Algorithm for generating permutations without repetition
Hy,
I want to generate all the permutations for a list of n objects and I do that by making a tree having the children the elements in the list, and the children for the first node the elements in the list except that node etc.. After that I make a depth-first search and make a list with all the paths to the nodes situated at a certain level p. The algorithm works fine if n = 10 and p =3 for example because there are 10!/7! paths generated. Unfortunately the list of objects can have 30 elements and the level p can be 15.. so the algorithm doesn't help me in this case. I want to ask if anyone has any ideas about how can the algorithm be improved to work better (less than 1 minute) for the later case.
Thanks
bluedragonfly wrote:
I want to know if there is another way to do things..and of course not the impossibleThe only thing you have asked for is 2.03e20 permutations. We don't know how you intend to process the permutations and even if we could guess what you are doing then just processing 2.03e20 anythings is just about impossible.
So, how do we know "if there is another way to do things" ?
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Collections.shuffle(list) without repetition.
Hello every one,
I am new to Collections. I have an String ArrayList having 3 strings in it. [A, B, C]. Now all i want is 6 permutations.
[A, B, C]
[A, C, B]
[B, A, C]
[B, C, A]
[C, A, B]
[C, B, A]
Here is the simple code i have written. I hard coded the number of permutations as i know the value to make it simple.
The problem is in those 6 permutations, few are being repeated which I dont want
Please help me out with this problem.
import java.util.*;
import java.util.Collections;
import java.util.List;
import java.util.Random;
public class sample {
public static void main(String args[]) {
ArrayList<String> list = new ArrayList<String>();
list.add("A");
list.add("B");
list.add("C");
int permutations = 6;
for ( int i = 0 ; i < permutations; i++)
Collections.shuffle(list);
System.out.println(list.toString());
Thanks in advance
aadyhi,
may be this can help you. i found this code in [http://compsci.ca/v3/viewtopic.php?t=5214|http://compsci.ca/v3/viewtopic.php?t=5214] under the user zylum.
the code is used for integers. but i managed to change is to Strings too and it shuffles without REPETITION. :) so i hope this helps:
int[] numbers = new int[52];
int randNumbers = new int[22];
for (int i = 0; i < 52; i ++) numbers[i] = i;
for (int i = 0; i < 52; i ++) {
int idx = Math.random()*52; //i forgot how to do random in java so im using the action script way which is prolly the same way
int temp = numbers[idx];
numbers[idx] = numbers;
numbers[i] = temp;
for (int i = 0; i < 22; i ++) randNumbers[i] = numbers[i];
Edited by: djmlog103 on Apr 2, 2008 9:53 AM
Edited by: djmlog103 on Apr 2, 2008 9:54 AM -
Generate Permutations of array recurssive function.
Producing consecutive permutations.Need to develop a method that lists one by one all permutations of the numbers 1, 2, �, n (n is a positive integer).
(a) Recursive method . Given a verbal description of the algorithm listing all permutations one by one, you are supposed to develop a recursive method with the following header:
public static boolean nextPermutation(int[] array)
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Here is a verbal description of the recursive algorithm you need to implement:
1. The first permutation is the permutation represented by the sequence (1, 2, �, n).
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the following procedure:
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(iii) Consecutively applying this algorithm to permutations starting from (1, 2, �, n),you will eventually list all n! possible permutations. The last one will be (n, �, 2, 1).For example, below is the sequence of permutations for n = 3 , listed by the described algorithm:
(0 1 2) ; (0 2 1) ; (2 0 1) ; (1 0 2) ; (1 2 0) ; (2 1 0)
Please help...i have done the iterative one..but unable to figure out the recursive..
Thank you in advance..
Please helpThanks for your reply...it would be great if a code is provided..i am try to do but this not working here is my code
* Permutations.java
* Created on April 25, 2006, 12:42 PM
package javaapplication1;
* @author ismail
public class Permutations {
// print N! permutation of the elements of array a (not in order)
public static boolean nextPermutation(int a[]) {
int N = a.length;
int[] t = new int[N];
/* for (int i = 0; i < N; i++)
t[i] = a;*/
return(perm2(a, N));
private static boolean perm2(int [] a, int n) {
int i=n-1;
// start from last but 2
for (; i>0; i--) if (a[i]<a[i+1]) break;
if (i <0) return false;
for (int j = 0; j < n; j++) {
swap(a, j, n-1);
perm2(a, n-1);
// swap(a, j, n-1);
return true;
// swap the characters at indices i and j
private static void swap(int[] a, int i, int j) {
int c;
c = a[i]; a[i] = a[j]; a[j] = c;
private static final int num = 3; // number of elements to permutate
private static int [] a = new int[num]; // permutations array
private static int qkCounter = 0; // Counter for Quick sort
private static int mrgCounter = 0; // Counter for Merge sort
private static int qkSum = 0; // Total sum for Quick sort
private static int mrgSum = 0; // Total sum for Merge sort
public static void main(String s[]){
int count=1;
int [] qkArray = new int[num]; // Quick Sort array
int [] mrgArray = new int[num]; // Merge Sort array
int avgQuick = 0;
int avgMerge = 0;
// Initialize array
for(int i=0;i<num;i++){
a[i]=i;
// call nextPermutation to generate possible permutations - iterative method
for (int j=0; nextPermutation(a);j++){
count++; // Count number of records
// print the permutation array
for(int i=0;i<num;i++){
System.out.print(a[i]);
not working means-- it keeps on printing 012 -
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Hi, is the "Seminumerical Algorithms" a book?What
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don't know, it may as well be randomly picked.As far as I know the numbers aren't randomly picked. You want the distribution to be good, and the length of the serie should also be long. Some numbers causes the cycle to be short, or the distribution to be bad. I think there are articles that lists 'good' numbers.
/Kaj -
Announce: Generic Algorithms for Java, 0.3 release
The latest release of jga: Generic Algorithms for Java is now available at
http://jga.sf.net/
Included in this release are a collection of algorithms adapted from STL that operate over generic collections and iterations. What's included are the algorithms that do not attempt to modify the collection/iteration that they're given: in other words, all of the find/search/count algorithms. Also included are the complete set of functors and predicates that were present in the last release, with a few modifications for easier use.
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Name TCURC is not in the namespace for generated BI meta objects
Hi,
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UNCAUGHT_EXCEPTION
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SY-MSGNO 019
SY-MSGV1 TCURC
which gives the message text "Name TCURC is not in the namespace for generated BI meta objects"
If I remove the mapping with Amount KF in the TRFN and load data it works fine without error but on mapping the KF with currency unit I get this error.It is something related to Currency but I'm unable to figure out the problem.
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regards
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There is no standard process to handle ARE 2 in SAP.
Where, rightly mention in one of the previous post that you can very much follow process flow for ARE 1 in SAP.
So, there will not be any combine process for
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- export under bond without payment
In SAP, you have to dealt them separately in ARE 1.
For further reference/understanding, refer following SDN threads links for:
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Take use of following TCodes:
TCode
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J1IA103
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J1IA104
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J1IANX18
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J1IBN01
- Create Excise Bond
J1IBN02
- Change Excise Bond
J1IBN03
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J1IBN04
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J1IBN05
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J1IBONSUM
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Regards
JP -
Help me to get the algorithm for this assingment.
I have tried to draw the x-y axis and the rectangle. I hava a problem to draw the rectangle inside of the lablled grid.
NEED EVERYONE'S HELP HERE!
Assignment #4
Goal:
� This lab gives students more experience in
� developing proper algorithms for real-life problems;
� using program constructs � sequential, selection, and repetition.
Algorithm. An algorithm is a description of steps to solve a problem that is unambiguous, executable, and terminating. In order to solve a problem by computer, you must know an algorithm for finding a solution.
Problem Description
Develop an algorithm and then write a java program that can accept commands from the user to draw a rectangle. The program should accept the xy-coordinate pair of the upper-left corner, and the width and height of a rectangle. It should then draw the rectangle pattern on a 40 X 20 labelled grid. For example, if the upper-left corner is (5, 10) with a width of 10 and height of 5, then the program would display:
^
20+
|
|
|
|
|
15+
|
|
|
|
10+
| ++++++++++
| ++++++++++
| ++++++++++
| ++++++++++
5+ ++++++++++
|
|
|
|
________________________________>
0 5 10 15 20 25 30 35 40
Upper Left (5,10) width:10, Height: 5
NOTE:
� Your display should include the labelled axis, as shown on the previous page, and also the lines that are drawn horizontally across the page.
� Make sure the rectangle will fit on the 40X20 grid.
