Create poem096.py

POEM_096/poem096.py

@@ -0,0 +1,81 @@
+#we went with anytree because we saw feasible start as a backtracking problem, and anytree seemed like it would be an effective
+#method of constructing a tree to backtrack through, well documented, high functionality, etc.
+from anytree import Node, RenderTree
+from sympy import symbols, diff
+import re
+
+def problem_to_tree(problem):
+    """
+    Based on this line of code from parabaloid example:
+        prob.model.add_subsystem('paraboloid', om.ExecComp('f = (x-3)**2 + x*y + (y+4)**2 - 3'))
+    
+    We assume 'problem' as it is passed is an OpenMDAO problem, and we constructing the tree from the equation stored in the subsystem
+
+
+    Level 1 is the initial problem (represented by a function) and the left hand sign of the equation, in this case single variable Y
+
+    1            Y = A*B + C*D
+               /   \
+              /     \
+    2        AB     CD
+            /  \    / \
+           /    \  /   \
+    3     A      B C    D
+
+        Gradients: the gradient list ends up being all leaf nodes at the end of the tree's formation
+
+        Partial Dervivative: Compute the partial derivative of each parent with respect to all it's children
+        eg, the PARTIAL of A would be partial_derivative(parent, child) or partial of AB with respect to A, which is B
+
+        All nodes link to both their parent and all children (lead nodes and root nodes only link 1 way)
+
+
+        per prob.model.add_subsystem('paraboloid', om.ExecComp('f = (x-3)**2 + x*y + (y+4)**2 - 3')),
+        subsystem was assumed to be tuple of name and om.ExecComp and that ExecComp was stored as a string
+    """
+    equation = problem.model.subsystem[1] #abstraction, unsure how to access the equation string stored in subsystem
+
+    equation = equation.split('=')
+
+    lhs = equation[0]
+    lhs = lhs.strip(' ')
+    root = Node(lhs, gradients=[]) #gradients starts empty, and is filled during back propagation
+
+    rhs = equation[1]
+    terms_list = re.split(r'\D', rhs) #rough idea, does not take into account full order of operations
+
+    for each term in terms_list:
+        """
+        construct an anytree node (as outlined for the root) and 
+        store which operation (+, -, /, *, etc.) took you from parent to child
+
+        Example Node AB would be (from level 2 in diagram above):
+           node_ab = Node(AB, parent=root, gradients=[], operation='*')
+
+            - AB is the node's name
+            - the parent node is the root of the tree in this case
+            - gradients are empty until back propagation
+            - the operation which forms AB's children is multiplication. Operation of leaf node is null
+        """
+
+
+    return RenderTree(root)
+
+
+def find_feasible_start(problem):
+    tree = problem_to_tree(problem)
+
+    equation = problem.model.subsystem[1] #abstraction, unsure how to access the equation string stored in subsystem
+    equation = equation.split('=')
+    rhs = equation[1]
+    exclude = {'*', '/', '+', '-', '0', '1', '2',
+               '3', '4', '5', '6', '7', '8', '9'} #characters we don't differentiate with respect to
+
+    symbols = [] #each symbol (eg variable) we want to differentiate with respect to
+    for char in rhs:
+        if char not in exclude:
+            symbols.append(char)
+
+
+
+