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Displaying Numbers in Certain Shapes

I just thought of this problem 15 minutes ago and even though it appears insanely easy I'm having a serious problem coming up with an answer.
Basically what I would like to do is based on a number (n) given by the user, I would like to draw a square shape.
Example: let's say the user gives the number 2, the result should be:
12
43
Now, suppose the user gives the number 3, the result should be:
123
894
765
etc..
Please don't give me the solution to this problem, I just want a clue or two to get me going.
I thought about doing it with a simple java class but I'm still struggling to get past the first condition:
public class DrawSquareWithNumbers {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter your number: ");
int number = scanner.nextInt();
for (int i = 0; i <= number; i++) {
if (i<number)
System.out.print(i);
if (i>=number) {
System.out.println("\n"+i);
}
}
}
}
Any tip? Thanks in advance.

So I think I just put way too much time in this but it's a fun challenge so I thought let's give it a go.
I implented a code version for this solution and it works quite well although it's probably not the cleanest as I approached the whole problem backwards.
Here is my solution to try out online (note it's severely unoptimized and by no ways good Java code. It's a quick and dirty implementation to be honest):
https://ideone.com/97JB7Y
So the idea is quite simple: We first calculate the correct value for each position in a matrix and then we print that matrix out.
Let's go over it in a bit more detail:
We start of by creating the Matrix for our values to print:
With a given size n this is
int[][] values = new int[n][n];
Now we want to calculate the correct value at each point. I chose to tackle it the "wrong way" around by not starting at the first point but at the center of the spiral.
Basically imagine this matrix with n = 3:
[1][2][3]
[8][9][4]
[7][6][5]
Instead of at 1 I just start at 9. Reasoning for this is that it's actually easier to calculate the position spiraling out from a point over spiraling in to a point.
So starting at this center point we spiral out from there in a circular fashion. For the matrix
[1][2]
[4][3]
this means we visit 4 -> 3 -> 2 -> 1. And then just save the correct value in the matrix.
Only problem with my approach is that for a matrix with uneven size (3, 5, 7, etc.) I still visit the points in spiraling order, for 3x3 the order of visiting is e.g. 9 -> 4 -> 3 -> 2 -> 1 -> 8 -> 7 -> 6 -> 5, as visualized in this perfect picture I totally drew in Paint:
This leads to the result matrix being inversed as such:
[5][6][7]
[4][9][8]
[3][2][1]
This small problem is easily fixed though by simply printing the matrix out inversed once more should n%2 != 0.
Hope I could help with maybe a different approach to the problem.

I think you want to make nxn matrix with user entered number. So you can check the input and then you can use loop as for(i=1; i<=n; i++) for rows and similarly for column(for j=0;j<=n;j++) and then you can print your desired shape. Since you have asked to give you idea only so I am not posting any code here. If in case you get stuck somewhere you can refer : https://www.google.com/amp/s/www.geeksforgeeks.org/print-a-given-matrix-in-spiral-form/amp/

Ok, let's give this a try. First let's assume you'll have to store the matrix before printing it and that there's no magic formula that allows you to print what you need in a single iteration.
Now, you've the NxN matrix, for example for 3 it'd be 3x3, 9 positions. Instead of solving it with a series of ifs in an ugly way, you could use direction vectors for a cleaner solution. Also assume for now that you've another NxN matrix filled with booleans, all set to false, that will represent the already printed positions in the NxN matrix that you will print in the end. When you write a number in the final NxN matrix, you put the same position's boolean to true in the boolean matrix.
So for example, you want to print the positions of the first row, 1 2 3. You are displacing to the right to print. This'd be the direction (1,0), aka the starting direction vector. You advance through the NxN matrix using this coordinates. When you go outside the matrix (in the example, your x position is 3) you decrease your x position by one and you "spin" your direction vector (this should be done in a separate function). (1,0) would spin to (0,-1). You keep using this vector to iterate your matrix, spinning as necesary. After the first whole circle, you will get to an already printed position before going outside the matrix. So after every print you've to check not only if you go outside the matrix, but also if that position has already a number on it. For this, you use the boolean matrix.
This is how I'd solve it, there are probably many other ways (and better ones). For starters you could use null, or a mark, in the final matrix and save yourself the booleans one.

Create the result matrix beforehand, and declare a variable for the current number, starting value = 1, the current co-ordinates, starting with (0,0), the stepping direction starting with "to the right"
Start loop
Calculate the co-ordinates of next step, and check it
If it is off the matrix, then change direction, and re calculate, re-check
If free, then put the number in the matrix, increment it, and loop
If it is not free, then end loop, print out result matrix

Coloring the Barnsley Fern fractal

While rendering the Barnsley fern fractal I come up with single color images or at most four color images i.e. the bottom left, bottom right, bottom stem and the rest of the leaves. Here is the image I get for example:
What I want however is to bring shades in the leaves and making stem thicker and of different color like:
I digged a bit about the algorithms that can be used, then I read in Draves's paper about fractal flames that during iteration of Iterated Function Systems a single point may be rendered many times if we use a single color which results in a loss of information so we need to create a histogram of how many times a point was to be rendered and then perform a rendering pass using the histogram with shades of colors log-density coloring.
I have brought myself to the point where I have the histogram but don't know how to use it to render the shades or using the log-density render technique. Can someone help me with such type of rendering or at least direct me to a source where I can read more about this with practical examples.
Here is what I have tried:
AffineTransformation f1 = new AffineTransformation(0,0,0,0.25,0,-0.4);
AffineTransformation f2 = new AffineTransformation(0.95,0.005,-0.005,0.93,-0.002,0.5);
AffineTransformation f3 = new AffineTransformation(0.035,-0.2,0.16,0.04,-0.09,0.02);
AffineTransformation f4 = new AffineTransformation(-0.04,0.2,0.16,0.04,0.083,0.12);
int N=Width*Height;
int pixelhistogram[] = new int[N];
for(int i=0;i< N*25;i++)
{
Point newpoint = new Point();
double probability = Math.random();
if(probability < 0.01)
{
newpoint = f1.transform(point);
}
else if(probability < 0.94)
{
newpoint = f2.transform(point);
}
else if(probability < 0.97)
{
newpoint = f3.transform(point);
}
else
{
newpoint = f4.transform(point);
}
point = newpoint;
// Translating the point to pixel in the image and
// incrementing that index in the pixelHistogram array by 1
// W and H are the Width and Height
int X=((int)(point.getX()*W/3)+W/2)/2 + W/4-1;
int Y=H-((int)(point.getY()*H/8) + H/9) -1;
pixelhistogram[W*Y+X]++;
}
// Now that I have the pixelhistogram
// I don't know how to render the shades using this
AffineTransformation is a simple class which performs Affine Transformation on a point. I omitted the code because otherwise the question would have become too lengthy.

A simple coloring would be to render pixel (X,Y) light green, green, or brown according to whether pixels[W*Y+X] is less than n1, between n1 and n2, or greater than n2. To determine n1 and n2, trial and error would probably be the simplest solution, but you could make an actual histogram of the log of the pixel counts that you have recorded to help judge where to put the cuts (or more generally you could use clustering algorithms to do it automatically).
PS: In the image that you show it looks like the stem is rendered with an L-system and the fronds are rendered using the three leaf transforms only (i.e. omit the fourth "stem-transform"); I would guess they are using the log pixel counts to shade the level of green but not to shade the stem.
Addition: I was asked, below, to discuss log-histograms. To avoid getting bogged down, I'd recommend first using a full featured data analysis software like R to see if this gets you what you want. Write out the pixels array to a text file with one number per line, then start R and run:
ct=scan('pixels_data.txt')
hist(log(ct))
If you see a a multimodal histogram (i.e. if it has clear peaks and valleys), that will suggest how to choose n1 and n2: put them in the valleys (i.e. if the valley on the plot is at y, set n1=exp(y)).
If you wind up plotting histograms in Java, it can apparently be done with the Jfreechart software. Just create an array with the logs of the values in the pixels array and create the histogram out of that.
At best I expect you to see only one valley in the histogram, if you use the standard 3-transform Barnsley fern, separating the really high stem values from the fronds. To color the fronds, if n is the cut between frond and stem, and pixels[W*Y+X] is less than n, you could color it using, say:
v=128.0*(log(n)-log(pixels[W*Y+X]))/log(n);
RGB=(v,255,v)
PS: Getting thick stems using the random iteration algorithm only is going to be a problem. If you change the 3rd transform to be less singular, your stems will wind up looking like thin ferns and not sticks. E.g.
{"title":"Thick Stem Fern","alist":[[
[0.11378443003074948,-0.005060836319767042,0.013131296101198788,0.21863066144310556,0.44540023470694723,0.01726296943557673],
[0.15415337683611596,-0.17449052243042712,0.23850452316465576,0.2090228040695959,0.3652068203134602,0.11052918709831461],
[-0.09216684947824424,0.20844742602316002,0.2262266208270773,0.22553569847678284,0.6389950926444947,-0.008256440681230735],
[0.8478159879190097,0.027115858923993468,-0.05918196850293869,0.8521840120809901,0.08189073762585078,0.1992198482087391]
]],"va":[1,0,0,1,0,0],"word_length":6,"level_max":40,"rect_size":1}
Is the json data to describe:

Why setSkew increase text size?

I'm using iText version 5.2.1.
To fully understand the setSkew method of Chunk object, i began to play with it with the following code:
for (int i=0; i <= 90; i+=5) {
Chunk c = new Chunk("A" + i);
c.setSkew((float)i, (float)-i);
document.add(c);
}
With my big surprise the text progressively gets bigger when approaching to 90 degree. I can't understand this behaviour: according to "itext in action 2nd ed." book, the first parameter of setSkew is the inclination of the baseline of the text, and the second the angle between characters and the (original) baseline.
So, what I'm missing?

What you're missing essentially is that you expect the skewing with your parameters to be merely something like a simple rotation which would keep sizes as they are. This is not the case for skewing. Instead skewing works like this:
(shamelessly copied out of "Figure 13 – Effects of Coordinate Transformations" in the PDF specification ISO 32000-1:2008)
What remains the same is the length of the projection of the skewed base line onto the regular base line of the text line:
Here x0 and x1 have the same length, but the skewed base line is longer. The steeper the angle is, the longer is the skewed base line and also the width of the glyphs there.
The analog goes for the y axis. Thus, your sample results in something like this:
If you prefer a mathematical reasoning, consider that the skewing transformation matrix has the values [1 tan(a) tan(b) 1 0 0]. So (1, 0) is skewed to (1, tan(a)), (0, 1) to (tan(b), 1), and everything else correspondignly according to linearity.

Fastest way to find if int array contains a number

This is an odd question. I have an integer array in Java, where each int represents a color. They will either be 0xFFFFFFFF or 0x0. What would be the FASTEST way to find if this array contains ANY values equal to 0xFFFFFFFF?
This is my current code:
int length = w * h;
for (int i = 0; i < length; i++) {
if (pixels[i] == 0xFFFFFFFF) {
return true;
}
}
I have no clue if there is a faster way to do this or not. I imagine you vets could have a trick or two though.
EDIT: Seeing as it is just a dumb array of pixels from Bitmap.getPixels(), there's no way it would be sorted or transformed to another storage structure. Thanks for the input, everyone, it seems like looping through is the best way in this case.

No, there is no faster way unless the array of integers is already sorted, which I doubt given it's an array of colours.
To scan through an unsorted array takes linear time "O(n)". That's what you do, and you exit the method as soon as a match is found which is good too.

Without switching to some other data structure, no, there is no better way to find whether the array contains that value. You have to look at all the array elements to see if it's there, since if you don't check some particular location you might miss the one copy of that pixel color.
That said, there are alternative ways that you could solve this problem. Here are a few thoughts on how to speed this up:
If every value is guaranteed to be either white or black, you could store two extra boolean values alongside the array representing whether there are white or black pixels. That way, once you've run the scan once, you could just read the booleans back. You could also store a count of the number of white and black pixels along with the array, and then whenever you write a pixel update the count by decrementing the number of pixels of the original color and incrementing the number of pixels of the new color. This would then give you the ability to check if a pixel of a given color exists in O(1) by just seeing if the correct counter is nonzero.
Alternatively, if you happen to know something about the image (perhaps where the white and black pixels ought to be), you could consider doing the iteration in a different order. For example, if the pixels you're looking for tend to be clustered in the center of the image, rewriting the loop to check there first might be a good idea since if there are any pixels of that type you'll find them more rapidly. This still has the same worst-case behavior, but for "realistic" images might be much faster.
If you have multiple threads available and the array is really huge (millions of elements), you could consider having multiple threads each search a part of the array for the value. This would only be feasible if you had a reason to suspect that most of the image was not white.
Since in most realistic images you might assume that the image is a mixture of colors and you're just looking for something of one color, then you might want to consider storing the image as a sparse array, where you store a list of the pixels that happen to be of one color (say, white) and then assume everything else is black. If you expect most images to be a solid color with a few outliers, this might be a very good representation. Additionally, it would give you constant-time lookup of whether any black or white pixels exist - just check if the list of set pixels is empty or consists of the entire image.
If the order doesn't matter, you could also store the elements in some container like a hash table, which could give you O(1) lookup of whether or not the element is there. You could also sort the array and then just check the endpoints.
As a microoptimization, you could consider always appending to the real image two values - one white pixel and one black pixel - so that you could always iterate until you find the value. This eliminates one of the comparisons from the loop (the check to see if you're in-bounds) and is recommended by some authors for very large arrays.
If you assume that most images are a nice mixture of white and black and are okay with getting the wrong answer a small fraction of the time, you could consider probing a few random locations and checking if any of them are the right color. If so, then clearly a pixel of the correct color exists and you're done. Otherwise, run the full linear scan. For images that are a nice blend of colors, this could save you an enormous amount of time, since you could probe some small number of locations (say, O(log n) of them) and end up avoiding a huge linear scan in many cases. This is exponentially faster than before.
If every value is either white or black, you could also consider storing the image in a bitvector. This would compress the size of the array by a factor of the machine word size (probably between 32-128x compression) You could then iterate across the compressed array and see if any value is not identically equal to 0 to see if any of the pixels are white. This also saves a huge amount of space, and I'd actually suggest doing this since it makes a lot of other operations easy as well.
Hope this helps!

It doesn't matter at the bytecode level, but at the native-code level,
if (pixels[i] != 0)
is likely to be a bit faster, given that you're sure only these two values can appear.

If your array is really big, it might be worth it to divide and conquer. That is, assign segments of the data to multiple threads (probably t threads where t is the number of available processor cores). With a sufficiently large data set, the parallelism may amortize the thread startup cost.

Here is the simple optimization that helps on large arrays: put the requested value at the end of the array and thus eliminate array bounds check. (templatetypedef has already mentioned this optimization.) This solution saves 25% of loop running time and it is good for large arrays:
tmp = a[n - 1]
a[n - 1] = 0xFFFFFFFF
pos = 0
while a[pos] != 0xFFFFFFFF
pos = pos + 1
a[n - 1] = tmp
if a[pos] = 0xFFFFFFFF then
return pos
return -1
There is the C# implementation with running time analysis on this address.

The only scope for improving the performance is the comparison. I feel bitwise operator would be a bit faster than the conditional operator.
You could do this
int length = w * h;
for (int i = 0; i < length; i++) {
if (pixels[i] & 0xFFFFFFFF) {
return true;
}
}

Can't you check when you insert the color into the array? If so, you could store the index of the array's element which contains the 0xFFFFFFFF color. Since you want "ANY" entry that has such value, this should do the trick :D
If not, your answer has the complexity of O(n) which is the best it could be, since the array isn't (and cannot be, as you say) ordered.

using the build-in foreach is a tad faster than the indexed for as id eliminates a bound check
for(int pix:pixels){
if(pix!=0)
return true;
}

Arrays.asList(...).contains(...)

Java Array Manipulation

I have a function named resize, which takes a source array, and resizes to new widths and height. The method I'm using, I think, is inefficient. I heard there's a better way to do it. Anyway, the code below works when scale is an int. However, there's a second function called half, where it uses resize to shrink an image in half. So I made scale a double, and used a typecast to convert it back to an int. This method is not working, and I dont know what the error is (the teacher uses his own grading and tests on these functions, and its not passing it). Can you spot the error, or is there a more efficient way to make a resize function?
public static int[][] resize(int[][] source, int newWidth, int newHeight) {
int[][] newImage=new int[newWidth][newHeight];
double scale=newWidth/(source.length);
for(int i=0;i<newWidth/scale;i++)
for(int j=0;j<newHeight/scale;j++)
for (int s1=0;s1<scale;s1++)
for (int s2=0;s2<scale;s2++)
newImage[(int)(i*scale+s1)][(int)(j*scale+s2)] =source[i][j];
return newImage;
}
/**
* Half the size of the image. This method should be just one line! Just
* delegate the work to resize()!
*/
public static int[][] half(int[][] source) {
int[][] newImage=new int[source.length/2][source[0].length/2];
newImage=resize(source,source.length/2,source[0].length/2);
return newImage;
}

So one scheme for changing the size of an image is to resample it (technically this is really the only way, every variation is really just a different kind of resampling function).
Cutting an image in half is super easy, you want to read every other pixel in each direction, and then load that pixel into the new half sized array. The hard part is making sure your bookkeeping is strong.
static int[][] halfImage(int[][] orig){
int[][] hi = new int[orig.length/2][orig[0].length/2];
for(int r = 0, newr = 0; r < orig.length; r += 2, newr++){
for(int c = 0, newc = 0; c < orig[0].length; c += 2, newc++){
hi[newr][newc] = orig[r][c];
}
}
return hi;
}
In the code above I'm indexing into the original image reading every other pixel in every other row starting at the 0th row and 0th column (assuming images are row major, here). Thus, r tells us which row in the original image we're looking at, and c tells us which column in the original image we're looking at. orig[r][c] gives us the "current" pixel.
Similarly, newr and newc index into the "half-image" matrix designated hi. For each increment in newr or newc we increment r and c by 2, respectively. By doing this, we skip every other pixel as we iterate through the image.
Writing a generalized resize routine that doesn't operate on nice fractional quantities (like 1/2, 1/4, 1/8, etc.) is really pretty hard. You'd need to define a way to determine the value of a sub-pixel -- a point between pixels -- for more complicated factors, like 0.13243, for example. This is, of course, easy to do, and you can develop a very naive linear interpolation principle, where when you need the value between two pixels you simply take the surrounding pixels, construct a line between their values, then read the sub-pixel point from the line. More complicated versions of interpolation might be a sinc based interpolation...or one of many others in widely published literature.
Blowing up the size of the image involves something a little different than we've done here (and if you do in fact have to write a generalized resize function you might consider splitting your function to handle upscaling and downscaling differently). You need to somehow create more values than you have originally -- those interpolation functions work for that too. A trivial method might simply be to repeat a value between points until you have enough, and slight variations on this as well, where you might take so many values from the left and so many from the right for a particular position.
What I'd encourage you to think about -- and since this is homework I'll stay away from the implementation -- is treating the scaling factor as something that causes you to make observations on one image, and writes on the new image. When the scaling factor is less than 1 you generally sample from the original image to populate the new image and ignore some of the original image's pixels. When the scaling factor is greater than 1, you generally write more often to the new image and might need to read the same value several times from the old image. (I'm doing a poor job highlighting the difference here, hopefully you see the dualism I'm getting at.)

What you have is pretty understandable, and I think it IS an O(n^4) algorithm. Ouchies!
You can improve it slightly by pushing the i*scale and j*scale out of the inner two loops - they are invariant where they are now. The optimizer might be doing it for you, however. There are also some other similar optimizations.
Regarding the error, run it twice, once with an input array that's got an even length (6x6) and another that's odd (7x7). And 6x7 and 7x6 while you're at it.

Based on your other question, it seems like you may be having trouble with mixing of types - with numeric conversions. One way to do this, which can make your code more debuggable and more readable to others not familiar with the problem space, would be to split the problematic line into multiple lines. Each minor operation would be one line, until you reach the final value. For example,
newImage[(int)(i*scale+s1)][(int)(j*scale+s2)] =source[i][j];
would become
int x = i * scale;
x += s1;
int y = j* scale;
y +=s2;
newImage[x][y] = source[i][j];
Now, you can run the code in a debugger and look at the values of each item after each operation is performed. When a value doesn't match what you think it should be, look at it and figure out why.
Now, back to the suspected problem: I expect that you need to use doubles somewhere, not ints - in your other question you talked about scaling factors. Is the factor less than 1? If so, when it's converted to an int, it'll be 0, and you'll get the wrong result.