In my programming course, we were given to add two matrices together with recursion only. Apparently our professor intended us to make a recursive method but still use one "for" loop to solve it.
However, I'm still convinced that everything that can be done with a for loop can be done with recursion. So I was trying to do it:
The method should look like this:
public static int[][] addMatrix(int[][] matrix1, int[][] matrix2)
no additional variables may be passed with.
I'm having a real hard time trying to solve this, since this is a very easy exercise using for loop - I thought it was pretty doable with recursion.
Any help ?
--UPDATE
So far my course of thought was like this:
To always take the first matrix with its full size, and step by step breaking the second matrix by taking matrix2[matrix2.length] and calling the method on it, that way I would know where to store the values calculated by the method without using an index variable.
Illustration:
X X Y Y
X X Y Y
X X Y Y
X are the variables in martix1, Y for matrix2
X X
X X
X X Y Y
Take the last "row":
X X
X X
X X Y
And if matrix2 is only 1x1, add it to the appropriate index in matrix1.
This is the best that I could have come up with.
With "matrix2[matrix2.length]" you'd be going out of range by the way since indexes start from 0.
Here is some pseudo code in no particular language that will give you the idea:
function addMatrices(mat1, mat2, result_mat /*pass by reference*/, i=0, j=0){
if (j >= mat1[0].size)
return result_mat;
result_mat[i][j] = mat1[i][j] + mat2[i][j];
i ++;
if (i == mat1.size){
i = 0;
j ++;
}
return addMatrices(mat1, mat2, result_mat, i, j);
}
Related
I am using the example of calculating the length of the arc around a circle and the area under the arc around a circle based on the radius of the circle (r) and the angle of the the arc(theta). The area and the length are both based on r and theta, and you can calculate them simultaneously in python.
In python, I can assign two values at the same time by doing this.
from math import pi
def circle_set(r, theta):
return theta * r, .5*theta*r*r
arc_len, arc_area = circle_set(1, .5*pi)
Implementing the same structure in R gives me this.
circle_set <- function(r, theta){
return(theta * r, .5 * theta * r *r)
}
arc_len, arc_area <- circle_set(1, .5*3.14)
But returns this error.
arc_len, arc_area <- circle_set(1, .5*3.14)
Error: unexpected ',' in "arc_len,"
Is there a way to use the same structure in R?
No, you can't do that in R (at least, not in base or any packages I'm aware of).
The closest you could come would be to assign objects to different elements of a list. If you really wanted, you could then use list2env to put the list elements in an environment (e.g., the global environment), or use attach to make the list elements accessible, but I don't think you gain much from these approaches.
If you want a function to return more than one value, just put them in a list. See also r - Function returning more than one value.
You can assign multiple variables the same value as below. Even here, I think the code is unusual and less clear, I think this outweighs any benefits of brevity. (Though I suppose it makes it crystal clear that all of the variables are the same value... perhaps in the right context it makes sense.)
x <- y <- z <- 1
# the above is equivalent to
x <- 1
y <- 1
z <- 1
As Gregor said, there's no way to do it exactly as you said and his method is a good one, but you could also have a vector represent your two values like so:
# Function that adds one value and returns a vector of all the arguments.
plusOne <- function(vec) {
vec <- vec + 1
return(vec)
}
# Creating variables and applying the function.
x <- 1
y <- 2
z <- 3
vec <- c(x, y, z)
vec <- plusOne(vec)
So essentially you could make a vector and have your function return vectors, which is essentially filling 3 values at once. Again, not what you want exactly, just a suggestion.
We have to create function(K) that returns vector which has all items smaller than or equal to K from fibonacci sequence. We can assume K is fibonacci item. For example if K is 3 the function would return vector (1,1,2,3).
In general, a for loop is used when you know how many iterations you need to do, and a while loop is used when you want to keep going until a condition is met.
For this case, it sounds like you get an input K and you want to keep going until you find a Fibonacci term > K, so use a while loop.
ans <- function(n) {
x <- c(1,1)
while (length(x) <= n) {
position <- length(x)
new <- x[position] + x[position-1]
x <- c(x,new)
}
return(x[x<=n])
}
`
Tried many different loops, and this is closest I get. It works with every other number but ans(3) gives 1,1,2 even though it should give 1,1,2,3. Couldn't see what is wrong with this.
I've tried a couple ways of doing this problem but am having trouble with how to write it. I think I did the first three steps correctly, but now I have to fill the vector z with numbers from y that are divisible by four, not divisible by three, and have an odd number of digits. I know that I'm using the print function in the wrong way, I'm just at a loss on what else to use ...
This is different from that other question because I'm not using a while loop.
#Step 1: Generate 1,000,000 random, uniformly distributed numbers between 0
#and 1,000,000,000, and name as a vector x. With a seed of 1.
set.seed(1)
x=runif(1000000, min=0, max=1000000000)
#Step 2: Generate a rounded version of x with the name y
y=round(x,digits=0)
#Step 3: Empty vector named z
z=vector("numeric",length=0)
#Step 4: Create for loop that populates z vector with the numbers from y that are divisible by
#4, not divisible by 3, with an odd number of digits.
for(i in y) {
if(i%%4==0 && i%%3!=0 && nchar(i,type="chars",allowNA=FALSE,keepNA=NA)%%2!=0){
print(z,i)
}
}
NOTE: As per #BenBolker's comment, a loop is an inefficient way to solve your problem here. Generally, in R, try to avoid loops where possible to maximise the efficiency of your code. #SymbolixAU has provided an example of doing so here in the comments. Having said that, in aid of helping you learn the ins-and-outs of loops and vectors, here's a solution which only requires a change to one line of your code:
You've got the vector created before the loop, that's a good start. Now, inside your loop, you need to populate that vector. To do so, you've currently got print(z,i), which won't really do too much. What you need to to change the vector itself:
z <- c( z, i )
Should work for you (just replace that print line in your loop).
What's happening here is that we're taking the existing z vector, binding i to the end of it, and making that new vector z again. So every time a value is added, the vector gets a little longer, such that you'll end up with a complete vector.
where you have print put this instead:
z <- append(z, i)
I am normally a maple user currently working with R, and I have a problem with correctly indexing variables.
Say I want to define 2 vectors, v1 and v2, and I want to call the nth element in v1. In maple this is easily done:
v[1]:=some vector,
and the nth element is then called by the command
v[1][n].
How can this be done in R? The actual problem is as follows:
I have a sequence M (say of length 10, indexed by k) of simulated negbin variables. For each of these simulated variables I want to construct a vector X of length M[k] with entries given by some formula. So I should end up with 10 different vectors, each of different length. My incorrect code looks like this
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
for(k in 1:sims){
x[k]<-rep(NA,M[k])
X[k]<-rep(NA,M[k])
for(i in 1:M[k]){x[k][i]<-runif(1,min=0,max=1)
if(x[k][i]>=0 & x[i]<=0.1056379){
X[k][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[k][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
}
The error appears to be that x[k] is not a valid name for a variable. Any way to make this work?
Thanks a lot :)
I've edited your R script slightly to get it working and make it reproducible. To do this I had to assume that eks_2016_kasko was an integer value of 10.
require(MASS)
sims<-10
# Because you R is not zero indexed add one
M<-rnegbin(sims, 10*exp(-2.17173), 840.1746) + 1
# Create a list
x <- list()
X <- list()
for(k in 1:sims){
x[[k]]<-rep(NA,M[k])
X[[k]]<-rep(NA,M[k])
for(i in 1:M[k]){
x[[k]][i]<-runif(1,min=0,max=1)
if(x[[k]][i]>=0 & x[[k]][i]<=0.1056379){
X[[k]][i]<-rlnorm(1, 6.228244, 0.3565041)}
else{
X[[k]][i]<-rlnorm(1, 8.910837, 1.1890874)
}
}
This will work and I think is what you were trying to do, BUT is not great R code. I strongly recommend using the lapply family instead of for loops, learning to use data.table and parallelisation if you need to get things to scale. Additionally if you want to read more about indexing in R and subsetting Hadley Wickham has a comprehensive break down here.
Hope this helps!
Let me start with a few remarks and then show you, how your problem can be solved using R.
In R, there is most of the time no need to use a for loop in order to assign several values to a vector. So, for example, to fill a vector of length 100 with uniformly distributed random variables, you do something like:
set.seed(1234)
x1 <- rep(NA, 100)
for (i in 1:100) {
x1[i] <- runif(1, 0, 1)
}
(set.seed() is used to set the random seed, such that you get the same result each time.) It is much simpler (and also much faster) to do this instead:
x2 <- runif(100, 0, 1)
identical(x1, x2)
## [1] TRUE
As you see, results are identical.
The reason that x[k]<-rep(NA,M[k]) does not work is that indeed x[k] is not a valid variable name in R. [ is used for indexing, so x[k] extracts the element k from a vector x. Since you try to assign a vector of length larger than 1 to a single element, you get an error. What you probably want to use is a list, as you will see in the example below.
So here comes the code that I would use instead of what you proposed in your post. Note that I am not sure that I correctly understood what you intend to do, so I will also describe below what the code does. Let me know if this fits your intentions.
# define M
library(MASS)
eks_2016_kasko <- 486689.1
sims<-10
M<-rnegbin(sims, eks_2016_kasko*exp(-2.17173), 840.1746)
# define the function that calculates X for a single value from M
calculate_X <- function(m) {
x <- runif(m, min=0,max=1)
X <- ifelse(x > 0.1056379, rlnorm(m, 6.228244, 0.3565041),
rlnorm(m, 8.910837, 1.1890874))
}
# apply that function to each element of M
X <- lapply(M, calculate_X)
As you can see, there are no loops in that solution. I'll start to explain at the end:
lapply is used to apply a function (calculate_X) to each element of a list or vector (here it is the vector M). It returns a list. So, you can get, e.g. the third of the vectors with X[[3]] (note that [[ is used to extract elements from a list). And the contents of X[[3]] will be the result of calculate_X(M[3]).
The function calculate_X() does the following: It creates a vector of m uniformly distributed random values (remember that m runs over the elements of M) and stores that in x. Then it creates a vector X that contains log normally distributed random variables. The parameters of the distribution depend on the value x.
Here's a super newbie question. I'm working through The Art of R Programming and came across this simple code:
z <- NULL
for (i in 1:10) if (i %% 2 == 0) z <- c(z,i)
z
the output is c(2,4,6,8,10).
What I don't understand is why you need c(z,i) to populate the vector z. Why wouldn't c(i) work. When I do that, it skips straight to the last iteration and fills vector z with the last associated value. Is there some form of rule that governs populating vectors? Try as I might, I didn't find a clear answer anywhere, just a bunch of samples that, while they made sense, didn't help me solve the above.