Assessing Task Execution with Server Core Availability in Python

from functools import reduce

def can_execute_tasks(tasks, cores):
    return reduce(lambda acc, task_cores: acc - task_cores if acc >= task_cores else -1, tasks, cores) >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

In this snippet, reduce iteratively applies the lambda function to the list of tasks, with the initial value being the total cores. The lambda function subtracts each task’s core requirement only if there are enough cores available. If at any point there aren’t enough cores, the accumulator becomes -1. If the final result is non-negative, it means all tasks could be allocated cores, and we return True.

Method 3: Priority Queue for Larges Tasks First

This method utilizes a priority queue (or a min heap in Python) to allocate cores to the largest tasks first, ensuring that the tasks with the highest core requirements get prioritized. If the largest task can be executed, it’s likely smaller ones can too.

Here’s an example:

import heapq

def can_execute_tasks(tasks, cores):
    max_heap = [-x for x in tasks]
    heapq.heapify(max_heap)
    while max_heap and cores >= 0:
        cores += heapq.heappop(max_heap)
    return cores >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

This code turns the tasks list into a max heap by negating the core requirements. We then pop off the largest item (most negative), add its value back to cores, and validate if enough cores are remaining. The process repeats until the heap is empty or not enough cores are available.

Method 4: Early Exit with Sorted Task List

By sorting the tasks list in descending order, we can allocate cores to bigger tasks first and exit early as soon as we don’t have enough cores for a task, potentially saving computation time.

Here’s an example:

def can_execute_tasks(tasks, cores):
    for task_cores in sorted(tasks, reverse=True):
        cores -= task_cores
        if cores < 0:
            return False
    return True

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Sorting the task list in descending order makes sure we try to allocate the maximum core requiring task first. If at any juncture we fail to have enough cores for a task, we immediately return False. If the loop completes, all tasks are allocatable, and we return True.

Bonus One-Liner Method 5: Compact Lambda with All

A one-liner approach leverages the all function to check the cumulative core availability through a lambda expression, offering a compact solution.

Here’s an example:

tasks = [1, 2, 1, 3]
cores = 4
print(all((cores := cores - task) >= 0 for task in tasks))

Output: False

The one-liner uses a generator expression inside the all function. It relies on Python’s walrus operator to subtract the core requirement from the cumulative cores variable and immediately evaluate if it’s non-negative. It returns False if any task’s allocation drops cores below zero.

Summary/Discussion

Method 1: Iterative Core Allocation Check. This is the simplest and most straightforward method. The weakness is that it does not prioritize tasks based on their core requirements.
Method 2: Using Python’s reduce Function. It gives a more functional programming approach. However, it might be less readable for those not familiar with the reduce function or lambda expressions.
Method 3: Priority Queue for Largest Tasks First. This method can be more efficient if the largest task exceeds available cores, stopping early. But it requires converting the list to a heap, adding a bit of overhead.
Method 4: Early Exit with Sorted Task List. Sorting tasks can help in an early exit in case of insufficient cores for larger tasks. The downside is the computational cost of sorting. However, if tasks are added incrementally, this can be quite efficient.
Bonus Method 5: Compact Lambda with All. It provides a concise one-liner that is elegant. The weakness can be the legibility for those not familiar with the walrus operator or comprehensions.

import heapq

def can_execute_tasks(tasks, cores):
    max_heap = [-x for x in tasks]
    heapq.heapify(max_heap)
    while max_heap and cores >= 0:
        cores += heapq.heappop(max_heap)
    return cores >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Method 4: Early Exit with Sorted Task List

By sorting the tasks list in descending order, we can allocate cores to bigger tasks first and exit early as soon as we don’t have enough cores for a task, potentially saving computation time.

Here’s an example:

def can_execute_tasks(tasks, cores):
    for task_cores in sorted(tasks, reverse=True):
        cores -= task_cores
        if cores < 0:
            return False
    return True

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Bonus One-Liner Method 5: Compact Lambda with All

A one-liner approach leverages the all function to check the cumulative core availability through a lambda expression, offering a compact solution.

Here’s an example:

tasks = [1, 2, 1, 3]
cores = 4
print(all((cores := cores - task) >= 0 for task in tasks))

Output: False

Summary/Discussion

Method 1: Iterative Core Allocation Check. This is the simplest and most straightforward method. The weakness is that it does not prioritize tasks based on their core requirements.
Method 2: Using Python’s reduce Function. It gives a more functional programming approach. However, it might be less readable for those not familiar with the reduce function or lambda expressions.
Method 3: Priority Queue for Largest Tasks First. This method can be more efficient if the largest task exceeds available cores, stopping early. But it requires converting the list to a heap, adding a bit of overhead.
Method 4: Early Exit with Sorted Task List. Sorting tasks can help in an early exit in case of insufficient cores for larger tasks. The downside is the computational cost of sorting. However, if tasks are added incrementally, this can be quite efficient.
Bonus Method 5: Compact Lambda with All. It provides a concise one-liner that is elegant. The weakness can be the legibility for those not familiar with the walrus operator or comprehensions.

from functools import reduce

def can_execute_tasks(tasks, cores):
    return reduce(lambda acc, task_cores: acc - task_cores if acc >= task_cores else -1, tasks, cores) >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Method 3: Priority Queue for Larges Tasks First

Here’s an example:

import heapq

def can_execute_tasks(tasks, cores):
    max_heap = [-x for x in tasks]
    heapq.heapify(max_heap)
    while max_heap and cores >= 0:
        cores += heapq.heappop(max_heap)
    return cores >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Method 4: Early Exit with Sorted Task List

By sorting the tasks list in descending order, we can allocate cores to bigger tasks first and exit early as soon as we don’t have enough cores for a task, potentially saving computation time.

Here’s an example:

def can_execute_tasks(tasks, cores):
    for task_cores in sorted(tasks, reverse=True):
        cores -= task_cores
        if cores < 0:
            return False
    return True

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Bonus One-Liner Method 5: Compact Lambda with All

A one-liner approach leverages the all function to check the cumulative core availability through a lambda expression, offering a compact solution.

Here’s an example:

tasks = [1, 2, 1, 3]
cores = 4
print(all((cores := cores - task) >= 0 for task in tasks))

Output: False

Summary/Discussion

Method 1: Iterative Core Allocation Check. This is the simplest and most straightforward method. The weakness is that it does not prioritize tasks based on their core requirements.
Method 2: Using Python’s reduce Function. It gives a more functional programming approach. However, it might be less readable for those not familiar with the reduce function or lambda expressions.
Method 3: Priority Queue for Largest Tasks First. This method can be more efficient if the largest task exceeds available cores, stopping early. But it requires converting the list to a heap, adding a bit of overhead.
Method 4: Early Exit with Sorted Task List. Sorting tasks can help in an early exit in case of insufficient cores for larger tasks. The downside is the computational cost of sorting. However, if tasks are added incrementally, this can be quite efficient.
Bonus Method 5: Compact Lambda with All. It provides a concise one-liner that is elegant. The weakness can be the legibility for those not familiar with the walrus operator or comprehensions.

💡 Problem Formulation: You have a multi-core server and multiple tasks to run, each with a defined core requirement. The challenge is to write a Python program to determine if all tasks can be executed without exceeding the available server cores. For instance, given 4 server cores and tasks requiring 1,2,1, and 3 cores respectively, we want our program to assert whether all tasks can be scheduled (True) or not (False).

Method 1: Iterative Core Allocation Check

This method involves iterating over each task and deducting its core requirements from the total available cores. If at any point available cores drop below zero, we return False, indicating not all tasks can be executed; otherwise, we return True.

Here’s an example:

def can_execute_tasks(tasks, cores):
    for task_cores in tasks:
        cores -= task_cores
        if cores < 0:
            return False
    return True

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

This code defines a function that takes a list of task core requirements and the total available cores. It traverses the task list, deducting the core requirement of each task from the total cores and checks after each deduction to ensure sufficient cores remain. If the available cores become negative, the function returns False, indicating that not all tasks can be executed; otherwise, it returns True.

Method 2: Using Python’s reduce Function

The reduce function from the functools module applies a cumulative operation to items of an iterable. In this case, we’ll use it to subtract each task’s core requirement from the total cores and return the sufficiency result.

Here’s an example:

def can_execute_tasks(tasks, cores):
    for task_cores in sorted(tasks, reverse=True):
        cores -= task_cores
        if cores < 0:
            return False
    return True

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Bonus One-Liner Method 5: Compact Lambda with All

A one-liner approach leverages the all function to check the cumulative core availability through a lambda expression, offering a compact solution.

Here’s an example:

tasks = [1, 2, 1, 3]
cores = 4
print(all((cores := cores - task) >= 0 for task in tasks))

Output: False

Summary/Discussion

Method 1: Iterative Core Allocation Check. This is the simplest and most straightforward method. The weakness is that it does not prioritize tasks based on their core requirements.
Method 2: Using Python’s reduce Function. It gives a more functional programming approach. However, it might be less readable for those not familiar with the reduce function or lambda expressions.
Method 3: Priority Queue for Largest Tasks First. This method can be more efficient if the largest task exceeds available cores, stopping early. But it requires converting the list to a heap, adding a bit of overhead.
Method 4: Early Exit with Sorted Task List. Sorting tasks can help in an early exit in case of insufficient cores for larger tasks. The downside is the computational cost of sorting. However, if tasks are added incrementally, this can be quite efficient.
Bonus Method 5: Compact Lambda with All. It provides a concise one-liner that is elegant. The weakness can be the legibility for those not familiar with the walrus operator or comprehensions.

import heapq

def can_execute_tasks(tasks, cores):
    max_heap = [-x for x in tasks]
    heapq.heapify(max_heap)
    while max_heap and cores >= 0:
        cores += heapq.heappop(max_heap)
    return cores >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Method 4: Early Exit with Sorted Task List

By sorting the tasks list in descending order, we can allocate cores to bigger tasks first and exit early as soon as we don’t have enough cores for a task, potentially saving computation time.

Here’s an example:

def can_execute_tasks(tasks, cores):
    for task_cores in sorted(tasks, reverse=True):
        cores -= task_cores
        if cores < 0:
            return False
    return True

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Bonus One-Liner Method 5: Compact Lambda with All

A one-liner approach leverages the all function to check the cumulative core availability through a lambda expression, offering a compact solution.

Here’s an example:

tasks = [1, 2, 1, 3]
cores = 4
print(all((cores := cores - task) >= 0 for task in tasks))

Output: False

Summary/Discussion

Method 1: Iterative Core Allocation Check. This is the simplest and most straightforward method. The weakness is that it does not prioritize tasks based on their core requirements.
Method 2: Using Python’s reduce Function. It gives a more functional programming approach. However, it might be less readable for those not familiar with the reduce function or lambda expressions.
Method 3: Priority Queue for Largest Tasks First. This method can be more efficient if the largest task exceeds available cores, stopping early. But it requires converting the list to a heap, adding a bit of overhead.
Method 4: Early Exit with Sorted Task List. Sorting tasks can help in an early exit in case of insufficient cores for larger tasks. The downside is the computational cost of sorting. However, if tasks are added incrementally, this can be quite efficient.
Bonus Method 5: Compact Lambda with All. It provides a concise one-liner that is elegant. The weakness can be the legibility for those not familiar with the walrus operator or comprehensions.

from functools import reduce

def can_execute_tasks(tasks, cores):
    return reduce(lambda acc, task_cores: acc - task_cores if acc >= task_cores else -1, tasks, cores) >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Method 3: Priority Queue for Larges Tasks First

Here’s an example:

import heapq

def can_execute_tasks(tasks, cores):
    max_heap = [-x for x in tasks]
    heapq.heapify(max_heap)
    while max_heap and cores >= 0:
        cores += heapq.heappop(max_heap)
    return cores >= 0

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Method 4: Early Exit with Sorted Task List

By sorting the tasks list in descending order, we can allocate cores to bigger tasks first and exit early as soon as we don’t have enough cores for a task, potentially saving computation time.

Here’s an example:

def can_execute_tasks(tasks, cores):
    for task_cores in sorted(tasks, reverse=True):
        cores -= task_cores
        if cores < 0:
            return False
    return True

# Example usage:
print(can_execute_tasks([1, 2, 1, 3], 4))

Output: False

Bonus One-Liner Method 5: Compact Lambda with All

A one-liner approach leverages the all function to check the cumulative core availability through a lambda expression, offering a compact solution.

Here’s an example:

tasks = [1, 2, 1, 3]
cores = 4
print(all((cores := cores - task) >= 0 for task in tasks))

Output: False

Summary/Discussion

Method 1: Iterative Core Allocation Check. This is the simplest and most straightforward method. The weakness is that it does not prioritize tasks based on their core requirements.
Method 2: Using Python’s reduce Function. It gives a more functional programming approach. However, it might be less readable for those not familiar with the reduce function or lambda expressions.
Method 3: Priority Queue for Largest Tasks First. This method can be more efficient if the largest task exceeds available cores, stopping early. But it requires converting the list to a heap, adding a bit of overhead.
Method 4: Early Exit with Sorted Task List. Sorting tasks can help in an early exit in case of insufficient cores for larger tasks. The downside is the computational cost of sorting. However, if tasks are added incrementally, this can be quite efficient.
Bonus Method 5: Compact Lambda with All. It provides a concise one-liner that is elegant. The weakness can be the legibility for those not familiar with the walrus operator or comprehensions.