I'm missing something fundamental here but I can't seem to find out what from all my research.
I have imported a csv file, split the string into floats, and now wish to connect all points to all other points with a line. My code is as follows:
String [] data;
void setup () {
size(300, 300);
background(255);
data = loadStrings("points.csv");
for (int i = 0; i < data.length; i++) {
String [] fields = split(data[i], ',');
float t = float(fields[0]);
float n = float(fields[1]);
float x = float(fields[2]);
float y = float(fields[3]);
ellipse(x, y, 10, 10);
line(x, y, x[i], y[i]);
}
}
The error message is "The type of expression must be an array type but it resolved to float"
I'm sure this is extremely basic but, I dont understand why x[i] or y[i] are not seen as an array type.
I would really appreciate any help with this. Many thanks in advance.
Sam
*UPDATE***
An exract from the points.csv file is as follows:
219185750 rabih_takkoush 20.88521 19.49821
219185716 MoustaphaAjram 100.870896 59.515259
219185709 jinanejghosh 56.886441 35.489087
219185557 MoustaphaAjram 34.870904 78.515243
219185555 Mohammad8Itani 12.8946 49.48179
What I am trying to accomplish is plotting the various geolocations (whereby col 3 = x, col 4 = y) and then connecting all points with all other points with a line.
The following script works plotting all locations specified in the array within the script:
float[] x = { 50, 100, 150, 200,20,20 };
float[] y = { 10, 30, 20, 250,20,90 };
void setup () {
size(300, 300);
background(255);
}
void draw() {
for (int i = 0; i < x.length; i++) {
ellipse(x[i], y[i], 10, 10);
for (int j = 0; j < x.length; j++) {
line(x[j], y[j], x[i], y[i]);
}
}
}
What I wish to do is do the same, but reading columns 3 and 4 of the csv file.
You're splitting your data into iteration-scoped floats, then you try to access them as if they're both floats as well as arrays in your line() call. Try this:
String[] data;
float[] x, y, t, n;
void setup () {
size(300, 300);
data = loadStrings("points.csv");
int len = data.length;
x = new float[len];
x = new float[len];
t = new float[len];
n = new float[len];
for (int i=0; i<len; i++) {
String line = data[i];
String[] fields = split(line, ',');
t[i] = float(fields[0]),
n[i] = float(fields[1]),
x[i] = float(fields[2]),
y[i] = float(fields[3]);
}
// don't put draw instructions in setup,
// put them in draw. if you want to run once,
// issue a noLoop() so that happens.
noLoop();
}
void draw() {
float prevx = x[0], prevy = y[0];
for (int i=0, last=x.length; i<last; i++) {
ellipse(x[i], y[i], 10, 10);
line(prevx,prevy, x[i],y[i]);
prevx=x[i];
prevy=y[i];
}
}
Now we're storing the data from CVS in linked arrays that we can access throughout the sketch, rather than throwing them away after setup().
ok so if you go with the first code you made, you only need to change a few things, here is what you can do:
float[] x;
float[] y;
string[] data;
void setup () {
size(300, 300);
background(255);
data = loadStrings("points.csv");
x = new float[data.length];
y = new float[data.length];
for (int i = 0; i < data.length; i++) {
String [] fields = split(data[i], ',');
float t = float(fields[0]);
float n = float(fields[1]);
float x = float(fields[2]);
float y = float(fields[3]);
}
}
void draw() {
for (int i = 0; i < x.length; i++) {
ellipse(x[i], y[i], 10, 10);
for (int j = 0; j < x.length; j++) {
line(x[j], y[j], x[i], y[i]);
}
}
}
As you can see nothing really new, it's a mix between your initial code and the one you made for the csv.
And actually you mainly needed to declare your x and y variables as float[] instead of just float. But also there were some changes to make in the loop.
In this code you load the data in your arrays first (exactly like you did by declaring array's values in your code, but this time you read these values from your file), and then call your draw method as before.
hope it works now
Given a set {1,2,3,4,5...n} of n elements, we need to find all subsets of length k .
For example, if n = 4 and k = 2, the output would be {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}.
I am not even able to figure out how to start. We don't have to use the inbuilt library functions like next_permutation etc.
Need the algorithm and implementation in either C/C++ or Java.
Recursion is your friend for this task.
For each element - "guess" if it is in the current subset, and recursively invoke with the guess and a smaller superset you can select from. Doing so for both the "yes" and "no" guesses - will result in all possible subsets.
Restraining yourself to a certain length can be easily done in a stop clause.
Java code:
private static void getSubsets(List<Integer> superSet, int k, int idx, Set<Integer> current,List<Set<Integer>> solution) {
//successful stop clause
if (current.size() == k) {
solution.add(new HashSet<>(current));
return;
}
//unseccessful stop clause
if (idx == superSet.size()) return;
Integer x = superSet.get(idx);
current.add(x);
//"guess" x is in the subset
getSubsets(superSet, k, idx+1, current, solution);
current.remove(x);
//"guess" x is not in the subset
getSubsets(superSet, k, idx+1, current, solution);
}
public static List<Set<Integer>> getSubsets(List<Integer> superSet, int k) {
List<Set<Integer>> res = new ArrayList<>();
getSubsets(superSet, k, 0, new HashSet<Integer>(), res);
return res;
}
Invoking with:
List<Integer> superSet = new ArrayList<>();
superSet.add(1);
superSet.add(2);
superSet.add(3);
superSet.add(4);
System.out.println(getSubsets(superSet,2));
Will yield:
[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
Use a bit vector representation of the set, and use an algorithm similar to what std::next_permutation does on 0000.1111 (n-k zeroes, k ones). Each permutation corresponds to a subset of size k.
This is python. Sorry for the spanish ;)
from pprint import pprint
conjunto = [1,2,3,4, 5,6,7,8,9,10]
k = 3
lista = []
iteraciones = [0]
def subconjuntos(l, k):
if k == len(l):
if not l in lista:
lista.append(l)
return
for i in l:
aux = l[:]
aux.remove(i)
result = subconjuntos(aux, k)
iteraciones[0] += 1
if not result in lista and result:
lista.append( result)
subconjuntos(conjunto, k)
print (lista)
print ('cant iteraciones: ' + str(iteraciones[0]))
Check out my solution
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Set;
public class Subset_K {
public static void main(String[]args)
{
Set<String> x;
int n=4;
int k=2;
int arr[]={1,2,3,4};
StringBuilder sb=new StringBuilder();
for(int i=1;i<=(n-k);i++)
sb.append("0");
for(int i=1;i<=k;i++)
sb.append("1");
String bin=sb.toString();
x=generatePerm(bin);
Set<ArrayList <Integer>> outer=new HashSet<ArrayList <Integer>>();
for(String s:x){
int dec=Integer.parseInt(s,2);
ArrayList<Integer> inner=new ArrayList<Integer>();
for(int j=0;j<n;j++){
if((dec&(1<<j))>0)
inner.add(arr[j]);
}
outer.add(inner);
}
for(ArrayList<?> z:outer){
System.out.println(z);
}
}
public static Set<String> generatePerm(String input)
{
Set<String> set = new HashSet<String>();
if (input == "")
return set;
Character a = input.charAt(0);
if (input.length() > 1)
{
input = input.substring(1);
Set<String> permSet = generatePerm(input);
for (String x : permSet)
{
for (int i = 0; i <= x.length(); i++)
{
set.add(x.substring(0, i) + a + x.substring(i));
}
}
}
else
{
set.add(a + "");
}
return set;
}
}
I am working on a 4 element set for test purpose and using k=2. What I try to do is initially generate a binary string where k bits are set and n-k bits are not set. Now using this string I find all the possible permutations of this string. And then using these permutations I output the respective element in the set. Would be great if someone could tell me about the complexity of this problem.
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
vector<int> v;
vector<vector<int> > result;
void subset(int arr[],int k,int n,int idx){
if(idx==n)
return;
if(k==1){
for(int i=idx;i<n;i++)
{
v.push_back(arr[i]);
result.push_back(v);
v.pop_back();
}
}
for(int j=idx;j<n;j++) {
v.push_back(arr[j]);
subset(arr,k-1,n,j+1);
v.pop_back();
}
}
int main(){
int arr[] = {1,2,3,4,5,6,7};
int k = 4;
int n =sizeof(arr)/sizeof(arr[0]);
subset(arr,k,n,0);
for(int i = 0;i<result.size();i++)
{
for(int j = 0;j<result[i].size();j++)
{
cout << result[i][j] << " ";
}
cout << endl;
}
}
Another intresting solution.
#include<bits/stdc++.h>
using namespace std;
long factorial(int n) { return (n==1|| n==0|| n < 0) ? 1 : n *factorial(n-1) ;}
void printS(int set[],int n,int k)
{
long noofsubset = factorial(n) / (factorial(n-k)*factorial(k));
bitset<32> z ((1 << (k)) - 1);
string s = z.to_string();
int i = 0;
while(i<noofsubset)
{
for (int j = 0; j < n;j++)
{
if(s[(32-n)+j] == '1')
cout << set[j]<<" ";
}
cout << endl;
next_permutation(s.begin(),s.end());
i++;
}
}
void printSubsetsOfArray(int input[], int size) {
int k = 3;
printS(input,size,k) ;
}
Slight improvement for #amit top voted answer:
His code keep checking combinations even when there won't be any chance for them to reach the wanted length. We can stop creating combinations much earlier:
e.g. for [1,2,3,4,5,6,7,8,9,10] , length = 8 , the code will still try all combinations of length 7,6,5,4,3,2,1 although they will obviously just be thrown away, halting only when idx reaches the end of the list.
We can improve the running time by stopping earlier, when we already know the set we build + the optional remaining digits will still be too short.
change :
//unsuccessful stop clause
if (idx == superSet.size()) return;
into:
// unsuccessful stop clause
Integer maxFutureElements = superSet.size() - idx;
if (current.size() + maxFutureElements < length) return;
Please check my solution:-
private static void printPermutations(List<Integer> list, int subSetSize) {
List<Integer> prefixList = new ArrayList<Integer>();
printPermutations(prefixList, list, subSetSize);
}
private static void printPermutations(List<Integer> prefixList, List<Integer> list, int subSetSize) {
if (prefixList.size() == subSetSize) {
System.out.println(prefixList);
} else {
for (int i = 0; i < list.size(); i++) {
Integer removed = list.remove(i);
prefixList.add(removed);
printPermutations(prefixList, list, subSetSize);
prefixList.remove(removed);
list.add(i, removed);
}
}
}
This is similar to String permutations:-
private static void printPermutations(String str) {
printAllPermutations("", str);
}
private static void printAllPermutations(String prefix, String restOfTheString) {
int len = restOfTheString.length();
System.out.println(prefix);
for (int i = 0; i < len; i++) {
printAllPermutations(prefix + restOfTheString.charAt(i), restOfTheString.substring(0, i) + restOfTheString.substring(i + 1, len));
}
}
This is an implemation in F#:
// allSubsets: int -> int -> Set<Set<int>>
let rec allSubsets n k =
match n, k with
| _, 0 -> Set.empty.Add(Set.empty)
| 0, _ -> Set.empty
| n, k -> Set.union (Set.map (fun s -> Set.add n s) (allSubsets (n-1) (k-1)))
(allSubsets (n-1) k)
You can try it in the F# REPL:
> allSubsets 3 2;;
val it : Set<Set<int>> = set [set [1; 2]; set [1; 3]; set [2; 3]]
> allSubsets 4 2;;
val it : Set<Set<int>> = set [set [1; 2]; set [1; 3]; set [1; 4]; set [2; 3]; set [2; 4]; set [3; 4]]
This Java class implements the same algorithm:
import java.util.HashSet;
import java.util.Set;
public class AllSubsets {
public static Set<Set<Integer>> allSubsets(int setSize, int subsetSize) {
if (subsetSize == 0) {
HashSet<Set<Integer>> result = new HashSet<>();
result.add(new HashSet<>());
return result;
}
if (setSize == 0) {
return new HashSet<>();
}
Set<Set<Integer>> sets1 = allSubsets((setSize - 1), (subsetSize - 1));
for (Set<Integer> set : sets1) {
set.add(setSize);
}
Set<Set<Integer>> sets2 = allSubsets((setSize - 1), subsetSize);
sets1.addAll(sets2);
return sets1;
}
}
If you do not like F# or Java then visit this website. It lists solutions to your particular problem in various programming languages:
http://rosettacode.org/wiki/Combinations
JavaScript implementation:
var subsetArray = (function() {
return {
getResult: getResult
}
function getResult(array, n) {
function isBigEnough(value) {
return value.length === n;
}
var ps = [
[]
];
for (var i = 0; i < array.length; i++) {
for (var j = 0, len = ps.length; j < len; j++) {
ps.push(ps[j].concat(array[i]));
}
}
return ps.filter(isBigEnough);
}
})();
var arr = [1, 2, 3, 4,5,6,7,8,9];
console.log(subsetArray.getResult(arr,2));
Here is an iterative version in python. Essence of it is increment_counters() function which returns all possible combinations. We know it needs to be called C(n,r) times.
def nchooser(n,r):
"""Calculate the n choose r manual way"""
import math
f = math.factorial
return f(n) / f(n-r) / f(r)
def increment_counters(rc,r,n):
"""This is the essense of the algorithm. It generates all possible indexes.
Ex: for n = 4, r = 2, rc will have values (0,1),(0,2),(0,3),(1,2),(1,3),(2,3).
You may have better understanding if you print all possible 35 values for
n = 7, r = 3."""
rc[r-1] += 1 # first increment the least significant counter
if rc[r-1] < n: # if it does not overflow, return
return
# overflow at the last counter may cause some of previous counters to overflow
# find where it stops (ex: in n=7,r=3 case, 1,2,3 will follow 0,5,6)
for i in range(r-2,-1,-1): # from r-2 to 0 inclusive
if rc[i] < i+n-r:
break
# we found that rc[i] will not overflow. So, increment it and reset the
# counters right to it.
rc[i] += 1
for j in range(i+1,r):
rc[j] = rc[j-1] + 1
def combinations(lst, r):
"""Return all different sub-lists of size r"""
n = len(lst)
rc = [ i for i in range(r) ] # initialize counters
res = []
for i in range(nchooser(n,r)): # increment the counters max possible times
res.append(tuple(map(lambda k: lst[k],rc)))
increment_counters(rc,r,n)
return res
Here is a Java version of what I think Simple is talking about, using a binary representation of all sets in the power set. It's similar to how Abhiroop Sarkar did it, but I think a boolean array makes more sense than a string when you are just representing binary values.
private ArrayList<ArrayList<Object>> getSubsets(int m, Object[] objects){
// m = size of subset, objects = superset of objects
ArrayList<ArrayList<Object>> subsets = new ArrayList<>();
ArrayList<Integer> pot = new ArrayList<>();
int n = objects.length;
int p = 1;
if(m==0)
return subsets;
for(int i=0; i<=n; i++){
pot.add(p);
p*=2;
}
for(int i=1; i<p; i++){
boolean[] binArray = new boolean[n];
Arrays.fill(binArray, false);
int y = i;
int sum = 0;
for(int j = n-1; j>=0; j--){
int currentPot = pot.get(j);
if(y >= currentPot){
binArray[j] = true;
y -= currentPot;
sum++;
}
if(y<=0)
break;
}
if(sum==m){
ArrayList<Object> subsubset = new ArrayList<>();
for(int j=0; j < n; j++){
if(binArray[j]){
subsubset.add(objects[j]);
}
}
subsets.add(subsubset);
}
}
return subsets;
}
If you are looking for Iterator pattern answer then here you go.
public static <T> Iterable<List<T>> getList(final Iterable<? extends T> list) {
List<List<T>> listOfList = new ArrayList<>();
for (T t: list)
listOfList.add(Collections.singletonList(t));
return listOfList;
}
public static <T> Iterable<List<T>> getIterable(final Iterable<? extends T> list, final int size) {
final List<T> vals = new ArrayList<>();
int numElements = 0;
for (T t : list) {
vals.add(t);
numElements++;
}
if (size == 1) {
return getList(vals);
}
if (size == numElements) {
return Collections.singletonList(vals);
}
return new Iterable<List<T>>() {
#Override
public Iterator<List<T>> iterator() {
return new Iterator<List<T>>() {
int currPos = 0;
Iterator<List<T>> nextIterator = getIterable(
vals.subList(this.currPos + 1, vals.size()), size - 1).iterator();
#Override
public boolean hasNext() {
if ((this.currPos < vals.size()-2) && (this.currPos+size < vals.size()))
return true;
return false;
}
#Override
public List<T> next() {
if (!nextIterator.hasNext()) {
this.currPos++;
nextIterator = getIterable(vals.subList(this.currPos+1, vals.size()), size-1).iterator();
}
final List<T> ret = new ArrayList<>(nextIterator.next());
ret.add(0, vals.get(this.currPos));
return ret;
}
};
}
};
}
Here's a short python algorithm. I haven't used any predefined functions as such so I believe it could be easily translated to Java/C
def subs(l,n):
if(len(l)<k):
return []
elif(k==0):
return [[]]
else:
lis=[[l[0]]+b for b in (subs(l[1:],k-1))]
return (lis+subs(l[1:],k))
Here l is the list [1,2,...,m]
Here is a simple algorithm to enumerate all k-subsets of [n]={0,...,n-1} in lexicographic order. That is, the first of these subsets is S0=(0,1,2...,k-1), and the last is Slast=(n-k, n-k+1,...,n-1). For any k-subset S and for any 0 < j < k, we have S[j-1] < S[j] <= n+j-k.
For example, if n=10 and k=4, S0=(0,1,2,3) and Slast=(6,7,8,9). Notice, for example, that no combination can have S[1]>7 (in which case we'd have S[j]>n+j-k), since then there would be not enough values left to fill thr remaining positions j=2..3.
The idea of the algorithm is to start with the first combination S0, and then call next() repeatedly to generate the next k-subset each time. The function next() traverses the current k-subset backwards, starting from the last position j=k-1 down to 0, until it finds an entry S[j] that has not yet reached its maximum allowed value n+j-k and can thus be increased. Then it increases this position by one and fills the remaining positions, j+1..k-1 with consecutive values from S[j]+1. The algorithm stops as soon as no position can be further increased.
For example, suppose we have S=(3,7,8,9). Starting from j=3, we see that S[3],S[2],S[1] have reached their maximum values. Thus, the rightmost position that can still be increased is S[0]. This value is updated to S[0]+1=4, and the following positions are updated to 5,6,7. Hence the next k-subset will be S=(4,5,6,7).
#include <stdlib.h>
#include <stdbool.h>
#include <stdio.h>
bool next(int *S, int k, int n) {
int j = k-1;
while (j >= 0 && S[j] == n + j - k)
j--;
if (j < 0) return false;
S[j] += 1;
for (int i = j+1; i < k ; i++)
S[i] = S[i-1] + 1;
return true;
}
int main(int argc, char *argv[])
{
int n = 10;
int k = 4;
int *S = (int *)calloc(k, sizeof(int));
for (int j = 0; j < k; S[++j] = j); //first k-subset
int no = 0;
do {
printf("subset #%d: ",no++);
for (int j=0; j < k; j++) {
printf("%d ", S[j]);
}
printf("\n");
} while(next(S, k, n));
return 0;
}