i really wondering what exactly are Triangular Array and Sparse Array. i have been looking in google but there is no article i found. only discussion and they not talking about the basic. thanks you.
There are wikipedia pages on them.
Sparse array and Triangular array
A sparse array is one in which most of the cells of the array are initialized to null or 0. In such cases we waste a lot of memory. That is why, we generally replace a sparse array with a linked list and when a new cell gets initialized to a non-null value, we just create a node and add the corresponding links properly to reconstruct to linked list. A sparse array is one in which more than 70 percent of the array has null(0) values. When we know that an array is going to be sparse, we generally replace it with a more memory efficient linked list.
A Triangular array is a special kind of a 2-D array where the ith row contains i number of elements. The most simple example is a Pascal's triangle.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
[nCr = (n-1)C(r-1) + (n-1)C(r)].
So, a[i][j] = a[i-1][j-1] + a[i-1][j]. So a Triangular array helps in the construction of the Pascal's Triangle.
Hope this helps!!
Related
I first want to clarify if my logic in the following is correct:
In a 2 dimensional array, say A[x][y], x represents the number of braces and y the number of elements in each brace. So int A[2][3] is initialized as:
{
{1,2,3}, //from A[0][0] to A[0][2];
{4,5,6} //from A[1]{0] to A[1][2];
};
Second, I want to know what the similar correlations are in a 3 dimensional array, four dimensional, and so on, and how to know which dimension number correlates to what level for any multi-dimensional array.
For example, in A[3][4][2], does the 3 denote the number of 2d tables, or rows/columns in each table? And in A[2][3][4][5], does 2 represent the number of 2d tables, no. of 3d tables, no. of 1d tables, or no. of rows/columns per 1d table? Note:I'm getting my head around multi dimensional arrays for the first time, please explain as simplistically as possible.
Yes what you say it's correct. You can think that recursively.
Start from a 1D array (let's assume that it has 3 elements):
int 1darray[] = {0, 1, 2};
Now producing a 2D array simply says go inside every element of 1darray and put another 1D array, to produce a 2D one, like this:
int 2darray[] = {1darray_0, 1darray_1, 1darray_2};
where the 1darray_0, 1darray_1, 1darray_2 are 1D arrays, just like the 1darray we created in the start. So now this will form a 3x3 2D array.
Now the 3D array can be formed like this:
int 3darray[] = {2darray_0, 2darray_1, 2darray_2};
where the 2darray_0, 2darray_1, 2darray_2 are 2D arrays, just like the 2darray we created above. So now this will form a 3x3x3 3D array.
Your example:
A[3][4][2]
says that A has:
3 rows
4 columns
2 z-columns
In general however, I would advice you to have in mind the picture I have in my 2D dynamic array (C):
which describes in a nutshell what I tried to explain in the start.
As you increase your dimensions, you replace every element of the previous array with an array of the next dimension, while you reach the end.
I know mapping 2D array into 1D array has been asked many times, but I did not find a solution that would fit a where the column count varies.
So I want get a 1-dimensional index from this 2-dimensional array
Col> _0____1____2__
Row 0 |_0__|_1__|_2__|
V 1 |_3__|_4__|
2 |_5__|_6__|_7__|
3 |_8__|_9__|
4 |_10_|_11_|_12_|
5 |_13_|_14_|
The normal formula index = row * columns + column does not work, since after the 2nd row the index is out of place.
What is the correct formula here?
EDIT:
The specific issue is that I have a list of items in with the layout like in the grid, but a one dimensional array for the data. So while looping through the elements in the UI, I need to get the correct data, but can only get the row and column for that element. I need to find a way to turn a row/column value into an index for the data-array
Bad picture trying to explain it
A truly optimal answer (or even a provably correct one) will depend on the language you are using and how it lays out memory for such arrays.
However, taking your question simply at face value, you have to know what the actual length of each row is in order to calculate a 1D index.
So either the row length follows some pattern that can be inferred from the data, or you have (or can write) a rlen = rowLength( 2dTable, RowNumber) function.
Then, depending on how big the tables are and how fast you need to run, you can calculate a 1D index from the 2d table by adding all the previous row lengths until the current row length is less than the 2d column index.
or build a 1d table of the row lengths (or commulative rowlengths) so you can scan it and so only call your rowlength function for each row only once.
With a better description of your problem, you might get a better answer...
For your example which alternates between 3 and 2 columns you can construct a formula:
index = (row / 2) * (3 + 2) + (row % 2 ? 3 : 0) + column
(C-like syntax, assuming integer division)
In general though, the one and only way to implement what you're doing here, jagged arrays, is to make an array of arrays, a.k.a. an Iliffe vector. That means, use the row number as index into an array of pointers which point to the individual row arrays containing the actual data.
You can have an additional 1D array having the length of the columns say "length". Then your formula is index=sum {length(i)}+column. i runs from 0 to row.
I have a question in algorithm design about arrays, which should be implement in C language.
Suppose that we have an array which has n elements. For simplicity n is power of '2' like 1, 2, 4, 8, 16 , etc. I want to separate this to 2 parts with (n/2) elements. Condition of separating is lowest absolute difference between sum of all elements in two arrays for example if I have this array (9,2,5,3,6,1,4,7) it will be separate to these arrays (9,5,1,3) and (6,7,4,2) . summation of first array's elements is 18 and the summation of second array's elements is 19 and the difference is 1 and these two arrays are the answer but two arrays like (9,5,4,2) and (7,6,3,1) isn't the answer because the difference of element summation is 4 and we have found 1 . so 4 isn't the minimum difference. How to solve this?
Thank you.
This is the Partition Problem, which is unfortunately NP-Hard.
However, since your numbers are integers, if they are relatively low, there is a pseudo polynomial O(W*n^2) solution using Dynamic Programming (where W is sum of all elements).
The idea is to create the DP matrix of size (W/2+1)*(n+1)*(n/2+1), based on the following recursive formula:
D(0,i,0) = true
D(0,i,k) = false k != 0
D(x,i,k) = false x < 0
D(x,0,k) = false x > 0
D(x,i,0) = false x > 0
D(x,i,k) = D(x,i-1,k) OR D(x-arr[i], i-1,k-1)
The above gives a 3d matrix, where each entry D(x,i,k) says if there is a subset containing exactly k elements, that sums to x, and uses the first i elements as candidates.
Once you have this matrix, you just need to find the highest x (that is smaller than SUM/2) such that D(x,n,n/2) = true
Later, you can get the relevant subset by going back on the table and "retracing" your choices at each step. This thread deals with how it is done on a very similar problem.
For small sets, there is also the alternative of a naive brute force solution, which basically splits the array to all possible halves ((2n)!/(n!*n!) of those), and picks the best one out of them.
I currently have a 3 dimensional matrix and I want to extract a single row (into the third dimension) from it by index (say matrix(2,1,:)). I initially anticipated that the result of this would be a 1 dimensional matrix however what I got was a 1 by 1 by n matrix. Usually this wouldn't be a problem but some of the functions I'm using don't like 3D matrices. For example see the problem replicated below:
threeDeeMatrix=rand(3,3,3);
oneDeeAttempt=threeDeeMatrix(1,1,:);
norm(oneDeeAttempt)
Which returns the error message:
Error using norm
Input must be 2-D.
This is because oneDeeAttempt is
oneDeeAttempt(:,:,1) =
0.8400
oneDeeAttempt(:,:,2) =
0.0700
oneDeeAttempt(:,:,3) =
0.7663
rather than [0.8400 0.0700 0.7663]
How can I strip these extra dimensions? The only solution I can come up with is to use a loop to manually copy the values but that seems a little excessive.
Using permute to rearrange the matrix
The solution (which I found in the final stages of asking this) is to use permute which rearranges the order of the dimensions (similar to a=a' for 2D matrices). Once the unit dimensions are last they are stripped from the matrix and it becomes 1 dimensional.
oneDee=permute(oneDeeAttempt,[3 1 2]) %rearrange so the previous third dimension is now the first
%the matrix is now 3 by 1 by 1 which becomes 3
Using squeeze to remove leading singleton dimensions
As pointed out by Luis Mendo squeeze will very simply remove these leading singleton dimensions without having to worry about which dimensions are non singleton
oneDee=squeeze(oneDeeAttempt);
I am not 100% what the role of the 1: is here. At which index to start the copy? But then why not two such parameters for the rank 2 array?
To be a little more explicit:
In Fortran90 and up you access a single array value by giving a single index, and access a subarray (a window of the array) by giving a range of indices separated by a colon.
a(1) = 0.
a(2:5) = (/3.14,2.71,1.62,0./)
You can also give a step size.
a(1:5:2) = (/0,2.71,0./)
And finally, you can leave out values and a default will be inserted. So if a runs from index 1 to 5 then I could write the above as
a(::2) = (/0,2.71,0./)
and the 1 and 5 are implied. Obviously, you shouldn't leave these out if it makes the code unclear.
With a multidimensional array, you can mix and match these on each dimension, as in your example.
You're taking a slice of array2, namely the elements in the D'th column from row 1 to C and putting them in the slice of array1 which is elements 1 through A
So both slices are 1-dimensional arrays
Slice may not be the correct term in Fortran