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https://github.com/bitburner-official/bitburner-src.git
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@@ -1,6 +1,7 @@
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import { getRandomInt } from "../utils/helpers/getRandomInt";
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import { MinHeap } from "../utils/Heap";
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import { comprGenChar, comprLZGenerate, comprLZEncode, comprLZDecode } from "../utils/CompressionContracts";
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import { HammingEncode, HammingDecode } from "../utils/HammingCodeTools";
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/* tslint:disable:completed-docs no-magic-numbers arrow-return-shorthand */
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@@ -1307,4 +1308,305 @@ export const codingContractTypesMetadata: ICodingContractTypeMetadata[] = [
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return parseInt(ans, 10) === HammingDecode(data);
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},
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},
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{
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name: "Proper 2-Coloring of a Graph",
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difficulty: 7,
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numTries: 5,
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desc: (data: [number, [number, number][]]): string => {
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return [
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`You are given the following data, representing a graph:\n`,
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`${JSON.stringify(data)}\n`,
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`Note that "graph", as used here, refers to the field of graph theory, and has`,
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`no relation to statistics or plotting.`,
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`The first element of the data represents the number of vertices in the graph.`,
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`Each vertex is a unique number between 0 and ${data[0] - 1}.`,
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`The next element of the data represents the edges of the graph.`,
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`Two vertices u,v in a graph are said to be adjacent if there exists an edge [u,v].`,
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`Note that an edge [u,v] is the same as an edge [v,u], as order does not matter.`,
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`You must construct a 2-coloring of the graph, meaning that you have to assign each`,
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`vertex in the graph a "color", either 0 or 1, such that no two adjacent vertices have`,
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`the same color. Submit your answer in the form of an array, where element i`,
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`represents the color of vertex i. If it is impossible to construct a 2-coloring of`,
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`the given graph, instead submit an empty array.\n\n`,
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`Examples:\n\n`,
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`Input: [4, [[0, 2], [0, 3], [1, 2], [1, 3]]]\n`,
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`Output: [0, 0, 1, 1]\n\n`,
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`Input: [3, [[0, 1], [0, 2], [1, 2]]]\n`,
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`Output: []`,
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].join(" ");
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},
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gen: (): [number, [number, number][]] => {
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//Generate two partite sets
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const n = Math.floor(Math.random() * 5) + 3;
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const m = Math.floor(Math.random() * 5) + 3;
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//50% chance of spawning any given valid edge in the bipartite graph
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const edges: [number, number][] = [];
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for (let i = 0; i < n; i++) {
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for (let j = 0; j < m; j++) {
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if (Math.random() > 0.5) {
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edges.push([i, n + j]);
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}
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}
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}
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//Add an edge at random with no regard to partite sets
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let a = Math.floor(Math.random() * (n + m));
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let b = Math.floor(Math.random() * (n + m));
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if (a > b) [a, b] = [b, a]; //Enforce lower numbers come first
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if (a != b && !edges.includes([a, b])) {
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edges.push([a, b]);
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}
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//Randomize array in-place using Durstenfeld shuffle algorithm.
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function shuffle(array: any[]): void {
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for (let i = array.length - 1; i > 0; i--) {
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const j = Math.floor(Math.random() * (i + 1));
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[array[i], array[j]] = [array[j], array[i]];
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}
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}
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//Replace instances of the original vertex names in-place
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const vertexShuffler = Array.from(Array(n + m).keys());
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shuffle(vertexShuffler);
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for (let i = 0; i < edges.length; i++) {
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edges[i] = [vertexShuffler[edges[i][0]], vertexShuffler[edges[i][1]]];
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if (edges[i][0] > edges[i][1]) {
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//Enforce lower numbers come first
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[edges[i][0], edges[i][1]] = [edges[i][1], edges[i][0]];
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}
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}
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//Shuffle the order of the edges themselves, as well
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shuffle(edges);
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return [n + m, edges];
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},
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solver: (data: [number, [number, number][]], ans: string): boolean => {
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//Case where the player believes there is no solution
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if (ans == "[]") {
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//Helper function to get neighbourhood of a vertex
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function neighbourhood(vertex: number): number[] {
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const adjLeft = data[1].filter(([a, _]) => a == vertex).map(([_, b]) => b);
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const adjRight = data[1].filter(([_, b]) => b == vertex).map(([a, _]) => a);
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return adjLeft.concat(adjRight);
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}
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//Verify that there is no solution by attempting to create a proper 2-coloring.
