Generating all possible combinations question

i'm trying to generate all possible combinations for several single column tables each with one string column.
so - if table 1 has values (aa, ab, ac)
table 2 has values (zz, zx)
table 3 has values (qw, qe)
the result set would contain all possible combinations:
aa, ab, ac, zz, zx, qw, qe, aazz, aazx, aaqw, aaqe, aazzqw, aazzqe, aazxqw, aazxqe...etc.
I've tried cross joins - but that does not get the smaller combinations.
I've looked at some code examples but they seem to be focused on single letter or number combinations.
I need to do this using tsql.
code examples or links to such would be of much help.
Thanks.

Something like this might work:
with t1 as (
select val from table1
union all
select ''
), t2 as (
select val from table2
union all
select ''
, t3 as (
Select val from table3
union all
select ''
select t1.val + t2.val + t3.val
from t1 cross join t2 cross join t3
Russel Loski, MCT, MCSE Data Platform/Business Intelligence. Twitter: @sqlmovers; blog: www.sqlmovers.com

Similar Messages

  • Fill Array with all possible combinations of 0/1

    Hi,
    i'm looking for a fast way to fill an array, respectively an int[8], with all possible combinations of 0 and 1 (or -1 and 1, shouldn't make a difference).
    I kind of know how to do it using multiple loops but I assume there is a more elegant or at leaster "better" practice.
    Thanks,
    nikolaus
            static int cnt = 0;
         public static void main(String[] args) {
              int[] element = new int[]{1,1,1,1,1,1,1,1};
              Integer[] x = new Integer[2];
              x[0] = 1;
              x[1] = -1;
              for(int i7:x){
                   element[7] = i7;
                   for(int i6:x){
                        element[6] = i6;
                        for(int i5:x){
                             element[5] = i5;
                             for(int i4:x){
                                  element[4] = i4;
                                  for(int i3:x){
                                       element[3] = i3;
                                       for(int i2:x){
                                            element[2] = i2;
                                            for(int i1:x){
                                                 element[1] = i1;
                                                 for(int i0:x){
                                                      element[0] = i0;
                                                      cnt++;
              }Edited by: NikolausO on Oct 30, 2008 4:21 AM
    Edited by: NikolausO on Oct 30, 2008 4:22 AM

    pm_kirkham wrote:
    No I replied to message number 5. as the ' (In reply to #5 )' above my post indicates, which was in reply to (a reply) to Sabre150's post which wasn't using enhanced loops, nor has any obvious place where you could use that approach to fill the array.
    Though you could pass in an array of the values to fill the array with, and loop over those, instead of using 0 or 1, at which point Sabre's approach becomes the same as your OP, but without the manual unrolling:
    import java.util.Arrays;
    public class NaryCombinations {
    public interface CombinationHandler {
    void apply (int[] combination) ;
    public static void main(String[] args) {
    calculateCombinations(new int[]{-1, 0, 1}, 4, new CombinationHandler () {
    public void apply (int[] combination) {
    System.out.println(Arrays.toString(combination));
    public static void calculateCombinations (int[] values, int depth, CombinationHandler handler) {
    recursivelyCalculateCombinations(values, 0, depth, handler, new int[depth]);
    private static void recursivelyCalculateCombinations (int[] values, int index, int depth,
    CombinationHandler handler, int[] combination) {
    if (index == depth) {
    handler.apply(combination);
    } else {
    for (int value : values) {
    combination[index] = value;
    recursivelyCalculateCombinations(values, index + 1, depth, handler, combination);
    Which looks to use the same basic approach to the generalization I created shortly after posting the original.
    public class Scratch1
         * A 'callback' to be invoked with every combination
         * of the result.
        public interface Callback
             * Invoked for each combination.
             * <br>
             * Each call is passed an new instance of the array.
             * @param array the array containing a combination.
            void processArray(int[] array);
        public Scratch1(final int[] array, final Callback callback, final int... values)
            if (callback == null)
                throw new IllegalArgumentException("The 'callback' cannot be 'null'");
            if (array.length < 1)
                throw new IllegalArgumentException("The array length must be >= 1");
            if ((values == null) || (values.length < 1))
                throw new IllegalArgumentException("The 'values' must have be at least of length 2");
            callback_ = callback;
            values_ = values.clone();
            array_ = array;
        public Scratch1(final int order, final Callback callback, final int... values)
            this(new int[order], callback, values);
         * Generates every possible value and invokes the callback for each one.
        public void process()
            process(0);
         * Internal method with no logical external use.
        private void process(int n)
            if (n == array_.length)
                callback_.processArray(array_.clone());
            else
                for (int v : values_)
                    array_[n] = v;
                    process(n + 1);
        private final Callback callback_;
        private final int[] values_;
        private final int[] array_;
        public static void main(String[] args) throws Exception
            final Callback callback = new Callback()
                public void processArray(int[] array)
                    System.out.println(java.util.Arrays.toString(array));
            new Scratch1(6, callback, 2, 1, 0).process();

  • Creating a tree of all possible combinations of {a,b,c,d}

    I am an MSc conversion student who is struggling with the initial stages of coding his Association Rule Mining dissertation project.
    I have a TreeNode class, which is in the process of being written.
    The class is designed to create a tree that stores all possible combinations of a Vector alphabet of strings {a,b,c,d}
    Please could somebody explain what my TreeNode class is doing so far?
    I have had a lot of help with this and I can't understand why there are so many vectors and what they all do.
    A few pointers (no sarcastic comments please) would be much appreciated.
    The code so far:
    package treenode;
    import java.io.*;
    import java.util.*;
    public class TreeNode
    public static final String INFO_TEXT ="A program to store a vector of string items a,b,c,d in all their possible combinations into a TreeNode class";
    private Vector alphabet;
    private Vector data; // this IS TreeRootData
    private Vector children;
    public TreeNode(Vector inputAlphabet, Vector inputData)
    // the alphabet and data have already been parsed in.
    this.alphabet = inputAlphabet;
    this.data = inputData;
    System.out.println("Tree node constructed with data "+displayVectorOfStrings(this.data));
    children = new Vector(); // we now construct the Vector of children
    //createChildren();
    public void createChildren()
    if data.length =
    // first check to see whether the current data length is the same
    // as the length of the alphabet
    // if it is then stop here (return;)
    // we loop through the alphabet we have...
    // for each candidate in the alphabet, attempt to create a new child TreeNode
    // with a data Vector that is the same as the data in this TreeNode plus
    // the candidate we are looking at in the alphabet. i.e. add available LHS
    // NB: Do NOT add if the candidate is already in the data. {a,b,a} is wrong!
    // NB: We will check to see whether data is already in the tree at a later stage
    // NB: The new Child TreeNode MUST BE ADDED to the children VECTOR -
    // or it will be lost
    public static String displayVectorOfStrings(Vector vector) //Printing the Vector of Strings
    String vectorText = "{";
    for (int i=0; i<vector.size(); i++)
    vectorText += (String)vector.elementAt(i)+",";
    if (vector.size() > 0) vectorText = vectorText.substring(0, vectorText.length()-1);
    // now we chop off the last element, replacing it with a closing bracket,
    // provided that the Vector is not empty.
    vectorText += "}";
    return vectorText;
    public static void main(String[] args) //main method WITH THE ARGUMENTS PASSED IN.
    // This main class can be thought of as OUTSIDE the TreeNode class. Although
    // it is in the file called TreeNode.java it is only here for convenience.
    // This is the top level of the program where everything starts. It is here -
    // that the alphabet is hardcoded in and that the Tree root is constructed.
    displayHeading();
    Vector alphabet = new Vector(); // constructing an initial alphabet,
    alphabet.addElement("a"); // which will eventually come from the database
    alphabet.addElement("b");
    alphabet.addElement("c");
    alphabet.addElement("d");
    System.out.println("Alphabet set to be "+TreeNode.displayVectorOfStrings(alphabet));
    // the root's initial data is now made into an empty Vector...
    Vector initialData = new Vector();
    // the root can now be constructed
    TreeNode treeRoot = new TreeNode(alphabet, initialData); // the treeRoot constructor
    public static void displayHeading()
    System.out.println();
    System.out.println(INFO_TEXT);

