Context:
I have a hydraulic erosion algorithm that needs to receive an array of droplet starting positions. I also already have a pattern replicating algorithm, so I only need a good pattern to replicate.
The Requirements:
I need an algorism that produces a set of n^2 entries in a set of format (x,y) or [index] that describe cells in an nxn grid (where n = 2^i where i is any positive integer).
(as a set it means that every cell is mentioned in exactly one entry)
The pattern [created by the algorism ] should contain zero to none clustering of "visited" cells at any stage.
The cell (0,0) is as close to (n-1,n-1) as to (1,1), this relates to the definition of clustering
Note
I was/am trying to find solutions through fractal-like patterns built through recursion, but at the time of writing this, my solution is a lookup table of a checkerboard pattern(list of black cells + list of white cells) (which is bad, but yields fewer artifacts than an ordered list)
C, C++, C#, Java implementations (if any) are preferred
You can use a linear congruential generator to create an even distribution across your n×n space. For example, if you have a 64×64 grid, using a stride of 47 will create the pattern on the left below. (Run on jsbin) The cells are visited from light to dark.
That pattern does not cluster, but it is rather uniform. It uses a simple row-wide transformation where
k = (k + 47) mod (n * n)
x = k mod n
y = k div n
You can add a bit of randomness by making k the index of a space-filling curve such as the Hilbert curve. This will yield the pattern on the right. (Run on jsbin)
You can see the code in the jsbin links.
I have solved the problem myself and just sharing my solution:
here are my outputs for the i between 0 and 3:
power: 0
ordering:
0
matrix visit order:
0
power: 1
ordering:
0 3 2 1
matrix visit order:
0 3
2 1
power: 2
ordering:
0 10 8 2 5 15 13 7 4 14 12 6 1 11 9 3
matrix visit order:
0 12 3 15
8 4 11 7
2 14 1 13
10 6 9 5
power: 3
ordering:
0 36 32 4 18 54 50 22 16 52 48 20 2 38 34 6
9 45 41 13 27 63 59 31 25 61 57 29 11 47 43 15
8 44 40 12 26 62 58 30 24 60 56 28 10 46 42 14
1 37 33 5 19 55 51 23 17 53 49 21 3 39 35 7
matrix visit order:
0 48 12 60 3 51 15 63
32 16 44 28 35 19 47 31
8 56 4 52 11 59 7 55
40 24 36 20 43 27 39 23
2 50 14 62 1 49 13 61
34 18 46 30 33 17 45 29
10 58 6 54 9 57 5 53
42 26 38 22 41 25 37 21
the code:
public static int[] GetPattern(int power, int maxReturnSize = int.MaxValue)
{
int sideLength = 1 << power;
int cellsNumber = sideLength * sideLength;
int[] ret = new int[cellsNumber];
for ( int i = 0 ; i < cellsNumber && i < maxReturnSize ; i++ ) {
// this loop's body can be used for per-request computation
int x = 0;
int y = 0;
for ( int p = power - 1 ; p >= 0 ; p-- ) {
int temp = (i >> (p * 2)) % 4; //2 bits of the index starting from the begining
int a = temp % 2; // the first bit
int b = temp >> 1; // the second bit
x += a << power - 1 - p;
y += (a ^ b) << power - 1 - p;// ^ is XOR
// 00=>(0,0), 01 =>(1,1) 10 =>(0,1) 11 =>(1,0) scaled to 2^p where 0<=p
}
//to index
int index = y * sideLength + x;
ret[i] = index;
}
return ret;
}
I do admit that somewhere along the way the values got transposed, but it does not matter because of how it works.
After doing some optimization I came up with this loop body:
int x = 0;
int y = 0;
for ( int p = 0 ; p < power ; p++ ) {
int temp = ( i >> ( p * 2 ) ) & 3;
int a = temp & 1;
int b = temp >> 1;
x = ( x << 1 ) | a;
y = ( y << 1 ) | ( a ^ b );
}
int index = y * sideLength + x;
(the code assumes that c# optimizer, IL2CPP, and CPP compiler will optimize variables temp, a, b out)
I'm trying to implement the a block conjugate gradient algorithm that is not subject to breakdown from non invertible residual matrices; But I'm getting nonsensical results (in each iteration, the rank of Rcurrent should be getting smaller, not increasing). It is presented in the paper "A breakdown-free block conjugate gradient method" by Hao Ji and Yaohang Li.
