Posix Threads

In this post I would demonstrate the usage of Posix threads through a very simple problem. I would also demonstrate how we can synchronize Posix threads using mutex. Though there are many synchronizing primitives available, I would stick to mutex for simplicity. Please note that the code presented here is only for demonstration purpose and is inefficient.

Before starting on the problem, let me give a brief introduction of Posix threads. Historically, hardware vendors developed their own proprietary implementations of threads. These implementations differed substantially from each other. So, it was very difficult to develop multi-threaded portable applications. So, a standard specification was required to standardize the thread interface and this interface is Posix thread.

Problem :

A string is to be processed by a given number of threads. The processing in this case would be printing of the string. Each thread should do specific amount of work before passing control to the next thread. The work in our case is the printing of characters. The control is to be passed to the next thread in a round robin manner till the string is processed completely.

Program should take 3 command line arguments.
<Program name> <Input String> <Amount of work> <Number of threads>
Here is the desired output for a sample run.
ankit@ankit:~/problems/p1$ ./driver ABCDEFGHILKLMN 2 3
Thread 1 : AB
Thread 2 : CD
Thread 3 : EF
Thread 1 : GH
Thread 2 : IL
Thread 3 : KL
Thread 1 : MN
ankit@ankit:~/problems/p1$

Note that each thread is getting control in a round robin manner. In the beginning, thread 1 printed "AB" since the amount of work specified is 2. Then thread 2 printed "CD", then thread 3 printed "EF". After this, control again came to thread 1.

This program should be complied using -lpthread option. eg: gcc driver.c -lpthread -o driver

This problem throws light on many basic aspects of Posix threads like thread creation, thread synchronization. Ideally the sleeping part in the program is not efficient. Instead we should wait the thread and signal when its turn come. But this is for demonstration purpose.

Subset Problems

In this post, I would discuss some basic programming problems on sets.

Problem Statement:

Given a list of numbers, find all the possible subsets of the list of numbers. For example, if the set of numbers given is {1,2,3}, the output should be {},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}.

Solution:

The first solution is recursive is pretty straight forward. Please follow the Java code given below.

The second solution is iterative and is more efficient than the first solution. This solution works based on the principle that for a set of n elements, there would 2^n subsets. The binary representation of an integer i in the range 0 to (2^n - 1) is used to find the corresponding ith subset. For example, if there are 3 elements in the set, we enumerate the binary representation of all the numbers from 0 to 7. Each binary representation represents a subset. The set bit at ith index represents that take ith element from the set. Here are those :
000 -> Empty set
001 -> Only take 1st element
010 -> Only take 2nd element
011 -> Take 1st and 2nd element
100 -> Only take 3rd element
101 -> Take 1st and 3rd element
110 -> Take 2nd and 3rd element
111 -> Take all elements

If we tweek the above problem statement a little. Here is the new problem statement:

Problem Statement:

Given a list of numbers from 0-9, find the greatest number that can be formed from these numbers such that the number is divisible by three

Solution:

We can use the recursive approach above. One more property is needed : If a number is divisible by 3, then the sum of its digits is also divisible by three. Since we have to find the greatest number, we can start with the greatest number in the recursive solution above. Also the list need to be reverse  sorted so that the smallest number is removed first. Here is the code segment :

Finding Overlapping Rectangles in a given set of Axis aligned rectagles

Describe an algorithm that takes an unsorted array of axis-aligned rectangles and returns any pair of rectangles that overlaps, if there is such a pair. Axis-aligned means that all the rectangle sides are either parallel or perpendicular to the x- and y-axis. You can assume that each rectangle object has two variables in it: the x-y coordinates of the upper-left corner and the bottom-right corner.

Solution :

We can use Sweep Line algorithm for this. Create a sorted array of the x-coordinates of the left and the right edges of all the rectangles. The x-coordinates should be annotated with the rectangles they belong and whether they are the left or the right edge of the rectangle. Now, scan the sorted list from left to right, adding each rectangle to a balanced BST when you encounter the left edge of a rectangle and removing the rectangle from the balanced BST when you encounter the right edge of the rectangle. The solution requires the usage of one dimensional range search tree. A range search tree is a tree which is capable of performing range search queries efficiently. The time complexity of a range search query in a range search tree is O(R + logN), where N is the number of nodes in the tree and R is the number of elements in the result set.

First lets see how a search query works in a one dimensional range search tree:
1DRangeSearchTree(k1, k2, v):
    Input : Keys k1 and k2, and a node v of a binary search tree T
    Output : The elements whose keys are greater than or equal to k1 and less than or equal to k2.
    Algorithm :
        if key(v) >= k1 and key(v) <= k2
            ElementsInLeftSubTree = 1DRangeSearchTree(k1, k2, T.left(v))
            ElementsInRightSubTree = 1DRangeSearchTree(k1, k2, T.right(v))
            return ElementsInLeftSubTree U ElementsInRightSubTree  U {element(v)}
        else if key(v) < k1
            return 1DRangeSearchTree(k1, k2, T.right(v))
        else if key(v) > k2
            return 1DRangeSearchTree(k1, k2, T.left(v))

Now lets see the detailed algorithm of the original problem:

1) Sort all left and right rectange edges, according to their X value, into list L.

2) Create a new empty range search tree T, for Y ordering of rectangle tops/bottoms

3) Create a new empty result set RS of unique rectangle pairs

4) Traverse L in ascending order. For all V in L:

   If V.isRightEdge()

      T.remove(V.rectangle.top)

      T.remove(V.rectangle.bottom)

   else

       For all U in T.getRange(V.rectangle.top, V.rectangle.bottom)

         RS.add(<V.rectangle, U.rectangle>)

      T.add(V.rectangle.top)

      T.add(V.rectangle.bottom)

5) return RS

Time complexity is O(R + N log N) where R is the actual number of intersections.

Please note that this algorithm does not consider two rectangles as overlapping if one rectangle is completely within another rectangle. However this can be corrected with a minor tweek in the search algorithm for 1DRangeSearchTree.

Reference : http://stackoverflow.com/questions/3324880/axisaligned-rectangles-intersection

Java ConcurrentHashMap demystified

This post is in continuation of my previous post on Java HashMap. You may want to go through that post if you are not sure of Java HashMap internals.

As mentioned in the previous post, Java HashMap is not thread safe. If you want to use HashMap in a multi-threaded environment, you might look to HashTable or ConcurrentHashMap. HashTable has a bottleneck that it uses map wide lock which greatly slows down the application. ConcurrentHashMap uses several tricks to achieve a high level of concurrency and avoid locking, including using multiple write locks for hash buckets. It also exploits the uncertainties of the JMM to minimize the time that a lock is held or avoid locking at all.

Please note that the major barrier for HashTable is that it uses single map wide lock that is held for the entirety of insertion, retrieval, removal or traversal. ConcurrentHashMap uses a collection of 32 locks each of which guards a subset of hash buckets. This means that maximum of 32 threads can modify the map at one time. 32 is the theoretical concurrency limit and may not be achievable in practice. So, if 31 threads are writing to the map, it is very much possible that 32nd thread trying to modify the map will block.

Retrieval :

Unlike HashTable, ConcurrentHashMap may not acquire a lock while retrieving a value from the map. A very important point to note here is that synchronization offers much more than atomicity. It also guarantees ordering. Lets see how... Java Language Specification permits some memory operations not be made visible to other threads instantly because of the performance benefits of using per processor caches. Since ConcurrentHashMap may not acquire a lock during retrieval, it should be prepared to handle stale values. Instead of locking, the LinkedList used at any bucket location in ConcurrentHashMap is designed such that the implementation can detect when its view of the list has become inconsistent or stale. If it finds so, it acquires the appropriate lock on the bucket and searches again. This is one of the reasons why  ConcurrentHashMap performs badly in single threaded environment. This approach optimizes the lookup in the common case - where most retrievals are successful and retrievals outnumber insertions and removals.

Removal :

Removal operation on ConcurrentHashMap holds the bucket lock for the duration of its execution. To understand the removal, we need to know the data structure of an Entry object in ConcurrentHashMap. So, here it is :
Please note that all elements except the value are final, whereas value is volatile. I assume that the reader is aware of what a volatile variable means in Java. Since the next variable in Entry object is final, we can't add or remove an element to the LInkedList at a bucket location. So, how does removal work ??? When an entry object is removed from the LinkedList, its value field is assigned the value null. Since value field is volatile, it changed value is immediately visible to all threads. Once the value field is assigned the value null, the portion of the LinkedList from head to the removed element is cloned and joined to the remainder of the LinkedList following the removed element. So, if another thread is looking for the removed element in the stale chain, it would see null value and would retry with lock which would ensure that the chain is upto date. Eventually the discarded portion of the stale chain would be garbage collected.

Insertion :

A new Entry object is always added to the head of the LinkedList at a bucket. The same logic would apply as explained in Removal section. If another thread is trying to retrieve the inserted value in a stale chain, it would retry with synchronization since the inserted value won't be present in the stale chain.

Update :

Like removal operation, update operation on ConcurrentHashMap also holds the bucket lock for the duration of its execution.  This doesn't necessarily block the reader from executing since the retrieve operation doesn't always need a lock. Update operation just looks for the appropriate Entry object in the chain and updates its value field. Since the value field is volatile, its changed value is available to all the threads at once.