    I don't remember from which site I took this material.
    You read this to find all possible combination of a set {a,b,c,d}
    Consider three elements A, B, and C. It turns out that for n distinct elements there are n! permutations possible. For these three elements all of the possible permutations are listed below:
    ABC, ACB, BAC, BCA, CAB, CBA
    Since there are 3 elements there are 3! = 6 permutations possible.
    Suppose instead of A, B, and C you have a deck of 52 playing cards. There are 52! permutations possible (i.e., the number of different ways the cards could be shuffled). Since 52! = 8x10^67, you will need to play a lot of cards before you have seen every permutation of the deck! Since shuffling the deck is a random process (or it should be), each of these 52! permutations is equally likely to occur.
    Developing a non-recursive algorithm to generate all permutations of n elements is a non-trivial task. However, developing a recursive algorithm to do this requires only modest effort and this is what we will do next.
    1. Let E = {e1, e2, ..., en } denote the set of n elements whose permutations we are to generate.
    2. Let Ei be the set obtained by removing element i from E.
    3. Let perm(X) denoted the permutations of the elements in the set X.
    4. Let ei.perm(X) denote the permutation list obtained by prefixing each permutation in perm(X) with element ei.
    Consider the following example, which is provided to clarify the definitions given above:
    If E = {a, b, c}, then E1 = {b, c} since the element in position 1 has been removed from E to create E1. Perm(E1) = (bc, cb) since these are the only two possible permutations of the two elements which appear in E1. Finally, e1.perm(E1) = (abc, acb) which should be apparent from the value of e1 and perm(E1) since the element a is simply used as a prefix to each permutation in perm(E1).
    Given these definitions, we set the recursion base case when n = 1. Since only one permutation is possible for one element, perm(E) = (e) where e is the lone element in E. When n > 1, perm(E) is the list e1.perm(E1) followed by e2.perm(E2) followed by e3.perm(E3) � followed by en.perm(En). This recursive definition of perm(E) defines perm(E) in terms of n perm(X)s, each of which involves an X with n � 1 elements. Thus both the base component and the recursive component (of a recursive algorithm) have been established and we thus have a complete recursive technique to generate the permutations.
    The following Java method is an implementation of this recursive definition of perm(E). This method will output all permutations whose prefix is list[0:k-1] and whose suffixes are the permutations of list[k:m]. By invoking the method with Perm(list, 0, n-1) all n! permutations of list[0: n-1] will be produced. [This is because this invocation will set k = 0 and m = n � 1, so the prefix of the generated permutations is null and their suffixes are the permutations of list[0: n-1]. When k = m, there is only one suffix which is list[m] and list[0: m] defines a permutation that is to be produced. When k < m, the else clause is executed.
    In the algorithm, let E denote the elements in list[k: m] and let Ei be the set obtained by removing ei = list[i] from E. The first swapping sequence in the for-loop has the effect of setting list[k] = ei and list[k+1: m] = Ei. Therefore, next statement which is the call to perm computes ei.perm(Ei). The final swapping sequence restores list[k: m] to its state prior to the first swapping sequence (the state is was in before the recursive call occurred).
    public static void perm(Object [] list, int k, int m)
    {    //generate all permutations of list[k:m]
    int i; Object temp;
    if (k == m) //list has one permutation � so output it
    {      for (i = 0; i <= m; i++)
    System.out.print(list);
    System.out.println;
    else //list has more than one permutation � so generate them recursively
    for (i = k; i <= m; i++)
    {     temp = list[k]; //these next three lines are a simple swap function
    list[k] = list[i];
    list[i] = temp;
    perm(list, k+1, m); //the recursive call
    temp = list[k]; //reset the order in the list
    list[k] = list[i];
    list[i] = temp;
    }//end for-loop
    } //end method perm
    See the program pasted below that prints all possible combination of a set {a,b,c,d}
    public class Perm
    public Perm()
    public static void main(String[] args)
    String s = "abcd";
    char [] chars = s.toCharArray();
    perm(chars, 0);
    public static void perm(char [] list, final int k)
    int i;
    char temp;
    if (k == list.length-1) //list has one permutation, so output it
    for (i = 0; i <= list.length-1; i++)
    System.out.print(list[i]);
    System.out.println();
    else { //list has more than one permutation, so generate them recursively
    for (i = k; i <= list.length-1; i++)
    temp = list[k]; //these next three lines are a simple swap function
    list[k] = list[i];
    list[i] = temp;
    perm(list, k+1); //the recursive call
    temp = list[k]; //reset the order in the list
    list[k] = list[i];
    list[i] = temp;
    }//end for-loop

  • Printing all possible combinations of a given word.

    Well I'm supposed to print all possible combinations of a given word(I'm writing this for the benefit of those who might not have guessed this from the subject). I've got some code but its horribly wrong. Could anyone point out any mistakes.
    public class Try
        public void display(String s)
            char ch[]=s.toCharArray();
            for(int i=0;i<ch.length;i++)
                String word="";
                word+=ch;
    char c[]=new char[s.length()];
    for(int j=0;j<ch.length;j++)
    if(ch[j]!=ch[i])
    c[j]=ch[j];
    for(int j=0;j<ch.length;j++)
    for(int k=0;k<ch.length;k++)
    if(c[j]!=c[k])
    word+=c[k];
    word+=c[j];
    System.out.println(word);

    Me and Lucifer have been workin on this program for ages.... we cant quite understand your pseudo code.. could you elaborate a bit more on how it works ??
    Meanwhile this is the dysfunctional program weve managed to come up with so far:
    public class SHAM
        public void display(String s)
            char c[]=s.toCharArray();
            for(int i=0;i<c.length;i++)
                String word="";
                char c1[]=new char[c.length-1];
                int a=0;
                for(int l=0;l<c.length;l++)
                    if(i!=l)
                        c1[a++]=c[l];
                for(int j=0;j<c.length-1;j++)
                    char c2[]=c1;
                    word=c[i]+word;
                    if(j+1!=c.length)
                        char t=c2[j];
                        c2[j]=c2[c2.length-1];
                        c2[c2.length-1]=t;
                    for(int k=0;k<c2.length;k++)
                        word+=c2[k];
                    System.out.println(word);               
    }And this is my coding of your pseudo code:
    public class WordCombo2
        public void combo(String s)
            getCombos(s.toCharArray(), s.length());
        private void getCombos(char[] c,int n)
            if (n==1)
                print(c);
                System.out.println();
            else
                for (int i=0;i<c.length;i++)
                    if (i==0 && i<n)
                        c=c[n-1];
    n--;
    getCombos(c,n);
    c[i]=c[n-1];
    private void print(char c[])
    for (int i=0;i<c.length;i++)
    System.out.print(c[i]+" ");
    I really dont understand how this is supposed to work so i dont know whats wrong :( Im sorry this program is just totally baffling me, and on top of that recursion kind of confuses me.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           

  • All possible combinations of 23 columns in a table.

    Hi,
    We have a table as follows;
    CREATE TABLE ALL_PROD_COMB_TMP
    HDLM_ID NUMBER(10),
    MULTIPATH_ID NUMBER(10),
    TRUE_COPY_ID NUMBER(10),
    UVM_ID NUMBER(10),
    TUNING_ID NUMBER(10),
    CLUSTER_ID NUMBER(10),
    MIDDLEWARE_ID NUMBER(10),
    TAPE_ID NUMBER(10),
    THIRD_PARTY_ID NUMBER(10),
    NAS_ID NUMBER(10),
    HBA_ID NUMBER(10),
    BCM_ID NUMBER(10),
    VOLUME_MIGRATION_VERSION_ID NUMBER(10),
    TIERED_STO_MGR_VERSION_ID NUMBER(10),
    HDPS_ID NUMBER(10),
    SERVER_ID NUMBER(10),
    OS_ID NUMBER(10),
    DWDM_ID NUMBER(10),
    EXTENDER_ID NUMBER(10),
    SWITCH_ID NUMBER(10),
    INTERFACE_ID NUMBER(10),
    HUR_ID NUMBER(10),
    STORAGE_FAMILY_ID NUMBER(10)
    The data in the table is as follows;
    For formatting purpose the table data is shown in transposed form.
    Serial No. Column Name Data Value
    1     HDLM_ID                    377     586               
    2     MULTIPATH_ID               142     357               
    3     TRUE_COPY_ID               1     87     12          
    4     UVM_ID                    89     79               
    5     TUNING_ID               10     9     8          
    6     CLUSTER_ID               3     327     638          
    7     MIDDLEWARE_ID                         
    8     TAPE_ID                    3     47     54          
    9     THIRD_PARTY_ID               3                    
    10     NAS_ID                    12     91     1          
    11     HBA_ID                    300     400               
    12     BCM_ID                    14                    
    13     VOLUME_MIGRATION_VERSION_ID     13     12               
    14     TIERED_STO_MGR_VERSION_ID     1     2               
    15     HDPS_ID                    10     100               
    16     SERVER_ID               3     4               
    17     OS_ID                    2     7               
    18     DWDM_ID                    1                    
    19     EXTENDER_ID               520                    
    20     SWITCH_ID               4                    
    21     INTERFACE_ID               2     1     3     4     
    22     HUR_ID                    2     1     3     4     5
    23     STORAGE_FAMILY_ID          2
    Now we have another table named ALL_PROD_COMB having the same structure as that of ALL_PROD_COMB_TMP. In this ALL_PROD_COMB we need all the possible combinations of all the column values from ALL_PROD_COMB_TMP.
    We tried the following thing;
    insert into
         all_prod_comb (
         hdlm_id           
                                       ,multipath_id           
                                       ,true_copy_id           
                                       ,uvm_id                
                                       ,tuning_id           
                                       ,cluster_id           
                                       ,middleware_id
                                       ,tape_id
                                       ,third_party_id
                                       ,nas_id
                                       ,hba_id
                                       ,bcm_id
                                       ,volume_migration_version_id
                                       ,tiered_sto_mgr_version_id
                                       ,hdps_id
                                       ,server_id
                                       ,os_id                
                                       ,dwdm_id           
                                       ,extender_id           
                                       ,switch_id           
                                       ,interface_id           
                                       ,hur_id                
                                       ,storage_family_id
    select distinct
         a.hdlm_id           
                   ,b.multipath_id           
                   ,c.true_copy_id           
                   ,d.uvm_id                
                   ,e.tuning_id           
                   ,f.cluster_id           
                   ,g.middleware_id
                   ,h.tape_id
                   ,i.third_party_id
                   ,j.nas_id
                   ,k.hba_id
                   ,l.bcm_id
                   ,m.volume_migration_version_id
                   ,n.tiered_sto_mgr_version_id
                   ,o.hdps_id
                   ,p.server_id
                   ,q.os_id                
                   ,r.dwdm_id           
                   ,s.extender_id           
                   ,t.switch_id           
                   ,u.interface_id           
                   ,v.hur_id                
                   ,w.storage_family_id
         from (select hdlm_id from all_prod_comb_tmp) a
    ,(select multipath_id from all_prod_comb_tmp) b
    ,(select true_copy_id from all_prod_comb_tmp) c
                   ,(select uvm_id from all_prod_comb_tmp) d
                   ,(select tuning_id from all_prod_comb_tmp) e
                   ,(select cluster_id from all_prod_comb_tmp) f
                   ,(select middleware_id from all_prod_comb_tmp) g
                   ,(select tape_id from all_prod_comb_tmp) h
                   ,(select third_party_id from all_prod_comb_tmp) i
                   ,(select nas_id from all_prod_comb_tmp) j
                   ,(select hba_id from all_prod_comb_tmp) k
                   ,(select bcm_id from all_prod_comb_tmp) l
                   ,(select volume_migration_version_id from all_prod_comb_tmp) m
                   ,(select tiered_sto_mgr_version_id from all_prod_comb_tmp) n
                   ,(select hdps_id from all_prod_comb_tmp) o
                   ,(select server_id from all_prod_comb_tmp) p
                   ,(select os_id from all_prod_comb_tmp) q
                   ,(select dwdm_id from all_prod_comb_tmp) r
                   ,(select extender_id from all_prod_comb_tmp) s
                   ,(select switch_id from all_prod_comb_tmp) t
                   ,(select interface_id from all_prod_comb_tmp) u
                   ,(select hur_id from all_prod_comb_tmp) v
                   ,(select storage_family_id from all_prod_comb_tmp) w;
    But after looking at the explain plan we had to discard this. Simply not possible. Please help us with some other feasible solution.
    regards,
    Dipankar Kushari.

    Hi,
    That's what I have written while posting. Cartesian is no way possible. But is there any way so that I can find out all possible combination of the following values which is there in a table.
    1 HDLM_ID 377 586
    2 MULTIPATH_ID 142 357
    3 TRUE_COPY_ID 1 87 12
    4 UVM_ID 89 79
    5 TUNING_ID 10 9 8
    6 CLUSTER_ID 3 327 638
    7 MIDDLEWARE_ID
    8 TAPE_ID 3 47 54
    9 THIRD_PARTY_ID 3
    10 NAS_ID 12 91 1
    11 HBA_ID 300 400
    12 BCM_ID 14
    13 VOLUME_MIGRATION_VERSION_ID 13 12
    14 TIERED_STO_MGR_VERSION_ID 1 2
    15 HDPS_ID 10 100
    16 SERVER_ID 3 4
    17 OS_ID 2 7
    18 DWDM_ID 1
    19 EXTENDER_ID 520
    20 SWITCH_ID 4
    21 INTERFACE_ID 2 1 3 4
    22 HUR_ID 2 1 3 4 5
    23 STORAGE_FAMILY_ID 2
    regards,
    Dipankar.

  • All possible combinations of a column as applied to a unique entry?

    i am having trouble getting a loop to do what i want, if you
    can help me find out what i am doing wrong i would appreciate it:
    i have a database with 3 tables in it
    table 1 has a list of documents - each one having a primary
    key
    table 2 has a list of document properties, i will call them
    property A, B, and C - each one having a primary key
    table 3 is a "relational" table that has a column for a
    document's PK and a property's PK that that document has
    Example of table 3:
    docPK / propPK
    1 (Document 1) / 1 (A)
    2 (Document 2) / 1 (B)
    3 (Document 3) / 2 (B)
    1 (Document 1) / 2 (B)
    1 (Document 1) / 3 (C)
    i need to create a loop in ColdFusion that spits out the
    number of possible combinations that exist for each document
    so the correct output should look like:
    Property A
    Document 1
    Property B
    Document 1
    Document 2
    Document 3
    Property C
    Document 3
    Property A, Property B
    Document 1
    Proberty A, Property B, Property C
    Document 1
    this output displays all possible combinations of properties
    under the conditions of existing documents
    here is the loop i have so far that does not seem to be
    working for me,
    <cfoutput query="rsProperties">
    <ul>
    <cfloop from="0" to="# of properties per document"
    index="i">
    <li>
    <cfoutput group="propPK">
    <cfif i EQ "# of properties per document">
    [#Trim(propertyName)#]
    </cfif>
    </cfoutput>
    </li>
    <ul>
    <cfoutput group="docPK">
    <li><a href="">#Document
    Name#</a></li>
    </cfoutput>
    </ul>
    </cfloop>
    </ul>
    </cfoutput>
    my loop returns possible combinations, but it is not
    returning ALL possible combinations, for example it may only return
    Property A as a single instead of also returning singles Property B
    and C
    my query simply consists of a SELECT * FROM [the database]
    WHERE [the three tables are set equal (as a join)] AND WHERE
    [documents exist]
    i know this is all rather confusing but if you can help me
    make sense of it i will be grateful
    thanks for your time

    Read the cfoutput section of the cfml reference manual. If
    you don't have one, the internet does.
    Look for the example that shows you how to use the group
    attribute.

  • How to generate all possible keys for DES algorithm

    Hello every one,
    I want to generate all the possible key combinations for the DES algorithm 56-bit "actuallly 64 but the last 8 bits are just padding", so can anyone plz help how can i do that? or give me article or something that might help me?
    Thanks in advance,
    Amr M. Kamel.

    I know that but who said that it will be one
    processor it will be on distributed and clustered
    environment :). I just want to know how to generate
    the keys "or write binary".
    Thans again for your helpA cluster of 2285 machines will still take a year! The simplest algorithm is to just count from 0 to 2^56 .
    If you are thinking of cracking DES then there is a Book I have on my shelf 'Cracking DES' published by the 'Electronic Frontier Foundation' ISBN 1-56592-520-3 . It contains the C code and describes the hardware used. The hardware cost was about $100,000 and it cracks DES in about 3.5 days.
    Assuming Moore's law ( http://en.wikipedia.org/wiki/Moore's_law ) applies then it should now take about much less than a day and cost much less than $100,000. BUT - it will take a long time using basic computers.

  • Help with oracle sql to get all possible combinations in a table.

    Hello guys I have a small predicatement that has me a bit stumped. I have a table like the following.(This is a sample of my real table. I use this to explain since the original table has sensitive data.)
    CREATE TABLE TEST01(
    TUID VARCHAR2(50),
    FUND VARCHAR2(50),
    ORG  VARCHAR2(50));
    Insert into TEST01 (TUID,FUND,ORG) values ('9102416AB','1XXXXX','6XXXXX');
    Insert into TEST01 (TUID,FUND,ORG) values ('9102416CC','100000','67130');
    Insert into TEST01 (TUID,FUND,ORG) values ('955542224','1500XX','67150');
    Insert into TEST01 (TUID,FUND,ORG) values ('915522211','1000XX','67XXX');
    Insert into TEST01 (TUID,FUND,ORG) values ('566653456','xxxxxx','xxxxx');
    "TUID"                        "FUND"                        "ORG"                        
    "9102416AB"                   "1XXXXX"                      "6XXXXX"                     
    "9102416CC"                   "100000"                      "67130"                      
    "955542224"                   "1500XX"                      "67150"                      
    "915522211"                   "1000XX"                      "67XXX"                      
    "566653456"                   "xxxxxx"                      "xxxxx"                       The "X"'s are wild card elements*( I inherit this and i cannot change the table format)* i would like to make a query like the following
    select tuid from test01 where fund= '100000' and org= '67130'however what i really like to do is retrieve any records that have have those segements in them including 'X's
    in other words the expected output here would be
    "TUID"                        "FUND"                        "ORG"                        
    "9102416AB"                   "1XXXXX"                      "6XXXXX"                     
    "9102416CC"                   "100000"                      "67130"                      
    "915522211"                   "1000XX"                      "67XXX"                      
    "566653456"                   "xxxxxx"                      "xxxxx"  i have started to write a massive sql statement that would have like 12 like statement in it since i would have to compare the org and fund every possible way.
    This is where im headed. but im wondering if there is a better way.
    select * from test02
    where fund = '100000' and org = '67130'
    or fund like '1%' and org like '6%'
    or fund like '1%' and org like '67%'
    or fund like '1%' and org like '671%'
    or fund like '1%' and org like '6713%'
    or fund like '1%' and org like '67130'
    or fund like '10%' and org like '6%'...etc
    /*seems like there should be a better way..*/can anyone give me a hand coming up with this sql statement...

    mlov83 wrote:
    if i run this
    select tuid,fund, org
    from   test01
    where '100000' like translate(fund, 'xX','%%') and '67130' like translate(org, 'xX','%%');this is what i get
    "TUID"                        "FUND"                        "ORG"                        
    "9102416AB"                   "1XXXXX"                      "6XXXXX"                     
    "9102416CC"                   "100000"                      "67130"                      
    "915522211"                   "1000XX"                      "67XXX"                      
    "566653456"                   "xxxxxx"                      "xxxxx"                      
    "9148859fff"                  "1XXXXXX"                     "X6XXX"                       the last item should be excluded. The second digit in "org" is a "7" Fund is wrong, too. You're looking for 6 characters ('100000'), but fund on that row is 7 characters ('1XXXXXX').
    and this is sitll getting picked up.That's why you should use the _ wild-card, instead of %
    select  tuid, fund, org
    from    test01
    where  '100000' like translate (fund, 'xX', '__')
    and    '67130'  like translate (org,  'xX', '__')
    ;It's hard to see, but, in both calls to TRANSLATE, the 3rd argument is a string of 2 '_'s.

  • Possible combination of coins

    Hello, I'm just new here. I have some algorithm problem with algorithm that couldn't put it to a code. The problem is "For a given denomination of money (5, 10, 20, 50, 100), get all possible combinations that will not exceed 2000 and at least one of them is present".
    I know this is easy for others but for me, it's quite difficult. I already searched but I can't find something that is similar with this problem.
    Here's what I tried,
    Given:
    denominations = { 5, 10, 20, 50, 100};
    total = 2000;
    A=5, B=10, C=20, D=50, E=100
    A' = counts of A // A' >= 1
    B' = counts of B // B' >= 1
    C' = counts of C // C' >= 1
    D' = counts of D // D' >= 1
    E' = counts of E // E' >= 1
    A A' + BB' + C*C' + D*D' + E*E' <= total
    The problem is I can't find a way for a possible counts of the denomination.
    Any help is much appreciated.
    Thanks.

    For a brute force approach, consider a simpler problem:
    Can you solve the problem if denominations = {5} and maxValue = 1820?
    For each solution of the problem the above case, does it generate a smaller problem to be solved for the next higher denomination in {5, 10, 20, 50, 100}?
    Can you see how to get a solution to the bigger problem from the smaller ones?
    If you've done data structures in your course, you may also want to make it more efficient by memotisation so you don't recompute the combinations of 20,50,100 given 10 = 2 * 5 etc.

  • Easiest way to sum all possible (different) combinations in an array?

    Hey guys, 
    Say I have an array of sorted values (of attenuation), I want to generate an array of all possible sums of those values, sort it, then refer to it as a list of all possible attenuation values (lets say theyre bits on a discrete digital attenuator).  I will try to show you an example of what I mean with letters:
    Array of attens: [a, b, c, d]
    I want an array like this: [a+b,  a+c,  a+d,  a+b+c,  a+b+d,  a+c+d,  b+c,  b+d,  b+c+d]... I hope I'm not missing any combos here...
    I'm attaching my attempt at this idea but there's something missing I think, I am doing something wrong...
    Thanks for any ideas or help.
    Solved!
    Go to Solution.
    Attachments:
    Sum all possible combos.vi ‏18 KB

    Ben wrote:
    The challenge is getting all possible combos.
    Have you concidered looking at this as a variation on binary counting?
    The total number of combinations is 2^(NumberOfSettings) if you include "None".
    So generate a ramp from zero to the total number possible.
    Convert each number to a boolean array and then use the boolean to determine if its corresponding value gets add into the total.
    After processing all of the values from the ramp the final array should be in ascending order.
    I hope that outline helps,
    Ben
    Ben, that is actually quite brilliantly simple.  I like it.
    There are only two ways to tell somebody thanks: Kudos and Marked Solutions
    Unofficial Forum Rules and Guidelines

  • HT4623 Hello - I have a single question. Anyone who how to undo an update ? The update to iOS 7... was a failure. Is it at all possible to undo an update?

    Hello - I have a single question. Anyone who how to undo an update ? The update to iOS 7... was a failure. Is it at all possible to undo an update?

    It is possible to RE-do one but not UN-do unless you consider both to be the same. But you can not downgrade to an earlier version.

  • Show all the possible combinations of n elements in an array???

    If I have an array of n elements, how to show all the possible combinations of n elements.
    What is the java library for doing the combinations??
    For example,
    public class CombinationTest
    {     public static void main(String[] args)     
         {     ArrayList letters = new ArrayList();
              leters.add("s1");
              leters.add("s2");
              leters.add("s3");
    Find all the combinations of 3 elements: C(3,2). Here's the result:
    1) s1, s2
    2) s1, s3
    3) s2, s3
    }

    There isn't a built-in API method for this. Still, combinations are pretty easy. Pseudocode (borrowing notation from ML and GJ, and assuming immutable sets): Set<Set<Object>> getCombinations(int itemsInCombo, Set<Object> itemsToCombine)
        // Handle base case and exceptional cases
        if (itemsInCombo = 0) return Set.emptySet();
        if (itemsInCombo < 0) throw new Exception("?!");
        // Recursive case
        Object o = itemsToCombine.first(); // use iterator - may throw exception
        // Get combos with first element
        Set<Set<Object>> rv = map (x => x.add(o)) getCombinations(itemsInCombo - 1, itemsToCombine.remove(o));
        // Add combos without first element
        rv.add(getCombinations(itemsInCombo, itemsToCombine.remove(o)));
        return rv;
    }

  • Dependent Values read only or all possible values

    Hello
    I have read the following instructions.
    [How to Configure Predefined Properties with Dependent Values (NW7.0)]|http://www.sdn.sap.com/irj/scn/go/portal/prtroot/docs/library/uuid/30b1ec90-0201-0010-6192-d3de721f8ae4?quicklink=index&overridelayout=true]
    I need this scenario in the Portal search.
    So my question is:
    Is it possible, if Country is empty, that the field City is read only.
    In the How to guide you can see , if no city is selected the property city is a text field(not Dropdown) and you can enter a sting.
    Or it is possible, if no country is selected that the field city shows all possible cities, not restricted by countries.
    For example no country is selected and the field city shows all cities.
    or no country is selected and the field city is read only.
    The html code of dropdown list country is
    <option value="&lt;empty&gt;"><option value="Germany">Germany<option value="France">France
    Value of no selection
    &lt;empty&gt = <empty>
    If no country is selected the selected value is &lt;empty&gt or <empty>.
    I tried this:
    "dependon=Country,*=allcities,Germany=GermanyCity,France=FrenchCity" and so on.
    "dependon=Country,<empty>=allcities,Germany=GermanyCity,France=FrenchCity" and so on.
    "dependon=Country,&lt;empty&gt=allcities,Germany=GermanyCity,France=FrenchCity" and so on.
    without success
    Can some one help?
    Edited by: Dirk Wiegele on Jan 23, 2011 12:21 PM
    Edited by: Dirk Wiegele on Jan 23, 2011 12:22 PM

    Hi
    Yes, it is possible to generate an excel file in SAP. You can use any of the following FM's for the same:
    1. EXCEL_OLE_STANDARD_DAT
    2. GUI_DOWNLOAD
    3. XXL_FULL_API
    Hope it will help.
    Regards,
    Nikita

  • How to get table and column names thats being used in SQL , that's generating all my reports on SSRS.

    Good day,
    I searched through the forum and cant find anything.
    I have around 300 published reports on SSRS and we are busy migrating to a new system.
    They have already setup their tables on the new system and I need to provide them with a list of table names and column names that are being used currently to generate the 300 reports on SSRS.
    We use various tables and databases to generate these reports, and will take me forever to go through each query to get this info.
    Is it at all possible to write a query in SQL 2008 that will give me all the table names and columns being used?
    Your assistance is greatly appreciated.
    I thank you.
    Andre.

    There's no straightforward method for that I guess. There are couple of things you can use to get these details
    1. query the ReportServer.dbo.Catalog table
    for getting details
    you may use script below for that
    http://gallery.technet.microsoft.com/scriptcenter/42440a6b-c5b1-4acc-9632-d608d1c40a5c
    2. Another method is to run the reports and run sql profiler trace on background to retrieve queries used.
    But in some of these cases the report might be using a procedure and you will get only procedure. Then its upto you to get the other details from procedure like tables used, columns etc
    Please Mark This As Answer if it helps to solve the issue Visakh ---------------------------- http://visakhm.blogspot.com/ https://www.facebook.com/VmBlogs

  • Can I set up a review process in LiveCycle? Is what I want to do, at all possible? Anyone chime in.

    What is the best way to set up a review process, without violating the EULA and having a form that is editable by the end user AFTER it is submitted.
    Basically the process will go like this..
    1. A user will log onto our website and click a link where he or she downloads a dynamic form.
    2. The user will submit the form to multiple parties at once. (Conditional e-mail, depending on what sections in the form are filled out, different people will get the form)
    3. The form will get reviewed, and if any changes need to be made, the person that originally filled out the form will need to get it back and make the necessary changes. (The forms are long, so they can't start from scratch).
    4. After the changes have been made, the form will be re-submitted for final review.
    5. After the form has been finalized, the completed form will be distributed to about 80 people within the company (attached to an e-mail or something) and these people will use the data within the form to populate what they need to.
    Additional information:
    -I am the only one that has Acrobat 9 Standard OR Professional.
    -Everyone else has Reader only
    -The forms are dynamic
    -They have to be posted to a website
    Is this at all possible? I can't figure out how this can be done.
    I hope that was clear, please ask if you have questions.

    Hi,
    You can Reader Enable a form in Acrobat v9 Standard, however note the EULA with the restriction on 500 unique recipients. If it is going out to less than 500 you are not limited to the number of data processes that you undertake on the form.
    See discussion here on Reader Enabling a form: http://assure.ly/etkFNU.
    You would need to Reader Enable the form before you put it up on the web page.
    You can script a regular button to submit the form to user-entered email addresses. See here: http://assure.ly/eUR4wJ. The first email button could be used to send out the form for review. The last section in the form could be a text field to allow reviewers to add comments and a second email button that would issued the amended form back to the originator for further action.
    Both email buttons would get their email address(s) from completed fields in the form. In addition the email submission can be set up to return the PDF form and not just the data. IF the form is Reader Enabled then users with Reader will be able to complete this action.
    Hope that helps,
    Niall

Maybe you are looking for