# CodinGame: Derivative Time - Part 1, Recursion (Typescript)

Solution of CodinGame exercise. Simple recursion example with typescript. Formula representation inspired by lisp.

Now we will solve exercise DERIVATIVE TIME:

Our goal is, evaluate the partial derivative of a given formula.

For instance, given the function formula "(5*(x*(y^2)))" and "y x", the variables in respect with you must derive it

So here f(x,y) = 5xy² and you have to calculate:

   d²f(x,y)
----------
dxdy

it gives you the formula 10*y. At last "x 2 y 6" means x=2, y=6, gives you the values with which you must evaluate the obtained derivative. So the answer should be 60

Note
To simplify the task, only consider +, * and ^. And assume that +, * and ^ always take two arguments and that expressions are fully parenthesized.

Negative power has no parenthesis. e.g. (((18*(x^-1))+y)+z)

vars may be in other forms other than x, y, and z. Similar to identifiers in many programming languages, the var would be some letter followed by letters, numbers or underscore.

The rules needed here:

a'=0
(a*x)'=a
(x^a)'=a*x^(a-1) (when a is not 0)
(u+v)'=u'+v'
(u*v)'=u'*v+v'*u

# Theoretical introduction to derivatives

To compute derivative of function we have to represent function as object possible to manipulate. One of possibilities is representing it as an array with operation placed on the beginning and with arguments occupying rest of positions.

This concept is intensively used in some of languages as Lisp or Scheme. If you do not know Lisp you can easy learn it in y minutes on page:

a + b

in lisp you will write

(+ a b)

Our further solution will be heavily based on this concept of ordering. How advantages it have over a + b? Firstly we can decompose our expressions by

const [operation, ...arguments] = expression;

Secondly for associative operations we can use reduce function like

const res = arguments.reduce(fn, identity);

In this case identity is value that will not change result of operation, and fn is function assigned to operation, eg (a,b) => a+b for addition.

Having these basic concept in mind we can consider derivative as function that transform array representing mathematical formula to other array with other formula.

We know which transformations should be applied to addition, multiplication or powers. We also able to simplify result knowing derivatives of constant or identity function.

So lets divide our solution to four steps.

1. Setup project
2. Parsing input
3. Evaluating values
4. Computing derivative

## Setup NodeJS typescript project with jest

Lets init project

npm init -y
tsc --init

in generated tsconfigl.json we have to set

"target": "ESNext",

to be albe to use Map or other modern data structures or syntax.

In package.json we should set

"test": "jest"

in scripts section and install required packages.

