mathsteps/lib/solveEquation/stepThrough.js

Step by step math solutions

const ChangeTypes = require('../ChangeTypes');
const checks = require('../checks');
const Equation = require('../equation/Equation');
const EquationOperations = require('./EquationOperations');
const EquationStatus = require('../equation/Status');
const evaluate = require('../util/evaluate');
const factor = require('../factor/stepThrough');
const flattenOperands = require('../util/flattenOperands');
const Node = require('../node');
const removeUnnecessaryParens = require('../util/removeUnnecessaryParens');
const simplifyExpressionNode = require('../simplifyExpression/stepThrough');
const Symbols = require('../Symbols');

const COMPARATOR_TO_FUNCTION = {
  '=': function(left, right) { return left === right; },
  '>': function(left, right) { return left > right; },
  '>=': function(left, right) { return left >= right; },
  '<': function(left, right) { return left < right; },
  '<=': function(left, right) { return left <= right; },
};

// Given a leftNode, rightNode and a comparator, will return the steps to get
// the solution. Possible solutions include:
// - solving for a variable (e.g. 'x=3' for '3x=4+5')
// - the result of comparing values (e.g. 'true' for 3 = 3, 'false' for 3 < 2)
function stepThrough(leftNode, rightNode, comparator, debug=false) {
  let equation = new Equation(leftNode, rightNode, comparator);

  if (debug) {
    // eslint-disable-next-line
    console.log('\n\nSolving: ' + equation.ascii(false, true));
  }

  // we can't solve/find steps if there are any unsupported nodes
  if (checks.hasUnsupportedNodes(equation.leftNode) ||
      checks.hasUnsupportedNodes(equation.rightNode)) {
    return [];
  }

  const symbolSet = Symbols.getSymbolsInEquation(equation);

  if (symbolSet.size === 0) {
    return solveConstantEquation(equation, debug);
  }
  const symbolName = symbolSet.values().next().value;

  let equationStatus;
  let steps = [];

  const originalEquationStr = equation.ascii();
  const MAX_STEP_COUNT = 20;
  let iters = 0;

  // Remove unnecessary parentheses here, before doing the find roots check
  // 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);

  // Checks if there are roots in the original equation before we
  // do any simplification.
  // E.g. (33 + 89) (x - 99) = 0
  if (checks.canFindRoots(equation)) {
    steps.push(getRootsStatus(equation));
    return steps;
  }

  // Step through the math equation until nothing changes
  do {
    steps = addSimplificationSteps(steps, equation, debug);

    if (steps.length > 0) {
      const lastStep = steps[steps.length - 1];
      equation = Equation.createEquationFromString(
        lastStep.newEquation.ascii(), equation.comparator);
    }

    equation.leftNode = flattenOperands(equation.leftNode);
    equation.rightNode = flattenOperands(equation.rightNode);

    // at this point, the symbols might have cancelled out.
    if (Symbols.getSymbolsInEquation(equation).size === 0) {
      return solveConstantEquation(equation, debug, steps);
    }

    // The left side of the equation is either factored or simplified.
    // If it is factor and we can find roots, return them.
    // e.g. x^2 + 3x + 2 = 0 -> (x + 1) (x + 2) = 0 -> x = -1
    if (checks.canFindRoots(equation)) {
      steps.push(getRootsStatus(equation));
      return steps;
    }

    try {
      equationStatus = step(equation, symbolName);
    }
    catch (e) {
      // This error happens for some math that we don't support
      if (e.message.startsWith('No term with symbol: ')) {
        // eslint-disable-next-line
        console.error('Math error: ' + e.message + ', returning no steps');
        return [];
      }
      else {
        throw e; // bubble up
      }
    }

    if (equationStatus.hasChanged()) {
      if (equationStatus.newEquation.ascii().length > 300) {
        // eslint-disable-next-line
        throw Error('Math error: Potential infinite loop for equation ' +
                    originalEquationStr +  '. It reached over 300 characters '+
                    ' long, so returning no steps');
      }
      if (debug) {
        logSteps(equationStatus);
      }
      steps.push(equationStatus);
    }

    equation = EquationStatus.resetChangeGroups(equationStatus.newEquation);
    if (iters++ === MAX_STEP_COUNT) {
      // eslint-disable-next-line
      console.error('Math error: Potential infinite loop for equation: ' +
                    originalEquationStr + ', returning no steps');
      return [];
    }
  } while (equationStatus.hasChanged());

  return steps;
}

/*
  Helper function to return the roots to a factor of an equation
  e.g. (x + 2) (x - 2) = 0 -> getRootsStatus (x + 2 = 0) will return
  a new EquationStatus with x = -2 as the roots
  Similarly getRootsStatus(x - 2 = 0) will return x = 2
*/
function getRootsStatus(equation) {
  const [solutions, symbol] = getSolutionsAndSymbol(equation);

  let allSolutions;

  if (solutions.length > 1) {
    const flattenSolutionsList = [];
    solutions.forEach(s => s.items
                      ? flattenSolutionsList.push(...s.items)
                      : flattenSolutionsList.push(s));
    allSolutions =  Node.Creator.list(flattenSolutionsList);
  }
  else if (solutions.length === 1) {
    allSolutions = solutions[0];
  }
  else {
    allSolutions = Node.Creator.list([]);
  }

  const roots = new Equation(symbol, allSolutions, '=');

  return new EquationStatus(ChangeTypes.FIND_ROOTS, equation, roots);
}

/*
  Helper function that returns the roots and symbol of an input equation
  that has roots to be found. (We check if the equation has roots first)
  For every factor on the left hand side, solve a new equation that is factor = 0
  for the symbol.
  TODO: handle multiple variable solutions e.g (x + 2) (y + 2) = 0
*/

function getSolutionsAndSymbol (equation) {
  const leftNode = equation.leftNode;

  const solutions = [];

  let symbol, steps, factorsWithSymbols;

  // If left hand side is a power node and it does not resolve to a constant
  // then it is a factor. (e.g. 2^7 resolves to a constant so it is not a factor, but
  // x^2 would have factors x = 0)
  // If left hand side is a multiplication node, return a list of all the valid factors.
  if (Node.Type.isOperator(leftNode, '^') && !checks.resolvesToConstant(leftNode)){
    factorsWithSymbols = [leftNode];
  }
  else {
    factorsWithSymbols = equation.leftNode.args.filter(arg => !checks.resolvesToConstant(arg));
  }

  /*
    For each factor, solve the equation factor = 0 for the symbol
    If the factor is a power node, solve the equation base = 0 for the symbol
    Show duplicate roots when factor is a power node
    e.g. (x + 1)^2 -> x = [-1, -1]
  */

  for (var f in factorsWithSymbols) {
    let factor = factorsWithSymbols[f];
    let exponent = 1;

    if (Node.Type.isOperator(factor, '^')) {
      exponent = parseFloat(factor.args[1].value);
      factor = factor.args[0];
    }

    const leftNode = Node.Type.isParenthesis(factor)
          ? factor.content
          : factor;

    steps = stepThrough(leftNode, equation.rightNode, '=');

    if (steps.length === 0 && Node.Type.isSymbol(leftNode)) {
      // e.g. x = 0
      symbol = leftNode;

      // Push the solution multiple times when we have duplicate roots
      // e.g. (x + 1)^2 -> x = [-1, -1]
      solutions.push(...Array(exponent).fill(equation.rightNode));
    }
    else if (steps.length !== 0) {
      // Solving for the variable on one side may sometimes
      // result in more than one step
      // e.g. x - 2 = 0
      const lastStep = steps.slice(-1)[0];

      // Append to a list of roots
      if (Node.Type.isSymbol(lastStep.newEquation.leftNode)) {
        symbol = lastStep.newEquation.leftNode;
        solutions.push(...Array(exponent).fill(lastStep.newEquation.rightNode));
      }
    }
  }

  return [solutions, symbol];
}

// Given an equation of constants, will simplify both sides, returning
// the steps and the result of the equation e.g. 'True' or 'False'
function solveConstantEquation(equation, debug, steps=[]) {
  const compareFunction = COMPARATOR_TO_FUNCTION[equation.comparator];

  if (!compareFunction) {
    throw Error('Unexpected comparator');
  }

  steps = addSimplificationSteps(steps, equation, true, debug);
  if (steps.length > 0) {
    const lastStep = steps[steps.length - 1];
    equation = Equation.createEquationFromString(
      lastStep.newEquation.ascii(), equation.comparator);
  }

  // If the left or right side didn't have any steps, unnecessary parens
  // might not have been removed, so do that now.
  equation.leftNode = removeUnnecessaryParens(equation.leftNode);
  equation.rightNode = removeUnnecessaryParens(equation.rightNode);

  if (!Node.Type.isConstantOrConstantFraction(equation.leftNode, true) ||
      !Node.Type.isConstantOrConstantFraction(equation.rightNode, true)) {
    throw Error('Expected both nodes to be constants, instead got: ' +
                equation.ascii());
  }

  const leftValue = evaluate(equation.leftNode);
  const rightValue = evaluate(equation.rightNode);
  let changeType;
  if (compareFunction(leftValue, rightValue)) {
    changeType = ChangeTypes.STATEMENT_IS_TRUE;
  }
  else {
    changeType = ChangeTypes.STATEMENT_IS_FALSE;
  }

  // there's no oldEquation or change groups because nothing actually changes
  // here, it's just a final step that states the solution
  const equationStatus = new EquationStatus(changeType, null, equation);
  if (debug) {
    logSteps(equationStatus);
  }
  steps.push(equationStatus);
  return steps;
}

// Given a symbol and an equation, performs a single step to
// solve for the symbol. Returns an Status object.
function step(equation, symbolName) {
  const solveFunctions = [
    // ensure the symbol is always on the left node
    EquationOperations.ensureSymbolInLeftNode,
    // get rid of denominators that have the symbol
    EquationOperations.removeSymbolFromDenominator,
    // remove the symbol from the right side
    EquationOperations.removeSymbolFromRightSide,
    // isolate the symbol on the left side
    EquationOperations.isolateSymbolOnLeftSide,
  ];

  for (let i = 0; i < solveFunctions.length; i++) {
    const equationStatus = solveFunctions[i](equation, symbolName);

    if (equationStatus.hasChanged()) {
      return equationStatus;
    }
  }
  return EquationStatus.noChange(equation);
}

// Simplifies the equation and returns the simplification steps
function addSimplificationSteps(steps, equation, debug=false) {
  let oldEquation = equation.clone();

  /*
    1. On the left side, we should first simplify,
    and add all those simplify substeps to a list of leftSteps.
    2. Then we should factor the simplified equation, and add all
    those factoring substeps to the leftSteps list.
    3. On the right side, there should be no need to factor,
    because we always move everything to the left side first
    e.g. x^2 + 3x + 2 = 0 <- factor the left side
    e.g. x + 4 + 2 = 0 <- simplify the left side
    e.g. 0 = x^2 + 3x + 2 -> x^2 + 3x + 2 = 0 <- swap to the left side
  */
  const leftSimplifySteps = simplifyExpressionNode(equation.leftNode, false);
  const simplifiedLeftNode = leftSimplifySteps.length !== 0
        ? leftSimplifySteps.slice(-1)[0].newNode
        : equation.leftNode;
  const leftFactorSteps = factor(simplifiedLeftNode, false);

  const leftSubSteps = [];

  for (let i = 0; i < leftSimplifySteps.length; i++) {
    const step = leftSimplifySteps[i];
    leftSubSteps.push(EquationStatus.addLeftStep(equation, step));
  }

  for (let i = 0; i < leftFactorSteps.length; i++) {
    const step = leftFactorSteps[i];
    leftSubSteps.push(EquationStatus.addLeftStep(equation, step));
  }

  if (leftSubSteps.length === 1) {
    const step = leftSubSteps[0];
    if (debug) {
      logSteps(step);
    }
    steps.push(step);
  }
  else if (leftSubSteps.length > 1) {
    const lastStep = leftSubSteps[leftSubSteps.length - 1];
    const finalEquation = EquationStatus.resetChangeGroups(lastStep.newEquation);
    // no change groups are set here - too much is changing for it to be useful
    const simplifyStatus = new EquationStatus(
      ChangeTypes.SIMPLIFY_LEFT_SIDE,
      oldEquation, finalEquation, leftSubSteps);
    if (debug) {
      logSteps(simplifyStatus);
    }
    steps.push(simplifyStatus);
  }

  // update `equation` to have the new simplified left node
  if (steps.length > 0) {
    equation = EquationStatus.resetChangeGroups(
      steps[steps.length - 1].newEquation);
  }

  // the updated equation from simplifing the left side is the old equation
  // (ie the "before" of the before and after) for simplifying the right side.
  oldEquation = equation.clone();

  const rightSteps = simplifyExpressionNode(equation.rightNode, false);
  const rightSubSteps = [];

  for (let i = 0; i < rightSteps.length; i++) {
    const step = rightSteps[i];
    rightSubSteps.push(EquationStatus.addRightStep(equation, step));
  }

  if (rightSubSteps.length === 1) {
    const step = rightSubSteps[0];
    if (debug) {
      logSteps(step);
    }
    steps.push(step);
  }
  else if (rightSubSteps.length > 1) {
    const lastStep = rightSubSteps[rightSubSteps.length - 1];
    const finalEquation = EquationStatus.resetChangeGroups(lastStep.newEquation);
    // no change groups are set here - too much is changing for it to be useful
    const simplifyStatus = new EquationStatus(
      ChangeTypes.SIMPLIFY_RIGHT_SIDE,
      oldEquation, finalEquation, rightSubSteps);
    if (debug) {
      logSteps(simplifyStatus);
    }
    steps.push(simplifyStatus);
  }

  return steps;
}

function logSteps(equationStatus) {
  // eslint-disable-next-line
  console.log('\n' + equationStatus.changeType);
  // eslint-disable-next-line
  console.log(equationStatus.newEquation.ascii());
  if (equationStatus.substeps.length > 0) {
    // eslint-disable-next-line
    console.log('\n substeps: ');
    equationStatus.substeps.forEach(logSteps);
  }
}


module.exports = stepThrough;