私はGoogle Foobarチャレンジに取り組んでおり、レベル3のチャレンジであるDoomsday Fuelにいます。手順は次のとおりです。
LAMBCHOPの原子炉の炉心に燃料を作ることは、複雑な問題が関係しているため、トリッキーなプロセスです。原鉱として始まり、処理中にフォーム間でランダムに変化し始め、最終的に安定したフォームに到達します。サンプルが最終的に到達できる安定した形状が複数ある場合がありますが、それらのすべてが燃料として役立つわけではありません。
ラムダ司令官は、与えられた鉱石サンプルの最終状態を予測することにより、科学者が燃料生成効率を高めるのを助けるようにあなたに命じました。あなたは、鉱石が取ることができるさまざまな構造と、それが受ける遷移を注意深く研究しました。ランダムではありますが、各構造が変形する確率は固定されているようです。つまり、鉱石が1つの状態になるたびに、次の状態(同じ状態である可能性があります)に入る確率は同じになります。観測された遷移をマトリックスに記録しました。実験室の他の人たちは、鉱石がなり得るよりエキゾチックな形を仮定しましたが、あなたはそれらのすべてを見たことはありません。
その状態が次の状態に移行した回数を表す非負の整数の配列の配列を取り、各最終状態の正確な確率を与える各最終状態の整数の配列を返す関数solution(m)を記述します。各状態の分子、最後にそれらのすべての分母、最も単純な形式。行列は最大で10 x 10です。鉱石がどの状態にある場合でも、その状態から最終状態へのパスが存在することが保証されています。つまり、処理は常に最終的に安定した状態で終了します。鉱石は状態0から始まります。分数が定期的に単純化されている限り、分母は計算中に符号付き32ビット整数内に収まります。
>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].
Javaソリューションを提供するには、Solution.Javaを編集しますPythonソリューションを提供するには、solution.pyを編集します
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];
}
}
}
}
すべてのテストケースに合格しますが、表示されない2つの非表示のケースがあります。
私はコード内の障害を見つけることができる可能性があるすべてのテストケースを試しましたが、できません。
コードが失敗するテストケースはありますか?
問題は行にあります
double[] unsimplifiedAns = FRMatrix[0];
上記は、状態0が非終了の場合にのみ当てはまります。
そうでない場合、出力配列は、最初と最後の要素が「1」である場合を除き、すべて「0」になります。