� Appropriate error messages must be displayed for invalid data provided by the user.If you format your code, it looks like thispublic class Rectangle {
public static void main(String[] args) {
int maxYAxisPoint = 20;
int maxXAxisPoint = 40;
int xCoodinate = 5;
int yCoodinate = 10;
int width = 5;
int height = 10;
int leftLowPoint;
System.out.println("This program draw a rectangle pattern on a 40*20 labelled grid.");
System.out.println("Please enter the x-coordinate of the upper left point: ");
xCoodinate = MyIO.readLineInt();
System.out.println("Please enter the y-coordinate of the upper left point: ");
yCoodinate = MyIO.readLineInt();
System.out.println("Please enter the width of the rectangle: ");
width = MyIO.readLineInt();
System.out.print("Enter the height of the rectangle:");
height = MyIO.readLineInt();
if (yCoodinate < maxYAxisPoint && yCoodinate + width < 40 ||
xCoodinate + width < maxXAxisPoint &&
xCoodinate + height < maxYAxisPoint) {
System.out.println(" ^");
for (maxYAxisPoint = 20; maxYAxisPoint >= 0; maxYAxisPoint--) {
if ( (maxYAxisPoint % 5) == 0) {
System.out.println(maxYAxisPoint + "+");
for (leftLowPoint = (maxYAxisPoint - height) + 1;
leftLowPoint <= yCoodinate; leftLowPoint--) {
System.out.println("+");
//Draw the x-axis
for (maxXAxisPoint = 0; maxXAxisPoint <= 40; maxXAxisPoint++) {
System.out.print("+____");
System.out.println("+>");
for (maxXAxisPoint = 0; maxXAxisPoint <= 40; maxXAxisPoint++) {
System.out.print(maxXAxisPoint + " ");
} else {
System.out.print(" | \n" + " |\n" + " |\n" + " |");
} else {
System.out.print("You entered a wrong x-y coordinate of the rectangle,\n the rectangle is not in the labelled grid.");
System.out.print("Upper Left(" + xCoodinate + "," + yCoodinate + ")");
System.out.print("width:" + width);
System.out.println("height: " + height);
}You have nested your loops. I'm pretty sure you don't want to do this. Here's an example of the difference. First a nested looppublic class Test2 {
public static void main(String[] args) {
for (int i=0; i<3; i++) {
System.out.println("i="+i);
for (int j=0; j<4; j++) {
System.out.println(" j="+j);
Outputs this
i=0
j=0
j=1
j=2
j=3
i=1
j=0
j=1
j=2
j=3
i=2
j=0
j=1
j=2
j=3Now an unnested looppublic class Test2 {
public static void main(String[] args) {
for (int i=0; i<3; i++) {
System.out.println("i="+i);
for (int j=0; j<4; j++) {
System.out.println(" j="+j);
outputs this
i=0
i=1
i=2
j=0
j=1
j=2
j=3This means you are printing the x-axis for every point on the graph. -
Generate XML without header tags in Oracle 9i Release1
I am a novoice to Oracle. I am using Oracle 9i Release 1.
I have a table "Attribution" with the following structure.
SQL> Desc attribution
Name Null? Type
REPORT_ID NUMBER
LABEL VARCHAR2(20)
VALUE NUMBER(25,10)
COLOR VARCHAR2(10)
The inserted values in this table are:
SQL> select * from attribution;
REPORT_ID LABEL VALUE COLOR
24 Allocation .025172879 996699
24 Selection -.03198432 FF9933
24 Timing .000016592 8B5A13
24 Interaction -.0148997 99583D
24 Derivative Effect -.00026146 CDCD00
24 Active Return -.02195601 638B45
24 Active Risk .031745116 800000
I want to generate an xml without the XML declaration row i.e. <?xml version="1.0"?>
Also I do not want the <Rowset> tags in my xml.
My generated XML should be like:
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<set label="Selection" value="-4.3508000000" color="FF9933"/>
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<set label="Derivative Effect" value="-0.0184000000" color="CDCD00"/>
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Please help me in generating this xml.
Thanks in Advance
JaspreetThanks for your reply.
Essentially my requirement was to concatinate three sets of xml strings. The application which is parsing my xml requires the XML in the following format:
XMLStringA + GeneratedXMLfromMyQuery + XMLStringB
where XMLStringA and XMLStringB are static XML strings and do not change.
I have generated GeneratedXMLfromMyQuery as from the following query.
Declare
qryctx dbms_xmlgen.ctxhandle;
GeneratedXMLfromMyQuery CLOB;
Begin
qryctx := dbms_xmlgen.newcontext('SELECT label "@label", value*100 "@value", color "@color" FROM Attribution order by rownum');
dbms_xmlgen.setrowtag(qryctx,'set');
GeneratedXMLfromMyQuery := dbms_xmlgen.getxml(qryctx);
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hi,
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Alvon Sibaraniuse transaction mb0a it will work for the scenario.
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Greetings.
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