6.00.2x-pset1.py
def brute_force_cow_transport(cows, limit=10):
    """
    Finds the allocation of cows that minimizes the number of spaceship trips
    via brute force.  The brute force algorithm follows the following method:

    1. Enumerate all possible ways that the cows can be divided into separate
    trips.
    2. Select the allocation that minimizes the number of trips without making
    any trip that does not obey the weight limitation.

    Does not mutate the given dictionary of cows.

    Parameters:
    cows - a dictionary of name (string), weight (int) pairs
    limit - weight limit of the spaceship (an int)

    Returns:
    A list of lists, with each inner list containing the names of cows
    transported on a particular trip and the overall list containing all the
    trips
    """
    listCows = list(get_partitions(cows))
    numberOfTrips = len(cows.keys()) + 1
    result = []
    for i in range(len(listCows)):
        if len(listCows[i]) < numberOfTrips:
            for j in range(len(listCows[i])):
                totalWeight = 0
                weightExceeded = False
                for k in listCows[i][j]:
                    totalWeight += cows[k]
                    if totalWeight > limit:
                        weightExceeded = True
                        break
                if weightExceeded:
                    break
            if not weightExceeded:
                numberOfTrips = len(listCows[i])
                result = list(listCows[i])
    return result

    Finds the allocation of cows that minimizes the number of spaceship trips
    via brute force.  The brute force algorithm follows the following method:

    1. Enumerate all possible ways that the cows can be divided into separate
    trips.
    2. Select the allocation that minimizes the number of trips without making
    any trip that does not obey the weight limitation.

    Does not mutate the given dictionary of cows.

    Parameters:
    cows - a dictionary of name (string), weight (int) pairs
    limit - weight limit of the spaceship (an int)

    Returns:
    A list of lists, with each inner list containing the names of cows
    transported on a particular trip and the overall list containing all the
    trips
    
Finds the allocation of cows that minimizes the number of spaceship trips
via brute force.  The brute force algorithm follows the following method.

1. Enumerate all possible ways that the cows can be divided into separate
trips.
2. Select the allocation that minimizes the number of trips without making
any trip that does not obey the weight limitation.

Does not mutate the given dictionary of cows.

a dictionary of name (string), weight (int) pairs
limit - weight limit of the spaceship (an int)

A list of lists, with each inner list containing the names of cows
transported on a particular trip and the overall list containing all the
trips
Finds the allocation of cows that minimizes the number of spaceship trips via brute force . The brute force algorithm follows the following method . 1 . Enumerate all possible ways that the cows can be divided into separate trips . 2 . Select the allocation that minimizes the number of trips without making any trip that does not obey the weight limitation . Does not mutate the given dictionary of cows . a dictionary of name ( string ) weight ( int ) pairs limit - weight limit of the spaceship ( an int ) A list of lists with each inner list containing the names of cows transported on a particular trip and the overall list containing all the trips
def brute_force_cow_transport(cows, limit=10):
    listCows = list(get_partitions(cows))
    numberOfTrips = len(cows.keys()) + 1
    result = []
    for i in range(len(listCows)):
        if len(listCows[i]) < numberOfTrips:
            for j in range(len(listCows[i])):
                totalWeight = 0
                weightExceeded = False
                for k in listCows[i][j]:
                    totalWeight += cows[k]
                    if totalWeight > limit:
                        weightExceeded = True
                        break
                if weightExceeded:
                    break
            if not weightExceeded:
                numberOfTrips = len(listCows[i])
                result = list(listCows[i])
    return result
def brute_force_cow_transport ( cows , limit = 10 ) : listCows = list ( get_partitions ( cows ) ) numberOfTrips = len ( cows . keys ( ) ) + 1 result = [ ] for i in range ( len ( listCows ) ) : if len ( listCows [ i ] ) < numberOfTrips : for j in range ( len ( listCows [ i ] ) ) : totalWeight = 0 weightExceeded = False for k in listCows [ i ] [ j ] : totalWeight += cows [ k ] if totalWeight > limit : weightExceeded = True break if weightExceeded : break if not weightExceeded : numberOfTrips = len ( listCows [ i ] ) result = list ( listCows [ i ] ) return result
Finds the allocation of cows that minimizes the number of spaceship trips
via brute force.
Finds the allocation of cows that minimizes the number of spaceship trips via brute force .
"""
    Finds the allocation of cows that minimizes the number of spaceship trips
    via brute force.  The brute force algorithm follows the following method:

    1. Enumerate all possible ways that the cows can be divided into separate
    trips.
    2. Select the allocation that minimizes the number of trips without making
    any trip that does not obey the weight limitation.

    Does not mutate the given dictionary of cows.

    Parameters:
    cows - a dictionary of name (string), weight (int) pairs
    limit - weight limit of the spaceship (an int)

    Returns:
    A list of lists, with each inner list containing the names of cows
    transported on a particular trip and the overall list containing all the
    trips
    """
cows
limit