Chris Doyle Chris Doyle -4 years ago 85
Java Question

Java program to identify patterns in numbers

I am looking to create a program that would idetify certain patterns in numbers. I am not sure if this needs an algorithm or just carefully thought out programming. I am not looking for someone to provide source code, just some thought provoking ideas to start me in the right direction.

Numbers will be fixed length of 6 digits from 000000 to 999999. I guess each number would be stored as part of an array. I would then like to test the number against a pattern.

For example lets say the 3 patterns I am using are

A A A A A A - would match such examples as 111111 , 222222, 333333 etc where
A B A B A B - would match such examples as 121212 , 454545, 919191 etc
A (A+1) (A+2) B (B+1) (B+2) - would match such examples as 123345, 789123, 456234


I guess the part that I am stuck on is how to allocate each part of the integer array to a value such as A or B

My inital thoughts would be just to allocate each part as an individual letter. So if the array consisted of 1 3 5 4 6 8, then I would create a map like

A=1
B=3
C=5
D=4
E=6
F=8


Then some how take the first pattern,

AAAAAA


and test with something like if (AAAAAA = ABCDEF) then we have matched AAAAAAA

if not then try (ABABAB = ABCDEF) etc through all my patterns

In this scenario there is no reason why the value assigned to C cant be the same as the value assigend to F like in the number 234874.

I am not sure if this will even make sense to anyone but I guess I can refine my question based on feedback.

In summary, I am looking for ideas on how to have a program accept a 6 digit number and return back to us which pattern it matched.

SOLUTION

After the comments given put me on a good track below is what i created as a final solution.

package com.doyleisgod.number.pattern.finder;

import java.io.File;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;


public class FindPattern {
private final int[] numberArray; //Array that we will match patterns against.
private final Document patternTree = buildPatternTree(); //patternTree containing all the patterns
private final Map<String, Integer> patternisedNumberMap; //Map used to allocate ints in the array to a letter for pattern analysis
private int depth = 0; //current depth of the pattern tree

// take the int array passed to the constructor and store it in out numberArray variable then build the patternised map
public FindPattern (int[] numberArray){
this.numberArray = numberArray;
this.patternisedNumberMap = createPatternisedNumberMap();
}

//builds a map allocating numbers to letters. map is built from left to right of array and only if the number does not exist in the map does it get added
//with the next available letter. This enforces that the number assigned to A can never be the same as the number assigned to B etc
private Map<String, Integer> createPatternisedNumberMap() {
Map<String, Integer> numberPatternMap = new HashMap<String, Integer>();

ArrayList<String> patternisedListAllocations = new ArrayList<String>();
patternisedListAllocations.add("A");
patternisedListAllocations.add("B");
patternisedListAllocations.add("C");
patternisedListAllocations.add("D");
Iterator<String> patternisedKeyIterator = patternisedListAllocations.iterator();

for (int i = 0; i<numberArray.length; i++){
if (!numberPatternMap.containsValue(numberArray[i])) {
numberPatternMap.put(patternisedKeyIterator.next(), numberArray[i]);
}
}
return numberPatternMap;
}

//Loads an xml file containing all the patterns.
private Document buildPatternTree(){
Document document = null;
try {
File patternsXML = new File("c:\\Users\\echrdoy\\Desktop\\ALGO.xml");
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
document = db.parse(patternsXML);

} catch (Exception e){
e.printStackTrace();
System.out.println("Error building tree pattern");
}
return document;
}

//gets the rootnode of the xml pattern list then called the dfsnodesearch method to analyse the pattern against the int array. If a pattern is found a
//patternfound exception is thorwn. if the dfsNodeSearch method returns without the exception thrown then the int array didnt match any pattern
public void patternFinder() {
Node rootnode= patternTree.getFirstChild();
try {
dfsNodeSearch(rootnode);
System.out.println("Pattern not found");
} catch (PatternFoundException p) {
System.out.println(p.getPattern());
}
}

//takes a node of the xml. the node is checked to see if it matches a pattern (this would only be true if we reached the lowest depth so must have
//matched a pattern. if no pattern then analyse the node for an expression. if expression is found then test for a match. the int from the array to be tested
//will be based on the current depth of the pattern tree. as each depth represent an int such as depth 0 (i.e root) represent position 0 in the int array
//depth 1 represents position 1 in the int array etc.
private void dfsNodeSearch (Node node) throws PatternFoundException {
if (node instanceof Element){
Element nodeElement = (Element) node;
String nodeName = nodeElement.getNodeName();

//As this method calls its self for each child node in the pattern tree we need a mechanism to break out when we finally reach the bottom
// of the tree and identify a pattern. For this reason we throw pattern found exception allowing the process to stop and no further patterns.
// to be checked.
if (nodeName.equalsIgnoreCase("pattern")){
throw new PatternFoundException(nodeElement.getTextContent());
} else {
String logic = nodeElement.getAttribute("LOGIC");
String difference = nodeElement.getAttribute("DIFFERENCE");

if (!logic.equalsIgnoreCase("")&&!difference.equalsIgnoreCase("")){
if (matchPattern(nodeName, logic, difference)){
if (node.hasChildNodes()){
depth++;
NodeList childnodes = node.getChildNodes();
for (int i = 0; i<childnodes.getLength(); i++){
dfsNodeSearch(childnodes.item(i));
}
depth--;
}
}
}
}


}
}

//for each node at a current depth a test will be performed against the pattern, logic and difference to identify if we have a match.
private boolean matchPattern(String pattern, String logic, String difference) {
boolean matched = false;
int patternValue = patternisedNumberMap.get(pattern);

if (logic.equalsIgnoreCase("+")){
patternValue += Integer.parseInt(difference);
} else if (logic.equalsIgnoreCase("-")){
patternValue -= Integer.parseInt(difference);
}

if(patternValue == numberArray[depth]){
matched=true;
}

return matched;
}


}


the xml list of patterns looks like this

<?xml version="1.0"?>
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(A)(A)</pattern>
</A>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(B)(A)</pattern>
</B>
</A>
<B LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(B)(A)</pattern>
</A>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(A)(B)(B)</pattern>
</B>
</B>
</A>
<A LOGIC="+" DIFFERENCE="2">
<A LOGIC="+" DIFFERENCE="4">
<A LOGIC="+" DIFFERENCE="6">
<pattern>(A)(A+2)(A+4)(A+6)</pattern>
</A>
</A>
</A>
<B LOGIC="=" DIFFERENCE="0">
<A LOGIC="+" DIFFERENCE="1">
<B LOGIC="+" DIFFERENCE="1">
<pattern>(A)(B)(A+1)(B+1)</pattern>
</B>
</A>
<A LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(A)(A)</pattern>
</A>
<B LOGIC="+" DIFFERENCE="1">
<pattern>(A)(B)(A)(B+1)</pattern>
</B>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(A)(B)</pattern>
</B>
</A>
<B LOGIC="=" DIFFERENCE="0">
<A LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(B)(A)</pattern>
</A>
<B LOGIC="=" DIFFERENCE="0">
<pattern>(A)(B)(B)(B)</pattern>
</B>
</B>
<A LOGIC="-" DIFFERENCE="1">
<B LOGIC="-" DIFFERENCE="1">
<pattern>(A)(B)(A-1)(B-1)</pattern>
</B>
</A>
</B>
<A LOGIC="+" DIFFERENCE="1">
<A LOGIC="+" DIFFERENCE="2">
<A LOGIC="+" DIFFERENCE="3">
<pattern>(A)(A+1)(A+2)(A+3)</pattern>
</A>
</A>
</A>
<A LOGIC="-" DIFFERENCE="1">
<A LOGIC="-" DIFFERENCE="2">
<A LOGIC="-" DIFFERENCE="3">
<pattern>(A)(A-1)(A-2)(A-3)</pattern>
</A>
</A>
</A>
</A>


and my pattern found exception class looks like this

package com.doyleisgod.number.pattern.finder;

public class PatternFoundException extends Exception {
private static final long serialVersionUID = 1L;
private final String pattern;

public PatternFoundException(String pattern) {
this.pattern = pattern;
}

public String getPattern() {
return pattern;
}

}


Not sure if any of this will help anyone with a similar problem or if anyone has any comments on how this is working would be great to hear them.

Answer Source

I would suggest to build state machine:

A. Initialization with patterns:

  1. Normalize all patterns
  2. Build a tree with depth = 6 that represents all patterns starting from root and all possible choices on every depth.

B. Run state machine


A.1. Pattern normalization.

A A A A A A => A0 A0 A0 A0 A0 A0

C A C A C A => A0 B0 A0 B0 A0 B0 (always start with A, and then B, C, D, etc.)

B B+1 B+2 A A+1 A+2 => A0 A1 A2 B0 B1 B2

Thus, you always have normalized pattern start with A0.

A.2. Build a tree

1.       A0
       /  | \
2.   A0  B0  A1
      |   |   |
3.   A0  A0  A2
      |   |   |
4.   A0  B0  B0
      |   |   |
5.   A0  A0  B1
      |   |   |
6.   A0  B0  B2
      |   |   |
     p1  p2  p3

B. Run state machine

Use Depth-first search algorithm using recursion to find matched pattern.

Does it make sense?

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