Here is the algorithm:
This is my implementation in Julia:
function orth(M::Matrix)
matrixRank = rank(M)
Ufactor = svdfact(M)[:U]
return Ufactor[:,1:matrixRank]
end
function BFBCG(A::Matrix, Xcurrent::Matrix, M::Matrix, tol::Number, maxit::Number, Rcurrent::Matrix)
# initialization
#Rcurrent = B - A*Xcurrent;
Zcurrent = M*Rcurrent;
Pcurrent = orth(Zcurrent);
Xnext::Matrix = ones(size(Xcurrent))
# iterative method
for i = 0:maxit
Qcurrent = A*Pcurrent
acurrent = (Pcurrent' * Qcurrent)\(Pcurrent'*Rcurrent)
Xnext = Xcurrent+Pcurrent*acurrent
Rnext = Rcurrent-Qcurrent*acurrent
# if Residual norm of columns in Rcurrent < tol, stop
Znext = M*Rnext
bcurrent = -(Pcurrent' * Qcurrent)\ (Qcurrent'*Znext)
Pnext = orth(Znext+Pcurrent*bcurrent)
Xcurrent = Xnext
Zcurrent = Znext
Rcurrent = Rnext
Pcurrent = Pnext
#printf("\nRANK:\t%d",rank(Rcurrent))
#printf("\nNORM column1:\t%1.8f",vecnorm(Rcurrent[:,1]))
#printf("\nNORM column2:\t%1.8f\n=============",vecnorm(Rcurrent[:,2]))
end
return Xnext
end
The results of the paper for those inputs:
A = [15 5 4 3 2 1; 5 35 9 8 7 6; 4 9 46 12 11 10; 3 8 12 50 14 13; 2 7 11 14 19 15; 1 6 10 13 15 45]
M = eye(6)
guess = rand(6,2)
R0 = [1 0.537266261211281;2 0.043775211060964;3 0.964458562037146;4 0.622317517840541;5 0.552735938776748;6 0.023323943544997]
X = BFBCG(A,guess,M,tol,9,R0)
are a rank that reaches zero in the third iteration.
The algorithm works, and the rank goes to zero in the third iteration. The problem is numerical inaccuracies which would leave any matrix fully ranked. To get a better result, use rank(Rcurrent, tol) instead of rank(Rcurrent) which is a version which takes tolerance into account. After which, at least on my machine, the rank drops to zero.
julia> X = BFBCG(A,guess,M,tol,9,R0)
RANK: 2
NORM column1: 1.78951939
NORM column2: 0.41155080
=============
RANK: 2
NORM column1: 0.97949620
NORM column2: 0.16170799
=============
RANK: 0
NORM column1: 0.00000000
NORM column2: 0.00000000
=============
RANK: 0
NORM column1: 0.00000000
NORM column2: 0.00000000
=============
I'm working on an implementation of the JPEG compression algorithm in MATLAB. I've run into some issues when computing the discrete cosine transform(DCT) of the 8x8 image blocks(T = H * F * H_transposed, H is the matrix containing the DCT coefficients of an 8x8 matrix, generated with dctmtx(8) and F is an 8x8 image block). The code is bellow:
jpegCompress.m
function y = jpegCompress(x, quality)
% y = jpegCompress(x, quality) compresses an image X based on 8 x 8 DCT
% transforms, coefficient quantization and Huffman symbol coding. Input
% quality determines the amount of information that is lost and compression achieved. y is the encoding structure containing fields:
% y.size size of x
% y.numblocks number of 8 x 8 encoded blocks
% y.quality quality factor as percent
% y.huffman Huffman coding structure
narginchk(1, 2); % check number of input arguments
if ~ismatrix(x) || ~isreal(x) || ~ isnumeric(x) || ~ isa(x, 'uint8')
error('The input must be a uint8 image.');
end
if nargin < 2
quality = 1; % default value for quality
end
if quality <= 0
error('Input parameter QUALITY must be greater than zero.');
end
m = [16 11 10 16 24 40 51 61 % default JPEG normalizing array
12 12 14 19 26 58 60 55 % and zig-zag reordering pattern
14 13 16 24 40 57 69 56
14 17 22 29 51 87 80 62
18 22 37 56 68 109 103 77
24 35 55 64 81 104 113 92
49 64 78 87 103 121 120 101
72 92 95 98 112 100 103 99] * quality;
order = [1 9 2 3 10 17 25 18 11 4 5 12 19 26 33 ...
