Construct a model of an electric circuit in java - java

I was recently in a job interview for a position as Java Developer. I was given a task: To think about a good way to represent an electric circuit (such as the one in the figure below) in Java.
A circuit is a combination of logic gates XOR, AND, OR etc.. Each gate has two input ports and an output port. Each output is connected to an input of another gate, which goes all the way to a higher gate (as in the picture). Make the system simple, no cycles are allowed (although real life circuits can have them).
I was asked to think about a good way to represent this model in Java, using the following guidelines:
I am given a circuit and a list of values which should be provided to its inputs.
I need to create a model to represent the circuit in Java, i.e., I need to define classes and an API that could be used to represent the circuit.
According to the input values and how the gates are connected, I need to calculate the output that the represented circuit would produce.
I need to think about a way to represent the board, use abstract classes or interfaces, and show understanding of the model (and if needed to use pattern design).
I chose to design the system as a tree and the interviewer told me it was a good choice. Then I build those classes:
Node
public class gate_node {
gate_node right_c,left_c;
Oprtator op;
int value;
int right_v,left_v;
public gate_node(gate_node right,gate_node left,Oprtator op){
this.left_c=left;
this.right_c=right;
this.op=op;
right_v=left_v=0;
}
}
Tree
public class tree {
gate_node head;
tree(gate_node head) {
this.head = head;
}
void go_right() {
head = head.right_c;
}
void go_left() {
head = head.left_c;
}
static int arr[] = { 0, 0, 1, 0 };
static int counter=0;
static int compute(gate_node head) {
if ((head.left_c == null) && (head.right_c == null))
{
int ret=(head.op.calc(arr[counter], arr[counter+1]));
counter++;
counter++;
return ret;
}
return (head.op.calc(compute(head.left_c), compute(head.right_c)));
}
public static void main(String[] args) {
tree t = new tree(new gate_node(null, null, new and()));
t.head.left_c = new gate_node(null, null, new and());
t.head.right_c = new gate_node(null, null, new or());
System.out.println(tree.compute(t.head));
}
}
Oprtator class:
public abstract class Oprtator {
abstract int calc(int x, int y);
}
OR gate:
public class or extends Oprtator {
public int calc(int x, int y){
return (x|y);
}
}
In the above code, I implemented the board as a tree with its a current head (which can go down to the left/right children). Every node has 2 children (also node type), 2 entries (0/1), a value and an operator (abstract class, could be extended by OR/AND..).
I used a counter and an array to insert the values to the appropriate leafs of the tree (as mentioned in the code).
It works but still I got a feeling that my interviewer wanted something more. Is my code is correct? Does anyone have a better way of represent this electrical board and how to give a good output (in terms of complexity or using better connection from one class to another, design patterns, something else?)

It's not a "perfect" answer, but you can solve this using a few classes to hold the logical connection/evaluation and then recursively evaluate the circuit.
Create a base class LogicalNode and give it a list of inputs to manage. Give it a base class function to evaluate all the inputs and return an output. This gets overridden in derived classes. Each type of node (INPUT, NOT, AND, OR) gets a derived class with a special "ComputOutput" overridden version. If you compute the output at the output node, it should recurse up the tree, computing all inputs of inputs of inputs, etc., till it reaches the "INPUT" nodes, which are fixed/logical inputs to the system.
You can create new types fairly quickly (see code). Not a lot of comments here, but it should be somewhat self explanatory.
Something like this (in C#):
public class LogicalNode
{
private List<LogicalNode> _inputs = new List<LogicalNode>();
private String _name = "Not Set";
public override string ToString()
{
return String.Format("Node {0}", _name);
}
public void Reset()
{
_inputs.Clear();
}
public void SetName(String name)
{
_name = name;
}
protected List<LogicalNode> GetInputs()
{
return _inputs;
}
public void AddInput(LogicalNode node)
{
_inputs.Add(node);
}
protected virtual bool ComputeOutputInternal()
{
return false;
}
public bool ComputeOutput()
{
// Console.WriteLine("Computing output on {0}.", _name);
return ComputeOutputInternal();
}
}
public class LogicalInput : LogicalNode
{
private bool _state = true;
public void SetState(bool state)
{
_state = state;
}
public bool GetState() { return _state; }
protected override bool ComputeOutputInternal()
{
return _state;
}
}
public class LogicalAND : LogicalNode
{
protected override bool ComputeOutputInternal()
{
List<LogicalNode> inputs = GetInputs();
bool result = true;
for (Int32 idx = 0; idx < inputs.Count && result; idx++)
{
result = result && inputs[idx].ComputeOutput();
}
return result;
}
}
public class LogicalOR : LogicalNode
{
protected override bool ComputeOutputInternal()
{
List<LogicalNode> inputs = GetInputs();
bool result = false;
for (Int32 idx = 0; idx < inputs.Count; idx++)
{
result = inputs[idx].ComputeOutput();
if (result)
// If we get one true, that is enough.
break;
}
return result;
}
}
public class LogicalNOT : LogicalNode
{
protected override bool ComputeOutputInternal()
{
List<LogicalNode> inputs = GetInputs();
if (inputs.Count > 0)
{ // NOTE: This is not an optimal design for
// handling distinct different kinds of circuits.
//
// It it demonstrative only!!!!
return !inputs[0].ComputeOutput();
}
return false;
}
And then to (quickly) test it:
static void Main(string[] args)
{
// The test circuit
// !((A&&B) || C)
// A B C Out
// 1 1 1 0
// 1 1 0 0
// 1 0 1 0
// 1 0 0 1
// 0 1 1 0
// 0 1 0 1
// 0 0 1 0
// 0 0 0 1
//
//
//
/* ------- -------
* A - | | | |
* | AND |-----| | -------
* B - | (D) | | | | |
* ------- | OR |----| NOT |----
* | (E) | | (F) |
* C --------------| | | |
* ------- -------
*/
LogicalInput A = new LogicalInput();
LogicalInput B = new LogicalInput();
LogicalInput C = new LogicalInput();
LogicalAND D = new LogicalAND();
LogicalOR E = new LogicalOR();
LogicalNOT F = new LogicalNOT();
A.SetName("A");
B.SetName("B");
C.SetName("C");
D.SetName("D");
E.SetName("E");
F.SetName("F");
D.AddInput(A);
D.AddInput(B);
E.AddInput(D);
E.AddInput(C);
F.AddInput(E);
// Truth Table
bool[] states = new bool[]{ true, false };
for(int idxA = 0; idxA < 2; idxA++)
{
for(int idxB = 0; idxB < 2; idxB++)
{
for(int idxC = 0; idxC < 2; idxC++)
{
A.SetState(states[idxA]);
B.SetState(states[idxB]);
C.SetState(states[idxC]);
bool result = F.ComputeOutput();
Console.WriteLine("A = {0}, B = {1}, C = {2}, Output = {3}",
A.GetState(), B.GetState(), C.GetState(), result.ToString());
}
}
}
}
}
With output:
A = True, B = True, C = True, Output = False
A = True, B = True, C = False, Output = False
A = True, B = False, C = True, Output = False
A = True, B = False, C = False, Output = True
A = False, B = True, C = True, Output = False
A = False, B = True, C = False, Output = True
A = False, B = False, C = True, Output = False
A = False, B = False, C = False, Output = True
Was this helpful?

