问题
I'm working my way through the Google Foobar challenge and am now at the level 3 challenge Doomsday Fuel. The instructions are as follows:
Doomsday Fuel
Making fuel for the LAMBCHOP's reactor core is a tricky process because of the exotic matter involved. It starts as raw ore, then during processing, begins randomly changing between forms, eventually reaching a stable form. There may be multiple stable forms that a sample could ultimately reach, not all of which are useful as fuel.
Commander Lambda has tasked you to help the scientists increase fuel creation efficiency by predicting the end state of a given ore sample. You have carefully studied the different structures that the ore can take and which transitions it undergoes. It appears that, while random, the probability of each structure transforming is fixed. That is, each time the ore is in 1 state, it has the same probabilities of entering the next state (which might be the same state). You have recorded the observed transitions in a matrix. The others in the lab have hypothesized more exotic forms that the ore can become, but you haven't seen all of them.
Write a function solution(m) that takes an array of array of nonnegative ints representing how many times that state has gone to the next state and return an array of ints for each terminal state giving the exact probabilities of each terminal state, represented as the numerator for each state, then the denominator for all of them at the end and in simplest form. The matrix is at most 10 by 10. It is guaranteed that no matter which state the ore is in, there is a path from that state to a terminal state. That is, the processing will always eventually end in a stable state. The ore starts in state 0. The denominator will fit within a signed 32-bit integer during the calculation, as long as the fraction is simplified regularly.
>For example, consider the matrix m:
[
[0,1,0,0,0,1], # s0, the initial state, goes to s1 and s5 with equal probability
[4,0,0,3,2,0], # s1 can become s0, s3, or s4, but with different probabilities
[0,0,0,0,0,0], # s2 is terminal, and unreachable (never observed in practice)
[0,0,0,0,0,0], # s3 is terminal
[0,0,0,0,0,0], # s4 is terminal
[0,0,0,0,0,0], # s5 is terminal
]
So, we can consider different paths to terminal states, such as:
s0 -> s1 -> s3
s0 -> s1 -> s0 -> s1 -> s0 -> s1 -> s4
s0 -> s1 -> s0 -> s5
Tracing the probabilities of each, we find that
s2 has probability 0
s3 has probability 3/14
s4 has probability 1/7
s5 has probability 9/14
So, putting that together, and making a common denominator, gives an answer in the form of
[s2.numerator, s3.numerator, s4.numerator, s5.numerator, denominator] which is
[0, 3, 2, 9, 14].
Languages
To provide a Java solution, edit Solution.java To provide a Python solution, edit solution.py
Test cases
==========
>Your code should pass the following test cases.
Note that it may also be run against hidden test cases not shown here.
>-- Java cases --
Input:
Solution.solution({{0, 2, 1, 0, 0}, {0, 0, 0, 3, 4}, {0, 0, 0, 0, 0}, {0, 0, 0, 0,0}, {0, 0, 0, 0, 0}})
Output:
[7, 6, 8, 21]
>Input:
Solution.solution({{0, 1, 0, 0, 0, 1}, {4, 0, 0, 3, 2, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}})
Output:
[0, 3, 2, 9, 14]
>-- Python cases --
Input:
solution.solution([[0, 2, 1, 0, 0], [0, 0, 0, 3, 4], [0, 0, 0, 0, 0], [0, 0, 0, 0,0], [0, 0, 0, 0, 0]])
Output:
[7, 6, 8, 21]
>Input:
solution.solution([[0, 1, 0, 0, 0, 1], [4, 0, 0, 3, 2, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0]])
Output:
[0, 3, 2, 9, 14]
>Use verify [file] to test your solution and see how it does. When you are finished editing your code, use submit [file] to submit your answer. If your solution passes the test cases, it will be removed from your home folder.
I have written the following code to solve it:
import java.util.ArrayList;
public class Solution {
public static int[] solution(int[][] m) {
double[][] mDouble = toDouble(m);
//TODO: change the double back into an int
// GOAL ONE: find Q matrix :
// 1:seperate the input into two 2d arrays
ArrayList<double[]> ters = new ArrayList<double[]>();
ArrayList<double[]> nonTers = new ArrayList<double[]>();
for(int i = 0; i < mDouble.length; i++){
boolean isTerminal = true;
int sum = 0;
for(int j = 0; j < mDouble[0].length; j++){
sum += mDouble[i][j];
if(mDouble[i][j] != 0){
isTerminal = false;
}
}
if(isTerminal){
ters.add(mDouble[i]);
}else{
for(int j = 0; j < mDouble[0].length; j++){
mDouble[i][j] = mDouble[i][j]/sum;
}
nonTers.add(mDouble[i]);
}
}
double[][] terminalStates = new double[ters.size()][m.length];
double[][] nonTerminalStates = new double[nonTers.size()][m.length];
for(int i = 0; i < ters.size(); i++){
terminalStates[i] = ters.get(i);
}
for(int i = 0; i < nonTers.size(); i++){
nonTerminalStates[i] = nonTers.get(i);
}
// 2: Plug into a function that will create the 2d array
double[][] QMatrix = getQMatrix(nonTerminalStates);
// GOAL TWO: find I matrix
double[][] IMatrix = makeIMatrix(QMatrix.length);
//GOAL 3: Find F matrix
//1: subtract the q matrix from the I matrix
double[][] AMatrix = SubtractMatrices(IMatrix, QMatrix);
//2: find the inverse TODO WRITE FUNCTION
double[][] FMatrix = invert(AMatrix);
//GOAL 4: multiply by R Matrix
//1: find r Matrx
double[][] RMatrix = getRMatrix(nonTerminalStates, terminalStates.length);
//2: use multiply function to get FR Matrix
double[][] FRMatrix = multiplyMatrices(FMatrix, RMatrix);
//GOAL 5: find answer array
//1: get the one dimensional answer
double[] unsimplifiedAns = FRMatrix[0];
//2: get fractions for the answers
int[] ans = fractionAns(unsimplifiedAns);
return ans;
}
public static int[] fractionAns(double[] uAns){
int[] ans = new int[uAns.length + 1];
int[] denominators = new int[uAns.length];
int[] numerators = new int[uAns.length];
for(int i = 0; i < uAns.length; i++){
denominators[i] = (int)(convertDecimalToFraction(uAns[i])[1]);
numerators[i] = (int)(convertDecimalToFraction(uAns[i])[0]);
}
int lcm = (int) lcm_of_array_elements(denominators);
for(int i = 0; i < uAns.length; i++){
ans[i] = numerators[i]*(lcm/convertDecimalToFraction(uAns[i])[1]);
}
ans[ans.length-1] = lcm;
return ans;
}
static private int[] convertDecimalToFraction(double x){
double tolerance = 1.0E-10;
double h1=1; double h2=0;
double k1=0; double k2=1;
double b = x;
do {
double a = Math.floor(b);
double aux = h1; h1 = a*h1+h2; h2 = aux;
aux = k1; k1 = a*k1+k2; k2 = aux;
b = 1/(b-a);
} while (Math.abs(x-h1/k1) > x*tolerance);
return new int[]{(int)h1, (int)k1};
}
public static long lcm_of_array_elements(int[] element_array)
{
long lcm_of_array_elements = 1;
int divisor = 2;
while (true) {
int counter = 0;
boolean divisible = false;
for (int i = 0; i < element_array.length; i++) {
// lcm_of_array_elements (n1, n2, ... 0) = 0.
// For negative number we convert into
// positive and calculate lcm_of_array_elements.
if (element_array[i] == 0) {
return 0;
}
else if (element_array[i] < 0) {
element_array[i] = element_array[i] * (-1);
}
if (element_array[i] == 1) {
counter++;
}
// Divide element_array by devisor if complete
// division i.e. without remainder then replace
// number with quotient; used for find next factor
if (element_array[i] % divisor == 0) {
divisible = true;
element_array[i] = element_array[i] / divisor;
}
}
// If divisor able to completely divide any number
// from array multiply with lcm_of_array_elements
// and store into lcm_of_array_elements and continue
// to same divisor for next factor finding.
// else increment divisor
if (divisible) {
lcm_of_array_elements = lcm_of_array_elements * divisor;
}
else {
divisor++;
}
// Check if all element_array is 1 indicate
// we found all factors and terminate while loop.
if (counter == element_array.length) {
return lcm_of_array_elements;
}
}
}
public static double[][] toDouble(int[][] ma){
double[][] retArr = new double[ma.length][ma.length];
for(int i = 0; i < retArr.length; i++){
for(int j = 0; j < retArr[0].length; j++){
retArr[i][j] = (ma[i][j]);
}
}
return retArr;
}
public static double[][] getRMatrix(double[][] nonTerminals, int terminalLength){
double[][] retArr = new double[nonTerminals.length][terminalLength];
for(int i = 0; i < retArr.length; i++){
for(int j = nonTerminals.length; j < nonTerminals[0].length; j++){
retArr[i][j-nonTerminals.length] = (nonTerminals[i][j]);
}
}
return retArr;
}
public static double[][] multiplyMatrices(double[][] firstMatrix, double[][] secondMatrix){
int r1 = firstMatrix.length;
int c1 = firstMatrix[0].length;
int c2 = secondMatrix[0].length;
double[][] product = new double[r1][c2];
for(int i = 0; i < r1; i++) {
for (int j = 0; j < c2; j++) {
for (int k = 0; k < c1; k++) {
product[i][j] += firstMatrix[i][k] * secondMatrix[k][j];
}
}
}
return product;
}
public static double[][] inverseMatrix(double[][] Amatrix){
return null;
}
public static double[][] SubtractMatrices(double[][] I, double[][] Q){
double[][] retArr = new double[I.length][I.length];
for(int i = 0; i < retArr.length; i++){
for(int j = 0; j < retArr.length; j++){
retArr[i][j] = I[i][j]-Q[i][j];
}
}
return retArr;
}
public static double[][] getQMatrix(double[][] qArr){
int size = qArr.length;
double[][] retArr = new double[size][size];
for(int i = 0; i < size; i++){
for(int j = 0; j < size; j++){
retArr[i][j] = qArr[i][j];
}
}
return retArr;
}
public static double[][] makeIMatrix(int size){
double[][] retArr = new double[size][size];
for(int i = 0; i < size; i++){
for(int j = 0; j < size; j++){
if(i == j){
retArr[i][j] = 1;
}else{
retArr[i][j] = 0;
}
}
}
return retArr;
}
public static double[][] invert(double a[][])
{
int n = a.length;
double x[][] = new double[n][n];
double b[][] = new double[n][n];
int index[] = new int[n];
for (int i=0; i<n; ++i)
b[i][i] = 1;
// Transform the matrix into an upper triangle
gaussian(a, index);
// Update the matrix b[i][j] with the ratios stored
for (int i=0; i<n-1; ++i)
for (int j=i+1; j<n; ++j)
for (int k=0; k<n; ++k)
b[index[j]][k]
-= a[index[j]][i]*b[index[i]][k];
// Perform backward substitutions
for (int i=0; i<n; ++i)
{
x[n-1][i] = b[index[n-1]][i]/a[index[n-1]][n-1];
for (int j=n-2; j>=0; --j)
{
x[j][i] = b[index[j]][i];
for (int k=j+1; k<n; ++k)
{
x[j][i] -= a[index[j]][k]*x[k][i];
}
x[j][i] /= a[index[j]][j];
}
}
return x;
}
// Method to carry out the partial-pivoting Gaussian
// elimination. Here index[] stores pivoting order.
public static void gaussian(double a[][], int index[])
{
int n = index.length;
double c[] = new double[n];
// Initialize the index
for (int i=0; i<n; ++i)
index[i] = i;
// Find the rescaling factors, one from each row
for (int i=0; i<n; ++i)
{
double c1 = 0;
for (int j=0; j<n; ++j)
{
double c0 = Math.abs(a[i][j]);
if (c0 > c1) c1 = c0;
}
c[i] = c1;
}
// Search the pivoting element from each column
int k = 0;
for (int j=0; j<n-1; ++j)
{
double pi1 = 0;
for (int i=j; i<n; ++i)
{
double pi0 = Math.abs(a[index[i]][j]);
pi0 /= c[index[i]];
if (pi0 > pi1)
{
pi1 = pi0;
k = i;
}
}
// Interchange rows according to the pivoting order
int itmp = index[j];
index[j] = index[k];
index[k] = itmp;
for (int i=j+1; i<n; ++i)
{
double pj = a[index[i]][j]/a[index[j]][j];
// Record pivoting ratios below the diagonal
a[index[i]][j] = pj;
// Modify other elements accordingly
for (int l=j+1; l<n; ++l)
a[index[i]][l] -= pj*a[index[j]][l];
}
}
}
}
It passes all the test cases but two hidden ones I cannot see.
I've tried all the test cases I possibly could to find the fault in my code but I cannot.
Are there any test cases here where my code fails?
回答1:
The problem lies in the line
double[] unsimplifiedAns = FRMatrix[0];
The above is true only if state 0 is non-terminating.
Otherwise the output array will be all '0's except the first and last element as '1'.
来源：https://stackoverflow.com/questions/61627375/hidden-test-cases-not-passing-for-google-foobar-challenge-doomsday-fuel