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Better handling of Unnecessary Parentheses #252

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Better handling of Unnecessary Parentheses #252

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e7b41af
spec for parenthesis on both sides of equation
GabrielMahan Apr 10, 2020
c9593c4
leverage removeUnnecessaryParens root node recrusive search
GabrielMahan May 18, 2020
588370a
revert previous non-spec changes to account for unnecessary parens (7…
GabrielMahan May 18, 2020
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4 changes: 0 additions & 4 deletions lib/Symbols.js
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Original file line number Diff line number Diff line change
Expand Up @@ -39,10 +39,6 @@ Symbols.getLastSymbolTerm = function(node, symbolName) {
}
}
}
else if (Node.Type.isParenthesis(node)) {
return Symbols.getLastSymbolTerm(node.content, symbolName);
}

return null;
};

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10 changes: 2 additions & 8 deletions lib/solveEquation/EquationOperations.js
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Expand Up @@ -128,16 +128,10 @@ EquationOperations.removeSymbolFromRightSide = function(equation, symbolName) {
// or dividing all other symbols and constants from both sides appropriately
// TODO: support inverting functions e.g. sqrt, ^, log etc.
EquationOperations.isolateSymbolOnLeftSide = function(equation, symbolName) {
let leftNode = equation.leftNode;

if (Node.Type.isParenthesis(leftNode)) {
// if entire left node is a parenthesis, we can ignore the parenthesis
leftNode = leftNode.content;
}

const leftNode = equation.leftNode;
let nonSymbolTerm = Symbols.getLastNonSymbolTerm(leftNode, symbolName);
let inverseOp, inverseTerm, changeType;

let inverseOp, inverseTerm, changeType;
if (!nonSymbolTerm) {
return EquationStatus.noChange(equation);
}
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4 changes: 2 additions & 2 deletions lib/solveEquation/stepThrough.js
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Expand Up @@ -55,8 +55,8 @@ function stepThrough(leftNode, rightNode, comparator, debug=false) {
// If we have roots, we return early and do not go through simplification,
// so we can't rely on that flow for parentheses removal
// e.g. x^(2) = 0 -> x^2 = 0
equation.leftNode = removeUnnecessaryParens(equation.leftNode);
equation.rightNode = removeUnnecessaryParens(equation.rightNode);
equation.leftNode = removeUnnecessaryParens(equation.leftNode, true);
equation.rightNode = removeUnnecessaryParens(equation.rightNode, true);

// Checks if there are roots in the original equation before we
// do any simplification.
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3 changes: 2 additions & 1 deletion test/solveEquation/solveEquation.test.js
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Expand Up @@ -96,7 +96,8 @@ describe('solveEquation for =', function () {
['(3 + x) / (x^2 + 3) = 1', 'x = [0, 1]'],
['6/x + 8/(2x) = 10', 'x = 1'],
['(x+1)=4', 'x = 3'],
['((x)/(4))=4', 'x = 16']
['((x)/(4))=4', 'x = 16'],
['(2x-12)=(x+4)', 'x = 16']
// TODO: fix these cases, fail because lack of factoring support, for complex #s,
// for taking the sqrt of both sides, etc
// ['(x + y) (y + 2) = 0', 'y = -y'],
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