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A circle with number x means:
“This circle is connected (straight line) to x circles with a true statement
This is valid only if the statement in the circle is True.

If a circle is False, then it could be interpreted like this:
“This circle is NOT connected (straight line) to x circles with a true statement
Thus, the number of True statements connected to a False circle should be different than the number in the False circle itself.

Let’s say a true circle is a circle with a true statement.
So if the number is 3 it means the circle is connected by 3 true circles.
Some circles are true and some are not.
Each circle is connected to 5 other circles, except the middle circle.
The middle circle is connected to 10 other circles.
Create another star with Boolean (T/F) input, to show which circles are true, and which circles are false.


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This is a variation from this puzzle.

Only for those who are still confused:

A is connected to (B,C,D,E,F)
B is connected to (A,C,F,G,J)
C is connected to (A,B,D,F,G)
D is connected to (A,C,F,H,E)
E is connected to (A,D,F,H,K)
F is connected to (A,B,C,D,E,G,H,I,J,K)
G is connected to (B,C,F,I,J)
H is connected to (E,D,F,I,K)
I is connected to (J,G,F,H,K)
J is connected to (B,G,F,I,K)
K is connected to (J,I,F,H,E)

  • $\begingroup$ Why is the "0" circle connected to 3 True circles? Shouldn't that mean it must have had a different value than 0? $\endgroup$ – Ivanhoe Sep 28 '16 at 7:41
  • $\begingroup$ @Ivanhoe So the circle is False (F) $\endgroup$ – Jamal Senjaya Sep 28 '16 at 7:43
  • $\begingroup$ @JamalSenjaya I think it would be best if you add an explanation on how this type of puzzle actually works. Because it was confusing at first for me. And as I can see from the answers/comments - not only for me. $\endgroup$ – Maria Deleva Sep 28 '16 at 8:03
  • $\begingroup$ @MariaDeleva I have tried to explain clearly, but maybe still confusing for some persons. Anyone can edit my puzzle to make it more clear. $\endgroup$ – Jamal Senjaya Sep 28 '16 at 8:12
  • $\begingroup$ @JamalSenjaya no problem. :) $\endgroup$ – Maria Deleva Sep 28 '16 at 8:28

The following truth values should solve the puzzle:

The numbers in parentheses are the number of wanted/connected True nodes
A = True (3/3)
B = True (2/2)
C = False (3/4)
D = True (2/2)
E = True (2/2)
F = False (7/6)
G = True (2/2)
H = False (4/3)
I = True (1/1)
J = False (0/3)
K = False (3/2)

Steps I used to find the solution:

Assume J is true, this makes B, G, I, K false, this makes F and H false, which would make I true, a contradiction, thus J is false.
Assume I is true, this makes F, H, K false and thus G true. Now assume E is true, this makes A and D true. It follows that C must be false and B must be true. This gives a correct solution.

  • $\begingroup$ I appreciate the steps towards the solution. A lot easier for me to know where to start on a puzzle like this now. Thanks! $\endgroup$ – Brian J Sep 28 '16 at 12:48
  • 1
    $\begingroup$ Maybe a small comment on how I chose the nodes to start from. I looked for nodes with small numbers that had nodes with high numbers around them. Assuming those as true will lead to several implications that should lead to a contradiction or solution very fast. $\endgroup$ – w l Sep 28 '16 at 12:56

I know this has been answered but I worked on it for a while and I don't want it to go to waste. I would have answered sooner but there was a fire drill in the building and had to leave.

I thought about brute forcing this since there are only $2^{11}$ possible combinations.
The idea is to write a base 2 number that contains 0 on a position if the circle is false and 1 if it's true.
I used PHP to go through all the possible values.
This is in no way the optimal approach.

The code below returns the following result

[0] => Array
[a] => 1
[b] => 1
[c] => 0
[d] => 1
[e] => 1
[f] => 0
[g] => 1
[h] => 0
[i] => 1
[j] => 0
[k] => 0


class Star 
    protected $values;
    protected $around;
    protected $size;
    public function __construct($values, $around)
        $this->values = $values;
        $this->around = $around;
        $this->size = count($values);
    protected function generateTrueFalse($number)
        $value = base_convert($number, 10, 2);
        $value = str_pad($value, $this->size, '0', STR_PAD_LEFT);
        $array = array();
        for ($i = 0;$i<strlen($value);$i++) {
            $array[chr(ord('a')+$i)] = $value[$i];
        return $array;
    protected function sumAround($char, $trueFalse) 
        $sum = 0;
        foreach ($this->around[$char] as $near) {
            $sum += $trueFalse[$near];
        return $sum;

    public function run()
        $results = array();
        for ($i = 0; $i<pow(2, $this->size);$i++) {
            $base2Array = $this->generateTrueFalse($i);
            $valid = true;
            foreach ($this->values as $key => $value) {
                $sum = $this->sumAround($key, $base2Array);
                if (($sum != $value && $base2Array[$key] == 1) || ($sum == $value && $base2Array[$key] == 0)) {
                    $valid = false;
            if ($valid) {
                $results[] = $base2Array;
        return $results;
$values = array(
    'a' => 3,
    'b' => 2,
    'c' => 3,
    'd' => 2,
    'e' => 2,
    'f' => 7,
    'g' => 2,
    'h' => 4,
    'i' => 1,
    'j' => 0,
    'k' => 3

$around = array(
	'a' => array('b','c', 'd', 'e', 'f'),
	'b' => array('a','c', 'f', 'g', 'j'),
	'c' => array('a','b', 'd', 'f', 'g'),
	'd' => array('a','c', 'f', 'h', 'e'),
	'e' => array('a','d', 'f', 'h', 'k'),
	'f' => array('a','b', 'c', 'd', 'e', 'h', 'i', 'j', 'k'),
	'g' => array('b','c', 'f', 'i', 'j'),
	'h' => array('e','d', 'f', 'i', 'j'),
	'i' => array('j','g', 'f', 'h', 'k'),
	'j' => array('b','g', 'f', 'i', 'k'),
	'k' => array('e','f', 'h', 'i', 'j')
$star = new Star($values, $around);
echo "<pre>"; print_r($star->run());
  • $\begingroup$ So you have proved the question have a unique answer. $\endgroup$ – Jamal Senjaya Sep 28 '16 at 10:11
  • $\begingroup$ No because I stop the execution when i find the first solution. I will change it to run through all the possible combinations. $\endgroup$ – Marius Sep 28 '16 at 10:22
  • 1
    $\begingroup$ @JamalSenjaya. I've modified my class to return all the possible results. And indeed it returns only one combination. $\endgroup$ – Marius Sep 28 '16 at 11:09

My answer below is found WRONG. see comments.

It seems like there are multiple solutions.
Below is the one I found.


  • 1
    $\begingroup$ you solution is contradictory - H states 4. You put a True value inside it, yet none of the circles surrounding it contain True value (if it is T, and the statement is 4, then there should be 4 circles linked to it with T). $\endgroup$ – Maria Deleva Sep 28 '16 at 7:58

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