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/* | |
Countdown rules: | |
Players select 6 number cards at random from two different decks. | |
One deck has 25, 50, 75, 100. | |
Another has the numbers 1-10, each twice. | |
A random number, the target, is picked between 100-999. | |
Players must find a way to the target using the number cards. | |
Numbers can be added, deducted, multiplied, or divided. | |
A card can only be used once in a given solution. | |
Negative numbers are not allowed. | |
Non-integers are not allowed. | |
*/ | |
function pickRandom<T>(arr: T[]): T { | |
const index = Math.floor(Math.random() * arr.length); | |
return arr.splice(index, 1)[0]; | |
} | |
const largeNums = [25, 50, 75, 100]; | |
// 1-10, twice | |
const smallNums = Array.from({ length: 20 }, (_, i) => (i % 10) + 1); | |
const ops = ['+', '-', '*', '/']; | |
// 100-999 | |
const target = Math.floor(Math.random() * 900) + 100; | |
const numbers = [ | |
pickRandom(largeNums), | |
pickRandom(largeNums), | |
pickRandom(smallNums), | |
pickRandom(smallNums), | |
pickRandom(smallNums), | |
pickRandom(smallNums), | |
]; | |
console.log('Target', target); | |
console.log('Numbers', numbers); | |
let solutionSteps = Infinity; | |
function solveBranch(numbers: number[], steps: number = 0): string | undefined { | |
let branchSolution: string | undefined = undefined; | |
// The solution is not allowed to dip into negative numbers at any point. | |
// So. sort the numbers descending, to avoid negatives and fractions < 1. | |
numbers.sort((a, b) => b - a); | |
for (let i = 0; i < numbers.length; i++) { | |
// We only need to look forward from the current position, because: | |
// Order doesn't matter with addition. | |
// Order matters with deduction, but the other way around would be < 0. | |
// Order doesn't matter with multiplication. | |
// Order matters with division, but the other way around would be < 1. | |
for (let j = i + 1; j < numbers.length; j++) { | |
for (const op of ops) { | |
let subResult: number; | |
switch (op) { | |
case '+': | |
subResult = numbers[i] + numbers[j]; | |
break; | |
case '-': | |
subResult = numbers[i] - numbers[j]; | |
// Zero doesn't help, and might land us with div-by-zero later. | |
if (subResult === 0) continue; | |
break; | |
case '*': | |
// Multiplying by 1 is a waste of a move. | |
if (numbers[i] === 1 || numbers[j] === 1) continue; | |
subResult = numbers[i] * numbers[j]; | |
break; | |
default: | |
// '/' | |
// Dividing by 1 is a waste of a move. | |
// Fractions are not allowed. | |
if (numbers[j] === 1 || numbers[i] % numbers[j]) continue; | |
subResult = numbers[i] / numbers[j]; | |
break; | |
} | |
if (subResult === target) { | |
solutionSteps = steps + 1; | |
return `${numbers[i]} ${op} ${numbers[j]} = ${subResult}`; | |
} | |
// We've already found a simpler solution. Bail. | |
if (steps + 1 === solutionSteps) return; | |
// Create a new set of numbers, removing the cards we've used, | |
// and replacing them with a made-up card of our intermediate result. | |
const newNumbers = [ | |
...numbers.slice(0, i), | |
...numbers.slice(i + 1, j), | |
...numbers.slice(j + 1), | |
subResult, | |
]; | |
const subSolution = solveBranch(newNumbers, steps + 1); | |
if (!subSolution) continue; | |
// Document the solution, but a shorter solution may come up later. | |
branchSolution = `${numbers[i]} ${op} ${numbers[j]} = ${subResult}\n${subSolution}`; | |
} | |
} | |
} | |
return branchSolution; | |
} | |
console.log(solveBranch(numbers) || 'Could not solve'); |
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/* | |
Countdown rules: | |
Players select 6 number cards at random from two different decks. | |
One deck has 25, 50, 75, 100. | |
Another has the numbers 1-10, each twice. | |
A random number, the target, is picked between 100-999. | |
Players must find a way to the target using the number cards. | |
Numbers can be added, deducted, multiplied, or divided. | |
A card can only be used once in a given solution. | |
Negative numbers are not allowed. | |
Non-integers are not allowed. | |
*/ | |
function pickRandom<T>(arr: T[]): T { | |
const index = Math.floor(Math.random() * arr.length); | |
return arr.splice(index, 1)[0]; | |
} | |
const largeNums = [25, 50, 75, 100]; | |
// 1-10, twice | |
const smallNums = Array.from({ length: 20 }, (_, i) => (i % 10) + 1); | |
const ops = ['+', '-', '*', '/']; | |
// 100-999 | |
const target = Math.floor(Math.random() * 900) + 100; | |
const numbers = [ | |
pickRandom(largeNums), | |
pickRandom(largeNums), | |
pickRandom(smallNums), | |
pickRandom(smallNums), | |
pickRandom(smallNums), | |
pickRandom(smallNums), | |
]; | |
console.log('Target', target); | |
console.log('Numbers', numbers); | |
let solutionDiff = Infinity; | |
let currentSolution: string[] | undefined = undefined; | |
function solveBranch(numbers: number[], steps: string[] = []): void { | |
// The solution is not allowed to dip into negative numbers at any point. | |
// So, sort the numbers descending, to avoid negatives and fractions < 1. | |
numbers.sort((a, b) => b - a); | |
for (let i = 0; i < numbers.length; i++) { | |
// We only need to look forward from the current position, because: | |
// Order doesn't matter with addition. | |
// Order matters with deduction, but the other way around would be < 0. | |
// Order doesn't matter with multiplication. | |
// Order matters with division, but the other way around would be < 1. | |
for (let j = i + 1; j < numbers.length; j++) { | |
for (const op of ops) { | |
let subResult: number; | |
switch (op) { | |
case '+': | |
subResult = numbers[i] + numbers[j]; | |
break; | |
case '-': | |
subResult = numbers[i] - numbers[j]; | |
// Zero doesn't help, and might land us with div-by-zero later. | |
if (subResult === 0) continue; | |
break; | |
case '*': | |
// Multiplying by 1 is a waste of a move. | |
if (numbers[i] === 1 || numbers[j] === 1) continue; | |
subResult = numbers[i] * numbers[j]; | |
break; | |
default: | |
// '/' | |
// Dividing by 1 is a waste of a move. | |
// Fractions are not allowed. | |
if (numbers[j] === 1 || numbers[i] % numbers[j]) continue; | |
subResult = numbers[i] / numbers[j]; | |
break; | |
} | |
const diff = Math.abs(target - subResult); | |
const step = `${numbers[i]} ${op} ${numbers[j]} = ${subResult}`; | |
if ( | |
// A better solution, or our first solution | |
diff < solutionDiff || | |
// Or a same solution in fewer steps | |
(diff === solutionDiff && steps.length + 1 < currentSolution!.length) | |
) { | |
solutionDiff = diff; | |
currentSolution = [...steps, step]; | |
} | |
// If we've got an exact answer, and we're about to perform more steps, bail. | |
if (solutionDiff === 0 && steps.length + 1 >= currentSolution!.length) { | |
return; | |
} | |
// Create a new set of numbers, removing the cards we've used, | |
// and replacing them with a made-up card of our intermediate result. | |
const newNumbers = [ | |
...numbers.slice(0, i), | |
...numbers.slice(i + 1, j), | |
...numbers.slice(j + 1), | |
subResult, | |
]; | |
solveBranch(newNumbers, [...steps, step]); | |
} | |
} | |
} | |
} | |
solveBranch(numbers); | |
console.log(currentSolution!.join('\n')); | |
if (solutionDiff) console.log(solutionDiff, 'out'); |
I hope this is the right place to post other solutions, here is mine
const randomInRange = (from, to) => Math.floor(Math.random() * (to - from)) + from;
const pickRandom = arr => arr.splice(Math.floor(Math.random() * arr.length), 1)[0];
const operations = [{
name: "+",
fn: (a, b) => a + b,
},
{
name: "-",
fn: (a, b) => a - b,
},
{
name: "*",
fn: (a, b) => a * b,
},
{
name: "/",
fn: (a, b) => a / b,
},
];
const target = randomInRange(100, 1000);
const smallNumbers = [25, 50, 75, 100];
const largeNumbers = [...Array(20).keys()].map(i => (i % 10) + 1);
const numbers = [
pickRandom(largeNumbers),
pickRandom(largeNumbers),
pickRandom(smallNumbers),
pickRandom(smallNumbers),
pickRandom(smallNumbers),
pickRandom(smallNumbers),
];
const solve = (numbers, target) => {
let permutations = numbers.map(permutation => ({ permutation, path: [permutation], remaining: numbers.filter(number => number !== permutation) }));
// every round we increase the number of numbers used by one, using the result of the previous round
// therefore the solution should be minimal
while (permutations[0].remaining.length !== 0) {
const foundTarget = permutations.find(({ permutation }) => target === permutation);
if (foundTarget) return foundTarget;
// we take the previous result, an operation and a not used number and calculate the new result
permutations = permutations.flatMap(({ permutation, path, remaining }) => remaining.flatMap(number => operations.map(({ fn, name }) => ({
permutation: fn(permutation, number),
path: [...path, name, number],
remaining: remaining.filter((numberA, i, self) => numberA !== number || self.indexOf(numberA) !== i),
}))))
.filter(({ permutation }) => permutation % 1 === 0 && permutation >= 0);
}
};
console.log(`looking for target ${target}`);
console.log(`${new Intl.ListFormat().format(numbers.map(String))} are given`);
const res = solve(numbers, target);
if (res) {
console.log(`\none of the shortest calculations is ${res.path.join(" ")}`);
} else {
console.log("\nthere is no solution");
}
Does this work if the set of numbers contains duplicates? Looking at those filters, it feels like it might remove too many.
True, remaining: remaining.filter(numberA => numberA !== number),
removes every number, event though it should only remove the first one. I fixed it by checking if the number to remove is the first occurrence in the array remaining: remaining.filter((numberA, i, self) => numberA !== number || self.indexOf(numberA) !== i),
.
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Running example: https://jsbin.com/yimejin/4/edit?js,console