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# Finding all permutations that match a set of rules

By : Bala
Date : October 21 2020, 11:12 PM
With these it helps Just as a hint.
You can treat your set of rules as a graph. Each index is a vertex, each rule is a directed edge.
code :

Share :

## Finding a set of permutations, with a constraint

By : Vanamali
Date : March 29 2020, 07:55 AM
I wish did fix the issue. What you want is a combinatorial block design. Using the nomenclature on the linked page, you want designs of size (n^2, n, 1) for maximum k. This will give you n(n+1) permutations, using your nomenclature. This is the maximum theoretically possible by a counting argument (see the explanation in the article for the derivation of b from v, k, and lambda). Such designs exist for n = p^k for some prime p and integer k, using an affine plane. It is conjectured that the only affine planes that exist are of this size. Therefore, if you can select n, maybe this answer will suffice.
However, if instead of the maximum theoretically possible number of permutations, you just want to find a large number (the most you can for a given n^2), I am not sure what the study of these objects is called.

## In ANTLR, is there a shortcut notation for expressing alternation of all the permutations of some set of rules?

By : Rita li
Date : March 29 2020, 07:55 AM
Hope this helps In ANTLR I want to define a rule like this: , I would just match any a, b or c once or more:
code :
``````rule
: ( a | b | c )+
;
``````
``````grammar Permutation;

parse
:  permutation[5] {System.out.println("parsed: " + \$permutation.text);} EOF
;

permutation[final int n]
@init{
java.util.Set set = new java.util.HashSet();
int counter = n;
}
:  (
{counter > 0}?=> token   // keep matching a `token` as long as `counter > 0`
{                        //
counter--;             // decrease `counter`
}                        //
)+
{set.size() == n}?         // if `set.size() != n`, an exception is thrown
;

token
:  A
|  B
|  C
|  D
|  E
;

A : 'A';
B : 'B';
C : 'C';
D : 'D';
E : 'E';

Space : ' ' {skip();};
``````
``````import org.antlr.runtime.*;

public class Main {
public static void main(String[] args) throws Exception {
PermutationLexer lexer = new PermutationLexer(new ANTLRStringStream(args[0]));
PermutationParser parser = new PermutationParser(new CommonTokenStream(lexer));
parser.parse();
}
}
``````
``````java -cp antlr-3.3.jar org.antlr.Tool Permutation.g
javac -cp antlr-3.3.jar *.java

java -cp .:antlr-3.3.jar Main "A B C D E"
parsed: ABCDE

java -cp .:antlr-3.3.jar Main "B D C E A"
parsed: BDCEA

java -cp .:antlr-3.3.jar Main "A B C D B"
line 1:9 rule permutation failed predicate: {set.size() == n}?
parsed: null
``````

## Finding permutations with foldl/map?

By : Gauri
Date : March 29 2020, 07:55 AM
code :
``````(define (insert l n e)
(if (= 0 n)
(cons e l)
(cons (car l)
(insert (cdr l) (- n 1) e))))

(define (seq start end)
(if (= start end)
(list end)
(cons start (seq (+ start 1) end))))

(define (permute l)
(if (null? l)
'(())
(apply append (map (lambda (p)
(map (lambda (n)
(insert p n (car l)))
(seq 0 (length p))))
(permute (cdr l))))))
``````

## WEKA FP-growth association rules not finding rules

By : Teodor Stelian Iones
Date : March 29 2020, 07:55 AM
around this issue one reason could be that your support and/or confidence value are too high. try low ones. e.g. a support and confidence level of 0.001%. another reason could be that your data set just doesn't contain any association rules. try another data set which certainly contains association rules from a set minimum support and confidence value.

## Finding even permutations using Haskell

By : İhsan Soydaş
Date : March 29 2020, 07:55 AM
wish help you to fix your issue To take a bird’s-eye view, what we want to do is count how many swaps we would need to sort the list in place. (Equivalently, in the language of group theory, to decompose the permutation into transpositions.) Which sorting algorithm we use to count swaps doesn’t matter for correctness: an even permutation will generate an even number of swaps and an odd permutation an odd number of swaps regardless of how we sort.
Let’s look first at: