I'm trying to get the sum of the first 120 terms of a series:
a(n) = (10+a(n-1))(1.005)^n
I've tried two approaches, but I don't know which mistakes I'm making.
Approach #1:
nval = 1
mval = zeros(120,1)
for nval=1:120
if nval <= 120 %counting up from 1 to 120
if nval == 1 %modifying first term to avoid 0 index
mval(1) = (10)*(1.005)
nval = nval +1;
else
mval(nval) = (10+mval(nval-1))*(1.005)^nval; %Assigning
nval = nval +1;
end
end
end
which results in a column vector with 10.05 at the top and zeroes for the rest. If if omit the opening definition of mval as zeroes, I get the same result as
Approach #2:
a(1)=10.05; %defining first term to avoid 0 index
for n=2:120, a(n)= (10+a(n-1))*(1.005)^n; end
which results in something far too large.
(The correct value for a(120) is ~1646.99)
Edit: my mistake and apologies - my series was wrong; the (1.005)^n is NOT to the nth power, but merely 1.005. Using this, both methods work.
Thanks to one and all for answers and suggestions
Related
I've written out a potential solution to a Leetcode problem but I get this error involving maximum recursion depth. I'm really unsure what I'm doing wrong. Here's what I've tried writing:
def orangesRotting(grid):
R,C = len(grid), len(grid[0])
seen = set()
min_time = 0
def fresh_search(r,c, time):
if ((r,c,time) in seen or r < 0 or c < 0 or r >= R or c >= C or grid[r][c] == 0):
return
elif grid[r][c] == 2:
seen.add((r,c,0))
elif grid[r][c] == 1:
seen.add((r,c, time + 1))
fresh_search(r+1,c,time+1)
fresh_search(r-1,c,time+1)
fresh_search(r,c+1,time+1)
fresh_search(r,c-1,time+1)
for i in range(R):
for j in range(C):
if grid[i][j] == 2:
fresh_search(i,j,0)
for _,_,t in list(seen):
min_time = max(min_time,t)
return min_time
Even on a simple input like grid = [[2,1,1], [1,1,0], [0,1,1]]. The offending line always appears to be at the if statement
if ((r,c,time) in seen or r < 0 or c < 0 or r >= R or c >= C or grid[r][c] == 0):
Please note, I'm not looking for help in solving the problem, just understanding why I'm running into this massive recursion issue. For reference, here is the link to the problem. Any help would be appreciated.
So let's trace through what you are doing here. You iterate through the entire grid and if the value for that cell is 2 you call fresh_search for that cell. We'll start with [0,0]
In fresh_search you then add the cell with times seen = 0 to your set.
Now for all neighboring cells you call fresh_search so we'll just look at r+1. For r+1 your method fresh_search adds the cell to your set with times seen = 1 and then calls fresh_search again with all neighboring cells.
Next we'll just look at r-1 which is our origin and now fresh_search is being called with this cell and times seen = 2. Now this value isn't in the set yet because (0,0,0) != (0,0,2) so it adds it to the set and again calls fresh_search with the r+1 cell but now times seen = 3
and so on and so forth until max recursion.
I am trying to find the duplicate value which occours the first in a list or array. I created the below code which worked in Python and trying to replicate it in Julia, but not getting the desired result. Can you please help?
def firstDuplicateValue(array):
# Write your code here.
index = float('inf')
for x, i in enumerate(array):
for j, k in enumerate(array):
if i == k and x != j and j > x:
if j < index:
index = j
if index == float('inf'):
return -1
else:
return array[index]
I have created the same code in Julia as below but I am getting the desired result -
function firstDuplicateValue(array)
index = Inf
for (ind_1, value_1) in enumerate(array)
for (ind_2, value_2) in enumerate(array)
if value_1 == value_2 && ind_2 > ind_1
if ind_2 < index
index = ind_2
end
end
end
end
if index == Inf
return -1
else return array[index]
end
end
I know there are other optimum ways of doing it, but this is just to learn basics of Julia.
added from an answer by the OP: It worked for me, I was providing the wrong input to the function, it was a typo.
It worked for me, I was providing the wrong input to the function, it was a typo.
I have an array 'A'. I want to do a loop over all the elements of A, checking to see if any are greater than or equal to 1. If they are, I would like to assign a '1' to a new array 'B' in the same element index of A.
How would I go about implementing this?
I have the cumbersome idea of:
for i in 1:end
for j in 1:end
if A[i,j] >= 1
B[i,j] = 1
else
B[i,j] = 0
end
end
end
but I would prefer something more succinct.
Just use broadcast:
B = A .≥ 1
You can certainly use broadcasting as Oscar suggested (e.g. B = A .>= 1), but there's also nothing wrong with loops, since loops are fast and avoid excess allocations. You really only need one loop though, and the if statement is slightly superfluous, so:
B = similar(A, Int64) # If B doesn't already exist, otherwise omit this line
#inbounds for i in eachindex(A)
B[i] = A[i] >= 1
end
The #inbounds is optional, but improves speed.
I have the following:
include("as_mod.jl")
solvetimes = 50:200
timevector = Array{Float64}(undef,length(solvetimes))
for i in solvetimes
global T
T = i
include("as_dat_large.jl")
m, x, z = build_model(true,true)
setsolver(m, GurobiSolver(MIPGap = 2e-2, TimeLimit = 3600))
solve(m)
timevector[i-49] = getsolvetime(m)
end
plot(solvetimes,log.(timevector),
title = "solvetimes vs T", xlabel = "T", ylabel = "log(t)")
And this works great as long as my solvetimes vector is incremented by only 1. However, I'm interested in an 30-increment and it obviously does not work then since my timevector then goes out of bounds. Is there any way of solving this issue? I read about and attempted to use the push! function but to no avail.
I apologize if my question is not good but I don't see how to improve it. The question is essentially about for loops where the index does NOT start at 1 and is only incremented with 1 up to an upper bound, but rather a non-one increment and a start different from 0 or one, if that makes sense.
The : syntax in 50:200 or 50:30:200 creates a range object in Julia. These range objects are not only iterable but also implement the method getindex which means that you can simply access the steps in the range with a[index] syntax as if it is an array.
julia> solvetimes = 50:30:200 # 50, 80, 110, 140, ...
50:30:200
julia> solvetimes[3]
110
You can solve your problem in several ways.
First, you can introduce an itercount variable to count the number of iterations and know at which index of timevector you will put the solve-time.
solvetimes = 50:30:200 # increment by 30
timevector = Vector{Float64}(undef,length(solvetimes))
itercount = 1
for i in solvetimes
...
timevector[itercount] = getsolvetime(m)
global itercount
itercount += 1
end
Other way would be to create an empty timevector and push!.
solvetimes = 50:30:200 # increment by 30
timevector = Float64[] # an empty Float64 vector
for i in solvetimes
...
push!(timevector, getsolvetime(m)) # push the value `getsolvetime(m)` into `timevector`
end
push! operation may require julia to allocate memory and copy data to compensate increasing array size, hence might not be very efficient, although it does not really matter in your problem.
Another way would be to iterate from 1 to length of solvetimes. Your loop control variable is still incremented one-by-one but now it represents the index in solvetimes rather than the time point.
solvetimes = 50:30:200 # increment by 30
len = length(solvetimes)
timevector = Vector{Float64}(undef, len)
for i in 1:len
global T
T = solvetimes[i]
...
timevector[i] = getsolvetime(m)
end
With these modifications, kth value in timevector, timevector[k] stands for the solve-time for solvetime[k].
You might also find other ways to solve the issue, like using Dicts etc.
got a while loop going, and that's working fine.
However I also need to add another condition.
I need the loop to keep going until it satisfies the while loop, but then I also need to add that this can only get repeated x times.
I think you would have to make a for loop to do x times, is it possible to put a while loop in this?
Basically how can I make a loop either reach the goal or stop after x loops??
The expression in while needs to be TRUE for the loop to continue. You can use | or & to add extra conditions. This loop runs 99 times or until sum of random variables is less than 100.
counter <- 0
result <- 0
while(counter < 100 | sum(result) < 100) {
result <- sum(result, rnorm(1, mean = 5, sd = 1))
counter <- sum(counter, 1)
}
> result
[1] 101.5264
> counter
[1] 21
Just pass the current iterator value as an argument to your function. That way you can break the recursion if that reaches a particular value.
But why do you have a while loop if you use recursion, for example:
add_one_recursive = function(number) {
number = number + 1
cat("New number = ", number, "\n")
if(number > 10) {
return(number)
} else {
return(add_one_recursive(number))
}
}
final_number = add_one_recursive(0)
New number = 1
New number = 2
New number = 3
New number = 4
New number = 5
New number = 6
New number = 7
New number = 8
New number = 9
New number = 10
New number = 11
Does not require an explicit loop at all.