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This is quite an interesting problem.
In consistent hashing, if one server goes down, all the load on that server goes to the next server. Let’s say server 3 went down due to some reason, now all responsibility falls on server 2. Server 2 could not handle this extra traffic and went down by itself! Now the responsibility of server 3 and 2 fell on top of 1, 1
The load on it will be more! In this way, if 1 server is down, it can collapse like dominoes one after another while handling the traffic.
An excellent solution to both of these problems is to map servers to multiple locations on the circle.
Earlier some server was getting more traffic, some server was getting less traffic. Now that problem has reduced a lot. Mapping to 3 locations instead of 2 will reduce the gap, but it will take more time to find out the location of the server for any key. As a developer you have to decide what is important to you.
The domino effect is no longer a problem. Now if server 3
If it goes down, the responsibility will not fall on one person, but on servers 1 and 0
Share that responsibility.
Service Discovery
If a cache server is down or a new server is added, the video server needs to know that, otherwise how will it know which cache server to fetch data from? For this we will use a coordinator service. The coordinator’s job is to ping periodically to see which servers are alive. The coordinator will keep that information to himself and inform the video service. When he gets the list of video service servers, he can hash it and figure out which server circle is mapped to where.
Login Form Using HTML and CSS Free Download Tutorial – Log In Form Design.
Coordinator service is very common in distributed systems, there are many tools dedicated to implement it, such as zookeeper and etcd. The question may be what happens if the coordinator service crashes? The coordinator itself is a distributed system, deployed on multiple nodes, so the chance of the whole thing crashing is very low. Even if it crashes, the video server can maintain a list of cache servers itself, if the coordinator crashes it will send requests from that list, but at that time it will not find new servers or down servers.
Bloom filter
As mentioned earlier it takes a bit more time to find data in LSM Tree than B+ Tree. So we want to minimize the search as much as possible. Using Bloom filter we can tell very quickly whether a data is present or not. The problem is that the Bloom filter is a probabilistic data structure, if there is no data it can definitely say ‘no’, but if it says there is data then there remains some probability that there is no data.
I have a write up on how it works, you can check it out later. For now it is enough to know that the Bloom filter takes up very little space in memory, 1 million strings can be inserted into a Bloom filter of less than 1 megabyte. Data can be read from a Bloom filter in O(1) complexity.
As we write the data to the memtable, we can also insert it into the Bloom filter. When I get a read request, I will first ask the bloom filter whether there is data or not. If he doesn’t say no, then there’s no need to look, if he does, we’ll look in the memtable and database to make sure.
last word
This time I have designed a fairly good database. A popular database created by Google is Level-DB which is built with LSM-Tree. Later, Facebook forked it and created another more advanced database called Rocks-DB. Also many other no-sql databases use close type structures.
Note one thing, in the above design, even if we write data once to the database, actually that data is going to at least 4 places (Index, WAL, SSTable, Bloom Filter), that is, Write Amplification is happening. Of course, this is not only a problem with LSM-Tree, but this is much more common in databases built with B+ trees.
LSM-tree is good for fast writing of data, but can be counterproductive if the compaction process is not configured properly because writes are stopped for a while when merging data.
After that, your task will be to study B+ tree and compare the two types of databases, then your knowledge will be stronger. For example, a B+ tree can perform a variety of join operations that cannot be performed in an LSM-tree.
Then we get that our function is f(u,guard)
where guard is a Boolean value representing u
Does it have a guard or not?
We can start deploying guards from any node, even though the image looks like 0
first node. We actually imagine a node as root for ease of understanding, in this case 0 is taken as root. By defining the root we can direct the edges of the tree.
If you think like this, the solution will be much easier. You will always think of the nodes and edges below the node you are currently at, and assume that all other nodes and edges have been magically resolved.
Now if 1 is not guarded then we have to solve f(1,false) problem. That is what is shown in the picture above left. Since 1 is unguarded, 1 and 2 must be guarded to guard the following paths, and f(3,true) and f(4,true)
Two subproblems are to be solved:
f(1,false)=f(3,true)+f(4,true)
Again if 1
3 and 4 if this is guarded
This guard placement is optional. We will solve all the problems with one sitting and one without sitting to see which is optimal:
f(1,true)=1+min(f(3,true),f(3,false))+min(f(4,true)+f(4,false))
Extra 1 this time
Added because of guard placement.
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We then understand how to move from one subproblem to another. 0
Taking the first node gives us the solution to the min(f(0,true),f(0,false)) problem. have to do
Note one thing, we are taking the input of the tree edges as bidirectional because the input can be given in any order. Only the direction is fixed after imagining the route to 0. We are keeping track of a node’s parent node in an array because we won’t be doing anything with the node above it.
Complexity
We have subproblem n∗2
Because we have n nodes and in each node we have two options. Running a loop over the edges again in each subproblem. Total complexity will be O(n2)
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