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	assymetric queueing logic
allow tagged job execution on an untagged node, but not vice-versa. This allows the admin to assign tags for UI grouping without worrying about nodes at all
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				| @ -529,16 +529,16 @@ bool Laminar::nodeCanQueue(const Node& node, const Run& run) const { | ||||
|     if(node.busyExecutors >= node.numExecutors) | ||||
|         return false; | ||||
| 
 | ||||
|     auto it = jobTags.find(run.name); | ||||
|     // if both nodes have no tags, it's OK
 | ||||
|     if(it == jobTags.end() && node.tags.size() == 0) | ||||
|     // if the node has no tags, allow the build
 | ||||
|     if(node.tags.size() == 0) | ||||
|         return true; | ||||
| 
 | ||||
|     // but if just one of them does, don't allow the build
 | ||||
|     if(it == jobTags.end() || node.tags.size() == 0) | ||||
|     auto it = jobTags.find(run.name); | ||||
|     // if the job has no tags, it cannot be run on this node
 | ||||
|     if(it == jobTags.end()) | ||||
|         return false; | ||||
| 
 | ||||
|     // in other cases, allow the build if they have a tag in common
 | ||||
|     // otherwise, allow the build if job and node have a tag in common
 | ||||
|     for(const std::string& tag : it->second) { | ||||
|         if(node.tags.find(tag) != node.tags.end()) | ||||
|             return true; | ||||
|  | ||||
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