I'm looking to find a way to take an array in ruby, two indices in that array and return an enumerable object which will yield, in order, all the elements between and including the two indices. But for performance reasons, I want to do this subject to the following two conditions:
This slice to enum does not create a copy of the subarray I want a return an enum over. This rules out array[i..j].to_enum, for example because array[i..j] is creating a new array.
It's not necessary to loop over the entire array to create the enum.
I'm wondering if there's a way to do this using the standard library's enumerable or array functionality without having to explicitly create my own custom enumerator.
What I'm looking for is a cleaner way to create the below enumerator:
def enum_slice(array, i, j)
Enumerator.new do |y|
while i <= j
y << array[i] # this is confusing syntax for yield (see here: https://ruby-doc.org/core-2.6/Enumerator.html#method-c-new)
i += 1
end
end
end
That seems pretty reasonable, and could even be turned into an extension to Array itself:
module EnumSlice
def enum_slice(i, j)
Enumerator.new do |y|
while i <= j
y << self[i]
i += 1
end
end
end
end
Now within the Enumerator block, y represents a Proc you call when you have more data. If that block ends it's presumed you're done enumerating. There's no requirement to ever terminate, an infinite Enumerator is allowed, and in that case it's up to the caller to stop iterating.
So in other words, the y block argument can be called zero or more times, and each time it's called output is "emitted" from the enumerator. When that block exits the enumerator is considered done and is closed out, y is invalid at that point.
All y << x does is call the << method on Enumerator::Yielder, which is a bit of syntactical sugar to avoid having to do y.call(x) or y[x], both of which look kind of ugly.
Now you can add this to Array:
Array.include(EnumSlice)
Where now you can do stuff like this:
[ 1, 2, 3, 4, 5, 6 ].enum_slice(2, 4).each do |v|
p v
end
Giving you the correct output.
It's worth noting that despite having gone through all this work, this really doesn't save you any time. There's already built-in methods for this. Your enum_slice(a, i, j) method is equivalent to:
a.drop(i).take(j)
Is that close in terms of performance? A a quick benchmark can help test that theory:
require 'benchmark'
Benchmark.bm do |bm|
count = 10000
a = (0..100_000).to_a
bm.report(:enum_slice) do
count.times do
a.enum_slice(50_000, 25_000).each do
end
end
end
bm.report(:drop_take) do
count.times do
a.drop(50_000).take(25_000).each do
end
end
end
end
The results are:
user system total real
enum_slice 0.020536 0.000200 0.020736 ( 0.020751)
drop_take 7.682218 0.019815 7.702033 ( 7.720876)
So your approach is about 374x faster. Not bad!
When I try to compile my code using -fcheck=all I get a runtime error since it seems I step out of bounds of my array dimension size. It comes from the part of my code shown below. I think it is because my loops over i,j only run from -ny to ny, -nx to nx but I try to use points at i+1,j+1,i-1,j-1 which takes me out of bounds in my arrays. When the loop over j starts at -ny, it needs j-1, so it immediately takes me out of bounds since I'm trying to access -ny-1. Similarly when j=ny, i=-nx,nx.
My question is, how can I fix this problem efficiently using minimal code?
I need the array grad(1,i,j) correctly defined on the boundary, and it needs to be defined exactly as on the right hand side of the equality below, I just don't know an efficient way of doing this. I can explicitly define grad(1,nx,j), grad(1,-nx,j), etc, separately and only loop over i=-nx+1,nx-1,j=-ny+1,ny-1 but this causes lots of duplicated code and I have many of these arrays so I don't think this is the logical/efficient approach. If I do this, I just end up with hundreds of lines of duplicated code that makes it very hard to debug. Thanks.
integer :: i,j
integer, parameter :: nx = 50, ny = 50
complex, dimension (3,-nx:nx,-ny:ny) :: grad,psi
real, parameter :: h = 0.1
do j = -ny,ny
do i = -nx,nx
psi(1,i,j) = sin(i*h)+sin(j*h)
psi(2,i,j) = sin(i*h)+sin(j*h)
psi(3,i,j) = sin(i*h)+sin(j*h)
end do
end do
do j = -ny,ny
do i = -nx,nx
grad(1,i,j) = (psi(1,i+1,j)+psi(1,i-1,j)+psi(1,i,j+1)+psi(1,i,j-1)-4*psi(1,i,j))/h**2 &
- (psi(2,i+1,j)-psi(2,i,j))*psi(1,i,j)/h &
- (psi(3,i,j+1)-psi(3,i,j))*psi(1,i,j)/h &
- psi(2,i,j)*(psi(1,i+1,j)-psi(1,i,j))/h &
- psi(3,i,j)*(psi(1,i,j+1)-psi(1,i,j))/h
end do
end do
If I was to do this directly for grad(1,nx,j), grad(1,-nx,j), it would be given by
do j = -ny+1,ny-1
grad(1,nx,j) = (psi(1,nx,j)+psi(1,nx-2,j)+psi(1,nx,j+1)+psi(1,nx,j-1)-2*psi(1,nx-1,j)-2*psi(1,nx,j))/h**2 &
- (psi(2,nx,j)-psi(2,nx-1,j))*psi(1,nx,j)/h &
- (psi(3,nx,j+1)-psi(3,nx,j))*psi(1,nx,j)/h &
- psi(2,nx,j)*(psi(1,nx,j)-psi(1,nx-1,j))/h &
- psi(3,nx,j)*(psi(1,nx,j+1)-psi(1,nx,j))/h
grad(1,-nx,j) = (psi(1,-nx+2,j)+psi(1,-nx,j)+psi(1,-nx,j+1)+psi(1,-nx,j-1)-2*psi(1,-nx+1,j)-2*psi(1,-nx,j))/h**2 &
- (psi(2,-nx+1,j)-psi(2,-nx,j))*psi(1,-nx,j)/h &
- (psi(3,-nx,j+1)-psi(3,-nx,j))*psi(1,-nx,j)/h &
- psi(2,-nx,j)*(psi(1,-nx+1,j)-psi(1,-nx,j))/h &
- psi(3,-nx,j)*(psi(1,-nx,j+1)-psi(1,-nx,j))/h
end do
One possible way for you could be using an additional index variable for the boundaries, modified from the original index to avoid getting out-of-bounds. I mean something like this:
do j = -ny,ny
jj = max(min(j, ny-1), -ny+1)
do i = -nx,nx
ii = max(min(i, nx-1), -nx+1)
grad(1,i,j) = (psi(1,ii+1,j)+psi(1,ii-1,j)+psi(1,i,jj+1)+psi(1,i,jj-1)-4*psi(1,i,j))/h**2 &
- (psi(2,ii+1,j)-psi(2,ii,j))*psi(1,i,j)/h &
- (psi(3,i,jj+1)-psi(3,i,jj))*psi(1,i,j)/h &
- psi(2,i,j)*(psi(1,ii+1,j)-psi(1,ii,j))/h &
- psi(3,i,j)*(psi(1,i,jj+1)-psi(1,i,jj))/h
end do
end do
It's hard for me to write a proper code because it seems you trimmed part of the original expression in the code you presented in the question, but I hope you understand the idea and apply it correctly for your logic.
Opinions:
Even though this is what you are asking for (as far as I understand), I would not recommend doing this before profiling and checking if assigning the boundary conditions manually after a whole array operation wouldn't be more efficient, instead. Maybe those extra calculations on the indices on each iteration could impact on performance (arguably less than if conditionals or function calls). Using "ghost cells", as suggested by #evets, could be even more performant. You should profile and compare.
I'd recommend you declaring your arrays as dimension(-nx:nx,-ny:ny,3) instead. Fortran stores arrays in column-major order and, as you are accessing values on the neighborhood of the "x" and "y", they would be non-contiguous memory locations for a fixed "other" dimension is the leftest, and that could mean less cache-hits.
In somewhat pseudo-code, you can do
do j = -ny, ny
if (j == -ny) then
p1jm1 = XXXXX ! Some boundary condition
else
p1jm1 = psi(1,i,j-1)
end if
if (j == ny) then
p1jp1 = YYYYY ! Some other boundary condition
else
p1jp1 = psi(1,i,j+1)
end if
do i = -nx, ny
grad(1,i,j) = ... term involving p1jm1 ... term involving p1jp1 ...
...
end do
end do
The j-loop isn't bad in that you are adding 2*2*ny conditionals. The inner i-loop is adding 2*2*nx conditionals for each j iteration (or 2*2*ny * 2*2*nx conditional). Note, you need a temporary for each psi with the triplet indices are unique, ie., psi(1,i,j+1), psi(1,i,j-1), and psi(3,i,j+1).
An exercise in Coderbyte is supposed to determine if some subset of integers in an array sum to the largest number in the array.
The following code seems to work on my computer, but when I submit it online, it seems to cause an endless loop. (There's never any output, regardless of the argument passed).
def arr_add?(arr)
a = arr.sort
lgst = a.pop
size = a.size
result = false
while size > 1
a.combination(size) {|c| result |= (c.inject {|r, a| r + a} == lgst)}
size -= 1
end
result.to_s
end
arr_add?([1, 2, 3, 4, 10, 14])
Any ideas why this might be the case?
I suspect that you are actually not running into an endless loop, but rather just take a really long time, because of the inefficiency of your algorithm.
def ArrayAdditionI(arr)
arr_size = arr.size
ary = arr.sort
largest = ary.pop
ary_size = arr_size - 1
combination_size = ary_size
result = false
while combination_size > 1
ary.combination(combination_size) {|combination|
result |= (combination.inject(
:+
) == largest)
}
combination_size -= 1
end
result.to_s
end
I introduced a new variable and renamed some others, so that it becomes easier to talk about the algorithm. I also reformatted it, to make the three nested "loops" more obvious.
Let's take a look at the algorithm.
The outer while loop is executed ary_size - 1 == arr_size - 2 times, with combination_size ranging from 2 to ary_size == arr_size - 1.
The combination "loop" is executed ary_size choose combination_size times, that's … well, a very quickly growing number.
The innermost "loop" (the operation performed by combination.inject) is executed combination_size - 1 times.
This gives a total execution count for the innermost operation of:
The Sum from 2 to arr_size - 1 of
arr_size - 1 choose combination_size times
combination_size - 1
In Wolfram Language, that's Sum[Binomial[a-1, c]*(c-1), c, 2, a-1], which Wolfram Alpha tells us is 2^(a-2) (a-3)+1, which is in O(2^n).
Playing around with the numbers a bit:
for 10 items, we have 1793 executions of the inject operation
for 15 items, we already have 98 305
for 20 items, we have 4 456 449
at 28 items, we cross the threshold to a billion operations: 1 677 721 601
for 1000 items, which I suspect is a somewhat reasonable input size CoderBytes might use, we have 2 670 735 203 411 771 297 463 949 434 782 054 512 824 301 493 176 042 516 553 547 843 013 099 994 928 903 285 314 296 959 198 121 926 383 029 722 247 001 218 461 778 959 624 588 092 753 669 155 960 493 619 769 880 691 017 874 939 573 116 202 845 311 796 007 113 080 079 901 646 833 889 657 798 860 899 142 814 122 011 828 559 707 931 456 870 722 063 370 635 289 362 135 539 416 628 419 173 512 766 291 969 operations. Oops.
Try your algorithm with arrays of length 5, 10, 15 (all instantaneous), 20 (a noticeable pause), and then 23, 24, 25 to get a feel for just how quickly the runtime grows.
Assuming that you could build a CPU which can execute the inner loop in a single instruction. Further assuming that a single instruction takes only a Planck time unit (i.e. the CPU has a frequency of roughly 20 000 000 000 000 000 000 000 000 000 000 THz). Further assuming that every single particle in the observable universe was such a CPU. It will still take more than the current age of the universe to execute your algorithm for an array of not even 500 items.
Note that with most of these programming puzzles, they are not actually programming puzzles, they are mathematics puzzles. They usually require a mathematical insight, in order to be able to solve them efficiently. Or, in this case, recognizing that it is the Subset sum problem, which is known to be NP-complete.
By the way, as a matter of style, here is (a slight variation of) your algorithm written in idiomatic Ruby style. As you can see, in idiomatic Ruby, it almost becomes a 1:1 translation of the English problem statement into code.
While it is asymptotically just as inefficient as your algorithm, it breaks early, as soon as the answer is true (unlike yours, will just keep running even if it already found a solution). (any? will do that for you automatically.)
def ArrayAdditionI(arr)
largest = arr.delete_at(arr.index(arr.max))
1.upto(arr.size).any? {|combination_size|
arr.combination(combination_size).any? {|combination|
combination.inject(:+) == largest
}
}.to_s
end
This is an alternative interpretation of the (unclear) problem statement:
def ArrayAdditionI(arr)
2.upto(arr.size).any? {|combination_size|
arr.combination(combination_size).any? {|combination|
combination.inject(:+) == arr.max
}
}.to_s
end
The code above is valid ruby code.
The result is "true".
It is a bit unusual perhaps in that while loops are somewhat rare to see/have, but since it is valid ruby code, it should work on that remote site too.
Contact whoever runs the online ruby interpreter at Coderbyte - their version appears to be incompatible with the MRI ruby.
Your code seems to count down; perhaps have a look at 10.downto(1) - replace with variables as appropriate.
In my code:
DO i=1,numJog,1
IF(val(i) .EQV. .TRUE.)THEN
DO j=1,contVenc,1
result(j) = i
END DO
END IF
END DO
Where val is a logical array, and result is a integer array.
For example, if val is:
F
T
F
T
Then, i=2 and i=4.
But the result array just write 4 twice. For example:
DO i=1,contVenc,1
WRITE(*,*) result(i)
END DO
The result is:
4
4
Instead of
2
4
If I make some changes in my code like:
DO i=1,numJog,1
IF(val(i) .EQV. .TRUE.)THEN
WRITE(*,*) i
END IF
END DO
The result is:
2
4
As I wanted.
Conclusion, I think this second loop is causing this problem.
Yes, your second loop is at fault here. You haven't said what contVenc is, but it crucially doesn't change at any point in the fragment you have there. That just means that the same elements of result are being assigned to whenever you have a .TRUE. in val.
In your case they are both set to 2 for the first .TRUE. and are then both set to 4 for the second.
You are more likely to mean something like (with extra tidying):
j = 0
DO i=1,numJog
IF (val(i)) THEN
j = j+1 ! Test this as a bound
result(j) = i
END IF
END DO
But then, I'd just use PACK. Your intended loop has the same effect as
result(1:COUNT(val(1:numJog))) = PACK([(i,i=1,numJog)], val(1:numJog))
Again hoping that result is large enough.
That said, if numJog is just the size of the array val (that is, you aren't just doing this on a sub-array) then, as High Performance Mark comments,
result(1:COUNT(val)) = PACK([(i,i=1,SIZE(val))], val)
avoids tracking this size separately.
Finally, with result an allocatable (Fortran 2003) array you needn't even (but still can) worry about counting the number of wanted indices and that the array is sufficiently large:
result = PACK([(i,i=1,SIZE(val))], val)