41 34 27 20 13 6 7 14 21 28 35 42 49 57 50 ...
43 36 29 22 15 8 16 23 30 37 44 51 58 59 52 ...
45 38 31 24 32 39 46 53 60 61 54 47 40 48 55 ...
62 63 56 64];
[xm, xn] = size(x); % retrieve size of input image
x = double(x) - 128; % level shift input
t = dctmtx(8); % compute 8 x 8 DCT matrix
% Compute DCTs pf 8 x 8 blocks and quantize coefficients
y = blkproc(x, [8 8], 'P1 * x * P2', t, t');
y = blkproc(y, [8 8], 'round(x ./ P1)', m); % <== nearly all elements from y are zero after this step
y = im2col(y, [8 8], 'distinct'); % break 8 x 8 blocks into columns
xb = size(y, 2); % get number of blocks
y = y(order, :); % reorder column elements
eob = max(x(:)) + 1; % create end-of-block symbol
r = zeros(numel(y) + size(y, 2), 1);
count = 0;
for j = 1:xb % process one block(one column) at a time
i = find(y(:, j), 1, 'last'); % find last non-zero element
if isempty(i) % check if there are no non-zero values
i = 0;
end
p = count + 1;
q = p + i;
r(p:q) = [y(1:i, j); eob]; % truncate trailing zeros, add eob
count = count + i + 1; % and add to output vector
end
r((count + 1):end) = []; % delete unused portion of r
y = struct;
y.size = uint16([xm xn]);
y.numblocks = uint16(xb);
y.quality = uint16(quality * 100);
y.huffman = mat2huff(r);
mat2huff is implemented as:
mat2huff.m
function y = mat2huff(x)
%MAT2HUFF Huffman encodes a matrix.
% Y = mat2huff(X) Huffman encodes matrix X using symbol
% probabilities in unit-width histogram bins between X's minimum
% and maximum value s. The encoded data is returned as a structure
% Y :
% Y.code the Huffman - encoded values of X, stored in
% a uint16 vector. The other fields of Y contain
% additional decoding information , including :
% Y.min the minimum value of X plus 32768
% Y.size the size of X
% Y.hist the histogram of X
%
% If X is logical, uintB, uint16 ,uint32 ,intB ,int16, or double,
% with integer values, it can be input directly to MAT2HUF F. The
% minimum value of X must be representable as an int16.
%
% If X is double with non - integer values --- for example, an image
% with values between O and 1 --- first scale X to an appropriate
% integer range before the call.For example, use Y
% MAT2HUFF (255 * X) for 256 gray level encoding.
%
% NOTE : The number of Huffman code words is round(max(X(:)))
% round (min(X(:)))+1. You may need to scale input X to generate
% codes of reasonable length. The maximum row or column dimension
% of X is 65535.
if ~ismatrix(x) || ~isreal(x) || (~isnumeric(x) && ~islogical(x))
error('X must be a 2-D real numeric or logical matrix.');
end
% Store the size of input x.
y.size = uint32(size(x));
% Find the range of x values
% by +32768 as a uint16.
x = round(double(x));
xmin = min(x(:));
xmax = max(x(:));
pmin = double(int16(xmin));
pmin = uint16(pmin+32768);
y.min = pmin;
% Compute the input histogram between xmin and xmax with unit
% width bins , scale to uint16 , and store.
x = x(:)';
h = histc(x, xmin:xmax);
if max(h) > 65535
h = 65535 * h / max(h);
end
h = uint16(h);
y.hist = h;
% Code the input mat rix and store t h e r e s u lt .
