Dynamically allocating 2d array with adjacent rows in memory

In a previous post “Allocating multidimentional array at runtime in C” I have explained a technique to allocate multidimensional arrays on runtime. While playing around with OpenMPI, I came to know that while sending/receiving a buffer, it requires the elements of a 2d matrix (or any dimension) to be in adjacent. Basically it does not care what you send, or receive, what it cares is the number of elements to be send should be adjacent to one another. In the previous post, the process will allocate the 2d or n-d matrix, but the rows of the matrix may not be adjacent to each other, as each row was allocated separately with malloc and then inserted into another array of pointers, each of which location points to the base addresses of these memory block. Read the post for details.

In C language the 2d array/matrix are stored in a row-major order, that is the elements of the matrix are stored adjacent to each other in the memory row wise. The first row comes first then just after the first row the second row starts, and so on. In the previous method the rows of the matrix may be scattered throughout the memory, as they are allocated with seperate malloc calls, but each of these returned addresses to the memory blocks (used as rows) are assigned to another array of pointers, which holds the rows together, and allows the mat[i][j] syntax to work.

For the applications in which, we might need to allocate the matrix dynamically at runtime, also have the rows of the matrix requires to be adjacent in the memory, and also make the mat[i][j] syntax work can be fulfilled by the following approach.
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Journey of a simple recursive code

Some one asked the following question in StackOverflow

Implement a function with prototype char *repeat(char *s, int n) so that it creates and returns a string which consists of n repetitions of the input string s. For example: if the input is “Hello” and 3, the output is “HelloHelloHello”. Use only recursive constructs.

I immediately started compiling the answer as it was a pretty straight forward question. But it was not only me who was posting the answer. So after posting, very quickly other answers started appearing. It was a general everyday answer, until a person (who also answered the question) pointed out a bug in my code. Bugs disturb the mind. So I dug into the code and fixed it. The things started to get challenging when others started to get more compact code, which didn’t let me sit idle.
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Implement stack using a queue

The puzzle is to implement basic stack operations with using only basic queue operations. That is, a queue object is given, we need construct a wrapper for the stack functions, push, pop, which will only use the queue object as its storage, and naturally will have to use the queue operations.

This can be done using only one queue object. Although either the push or the pop complexity will no more be O(1).
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Detect Endianness of a System

In a previous post “Little and Big Endian conversion” i have briefly discussed about the Big-Endian and the Little-Endian representations. It is the ordering of the bytes (elementary addressable elements) within the representation of a larger basic data type. These are the two byte orderings in the memory within a larger datatype. For example if the size of integer is 4 bytes in a system, then will the least significant byte be stored in the lower memory address or in the higher memory address. In short: if the most significant byte is stored in higher memory address then it is called the Little Endian representation, and if the most significant byte is stored in lower memory address than the least significant byte then it is called the Big-Endian representation. For a diagram see the post “Little and Big Endian conversion”. Also refer the Wikipedia Entry on Endianness.

So how to determine if your system a Little Endian system or a Big Endian system by running a piece of code? Continue reading “Detect Endianness of a System”

Get sorted index orderting of an array

Yesterday i was translating some code i wrote in R to C++. I had some calculation to do which required the list of index of the top n values in a list. Therefore what i needed was not a the list sorted itself or get a sorted copy of a given list, but actually the sorted order of the index into the actual array. To describe the problem, for example consider the following list:

arr = 20 50 40 80 10

The sorted (non-decreasing) ordering of the values is obviously

sorted_arr = 10 20 40 50 80

The sorted (non-decreasing) ordering of the index into the original array is (index starts at 1 in this example)

sorted_index_into_arr = 5 1 3 2 4

Therefore the smallest value in the list arr could be found by indexing into arr using the first value of sorted_index_into_arr, which stores the index of the array arr holding the smallest value.

arr[ sorted_index_into_arr[1] ]

The sorted index ordering is very easy to get in languages like R and Octave or Matlab. For example in R we can do the following to get the sorted index order using the order function:

> arr <- c (20, 50, 40, 80, 10)
> arr
[1] 20 50 40 80 10
> order (arr)
[1] 5 1 3 2 4

In the case of Octave or Matlab you can get the index order using the sort function, which will return two lists (1xn matrix), the first list is the sorted array itself, and the next one is the sorted index order into the original array.

octave:1> a = [20 50 40 80 10]
a =

   20   50   40   80   10

octave:2> [sorted index] = sort (a)
sorted =

   10   20   40   50   80

index =

   5   1   3   2   4

Although these functional languages provides gives these features in C/C++ it is not immediately available using the builtin sort library functions. But the sorting routines accepts a function pointer to a comparison function, writing appropriate code for which will do the trick.
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Find Sum (i … j) in a list

The problem is to sum the ith to jth element of a given list. The easiest solution runs in \mathcal{O}(n) time, which starts are the ith index and adds up the numbers in the list until it reaches the jth index. In this post i will show the process of calculating the sum of elements i through j both inclusive (where i \le j), which takes \mathcal{O}(1) time for finding the sum, and \mathcal{O}(n) time and space for preprocessing the given list of length n.
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Find a String in an Alphabet Grid

Everybody of us has definitely played the word search game in which a grid of characters are given, and a list of words are given which are to be found in the grid horizontally, vertically or diagonally. In this post i will show a simple solution to a more generalized version of the search which finds the strings in the grid with backtracking.
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Towers of Hanoi Iterative process: simulating recursion

The last post Recursive Solution to Towers of Hanoi described the well-known recursive definition and implementation of the Towers of Hanoi problem. This post is an extension presenting the same problem iteratively by simulating the recursion stack. This implementation will simply to simulate the recursion presented on the previous post by using an explicit manual stack.

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Recursive Solution to Towers of Hanoi

Towers of Hanoi is a mathematical game or a puzzle in which there are three pegs, and some disks (originally 8) of different radius placed on top of one another such that no larger disk is placed on a smaller disk. The task is to transfer such a column of disks from a source peg to another destination peg. The constraints are we can move only one disk at a time, and we may use the third peg as a temporary storage for the disks, and a larger disk cannot be placed on top of a smaller disk.

The puzzle was invented by the French mathematician Édouard Lucas in 1883. There is a legend about an Indian temple which contains a large room with three time-worn posts in it surrounded by 64 golden disks. Brahmin priests, acting out the command of an ancient prophecy, have been moving these disks, in accordance with the rules of the puzzle, since that time. The puzzle is therefore also known as the Tower of Brahma puzzle. According to the legend, when the last move of the puzzle is completed, the world will end.[2] It is not clear whether Lucas invented this legend or was inspired by it.


In this post I will describe the basic recursive solution to the Towers of Hanoi
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