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POAlgorithm to detect when graph re-converges (similar to common subtree?)
 

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  1. POAlgorithm to detect when graph re-converges (similar to common subtree?)
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    <p>I've been working on this problem all day, I'm re-writing one of our legacy products and I'm having a hard time determining how to find a specific node in my flow chart. The problem reminds me of University, but for the life of me I can't come up with an algorithm to solve this.</p> <p>I've attached 3 screen shots to help explain this, but the basic problem is, given a YES/NO? decision node, locate the closest child node that terminates the branch. </p> <p>I'm working in C# .NET and JSON. In JSON I've got an object that gives each node a unique identifier, and also identifies each "link" from one node to the next. I would hope to write a function (or several) to determine the first "end node" given a branched node in C#. Currently I've built out the jSON into XML in C#.</p> <p>Any and all ideas encouraged, not really looking for code but an approach/algorithm. </p> <p><img src="https://i.stack.imgur.com/frNUZ.jpg" alt="enter image description here"> <img src="https://i.stack.imgur.com/tW1jg.jpg" alt="enter image description here"></p> <p><img src="https://i.stack.imgur.com/tcjEj.jpg" alt="given the yes/no find the delay block.. first node that all child nodes traverse to"></p> <p>Attached is the output in jSON from the diagram:</p> <pre><code>{ "class": "go.GraphLinksModel", "linkFromPortIdProperty": "fromPort", "linkToPortIdProperty": "toPort", "nodeDataArray": [ {"key":-1, "category":"Start", "loc":"169 288", "text":"Start"}, {"key":-2, "category":"End", "loc":"855 394", "text":"End"}, {"category":"Branch", "text":"Yes or No", "key":-4, "loc":"284.8837209302326 285.7848837209302"}, {"category":"DelayNode", "text":"Delay", "key":-3, "loc":"365.8837209302326 215.52345997177622"}, {"category":"Branch", "text":"Yes or No", "key":-5, "loc":"478.8837209302326 214.52345997177622"}, {"category":"DelayNode", "text":"Delay", "key":-6, "loc":"568.8837209302326 151.52345997177622"}, {"category":"DelayNode", "text":"Delay", "key":-7, "loc":"573.8837209302326 268.5234599717762"}, {"category":"DelayNode", "text":"Delay", "key":-8, "loc":"653.8837209302326 215.52345997177622"}, {"category":"Branch", "text":"Yes or No", "key":-9, "loc":"392.8837209302326 392.5234599717762"}, {"category":"DelayNode", "text":"Delay", "key":-10, "loc":"454.8837209302326 317.5234599717762"}, {"category":"DelayNode", "text":"Delay", "key":-11, "loc":"550.8837209302326 473.5234599717762"}, {"category":"DelayNode", "text":"Delay", "key":-12, "loc":"549.8837209302326 317.5234599717762"}, {"category":"DelayNode", "text":"Delay", "key":-13, "loc":"711.8837209302326 343.5234599717762"}, {"category":"Branch", "text":"Yes or No", "key":-14, "loc":"434.8837209302326 487.5234599717762"} ], "linkDataArray": [ {"from":-4, "to":-3, "fromPort":"T", "toPort":"L", "visible":true}, {"from":-1, "to":-4, "fromPort":"R", "toPort":"L"}, {"from":-3, "to":-5, "fromPort":"R", "toPort":"L"}, {"from":-5, "to":-6, "fromPort":"T", "toPort":"L", "visible":true}, {"from":-5, "to":-7, "fromPort":"B", "toPort":"L", "visible":true, "text":"NO"}, {"from":-6, "to":-8, "fromPort":"R", "toPort":"L"}, {"from":-7, "to":-8, "fromPort":"R", "toPort":"L"}, {"from":-4, "to":-9, "fromPort":"B", "toPort":"L", "visible":true, "text":"NO"}, {"from":-9, "to":-10, "fromPort":"T", "toPort":"L", "visible":true}, {"from":-10, "to":-12, "fromPort":"R", "toPort":"L"}, {"from":-11, "to":-13, "fromPort":"R", "toPort":"L"}, {"from":-12, "to":-13, "fromPort":"R", "toPort":"L"}, {"from":-8, "to":-13, "fromPort":"R", "toPort":"L"}, {"from":-13, "to":-2, "fromPort":"R", "toPort":"L"}, {"from":-9, "to":-14, "fromPort":"B", "toPort":"L", "visible":true, "text":"NO"}, {"from":-14, "to":-11, "fromPort":"T", "toPort":"L", "visible":true}, {"from":-14, "to":-11, "fromPort":"B", "toPort":"L", "visible":true, "text":"NO"} ]} </code></pre>
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    1. COIn all of your examples, once you take one branch or the other, you always eventually end up at "end". You never loop back around and possibly enter the decision node again. Does that property hold in general, or do you need to consider situations where there are loops? The solution is straightforward if there are no such loops; if loops can exist then it's slightly trickier.
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    2. CO@EricLippert you are correct in assuming that there are no loops back, everything moves forward. I think my comment on the answer below is actually the solution I'm looking for.
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    3. COExcellent. Note also that you need to have the property that there is exactly one "end", or at least that if there is more than one "end" node, they compare as equal.
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