So I have the following code for a Merge Sort in Ruby.
class MergeSort
def sort(array)
if array.length == 1 || array.length == 0
return array
end
firstHalf = array[0..array.length / 2]
secondHalf = array[(array.length / 2) + 1..array.length]
firstHalf = sort(firstHalf)
secondHalf = sort(secondHalf)
b = 0
c = 0
for i in (0..(firstHalf.length - 1))
while b < secondHalf.length && firstHalf[i] >= secondHalf[b]
array[c] = secondHalf[b]
b = b + 1
c = c + 1
array[c] = firstHalf[i]
c = c + 1
end
return array
end
end
array = [1,4,9,14,20,25]
puts MergeSort::new.sort(array)
When I run the code, I get a SystemStackError. Can someone tell me why this is happening? Thanks.
At a guess, once the array length gets to 3 (i.e. elements [0..2]), the call
firstHalf = array[0..array.length / 2]
evaluates to
0..1.5 and if 1.5 is rounded up to 2
which then calls sort [0..2] again
and eventually you get a stack overflow?
In order to call .new you have to have an initialize method in your class.
What you probably wanted to do was calling .sort on the class itself, in which case you have to prefix it with self, so:
class MergeSort
def self.sort(array)
...
Afterwards you can call it like this:
MergeSort.sort(array)
Related
How can I get the number of iterations/steps that this method takes to find an answer?
def binary_search(array, n)
min = 0
max = (array.length) - 1
while min <= max
middle = (min + max) / 2
if array[middle] == n
return middle
elsif array[middle] > n
max = middle - 1
elsif array[middle] < n
min = middle + 1
end
end
"#{n} not found in this array"
end
One option to use instead of a counter is the .with_index keyword. To use this you'll need to use loop instead of while, but it should work the same. Here's a basic example with output.
arr = [1,2,3,4,5,6,7,8]
loop.with_index do |_, index| # The underscore is to ignore the first variable as it's not used
if (arr[index] % 2).zero?
puts "even: #{arr[index]}"
else
puts "odd: #{arr[index]}"
end
break if index.eql?(arr.length - 1)
end
=>
odd: 1
even: 2
odd: 3
even: 4
odd: 5
even: 6
odd: 7
even: 8
Just count the number of iterations.
Set a variable to 0 outside the loop
Add 1 to it inside the loop
When you return the index, return the count with it (return [middle, count]).
I assume the code to count numbers of interations required by binary_search is to be used for testing or optimization. If so, the method binary_search should be modified in such a way that to produce production code it is only necessary to remove (or comment out) lines of code, as opposed to modifying statements. Here is one way that might be done.
def binary_search(array, n)
# remove from production code lines marked -> #******
_bin_srch_iters = 0 #******
begin #******
min = 0
max = (array.length) - 1
loop do
_bin_srch_iters += 1 #******
middle = (min + max) / 2
break middle if array[middle] == n
break nil if min == max
if array[middle] > n
max = middle - 1
else # array[middle] < n
min = middle + 1
end
end
ensure #******
puts "binary_search reqd #{_bin_srch_iters} interations" #******
end #******
end
x = binary_search([1,3,6,7,9,11], 3)
# binary_search reqd 3 interations
#=> 1
binary_search([1,3,6,7,9,11], 5)
# binary_search reqd 3 interations
#=> nil
The code below returns "Arithmetic", "Geometric" if the input array is an arithmetic and geometric series respectively and -1 if it is neither.
Although the code works fine, when I change
if s = arr.length - 1
to
if s == arr.length - 1
in the while loop, the code is not working properly anymore.
I do not understand why. Shouldn't == work instead of =?
def ArithGeo(arr)
# code goes here
len = arr.length
difference = arr[len-1] - arr[len-2]
ratio = arr[len-1]/arr[len-2]
k = 0
s = k + 1
while (arr[s] - arr[k]) == difference && s < arr.length
if s = arr.length - 1
return "Arithmetic"
end
k += 1
end
k = 0
while arr[s] / arr[k] == ratio && s < arr.length
if s = arr.length - 1
return "Geometric"
end
k += 1
end
return -1
end
You're never changing the value of s which I think you want to do. You should do that at the point that you increment k
k += 1
s = k + 1
Also, at the point where you reinitialize k for the geometric test, you want to reset s as well...
k = 0
s = k + 1
You could also get rid of the variable s completely and make it a method... add these three lines at the top of the code
def s(k)
k + 1
end
And remove all the lines where you assign a value to s and use s(k)... s(k) will be a method that always returns the next higher value to k
The difference between those two statements is that variable s is set for the first statement but not for the second. The first if statement has thus a side effect of setting s to arr.length - 1
if s = arr.length - 1 # s => arr.length - 1
if s == arr.length - 1 # s => undefined
Because the if statement is inside a while loop which uses s in its expression the change of the statement changes the behavior of the programm.
If you put == the statement will try to check if they are equals , with just = the statement work properly because your are only setting the value to a value , so this is always true.
If it's different compare something to equals than just set a variable , that can be always true.
I'm trying to implement Smith-Waterman alignment in parallel using Julia (see: Figure 1 of http://www.cs.virginia.edu/~rl6sf/paper_dump/2011:12:33:22.pdf), but the algorithm is running much slower in Julia than the serial version. I'm using shared arrays to do this and figure I am doing something silly that is making the code run slow. Could someone take a look and see if my code is optimized as possible? The parallel version should run faster than in serial….
The basic concept of it is to compute the anti-diagonal elements of a matrix in parallel from the upper left to lower right corner and to update them. I'm trying to use 32 cores on a shared array machine to do this. I have a SharedArray matrix that I am using to do this and am computing the elements of each anti-diagonal in parallel as shown below. The while loops in the spSW function submit tasks to workers in sync for each anti-diagonal using the helper function shared_get_score(). The main goal of this function is to fill in each element in the shared arrays "matrix" and "path".
function spSW(seq1,seq2,p)
indel = -1
match = 2
seq1 = "^$seq1"
seq2 = "^$seq2"
col = length(seq1)
row = length(seq2)
wl = workers()
matrix,path = shared_initialize_path(seq1,seq2)
for j = 2:col
jcol = j
irow = 2
#sync begin
count = 0
while jcol > 1 && irow < row + 1
#println(j," ",irow," ",jcol)
if seq1[jcol] == seq2[irow]
equal = true
else
equal = false
end
w = wl[(count % p) + 1]
#async remotecall_wait(w,shared_get_score!,matrix,path,equal,indel,match,irow,jcol)
jcol -= 1
irow += 1
count += 1
end
end
end
for i = 3:row
jcol = col
irow = i
#sync begin
count = 0
while irow < row+1 && jcol > 1
#println(j," ",irow," ",jcol)
if seq1[jcol] == seq2[irow]
equal = true
else
equal = false
end
w = wl[(count % p) + 1]
#async remotecall_wait(w,shared_get_score!,matrix,path,equal,indel,match,irow,jcol)
jcol -= 1
irow += 1
count += 1
end
end
end
return matrix,path
end
The other helper functions are:
function shared_initialize_path(seq1,seq2)
col = length(seq1)
row = length(seq2)
matrix = convert(SharedArray,fill(0,(row,col)))
path = convert(SharedArray,fill(0,(row,col)))
return matrix,path
end
#everywhere function shared_get_score!(matrix,path,equal,indel,match,i,j)
pathvalscode = ["-","|","M"]
pathvals = [1,2,3]
scores = []
push!(scores,matrix[i,j-1]+indel)
push!(scores,matrix[i-1,j]+indel)
if equal
push!(scores,matrix[i-1,j-1]+match)
else
push!(scores,matrix[i-1,j-1]+indel)
end
val,ind = findmax(scores)
if val < 0
matrix[i,j] = 0
else
matrix[i,j] = val
end
path[i,j] = pathvals[ind]
end
Does anyone see an obvious way to make this run faster? Right now it's about 10 times slower than the serial version.
I need to initialize a 3D tensor with an index-dependent function in torch7, i.e.
func = function(i,j,k) --i, j is the index of an element in the tensor
return i*j*k --do operations within func which're dependent of i, j
end
then I initialize a 3D tensor A like this:
for i=1,A:size(1) do
for j=1,A:size(2) do
for k=1,A:size(3) do
A[{i,j,k}] = func(i,j,k)
end
end
end
But this code runs very slow, and I found it takes up 92% of total running time. Are there any more efficient ways to initialize a 3D tensor in torch7?
See the documentation for the Tensor:apply
These functions apply a function to each element of the tensor on
which the method is called (self). These methods are much faster than
using a for loop in Lua.
The example in the docs initializes a 2D array based on its index i (in memory). Below is an extended example for 3 dimensions and below that one for N-D tensors. Using the apply method is much, much faster on my machine:
require 'torch'
A = torch.Tensor(100, 100, 1000)
B = torch.Tensor(100, 100, 1000)
function func(i,j,k)
return i*j*k
end
t = os.clock()
for i=1,A:size(1) do
for j=1,A:size(2) do
for k=1,A:size(3) do
A[{i, j, k}] = i * j * k
end
end
end
print("Original time:", os.difftime(os.clock(), t))
t = os.clock()
function forindices(A, func)
local i = 1
local j = 1
local k = 0
local d3 = A:size(3)
local d2 = A:size(2)
return function()
k = k + 1
if k > d3 then
k = 1
j = j + 1
if j > d2 then
j = 1
i = i + 1
end
end
return func(i, j, k)
end
end
B:apply(forindices(A, func))
print("Apply method:", os.difftime(os.clock(), t))
EDIT
This will work for any Tensor object:
function tabulate(A, f)
local idx = {}
local ndims = A:dim()
local dim = A:size()
idx[ndims] = 0
for i=1, (ndims - 1) do
idx[i] = 1
end
return A:apply(function()
for i=ndims, 0, -1 do
idx[i] = idx[i] + 1
if idx[i] <= dim[i] then
break
end
idx[i] = 1
end
return f(unpack(idx))
end)
end
-- usage for 3D case.
tabulate(A, function(i, j, k) return i * j * k end)
I am using MATLAB to find all of the possible combinations of k elements out of n possible elements. I stumbled across this question, but unfortunately it does not solve my problem. Of course, neither does nchoosek as my n is around 100.
Truth is, I don't need all of the possible combinations at the same time. I will explain what I need, as there might be an easier way to achieve the desired result. I have a matrix M of 100 rows and 25 columns.
Think of a submatrix of M as a matrix formed by ALL columns of M and only a subset of the rows. I have a function f that can be applied to any matrix which gives a result of either -1 or 1. For example, you can think of the function as sign(det(A)) where A is any matrix (the exact function is irrelevant for this part of the question).
I want to know what is the biggest number of rows of M for which the submatrix A formed by these rows is such that f(A) = 1. Notice that if f(M) = 1, I am done. However, if this is not the case then I need to start combining rows, starting of all combinations with 99 rows, then taking the ones with 98 rows, and so on.
Up to this point, my implementation had to do with nchoosek which worked when M had only a few rows. However, now that I am working with a relatively bigger dataset, things get stuck. Do any of you guys think of a way to implement this without having to use the above function? Any help would be gladly appreciated.
Here is my minimal working example, it works for small obs_tot but fails when I try to use bigger numbers:
value = -1; obs_tot = 100; n_rows = 25;
mat = randi(obs_tot,n_rows);
while value == -1
posibles = nchoosek(1:obs_tot,i);
[num_tries,num_obs] = size(possibles);
num_try = 1;
while value == 0 && num_try <= num_tries
check = mat(possibles(num_try,:),:);
value = sign(det(check));
num_try = num_try + 1;
end
i = i - 1;
end
obs_used = possibles(num_try-1,:)';
Preamble
As yourself noticed in your question, it would be nice not to have nchoosek to return all possible combinations at the same time but rather to enumerate them one by one in order not to explode memory when n becomes large. So something like:
enumerator = CombinationEnumerator(k, n);
while(enumerator.MoveNext())
currentCombination = enumerator.Current;
...
end
Here is an implementation of such enumerator as a Matlab class. It is based on classic IEnumerator<T> interface in C# / .NET and mimics the subfunction combs in nchoosek (the unrolled way):
%
% PURPOSE:
%
% Enumerates all combinations of length 'k' in a set of length 'n'.
%
% USAGE:
%
% enumerator = CombinaisonEnumerator(k, n);
% while(enumerator.MoveNext())
% currentCombination = enumerator.Current;
% ...
% end
%
%% ---
classdef CombinaisonEnumerator < handle
properties (Dependent) % NB: Matlab R2013b bug => Dependent must be declared before their get/set !
Current; % Gets the current element.
end
methods
function [enumerator] = CombinaisonEnumerator(k, n)
% Creates a new combinations enumerator.
if (~isscalar(n) || (n < 1) || (~isreal(n)) || (n ~= round(n))), error('`n` must be a scalar positive integer.'); end
if (~isscalar(k) || (k < 0) || (~isreal(k)) || (k ~= round(k))), error('`k` must be a scalar positive or null integer.'); end
if (k > n), error('`k` must be less or equal than `n`'); end
enumerator.k = k;
enumerator.n = n;
enumerator.v = 1:n;
enumerator.Reset();
end
function [b] = MoveNext(enumerator)
% Advances the enumerator to the next element of the collection.
if (~enumerator.isOkNext),
b = false; return;
end
if (enumerator.isInVoid)
if (enumerator.k == enumerator.n),
enumerator.isInVoid = false;
enumerator.current = enumerator.v;
elseif (enumerator.k == 1)
enumerator.isInVoid = false;
enumerator.index = 1;
enumerator.current = enumerator.v(enumerator.index);
else
enumerator.isInVoid = false;
enumerator.index = 1;
enumerator.recursion = CombinaisonEnumerator(enumerator.k - 1, enumerator.n - enumerator.index);
enumerator.recursion.v = enumerator.v((enumerator.index + 1):end); % adapt v (todo: should use private constructor)
enumerator.recursion.MoveNext();
enumerator.current = [enumerator.v(enumerator.index) enumerator.recursion.Current];
end
else
if (enumerator.k == enumerator.n),
enumerator.isInVoid = true;
enumerator.isOkNext = false;
elseif (enumerator.k == 1)
enumerator.index = enumerator.index + 1;
if (enumerator.index <= enumerator.n)
enumerator.current = enumerator.v(enumerator.index);
else
enumerator.isInVoid = true;
enumerator.isOkNext = false;
end
else
if (enumerator.recursion.MoveNext())
enumerator.current = [enumerator.v(enumerator.index) enumerator.recursion.Current];
else
enumerator.index = enumerator.index + 1;
if (enumerator.index <= (enumerator.n - enumerator.k + 1))
enumerator.recursion = CombinaisonEnumerator(enumerator.k - 1, enumerator.n - enumerator.index);
enumerator.recursion.v = enumerator.v((enumerator.index + 1):end); % adapt v (todo: should use private constructor)
enumerator.recursion.MoveNext();
enumerator.current = [enumerator.v(enumerator.index) enumerator.recursion.Current];
else
enumerator.isInVoid = true;
enumerator.isOkNext = false;
end
end
end
end
b = enumerator.isOkNext;
end
function [] = Reset(enumerator)
% Sets the enumerator to its initial position, which is before the first element.
enumerator.isInVoid = true;
enumerator.isOkNext = (enumerator.k > 0);
end
function [c] = get.Current(enumerator)
if (enumerator.isInVoid), error('Enumerator is positioned (before/after) the (first/last) element.'); end
c = enumerator.current;
end
end
properties (GetAccess=private, SetAccess=private)
k = [];
n = [];
v = [];
index = [];
recursion = [];
current = [];
isOkNext = false;
isInVoid = true;
end
end
We can test implementation is ok from command window like this:
>> e = CombinaisonEnumerator(3, 6);
>> while(e.MoveNext()), fprintf(1, '%s\n', num2str(e.Current)); end
Which returns as expected the following n!/(k!*(n-k)!) combinations:
1 2 3
1 2 4
1 2 5
1 2 6
1 3 4
1 3 5
1 3 6
1 4 5
1 4 6
1 5 6
2 3 4
2 3 5
2 3 6
2 4 5
2 4 6
2 5 6
3 4 5
3 4 6
3 5 6
4 5 6
Implementation of this enumerator may be further optimized for speed, or by enumerating combinations in an order more appropriate for your case (e.g., test some combinations first rather than others) ... Well, at least it works! :)
Problem solving
Now solving your problem is really easy:
n = 100;
m = 25;
matrix = rand(n, m);
k = n;
cont = true;
while(cont && (k >= 1))
e = CombinationEnumerator(k, n);
while(cont && e.MoveNext());
cont = f(matrix(e.Current(:), :)) ~= 1;
end
if (cont), k = k - 1; end
end