Merge pull request #205 from lambda-feedback/tr173-changed-preview-code-gen-for-latex-response-with-pm

Found a way to solve the original problem without generating code for…

Author: KarlLundengaard

Diff for: app/preview_utilities.py

@@ -89,17 +89,13 @@ def parse_latex(response: str, symbols: SymbolDict, simplify : bool, parameters=
         raise ValueError(str(e))
 
     if (pm_placeholder is not None) or (mp_placeholder is not None):
-        result_str_set = set()
         result_str = str(expression)
         for ph in [(pm_placeholder, "plus_minus"), (mp_placeholder, "minus_plus")]:
             if ph[0] is not None:
                 result_str = result_str.replace("*"+ph[0]+"*", " "+ph[1]+" ")
                 result_str = result_str.replace(ph[0]+"*", " "+ph[1]+" ")
                 result_str = result_str.replace("*"+ph[0], " "+ph[1]+" ")
                 result_str = result_str.replace(ph[0], " "+ph[1]+" ")
-        for expr in create_expression_set(result_str, parameters):
-            result_str_set.add(expr)
-        result_str = '{'+', '.join(result_str_set)+'}'
     else:
         result_str = str(expression.xreplace(substitutions))
 

Diff for: app/symbolic_comparison_preview_tests.py

@@ -96,7 +96,7 @@ def test_latex_with_plus_minus(self):
         )
         result = preview_function(response, params)
         preview = result["preview"]
-        assert preview.get("sympy") in ['{3*i/sqrt(5), - 3*i/sqrt(5)}', '{- 3*i/sqrt(5), 3*i/sqrt(5)}']
+        assert preview.get("sympy") == ' plus_minus 3*i/sqrt(5)'
         assert preview.get("latex") == r'\pm \frac{3}{\sqrt{5}} i'
 
     def test_latex_conversion_preserves_default_symbols(self):