c# select smaller 3d array from larger 3d array - c#-4.0

I have an array (object[,,]), let us assume for arguments sake the array is x1000, y1000,z*1000 and represents a matrix of points on an x,y,z plane.
at position: x50,y10,z199, for example I want to extract another object[,,] containing a smaller cube, say a submatrix of 100 cubed(or whatever is available, nulls if empty?) from the parent array using the reference point as a centerpoint, is this possible, I was hoping I could do it in linq but got hopelessly lost.. how would you/ should I tackle this.. one though was to do the following:
1.Create a new 3d array with the size of the amount of items i want to retrieve (xyz).
2.Iterate over each axis (x, y, z).
3.Copy the value from the source array to the target array (offsetX+x, offsetY+y, offsetZ+z).
4.Return new array.
but if this is being called a lot, I see it being quite a bottleneck, ideas anyone?

Depending on your usage of the smaller array, this may or may not suit your needs.
To represent a sub section (chunk) of array, instead of creating a new array or doing any copying, you could write your own class which serves as a view to that chunk of array.
Note that this example has the following properties:
No guard clauses in constructor
Chunk is always cube (x, y, z length are equal)
Chunk length is always odd (since we are expanding out from point of reference)
public class ArrayChunk<T>
{
// Array this chunk is from.
private readonly T[,,] _parentArray;
// Point of reference.
private readonly int _x, _y, _z;
// How many elements to move outwards in each direction from point of reference.
private readonly int _numToExpand;
public ArrayChunk(T[,,] parentArray, int x, int y, int z, int numToExpand)
{
_parentArray = parentArray;
_x = x;
_y = y;
_z = z;
_numToExpand = numToExpand;
}
public int Length => _numToExpand*2 + 1;
public T this[int x, int y, int z]
{
get
{
// Make sure index is within chunk range.
EnsureInChunkRange(x, y, z);
// Map chunk index to parent array index.
int parentX = MapToParent(_x, x),
parentY = MapToParent(_y, y),
parentZ = MapToParent(_z, z);
// If parent array index is in parent array range, return element from parent array.
if (IsInRangeOfParent(parentX, parentY, parentZ))
return _parentArray[parentX, parentY, parentZ];
// Otherwise return default element for type T.
return default(T);
}
set
{
EnsureInChunkRange(x, y, z);
int parentX = MapToParent(_x, x),
parentY = MapToParent(_y, y),
parentZ = MapToParent(_z, z);
if (IsInRangeOfParent(parentX, parentY, parentZ))
_parentArray[parentX, parentY, parentZ] = value;
else
throw new InvalidOperationException();
}
}
private void EnsureInChunkRange(int x, int y, int z)
{
if (x < 0 || y < 0 || z < 0 ||
x >= Length || y >= Length || z >= Length)
{
throw new IndexOutOfRangeException();
}
}
private int MapToParent(int referenceIndex, int index)
{
return referenceIndex - _numToExpand + index;
}
private bool IsInRangeOfParent(int parentX, int parentY, int parentZ)
{
return
parentX >= 0 &&
parentY >= 0 &&
parentZ >= 0 &&
parentX < _parentArray.GetLength(0) &&
parentY < _parentArray.GetLength(1) &&
parentZ < _parentArray.GetLength(2);
}
}
To easily get chunk from array, you could declare an extension method:
public static class ArrayChunkExtensions
{
public static ArrayChunk<T> GetChunk<T>(this T[,,] array, int x, int y, int z, int numToExpand)
{
return new ArrayChunk<T>(array, x, y, z, numToExpand);
}
}
Here's a sample usage:
Action<int, Action<int, int, int>> iterate = (length, action) =>
{
for (int x = 0; x < length; x++)
for (int y = 0; y < length; y++)
for (int z = 0; z < length; z++)
action(x, y, z);
};
// Create 5x5x5 parent array.
const int size = 5;
var array = new string[size, size, size];
iterate(size, (x, y, z) => array[x, y, z] = $"x:{x} y:{y} z:{z}");
// Take 3x3x3 chunk from parent array center.
const int indexOfReference = 2;
const int numToExpand = 1;
ArrayChunk<string> chunk = array.GetChunk(indexOfReference, indexOfReference, indexOfReference, numToExpand);
iterate(chunk.Length, (x, y, z) => Console.WriteLine(chunk[x, y, z]));

Related

checking edge for circular movement

I want to make my dot program turn around when they reach edge
so basically i just simply calculate
x = width/2+cos(a)*20;
y = height/2+sin(a)*20;
it's make circular movement. so i want to make this turn around by checking the edge. i also already make sure that y reach the if condition using println command
class particles {
float x, y, a, r, cosx, siny;
particles() {
x = width/2; y = height/2; a = 0; r = 20;
}
void display() {
ellipse(x, y, 20, 20);
}
void explode() {
a = a + 0.1;
cosx = cos(a)*r;
siny = sin(a)*r;
x = x + cosx;
y = y + siny;
}
void edge() {
if (x>width||x<0) cosx*=-1;
if (y>height||y<0) siny*=-1;
}
}
//setup() and draw() function
particles part;
void setup(){
size (600,400);
part = new particles();
}
void draw(){
background(40);
part.display();
part.explode();
part.edge();
}
they just ignore the if condition
There is no problem with your check, the problem is with the fact that presumably the very next time through draw() you ignore what you did in response to the check by resetting the values of cosx and siny.
I recommend creating two new variables, dx and dy ("d" for "direction") which will always be either +1 and -1 and change these variables in response to your edge check. Here is a minimal example:
float a,x,y,cosx,siny;
float dx,dy;
void setup(){
size(400,400);
background(0);
stroke(255);
noFill();
x = width/2;
y = height/2;
dx = 1;
dy = 1;
a = 0;
}
void draw(){
ellipse(x,y,10,10);
cosx = dx*20*cos(a);
siny = dy*20*sin(a);
a += 0.1;
x += cosx;
y += siny;
if (x > width || x < 0)
dx = -1*dx;
if (y > height || y < 0)
dy = -1*dy;
}
When you run this code you will observe the circles bouncing off the edges:

NxN matrix is given and we have to find

N things to select for N people, you were given a NxN matrix and cost at each element, you needed to find the one combination with max total weight, such that each person gets exactly one thing.
I found difficulty in making its dp state.
please help me and if possible then also write code for it
C++ style code:
double max_rec(int n, int r, int* c, double** m, bool* f)
{
if (r < n)
{
double max_v = 0.0;
int max_i = -1;
for (int i = 0; i < n; i++)
{
if (f[i] == false)
{
f[i] = true;
double value = m[r][i] + max_rec(n, r + 1, c, m, f);
if (value > max_v)
{
max_v = value;
max_i = i;
}
f[i] = false;
}
}
c[i] = max_i;
return max_v;
}
return 0.0;
}
int* max_comb(int n, double** m)
{
bool* f = new bool[n];
int* c = new int[n];
max_rec(n, 0, c, m, f);
delete [] f;
return c;
}
Call max_comb with N and your NxN matrix (2d array). Returns the column indices of the maximum combination.
Time complexity: O(N!)
I know this is bad but the problem does not have a greedy structure.
And as #mszalbach said, try to attempt the problem yourself before asking.
EDIT: can reduce to polynomial time by memoizing.

MPI initialize array only on root

I have a working Wafefront program using MPI express. What happens in this program is that for a matrix of n x m there are n processes. Each process is assigned a row. Each process does the following:
for column = 0 to matrix_width do:
1) x = get the value of this column from the row above (rank - 1 process)
2) y = Get the value left of us (our row, column-1)
3) Add to our current column value: (x + y)
So on the master process I will declare an array of n x m elements. Each slave process should thus allocate an array of length m. But as it stands in my solution each process has to allocate an array of n x m for the scatter operation to work, otherwise I get a nullpointer (if I assign it null) or an out of bounds exception (if I instantiate it with new int[1]). I'm sure there has to be a solution to this, otherwise each process would require as much memory as the root.
I think I need something like allocatable in C.
Below the important part is the one marked "MASTER". Normally I would pull the initialization into the if(rank == 0) test and initialize the array with null (not allocating the memory) in the else branch but that does not work.
package be.ac.vub.ir.mpi;
import mpi.MPI;
// Execute: mpjrun.sh -np 2 -jar parsym-java.jar
/**
* Parallel and sequential implementation of a prime number counter
*/
public class WaveFront
{
// Default program parameters
final static int size = 4;
private static int rank;
private static int world_size;
private static void log(String message)
{
if (rank == 0)
System.out.println(message);
}
////////////////////////////////////////////////////////////////////////////
//// MAIN //////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////
public static void main(String[] args) throws InterruptedException
{
// MPI variables
int[] matrix; // matrix stored at process 0
int[] row; // each process keeps its row
int[] receiveBuffer; // to receive a value from ``row - 1''
int[] sendBuffer; // to send a value to ``row + 1''
/////////////////
/// INIT ////////
/////////////////
MPI.Init(args);
rank = MPI.COMM_WORLD.Rank();
world_size = MPI.COMM_WORLD.Size();
/////////////////
/// ALL PCS /////
/////////////////
// initialize data structures
receiveBuffer = new int[1];
sendBuffer = new int[1];
row = new int[size];
/////////////////
/// MASTER //////
/////////////////
matrix = new int[size * size];
if (rank == 0)
{
// Initialize matrix
for (int idx = 0; idx < size * size; idx++)
matrix[idx] = 0;
matrix[0] = 1;
receiveBuffer[0] = 0;
}
/////////////////
/// PROGRAM /////
/////////////////
// distribute the rows of the matrix to the appropriate processes
int startOfRow = rank * size;
MPI.COMM_WORLD.Scatter(matrix, startOfRow, size, MPI.INT, row, 0, size, MPI.INT, 0);
// For each column each process will calculate it's new values.
for (int col_idx = 0; col_idx < size; col_idx++)
{
// Get Y from row above us (rank - 1).
if (rank > 0)
MPI.COMM_WORLD.Recv(receiveBuffer, 0, 1, MPI.INT, rank - 1, 0);
// Get the X value (left from current column).
int x = col_idx == 0 ? 0 : row[col_idx - 1];
// Assign the new Z value.
row[col_idx] = row[col_idx] + x + receiveBuffer[0];
// Wait for other process to ask us for this value.
sendBuffer[0] = row[col_idx];
if (rank + 1 < size)
MPI.COMM_WORLD.Send(sendBuffer, 0, 1, MPI.INT, rank + 1, 0);
}
// At this point each process should be done so we call gather.
MPI.COMM_WORLD.Gather(row, 0, size, MPI.INT, matrix, startOfRow, size, MPI.INT, 0);
// Let the master show the result.
if (rank == 0)
for (int row_idx = 0; row_idx < size; ++row_idx)
{
for (int col_idx = 0; col_idx < size; ++col_idx)
System.out.print(matrix[size * row_idx + col_idx] + " ");
System.out.println();
}
MPI.Finalize(); // Don't forget!!
}
}

How do you flatten image coordinates into a 1D array?

My source code is from Heterogeneous Computing with OpenCL Chapter 4 Basic OpenCL Examples > Image Rotation. The book leaves out several critical details.
My major problem is that I don't know how to initialize the array that I supply to their kernel (they don't tell you how). What I have is:
int W = inImage.width();
int H = inImage.height();
float *myImage = new float[W*H];
for(int row = 0; row < H; row++)
for(int col = 0; col < W; col++)
myImage[row*W+col] = col;
which I supply to this kernel:
__kernel void img_rotate(__global float* dest_data, __global float* src_data, int W, int H, float sinTheta, float cosTheta)
{
const int ix = get_global_id(0);
const int iy = get_global_id(1);
float x0 = W/2.0f;
float y0 = H/2.0f;
float xoff = ix-x0;
float yoff = iy-y0;
int xpos = (int)(xoff*cosTheta + yoff*sinTheta + x0);
int ypos = (int)(yoff*cosTheta - xoff*sinTheta + y0);
if(((int)xpos>=0) && ((int)xpos < W) && ((int)ypos>=0) && ((int)ypos<H))
{
dest_data[iy*W+ix] = src_data[ypos*W+xpos];
//dest_data[iy*W+ix] = src_data[iy*W+ix];
}
}
I'm having trouble finding the right value for theta too. An integer would be an appropriate value for theta, right?
float theta = 45; // 45 degrees, right?
float cos_theta = cos(theta);
float sin_theta = sin(theta);
When writing my OpenCL code, I always treat each kernel as reading a 3D set of data, regardless if the data is 1D, 2D, or 3D:
__kernel void TestKernel(__global float *Data){
k = get_global_id(0); //also z
j = get_global_id(1); //also y
i = get_global_id(2); //also x
//Convert 3D to 1D
int linear_coord = i + get_global_size(0)*j + get_global_size(0)*get_global_size(1)*k;
//do stuff
}
When doing the clEnqueueNDKernelRange(...), just set the dimension to be:
int X = 500;
int Y = 300;
int Z = 1;
size_t GlobalDim = {Z, Y, X};
This let's all of my kernels work easily in all dimensions.

Generate list of all possible permutations of a string

How would I go about generating a list of all possible permutations of a string between x and y characters in length, containing a variable list of characters.
Any language would work, but it should be portable.
There are several ways to do this. Common methods use recursion, memoization, or dynamic programming. The basic idea is that you produce a list of all strings of length 1, then in each iteration, for all strings produced in the last iteration, add that string concatenated with each character in the string individually. (the variable index in the code below keeps track of the start of the last and the next iteration)
Some pseudocode:
list = originalString.split('')
index = (0,0)
list = [""]
for iteration n in 1 to y:
index = (index[1], len(list))
for string s in list.subset(index[0] to end):
for character c in originalString:
list.add(s + c)
you'd then need to remove all strings less than x in length, they'll be the first (x-1) * len(originalString) entries in the list.
It's better to use backtracking
#include <stdio.h>
#include <string.h>
void swap(char *a, char *b) {
char temp;
temp = *a;
*a = *b;
*b = temp;
}
void print(char *a, int i, int n) {
int j;
if(i == n) {
printf("%s\n", a);
} else {
for(j = i; j <= n; j++) {
swap(a + i, a + j);
print(a, i + 1, n);
swap(a + i, a + j);
}
}
}
int main(void) {
char a[100];
gets(a);
print(a, 0, strlen(a) - 1);
return 0;
}
You are going to get a lot of strings, that's for sure...
Where x and y is how you define them and r is the number of characters we are selecting from --if I am understanding you correctly. You should definitely generate these as needed and not get sloppy and say, generate a powerset and then filter the length of strings.
The following definitely isn't the best way to generate these, but it's an interesting aside, none-the-less.
Knuth (volume 4, fascicle 2, 7.2.1.3) tells us that (s,t)-combination is equivalent to s+1 things taken t at a time with repetition -- an (s,t)-combination is notation used by Knuth that is equal to . We can figure this out by first generating each (s,t)-combination in binary form (so, of length (s+t)) and counting the number of 0's to the left of each 1.
10001000011101 --> becomes the permutation: {0, 3, 4, 4, 4, 1}
Non recursive solution according to Knuth, Python example:
def nextPermutation(perm):
k0 = None
for i in range(len(perm)-1):
if perm[i]<perm[i+1]:
k0=i
if k0 == None:
return None
l0 = k0+1
for i in range(k0+1, len(perm)):
if perm[k0] < perm[i]:
l0 = i
perm[k0], perm[l0] = perm[l0], perm[k0]
perm[k0+1:] = reversed(perm[k0+1:])
return perm
perm=list("12345")
while perm:
print perm
perm = nextPermutation(perm)
You might look at "Efficiently Enumerating the Subsets of a Set", which describes an algorithm to do part of what you want - quickly generate all subsets of N characters from length x to y. It contains an implementation in C.
For each subset, you'd still have to generate all the permutations. For instance if you wanted 3 characters from "abcde", this algorithm would give you "abc","abd", "abe"...
but you'd have to permute each one to get "acb", "bac", "bca", etc.
Some working Java code based on Sarp's answer:
public class permute {
static void permute(int level, String permuted,
boolean used[], String original) {
int length = original.length();
if (level == length) {
System.out.println(permuted);
} else {
for (int i = 0; i < length; i++) {
if (!used[i]) {
used[i] = true;
permute(level + 1, permuted + original.charAt(i),
used, original);
used[i] = false;
}
}
}
}
public static void main(String[] args) {
String s = "hello";
boolean used[] = {false, false, false, false, false};
permute(0, "", used, s);
}
}
Here is a simple solution in C#.
It generates only the distinct permutations of a given string.
static public IEnumerable<string> permute(string word)
{
if (word.Length > 1)
{
char character = word[0];
foreach (string subPermute in permute(word.Substring(1)))
{
for (int index = 0; index <= subPermute.Length; index++)
{
string pre = subPermute.Substring(0, index);
string post = subPermute.Substring(index);
if (post.Contains(character))
continue;
yield return pre + character + post;
}
}
}
else
{
yield return word;
}
}
There are a lot of good answers here. I also suggest a very simple recursive solution in C++.
#include <string>
#include <iostream>
template<typename Consume>
void permutations(std::string s, Consume consume, std::size_t start = 0) {
if (start == s.length()) consume(s);
for (std::size_t i = start; i < s.length(); i++) {
std::swap(s[start], s[i]);
permutations(s, consume, start + 1);
}
}
int main(void) {
std::string s = "abcd";
permutations(s, [](std::string s) {
std::cout << s << std::endl;
});
}
Note: strings with repeated characters will not produce unique permutations.
I just whipped this up quick in Ruby:
def perms(x, y, possible_characters)
all = [""]
current_array = all.clone
1.upto(y) { |iteration|
next_array = []
current_array.each { |string|
possible_characters.each { |c|
value = string + c
next_array.insert next_array.length, value
all.insert all.length, value
}
}
current_array = next_array
}
all.delete_if { |string| string.length < x }
end
You might look into language API for built in permutation type functions, and you might be able to write more optimized code, but if the numbers are all that high, I'm not sure there is much of a way around having a lot of results.
Anyways, the idea behind the code is start with string of length 0, then keep track of all the strings of length Z where Z is the current size in the iteration. Then, go through each string and append each character onto each string. Finally at the end, remove any that were below the x threshold and return the result.
I didn't test it with potentially meaningless input (null character list, weird values of x and y, etc).
This is a translation of Mike's Ruby version, into Common Lisp:
(defun perms (x y original-string)
(loop with all = (list "")
with current-array = (list "")
for iteration from 1 to y
do (loop with next-array = nil
for string in current-array
do (loop for c across original-string
for value = (concatenate 'string string (string c))
do (push value next-array)
(push value all))
(setf current-array (reverse next-array)))
finally (return (nreverse (delete-if #'(lambda (el) (< (length el) x)) all)))))
And another version, slightly shorter and using more loop facility features:
(defun perms (x y original-string)
(loop repeat y
collect (loop for string in (or (car (last sets)) (list ""))
append (loop for c across original-string
collect (concatenate 'string string (string c)))) into sets
finally (return (loop for set in sets
append (loop for el in set when (>= (length el) x) collect el)))))
Here is a simple word C# recursive solution:
Method:
public ArrayList CalculateWordPermutations(string[] letters, ArrayList words, int index)
{
bool finished = true;
ArrayList newWords = new ArrayList();
if (words.Count == 0)
{
foreach (string letter in letters)
{
words.Add(letter);
}
}
for(int j=index; j<words.Count; j++)
{
string word = (string)words[j];
for(int i =0; i<letters.Length; i++)
{
if(!word.Contains(letters[i]))
{
finished = false;
string newWord = (string)word.Clone();
newWord += letters[i];
newWords.Add(newWord);
}
}
}
foreach (string newWord in newWords)
{
words.Add(newWord);
}
if(finished == false)
{
CalculateWordPermutations(letters, words, words.Count - newWords.Count);
}
return words;
}
Calling:
string[] letters = new string[]{"a","b","c"};
ArrayList words = CalculateWordPermutations(letters, new ArrayList(), 0);
... and here is the C version:
void permute(const char *s, char *out, int *used, int len, int lev)
{
if (len == lev) {
out[lev] = '\0';
puts(out);
return;
}
int i;
for (i = 0; i < len; ++i) {
if (! used[i])
continue;
used[i] = 1;
out[lev] = s[i];
permute(s, out, used, len, lev + 1);
used[i] = 0;
}
return;
}
permute (ABC) -> A.perm(BC) -> A.perm[B.perm(C)] -> A.perm[(*BC), (CB*)] -> [(*ABC), (BAC), (BCA*), (*ACB), (CAB), (CBA*)]
To remove duplicates when inserting each alphabet check to see if previous string ends with the same alphabet (why? -exercise)
public static void main(String[] args) {
for (String str : permStr("ABBB")){
System.out.println(str);
}
}
static Vector<String> permStr(String str){
if (str.length() == 1){
Vector<String> ret = new Vector<String>();
ret.add(str);
return ret;
}
char start = str.charAt(0);
Vector<String> endStrs = permStr(str.substring(1));
Vector<String> newEndStrs = new Vector<String>();
for (String endStr : endStrs){
for (int j = 0; j <= endStr.length(); j++){
if (endStr.substring(0, j).endsWith(String.valueOf(start)))
break;
newEndStrs.add(endStr.substring(0, j) + String.valueOf(start) + endStr.substring(j));
}
}
return newEndStrs;
}
Prints all permutations sans duplicates
Recursive solution in C++
int main (int argc, char * const argv[]) {
string s = "sarp";
bool used [4];
permute(0, "", used, s);
}
void permute(int level, string permuted, bool used [], string &original) {
int length = original.length();
if(level == length) { // permutation complete, display
cout << permuted << endl;
} else {
for(int i=0; i<length; i++) { // try to add an unused character
if(!used[i]) {
used[i] = true;
permute(level+1, original[i] + permuted, used, original); // find the permutations starting with this string
used[i] = false;
}
}
}
In Perl, if you want to restrict yourself to the lowercase alphabet, you can do this:
my #result = ("a" .. "zzzz");
This gives all possible strings between 1 and 4 characters using lowercase characters. For uppercase, change "a" to "A" and "zzzz" to "ZZZZ".
For mixed-case it gets much harder, and probably not doable with one of Perl's builtin operators like that.
Ruby answer that works:
class String
def each_char_with_index
0.upto(size - 1) do |index|
yield(self[index..index], index)
end
end
def remove_char_at(index)
return self[1..-1] if index == 0
self[0..(index-1)] + self[(index+1)..-1]
end
end
def permute(str, prefix = '')
if str.size == 0
puts prefix
return
end
str.each_char_with_index do |char, index|
permute(str.remove_char_at(index), prefix + char)
end
end
# example
# permute("abc")
The following Java recursion prints all permutations of a given string:
//call it as permut("",str);
public void permut(String str1,String str2){
if(str2.length() != 0){
char ch = str2.charAt(0);
for(int i = 0; i <= str1.length();i++)
permut(str1.substring(0,i) + ch + str1.substring(i,str1.length()),
str2.substring(1,str2.length()));
}else{
System.out.println(str1);
}
}
Following is the updated version of above "permut" method which makes n! (n factorial) less recursive calls compared to the above method
//call it as permut("",str);
public void permut(String str1,String str2){
if(str2.length() > 1){
char ch = str2.charAt(0);
for(int i = 0; i <= str1.length();i++)
permut(str1.substring(0,i) + ch + str1.substring(i,str1.length()),
str2.substring(1,str2.length()));
}else{
char ch = str2.charAt(0);
for(int i = 0; i <= str1.length();i++)
System.out.println(str1.substring(0,i) + ch + str1.substring(i,str1.length()),
str2.substring(1,str2.length()));
}
}
import java.util.*;
public class all_subsets {
public static void main(String[] args) {
String a = "abcd";
for(String s: all_perm(a)) {
System.out.println(s);
}
}
public static Set<String> concat(String c, Set<String> lst) {
HashSet<String> ret_set = new HashSet<String>();
for(String s: lst) {
ret_set.add(c+s);
}
return ret_set;
}
public static HashSet<String> all_perm(String a) {
HashSet<String> set = new HashSet<String>();
if(a.length() == 1) {
set.add(a);
} else {
for(int i=0; i<a.length(); i++) {
set.addAll(concat(a.charAt(i)+"", all_perm(a.substring(0, i)+a.substring(i+1, a.length()))));
}
}
return set;
}
}
I'm not sure why you would want to do this in the first place. The resulting set for any moderately large values of x and y will be huge, and will grow exponentially as x and/or y get bigger.
Lets say your set of possible characters is the 26 lowercase letters of the alphabet, and you ask your application to generate all permutations where length = 5. Assuming you don't run out of memory you'll get 11,881,376 (i.e. 26 to the power of 5) strings back. Bump that length up to 6, and you'll get 308,915,776 strings back. These numbers get painfully large, very quickly.
Here's a solution I put together in Java. You'll need to provide two runtime arguments (corresponding to x and y). Have fun.
public class GeneratePermutations {
public static void main(String[] args) {
int lower = Integer.parseInt(args[0]);
int upper = Integer.parseInt(args[1]);
if (upper < lower || upper == 0 || lower == 0) {
System.exit(0);
}
for (int length = lower; length <= upper; length++) {
generate(length, "");
}
}
private static void generate(int length, String partial) {
if (length <= 0) {
System.out.println(partial);
} else {
for (char c = 'a'; c <= 'z'; c++) {
generate(length - 1, partial + c);
}
}
}
}
Here's a non-recursive version I came up with, in javascript.
It's not based on Knuth's non-recursive one above, although it has some similarities in element swapping.
I've verified its correctness for input arrays of up to 8 elements.
A quick optimization would be pre-flighting the out array and avoiding push().
The basic idea is:
Given a single source array, generate a first new set of arrays which swap the first element with each subsequent element in turn, each time leaving the other elements unperturbed.
eg: given 1234, generate 1234, 2134, 3214, 4231.
Use each array from the previous pass as the seed for a new pass,
but instead of swapping the first element, swap the second element with each subsequent element. Also, this time, don't include the original array in the output.
Repeat step 2 until done.
Here is the code sample:
function oxe_perm(src, depth, index)
{
var perm = src.slice(); // duplicates src.
perm = perm.split("");
perm[depth] = src[index];
perm[index] = src[depth];
perm = perm.join("");
return perm;
}
function oxe_permutations(src)
{
out = new Array();
out.push(src);
for (depth = 0; depth < src.length; depth++) {
var numInPreviousPass = out.length;
for (var m = 0; m < numInPreviousPass; ++m) {
for (var n = depth + 1; n < src.length; ++n) {
out.push(oxe_perm(out[m], depth, n));
}
}
}
return out;
}
In ruby:
str = "a"
100_000_000.times {puts str.next!}
It is quite fast, but it is going to take some time =). Of course, you can start at "aaaaaaaa" if the short strings aren't interesting to you.
I might have misinterpreted the actual question though - in one of the posts it sounded as if you just needed a bruteforce library of strings, but in the main question it sounds like you need to permutate a particular string.
Your problem is somewhat similar to this one: http://beust.com/weblog/archives/000491.html (list all integers in which none of the digits repeat themselves, which resulted in a whole lot of languages solving it, with the ocaml guy using permutations, and some java guy using yet another solution).
I needed this today, and although the answers already given pointed me in the right direction, they weren't quite what I wanted.
Here's an implementation using Heap's method. The length of the array must be at least 3 and for practical considerations not be bigger than 10 or so, depending on what you want to do, patience and clock speed.
Before you enter your loop, initialise Perm(1 To N) with the first permutation, Stack(3 To N) with zeroes*, and Level with 2**. At the end of the loop call NextPerm, which will return false when we're done.
* VB will do that for you.
** You can change NextPerm a little to make this unnecessary, but it's clearer like this.
Option Explicit
Function NextPerm(Perm() As Long, Stack() As Long, Level As Long) As Boolean
Dim N As Long
If Level = 2 Then
Swap Perm(1), Perm(2)
Level = 3
Else
While Stack(Level) = Level - 1
Stack(Level) = 0
If Level = UBound(Stack) Then Exit Function
Level = Level + 1
Wend
Stack(Level) = Stack(Level) + 1
If Level And 1 Then N = 1 Else N = Stack(Level)
Swap Perm(N), Perm(Level)
Level = 2
End If
NextPerm = True
End Function
Sub Swap(A As Long, B As Long)
A = A Xor B
B = A Xor B
A = A Xor B
End Sub
'This is just for testing.
Private Sub Form_Paint()
Const Max = 8
Dim A(1 To Max) As Long, I As Long
Dim S(3 To Max) As Long, J As Long
Dim Test As New Collection, T As String
For I = 1 To UBound(A)
A(I) = I
Next
Cls
ScaleLeft = 0
J = 2
Do
If CurrentY + TextHeight("0") > ScaleHeight Then
ScaleLeft = ScaleLeft - TextWidth(" 0 ") * (UBound(A) + 1)
CurrentY = 0
CurrentX = 0
End If
T = vbNullString
For I = 1 To UBound(A)
Print A(I);
T = T & Hex(A(I))
Next
Print
Test.Add Null, T
Loop While NextPerm(A, S, J)
J = 1
For I = 2 To UBound(A)
J = J * I
Next
If J <> Test.Count Then Stop
End Sub
Other methods are described by various authors. Knuth describes two, one gives lexical order, but is complex and slow, the other is known as the method of plain changes. Jie Gao and Dianjun Wang also wrote an interesting paper.
Here is a link that describes how to print permutations of a string.
http://nipun-linuxtips.blogspot.in/2012/11/print-all-permutations-of-characters-in.html
This code in python, when called with allowed_characters set to [0,1] and 4 character max, would generate 2^4 results:
['0000', '0001', '0010', '0011', '0100', '0101', '0110', '0111', '1000', '1001', '1010', '1011', '1100', '1101', '1110', '1111']
def generate_permutations(chars = 4) :
#modify if in need!
allowed_chars = [
'0',
'1',
]
status = []
for tmp in range(chars) :
status.append(0)
last_char = len(allowed_chars)
rows = []
for x in xrange(last_char ** chars) :
rows.append("")
for y in range(chars - 1 , -1, -1) :
key = status[y]
rows[x] = allowed_chars[key] + rows[x]
for pos in range(chars - 1, -1, -1) :
if(status[pos] == last_char - 1) :
status[pos] = 0
else :
status[pos] += 1
break;
return rows
import sys
print generate_permutations()
Hope this is of use to you. Works with any character, not only numbers
Many of the previous answers used backtracking. This is the asymptotically optimal way O(n*n!) of generating permutations after initial sorting
class Permutation {
/* runtime -O(n) for generating nextPermutaion
* and O(n*n!) for generating all n! permutations with increasing sorted array as start
* return true, if there exists next lexicographical sequence
* e.g [a,b,c],3-> true, modifies array to [a,c,b]
* e.g [c,b,a],3-> false, as it is largest lexicographic possible */
public static boolean nextPermutation(char[] seq, int len) {
// 1
if (len <= 1)
return false;// no more perm
// 2: Find last j such that seq[j] <= seq[j+1]. Terminate if no such j exists
int j = len - 2;
while (j >= 0 && seq[j] >= seq[j + 1]) {
--j;
}
if (j == -1)
return false;// no more perm
// 3: Find last l such that seq[j] <= seq[l], then exchange elements j and l
int l = len - 1;
while (seq[j] >= seq[l]) {
--l;
}
swap(seq, j, l);
// 4: Reverse elements j+1 ... count-1:
reverseSubArray(seq, j + 1, len - 1);
// return seq, add store next perm
return true;
}
private static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}
private static void reverseSubArray(char[] a, int lo, int hi) {
while (lo < hi) {
swap(a, lo, hi);
++lo;
--hi;
}
}
public static void main(String[] args) {
String str = "abcdefg";
char[] array = str.toCharArray();
Arrays.sort(array);
int cnt=0;
do {
System.out.println(new String(array));
cnt++;
}while(nextPermutation(array, array.length));
System.out.println(cnt);//5040=7!
}
//if we use "bab"-> "abb", "bab", "bba", 3(#permutations)
}
Recursive Approach
func StringPermutations(inputStr string) (permutations []string) {
for i := 0; i < len(inputStr); i++ {
inputStr = inputStr[1:] + inputStr[0:1]
if len(inputStr) <= 2 {
permutations = append(permutations, inputStr)
continue
}
leftPermutations := StringPermutations(inputStr[0 : len(inputStr)-1])
for _, leftPermutation := range leftPermutations {
permutations = append(permutations, leftPermutation+inputStr[len(inputStr)-1:])
}
}
return
}
Though this doesn't answer your question exactly, here's one way to generate every permutation of the letters from a number of strings of the same length: eg, if your words were "coffee", "joomla" and "moodle", you can expect output like "coodle", "joodee", "joffle", etc.
Basically, the number of combinations is the (number of words) to the power of (number of letters per word). So, choose a random number between 0 and the number of combinations - 1, convert that number to base (number of words), then use each digit of that number as the indicator for which word to take the next letter from.
eg: in the above example. 3 words, 6 letters = 729 combinations. Choose a random number: 465. Convert to base 3: 122020. Take the first letter from word 1, 2nd from word 2, 3rd from word 2, 4th from word 0... and you get... "joofle".
If you wanted all the permutations, just loop from 0 to 728. Of course, if you're just choosing one random value, a much simpler less-confusing way would be to loop over the letters. This method lets you avoid recursion, should you want all the permutations, plus it makes you look like you know Maths(tm)!
If the number of combinations is excessive, you can break it up into a series of smaller words and concatenate them at the end.
c# iterative:
public List<string> Permutations(char[] chars)
{
List<string> words = new List<string>();
words.Add(chars[0].ToString());
for (int i = 1; i < chars.Length; ++i)
{
int currLen = words.Count;
for (int j = 0; j < currLen; ++j)
{
var w = words[j];
for (int k = 0; k <= w.Length; ++k)
{
var nstr = w.Insert(k, chars[i].ToString());
if (k == 0)
words[j] = nstr;
else
words.Add(nstr);
}
}
}
return words;
}
def gen( x,y,list): #to generate all strings inserting y at different positions
list = []
list.append( y+x )
for i in range( len(x) ):
list.append( func(x,0,i) + y + func(x,i+1,len(x)-1) )
return list
def func( x,i,j ): #returns x[i..j]
z = ''
for i in range(i,j+1):
z = z+x[i]
return z
def perm( x , length , list ): #perm function
if length == 1 : # base case
list.append( x[len(x)-1] )
return list
else:
lists = perm( x , length-1 ,list )
lists_temp = lists #temporarily storing the list
lists = []
for i in range( len(lists_temp) ) :
list_temp = gen(lists_temp[i],x[length-2],lists)
lists += list_temp
return lists
def permutation(str)
posibilities = []
str.split('').each do |char|
if posibilities.size == 0
posibilities[0] = char.downcase
posibilities[1] = char.upcase
else
posibilities_count = posibilities.length
posibilities = posibilities + posibilities
posibilities_count.times do |i|
posibilities[i] += char.downcase
posibilities[i+posibilities_count] += char.upcase
end
end
end
posibilities
end
Here is my take on a non recursive version

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