addresses and do all comparisons in a single clock cycle. This is one of the reasons why 100MHz FPGA can sometimes outperform the 4GHz CPU :)
Here's some random experimental evidence that I haven't vetted at all showing that it's slower than binary search.
Almost all textbooks and websites on binary search trees do not really talk about binary trees! They show you ternary search trees! True binary trees store data in their leaves not internal nodes (except for keys to navigate). Some call these leaf trees and make the distinction between node trees shown in textbooks:
J. Nievergelt, C.-K. Wong: Upper Bounds for the Total Path Length of Binary Trees,
Journal ACM 20 (1973) 1–6.
The following about this is from Peter Brass's book on data structures.
2.1 Two Models of Search Trees
In the outline just given, we supressed an important point that at first seems
trivial, but indeed it leads to two different models of search trees, either of
which can be combined with much of the following material, but one of which
is strongly preferable.
If we compare in each node the query key with the key contained in the
node and follow the left branch if the query key is smaller and the right branch
if the query key is larger, then what happens if they are equal? The two models
of search trees are as follows:
Take left branch if query key is smaller than node key; otherwise take the
right branch, until you reach a leaf of the tree. The keys in the interior node
of the tree are only for comparison; all the objects are in the leaves.
Take left branch if query key is smaller than node key; take the right branch
if the query key is larger than the node key; and take the object contained
in the node if they are equal.
This minor point has a number of consequences:
{ In model 1, the underlying tree is a binary tree, whereas in model 2, each
tree node is really a ternary node with a special middle neighbor.
{ In model 1, each interior node has a left and a right subtree (each possibly a
leaf node of the tree), whereas in model 2, we have to allow incomplete
nodes, where left or right subtree might be missing, and only the
comparison object and key are guaranteed to exist.
So the structure of a search tree of model 1 is more regular than that of a tree
of model 2; this is, at least for the implementation, a clear advantage.
{ In model 1, traversing an interior node requires only one comparison,
whereas in model 2, we need two comparisons to check the three
possibilities.
Indeed, trees of the same height in models 1 and 2 contain at most approximately
the same number of objects, but one needs twice as many comparisons in model
2 to reach the deepest objects of the tree. Of course, in model 2, there are also
some objects that are reached much earlier; the object in the root is found
with only two comparisons, but almost all objects are on or near the deepest
level.
Theorem. A tree of height h and model 1 contains at most 2^h objects.
A tree of height h and model 2 contains at most 2^h+1 − 1 objects.
This is easily seen because the tree of height h has as left and right subtrees a
tree of height at most h − 1 each, and in model 2 one additional object between
them.
{ In model 1, keys in interior nodes serve only for comparisons and may
reappear in the leaves for the identification of the objects. In model 2, each
key appears only once, together with its object.
It is even possible in model 1 that there are keys used for comparison that
do not belong to any object, for example, if the object has been deleted. By
conceptually separating these functions of comparison and identification, this
is not surprising, and in later structures we might even need to define artificial
tests not corresponding to any object, just to get a good division of the search
space. All keys used for comparison are necessarily distinct because in a model
1 tree, each interior node has nonempty left and right subtrees. So each key
occurs at most twice, once as comparison key and once as identification key in
the leaf.
Model 2 became the preferred textbook version because in most textbooks
the distinction between object and its key is not made: the key is the object.
Then it becomes unnatural to duplicate the key in the tree structure. But in
all real applications, the distinction between key and object is quite important.
One almost never wishes to keep track of just a set of numbers; the numbers
are normally associated with some further information, which is often much
larger than the key itself.
You may have heard ternary search being used in those riddles that involve weighing things on scales. Those scales can return 3 answers: left is lighter, both are the same, or left is heavier. So in a ternary search, it only takes 1 comparison.
However, computers use boolean logic, which only has 2 answers. To do the ternary search, you'd actually have to do 2 comparisons instead of 1.
I guess there are some cases where this is still faster as earlier posters mentioned, but you can see that ternary search isn't always better, and it's more confusing and less natural to implement on a computer.
