Problems with fold in Rust - fold

Let's assume fn scan(int, int) -> int.
When using
fn count(x: int, y: int) -> int
{
scan(x - 1, y - 1) + scan(x - 1, y) + scan(x - 1, y + 1) + scan(x, y - 1)
+ scan(x, y + 1) + scan(x + 1, y - 1) + scan(x + 1, y) + scan(x + 1, y + 1)
}
I get correct results. I am trying to get the same results by folding the scan function over the given value ranges; however, I can't seem to get it right. My current attempt is
fn count_fold(x: int, y: int) -> int
{
std::iter::range_inclusive(-1, 1).zip(std::iter::range_inclusive(-1, 1)).fold(0, |a, (i, j)| { a + scan(x + i, y + j) })
}
which seems to return only a subset of the correct results. What am I doing wrong? TIA.

When you zip two iterators, you are not creating the "product" of the iterations, like you seem to be wanting to do. Rather, you are iterating both iterators at the same time and creating a pair with the iterated values. So in the count_fold version, the closure will only be called with the following pairs:
(-1, -1)
(0, 0)
(1, 1)
So your count_fold function is actually akin to
scan(x - 1, y - 1) + scan(x, y) + scan(x - 1, y + 1)
I can be wrong, but I don't think there is a function in std that create the product of two iterators.
Moreover, your count method do not use scan(x, y) in the sum, so it is not even really the product of the iterators; you have to be careful about that if you want to create your own product iterator and use it for that purpose.

Cyrille is correct in saying there is no iterator-product function. However, one can manually take the product of the two iterators by folding twice:
use std::iter::range_inclusive;
fn count_fold(x: int, y: int) -> int {
range_inclusive(-1, 1).fold(0,
|a, i| range_inclusive(-1, 1).fold(a, |b, j| b + scan(x + i, y + j)))
}
Although it looks like you'll need to filter the case when i == 0 && j == 0, i.e.
fn count_fold(x: int, y: int) -> int {
range_inclusive(-1, 1).fold(0,
|a, i| range_inclusive(-1, 1).fold(a,
|b, j| if i == 0 && j == 0 {b} else {b + scan(x + i, y + j)}))
}
or
fn count_fold(x: int, y: int) -> int {
range_inclusive(-1, 1).fold(0,
|a, i| range_inclusive(-1, 1)
.filter(|&j| !(i == 0 && j == 0))
.fold(a, |b, j| b + scan(x + i, y + j)))
}
However, I'd almost say that this is clearer imperatively:
fn count_fold(x: int, y: int) -> int {
let mut a = 0;
for i in range_inclusive(-1, 1) {
for j in range_inclusive(-1, 1) {
if !(i == 0 && j == 0) { a += scan(x + i, y + j) }
}
}
a
}

Related

C++ equations, not sure if i wrote it correctly

Are and correctly written?
if(x>0.1){z = 2*n+x}
if else(x <= 0.1){x(pow,n) - 1 / sqrt(n(pow,2)) + x(pow,2) * n}
x = n / 2 * n(pow,2) + 3 * n - 2
x is correctly translated to C++.
As for the rest of the code, it would be:
if(x > 0.1) {
z = 2 * n + x;
}
else {
z = pow(x, n) - 1 / sqrt(pow(n, 2) + pow(x, 2) * n);
}

How should I go about solving this recursion without trial and error

int sum_down(int x)
{
if (x >= 0)
{
x = x - 1;
int y = x + sum_down(x);
return y + sum_down(x);
}
else
{
return 1;
}
}
What is this smallest integer value of the parameter x, so that the returned value is greater than 1.000.000 ?
Right now I am just doing it by trial and error and since this question is asked via a paper format. I don't think I will have enough time to do trial and error. Question is, how do you guys visualise this quickly such that it can be solved easily. Thanks guys and I am new to programming so thanks in advance!
The recursion logic:
x = x - 1;
int y = x + sum_down(x);
return y + sum_down(x);
can be simplified to:
x = x - 1;
int y = x + sum_down(x) + sum_down(x);
return y;
which can be simplified to:
int y = (x-1) + sum_down(x-1) + sum_down(x-1);
return y;
which can be simplified to:
return (x-1) + 2*sum_down(x-1);
Put in mathematical form,
f(N) = (N-1) + 2*f(N-1)
with the recursion terminating when N is -1. f(-1) = 1.
Hence,
f(0) = -1 + 2*1 = 1
f(1) = 0 + 2*1 = 2
f(2) = 1 + 2*2 = 5
...
f(18) = 17 + 2*f(17) = 524269
f(19) = 18 + 2*524269 = 1048556
Your program can be written this way (sorry about c#):
public static void Main()
{
int i = 0;
int j = 0;
do
{
i++;
j = sum_down(i);
Console.Out.WriteLine("j:" + j);
} while (j < 1000000);
Console.Out.WriteLine("i:" + i);
}
static int sum_down(int x)
{
if (x >= 0)
{
return x - 1 + 2 * sum_down(x - 1);
}
else
{
return 1;
}
}
So at first iteration you'll get 2, then 5, then 12... So you can neglect the x-1 part since it'll stay little compared to the multiplication.
So we have:
i = 1 => sum_down ~= 4 (real is 2)
i = 2 => sum_down ~= 8 (real is 5)
i = 3 => sum_down ~= 16 (real is 12)
i = 4 => sum_down ~= 32 (real is 27)
i = 5 => sum_down ~= 64 (real is 58)
So we can say that sum_down(x) ~= 2^x+1. Then it's just basic math with 2^x+1 < 1 000 000 which is 19.
A bit late, but it's not that hard to get an exact non-recursive formula.
Write it up mathematically, as explained in other answers already:
f(-1) = 1
f(x) = 2*f(x-1) + x-1
This is the same as
f(-1) = 1
f(x+1) = 2*f(x) + x
(just switched from x and x-1 to x+1 and x, difference 1 in both cases)
The first few x and f(x) are:
x: -1 0 1 2 3 4
f(x): 1 1 2 5 12 27
And while there are many arbitrary complicated ways to transform this into a non-recursive formula, with easy ones it often helps to write up what the difference is between each two elements:
x: -1 0 1 2 3 4
f(x): 1 1 2 5 12 27
0 1 3 7 15
So, for some x
f(x+1) - f(x) = 2^(x+1) - 1
f(x+2) - f(x) = (f(x+2) - f(x+1)) + (f(x+1) - f(x)) = 2^(x+2) + 2^(x+1) - 2
f(x+n) - f(x) = sum[0<=i<n](2^(x+1+i)) - n
With eg. a x=0 inserted, to make f(x+n) to f(n):
f(x+n) - f(x) = sum[0<=i<n](2^(x+1+i)) - n
f(0+n) - f(0) = sum[0<=i<n](2^(0+1+i)) - n
f(n) - 1 = sum[0<=i<n](2^(i+1)) - n
f(n) = sum[0<=i<n](2^(i+1)) - n + 1
f(n) = sum[0<i<=n](2^i) - n + 1
f(n) = (2^(n+1) - 2) - n + 1
f(n) = 2^(n+1) - n - 1
No recursion anymore.
How about this :
int x = 0;
while (sum_down(x) <= 1000000)
{
x++;
}
The loop increments x until the result of sum_down(x) is superior to 1.000.000.
Edit : The result would be 19.
While trying to understand and simplify the recursion logic behind the sum_down() function is enlightening and informative, this snippet tend to be logical and pragmatic in that it does not try and solve the problem in terms of context, but in terms of results.
Two lines of Python code to answer your question:
>>> from itertools import * # no code but needed for dropwhile() and count()
Define the recursive function (See R Sahu's answer)
>>> f = lambda x: 1 if x<0 else (x-1) + 2*f(x-1)
Then use the dropwhile() function to remove elements from the list [0, 1, 2, 3, ....] for which f(x)<=1000000, resulting in a list of integers for which f(x) > 1000000. Note: count() returns an infinite "list" of [0, 1, 2, ....]
The dropwhile() function returns a Python generator so we use next() to get the first value of the list:
>>> next(dropwhile(lambda x: f(x)<=1000000, count()))
19

shell sort sequence implementation in C++

I am reading about shell sort in Algorithms in C++ by Robert Sedwick.
Here outer loop to change the increments leads to this compact shellsort implementation, which uses the increment sequence 1 4 13 40 121 364 1093 3280 9841 . . . .
template <class Item>
void shellsort(Item a[], int l, int r)
{
int h;
for (h = 1; h <= (r - l) / 9; h = 3 * h + 1);
for (; h > 0; h = h / 3)
{
for (int i = l + h; i <= r; i++)
{
int j = i; Item v = a[i];
while (j >= l + h && v < a[j - h])
{
a[j] = a[j - h]; j -= h;
}
a[j] = v;
}
}
}
My question under what basis author is checking for condition h <= (r-l)/9, and why author is dividing by 9.
The loop:
for (h = 1; h <= (r - l) / 9; h = 3 * h + 1);
calculates the initial value of h. This value must be smaller than the range it will be used in:
h <= (r - l)
Everytime this condition passes, h gets updated to 3 * h + 1, which means that even though h is smaller than (r-l), the updated value might be larger. To prevent this, we could check if the next value of h would surpass the largest index:
(h * 3) + 1 <= (r - l)
This will make sure h is smaller than range of the array.
For example: say we have an array of size 42, which means indices go from 0 to 41. Using the condition as described above:
h = 1, is (3 * 1 + 1) <= (41 - 0) ? yes! -> update h to 4
h = 4, is (3 * 4 + 1) <= (41 - 0) ? yes! -> update h to 13
h = 13, is (3 * 13 + 1) <= (41 - 0) ? yes! -> update h to 40
h = 40, is (3 * 40 + 1) <= (41 - 0) ? no! => h will begin at 40
This means our initial h is 40, because h is only marginally smaller than the range of the array, very little work will be done, the algorithm will only check the following:
Does array[40] needs to be swapped with array[0] ?
Does array[41] needs to be swapped with array[1] ?
This is a bit useless, the first iteration only performs two checks. A smaller initial value of h means more work will get done in the first iteration.
Using:
h <= (r - l) / 9
ensures the initial value of h to be sufficiently small to allow the first iteration to do useful work. As an extra advantage, it also looks cleaner than the previous condition.
You could replace 9 by any value greater than 3. Why greater than 3? To ensure (h * 3) + 1 <= (r - l) is still true!
But do remember to not make the initial h too small: Shell Sort is based on Insertion Sort, which only performs well on small or nearly sorted arrays. Personally, I would not exceed h <= (r - l) / 15.

C++ Points of Vertices in Cuboid (Bitwise AND)

I'm trying to calculate the points in a cuboid given its centre (which is a Vector3) and the lengths of the sides along the x, y and z axis. I found the following on math.stackexchange.com: https://math.stackexchange.com/questions/107778/simplest-equation-for-drawing-a-cube-based-on-its-center-and-or-other-vertices which says I can use the following formulae:
The constructor for the World class is:
World::World(Vector3 o, float d1, float d2, float d3) : origin(o)
{
// If we consider an edge length to be d, we need to find r such that
// 2r = d in order to calculate the positions of each vertex in the world.
float r1 = d1 / 2,
r2 = d2 / 2,
r3 = d3 / 2;
for (int i = 0; i < 8; i++)
{
/* Sets up the vertices of the cube.
*
* #see http://bit.ly/1cc2RPG
*/
float x = o.getX() + (std::pow(-1, i&1) * r1),
y = o.getY() + (std::pow(-1, i&2) * r2),
z = o.getZ() + (std::pow(-1, i&4) * r3);
points[i] = Vector3(x, y, z);
std::cout << points[i] << "\n";
}
}
And I passing the following parameters to the constructor:
Vector3 o(0, 0, 0);
World w(o, 100.f, 100.f, 100.f);
The coordinates being output for all 8 vertices are:
(50, 50, 50)
(-50, 50, 50)
(50, 50, 50)
(-50, 50, 50)
(50, 50, 50)
(-50, 50, 50)
(50, 50, 50)
(-50, 50, 50)
Which cannot be correct. Any guidance would be very much appreciated!
The problem lies in the bitwise & inside your pow calls:
In the y and z components, they always return 0 and 2 or 4, respectively. -1^2 = -1^4 = 1, which is why the sign of these components is always positive. You could try (i&2)!=0 or (i&2) >> 1 for the y component instead. The same goes for the z component.
Change this:
float x = o.getX() + (std::pow(-1, i&1) * r1),
y = o.getY() + (std::pow(-1, i&2) * r2),
z = o.getZ() + (std::pow(-1, i&4) * r3);
To this:
float x = o.getX() + (std::pow(-1, (i ) & 1) * r1), // pow(-1, 0) == 1, pow(-1, 1) == -1
y = o.getY() + (std::pow(-1, (i >> 1) & 1) * r2), // pow(-1, 0) == 1, pow(-1, 1) == -1
z = o.getZ() + (std::pow(-1, (i >> 2) & 1) * r3); // pow(-1, 0) == 1, pow(-1, 1) == -1
Or even to this:
float x = o.getX() + (std::pow(-1, (i )) * r1), // pow(-1, {0, 2, 4, 6}) == 1, pow(-1, {1, 3, 5, 7}) == -1
y = o.getY() + (std::pow(-1, (i >> 1)) * r2), // pow(-1, {0, 2}) == 1, pow(-1, {1, 3}) == -1
z = o.getZ() + (std::pow(-1, (i >> 2)) * r3); // pow(-1, 0) == 1, pow(-1, 1) == -1
The problem is that as written even though the values you mask out identify weather or not the lengths need to be negated. They are not in the correct place value to get the desired properties from the exponentiation of -1.
Rewriting the code as I have above will solve this issue, however it would be more readable and in general more permanent just to unroll the loop and manually write if each one is an addition or subtraction without using the pow function.

Finding Neighbourhood in Matrices [duplicate]

This question already has an answer here:
Conway's Game of Life, counting neighbors [closed]
(1 answer)
Closed 9 years ago.
I am working on project containing cellular automat methods. What I am trying to figure is how to write function helping to find all the neighbours in a 2d array.
for example i ve got size x size 2d array [size = 4 here]
[x][n][ ][n]
[n][n][ ][n]
[ ][ ][ ][ ]
[n][n][ ][n]
Field marked as x [0,0 index] has neighbours marked as [n] -> 8 neighbours. What Im trying to do is to write a function which can find neighbours wo writting tousands of if statements
Does anybody have an idea how to do it ?
thanks
For the neighbours of element (i,j) in NxM matrix:
int above = (i-1) % N;
int below = (i+1) % N;
int left = (j-1) % M;
int right = (j+1) % M;
decltype(matrix[0][0]) *indices[8];
indices[0] = & matrix[above][left];
indices[1] = & matrix[above][j];
indices[2] = & matrix[above][right];
indices[3] = & matrix[i][left];
// Skip matrix[i][j]
indices[4] = & matrix[i][right];
indices[5] = & matrix[below][left];
indices[6] = & matrix[below][j];
indices[7] = & matrix[below][right];
Suppose you are in cell (i, j). Then, on an infinite grid, your neighbors should be [(i-1, j-1), (i-1,j), (i-1, j+1), (i, j-1), (i, j+1), (i+1, j-1), (i+1, j), (i+1, j+1)].
However, since the grid is finite some of the above values will get outside the bounds. But we know modular arithmetic: 4 % 3 = 1 and -1 % 3 = 2. So, if the grid is of size n, m you only need to apply %n, %m on the above list to get the proper list of neighbors: [((i-1) % n, (j-1) % m), ((i-1) % n,j), ((i-1) % n, (j+1) % m), (i, (j-1) % m), (i, (j+1) % m), ((i+1) % n, (j-1) % m), ((i+1) % n, j), ((i+1) % n, (j+1) % m)]
That works if your coordinates are between 0 and n and between 0 and m. If you start with 1 then you need to tweak the above by doing a -1 and a +1 somewhere.
For your case n=m=4 and (i, j) = (0, 0). The first list is [(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 1), (1, -1), (1, 0), (1, 1)]. Applying the modulus operations you get to [(3, 3), (3, 0), (3, 1), (0, 3), (0, 1), (1, 3), (1, 0), (1, 1)] which are exactly the squares marked [n] in your picture.
Add and subtract one from the coordinates, in all possible permutations. Results outside the boundaries wrap around (e.g. -1 becomes 3 and 4 becomes 0). Just a couple of simple loops needed basically.
Something like
// Find the closest neighbours (one step) from the coordinates [x,y]
// The max coordinates is max_x,max_y
// Note: Does not contain any error checking (for valid coordinates)
std::vector<std::pair<int, int>> getNeighbours(int x, int y, int max_x, int max_y)
{
std::vector<std::pair<int, int>> neighbours;
for (int dx = -1; dx <= 1; ++dx)
{
for (int dy = -1; dy <= 1; ++dy)
{
// Skip the coordinates [x,y]
if (dx == 0 && dy == 0)
continue;
int nx = x + dx;
int ny = y + dy;
// If the new coordinates goes out of bounds, wrap them around
if (nx < 0)
nx = max_x;
else if (nx > max_x)
nx = 0;
if (ny < 0)
ny = max_y;
else if (ny > max_y)
ny = 0;
// Add neighbouring coordinates to result
neighbours.push_back(std::make_pair(nx, ny));
}
}
return neighbours;
}
Example use for you:
auto n = getNeighbours(0, 0, 3, 3);
for (const auto& p : n)
std::cout << '[' << p.first << ',' << p.second << "]\n";
Prints out
[3,3]
[3,0]
[3,1]
[0,3]
[0,1]
[1,3]
[1,0]
[1,1]
which is the correct answer.