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const coloring: (number | undefined)[] = Array(data[0]).fill(undefined);
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while (coloring.some((val) => val === undefined)) {
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//Color a vertex in the graph
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const initialVertex: number = coloring.findIndex((val) => val === undefined);
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coloring[initialVertex] = 0;
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const frontier: number[] = [initialVertex];
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//Propogate the coloring throughout the component containing v greedily
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while (frontier.length > 0) {
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const v: number = frontier.pop() || 0;
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const neighbors: number[] = neighbourhood(v);
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//For each vertex u adjacent to v
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for (const id in neighbors) {
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const u: number = neighbors[id];
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//Set the color of u to the opposite of v's color if it is new,
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//then add u to the frontier to continue the algorithm.
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if (coloring[u] === undefined) {
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if (coloring[v] === 0) coloring[u] = 1;
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else coloring[u] = 0;
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frontier.push(u);
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}
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//Assert u,v do not have the same color
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else if (coloring[u] === coloring[v]) {
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//If u,v do have the same color, no proper 2-coloring exists, meaning
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//the player was correct to say there is no proper 2-coloring of the graph.
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return true;
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}
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}
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}
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}
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//If this code is reached, there exists a proper 2-coloring of the input
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//graph, and thus the player was incorrect in submitting no answer.
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return false;
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}
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//Sanitize player input
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const sanitizedPlayerAns: string = removeBracketsFromArrayString(ans);
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const sanitizedPlayerAnsArr: string[] = sanitizedPlayerAns.split(",");
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const coloring: number[] = sanitizedPlayerAnsArr.map((val) => parseInt(val));
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//Solution provided case
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if (coloring.length == data[0]) {
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const edges = data[1];
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const validColors = [0, 1];
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//Check that the provided solution is a proper 2-coloring
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return edges.every(([a, b]) => {
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const aColor = coloring[a];
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const bColor = coloring[b];
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return (
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validColors.includes(aColor) && //Enforce the first endpoint is color 0 or 1
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validColors.includes(bColor) && //Enforce the second endpoint is color 0 or 1
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aColor != bColor //Enforce the endpoints are different colors
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);
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});
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}
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//Return false if the coloring is the wrong size
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else return false;
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},
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},
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{
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name: "Compression I: RLE Compression",
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difficulty: 2,
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numTries: 10,
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desc: (plaintext: string): string => {
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return [
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"Run-length encoding (RLE) is a data compression technique which encodes data as a series of runs of",
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"a repeated single character. Runs are encoded as a length, followed by the character itself. Lengths",
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"are encoded as a single ASCII digit; runs of 10 characters or more are encoded by splitting them",
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"into multiple runs.\n\n",
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"You are given the following input string:\n",
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` ${plaintext}\n`,
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"Encode it using run-length encoding with the minimum possible output length.\n\n",
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"Examples:\n",
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" aaaaabccc -> 5a1b3c\n",
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" aAaAaA -> 1a1A1a1A1a1A\n",
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" 111112333 -> 511233\n",
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" zzzzzzzzzzzzzzzzzzz -> 9z9z1z (or 9z8z2z, etc.)\n",
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].join(" ");
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},
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gen: (): string => {
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const length = 50 + Math.floor(25 * (Math.random() + Math.random()));
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let plain = "";
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while (plain.length < length) {
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const r = Math.random();
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let n = 1;
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if (r < 0.3) {
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n = 1;
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} else if (r < 0.6) {
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n = 2;
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} else if (r < 0.9) {
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n = Math.floor(10 * Math.random());
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} else {
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n = 10 + Math.floor(5 * Math.random());
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}
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const c = comprGenChar();
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plain += c.repeat(n);
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}
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return plain.substring(0, length);
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},
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solver: (plain: string, ans: string): boolean => {
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if (ans.length % 2 !== 0) {
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return false;
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}
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let ans_plain = "";
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for (let i = 0; i + 1 < ans.length; i += 2) {
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const length = ans.charCodeAt(i) - 0x30;
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if (length < 0 || length > 9) {
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return false;
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}
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ans_plain += ans[i + 1].repeat(length);
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}
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if (ans_plain !== plain) {
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return false;
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}
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let length = 0;
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for (let i = 0; i < plain.length; ) {
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let run_length = 1;
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while (i + run_length < plain.length && plain[i + run_length] === plain[i]) {
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++run_length;
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}
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i += run_length;
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while (run_length > 0) {
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run_length -= 9;
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length += 2;
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}
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}
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return ans.length === length;
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},
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},
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{
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name: "Compression II: LZ Decompression",
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difficulty: 4,
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numTries: 10,
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desc: (compressed: string): string => {
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return [
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"Lempel-Ziv (LZ) compression is a data compression technique which encodes data using references to",
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"earlier parts of the data. In this variant of LZ, data is encoded in two types of chunk. Each chunk",
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"begins with a length L, encoded as a single ASCII digit from 1 - 9, followed by the chunk data,",
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"which is either:\n\n",
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"1. Exactly L characters, which are to be copied directly into the uncompressed data.\n",
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"2. A reference to an earlier part of the uncompressed data. To do this, the length is followed",
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"by a second ASCII digit X: each of the L output characters is a copy of the character X",
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"places before it in the uncompressed data.\n\n",
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"For both chunk types, a length of 0 instead means the chunk ends immediately, and the next character",
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"is the start of a new chunk. The two chunk types alternate, starting with type 1, and the final",
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"chunk may be of either type.\n\n",
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"You are given the following LZ-encoded string:\n",
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` ${compressed}\n`,
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"Decode it and output the original string.\n\n",
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"Example: decoding '5aaabc340533bca' chunk-by-chunk\n",
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" 5aaabc -> aaabc\n",
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" 5aaabc34 -> aaabcaab\n",
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" 5aaabc340 -> aaabcaab\n",
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" 5aaabc34053 -> aaabcaabaabaa\n",
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" 5aaabc340533bca -> aaabcaabaabaabca",
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].join(" ");
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},
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gen: (): string => {
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return comprLZEncode(comprLZGenerate());
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},
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solver: (compr: string, ans: string): boolean => {
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return ans === comprLZDecode(compr);
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},
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},
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{
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name: "Compression III: LZ Compression",
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difficulty: 10,
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numTries: 10,
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desc: (plaintext: string): string => {
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return [
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"Lempel-Ziv (LZ) compression is a data compression technique which encodes data using references to",
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"earlier parts of the data. In this variant of LZ, data is encoded in two types of chunk. Each chunk",
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"begins with a length L, encoded as a single ASCII digit from 1 - 9, followed by the chunk data,",
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"which is either:\n\n",
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"1. Exactly L characters, which are to be copied directly into the uncompressed data.\n",
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"2. A reference to an earlier part of the uncompressed data. To do this, the length is followed",
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"by a second ASCII digit X: each of the L output characters is a copy of the character X",
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"places before it in the uncompressed data.\n\n",
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"For both chunk types, a length of 0 instead means the chunk ends immediately, and the next character",
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"is the start of a new chunk. The two chunk types alternate, starting with type 1, and the final",
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"chunk may be of either type.\n\n",
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"You are given the following input string:\n",
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` ${plaintext}\n`,
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"Encode it using Lempel-Ziv encoding with the minimum possible output length.\n\n",
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"Examples (some have other possible encodings of minimal length):\n",
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" abracadabra -> 7abracad47\n",
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" mississippi -> 4miss433ppi\n",
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" aAAaAAaAaAA -> 3aAA53035\n",
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" 2718281828 -> 627182844\n",
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" abcdefghijk -> 9abcdefghi02jk\n",
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" aaaaaaaaaaa -> 1a911a\n",
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" aaaaaaaaaaaa -> 1a912aa\n",
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" aaaaaaaaaaaaa -> 1a91031",
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].join(" ");
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},
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gen: (): string => {
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return comprLZGenerate();
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},
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solver: (plain: string, ans: string): boolean => {
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return comprLZDecode(ans) === plain && ans.length === comprLZEncode(plain).length;
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},
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},
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];
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Reference in New Issue
Block a user