For N=1, I would use std.array : empty to check for length being at least N, and avoid having to go through the whole input.
For N>1 (or all N), what is the idiomatic way in D language?
I've tried to use the std.range : take which "Lazily takes only up to n elements of a range.". It works for arrays but not for ranges (unless of course I make a sub-range into an array):
#!/usr/bin/env rdmd
module test_ranges;
void main()
{
import std.container.dlist : DList;
assert(lengthAtLeast([1, 2, 3], 2) == true);
// assert(lengthAtLeast(DList!int(1, 2, 3)[], 2) == true);
/*
test_ranges.d(64): Error: no property length for type Take!(Range)
test_ranges.d(10): Error: template instance `test_ranges.lengthAtLeast!(Range)` error instantiating
Failed: ["/usr/bin/dmd", "-v", "-o-", "test_ranges.d", "-I."]
*/
}
bool lengthAtLeast(R)(R input, size_t n)
{
import std.range : take;
return input.take(n).length == n;
// this makes it work for arrays and ranges alike, but is not nice, is it?
// import std.array : array;
// return input.take(n).array.length == n;
}
walkLength does what you want:
bool lengthAtLeast(R)(R input, size_t n)
{
import std.range.primitives : walkLength;
return input.walkLength(n) >= n; // walks upTo n or returns the length if known
}
Related
So I was asked to write a function that changes array's values in a way that:
All of the values that are the smallest aren't changed
if, let's assume, the smallest number is 2 and there is no 3's and 4's then all 5's are changed for 3's etc.
for example, for an array = [2, 5, 7, 5] we would get [2, 3, 4, 3], which generalizes to getting a minimal value of an array which remains unchanged, and every other minimum (not including the first one) is changed depending on which minimum it is. On our example - 5 is the first minimum (besides 2), so it is 2 (first minimum) + 1 = 3, 7 is 2nd smallest after 2, so it is 2+2(as it is 2nd smallest).
I've come up with something like this:
int fillGaps(int arr[], size_t sz){
int min = *min_element(arr, arr+sz);
int w = 1;
for (int i = 0; i<sz; i++){
if (arr[i] == min) {continue;}
else{
int mini = *min_element(arr+i, arr+sz);
for (int j = 0; j<sz; j++){
if (arr[j] == mini){arr[j] = min+w;}
}
w++;}
}
return arr[sz-1];
}
However it works fine only for the 0th and 1st value, it doesnt affect any further items. Could anyone please help me with that?
I don't quite follow the logic of your function, so can't quite comment on that.
Here's how I interpret what needs to be done. Note that my example implementation is written to be as understandable as possible. There might be ways to make it faster.
Note that I'm also using an std::vector, to make things more readable and C++-like. You really shouldn't be passing raw pointers and sizes, that's super error prone. At the very least bundle them in a struct.
#include <algorithm>
#include <set>
#include <unordered_map>
#include <vector>
int fillGaps (std::vector<int> & data) {
// Make sure we don't have to worry about edge cases in the code below.
if (data.empty()) { return 0; }
/* The minimum number of times we need to loop over the data is two.
* First to check which values are in there, which lets us decide
* what each original value should be replaced with. Second to do the
* actual replacing.
*
* So let's trade some memory for speed and start by creating a lookup table.
* Each entry will map an existing value to its new value. Let's use the
* "define lambda and immediately invoke it" to make the scope of variables
* used to calculate all this as small as possible.
*/
auto const valueMapping = [&data] {
// Use an std::set so we get all unique values in sorted order.
std::set<int> values;
for (int e : data) { values.insert(e); }
std::unordered_map<int, int> result;
result.reserve(values.size());
// Map minimum value to itself, and increase replacement value by one for
// each subsequent value present in the data vector.
int replacement = *values.begin();
for (auto e : values) { result.emplace(e, replacement++); }
return result;
}();
// Now the actual algorithm is trivial: loop over the data and replace each
// element with its replacement value.
for (auto & e : data) { e = valueMapping.at(e); }
return data.back();
}
I need to put the first value of a loop = 0, and then use a range to start the loop.
In MatLab this is possible : x = [0 -range:range] (range is a integer)
This will give a value of [0, -range, -range+1, -range+2, .... , range-1, range]
The problem is I need to do this in C++, I tried to do by an array and then put in like the value on the loop without success.
//After loading 2 images, put it into matrix values and then trying to compare each one.
for r=1:bRows
for c=1:bCols
rb=r*blockSize;
cb=c*blockSize;
%%for each block search in the near position(1.5 block size)
search=blockSize*1.5;
for dr= [0 -search:search] //Here's the problem.
for dc= [0 -search:search]
%%check if it is inside the image
if(rb+dr-blockSize+1>0 && rb+dr<=rows && cb+dc-blockSize+1>0 && cb+dc<=cols)
%compute the error and check if it is lower then the previous or not
block=I1(rb+dr-blockSize+1:rb+dr,cb+dc-blockSize+1:cb+dc,1);
TE=sum( sum( abs( block - cell2mat(B2(r,c)) ) ) );
if(TE<E)
M(r,c,:)=[dr dc]; %store the motion vector
Err(r,c,:)=TE; %store th error
E=TE;
end
end
end
end
%reset the error for the next search
E=255*blockSize^2;
end
end
C++ doesn't natively support ranges of the kind you know from MatLab, although external solutions are available, if somewhat of an overkill for your use case. However, C++ allows you to implement them easily (and efficiently) using the primitives provided by the language, such as for loops and resizable arrays. For example:
// Return a vector consisting of
// {0, -limit, -limit+1, ..., limit-1, limit}.
std::vector<int> build_range0(int limit)
{
std::vector<int> ret{0};
for (auto i = -limit; i <= limit; i++)
ret.push_back(i);
return ret;
}
The resulting vector can be easily used for iteration:
for (int dr: build_range0(search)) {
for (int dc: build_range0(search)) {
if (rb + dr - blockSize + 1 > 0 && ...)
...
}
}
The above of course wastes some space to create a temporary vector, only to throw it away (which I suspect happens in your MatLab example as well). If you want to just iterate over the values, you will need to incorporate the loop such as the one in build_range0 directly in your function. This has the potential to reduce readability and introduce repetition. To keep the code maintainable, you can abstract the loop into a generic function that accepts a callback with the loop body:
// Call fn(0), fn(-limit), fn(-limit+1), ..., fn(limit-1), and fn(limit)
template<typename F>
void for_range0(int limit, F fn) {
fn(0);
for (auto i = -limit; i <= limit; i++)
fn(i);
}
The above function can be used to implement iteration by providing the loop body as an anonymous function:
for_range0(search, [&](int dr) {
for_range0(search, [&](int dc) {
if (rb + dr - blockSize + 1 > 0 && ...)
...
});
});
(Note that both anonymous functions capture enclosing variables by reference in order to be able to mutate them.)
Reading your comment, you could do something like this
for (int i = 0, bool zero = false; i < 5; i++)
{
cout << "hi" << endl;
if (zero)
{
i = 3;
zero = false;
}
}
This would start at it 0, then after doing what I want it to do, assign i the value 3, and then continue adding to it each iteration.
Working on below problem as an algorithm puzzle. Referred a few similar solutions (and post one of them below), tried and they worked. The question is, for the line "swap(num[i], num[k]);", how do we ensure we could always swap to a number which never tried before (e.g. suppose we swap 1 with 2 in current iteration of the for loop, then it is possible later we swap 2 back with 1 in next iterations of the same for loop of the same level/layer of recursive call)? I have the confusion since we pass num by reference, and it is very possible later (lower level/layer) recursive calls modify content of num, which cause numbers we already evaluated swap back. However, I tried and it works for all of my test cases. Wondering if below solution is 100% correct, or happened to pass my test cases? :)
Here are detailed problem statement and code I am debugging,
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1]
class Solution {
public:
void recursion(vector<int> num, int i, int j, vector<vector<int> > &res) {
if (i == j-1) {
res.push_back(num);
return;
}
for (int k = i; k < j; k++) {
if (i != k && num[i] == num[k]) continue;
swap(num[i], num[k]);
recursion(num, i+1, j, res);
}
}
vector<vector<int> > permuteUnique(vector<int> &num) {
sort(num.begin(), num.end());
vector<vector<int> >res;
recursion(num, 0, num.size(), res);
return res;
}
};
thanks in advance,
Lin
As #notmyfriend said in the comments, num is actually copied each function call.
So now it boils down to:
Of all array values, select one to be the first one and place it there.
That in a loop for each value one time, and then recursively:
Of all values after the first one, select one to be the first and place it there...
...and so on, combined with a check to filter out swaps where nothing changes, ie. filter out duplicates.
If num were a real reference, it won't work anymore (at least not without additional steps).
Eg. 1 1 2 is an easy conterexample, it would give the results:
112, 121, 211, 112, 121
ie. there are duplicates despite the check (and probably there
are examples where some permutations are not generated at all, too).
About the comment:
Per default, every normal function parameter in C++ is copied
(normal = without explicit reference symbol '&' etc.).
Maybe you're thinking of C-style arrays: Essentially, what is passed there is a pointer (to the first value). The pointer is copied, but both original and copied pointer point to the same memory location.
While the purpose of std::vector is (too) to contain an array, the vector itself is a single class object (which contains a pointer to the values somewhere). A class can define itself how it should be copied (with a copy constructor).
Technically, the vector class could implement copying as pointer copying, then it would have the same effect as passing the whole vector as reference; but the C++ creators wanted to keep the copy semantics, ie. that copying a container class should make a real copy with all values duplicated.
For non-copying, there are references already...
Below you can find a solution written in Java. Sorry for not providing a solution in C++, I'm not using it for a long time. But the syntax would be similar.
Solution is using Backtracking (https://en.wikipedia.org/wiki/Backtracking)
Also I'm using hashset to check uniqueness, may be there is a solution which does not use any hashset type data structure, becase my solution is using extra memory in order to provide unique solutions.
Sample input and output;
input : [1, 1, 2]
output : [1, 1, 2]
[1, 2, 1]
[2, 1, 1]
And the solution is;
public class permutation {
public static void main(String[] args) {
permutation p = new permutation();
p.permute(new int[] { 1, 1, 2 });
}
HashSet<String> set = new HashSet<String>();
private void permute(int[] arr) {
set.clear();
this.permute(arr, 0, arr.length - 1);
}
private void permute(int[] arr, int l, int r) {
if (l == r) {
String key = Arrays.toString(arr);
if (set.contains(key))
return;
set.add(key);
System.out.println(key);
} else {
for (int i = l; i <= r; i++) {
swap(arr, l, i);
permute(arr, l + 1, r);
swap(arr, i, l);
}
}
}
private void swap(int[] arr, int l, int r) {
int tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
}
}
I've never programmed in C++ before and I'm trying to figure out how to recursively pass segments of an array in a C++ method. I am trying to convert the following pseudo code into C++.
SlowSort(A[1...n])
if n = 2
if A[1] > A[2]
Swap(A[1], A[2])
else if n > 2
SlowSort(A[1...(2n/3)])
SlowSort(A[(n/3+1)... n])
SlowSort(A[1...(2n/3)])
The recursive calls are the bits I'm having a problem with. I was thinking about creating two new arrays that point to the wanted locations but don't know how to go about that, specifically doing that and defining the length of the array. I've tried googling it and searching this site, but there doesn't seem to be anything, that I understand, on it. Also, in case I fudged up somewhere in my code, here's what I have for the first bit.
int SlowSort(int A[])
{
int length = (sizeof(A)/sizeof(*A));
if(length ==2)
{
if(A[0] > A[1])
{
int temp = A[0];
A[0] = A[1];
A[1] = temp;
}
}
In short, how do In covert the else if statement into C++? Explanation would be nice too.
Thanks
You will want to pass indices into the array instead, and use those.
void SlowSort(int A[], int left, int right)
{
if (right - left == 2)
if (A[left] > A[right])
Swap(A[left], A[right]);
else
{
int n = right - left + 1;
SlowSort(A, left, 2 * n / 3);
SlowSort(A, left + n / 3 + 1, right);
SlowSort(A, left, left + 2* n / 3);
}
The above code might not be correct regarding what the algorithm is supposed to do, but you get the idea I'm trying to describe. The thing is: you don't make a copy of the array. Instead, pass the same array always and the range (i.e. the indices) you are sorting.
You simply pass required pointer using the pointer arithmetic. For example the following pseudo code
SlowSort(A[(n/3+1)... n])
could be written as
SlowSort( A + n/3+1, n - n/3 - 1 );
So the function could be declared as
void SlowSort( int A[], size_t n );
As for this code snippet
int SlowSort(int A[])
{
int length = (sizeof(A)/sizeof(*A));
then it is invalid because array is implicitly converted to a ponter to its first element when it is passed as an argument to a function seclared such a way. So the value of length will not be equal to the number of elements.
This is pretty simple. Since arrays are just consecutive pointers. If you have a method:
Your code would look like this:
void slowSort(int[] array, int length)
{
if(length == 2)
{
if(array[0] > array[1])
{
int temp = array[0];
array[0] = array[1];
array[1] = temp;
}
}
else
{
slowSort(&array[0], (2 * length) / 3 - 1);
slowSort(&array[length / 3], length - (length / 3 - 1));
slowSort(&array[0], (2 * length) / 3 - 1);
}
}
The trick I use here is that I pass the pointer of the element I want to start with and the pass the end point.
This works because when you pass an array in C++ you just pass the pointer of the first element. Here I pass a custom pointer of the array.
The modern C++ way to do this would be to pass iterators to the beginning and one-past-the-end of the range. In this case, the iterators are pointers.
void SlowSort(int* begin, int* end)
{
unsigned length = end-begin;
if(length == 2)
{
if(begin[0] > begin[1])
{
std::swap( begin[0], begin[1] );
}
} else if(length>2) {
SlowSort(begin, begin+2*length/3);
SlowSort(begin+length/3, end);
SlowSort(begin, begin+2*length/3);
}
}
then, for the case of working with an entire array:
template<unsigned N>
void SlowSort( int(&Arr)[N] ) {
return SlowSort( Arr, Arr+N );
}
we dispatch it to the iterator version, relying on decaying of array-to-pointer. This has to be a template function, as we want it to work with multiple different array sizes.
Note that an int Arr[] is not an array. It is a different way to say int* Arr, left over as a legacy from C. In fact, as a parameter to a function, saying void foo( int A[27] ) results in void foo( int* A ): function parameters cannot be arrays.
They can, however, be references-to-arrays, which is what the above template function uses.
I am trying to make a function that can return the prime factors of a given number in an array (or multi-set, but I'm trying to use an array).
For example, if I put in 12, I want to get 2, 2, and 3, not 2, and 3 like with a set. This is so that I can use these to see if it is a Smith number or not, so I need the numbers seperately.
Also, I am taking a recursive approach.
I have tried (to no avail) to return the array many ways, including passing an initial pointer into the code which points to a space to store the array.
I've tried just initializing the array in the function and then returning it.
From what I can tell, I can get the array back from the base case iteration and then when trying to construct a new array with size oldArray+1 to copy values to, things get messy. This is where I get lost.
From what I've read, although this isn't the most efficient implementation, I should be able to make it work.
I have a function, nextPrime(int n), which given n will give back the next prime up from that number.
See source below:
int* find(int n, int p) {
int root = (int) floor(sqrt(n));
if (p > root) {
// Base case, array gets initialized and returned
// depending on value of n and p.
if (n > 1) {
factors = new int[1];
factors[0] = n;
return factors;
}
else {
factors = new int[0];
return factors;
}
}
else
if (n%p == 0){
// Inductive step if p is a factor
int newFloor = (int) floor(n/p);
factors = find(newFloor, p);
// Initialize new array.
int* newFactors;
newFactors = new int[(sizeof(factors) / sizeof(int)) + 1];
// Add p to first slot, fill rest with contents of factors.
factors[0] = p;
for (int i = 0; i < (sizeof(factors) / sizeof(int)); i++) {
newFactors[i+1] = factors[i];
}
return newFactors;
}
else {
// Inductive step p isn't a factor of n
factors = find(n, factors, nextPrime(p));
return factors;
}
}
As I say, the error is with returning the array and using its value, but why does it seem to return OK from the first iteration?
Something like this could work. Not terribly efficient !!
void FindFactors( int number , std::vector<int>& factors )
{
for ( int i = 2; i <= number; ++i )
{
if ( number % i == 0 )
{
factors.push_back( i );
FindFactors( number / i , factors);
break;
}
}
}
int main()
{
std::vector<int> factors;
FindFactors( 121 , factors );
return 0;
}
After you call the function factors will contain only the prime factors.
You should be using std::vector for this. The main problem you have is that a pointer to an array has no way of knowing the number of items the array contains. Concretely, the part where you say sizeof(factors) is wrong. As I understand, you're expecting that to give you the number of items in the array pointed to by factors, but it really gives you the number of bytes needed to store a pointer to int.
You should be either returning a vector<int> or passing it in as a reference and updating it each time you find a factor.