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Count all lists of adjacent nodes stored in an array


Count all lists of adjacent nodes stored in an array

By : Nadine Wickord
Date : November 19 2020, 07:01 PM
help you fix your problem I think this O(n) algorithm does what you want. I doubt you can do this faster since you have to analyse each element.
code :


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Determining valid adjacent cells of a square stored as an array

Determining valid adjacent cells of a square stored as an array


By : Brendan Tuppack
Date : March 29 2020, 07:55 AM
wish helps you This is easier if you used 0-based indexing. These rules work if you subtract 1 from all your indexes:
Two indexes are in the same square if (a/9) == (b/9) and a >= 0 and b >= 0. Two indexes are in the same row if they are in the same square and (a/3) == (b/3). Two indexes are in the same column if they are in the same square and (a%3) == (b%3).
Count elements in adjacent array

Count elements in adjacent array


By : user3201912
Date : March 29 2020, 07:55 AM
With these it helps I would just count the numbers.
This means each item of the base key plus the same for it's children.
code :
$count = array();
$stack[] = array(-1, $arr);
while($item = array_shift($stack)) {
    list($level, $node) = $item;
    $count[$level] = isset($count[$level]) ? $count[$level]+1 : 1;
    if (!isset($node['_children'])) 
        continue;
    foreach($node['_children'] as $item)
        $stack[] = array($level+1, $item);
}
var_dump($count);
array(4) {
  [-1]=> int(1)
  [0] => int(2)
  [1] => int(1)
  [2] => int(1)
}
array_shift($count);
$count = array();
$nodes = array(); // nodes per level go here.
$stack[] = array(-1, $arr);
while($item = array_shift($stack)) {
    list($level, $node) = $item;
    $count[$level] = isset($count[$level]) ? $count[$level]+1 : 1;
    $nodes[$level][] = $node; // set here!
    if (!isset($node['_children'])) 
        continue;
    foreach($node['_children'] as $item)
        $stack[] = array($level+1, $item);
}
var_dump($count, $nodes);
/**
 * class implementation of a recursive child counter
 * 
 * Usage:
 * 
 *   Object use:
 * 
 *    $counter = new LevelChildCounter($arr, '_children');
 *    $count = $counter->countPerLevel();
 * 
 *  Functional use:
 * 
 *    $count = LevelChildCounter::count($arr, '_children');
 */
class LevelChildCounter {
    private $tree;
    private $childKey;
    private $counter;
    public function __construct(array $tree, $childKey) {
        $this->tree = $tree;
        $this->childKey = $childKey;
    }
    /**
     * functional interface of countPerLevel
     */
    public static function count(array $tree, $childKey) {
        $counter = new self($tree, $childKey);
        return $counter->countPerLevel();
    }
    private function countUp($level) {
        isset($this->counter[$level])
          ? $this->counter[$level]++
          : $this->counter[$level] = 1;
    }
    private function countNode($node, $level) {
        // count node
        $this->countUp($level);

        // children to handle?        
        if (!isset($node[$this->childKey]))
            return;

        // recursively count child nodes            
        foreach($node[$this->childKey] as $childNode)
                $this->countNode($childNode, $level+1)
                ;
    }
    /**
     * Count all nodes per level
     * 
     * @return array
     */
    public function countPerLevel() {
        $this->counter = array();
        $this->countNode($this->tree, -1);
        return $this->counter;
    }
}

$count = LevelChildCounter::count($arr, '_children');
array_shift($count);

var_dump($count);
array(3) {
  [0]=> int(2)
  [1]=> int(1)
  [2]=> int(1)
}
iterate all nodes in a path and comparing the property in adjacent nodes to output only the ones that match a certain co

iterate all nodes in a path and comparing the property in adjacent nodes to output only the ones that match a certain co


By : Prasit Lee
Date : March 29 2020, 07:55 AM
I wish this help you You can first get the start node and the end node of each relationship, and then use "Where" to return the end nodes of those relationships whose end nodes satisfy the difference condition, something like this,
code :
Match p = a-[:Rel*]->e
Where a.name?='A'
With endNode(last(relationships(p))) as second, startNode(last(relationships(p))) as first
Where second.wt - first.wt >= 20
Return second
R: Collect igraph nodes adjacent to selected nodes in a list

R: Collect igraph nodes adjacent to selected nodes in a list


By : Santiago Frias
Date : March 29 2020, 07:55 AM
I hope this helps you . You can just unlist(adj) and use that to index into V(g) I will repeat some code from the question and set the random seed to make the result repeatable,
code :
set.seed(26)
select = V(g)[sample(1:5,2)]
select
+ 2/5 vertices, named, from e2f7066:
[1] John Jim 

adj = adjacent_vertices(g,select,mode="all")
N = unlist(adj)
V(g)[N]
+ 5/5 vertices, named, from e2f7066:
[1] Jim  Jill John Jill Bob 
Find adjacent cells in a grid stored in array (not multidimensional)

Find adjacent cells in a grid stored in array (not multidimensional)


By : Kostas Drosatos
Date : March 29 2020, 07:55 AM
seems to work fine If anyone is interested, I figured it out (though I don't know if it's the best way):
code :
function getAdjacents(i, gridSize){
  var mod = i%gridSize;
  var adjacents = {
    topLeft: mod == 0 || i < gridSize ? null : i- (gridSize + 1),
    top: i < gridSize ? null : i-gridSize,
    topRight: i < gridSize ||  mod == 4 ? null : i- (gridSize - 1),
    left: mod == 0 ? null : i-1,
    right: mod == 4 ? null : i+1,
    bottomLeft: mod == 0 || i >= (gridSize*gridSize-gridSize) ? null :  i + (gridSize-1),
    bottom: i >= (gridSize*gridSize-gridSize) ? null : i+gridSize,
    bottomRight: mod == 4 || i >= (gridSize*gridSize-gridSize) ? null :  i + (gridSize+1)
  }
  return adjacents;
}
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