等しい確率で 0 または 1 を返す関数 rand2() が与えられた場合、等しい確率で 0 1 または 2 を返す rand2() を使用して rand3() を実装します。 rand2() メソッドの呼び出し数を最小限に抑えます。また、他のライブラリ関数や浮動小数点演算の使用も許可されません。
春のjpa
アイデアは表現を使用することです 2 * rand2() + rand2() 。等しい確率で 0 1 2 3 を返します。同じ確率で 0 1 2 を返すようにするには、望ましくないイベント 3 を削除します。
以下は上記のアイデアの実装です –
C++
// C++ Program to print 0 1 or 2 with equal // probability #include
// Java Program to print 0 1 or 2 with equal // probability import java.util.Random; class GFG { // Random Function to that returns 0 or 1 with // equal probability static int rand2() { // rand() function will generate odd or even // number with equal probability. If rand() // generates odd number the function will // return 1 else it will return 0. Random rand = new Random(); return (rand.nextInt() & 1); } // Random Function to that returns 0 1 or 2 with // equal probability 1 with 75% static int rand3() { // returns 0 1 2 or 3 with 25% probability int r = 2 * rand2() + rand2(); if (r < 3) return r; return rand3(); } // Driver code public static void main(String[] args) { for(int i = 0; i < 100; i++) System.out.print(rand3()); } } // This code is contributed by divyesh072019.
# Python3 Program to print 0 1 or 2 with equal # Probability import random # Random Function to that returns 0 or 1 with # equal probability def rand2(): # randint(0100) function will generate odd or even # number [1100] with equal probability. If rand() # generates odd number the function will # return 1 else it will return 0 tmp=random.randint(1100) return tmp%2 # Random Function to that returns 0 1 or 2 with # equal probability 1 with 75% def rand3(): # returns 0 1 2 or 3 with 25% probability r = 2 * rand2() + rand2() if r<3: return r return rand3() # Driver code to test above functions if __name__=='__main__': for i in range(100): print(rand3()end='') #This code is contributed by sahilshelangia
// C# Program to print 0 1 or 2 with equal // probability using System; class GFG { // Random Function to that returns 0 or 1 with // equal probability static int rand2() { // rand() function will generate odd or even // number with equal probability. If rand() // generates odd number the function will // return 1 else it will return 0. Random rand = new Random(); return (rand.Next() & 1); } // Random Function to that returns 0 1 or 2 with // equal probability 1 with 75% static int rand3() { // returns 0 1 2 or 3 with 25% probability int r = 2 * rand2() + rand2(); if (r < 3) return r; return rand3(); } // Driver code static void Main() { for(int i = 0; i < 100; i++) Console.Write(rand3()); } } // This code is contributed by divyeshrabadiya07.
// PHP Program to print 0 1 or // 2 with equal probability // Random Function to that // returns 0 or 1 with // equal probability function rand2() { // rand() function will generate // odd or even number with equal // probability. If rand() generates // odd number the function will // return 1 else it will return 0. return rand() & 1; } // Random Function to that // returns 0 1 or 2 with // equal probability 1 with 75% function rand3() { // returns 0 1 2 or 3 // with 25% probability $r = 2 * rand2() + rand2(); if ($r < 3) return $r; return rand3(); } // Driver Code // Initialize random // number generator srand(time(NULL)); for($i = 0; $i < 100; $i++) echo rand3(); // This code is contributed by aj_36 ?> <script> // Javascript program to print 0 1 or 2 with equal // probability // Random Function to that returns 0 or 1 with // equal probability function rand2() { // Math.random()*2 function generates // 0 and 1 with equal probability return Math.floor(Math.random()*2); } // Random Function to that returns 0 1 or 2 with // equal probability 1 with 75% function rand3() { // returns 0 1 2 or 3 with 25% probability var r = 2 * rand2() + rand2(); if (r < 3) return r; return rand3(); } var ans = ''; //to store the output for(var i = 0; i < 100; i++) ans += rand3(); document.write(ans); // This code is contributed by shruti456rawal </script>
出力:
2111011101112002111002020210112022022022211100100121202021102100010200121121210122011022111020
別の解決策 -
x = rand2() および y = rand2() の場合、x + y は 25% の確率で 0 と 2 を返し、50% の確率で 1 を返します。 1 の確率を 0 と 2 の確率、つまり 25% に等しくするには、x + y = 1、つまり (x = 1 y = 0) または (x = 0 y = 1) をもたらす望ましくないイベントを 1 つ削除します。
int rand3() { int x y; do { x = rand2(); y = rand2(); } while (x == 0 && y == 1); return x + y; }
上記のソリューションは実行するたびに異なる結果が生成されることに注意してください。