Filling a grid with boxes with numbers with decreasing size - math

There is this interesting game, that has numbers in a grid, where each number has progressively smaller font. The player's task is to click on the numbers in succession.
I'm interested in the algorithm that creates the boxes for the numbers, I cannot think of a way it works.
Apparently the grid has N numbers (apart from the 1.88 in the picture), number 1 has the biggest font and succesively the font size decreases. Then the numbers are somehow placed on the grid and boxes grow around them. But it doesn't seem totally random, as there are some horizotal lines that go across the whole grid.
Do you have an idea on how this might work?

It looks to me as though the boxes have been generated by successive division. That is, starting with the full rectangle, a dividing line (horizontal or vertical) was placed, and then the two resulting rectangles were subdivided in turn, until there were enough rectangles for the game.
Here's a sketch of the first algorithm I would try. N is the number of rectangles I want to divide the original rectangle into, and A is a critical aspect ratio used to stop the small rectangles getting too narrow. (Perhaps A = 1.5 would be a good start.)
Create an empty priority queue and add the full rectangle to it.
If the length of the priority queue is greater than or equal to N, stop.
Remove the largest rectangle, R, from the priority queue.
Choose whether to divide it horizontally or vertically: if its aspect ratio (width/height) is greater than A, divide it vertically; if less than 1/A, divide it horizontally, otherwise choose at random.
Decide where to put the dividing line. (Perhaps randomly between 40% and 60% along the chosen dimension.)
This divides R into two smaller rectangles. Add both of them to the priority queue. Go to step 2.
When this algorithm completes, there are N rectangles in the queue. Put the number 1 in the largest of them, the number 2 in the second largest, and so on.
It turns out that putting the numbers into the boxes is not quite as straightforward as I assumed in my first attempt. The area metric works well for subdividing the rectangles nicely, but it doesn't work for putting the numbers into the boxes, because for fitting text into a box, the height and width both have to be taken into account (the area is not so useful).
Instead of explaining the algorithm for putting numbers into the boxes, I will just give you some sample code in Python and let you reverse-engineer it!
import heapq, itertools, random
import Image, ImageDraw, ImageFont
# ALGORITHM PARAMETERS
aspect_max = 1.5 # Above this ratio, always divide vertically
aspect_min = 1.0 # Below this ratio, always divide horizontally
div_min = 0.4 # Minimum position for dividing line
div_max = 0.6 # Maximum position for dividing line
digit_ratio = 0.7 # Aspect ratio of widest digit in font
label_margin = 2 # Margin around label (pixels)
class Rectangle(object):
def __init__(self, x, y, w, h):
self.x = x
self.y = y
self.w = w
self.h = h
self.area = self.w * self.h
self.aspect = float(self.w) / self.h
def __le__(self, other):
# The sense of this comparison is inverted so we can put
# Rectangles into a min-heap and be able to pop the largest.
return other.area <= self.area
def __repr__(self):
return 'Rectangle({0.x}, {0.y}, {0.w}, {0.h})'.format(self)
def divide(self, n):
"""
Divide this rectangle into `n` smaller rectangles and yield
them in order by area (largest first).
"""
def division(l):
return random.randrange(int(l * div_min), int(l * div_max))
queue = [self]
while len(queue) < n:
r = heapq.heappop(queue)
if (r.aspect > aspect_max
or r.aspect > aspect_min
and random.random() < 0.5):
# Vertical division
w = division(r.w)
heapq.heappush(queue, Rectangle(r.x, r.y, w, r.h))
heapq.heappush(queue, Rectangle(r.x + w, r.y, r.w - w, r.h))
else:
# Horizontal division
h = division(r.h)
heapq.heappush(queue, Rectangle(r.x, r.y, r.w, h))
heapq.heappush(queue, Rectangle(r.x, r.y + h, r.w, r.h - h))
while queue:
yield heapq.heappop(queue)
def font_height(self, n):
"""
Return the largest font height such that we can draw `n`
digits in this rectangle.
"""
return min(int((self.w - label_margin * 2) / (digit_ratio * n)),
self.h - label_margin * 2)
def draw_rectangles(rectangles, fontfile):
"""
Create and return an Image containing `rectangles`. Label each
rectangle with a number using the TrueType font in `fontfile`.
"""
rectangles = list(rectangles)
im = Image.new('RGBA', (1 + max(r.x + r.w for r in rectangles),
1 + max(r.y + r.h for r in rectangles)))
draw = ImageDraw.Draw(im)
for digits in itertools.count(1):
rectangles = sorted(rectangles,
key = lambda r: r.font_height(digits),
reverse = True)
i_min = 10 ** (digits - 1)
i_max = 10 ** digits
i_range = i_max - i_min
for i in xrange(i_range):
if i >= len(rectangles): return im
r = rectangles[i]
draw.line((r.x, r.y, r.x + r.w, r.y, r.x + r.w, r.y + r.h,
r.x, r.y + r.h, r.x, r.y),
fill = 'black', width = 1)
label = str(i + i_min)
font = ImageFont.truetype(fontfile, r.font_height(digits))
lw, lh = font.getsize(label)
draw.text((r.x + (r.w - lw) // 2, r.y + (r.h - lh) // 2),
label, fill = 'black', font = font)
rectangles = rectangles[i_range:]
Here's a sample run:
>>> R = Rectangle(0, 0, 400, 400)
>>> draw_rectangles(R.divide(30), '/Library/Fonts/Verdana.ttf').save('q10009528.png')

The pattern of the cuts looks recursive. That is, the process of dividing the region into rectangles consists of cutting a rectangle in two, over and over. There are two cuts that divide the whole rectangular region (the horizontal cuts above and below 1), so we can't tell which cut came first, but we see the cuts as a kind of tree: the cut that separates 1 from 10 produced a large rectangle below it (20, 21, 4, 10, etc.), which was then divided by the vertical cut between 21 and 4, the rectangle containing 4 was later divided by the cut that separates 4 and 14, and so on. There are N cuts that produce N regions plus one leftover ("1.88") which is not necessary but which might give us a clue.
Now we just have to figure out the order of the cuts, the choice of proportion and the choice of orientation.
Consecutive numbers are rarely neighbors, so it looks as if numbers are not assigned as the cutting progresses. Instead, the region is chopped into rectangles, the rectangles are sorted by size and then numbers are assigned (notice that 20 and 21 are neighbors, but they were formed by other cuts after the one that divides them).
A plausible hypothesis for the order of the cuts is that the algorithm always cuts the largest rectangle. If that were not true, we might see, e.g., 14 bigger than 15 and 18 combined, and I see no example of that.
Proportion... With careful measurement we could see the actual distribution of proportions, but I don't feel like doing that much work. We see no very long, thin rectangles and no 50/50 cuts, so at a guess I'd say the algorithm chooses randomly, in some range like [0.6, 0.8]. Maybe it tries to avoid making a rectangle very close to the size of a rectangle that already exists. After all the cuts, the rectangle chosen to be left over ("1.88") is neither the biggest nor the smallest; maybe it's random, maybe it's the second-biggest, maybe it's something else-- more examples would be useful.
The orientation seems to be strongly biased towards cutting rectangles across their narrow width, rather than "lengthwise". This has the effect of producing rectangles more like squares and less like books on a shelf. The only possible exception I can see is the 1-9 cut, which might divide the block whose lower-right number is 1 lengthwise. But that depends on the order of cuts above and below 1, so it leads to a hypothesis: the algorithm always cuts a rectangle along its shorter dimension, and the 1-9 cut was actually the first.
That's about as far as I can go, short of breaking out a ruler and calculator.

Related

How to subdivide the space inside a small multiple

I need to create small multiples (like the one shown in the picture) using ggplot2 where each circle occupies 70% of the total size of it's own small multiple and the remaining 30% is empty space.
I know the center and the radius of the circle.
Question: Is it possible to do this, and if so, how?
Area of ​​a circle = C = π * r²
Area of your square = S = width * length = side²
So just calculate C, so you can calculate how big S must be to fullfill your criteria. When you know S you know the side, which translates to how you have to set the axis limits for x and y.

How can I seamlessly wrap map tiles around cylindrically?

I'm creating a game that takes place on a map, and the player should be able to scroll around the map. I'm using real-world data from NASA as a 5700 by 2700 pixel image split into 4 smaller ones, each corresponding to a hemisphere:
How I split up the image:
The player will be viewing the world through a camera, which is currently in a 4:3 aspect ratio, which can be moved around. Its height and width can be described as two variables x and y, currently at 480 and 360 respectively.
Model of the camera:
In practice, the camera is "fixed" and instead the tiles move. The camera's center is described as two variables: xcam and ycam.
Currently, the 4 tiles move and hide flawlessly. The problem arises when the camera passes over the "edge" at 180 degrees latitude. What should happen is that the tiles on one side should show and move as if the world was a cylinder without any noticeable gaps. I update xcam by doing this equation to it:
xcam = ((xcam + (2700 - x) mod (5400 - x)) - (2700 - x)
And the tiles' centers update according to these equations (I will focus only on tiles 1 and 2 for simplicity):
tile1_x = xcam - 1350
tile1_y = ycam + 650
tile2_x = xcam + 1350
tile2_y = ycam + 650
Using this, whenever the camera moves past the leftmost edge of tile 1, it "skips" and instead of tile 1 still being visible with tile 2 in view, it moves enough so that tile 2's rightmost edge is in the camera's rightmost edge.
Here's what happens in reality: ,
and here's what I want to happen: .
So, is there any way to update the equations I'm using (or even completely redo everything) so that I can get smooth wrapping?
I think you unnecessarily hard-code a number of tiles and their sizes, and thus bind your code to those data. In my opinion it would be better to store them in some variables, so that they can be easily modified in one place if data ever changes. This also allows us to write a more flexible code.
So, let's assume we have variables:
// logical size of the whole Earth's map,
// currently 2 and 2
int ncols, nrows;
// single tile's size, currently 2700 and 1350
int wtile, htile;
// the whole Earth map's size
// always ncols*wtile and nrows*htile
int wmap, hmap;
Tile tiles[nrows][ncols];
// viewport's center in map coordinates
int xcam, ycam;
// viewport's size in map coordinates, currently 480 and 360
int wcam, hcam;
Whenever we update the player's position, we need to make sure the position falls within an allowed range. But, we need to establish the coordinates system first in order to define the allowed range. For example, if x values span from 0 to wmap-1, increasing rightwards (towards East), and y values span from 0 to hmap-1, increasing downwards (toward South), then:
// player's displacement
int dx, dy;
xcam = (xcam + dx) mod wmap
ycam = (ycam + dy) mod hmap
assures the camera position is always within the map. (Assumed the mod operator always returns non-negative value. Should it work like the C language % operator, which returns negative result for negative dividend, one needs to add a divisor first to make sure the first argument is non-negative: xcam = (xcam + dx + wmap) mod wmap, etc.)
If you'd rather like to have xcam,ycam = 0,0 at the center of a map (that is, at the Greenwich meridian and the equator), then the allowed range would be -wmap/2 through wmap/2-1 for x and -hmap/2 through hmap/2 - 1 for y. Then:
xcam = (xcam + dx + wmap/2) mod wmap - wmap/2
ycam = (ycam + dy + hmap/2) mod hmap - hmap/2
More generally, let x0, y0 denote the 'zero' position of camera relative to the upper-left corner of the map. Then we can update the camera position by transforming it to the map's coordinates, then shifting and wrapping, and finally transforming back to camera's coordinates:
xmap = xcam + x0
ymap = ycam + y0
xmap = (xmap + dx) mod wmap
ymap = (ymap + dy) mod hmap
xcam = xmap - x0
ycam = ymap - y0
or, more compactly:
xcam = (xcam + dx + x0) mod wmap - x0
ycam = (ycam + dy + y0) mod hmap - y0
Now, when we know the position of the viewport (camera) relative to the map, we need to fill it with the map tiles. And a new decision must be made here.
When we travel from Anchorage, Alaska (western hemisphere) to the North, we eventually reach the North Pole and then we'll find ourselves in the eastern hemisphere, headin South. If we proceed in the same direction, we'll get to Kuusamo, Norway, then Sankt Petersburg, Russia, then Kiev, Ukraine... But that would be a travel to the South! We usually do not describe it as a next part of the initial North route. Consequently, we do not show the part 'past the pole' as an upside-down extension of a map. Hence the map should never show tiles above row number 0 or below row nrows-1.
On the other hand, when we travel along circles of latitude, we smoothly cross the 0 and 180 meridians and switch between the eastern and western hemisphere. So if the camera view covers area on both sides of the left or right edge of the map, we need to continue filling the view with tiles from the other end of the tiles array. If we use a map scaled down, so that it is smaller than the viewport, we may even need to iterate that more than once!
The left edge of a camera view corresponds to the 'longitude' of xleft = xcam - wcam/2 and the right one to xrght = xcam + wcam/2. So we can step across the viewport by the tile's width to find out appropriate columns and show them:
x = xleft
repeat
show a column at x
x = x + wtile
until x >= xrght
The 'show a column at x' part requires finding appropriate column, then iterating across the column to show corresponding tiles. Let's find out which tiles fit the camera view:
ytop = ycam - hcam/2
ybot = ycam + hcam/2
y=ytop
repeat
show a tile at x,y
y = y + htile
until y >= ybot
To show the tile we need to locate appropriate tile and then send it to appropriate position in the camera view.
However, we treat column number differently from the row number: columns wrap while rows do not:
row = y/htile
if (0 <= row) and (row < nrows) then
col = (x/wtile) mod ncols
xtile = x - (x mod wtile)
ytile = y - (y mod htile)
display tile[row][col] at xtile,ytile
endif
Of course xtile and ytile are our map-scale longitude and latitude, so the 'display tile at' routine must transform them to the camera view coordinates by subtracting the camera position from them:
xinwiev = xtile - xcam
yinview = ytile - ycam
and then apply the resulting values relative to the camera view's center at the displaying device (screen).
Another level of complication will appear if you want to implement zooming in and out the view, that is dynamic scaling of the map, but I'm sure you'll find out yourself which calculations will need applying the zoom factor for correct results. :)

Weighted random coordinates

This may be more of a search for a term, but solutions are also welcome. I'm looking to create n amount of random x,y coordinates. The issue I am having is that I would like the coordinates to be "weighted" or have more of a chance of falling closer to a specific point. I've created something close by using this pseudo code:
x = rand(100) //random integer between 0 and 100
x = rand(x) //random number between 0 and the previous rand value
//randomize x to positive or negative
//repeat for y
This works to pull objects toward 0,0 - however if you create enough points, you can see a pattern of the x and y axis. This is because the even if x manages to get to 100, the chances are high that y will then be closer to.
I'm looking to avoid the formation of this x,y line. Bonus points if there is a way to throw in multiple "weighted coordinates" that the random coordinates would sort of gravitate to, instead of statically to 0,0.
This is easier in polar coordinates. All you have to do is to generate a uniform random angle and a power distributed distance. Here's an example in Python:
import math
from random import random
def randomPoint(aroundX, aroundY, scale, density):
angle = random()*2*math.pi
x = random()
if x == 0:
x = 0.0000001
distance = scale * (pow(x, -1.0/density) - 1)
return (aroundX + distance * math.sin(angle),
aroundY + distance * math.cos(angle))
Here's the distribution of randomPoint(0, 0, 1, 1):
We can shift it to center around another point like 1,2 with randomPoint(1, 2, 1, 1):
We can spread across a larger area by increasing the scale. Here's randomPoint(0, 0, 3, 1):
And we can change the shape, that is the tendency to flock together, by changing the density. randomPoint(0, 0, 1, 3):

Draw a circle with given curvature

I want to draw a circe with given curvature k.
I just need to know the y-coordinate for a given x-coordinate. So i.e. z = 1/k + sqrt(1/k^2 - x^2) is what I would normally use.
The problem is that my k is allowed to become zero. Which means that my circle becomes a line. For a mathematican thats no problem. But for my computer it is. For example when k is minimum double value, y will be infinity, for k == 0 I receive nan for y.
Are there any ways to get this done?
Given such border cases, I would just test the input parameters to see if one of them applies and use separate logic to just draw a horizontal or vertical line as appropriate if a border case applies.
That is a fairly common approach and computationally quite efficient.
When testing for border cases, test k to ensure that:
- k^2 will not overflow the data type in use
- k is not so small that 1/k^2 will underflow the data type in use
In either case, use the appropriate border case logic. Thanks #Godeke for pointing that out.
You gave the formula
y1 = 1/k + sqrt(1/k^2 - x^2) // (1)
which describes the upper half of the circle with radius 1/k and center (0, 1/k). Now for small k these values become very large and will eventually be outside of your drawing are.
The lower half of the circle is given by
y2 = 1/k - sqrt(1/k^2 - x^2) // (2)
For k approaching zero, these values "approach" the line y = 0. But for small values of k, (2) computes the difference of two large numbers. This causes a loss of precision and possible overflow.
But you can rewrite the formula (2) into the equivalent form
y2 = k * x^2 / (1 + sqrt(1 - k^2 * x^2)) // (2a)
Now you can compute the lower half of the circle for small values of k and even for k = 0 without any overflow or precision loss.
For the upper half you always have y1 >= 1/k. So if 1/k is larger than the boundary of your drawing area, you can ignore the upper value. Otherwise you can compute y1 via
y1 = 2/k - y2