Theoretically the minimum of k/ln(k) is achieved at e and since 3 is closer to e than 2 it requires less comparisons. You can check that 3/ln(3) = 2.73.. and 2/ln(2) = 2.88.. The reason why binary search could be faster is that the code for it will have less branches and will run faster on modern CPUs.
I have just posted a blog about the ternary search and I have shown some results. I have also provided some initial level implementations on my git repo I totally agree with every one about the theory part of the ternary search but why not give it a try? As per the implementation that part is easy enough if you have three years of coding experience.
I found that if you have huge data set and you need to search it many times ternary search has an advantage.
If you think you can do better with a ternary search go for it.
Although you get the same big-O complexity (ln n) in both search trees, the difference is in the constants. You have to do more comparisons for a ternary search tree at each level. So the difference boils down to k/ln(k) for a k-ary search tree. This has a minimum value at e=2.7 and k=2 provides the optimal result.
I remember learning a data structure that stored a set of integers as ranges in a tree, but it's been 10 years and I can't remember the name of the data structure, and I'm a bit fuzzy on the details. If it helps, it's a functional data structure that was taught at CMU, I believe in 15-212 (Principles of Programming) in 2002.
Basically, I want to store a set of integers, most of which are consecutive. I want to be able to query for set membership efficiently, add a range of integers efficiently, and remove a range of integers efficiently. In particular, I don't care to preserve what the original ranges are. It's better if adjacent ranges are coalesced into a single larger range.
A naive implementation would be to simply use a generic set data structure such as a HashSet or TreeSet, and add all integers in a range when adding a range, or remove all integers in a range when removing a range. But of course, that would waste a lot of memory in addition to making add and remove slow.
I'm thinking of a purely functional data structure, but for my current use I don't need it to be. IIRC, lookup, insertion, and deletion were all O(log N), where N was the number of ranges in the set.
So, can you tell me the name of the data structure I'm trying to remember, or a suitable alternative?
I found the old homework and the data structure I had in mind were Discrete Interval Encoding Trees or diets for short. They are described in detail in Diets for Fat Sets, Martin Erwig. Journal of Functional Programming, Vol. 8, No. 6, 627-632, 1998. It is basically a tree of intervals with the invariant that all of the intervals are non-overlapping and non-touching. There is a Haskell implementation in Hackage. I was hoping there would be an existing implementation for Scala, but I'm not seeing any.
The homework also included another data structure they called a Recursive Interval-Occluding Tree (RIOT), which rather than keeping only an interval at each node keeps an interval and another (possibly empty) RIOT of things removed from the interval. The assignment included benchmarks showing it did better than diets for random insertions and deletions. AFAICT it is simply something the TAs made up and never published as it no longer seems to exist anywhere on the Internets, at least not under that name.
You probably are looking for segment trees. This might be helpful: http://www.topcoder.com/tc?d1=tutorials&d2=lowestCommonAncestor&module=Static
You can also use binary search trees for the same, for which each node will have two data fields: min_val and max_val.
During insertion algorithm, you just need to call another merging operation to check if the left-child,parent,right-child create a sequence, so as to club them into a single node. This will take O(log n) time.
Other operations like deletion and look-up will take O(log n) time as usual, but special measures need to be taken while deletion.
I am looking for a method to do fast nearest neighbour (hopefully O(log n)) for high dimensional points (typically ~11-13 dimensional). I would like it to behave optimally during insertions after having initialized the structure. KD tree came to my mind but if you do not do bulk loading but do dynamic insertions, then kd tree ceases to be balanced and afaik balancing is an expensive operation.
So, I wanted to know what data structures would you prefer for such kind of setting. You have high dimensional points and you would like to do insertions and query for nearest neighbour.
Another data structure that comes to mind is the cover tree. Unlike KD trees which were originally developed to answer range queries, this data structure is optimal for nearest neighbor queries. It has been used in n-body problems that involve computing the k nearest neighbors of all the data points. Such problems also occur in density estimation schemes (Parzen windows).
I don't know enough about your specific problem, but I do know that there are online versions of this data structure. Check out Alexander Gray's page and this link
The Curse of Dimensionality gets in the way here. You might consider applying Principal Component Analysis (PCA) to reduce the dimensionality, but as far as I know, nobody has a great answer for this.
I have dealt with this type of problem before (in audio and video fingerprinting), sometimes with up to 30 dimensions. Analysis usually revealed that some of the dimensions did not contain relevant information for searches (actually fuzzy searches, my main goal), so I omitted them from the index structures used to access the data, but included them in the logic to determine matches from a list of candidates found during the search. This effectively reduced the dimensionality to a tractable level.
I simplified things further by quantizing the remaining dimensions severely, such that the entire multidimensional space was mapped into a 32-bit integer. I used this as the key in an STL map (a red-black tree), though I could have used a hash table. I was able to add millions of records dynamically to such a structure (RAM-based, of course) in about a minute or two, and searches took about a millisecond on average, though the data was by no means evenly distributed. Searches required careful enumeration of values in the dimensions that were mapped into the 32-bit key, but were reliable enough to use in a commercial product. I believe it is used to this day in iTunes Match, if my sources are correct. :)
The bottom line is that I recommend you take a look at your data and do something custom that exploits features in it to make for fast indexing and searching. Find the dimensions that vary the most and are the most independent of each other. Quantize those and use them as the key in an index. Each bucket in the index contains all items that share that key (there will likely be more than one). To find nearest neighbors, look at "nearby" keys and within each bucket, look for nearby values. Good luck.
p.s. I wrote a paper on my technique, available here. Sorry about the paywall. Perhaps you can find a free copy elsewhere. Let me know if you have any questions about it.
If you use a Bucket Kd-Tree with a reasonably large bucket size it lets the tree get better idea of where to split when the leaves get too full. The guys in Robocode do this under extremely harsh time-constraints, with random insertions happening on the fly and kNN with k>80, d > 10 and n > 30k in under 1ms. Check out this kD-Tree Tutorial which explains a bunch of kD-Tree enhancements and how to implement them.
In my experience, 11-13 dimensions is not too bad -- if you bulk-load. Both bulk-loaded R-trees (in contrast to k-d-trees these remain balanced!) and k-d-trees should still work much better than linear scanning.
Once you go fully dynamic, my experiences are much worse. Roughly: with bulk loaded trees I'm seeing 20x speedups, with incrementally built R-trees just 7x. So it does pay off to frequently rebuild the tree. And depending on how you organize your data, it may be much faster than you think. The bulk load for the k-d-tree that I'm using is O(n log n), and I read that there is a O(n log log n) variant, too. With a low constant factor. For the R-tree, Sort-Tile-Recursive is the best bulk load I have seen so far, and also O(n log n) with a low constant factor.
So yes, in high-dimensionality I would consider to just reload the tree from time to time.
I'm just starting to learn about parallel programming, and I'm looking at binary search.
This can't really be optimized by throwing more processors at it right? I know it's supposedly dividing and conquering, but you're really "decreasing and conquering" (from Wikipedia).
Or could you possibly parallelize the comparisons? (if X is less than array[mid], search from low to mid - 1; else if X is greater than array[mid] search from mid + 1 to high, else return mid, the index of X)
Or how about you give half of the array to one processor to do binary search on, and the other half to another? Wouldn't that be wasteful though? Because it's decreasing and conquering rather than simply dividing and conquering? Thoughts?
You can easily use parallelism.
For k is less than n processors, split the array into n/k groups and assign a processor to each group.
Run binary search on that group.
Now the time is log(n/k).
There is also a crew method that is logn/log(k+1).
I would think it certainly qualifies for parallelisation. At least, across two threads. Have one thread do a depth-first search, and the other do a breadth-first search. The winner is the algorithm that performs the fastest, which may be different from data-set to data-set.
I don't have much experience in parallel programming, but I doubt this is a good candidate for parallel processing. Each step of the algorithm depends on performing one comparison, and then proceeding down a set "path" based on this comparison (you either found your value, or now have to keep searching in a set "direction" based on the comparison). Two separate threads performing the same comparison won't get you anywhere any faster, and separate threads will both need to rely on the same comparison to decide what to do next, so they can't really do any useful, divided work on their own.
As far as your idea of splitting the array, I think you are just negating the benefit of binary search in this case. Your value (assuming it's in your array), will either be in the top or the bottom half of your array. The first comparison (at the midpoint) in a binary search is going to tell you which half you should be looking in. If you take that even further, consider breaking an array of N elements into N different binary searches (a naive attempt to parallel-ize). You are now doing N comparisons, when you don't need to. You are losing the power of binary search, in that each comparison will narrow down your search to the appropriate subset.
Hope that helps. Comments welcome.
Yes, in the classical sense of parallelization (multi-core), binary search and BST are not much better.
There are techniques like having multiple copies of the BST on L1 cache for each processor. Only one processor is active but the gains from having multiple L1 caches can be great (4 cycles for L1 vs 14 cycles for L2).
In real world problems you are often searching multiple keys at the same time.
Now, there is another kind of parallelization that can help: SIMD! Check out "Fast architecture sensitive tree search on modern CPUs and GPUs" by a team from Intel/UCSC/Oracle (SIGMOD 2010). It's very cool. BTW I'm basing my current research project on this very paper.
Parallel implementation can speed up a binary search, but the improvement is not particularly significant. Worst case, the time required for a binary search is log_2(n) where n is the number of elements in the list. A simple parallel implementation breaks the master list into k sub-lists to be bin-searched by parallel threads. The resulting worst-case time for the binary search is log_2(n/k) realizing a theoretical decrease in the search time.
Example:
A list of 1024 entries takes as many as 10 cycles to binary search using a single thread. Using 4 threads, each thread only would only take 8 cycles to complete the search. And using 8 threads, each thread takes 7 cycles. Thus, an 8 threaded parallel binary search could be up to 30% faster than the single threaded model.
However, his speed-up should not be confused with a improvement in efficiency: The 8 threaded model actually executes 8 * 7 = 56 comparisons to complete the search compared to the 10 comparisons executed by the single threaded binary search. It is up to the discretion of the programmer if the marginal gain in speed of a parallel application of binary search is appropriate or advantageous for their application.
I am pretty sure binary search can be speed up by a factor of log (M) where M is the number of processors. log(n/M) = log(n) - log(M) > log(n)/ log(M) for a constant M. I do not have a proof for a tight lower bound, but if M=n, the execution time is O(1), which cannot be any better. An algorithm sketch follows.
Parallel_Binary_Search(sorted_arraylist)
Divide your sorted_arraylist into M chunks of size n/M.
Apply one step of comparison to the middle element of each chunk.
If a comparator signals equality, return the address and terminate.
Otherwise, identify both adjacent chunks where comparators signaled (>) and (<), respectively.
Form a new Chunk starting from the element following the one that signaled (>) and ending at the element preceding the one that signaled (<).
If they are the same element, return fail and terminate.
Otherwise, Parallel_Binary_Search(Chunk)
I recently heard about ternary search in which we divide an array into 3 parts and compare. Here there will be two comparisons but it reduces the array to n/3. Why don't people use this much?
Actually, people do use k-ary trees for arbitrary k.
This is, however, a tradeoff.
To find an element in a k-ary tree, you need around k*ln(N)/ln(k) operations (remember the change-of-base formula). The larger your k is, the more overall operations you need.
The logical extension of what you are saying is "why don't people use an N-ary tree for N data elements?". Which, of course, would be an array.
A ternary search will still give you the same asymptotic complexity O(log N) search time, and adds complexity to the implementation.
The same argument can be said for why you would not want a quad search or any other higher order.
Searching 1 billion (a US billion - 1,000,000,000) sorted items would take an average of about 15 compares with binary search and about 9 compares with a ternary search - not a huge advantage. And note that each 'ternary compare' might involve 2 actual comparisons.
Wow. The top voted answers miss the boat on this one, I think.
Your CPU doesn't support ternary logic as a single operation; it breaks ternary logic into several steps of binary logic. The most optimal code for the CPU is binary logic. If chips were common that supported ternary logic as a single operation, you'd be right.
B-Trees can have multiple branches at each node; a order-3 B-tree is ternary logic. Each step down the tree will take two comparisons instead of one, and this will probably cause it to be slower in CPU time.
B-Trees, however, are pretty common. If you assume that every node in the tree will be stored somewhere separately on disk, you're going to spend most of your time reading from disk... and the CPU won't be a bottleneck, but the disk will be. So you take a B-tree with 100,000 children per node, or whatever else will barely fit into one block of memory. B-trees with that kind of branching factor would rarely be more than three nodes high, and you'd only have three disk reads - three stops at a bottleneck - to search an enormous, enormous dataset.
Reviewing:
Ternary trees aren't supported by hardware, so they run less quickly.
B-tress with orders much, much, much higher than 3 are common for disk-optimization of large datasets; once you've gone past 2, go higher than 3.
The only way a ternary search can be faster than a binary search is if a 3-way partition determination can be done for less than about 1.55 times the cost of a 2-way comparison. If the items are stored in a sorted array, the 3-way determination will on average be 1.66 times as expensive as a 2-way determination. If information is stored in a tree, however, the cost to fetch information is high relative to the cost of actually comparing, and cache locality means the cost of randomly fetching a pair of related data is not much worse than the cost of fetching a single datum, a ternary or n-way tree may improve efficiency greatly.
What makes you think Ternary search should be faster?
Average number of comparisons:
in ternary search = ((1/3)*1 + (2/3)*2) * ln(n)/ln(3) ~ 1.517*ln(n)
in binary search = 1 * ln(n)/ln(2) ~ 1.443*ln(n).
Worst number of comparisons:
in ternary search = 2 * ln(n)/ln(3) ~ 1.820*ln(n)
in binary search = 1 * ln(n)/ln(2) ~ 1.443*ln(n).
So it looks like ternary search is worse.
Also, note that this sequence generalizes to linear search if we go on
Binary search
Ternary search
...
...
n-ary search ≡ linear search
So, in an n-ary search, we will have "one only COMPARE" which might take upto n actual comparisons.
"Terinary" (ternary?) search is more efficient in the best case, which would involve searching for the first element (or perhaps the last, depending on which comparison you do first). For elements farther from the end you're checking first, while two comparisons would narrow the array by 2/3 each time, the same two comparisons with binary search would narrow the search space by 3/4.
Add to that, binary search is simpler. You just compare and get one half or the other, rather than compare, if less than get the first third, else compare, if less than get the second third, else get the last third.
Ternary search can be effectively used on parallel architectures - FPGAs and ASICs. For example if internal FPGA memory required for search is less than half of the FPGA resource, you can make a duplicate memory block. This would allow to simultaneously access two different memory addresses and do all comparisons in a single clock cycle. This is one of the reasons why 100MHz FPGA can sometimes outperform the 4GHz CPU :)
Here's some random experimental evidence that I haven't vetted at all showing that it's slower than binary search.
Almost all textbooks and websites on binary search trees do not really talk about binary trees! They show you ternary search trees! True binary trees store data in their leaves not internal nodes (except for keys to navigate). Some call these leaf trees and make the distinction between node trees shown in textbooks:
J. Nievergelt, C.-K. Wong: Upper Bounds for the Total Path Length of Binary Trees,
Journal ACM 20 (1973) 1–6.
The following about this is from Peter Brass's book on data structures.
2.1 Two Models of Search Trees
In the outline just given, we supressed an important point that at first seems
trivial, but indeed it leads to two different models of search trees, either of
which can be combined with much of the following material, but one of which
is strongly preferable.
If we compare in each node the query key with the key contained in the
node and follow the left branch if the query key is smaller and the right branch
if the query key is larger, then what happens if they are equal? The two models
of search trees are as follows:
Take left branch if query key is smaller than node key; otherwise take the
right branch, until you reach a leaf of the tree. The keys in the interior node
of the tree are only for comparison; all the objects are in the leaves.
Take left branch if query key is smaller than node key; take the right branch
if the query key is larger than the node key; and take the object contained
in the node if they are equal.
This minor point has a number of consequences:
{ In model 1, the underlying tree is a binary tree, whereas in model 2, each
tree node is really a ternary node with a special middle neighbor.
{ In model 1, each interior node has a left and a right subtree (each possibly a
leaf node of the tree), whereas in model 2, we have to allow incomplete
nodes, where left or right subtree might be missing, and only the
comparison object and key are guaranteed to exist.
So the structure of a search tree of model 1 is more regular than that of a tree
of model 2; this is, at least for the implementation, a clear advantage.
{ In model 1, traversing an interior node requires only one comparison,
whereas in model 2, we need two comparisons to check the three
possibilities.
Indeed, trees of the same height in models 1 and 2 contain at most approximately
the same number of objects, but one needs twice as many comparisons in model
2 to reach the deepest objects of the tree. Of course, in model 2, there are also
some objects that are reached much earlier; the object in the root is found
with only two comparisons, but almost all objects are on or near the deepest
level.
Theorem. A tree of height h and model 1 contains at most 2^h objects.
A tree of height h and model 2 contains at most 2^h+1 − 1 objects.
This is easily seen because the tree of height h has as left and right subtrees a
tree of height at most h − 1 each, and in model 2 one additional object between
them.
{ In model 1, keys in interior nodes serve only for comparisons and may
reappear in the leaves for the identification of the objects. In model 2, each
key appears only once, together with its object.
It is even possible in model 1 that there are keys used for comparison that
do not belong to any object, for example, if the object has been deleted. By
conceptually separating these functions of comparison and identification, this
is not surprising, and in later structures we might even need to define artificial
tests not corresponding to any object, just to get a good division of the search
space. All keys used for comparison are necessarily distinct because in a model
1 tree, each interior node has nonempty left and right subtrees. So each key
occurs at most twice, once as comparison key and once as identification key in
the leaf.
Model 2 became the preferred textbook version because in most textbooks
the distinction between object and its key is not made: the key is the object.
Then it becomes unnatural to duplicate the key in the tree structure. But in
all real applications, the distinction between key and object is quite important.
One almost never wishes to keep track of just a set of numbers; the numbers
are normally associated with some further information, which is often much
larger than the key itself.
You may have heard ternary search being used in those riddles that involve weighing things on scales. Those scales can return 3 answers: left is lighter, both are the same, or left is heavier. So in a ternary search, it only takes 1 comparison.
However, computers use boolean logic, which only has 2 answers. To do the ternary search, you'd actually have to do 2 comparisons instead of 1.
I guess there are some cases where this is still faster as earlier posters mentioned, but you can see that ternary search isn't always better, and it's more confusing and less natural to implement on a computer.
Theoretically the minimum of k/ln(k) is achieved at e and since 3 is closer to e than 2 it requires less comparisons. You can check that 3/ln(3) = 2.73.. and 2/ln(2) = 2.88.. The reason why binary search could be faster is that the code for it will have less branches and will run faster on modern CPUs.
I have just posted a blog about the ternary search and I have shown some results. I have also provided some initial level implementations on my git repo I totally agree with every one about the theory part of the ternary search but why not give it a try? As per the implementation that part is easy enough if you have three years of coding experience.
I found that if you have huge data set and you need to search it many times ternary search has an advantage.
If you think you can do better with a ternary search go for it.
Although you get the same big-O complexity (ln n) in both search trees, the difference is in the constants. You have to do more comparisons for a ternary search tree at each level. So the difference boils down to k/ln(k) for a k-ary search tree. This has a minimum value at e=2.7 and k=2 provides the optimal result.