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  1. POAdvantage and disadvantage of spanning tree with even distance
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    <p>It's new year day and still can't solve my problem about a spanning tree algorithm. I can't insert picture yet so I have to try to explain the enviroment with words.</p> <p>It's 36 nodes and the distance to every nodes is even. The question is if the distance is even, it doesn't matter which way to pass message from node with ID 1 (the root) to the last node with ID 36. Because the distance is even there's no time saving, energy saving or message saving algorithm right? I hope someone understand my question</p> <p>edited:</p> <ol> <li><p>Enviroment</p> <pre><code>1 - 2 - 3 - 4 - 5 - 6 | | | | | | 7 8 9 10 11 12 | | | | | | 13 14 15 16 17 18 | | | | | | 19 20 21 22 23 24 | | | | | | 25 26 27 28 29 30 | | | | | | 31 32 33 34 35 36 </code></pre></li> </ol> <p>This is my choice of spanning tree. Node with ID 36 send it information thru 30,24,18,12,6,5,4,3,2,1 (one is the root) and then node 1 send information to the base station. Because it doesn't have any cost it doesn't really matter which path I choose to send the information from node 36 to node 1 because the cost will still be the same.</p> <ol> <li><p>My Spanning tree Algorithm</p> <ul> <li>When start, only the root is marked.</li> <li>The root send search message to it neighbor</li> <li>If a node is not marked, when it recieves search messages from other nodes:</li> <li>it mark itself </li> <li>Select the nodes with lowest ID as a "parent" and reply "non-parent" to the other nodes</li> <li>If the node is already mark, it replies "non-parent"</li> <li>If a node is already marked and recieve a parent message it marks the sender as a child</li> </ul></li> <li><p>I can't show you guys the flowchart because I don't have the privilege to insert images.</p></li> <li><p>Pseudo Code (haven't done it)</p></li> <li><p>Conclusion - Here I should write down the advantage and disadvantage of my algorithm, but right now I can't think of any advantage and disadvantage</p></li> </ol>
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