Finding Whether A Point is Within a Right-Angle Triangle

I've always wondered the easiest way to figure out whether or not a point lies within a triangle, or in this instance, a rectangle cut into half diagonally.
Let's say I have a rectangle that is 64x64 pixels. With this rectangle, I want to return a TRUE value if a passed point is within the upper-left corner of the rectangle, and FALSE if it isn't.
-----
| /|
| / |
|<__|
Horray for bad ASCII art.
Anyway, the hypothetical points for this triangle that would return TRUE would be (0,0) and (63,0) and (0, 63). If a point lands on a line (e.g., 50,0) it would return TRUE as well.
Assuming 0,0 is in the upper-left corner and increases downwards...
I've had a possible solution in my head, but it seems more complicated than it should be - taking the passed Y value, determining where it would be in the rectangle, and figuring out manually where the line would cut at that Y value. E.g, a passed Y value of 16 would be quarter height of the rectangle. And thus, depending on what side you were checking (left or right), the line would either be at 16px or 48px, depending on the direction of the line. In the example above, since we're testing the upper-left corner, at 16px height, the line would be at 48px width
There has to be a better way.
EDIT:
The rectangle could also look like this as well
-----
|\ |
| \ |
|__>|
But I'm figuring in most cases the current answers already provided should still hold up...
Top-left/bottom-right triangles: For all points in the top-left triangle, x+y<=64. Points in the bottom-right triangle have x+y>64.
(for a rectangle of size (w,h) use w*y+h*x-w*h<0)
Top-right/bottom-left triangles: For all points in the bottom-left triangle, x<=y. Points in the top-right triangle have x>y.
(for a rectangle of size (w,h) use h*x-w*y<0)
How did we get there?
For a rectangle of dimensions (w,h) and TL/BR triangles, the equation of the diagonal is (try it out! assign x=0 and check that you get y==h, and assign y=0 and check that x==w)
h*x + w*y - w*h = 0
Points on one side of that line will have
h*x + w*y - w*h > 0
While points on the other will have
h*x + w*y - w*h < 0
Inserting 64 for both w and h, we get:
64x + 64y - 64*64 < 0
Dividing by 64 gets us:
x+y < 64
For TR/BL triangles, the line equation and the resulting inequalities are:
h*x - w*y = 0
h*x - w*y < 0
h*x - w*y > 0
Inserting 64 for w and h, we get
64x-64y < 0
=> x<y
you can represent the triangle with three affine functions
take the unit triangle with corners at (0, 0), (1, 0) and (1, 1). the sides are represented by the three lines
y = 0
x = 1
y = x
So the interior and boundry of the triangle are given as the intersection of the sets
x >= 1
y >= 0
y <= x
so given a point, (x, y), you just need to verify that it satisfies those three inequalities.
You can of course generalize this to any triangle using the fact that any affine function (representing a line) can be written in the form y = mx + b.
The equation for the line looks like this :
y = mx + b
So, if you insert your x and y-Values into that equation, it will probably not hold anymore. Let's reformulate it:
mx + b - y = 0
Same thing, different look. Again, the result is probably not zero. But, the result will now tell you whether it's on the one side of the line or the other.
Now you just have to find out whether the point is inside your rectangle.
Lets assume your right angled triangle has one corner at 0,0 and the diagonal corner at a,b.
So y=mx+c c=0 as we start at the origin.
m=b/a
So y=bx/a
To know which half of the triangle your point (c,d) falls in
if (d<=(bc/a)) {//point is in bottom half}
if (d>(bc/a)) {//point is in top half}
I think...
A simple option is to use a ray casting algorithm. Whilst perhaps a little overkill for what you need, it does have the advantage that it will work with more complex triangles and polygons.
Loosely, the algorithm takes an imaginary point in a direction (infinitely off to the left, for example) and casts a ray to your test point; you then calculate whether each line of your triangle crosses that infinitely long line. If you get an odd number of crossings, your point is inside your triangle; even and you're out of your triangle

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