map = huffman(double(h)); % Make Huffman code map
hx = map(x(:) - xmin + 1); % Map image
hx = char(hx)'; % Convert to char array
hx = hx(:)';
hx(hx == ' ') = [ ]; % Remove blanks
ysize = ceil(length(hx) / 16); % Compute encoded size
hx16 = repmat('0', 1, ysize * 16); % Pre-allocate modulo-16 vector
hx16(1:length(hx)) = hx; % Make hx modulo-16 in length
hx16 = reshape(hx16, 16, ysize); % Reshape to 16-character words
hx16 = hx16' - '0'; % Convert binary string to decimal
twos = pow2(15 : - 1 : 0);
y.code = uint16(sum(hx16 .* twos(ones(ysize ,1), :), 2))';
Why is the block processing step generating mostly null values?
It is likely that multiplying the Quantization values you have by four is causing the DCT coefficients to go to zero.
I'm interested in finding the nth row of pascal triangle (not a specific element but the whole row itself). What would be the most efficient way to do it?
I thought about the conventional way to construct the triangle by summing up the corresponding elements in the row above which would take:
1 + 2 + .. + n = O(n^2)
Another way could be using the combination formula of a specific element:
c(n, k) = n! / (k!(n-k)!)
for each element in the row which I guess would take more time the the former method depending on the way to calculate the combination. Any ideas?
>>> def pascal(n):
... line = [1]
... for k in range(n):
... line.append(line[k] * (n-k) / (k+1))
... return line
...
>>> pascal(9)
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
This uses the following identity:
C(n,k+1) = C(n,k) * (n-k) / (k+1)
So you can start with C(n,0) = 1 and then calculate the rest of the line using this identity, each time multiplying the previous element by (n-k) / (k+1).
A single row can be calculated as follows:
First compute 1. -> N choose 0
Then N/1 -> N choose 1
Then N*(N-1)/1*2 -> N choose 2
Then N*(N-1)*(N-2)/1*2*3 -> N choose 3
.....
Notice that you can compute the next value from the previous value, by just multipyling by a single number and then dividing by another number.
This can be done in a single loop. Sample python.
def comb_row(n):
r = 0
num = n
cur = 1
yield cur
while r <= n:
r += 1
cur = (cur* num)/r
yield cur
num -= 1
The most efficient approach would be:
std::vector<int> pascal_row(int n){
std::vector<int> row(n+1);
row[0] = 1; //First element is always 1
for(int i=1; i<n/2+1; i++){ //Progress up, until reaching the middle value
row[i] = row[i-1] * (n-i+1)/i;
}
for(int i=n/2+1; i<=n; i++){ //Copy the inverse of the first part
row[i] = row[n-i];
}
return row;
}
here is a fast example implemented in go-lang that calculates from the outer edges of a row and works it's way to the middle assigning two values with a single calculation...
package main
import "fmt"
func calcRow(n int) []int {
// row always has n + 1 elements
row := make( []int, n + 1, n + 1 )
// set the edges
row[0], row[n] = 1, 1
// calculate values for the next n-1 columns
for i := 0; i < int(n / 2) ; i++ {
x := row[ i ] * (n - i) / (i + 1)
row[ i + 1 ], row[ n - 1 - i ] = x, x
}
return row
}
func main() {
for n := 0; n < 20; n++ {
fmt.Printf("n = %d, row = %v\n", n, calcRow( n ))
}
}
the output for 20 iterations takes about 1/4 millisecond to run...
n = 0, row = [1]
n = 1, row = [1 1]
n = 2, row = [1 2 1]
n = 3, row = [1 3 3 1]
n = 4, row = [1 4 6 4 1]
n = 5, row = [1 5 10 10 5 1]
n = 6, row = [1 6 15 20 15 6 1]
n = 7, row = [1 7 21 35 35 21 7 1]
n = 8, row = [1 8 28 56 70 56 28 8 1]
n = 9, row = [1 9 36 84 126 126 84 36 9 1]
n = 10, row = [1 10 45 120 210 252 210 120 45 10 1]
n = 11, row = [1 11 55 165 330 462 462 330 165 55 11 1]
n = 12, row = [1 12 66 220 495 792 924 792 495 220 66 12 1]
n = 13, row = [1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1]
n = 14, row = [1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1]
n = 15, row = [1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1]
n = 16, row = [1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1]
n = 17, row = [1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1]
n = 18, row = [1 18 153 816 3060 8568 18564 31824 43758 48620 43758 31824 18564 8568 3060 816 153 18 1]
n = 19, row = [1 19 171 969 3876 11628 27132 50388 75582 92378 92378 75582 50388 27132 11628 3876 969 171 19 1]
An easy way to calculate it is by noticing that the element of the next row can be calculated as a sum of two consecutive elements in the previous row.
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
For example 6 = 5 + 1, 15 = 5 + 10, 1 = 1 + 0 and 20 = 10 + 10. This gives a simple algorithm to calculate the next row from the previous one.
def pascal(n):
row = [1]
for x in xrange(n):
row = [l + r for l, r in zip(row + [0], [0] + row)]
# print row
return row
print pascal(10)
In Scala Programming: i would have done it as simple as this:
def pascal(c: Int, r: Int): Int = c match {
case 0 => 1
case `c` if c >= r => 1
case _ => pascal(c-1, r-1)+pascal(c, r-1)
}
I would call it inside this:
for (row <- 0 to 10) {
for (col <- 0 to row)
print(pascal(col, row) + " ")
println()
}
resulting to:
.
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
To explain step by step:
Step 1: We make sure that if our column is the first one we always return figure 1.
Step 2: Since each X-th row there are X number of columns. So we say that; the last column X is greater than or equal to X-th row, then the return figure 1.
Step 3: Otherwise, we get the sum of the repeated pascal of the column just before the current one and the row just before the current one ; and the pascal of that column and the row just before the current one.
Good Luck.
Let me build upon Shane's excellent work for an R solution. (Thank you, Shane!. His code for generating the triangle:
pascalTriangle <- function(h) {
lapply(0:h, function(i) choose(i, 0:i))
}
This will allow one to store the triangle as a list. We can then index whatever row desired. But please add 1 when indexing! For example, I'll grab the bottom row:
pt_with_24_rows <- pascalTriangle(24)
row_24 <- pt_with_24_rows[25] # add one
row_24[[1]] # prints the row
So, finally, make-believe I have a Galton Board problem. I have the arbitrary challenge of finding out percentage of beans have clustered in the center: say, bins 10 to 15 (out of 25).
sum(row_24[[1]][10:15])/sum(row_24[[1]])
Which turns out to be 0.7704771. All good!
In Ruby, the following code will print out the specific row of Pascals Triangle that you want:
def row(n)
pascal = [1]
if n < 1
p pascal
return pascal
else
n.times do |num|
nextNum = ((n - num)/(num.to_f + 1)) * pascal[num]
pascal << nextNum.to_i
end
end
p pascal
end
Where calling row(0) returns [1] and row(5) returns [1, 5, 10, 10, 5, 1]
Here is the another best and simple way to design a Pascal Triangle dynamically using VBA.
`1
11
121
1331
14641`
`Sub pascal()
Dim book As Excel.Workbook
Dim sht As Worksheet
Set book = ThisWorkbook
Set sht = book.Worksheets("sheet1")
a = InputBox("Enter the Number", "Fill")
For i = 1 To a
For k = 1 To i
If i >= 2 And k >= 2 Then
sht.Cells(i, k).Value = sht.Cells(i - 1, k - 1) + sht.Cell(i- 1, k)
Else
sht.Cells(i, k).Value = 1
End If
Next k
Next i
End Sub`
I used Ti-84 Plus CE
The use of –> in line 6 is the store value button
Forloop syntax is
:For(variable, beginning, end [, increment])
:Commands
:End
nCr syntax is
:valueA nCr valueB
List indexes start at 1 so that's why i set it to R+1
N= row
R= column
PROGRAM: PASCAL
:ClrHome
:ClrList L1
:Disp "ROW
:Input N
:For(R,0,N,1)
:N nCr R–>L1(R+1)
:End
:Disp L1
This is the fastest way I can think of to do this in programming (with a ti 84) but if you mean to be able to calculate the row using pen and paper then just draw out the triangle cause doing factorals are a pain!
Here's an O(n) space-complexity solution in Python:
def generate_pascal_nth_row(n):
result=[1]*n
for i in range(n):
previous_res = result.copy()
for j in range(1,i):
result[j] = previous_res[j-1] + previous_res[j]
return result
print(generate_pascal_nth_row(6))
class Solution{
public:
int comb(int n,int r){
long long c=1;
for(int i=1;i<=r;i++) { //calculates n!/(n-r)!
c=((c*n))/i; n--;
}
return c;
}
vector<int> getRow(int n) {
vector<int> v;
for (int i = 0; i < n; ++i)
v.push_back(comb(n,i));
return v;
}
};
faster than 100% submissions on leet code https://leetcode.com/submissions/detail/406399031/
The most efficient way to calculate a row in pascal's triangle is through convolution. First we chose the second row (1,1) to be a kernel and then in order to get the next row we only need to convolve curent row with the kernel.
So convolution of the kernel with second row gives third row [1 1]*[1 1] = [1 2 1], convolution with the third row gives fourth [1 2 1]*[1 1] = [1 3 3 1] and so on
This is a function in julia-lang (very simular to matlab):
function binomRow(n::Int64)
baseVector = [1] #the first row is equal to 1.
kernel = [1,1] #This is the second row and a kernel.
row = zeros(n)
for i = 1 : n
row = baseVector
baseVector = conv(baseVector, kernel) #convoltion with kernel
end
return row::Array{Int64,1}
end
To find nth row -
int res[] = new int[n+1];
res[0] = 1;
for(int i = 1; i <= n; i++)
for(int j = i; j > 0; j++)
res[j] += res[j-1];
I want to randomly generate points. Well at least there should be a limitation on the y-axis. Later I connect the points to a line which should proceed in a simple animation. You can imagine this as a random walk of a drunken person, going uphill and downhill.
This sounds very simple. I searched around the web and found that this could be accomplished using the markov chain. I think this idea is really interesting.
You can create the first state of your scene by yourself and pass this state as input to the markov chain algorithm. The algorithm randomly changes this state and creates a walk.
However I cannot find any example of that algorithm and no source code. I just found an applet that demonstrates the markov chain algorithm: http://www.probability.ca/jeff/java/unif.html
Please suggest some code. Any other ideas how to accomplish this are appreciated too.
I painted an example
So I want the line to proceed in a similar way. There are valleys, slopes ... they are random but the randomness still apply to the initial state of the line. This is why I found makrov chain so interesting here: http://www.suite101.com/content/implementing-markov-chains-a24146
Here's some code in Lua:
absstepmax = 25
ymin = -100
ymax = 100
x = 0
y = 5
for i = 1, 20 do
y = y + (math.random(2*absstepmax) - absstepmax - 1)
y = math.max(ymin, math.min(ymax, y))
x = x + 5
print (x,y)
end
absstepmax limits the size of a y step per iteration
ymin and ymax limit the extent of y
There is no bias in the example, i.e., y can change symmetrically up or down. If you want your "drunk" tending more "downhill" you can change the offset after the call to random from absstepmax - 1 to absstepmax - 5 or whatever bias you like.
In this example, the x step is fixed. You may make this random as well using the same mechanisms.
Here are some sample runs:
> absstepmax = 25
> ymin = -100
> ymax = 100
> x = 0
> y = 5
> for i = 1, 20 do
>> y = y + (math.random(2*absstepmax) - absstepmax - 1)
>> y = math.max(ymin, math.min(ymax, y))
>> x = x + 5
>> print (x,y)
>> end
5 4
10 22
15 37
20 39
25 50
30 40
35 21
40 22
45 12
50 16
55 16
60 12
65 -1
70 -8
75 -14
80 -17
85 -19
90 -25
95 -37
100 -59
> absstepmax = 25
> ymin = -100
> ymax = 100
> x = 0
> y = 5
> for i = 1, 20 do
>> y = y + (math.random(2*absstepmax) - absstepmax - 1)
>> y = math.max(ymin, math.min(ymax, y))
>> x = x + 5
>> print (x,y)
>> end
5 -2
10 -15
15 -7
20 1
25 1
30 12
35 23
40 45
45 43
50 65
55 56
60 54
65 54
70 62
75 57
80 62
85 86
90 68
95 76
100 68
>
Painted result added from OP: