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Posts published in “Concurrent”

花花酱 LeetCode 1195. Fizz Buzz Multithreaded

Write a program that outputs the string representation of numbers from 1 to n, however:

  • If the number is divisible by 3, output “fizz”.
  • If the number is divisible by 5, output “buzz”.
  • If the number is divisible by both 3 and 5, output “fizzbuzz”.

For example, for n = 15, we output: 1, 2, fizz, 4, buzz, fizz, 7, 8, fizz, buzz, 11, fizz, 13, 14, fizzbuzz.

Suppose you are given the following code:

class FizzBuzz {
  public FizzBuzz(int n) { ... }               // constructor
  public void fizz(printFizz) { ... }          // only output "fizz"
  public void buzz(printBuzz) { ... }          // only output "buzz"
  public void fizzbuzz(printFizzBuzz) { ... }  // only output "fizzbuzz"
  public void number(printNumber) { ... }      // only output the numbers
}

Implement a multithreaded version of FizzBuzz with four threads. The same instance of FizzBuzz will be passed to four different threads:

  1. Thread A will call fizz() to check for divisibility of 3 and outputs fizz.
  2. Thread B will call buzz() to check for divisibility of 5 and outputs buzz.
  3. Thread C will call fizzbuzz() to check for divisibility of 3 and 5 and outputs fizzbuzz.
  4. Thread D will call number() which should only output the numbers.

Solution:

4 Semaphores

C++

花花酱 LeetCode 1117. Building H2O

There are two kinds of threads, oxygen and hydrogen. Your goal is to group these threads to form water molecules. There is a barrier where each thread has to wait until a complete molecule can be formed. Hydrogen and oxygen threads will be given a releaseHydrogen and releaseOxygen method respectfully, which will allow them to pass the barrier. These threads should pass the barrier in groups of three, and they must be able to immediately bond with each other to form a water molecule. You must guarantee that all the threads from one molecule bond before any other threads from the next molecule do.

In other words:

  • If an oxygen thread arrives at the barrier when no hydrogen threads are present, it has to wait for two hydrogen threads.
  • If a hydrogen thread arrives at the barrier when no other threads are present, it has to wait for an oxygen thread and another hydrogen thread.

We don’t have to worry about matching the threads up explicitly; that is, the threads do not necessarily know which other threads they are paired up with. The key is just that threads pass the barrier in complete sets; thus, if we examine the sequence of threads that bond and divide them into groups of three, each group should contain one oxygen and two hydrogen threads.

Write synchronization code for oxygen and hydrogen molecules that enforces these constraints.

Example 1:

Input: "HOH"
Output: "HHO"
Explanation: "HOH" and "OHH" are also valid answers.

Example 2:

Input: "OOHHHH"
Output: "HHOHHO"
Explanation: "HOHHHO", "OHHHHO", "HHOHOH", "HOHHOH", "OHHHOH", "HHOOHH", "HOHOHH" and "OHHOHH" are also valid answers.

Constraints:

  • Total length of input string will be 3n, where 1 ≤ n ≤ 30.
  • Total number of H will be 2n in the input string.
  • Total number of O will be n in the input string.

Solution: Semaphore

C++

花花酱 LeetCode 1116. Print Zero Even Odd

Suppose you are given the following code:

class ZeroEvenOdd {
  public ZeroEvenOdd(int n) { ... }      // constructor
  public void zero(printNumber) { ... }  // only output 0's
  public void even(printNumber) { ... }  // only output even numbers
  public void odd(printNumber) { ... }   // only output odd numbers
}

The same instance of ZeroEvenOdd will be passed to three different threads:

  1. Thread A will call zero() which should only output 0’s.
  2. Thread B will call even() which should only ouput even numbers.
  3. Thread C will call odd() which should only output odd numbers.

Each of the thread is given a printNumber method to output an integer. Modify the given program to output the series 010203040506… where the length of the series must be 2n.

Example 1:

Input: n = 2
Output: "0102"
Explanation: There are three threads being fired asynchronously. One of them calls zero(), the other calls even(), and the last one calls odd(). "0102" is the correct output.

Example 2:

Input: n = 5
Output: "0102030405"

Solution: Mutex

C++

花花酱 LeetCode 1115. Print FooBar Alternately

Suppose you are given the following code:

class FooBar {
  public void foo() {
    for (int i = 0; i < n; i++) {
      print("foo");
    }
  }

  public void bar() {
    for (int i = 0; i < n; i++) {
      print("bar");
    }
  }
}

The same instance of FooBar will be passed to two different threads. Thread A will call foo() while thread B will call bar(). Modify the given program to output “foobar” n times.

Example 1:

Input: n = 1
Output: "foobar"
Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar(). "foobar" is being output 1 time.

Example 2:

Input: n = 2
Output: "foobarfoobar"
Explanation: "foobar" is being output 2 times.

Solution: Mutex

C++

花花酱 LeetCode 1114. Print in Order

Suppose we have a class:

The same instance of Foo will be passed to three different threads. Thread A will call first(), thread B will call second(), and thread C will call third(). Design a mechanism and modify the program to ensure that second() is executed after first(), and third() is executed after second().

Example 1:

Input: [1,2,3]
Output: "firstsecondthird"
Explanation: There are three threads being fired asynchronously. The input [1,2,3] means thread A calls first(), thread B calls second(), and thread C calls third(). "firstsecondthird" is the correct output.

Example 2:

Input: [1,3,2]
Output: "firstsecondthird"
Explanation: The input [1,3,2] means thread A calls first(), thread B calls third(), and thread C calls second(). "firstsecondthird" is the correct output.

Note:

We do not know how the threads will be scheduled in the operating system, even though the numbers in the input seems to imply the ordering. The input format you see is mainly to ensure our tests’ comprehensiveness.

Solution 1: Mutex

C++

Solution 2: Busy waiting

C++