npm i -D @types/jest @types/node esbuild-jest jest ts-node typescript

Now lets create jest config jest.config.ts

module.exports = {
roots: ['<rootDir>'],
testMatch: ['**/__tests__/**/*.+(ts|tsx)', '**/?(*.)+(spec|test).+(ts|tsx)'],
transform: {
'^.+\\.(ts|tsx)$': 'esbuild-jest', }, setupFilesAfterEnv: [], testEnvironment: 'node', }  Finally we can create three files: lib.ts with: export function run(input: string): string { return '' }  Test test/lib.test.ts with import {run} from "../lib"; describe('e2e', () => { it('temporary', () => { expect(run('')).toEqual('') }) }); and index.ts that is responsible for input/output operations import {run} from "./lib"; process.stdin.on('data', (buff) => { const input = buff.toString(); process.stdout.write(run(input)); })  Now we should be able to run program using echo '' | ts-node index.ts and test by command npm run test ## Parsing Input and building objects Lets see on input From content of this exercise we can read that input contains 3 lines: Line 1: formula Line 2: list of vars for partial derivative, separated by space, length of the list will be 1, 2 or 3. Line 3: vars' values, paired and separated by space Output is the result (always an integer). So for example input (5*(x*y)) x x 2 y 6 we will have derivative by x giving (5*y), that in point {x:2, y:6} is 5*6 or simply 30 Now we will introduce three objects that will refer to these lines: • Formula • DiffOperator • Coordinate Generally any of these object should be created basing on his line. Only exception is Formula that require variables from coordinate to recognize on arguments and treat them in a special way. Now run function from lib.ts can be rewritten as export function run(input: string): string { const [F, vs, dict] = input.split('\n'); const point = new Coordinate(dict); const diff = new DiffOperator(vs); const formula = new Formula(F, point.variables); return diff.actOn(formula).evaluate(point).toString(); }  What can be explained as acting derivative operator on function and evaluating it at given point. It is exactly what we need to do in this task. Now we will implement these classes and write some tests for them ### Coordinate implementation Lets start from coordinate export class Coordinate { map: Map<string, number> = new Map(); constructor(input: string) { const pairs = input.split(' '); for (let i = 0; i < pairs.length / 2; i++) { const [key, value] = [pairs[i * 2], Number.parseFloat(pairs[i * 2 + 1])]; this.map.set(key, value); } } get variables(): string[] { return [...this.map.keys()] } get(name): number { return this.map.get(name) ?? 0; } } It simply splits his string x 2 y 6 by spaces and move by pairs setting them as keys and values of internal Map. It can return both list of keys to help in Formula initialization and get singe value of key that is useful in evaluating. tests for coordinate are the following describe('coordinate', () => { it('parsing', () => { const point = new Coordinate('x 2 y 6'); expect(point.variables).toEqual(['x', 'y']); expect(point.get('x')).toEqual(2); expect(point.get('y')).toEqual(6); expect(point.get('undefined')).toEqual(0); }) }) ### Formula initialization Next object is formula. We need two additional types type Operator = '*' | '^' | '+' | '-'; type SubFormula = Array<string | number | SubFormula> | string | number and can write our formula class export class Formula { f: SubFormula; v: string[] = []; } following functions will be placed inside of this class. We need constructor that will get formula string and list of variables. To operate on string form of formula we will define two functions: • decomposeInput(input: string): SubFormula • isDecomposable(input: string): boolean Firs will do parsing, second will check if should we expand input to elements or should we left it as atomic part of formula. In our case we need to decompose if string starts and ends with brackets and contains operator  isDecomposable(input: string): boolean { return input.startsWith('(') && input.endsWith(')') && /[*^+-]/.test(input) } Decomposition contains part that prepare three elements p like previous, n like next, and operator. decomposeInput(input: string): SubFormula { let p: SubFormula = '', operator: Operator | '' = '', n: SubFormula = '', level: number = 0; if (!this.isDecomposable(input)) return [0]; for (let i = 1; i < input.length - 1; i++) { if (level === 0 && !operator && /[*^+-]/.test(input[i])) { operator = input[i] as Operator; continue; } else if (input[i] === '(') { level++; } else if (input[i] === ')') { level--; } if (operator) { n += input[i]; } else { p += input[i]; } } we are using level to determine which sub bracket level we reaching during loop. After loop over all characters and completing p, n, and operator we have to check if should we decompose or parse these elements using code  if (this.isDecomposable(p)) { p = this.decomposeInput(p); } else { if (!this.v.includes(p)) { p = Number.parseFloat(p); } } if (this.isDecomposable(n)) { n = this.decomposeInput(n); } else { if (!this.v.includes(n)) { n = Number.parseFloat(n); } } return [operator, p, n]; } after defining these functions we can write constructor  constructor(input: string, variables: string[]) { this.v = variables; this.f = this.decomposeInput(input); } and test for them describe('formula', () => { const f = new Formula('(5*(x*y))', ['x', 'y']); it('parse', () => { expect(f.f).toEqual(['*', 5, ['*', 'x', 'y']]); expect(f.v).toEqual(['x', 'y']); }); it('parse with many brackets', () => { const f = new Formula('((x^3)+(x^2))', ['x']); expect(f.f).toEqual(['+', ['^', 'x', 3], ['^', 'x', 2]]) }) }) ### DiffOperator (derivateve) initialization Fortunatelly derivative is super easy to construct export class DiffOperator { d: string[] = []; constructor(input: string) { this.d = input.split(' '); } } inside we want to store array of variables names that we will use seqentially acting on given function. We can skip tests for it and go to next part - Evaluating values ## Evaluating values Now we still will write code in Formula class. Our computing function will be able to call themself many times so to simply their interface we will split computation to: static compute(f: SubFormula | string | number, point: Coordinate): number and evaluate(point: Coordinate): number  first one will be used to operate directly on SubFormula, but second will be layer that was applied in run in expression diff.actOn(formula).evaluate(point) evaluate can be defined as  evaluate(point: Coordinate): number { return Formula.compute(this.f, point); } But compute should start from classification of his argument. It can be expression, then Array.isArray(f) will be true, number, then this number should be returned od name of coordinate, then coordinate saved in point can be returned. Our code will be the following  static compute(f: SubFormula | string | number, point: Coordinate): number { if (Array.isArray(f)) { // ... TO BE DEFINED IN NEXT STEP } else if (typeof f === 'number') { return f; } else { return point.get(f); } } Most interesting is case when, f is array. Then using theory from introduction we should decompose it to operator and arguments; const [op, ...rest] = f; and in next step in dependence from operator arguments can be processed by different functions: switch (op) { case '+': return <number>rest.reduce((p: number, n: SubFormula): number => p + this.compute(n, point), 0); case '*': return <number>rest.reduce((p: number, n: SubFormula): number => p * this.compute(n, point), 1); case '^': { const [p, n] = rest; return Math.pow(this.compute(p, point), this.compute(n, point)); } case '-': { const [p, n] = rest; return this.compute(p,point) - this.compute(n, point); } default: return 0; } For associative operations like * and + with existing identity value we used reduce. But for rest cases we assumed that there are two arguments and used destructuring. To prove that it works we can add the following test to formula section  it('evaluate', () => { expect(f.evaluate(new Coordinate('x 1 y 2'))).toStrictEqual(10); }); our formula defined ealier was (5*(x*y)) so we expect result 10. If you following all these concepts to this point, then next step will be easy logical continuation for you. ## Computing derivative Computing derivative will be done in the same style like evaluating value of function. We will split it on two methods Dynamic actOn with interface actOn(f: Formula): Formula and static internal function static derivative(s: SubFormula, dir: string): SubFormula First one will iterate over all directions saved in d property. Second is first order derivative in direction passed by second argument. These functions works also on other levels: actOn will use Formula abstraction, but derivative need and return SubFormula to operate on more low level structure. We will introduce also one more static function static simplify(s: SubFormula): SubFormula  that will not need direction and use simple algebra rules to make computation partial results more human readable and easy to process. Simplify will return the same mathematical formula but written in most simple possible manner removing unnecessary additions of 0 or multiplications by 1. Following functions are defined as methods of DiffOperator. ### Simplification of formula Lets start from simplify because of I started to explain it. To simplify formulas we will apply the folowing rules: • all subarguments should be simplified (recursive) • if all subarguments are numbers, they should be computed • if one of argument is 0 or 1 special treatment of second argument should be applied in dependence from operator Code that implement described behavior is  static simplify(s: SubFormula): SubFormula { if (Array.isArray(s)) { const [op, ...rest] = s; for (let i = 0; i < rest.length; i++) { rest[i] = this.simplify(rest[i]); } if (rest.every((r) => typeof r === 'number')) { return Formula.compute(s, new Coordinate('')); } if (rest[0] === 0) { if (op === '+') return rest[1] if (op === '*') return 0 if (op === '^') return 0 } if (rest[0] === 1) { if (op === '*') return rest[1] if (op === '^') return 1 } if (rest[1] === 0) { if (op === '+') return rest[0] if (op === '*') return 0 if (op === '^') return 1 } if (rest[1] === 1) { if (op === '*') return rest[0] if (op === '^') return rest[0] } return [op, ...rest]; } else { return s; } } ### Derivative of formula To define derivative of first order on given direction: static derivative(s: SubFormula, dir: string): SubFormula we have to assume that we can compute it on • expression • our variable • constant If it is executed on variable or constant then result will be 1 or 0, but on expression we have to check operator and decide which rule should we apply. This part is exactly the same concept as with evaluation, but we returns arrays with operators that represents functions instead of numbers. Code below:  static derivative(s: SubFormula, dir: string): SubFormula { if (Array.isArray(s)) { const [op, p, n] = s; switch (op) { case '*': return ['+', ['*', p, this.derivative(n, dir)], ['*', this.derivative(p, dir), n]]; case '+': return ['+', this.derivative(p, dir), this.derivative(n, dir)]; case '-': return ['-', this.derivative(p, dir), this.derivative(n, dir)]; case '^': { if (p === dir) { if (n === 0) return 0; return typeof n === 'number' ? ['*', n, ['^', p, n - 1]] : ['*', n, ['^', p, ['+', n, -1]]] } else { return 0; } } } } else if (typeof s === 'string' && s === dir) { return 1 } else { return 0; } } we can see that only with powers there is a little confuction, because of I decided to evaluate subtraction 1 on power if it is number. I decided to skip support for a^x because of it is exponential function that is out of scope of first part of this exercise. Finally to implement actOn we will need clone of formula, so on formula we should implement  clone(): Formula { const o = new Formula('', []); o.f = JSON.parse(JSON.stringify(this.f)); o.v = JSON.parse(JSON.stringify(this.f)); return o; } and then on DiffOperator  actOn(f: Formula): Formula { const o = f.clone(); return this.d.reduce((p, n) => { o.f = DiffOperator.simplify(DiffOperator.derivative(p.f, n)); return o; }, o); } to write stable, editable and working code we should add tests describe('diff operation', () => { it('simplify', () => { expect(DiffOperator.simplify(["+", ["*", 1, 0], ["*", 0, 1]])).toEqual(0); expect(DiffOperator.simplify(['^', 'x', 1])).toEqual('x'); expect(DiffOperator.simplify(['*', 2, ['^', 'x', 1]])).toEqual(['*', 2, 'x']); expect(DiffOperator.simplify(["*", 5, ["+", ["*", "x", 0], ["*", 1, "y"]]])).toEqual(['*', 5, 'y']) expect(DiffOperator.simplify(["+", ["*", 5, ["+", ["*", "x", 0], ["*", 1, "y"]]], ["*", 0, ["*", "x", "y"]]])).toEqual(['*', 5, 'y']) }); it('derivative', () => { expect(DiffOperator.simplify(DiffOperator.derivative(['*', 2, 'x'], 'x'))).toEqual(2); expect(DiffOperator.simplify(DiffOperator.derivative(['*', 5, ['*', 'x', 'y']], 'x'))).toEqual(['*', 5, 'y']) }) it("a'=0", () => { const f = new Formula('(1*1)', ['x']); const