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General Approach while designing such type of system:-The general approach is as follows:
Imagine we’re designing a hypothetical system X for millions of items (users, files, megabytes, etc):How would you solve it for a small number of items? Develop an algorithm for this case, which is often pretty straight-forward.What happens when you try to implement that algorithm with millions of items? It’s likely that you haverun out of space on the computer. So, divide up the files across many computers2.1) How do you divide up data across many machines? That is, do the first 100 items appear on the same computer? Or all items with the same hash value mod 100?2.2) About how many computers will you need? To estimate this, ask how big each item is and take a guess at how much space a typical computer hasNow, fix the problems that occur when you are using many computers. Make sure to answer the following questions:3.1) How does one machine know which machine it should access to look up data?
Can data get out of sync across computers? How do you handle that?3.2) How can you minimize expensive reads across computers?So this is the basic approach.Now come to the our task.Designing a data structure of a very large social network (eg. Facebook,LinkedList etc.)
Description and designing an algorithm to show the connection or path between two people. (eg. me->ram->naraian->vivek->you) :-Approach:Forget that we’re dealing with millions of users at first.Design this for the simple case
We can construct a graph by assuming every person is a node and if there is an edge between two nodes, then the two people are friends with each otherclass Person {
Person[] friends;
// Other info
}If I want to find the connection between two people, I would start with one person and do a
simple breadth first search.But here users are in Millions…… :(When we deal with a service the size of LinkedIn or Facebook, we cannot possibly keep all of our
data on one machine. That means that our simple Person data structure from above doesn’t
quite work—our friends may not live on the same machine as us. Instead, we can replace our
list of friends with a list of their IDs, and traverse as follows:1. For each friend ID: int machine_index = lookupMachineForUserID(id);2. Go to machine machine_index3. Person friend = lookupFriend(machine_index);There are more optimizations and follow up questions here than we could possibly discuss,
but here are just a few thoughts.Optimization: Reduce Machine JumpsJumping from one machine to another is expensive. Instead of randomly jumping from machine to machine with each friend, try to batch these jumps—
eg. if 5 of my friends live on one machine, I should look them up all at once.Optimization: Smart Division of People and MachinesPeople are much more likely to be friends with people who live in the same country as them
Rather than randomly dividing people up across machines, try to divide them up by country,
city, state, etc.
This will reduce the number of jumpsQuestion:
Breadth First Search usually requires “marking” a node as visited. How do you do that in this case?Usually, in BFS, we mark a node as visited by setting a flag visited in its node class. Here, we don’t want to do that (there could be multiple searches going on at the same time, so it’s bad to just edit our data)
In this case, we could mimic the marking of nodes with a hash table to lookup a node id and whether or not it’s been visited.Other Follow-Up Questions:1) In the real world, servers fail How does this affect you?2) How could you take advantage of caching?3) Do you search until the end of the graph (infinite)? How do you decide when to give up?4) In real life, some people have more friends of friends than others, and are therefore
more likely to make a path between you and someone else.How could you use this data
to pick where you start traversing?The following code demonstrates our algorithm:public class Server {

ArrayList<Machine> machines = new ArrayList<Machine>();

}

public class Machine {

public ArrayList<Person> persons = new ArrayList<Person>();

public int machineID;

}

public class Person {

private ArrayList<Integer> friends;

private int ID;

private int machineID;

private String info;

private Server server = new Server();

public String getInfo() { return info; }

public void setInfo(String info) {

this.info = info;

}

public int[] getFriends() {

int[] temp = new int[friends.size()];

for (int i = 0; i < temp.length; i++) {

temp[i] = friends.get(i);

}

return temp;

}

public int getID() { return ID; }

public int getMachineID() { return machineID; }

// Look up a person given their ID and Machine ID

public Person lookUpFriend(int machineID, int ID) {

for (Machine m : server.machines) {

if (m.machineID == machineID) {

for (Person p : m.persons) {

if (p.ID == ID){

return p;

}

}

}

}

return null;

}

// Look up a machine given the machine ID

public Machine lookUpMachine(int machineID) {

for (Machine m:server.machines) {

if (m.machineID == machineID)

return m;

}

return null;

}

public Person(int iD, int machineID) {

ID = iD;

this.machineID = machineID;

}
}

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