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Solution to Wooded Path

Reference Point

Answer: PINE LOG
by April and/or Eric Pinnick

For each scientific plant name, use the actual traits of each plant to find it in the idiosyncratic trinary search tree. With each plant is an associated rooted tree graph consisting of lines and nodes. The gray problems are example problems and result in 0/null/empty graph. The non-example (green) puzzle problems are used to get the answer.

Math is explained below. Complete each math problem. Each math problem results in a 6node_5line graph. This graph is used as a path in the trinary search tree to locate a new scientific name of a plant. Index into this plant name by the number in (). The answer is read off in order of the green problems.

Math explanation

Let the dark node be considered the ‘root’ and be located at the bottom of the graph.

Consider the following examples:

left_graph operator right_graph = new_graph


old_graph^double_dagger = new_graph


Addition: Like concatenation. The dark node of the right_graph overlaps the top node of the left_graph. A 3node_2line graph plus a 3node_2line graph results in a 5node_4line graph.

Subtraction: As addition, but in reverse. The right_graph is removed from the top of the left_graph.

Multiplication: Each instance of a node on the left_graph becomes the entirety of the right_graph. (The right_graph will be repeated several times with lines of the left_graph separating each instance). Thus, a 3node_2line graph multiplied by a 3node_2line graph will be a 9node_8line graph.

Division: As multiplication, but in reverse. The right_graph as a whole unit is removed multiple times from the left_graph. Each instance of the right_graph will be separated from the next instance of the right_graph by one line. The lines that remain (with nodes in between) will be the new_graph.

Transpose (double dagger): Each line in a graph is mirrored individually. Left lines become right lines, right lines become left lines, and middle/vertical lines are unchanged. Number of nodes and lines in the new_graph is unchanged from the old_graph