d = new DiffOperator('x'); expect(d.actOn(f).f).toEqual(0); }); it("(a*x)'=a", () => { const f = new Formula('(4*x)', ['x']); const d = new DiffOperator('x'); expect(d.actOn(f).f).toEqual(4); }); it("(x^a)'=a*x^(a-1) (when a is not 0)", () => { const cases = [ {in: '(x^5)', out: ['*', 5, ['^', 'x', 4]]}, {in: '(x^3)', out: ['*', 3, ['^', 'x', 2]]}, {in: '(x^2)', out: ['*', 2, 'x']}, ]; for (const c of cases) { const f = new Formula(c.in, ['x']); const d = new DiffOperator('x'); expect(d.actOn(f).f).toEqual(c.out); } }); it("(u+v)'=u'+v for (x+(x^2))'", () => { const f = new Formula('(x+(x^2))', ['x']); const d = new DiffOperator('x'); expect(d.actOn(f).f).toEqual(['+', 1, ['*', 2, 'x']]); }); it("(u+v)'=u'+v'", () => { const f = new Formula('((x^3)+(x^2))', ['x']); const d = new DiffOperator('x'); console.log("f", f); expect(d.actOn(f).f).toEqual(['+', ['*', 3, ['^', 'x', 2]], ['*', 2, 'x']]); }); it("(u*v)'=u'*v+v'*u", () => { const f = new Formula('(x*x)', ['x']); const d = new DiffOperator('x'); console.log("f", f); expect(d.actOn(f).f).toEqual(['+', 'x', 'x']); }); it('d(8*(y^x))/dy', () => { const f = new Formula('(8*(y^x))', ['x', 'y']); expect(f.f).toEqual(['*', 8, ['^', 'y', 'x']]); expect(DiffOperator.simplify(DiffOperator.derivative(['*', 8, ['^', 'y', 'x']], 'y'))).toEqual(['*', 8, ['*', 'x', ['^', 'y', ['+', 'x', -1]]]]); }) it('(18*(x^-1))',() => { const f = new Formula('(18*(x^-1))', ['x']); expect(f.f).toEqual(['*', 18, ['^', 'x', -1]]); expect(DiffOperator.simplify(DiffOperator.derivative(f.f, 'x'))).toEqual( ['*', 18, ['*', -1, ['^', 'x', -2]]]); }); it('d(((x^2)+(2*(z^5)))+(((18*(x^-1))+y)+z))/dx', () => { const f = new Formula('(((x^2)+(2*(z^5)))+(((18*(x^-1))+y)+z))', ['x', 'y', 'z']); expect(DiffOperator.simplify(DiffOperator.derivative(f.f, 'x'))).toEqual(['+', ['*', 2, 'x'], ['*', 18, ['*', -1, ['^', 'x', -2]]]]); }) }) Finally all elements of program works so we can add also e2e tests on run function describe('e2e', () => { it('easy multiply', () => { expect(run('(5*(x*y))\nx\nx 2 y 6')).toEqual('30') }) it("second derivative", () => { expect(run('(5*((x^4)*(y^2)))\nx x\nx 2 y 6')).toEqual('8640') }) it("second derivative mix", () => { expect(run('(5*(x*(y^2)))\ny x\nx 2 y 6')).toEqual('60') }) it("power with number", () => { expect(run('((x^2)+(9*(x+y)))\nx\nx 1 y 2')).toEqual('11') }) it("power with variable", () => { expect(run('(8*(y^x))\ny y\nx -1 y 2')).toEqual('2') }) it("3 variables", () => { expect(run('(((x^2)+(2*(z^5)))+((x+y)+z))\nz\nx 2 y 3 z 4')).toEqual('2561') }) it("fraction", () => { expect(run('(((x^2)+(2*(z^5)))+(((18*(x^-1))+y)+z))\nx\nx 3 y 4 z 1')).toEqual('4') }) it("longer multiply", () => { expect(run('(((x^2)*(2*(z^5)))*((x+y)+z))\nz\nx 1 y 1 z 1')).toEqual('32') }) it("3rd derivative", () => { expect(run('(((y^6)*(z^5))*(((3*(x^4))+y)+z))\ny y z\nx 1 y 1 z 2')).toEqual('16320') }) it("some Greek ;)", () => { expect(run('(((Beta^6)*(Gamma^5))*(((3*(Alpha^4))+Beta)+Gamma))\nBeta Beta Gamma\nAlpha 1 Beta 1 Gamma 2')).toEqual('16320') }) it("maybe not xyz ;)", () => { expect(run('(((x2^6)*(x3^5))*(((3*(x1^4))+x2)+x3))\nx2 x2 x3\nx1 1 x2 1 x3 2')).toEqual('16320') }) it("some Vars ;)))", () => { expect(run('(((Var_2^6)*(Var_3^5))*(((3*(Var_1^4))+Var_2)+Var_3))\nVar_2 Var_2 Var_3\nVar_1 1 Var_2 1 Var_3 2')).toEqual('16320') }) it("bigger constants", () => { expect(run('(50*((x^40)*(y^20)))\nx x\nx 1 y 1')).toEqual('78000') }) it("bigger power", () => { expect(run('(x^(y^10))\nx x\nx 1 y 2')).toEqual('1047552') }) it("cannot find", () => { expect(run('(5*(x*(y^2)))\nz\nx 2 y 6')).toEqual('0') }) }) and proper github workflow name: Node.js CI on: push: branches: [ "main" ] jobs: build: runs-on: ubuntu-latest strategy: matrix: node-version: [14.x, 16.x, 18.x] steps: - uses: actions/checkout@v3 - name: Use Node.js${{ matrix.node-version }}
uses: actions/setup-node@v3
with:
node-version: \${{ matrix.node-version }}
cache: 'npm'
- run: npm ci
- run: npm run build --if-present
- run: npm test


## Further steps

In codingame there is also second part of this exercise:

that will include trigonometric functions, logarithm, exponent and even the chain rule.

You can invite me to your friends on this platform using link:

https://www.codingame.com/servlet/urlinvite?u=5287657

You can find all presented code on my github: