I can at most read the data of a 3-way factorial design in R. But when the number of factors is more than 3, i can't read the data in R.
#2^3 design
trt=c("000","100","010","110","001","101","011","111")
m=array(trt,dim=c(2,2,2))
m
How can i read the data of 2^4 Factorial Design in R ?
Of course it depends how you will proceed with your data afterwards, but from my point of view a simple 2 dimensional data.frame should be very convenient to process afterwards plus you can generate the factorial designs rather easily:
expand.grid(a = 0:1, b = 0:1, c = 0:1, d = 0:1)
# a b c d
# 1 0 0 0 0
# 2 1 0 0 0
# 3 0 1 0 0
# 4 1 1 0 0
# 5 0 0 1 0
# 6 1 0 1 0
# 7 0 1 1 0
# 8 1 1 1 0
# 9 0 0 0 1
# 10 1 0 0 1
# 11 0 1 0 1
# 12 1 1 0 1
# 13 0 0 1 1
# 14 1 0 1 1
# 15 0 1 1 1
# 16 1 1 1 1
Related
I am trying to display some value of n consecutive numbers of a vector (in this example, vector x).
x = [1
1
1
1
1
1
0
0
0
0
1
1
0
0
0
0
0
0
1
0
1
0
1
0
1
0
1
1
0
0
1
1
1
1
1
1
0
0
1
1
1
1
0
0
1
1
1
1
1
1
0
0
0
0
0
1
0
0
1
0
0
0
0
1
0
0
1
0
1
0
0
0
1
0
0
0
0
1
0
1
0
0
1
0
0
0
0
1
1
0
0
0
1
0
0
0
0
0
0
1
0
1
0
0
0
0
1
0
1
0
0
0
1
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
1
1
1
1
0
0
1
1
0
0
1
0
1
0
0
0
0
0
0
1
0
1
0
0];
For example, I may want the first 3 values of 4 consecutive numbers which would be (3 values of 4 bits each):
1111
1100
0011
I may want the first 4 values of 2 consecutive numbers which would be (4 values of 2 bits each):
11
11
11
00
x is a double array. What would be an easy way to achieve this?
The simplest way is to reshape x to a matrix with your desired number of bits per value as the number of rows (MATLAB is column-major). For example, to get 4-digit values:
t = reshape(x,4,[]);
This will only work if the length of x evenly divides into 4. You could first crop x to the right number of elements to avoid errors where the length is not evenly divisible:
t = reshape(x(1:4*3),4,[]);
Now the transposed matrix t, converted to a string, looks like your desired output:
c = char(t.' + '0');
The output is a 3x4 char array:
'1111'
'1100'
'0011'
You can convert these binary representations back to numbers with bin2dec:
b = bin2dec(c);
The output is a 3-element double vector:
15
12
3
This is what I already implemented:
function B = bell(n)
B(1,1) = 1;
for i=2:n
B(i,1) = B(1,end);
for j = 1:i-1
B(i-j,j+1) = B(i-1,j)+B(i,j);
end
end
end
When n=3, i got:
1 2 3
1 3 0
2 0 0
instead of:
1 2 5
1 3 0
2 0 0
You need to change your line
B(i-j,j+1) = B(i-1,j)+B(i,j);
into
B(i-j,j+1) = B(i-j+1,j)+B(i-j,j);
Example: for n=5,
B =
1 2 5 15 52
1 3 10 37 0
2 7 27 0 0
5 20 0 0 0
15 0 0 0 0
in accordance with this.
I'm looking for a vectorized solution for this problem :
Let A a vector (great size : > 10000) of 0 and 1.
Ex :
A = [0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 1 etc]
I want to replace the 0 between the 1's (of odd ranks) by 2
i.e. to produce :
B = [0 0 0 1 2 2 2 2 2 1 0 0 0 1 2 2 1 0 0 1 2 1 etc]
Thanks for your help
It can be done quite easily with cumsum and mod:
A = [0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 1]
Short answer
A( mod(cumsum(A),2) & ~A ) = 2
A =
0 0 0 1 2 2 2 2 2 1 0 0 0 1 2 2 1 0 0 1 2 1
You requested to fill the islands of odd rank, but by changing mod(... to ~mod(... you can easily fill also the islands of even rank.
Explanation/Old answer:
mask1 = logical(A);
mask2 = logical(mod(cumsum(A),2))
out = zeros(size(A));
out(mask2) = 2
out(mask1) = 1
try using cumsum
cs = cumsum( A );
B = 2*( mod(cs,2)== 1 );
B(A==1) = 1;
I am working on a minigame called 'Pogo Painter', and I need some mathematical solutions. Below is an image (made with Paint) to illustrate a bit what it's all about.
Four players, each of different color, must claim squares to gain points. The minigame will be similar to this: http://www.youtube.com/watch?v=rKCQfAlaRrc, but slightly different. The players will be allowed to run around the playground and claim any of the squares, and points are gathered when a pattern is closed. For example, claiming blue square on A3 will create a closed blue pattern.
What kind of variables should I declare and how do I check if the pattern is closed?
Please answer if you have a solution :)
Here’s another (Discrete Optimization) way to model your problem.
Notation
View your grid as a ‘graph’ with n^2 nodes, and edges of length 1 (Edges connect two neighboring nodes.) Let the nodes be numbered 1:n^2. (For ease of notation, you can use a double array (x,y) to denote each node if you prefer.)
Decision Variables
There are k colors, one for each player (1 through 4). 0 is an unclaimed cell (white)
X_ik = 1 if player k has claimed node i. 0 otherwise.
To start out
X_i0 = 1 for all nodes i.
All nodes start out as white (0).
Neighboring sets: Two nodes i and j are ‘neighbors’ if they are adjacent to each other. (Any given node i can have at most 4 neighbors: Up down right and left.)
Edge variables:
We can now define a new set of edge variables Y_ijk that connect two adjacent nodes (i and j) with a common color k.
Y_ijk = 1 if neighboring nodes i and j are both of color k. 0 Otherwise.
(That is, X_ik = X_jk) for non-zero k.
We now have an undirected graph. Checking for ‘closed patterns’ is the same as detecting cycles.
Detecting Cycles:
A simple DFS search will do, since we have undirected cycles. Start with each colored node i, and check for cycles. If a path leads you back to a visited node, cycles exist. You can award points accordingly.
Finally, one suggestion as you design the game. You can reward points according to the “longest cycle” you detect. The shortest cycle gets 4 points, one point for each edge (or one point for each node in the cycle) whichever works best for you.
1 1
1 1 scores 4 points
1 1 1
1 1 1 scores 6 points
1 1 1
1 1 1
1 1 scores 8 points
Hope that helps.
Okay,
This is plenty of text, but it's simple.
An N-by-N square will satisfy as the game-board.
Each time a player claims a square,
If the square is not attached to any square of that player, then you must give that square a unique ID.
If the square is attached,
Count how many neighbours of each ID it has.
( See the demos I put below, to see what this means)
For each group
patterns_count += group_size - 1
If the number of groups is more than 1
Change the ID of that group as well as every other square connected to it so they all share the same ID
You must remember which IDs belong to which players.
This is what you have in your example
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 0 8 8 8
And this is what it would turn out like after blue grabs A-3
1 1 1 0 0 0 0 2 2
1 0 0 0 1 3 3 0 0
1 1 0 0 3 3 0 0 0
0 1 0 0 4 5 0 0 0
0 0 0 6 4 0 0 0 0
7 7 0 0 0 0 8 8 8
0 7 7 0 9 8 8 0 8
A A 7 0 9 8 0 0 8
A 0 7 0 0 8 8 8 8
More examples of the algorithm in use
1 1 1 0
1 0 1 0
1 1 0
0 0 0 0
2 neighbours. 2x'1'
1x closed pattern.
1 1 1 0
1 0 1 0
1 1 1 0
0 0 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 0 0
1 0 1 0 0
1 1 1 0 0
3 neighbours: 3x'1'
2x closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
1 0 1 0 0
1 1 1 0 0
--
1 1 1 0 0
1 0 1 0 0
1 1 2 2
0 0 2 0 2
0 0 2 2 2
4 neighbours: 2x'1', 2x'2'
2 Closed patterns
1 1 1 0 0
1 0 1 0 0
1 1 1 1 1
0 0 1 0 1
0 0 1 1 1
But I also consider these a closed pattern. You haven't given any description as to what should be considered one and what shouldn't be.
1 1 0
1 1 0
0 0 0
1 1 1
1 1 1
0 0 0
1 1 1
1 1 1
1 1
I have an array:
1 1 1 0 0
1 2 2 0 0
1 2 3 0 0
0 0 0 0 0
0 0 0 0 0
I want to make it
1 1 1 1 1
1 2 2 2 1
1 2 3 2 1
1 2 2 2 1
1 1 1 1 1
It is like rotating 1/4 piece of pie 270 degrees to fill out the remaining parts of the pie to make a full circle. Essentially mirroring the entire corner in all directions. I don't want to use any in built matlab features if possible - just some vector tricks if possible. Thanks.
EDIT:
This is embedded within an matrix of zeros of arbitrary size. I want it to work in both the above example and say this example:
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 1 1 1 0 0 0 0 0 0 0 0 0
0 0 1 2 2 0 0 0 0 0 0 0 0 0
0 0 1 2 3 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0
Ideally, I want to have a vector say [1,2,3.. N] which can be rotated circularly about the highest value in the array (N) centered about some point xc,yc in the grid. Or if this isn't possible, take an base array [1 1 1, 1 2 2, 1 2 3] and rotate it such that 3 is in the centre and you fill a circle as in the 2nd matrix above.
EDIT:
I found rot90(M,k) rotates matrix M k times but this produces:
Mrot = M + rot90(M,1) + rot90(M,2) + rot90(M,3)
Mrot =
1 1 2 1 1
1 2 4 2 1
2 4 12 4 2
1 2 4 2 1
1 1 2 1 1
This stacks it in the x,y directions which isn't correct.
Assuming the corner you want to replicate is symmetric about the diagonal (as in your example), then you can do this in one indexing step. Given a matrix M containing your sample 5-by-5 matrix, here's how to do it:
>> index = [1 2 3 2 1];
>> M = M(index, index)
M =
1 1 1 1 1
1 2 2 2 1
1 2 3 2 1
1 2 2 2 1
1 1 1 1 1