While I obviously can't say exactly what the interviewer was looking for, if I were interviewing you I would probably press you to make your compute method a non-static member of your gate_node class. This way, your networks do not have to be "balanced" (they can be deeper, have more inputs) on one side or the other.
Put another way, looking at your compute code, I do not believe it would work for general circuits.
Perhaps something like the following (in gate_node):
int compute() {
/* The following use of a static sInputCounter assumes that the static/global
* input array is ordered from left to right, irrespective of "depth". */
final int left = (null != left_c ? left_c.compute() : sInput[sInputCounter++]);
final int right = (null != right_c ? right_c.compute() : sInput[sInputCounter++]);
return op.calc(left, right);
}
This way, the "tree" can just be represented by the head/root gate_node (although you still probably want a class like your tree class -- I might call it "network" just to avoid confusion, with static methods for constructing the tree, setting up the inputs, etc.) and you trigger the network evaluation by calling head.compute().
Of course, you are still left with the non-trivial problem of constructing a network from some external specification. I imagine that another issue for the interviewer could have been that this aspect was not well-specified in your solution. (Nor in mine here either, I'm sorry.)

Related

Print Tree with 4 nodes (simple forest) for checking a benchmark

I implemented an experimental OOP language and now benchmark garbage collection using a Storage benchmark. Now I want to check/print the following benchmark for small depths (n=2, 3, 4,..).
The tree (forest with 4 subnode) is generated by the buildTreeDepth method. The code is as follows:
import java.util.Arrays;
public final class StorageSimple {
private int count;
private int seed = 74755;
public int randomNext() {
seed = ((seed * 1309) + 13849) & 65535;
return seed;
}
private Object buildTreeDepth(final int depth) {
count++;
if (depth == 1) {
return new Object[randomNext() % 10 + 1];
} else {
Object[] arr = new Object[4];
Arrays.setAll(arr, v -> buildTreeDepth(depth - 1));
return arr;
}
}
public Object benchmark() {
count = 0;
buildTreeDepth(7);
return count;
}
public boolean verifyResult(final Object result) {
return 5461 == (int) result;
}
public static void main(String[] args) {
StorageSimple store = new StorageSimple();
System.out.println("Result: " + store.verifyResult(store.benchmark()));
}
}
Is there a somewhat simple/straight forward way to print the tree generated by buildTreeDepth? Just the short trees of n=3, 4, 5.
As other has already suggested, you may choose some lib to do so. But if you just want a simple algo to test in command line, you may do the following, which I always use when printing tree in command line (write by handle, may have some bug. Believe you can get what this BFS algo works):
queue.add(root);
queue.add(empty);
int count = 1;
while (queue.size() != 1) {
Node poll = queue.poll();
if (poll == empty) {
count = 1;
queue.add(empty);
}
for (Node n : poll.getChildNodes()) {
n.setNodeName(poll.getNodeName(), count++);
queue.add(n);
}
System.out.println(poll.getNodeName());
}
Sample output:
1
1-1 1-2 1-3 1-4
1-1-1 1-1-2 1-1-3 1-2-1 1-2-2 1-3-1 1-3-2 1-4-1
...
And in your case you use array, which seems even easier to print.
Instead of using object arrays, use a List implementation like ArrayList. For an improved better result subclass ArrayList to also hold a 'level' value and add indentation to the toString() method.

Having 3 containers (2 full and 1 empty) and trying to create x amount from them

Note: I came across the question below and I wanted to generalize the problem and implement it, but it turns out that it is not easy. This question is making me crazy. This is NOT a Homework question just a curiosity.
Question
There are three containers whose sizes are 10 pints, 7 pints and 4 pints respectively. The 7-pint and 4-pint containers start out full of water, but the 10-pint container is initially empty.
Since there are no marks on the containers, you can pour the contents of one container into another and stop under the following conditions:
the source container is empty
the destination container is full
What sequence of moves should you make if you want to isolate exactly 2 pints of water?
Source: page 102, Question 3.8
Solution
Answering that question is easy using directed graph data structure where nodes contain tuples to represent a certain state.
We start from the initial state (node), and then we create a node representing a possible next state, and we connect it to initial node, and then run BFS to find the shortest path.
Pattern of each state (or node): <10-pint container, 7-pint container, 4-pint container>
Initial state or node: <0, 7, 4>.
The nodes connected to the initial state (or node): <7, 0, 4>, <4, 7, 0>, as you can see from the picture.
Generalised Question
But suppose if want to generalized the problem, suppose we have three containers whose sizes are x, y and z pints respectively such that x >= y >= z.
The y-pint and z-pint containers start out full of water but the x-pint container is initially empty.
What sequence of moves should you make if you want to isolate exactly a pints of water?
Suggesting a Solution to Generalised Version
Here (DropBox, GitHub) is my source code so far.
Here are two important method in main class. They fill the graph based on all possibilities, and it also makes sure there is no duplicate node.
public static void fillGraph(int x, int y, int z) {
TupleContainer initialState = new TupleContainer(x, y, z);
TupleContainer currentState = initialState;
Iterator<TupleContainer> it, it_1, it_2, it_3;
Graph.addNode(initialState);
it = addAdjacentEdgesToTuple(currentState).iterator();
while (it.hasNext()) {
it_1 = addAdjacentEdgesToTuple(it.next()).iterator();
while (it_1.hasNext()) {
it_2 = addAdjacentEdgesToTuple(it.next()).iterator();
while (it_2.hasNext()) {
it_3 = addAdjacentEdgesToTuple(it.next()).iterator();
while (it_3.hasNext()) {
addAdjacentEdgesToTuple(it.next()).iterator();
}
}
}
}
public static Collection<TupleContainer> addAdjacentEdgesToTuple(
TupleContainer currentState) {
TupleContainer tempTupleContainer;
Collection<TupleContainer> CollectionLevel;
Iterator<TupleContainer> it;
CollectionLevel = currentState.MixUpContainers();
it = CollectionLevel.iterator();
while (it.hasNext()) {
tempTupleContainer = it.next();
if (graphContains(tempTupleContainer) != null)
Graph.addNode(tempTupleContainer);
else
tempTupleContainer = graphContains(tempTupleContainer);
Graph.addEdge(currentState, tempTupleContainer);
}
return CollectionLevel;
}
My Question
My code just fills the graph to depth of 4, but how can I set depth and make it run recursively or how make it run until all possibilities are taken to consideration without going into infinite loop. What is the algorithm to this generalized question?
Hm ... there may be better algorithms, but if you simply want arbitrarily deep recursion without going into endless loops, you can use a breadth first search that visits each node only once, i.e. if it hasn't already been visited:
public class Test {
public static void main(String[] args) throws Exception {
State initialState = new State(null, 0, 7, 4);
Set<State> reached = new HashSet<>();
Queue<State> pending = new ArrayDeque<>();
pending.add(initialState);
while (!pending.isEmpty()) {
State state = pending.remove();
if (isGoal(state)) {
printPathTo(state);
return;
}
for (State s : state.adjacentStates()) {
if (!reached.contains(s)) {
reached.add(s);
pending.add(s);
}
}
}
System.out.println("There appears to be no solution.");
}
private static boolean isGoal(State state) {
for (int a : state.content) {
if (a == 2) {
return true;
}
}
return false;
}
private static void printPathTo(State state) {
if (state != null) {
printPathTo(state.previous);
System.out.println(state);
}
}
}
class State {
final static int[] capacity = { 10, 7, 4 };
final int[] content;
final State previous;
public State(State previous, int... content) {
this.content = content;
this.previous = previous;
}
Iterable<State> adjacentStates() {
List<State> result = new ArrayList<>();
for (int i = 0; i < content.length; i++) {
for (int j = 0; j < content.length; j++) {
if (i != j) {
int[] newContent = Arrays.copyOf(content, content.length);
int movedQuantity = Math.min(content[i], capacity[j] - content[j]);
newContent[i] -= movedQuantity;
newContent[j] += movedQuantity;
result.add(new State(this, newContent));
}
}
}
return result;
}
#Override
public int hashCode() {
return Arrays.hashCode(content);
}
#Override
public boolean equals(Object obj) {
return Arrays.equals(content, ((State) obj).content);
}
#Override
public String toString() {
return Arrays.toString(content);
}
}
You can also try iterative deepening search, we were given a demonstration of it working on the same problem in uni and it worked well.

JAVA Wrong output in object array program (beginner)

I have an assignment to calculate the amount of carbon dioxide produced in a year from
a household and compare how recycling can reduce its CO2 Footprint.
There are two classes in the program, CO2FromWaste, and CO2FromWasteTester.
The first class, CO2FromWaste is:
public class CO2FromWaste
{
//declaration of private instance variables
private int numPeople;
private boolean Paper, Plastic, Glass, Cans;
private double grossWasteEmission, wasteReduction, netWasteReduction;
//constructor
CO2FromWaste(int people, boolean paper, boolean plastic, boolean glass, boolean cans){
numPeople = people;
Paper = paper;
Plastic = plastic;
Glass = glass;
Cans = cans;
grossWasteEmission = 0.0;
wasteReduction = 0.0;
netWasteReduction = 0.0;
}
public void calcGrossWasteEmission(){
grossWasteEmission = numPeople * 1018;
}
public double getGrossWasteEmission(){
return grossWasteEmission;
}
public void calcWasteReduction(){
if (Paper == true){
wasteReduction = numPeople * 184;
}
if (Plastic == true){
wasteReduction += (numPeople * 25.6);
}
if (Glass == true){
wasteReduction+= (numPeople*46.6);
}
if (Cans == true){
wasteReduction+=(numPeople*165.8);
}
}
public double getWasteReduction()
{
return wasteReduction;
}
public void calcNetWasteReduction(){
netWasteReduction = grossWasteEmission = wasteReduction;
}
public double getNetWasteReduction(){
return netWasteReduction;
}
public int getNumPeople(){
return numPeople;
}
public boolean getPaper(){
return Paper;
}
public boolean getPlastic(){
return Plastic;
}
public boolean getGlass(){
return Glass;
}
public boolean getCans(){
return Cans;
}
}
The second class, CO2FromWasteTester is:
import java.util.ArrayList;
public class CO2FromWasteTester
{
public static void main(String[]args){
ArrayList<CO2FromWaste> waste = new ArrayList<CO2FromWaste>();
waste.add(new CO2FromWaste(1, true, true, true, true));
waste.add(new CO2FromWaste(3, true, false, true, true));
waste.add(new CO2FromWaste(4, false, false, false, false));
waste.add(new CO2FromWaste(1, true, true, true, true));
waste.add(new CO2FromWaste(1, true, true, true, true));
CO2FromWaste wasteRecord;
for (int index = 0; index < waste.size(); index++){
wasteRecord = waste.get(index);
wasteRecord.calcGrossWasteEmission();
wasteRecord.calcWasteReduction();
wasteRecord.calcNetWasteReduction();
}
System.out.println(" Household Waste Recycled Total Pounds of CO2 Net");
System.out.println(" Index People Paper Plastic Glass Cans Emission Reduction Emission ");
for (int index = 0; index < waste.size(); index ++)
{
wasteRecord = waste.get(index);
System.out.printf("%3d %9d %10s %10s %10s %10s %12.2f %10.2f %10.2f%n",index,wasteRecord.getNumPeople(),wasteRecord.getPaper(), wasteRecord.getPlastic(), wasteRecord.getGlass(),wasteRecord.getCans(),wasteRecord.getGrossWasteEmission(),wasteRecord.getWasteReduction(),wasteRecord.getNetWasteReduction());
}
}
}
The output should read like a table, with the correct data under the headers.
For the first line, the output should be
0 1 true true true true 1018.00 422.00 596.00
but it reads
0 1 true true true true 422.00 422.00 422.00
There is something wrong starting the the gross emission part, and that part should be fairly simple because all it needs to do is to multiply the number of people given by 1018.
I am not sure what to do from here and would appreciate some help.
This is the problem:
public void calcNetWasteReduction(){
netWasteReduction = grossWasteEmission = wasteReduction;
}
That's equivalent to:
public void calcNetWasteReduction(){
grossWasteEmission = wasteReduction;
netWasteReduction = grossWasteEmission;
}
In other words, it's modifying grossWasteEmission when it shouldn't be. I suspect you wanted:
public void calcNetWasteReduction(){
netWasteReduction = grossWasteEmission - wasteReduction;
}
In other words, making the second = a - instead.
It's not clear why you have the three separate methods at all, to be honest - why not perform the calculations in the constructor? Or in the getter methods (removing the intermediate fields altogether)?
Additionally, consider making an enum of waste reduction types (PAPER, GLASS, PLASTIC) and taking an EnumSet of them. If you do keep them as individual parameters, I would strongly advise you to change the names to be more conventional Java (paper instead of Paper etc) and use if (paper) rather than if (paper == true) etc. In general, avoid explicit checking against Boolean constants.
The problem is here:
for (int index = 0; index < waste.size(); index++){
wasteRecord = waste.get(index);
wasteRecord.calcGrossWasteEmission();
wasteRecord.calcWasteReduction();
wasteRecord.calcNetWasteReduction(); // <== netWasteReduction = grossWasteEmission = wasteReduction;
}
You just need to replace the '=' sign below. Its because of this line that you are getting all the three values equal. :-)
public void calcNetWasteReduction(){
netWasteReduction = grossWasteEmission = wasteReduction;
Hope this helps

Most optimal way to combine unique groups of size k from n that meet specific requirements

I have been working on something the past few days that seems to be working as intended, however I am looking for ways to improve it. I have a set of n items, and I need to put together groups of these items that MUST meet ALL of the following requirements:
2 items from Category A
2 items from Category B
2 items from Category C
2 Items from Category D
1 item from Category E
I am currently using the following recursive method to put my groups together and the isValid() method is being used to determine if the group meets the criteria.
void getGroups(String[] arr, int len, int startPosition, String[] result) {
if(len == 0) {
Group group = new Group(Arrays.asList(result));
if(group.isValid()) {
validGroups.add(group);
group.printGroup();
}
return;
}
for(int i = startPosition; i <= arr.length - len; i++) {
result[result.length - len] = arr[i];
getGroups(arr, len - 1, i + 1, result);
}
}
I am able to see valid results get printed as the program runs, however the original size of items that I am working with can be well over 100 items. This means there is a very large number of total possible groups that will be iterated through and a lot of times the program never actually completes.
I know that there are currently a bunch of wasted iterations, for example if at some point I detect a group is invalid because it has 3 items from Category A, I should be able to move on. I am not sure if my current method with a few tweaks is the best way to go about this, or if I should separate the items into their respective groups first, and then from their put only valid combinations together. Any help would be appreciated. Thanks.
EDIT: I tried to make the method a bit more simpler than my actual method. My actual method takes in an array of Objects that I've created that contain their value along with their category. I guess for the example we can assume that each category is represented by a list of Strings that it contains. The method can be called like:
String[] items = {"test1", "test2", "test3", "test4", "test5", "test6", "test7",
"test8", "test9", "test10", "test11", "test12", "test13",
"test14", "test15", "test16", "test17", "test18"};
getGroups(items, 9, 0, new String[9]);
EDIT2:
List<String> catA = new ArrayList<String>();
catA.add("test1");
catA.add("test2");
catA.add("test3");
catA.add("test4");
List<String> catB = new ArrayList<String>();
catB.add("test5");
catB.add("test6");
catB.add("test7");
catB.add("test8");
List<String> catC = new ArrayList<String>();
catC.add("test9");
catC.add("test10");
catC.add("test11");
catC.add("test12");
List<String> catS = new ArrayList<String>();
catD.add("test13");
catD.add("test14");
catD.add("test15");
catD.add("test16");
List<String> catE = new ArrayList<String>();
catE.add("test17");
catE.add("test18");
Output:
{"test1", "test2", "test5", "test6", "test9", "test10", "test13", "test14", "test17"}
{"test1", "test2", "test5", "test6", "test9", "test10", "test13", "test14", "test18"}
{"test1", "test2", "test5", "test6", "test9", "test10", "test13", "test16", "test17"}
{"test1", "test2", "test5", "test6", "test9", "test10", "test13", "test15", "test17"}
{"test1", "test2", "test5", "test6", "test9", "test10", "test14", "test15", "test17"}
etc...
This seems to work.
I use a BitPattern iterator I wrote a while ago that walks all n-bit numbers containing just k set bits and uses that to select from your categories.
Note that much of this code is building the test data to reflect your requirements.
I hold a List of Iterables which are the BitPatterns. A list of Iterators which are the currently in-use Iterators from the BitPatterns (they must be renewed every time they complete) and a List of BigIntgers that are the current values to explode into selections from the data.
public class Test {
enum Category {
A(2), B(2), C(2), D(2), E(1);
public final int required;
Category(int required) {
this.required = required;
}
}
private static final Category[] categories = Category.values();
static class Categorised {
final String name;
final Category category;
Categorised(String name, Category category) {
this.name = name;
this.category = category;
}
#Override
public String toString() {
return category.name() + ":" + name;
}
}
static final List<Categorised> data = new ArrayList<>();
static {
data.add(new Categorised("A-1", Category.A));
data.add(new Categorised("A-2", Category.A));
data.add(new Categorised("A-3", Category.A));
data.add(new Categorised("B-1", Category.B));
data.add(new Categorised("B-2", Category.B));
data.add(new Categorised("B-3", Category.B));
data.add(new Categorised("C-1", Category.C));
data.add(new Categorised("C-2", Category.C));
data.add(new Categorised("C-3", Category.C));
data.add(new Categorised("D-1", Category.D));
data.add(new Categorised("D-2", Category.D));
data.add(new Categorised("D-3", Category.D));
data.add(new Categorised("E-1", Category.E));
data.add(new Categorised("E-2", Category.E));
data.add(new Categorised("E-3", Category.E));
}
// Categorise the data.
private Map<Category, List<Categorised>> categorise(List<Categorised> data) {
Map<Category, List<Categorised>> categorised = new EnumMap<>(Category.class);
for (Categorised d : data) {
List<Categorised> existing = categorised.get(d.category);
if (existing == null) {
existing = new ArrayList<>();
categorised.put(d.category, existing);
}
existing.add(d);
}
return categorised;
}
public void test() {
// Categorise the data.
Map<Category, List<Categorised>> categorised = categorise(data);
// Build my lists.
// A source of Iteratprs.
List<BitPattern> is = new ArrayList<>(categories.length);
// The Iterators.
List<Iterator<BigInteger>> its = new ArrayList<>(categories.length);
// The current it patterns to use to select.
List<BigInteger> next = new ArrayList<>(categories.length);
for (Category c : categories) {
int k = c.required;
List<Categorised> from = categorised.get(c);
// ToDo - Make sure there are enough.
int n = from.size();
// Make my iterable.
BitPattern p = new BitPattern(k, n);
is.add(p);
// Gather an Iterator.
Iterator<BigInteger> it = p.iterator();
// Store it.
its.add(it);
// Prime it.
next.add(it.next());
}
// Walk the lists.
boolean stepped;
do {
// Interpret the current numbers.
List<Categorised> candidates = new ArrayList<>();
for ( int c = 0; c < categories.length; c++ ) {
BigInteger b = next.get(c);
List<Categorised> category = categorised.get(categories[c]);
// Step through the bits in the number.
BitSet bs = BitSet.valueOf(b.toByteArray());
for (int i = bs.nextSetBit(0); i >= 0; i = bs.nextSetBit(i + 1)) {
// Pull those entries from the categorised list.
candidates.add(category.get(i));
}
}
// Print it for now.
System.out.println(candidates);
// Step again.
stepped = step(is, its, next);
} while (stepped);
}
// Take one step.
private boolean step(List<BitPattern> is, List<Iterator<BigInteger>> its, List<BigInteger> next) {
boolean stepped = false;
// Step each one until we make one successful step.
for (int i = 0; i < is.size() && !stepped; i++) {
Iterator<BigInteger> it = its.get(i);
if (it.hasNext()) {
// Done here!
stepped = true;
} else {
// Exhausted - Reset it.
its.set(i, it = is.get(i).iterator());
}
// Pull that one.
next.set(i, it.next());
}
return stepped;
}
public static void main(String args[]) {
new Test().test();
}
}
This is the BitPattern iterator.
/**
* Iterates all bit patterns containing the specified number of bits.
*
* See "Compute the lexicographically next bit permutation"
* http://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation
*
* #author OldCurmudgeon
*/
public class BitPattern implements Iterable<BigInteger> {
// Useful stuff.
private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = ONE.add(ONE);
// How many bits to work with.
private final int bits;
// Value to stop at. 2^max_bits.
private final BigInteger stop;
// Should we invert the output.
private final boolean not;
// All patterns of that many bits up to the specified number of bits - invberting if required.
public BitPattern(int bits, int max, boolean not) {
this.bits = bits;
this.stop = TWO.pow(max);
this.not = not;
}
// All patterns of that many bits up to the specified number of bits.
public BitPattern(int bits, int max) {
this(bits, max, false);
}
#Override
public Iterator<BigInteger> iterator() {
return new BitPatternIterator();
}
/*
* From the link:
*
* Suppose we have a pattern of N bits set to 1 in an integer and
* we want the next permutation of N 1 bits in a lexicographical sense.
*
* For example, if N is 3 and the bit pattern is 00010011, the next patterns would be
* 00010101, 00010110, 00011001,
* 00011010, 00011100, 00100011,
* and so forth.
*
* The following is a fast way to compute the next permutation.
*/
private class BitPatternIterator implements Iterator<BigInteger> {
// Next to deliver - initially 2^n - 1
BigInteger next = TWO.pow(bits).subtract(ONE);
// The last one we delivered.
BigInteger last;
#Override
public boolean hasNext() {
if (next == null) {
// Next one!
// t gets v's least significant 0 bits set to 1
// unsigned int t = v | (v - 1);
BigInteger t = last.or(last.subtract(BigInteger.ONE));
// Silly optimisation.
BigInteger notT = t.not();
// Next set to 1 the most significant bit to change,
// set to 0 the least significant ones, and add the necessary 1 bits.
// w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
// The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the number of trailing zeros.
next = t.add(ONE).or(notT.and(notT.negate()).subtract(ONE).shiftRight(last.getLowestSetBit() + 1));
if (next.compareTo(stop) >= 0) {
// Dont go there.
next = null;
}
}
return next != null;
}
#Override
public BigInteger next() {
last = hasNext() ? next : null;
next = null;
return not ? last.not(): last;
}
#Override
public void remove() {
throw new UnsupportedOperationException("Not supported.");
}
#Override
public String toString () {
return next != null ? next.toString(2) : last != null ? last.toString(2): "";
}
}
}
I will not write code but will list a possible approach. I say possible because it will be running and storing all data in memory and is not the best in regard to algorithms. yet, it is an approach where you don't need to eliminate invalid options. I will use an example in order to make things more clear.
suppose you have categories A,B,C. Where K=2 for A,B and K=1 for C.
you also have the input items A1,B1,B2,A2,C1,A3
1- go over the items and divide them according to their category. so you prepare an array/list for each category that has all the items that belong to it.
so now you have arrays:
Category A = [A1,A2,A3] , Category B = [B1,B2] and Category C=[C1]
2- now after preparing the lists, prepare the various legal groups that you can have for picking K items from N items found in that list . here is a link that might help in doing that efficiently: How to iteratively generate k elements subsets from a set of size n in java?
now you have:
first group belonging to category A:
[A1,A2] , [A1,A3], [A2,A3] (3 elements)
second group belonging to category B:
[B1,B2] (1 element)
third group belonging to category C:
[C1] (1 element)
3- now, if you treat each such group as an item, the question transforms to how many different ways are there for picking exactly one element from each group. and that is supposed to be easier to program recursively and will not require eliminating options. and if the number of categories is constant, it will be nested loops over the sets of groups in second point above.
EDIT
the approach works well in eliminating the need to validate bad combinations.
yet, there will still be a problem in regard of time. Here is the code that I made to demonstrate. it makes a list of 100 items. then it does the steps mentioned.
Note that I commented out the code that prints the groups.
The calculation is very fast up to that point. I have added code that prints how many legal choices can be made from each group.
package tester;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Random;
/**
*
* #author
*/
public class Tester {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
//generate 100 random items belonging to categories
Random rand=new Random();
List<Item> items=new ArrayList<>();
int a=1,b=1,c=1,d=1,e=1;
for (int i=0;i<100;i++){
int randomNumber=rand.nextInt(5)+1;
CATEGORY_TYPE categoryType=null;
int num=0;
switch (randomNumber) {
case 1:
categoryType=CATEGORY_TYPE.A;
num=a++;
break;
case 2:
categoryType=CATEGORY_TYPE.B;
num=b++;
break;
case 3:
categoryType=CATEGORY_TYPE.C;
num=c++;
break;
case 4:
categoryType=CATEGORY_TYPE.D;
num=d++;
break;
case 5:
categoryType=CATEGORY_TYPE.E;
num=e++;
break;
}
String dummyData="Item "+categoryType.toString()+num;
Item item=new Item(dummyData,categoryType);
items.add(item);
}
//arrange the items in lists by category
List<Item> categoryAItemsList=new ArrayList<>();
List<Item> categoryBItemsList=new ArrayList<>();
List<Item> categoryCItemsList=new ArrayList<>();
List<Item> categoryDItemsList=new ArrayList<>();
List<Item> categoryEItemsList=new ArrayList<>();
for (Item item:items){
if (item.getCategoryType()==CATEGORY_TYPE.A)
categoryAItemsList.add(item);
else if (item.getCategoryType()==CATEGORY_TYPE.B)
categoryBItemsList.add(item);
else if (item.getCategoryType()==CATEGORY_TYPE.C)
categoryCItemsList.add(item);
else if (item.getCategoryType()==CATEGORY_TYPE.D)
categoryDItemsList.add(item);
else if (item.getCategoryType()==CATEGORY_TYPE.E)
categoryEItemsList.add(item);
}
//now we want to construct lists of possible groups of choosing from each category
List<Item[]> subsetStoringListA=new ArrayList<>();
List<Item[]> subsetStoringListB=new ArrayList<>();
List<Item[]> subsetStoringListC=new ArrayList<>();
List<Item[]> subsetStoringListD=new ArrayList<>();
List<Item[]> subsetStoringListE=new ArrayList<>();
processSubsets(categoryAItemsList.toArray(new Item[0]),2,subsetStoringListA);
processSubsets(categoryBItemsList.toArray(new Item[0]),2,subsetStoringListB);
processSubsets(categoryCItemsList.toArray(new Item[0]),2,subsetStoringListC);
processSubsets(categoryDItemsList.toArray(new Item[0]),2,subsetStoringListD);
processSubsets(categoryEItemsList.toArray(new Item[0]),1,subsetStoringListE);
System.out.println(" A groups number: "+subsetStoringListA.size());
System.out.println(" B groups number: "+subsetStoringListB.size());
System.out.println(" C groups number: "+subsetStoringListC.size());
System.out.println(" D groups number: "+subsetStoringListD.size());
System.out.println(" E groups number: "+subsetStoringListE.size());
//now we just print all possible combinations of picking a single group from each list.
//the group is an array with valid choices
// for (Item[] subsetA:subsetStoringListA){
// for (Item[] subsetB:subsetStoringListB){
// for (Item[] subsetC:subsetStoringListC){
// for (Item[] subsetD:subsetStoringListD){
// for (Item[] subsetE:subsetStoringListE){
// print(subsetA);
// print(subsetB);
// print(subsetC);
// print(subsetD);
// print(subsetE);
// System.out.println("\n");
// }
//
// }
// }
// }
// }
}
static void print(Item[] arr){
for (Item item:arr)
System.out.print(item.getDumyData()+" ");
}
static void processSubsets(Item[] set, int k,List<Item[]> subsetStoringList) {
Item[] subset = new Item[k];
processLargerSubsets(set, subset, 0, 0,subsetStoringList);
}
static void processLargerSubsets(Item[] set, Item[] subset, int subsetSize, int nextIndex,List<Item[]> subsetStoringList) {
if (subsetSize == subset.length) { //here we have a subset we need to store a copy from it
subsetStoringList.add(Arrays.copyOf(subset, subset.length));
} else {
for (int j = nextIndex; j < set.length; j++) {
subset[subsetSize] = set[j];
processLargerSubsets(set, subset, subsetSize + 1, j + 1,subsetStoringList);
}
}
}
public static enum CATEGORY_TYPE {A,B,C,D,E}
private static class Item{
private CATEGORY_TYPE categoryType;
private String dumyData;
public Item(String dumyData,CATEGORY_TYPE categoryType) {
this.dumyData = dumyData; //maybe bad name but i mean the object can have many other fields etc
this.categoryType = categoryType;
}
/**
* #return the categoryType
*/
public CATEGORY_TYPE getCategoryType() {
return categoryType;
}
/**
* #return the dumyData
*/
public String getDumyData() {
return dumyData;
}
}
}
in a specific run, it gave the following:
A groups number: 210
B groups number: 153
C groups number: 210
D groups number: 210
E groups number: 19
that means , if we had to print all possible choices of a single element (and here an elemnt is an array containing k choices from a category) from each of these, you will have : 210*153*210*210*19 = 26,921,727,000
now listing/printing over 26 billion variations will take time no matter what and I don't see how it will be minimized.
try setting the total items to 20 and uncomment the printing code to see that everything is working correctly. And see if you really need to list the possible combinations. please remember that every combination here is legal and there are no wasted iterations in all the parts of the code.
one final note: I did not treat edge cases like when there are no items in a category to complete K. that you can easily put in the code according to the desired behaviour in that case.
So this seems to be a constraint satisfaction problem. So maybe try backtracking?
I believe the following works, but plug in your own data to guaranteee.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Launch {
public static void main(String[] args) {
// Formulate the constraints.
int[] constraints = { 2, 1, 0, 1 };
// Create all the items.
List<boolean[]> items = new ArrayList<boolean[]>();
boolean[] i1 = { true, false, true, false };
boolean[] i2 = { true, false, false, false };
boolean[] i3 = { false, true, false, true };
boolean[] i4 = { false, false, false, true };
items.add(i1);
items.add(i2);
items.add(i3);
items.add(i4);
// Solve!
backtrack(constraints, items);
}
/**
* Recursively generate possible solutions but backtrack as soon as the constraints fail.
*/
private static void backtrack(int[] constraints, List<boolean[]> items) {
// We start with no items belonging to any categories.
List<List<boolean[]>> memberships = new ArrayList<List<boolean[]>>();
for (int i = 0; i < constraints.length; i++) {
memberships.add(new ArrayList<boolean[]>());
}
backtrack(constraints, items, memberships);
}
/**
* Recursively generate possible solutions but backtrack as soon as the constraints fail.
*/
private static void backtrack(int[] constraints, List<boolean[]> items,
List<List<boolean[]>> memberships) {
if (items.isEmpty() && !acceptable(constraints, memberships)) {
return;
} else if (acceptable(constraints, memberships)) {
display(memberships);
} else {
for (boolean[] item : items) {
int catIdx = 0;
for (boolean belongs : item) {
if (belongs) {
// The item and category were chosen so let's update
// memberships.
List<List<boolean[]>> newMemberships = new ArrayList<List<boolean[]>>();
for (List<boolean[]> old : memberships) {
newMemberships.add(new ArrayList<boolean[]>(old));
}
newMemberships.get(catIdx).add(item);
// We've placed the item so let's remove it from the
// possibilities.
List<boolean[]> newItems = new ArrayList<boolean[]>(
items);
newItems.remove(item);
// Now solve the sub-problem.
backtrack(constraints, newItems, newMemberships);
}
catIdx++;
}
}
}
}
/**
* A nice way to display the membership tables.
*/
private static void display(List<List<boolean[]>> memberships) {
System.out.println("---");
for (List<boolean[]> category : memberships) {
for (boolean[] item : category) {
System.out.print(Arrays.toString(item) + " ");
}
System.out.println();
}
}
/**
* Returns whether or not a list of memberships are accepted by the
* constraints.
*
* #param constraints
* - The number of items required per category.
* #param memberships
* - The current items per category.
*/
private static boolean acceptable(int[] constraints,
List<List<boolean[]>> memberships) {
boolean acceptable = memberships.size() == constraints.length;
for (int i = 0; i < memberships.size(); i++) {
acceptable = acceptable
&& constraints[i] == memberships.get(i).size();
}
return acceptable;
}
}

Java / JUnit - AssertionError testing a pre-defined polynomial

I have a Term class to define a polynomial:
public class Term
{
final private int coef;
final private int expo;
private static Term zero, unit;
static
{
try
{
zero = new Term(0, 0); // the number zero
unit = new Term(1, 0); // the number one
}
catch (Exception e)
{
// constructor will not throw an exception here anyway
}
}
/**
*
* #param c
* The coefficient of the new term
* #param e
* The exponent of the new term (must be non-negative)
* #throws NegativeExponent
*/
public Term(int c, int e) throws NegativeExponent
{
if (e < 0)
throw new NegativeExponent();
coef = c;
expo = (coef == 0) ? 1 : e;
}
final public static Term Zero = zero;
final public static Term Unit = unit;
public boolean isConstant()
{
boolean isConst = false;
if (this.expo == 0)
{
isConst = true;
}
return isConst;
}
}
And I have the JUnit test as follows:
/*
* const1 isConstant(zero) => true (0,0)
* const2 isConstant(unit) => true (1,0)
* const3 isConstant(0,5) => true
* const4 isConstant(5,1) => false
*/
#Test
public void const1() throws TError { assertTrue(Term.Zero.isConstant()); }
#Test
public void const2() throws TError { assertTrue(Term.Unit.isConstant()); }
#Test
public void const3() throws TError { assertTrue(new Term(0,5).isConstant()); }
#Test
public void const4() throws TError { assertFalse(new Term(5,1).isConstant()); }
Tests 2 and 4 pass as they should, but Tests 1 and 3 come up as failures and I cannot work out why, "Zero" is defining the polynomial as (0,0) and the other is defining it as (0,5). So by my thinking, the 1st one should give a green tick and the 3rd test should give a red cross as it has 5 as the exponent.
Any ideas?
Please review your constructor:
public Term(int c, int e) throws NegativeExponent{
if (e < 0) throw new NegativeExponent();
coef = c;
expo = (coef == 0) ? 1 : e;
}
When coef == 0, then exp is assigned with 1.
This makes zero as (0,1) not (0,0). this is the reason for test outcome as below:
const1 -> false as expo = 1-->failure as expecting true
const2 -> true as expo = 0 -->success as expecting true
const3 -> false as expo = 5.-->failure as expecting true
const4 -> false as expo = 1.-->success as expecting false
EDIT: To reverse fix your code to make the test cases pass, I think you may update your constructor as below:
public Term(int c, int e) throws NegativeExponent{
if (e < 0) throw new NegativeExponent();
coef = c;
expo = (c == 0 && e != 0) ? 0 : e;
}
Here, I have updated the constructor to set expo as 0, if coef is 0 as any expo for coef 0 is immaterial.
Yogendra covered why your first test fails. Your const3() fails (for me at least) because new Term(0, 5) has a expo of 5, which gets replaced with 1 b/c of the coef, but 1 is still not 0 so new Term(0, 5) is NOT constant.
Not that you asked, but I'll also give you some general Java pointers for the future...
1) Use all caps for static constants
public static final Term ZERO = new Term(0, 0);
public static final Term UNIT = new Term(1, 0);
this is just a convention that the Java community expects.
2) Give your test cases meaningful names. One powerful use of test cases is that they can double as documentation of your assumptions at the time. Rather than const1() which tells a fellow developer nothing, use a descriptive name like zeroShouldBeAConstant(). This tells the next person who looks at your code that Zero should be considered a constant in this system.
3) Whenever you are returning a boolean variable, consider just returning the statement. Compare the function below to yours and tell me which one is more readable:
public boolean isConstant()
{
return expo == 0;
}
It is less code and hopefully easier to read.
Good luck!

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