获取指定长度得所有序列

通过事件来表述这个序列，即n重伯努利实验（二项分布）的所有可能结果。例如时间a表示为： a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], 假设每次实验为从a中选择一个数字，那么进行n次实验，获得所有可能得序列。

比如，进行两次实验， n=2, 那么可能得结果有100个。这里因为每次实验都是相对独立的，所以每次实验的结果可能出现重复，也就是说在获得所有可能的序列中，可以存在重复得值。

递归实现，DFS（深度优先遍历）

def gen_all_sequence_dfs(outcomes, length):
    """
    generate all sequence by dfs

    outcomes: all the possible event, a list
    length: how many times does the sequence repeat, sequence length
    """

    res = []
    seq = []
    dfs_sequence(outcomes, length, seq, res)

    return res

def dfs_sequence(outcomes, length, seq, res):
    """
    deep first search
    """
    if 0 == length:
        res.append(tuple(seq[:]))
        return

    for key in outcomes:
        seq.append(key)
        dfs_sequence(outcomes, length - 1, seq, res)
        seq.pop()

def run_dfs_example1():
    """
    Example of all sequences
    """
    print ‘dfs gen all sequence‘
    outcomes = set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

    length = 2
    seq_outcomes = gen_all_sequence_dfs(outcomes, length)
    print "Computed", len(seq_outcomes), "sequences of", str(length), "outcomes"
    print "Sequences were", seq_outcomes

run_dfs_example1()

运行输出结果：

dfs gen all sequence
Computed 100 sequences of 2 outcomes
Sequences were [(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (5, 7), (5, 8), (5, 9), (6, 0), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (6, 7), (6, 8), (6, 9), (7, 0), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 7), (7, 8), (7, 9), (8, 0), (8, 1), (8, 2), (8, 3), (8, 4), (8, 5), (8, 6), (8, 7), (8, 8), (8, 9), (9, 0), (9, 1), (9, 2), (9, 3), (9, 4), (9, 5), (9, 6), (9, 7), (9, 8), (9, 9)]

非递归实现

利用动态规划的原理（这里我也不太熟悉是不是动态规划，暂且这么叫，如果有错误，请大家帮忙更正），动态的计算第k次实验后获得得所有得序列。根据第k-1次实验的所有得序列得结果，然后把每一次结果拿出来计算这一次结果再加上一次实验（即第k次实验）能够获得的结果。

def gen_all_sequences(outcomes, length):
    """
    Iterative function that enumerates the set of all sequences of
    outcomes of given length

    Permutation allow repeat
    """

    ans = set([()])
    for dummy_idx in range(length):
        temp = set()
        for seq in ans:
            for item in outcomes:
                new_seq = list(seq)
                new_seq.append(item)
                temp.add(tuple(new_seq))
        ans = temp
    return ans

# example for digits
def run_example1():
    """
    Example of all sequences
    """
    print ‘gen all sequence‘
    outcomes = set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
    #outcomes = set(["Red", "Green", "Blue"])
    #outcomes = set(["Sunday", "Mondy", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"])

    length = 2
    seq_outcomes = gen_all_sequences(outcomes, length)
    print "Computed", len(seq_outcomes), "sequences of", str(length), "outcomes"
    print "Sequences were", seq_outcomes


print ‘#############################################‘
run_example1()

代码运行结果：

gen all sequence
Computed 100 sequences of 2 outcomes
Sequences were set([(7, 3), (6, 9), (0, 7), (1, 6), (3, 7), (2, 5), (8, 5), (5, 8), (4, 0), (9, 0), (6, 7), (5, 5), (7, 6), (0, 4), (1, 1), (3, 2), (2, 6), (8, 2), (4, 5), (9, 3), (6, 0), (7, 5), (0, 1), (3, 1), (9, 9), (7, 8), (2, 1), (8, 9), (9, 4), (5, 1), (7, 2), (1, 5), (3, 6), (2, 2), (8, 6), (4, 1), (9, 7), (6, 4), (5, 4), (7, 1), (0, 5), (1, 0), (0, 8), (3, 5), (2, 7), (8, 3), (4, 6), (9, 2), (6, 1), (5, 7), (7, 4), (0, 2), (1, 3), (4, 8), (3, 0), (2, 8), (9, 8), (8, 0), (6, 2), (5, 0), (1, 4), (3, 9), (2, 3), (1, 9), (8, 7), (4, 2), (9, 6), (6, 5), (5, 3), (7, 0), (6, 8), (0, 6), (1, 7), (0, 9), (3, 4), (2, 4), (8, 4), (5, 9), (4, 7), (9, 1), (6, 6), (5, 6), (7, 7), (0, 3), (1, 2), (4, 9), (3, 3), (2, 9), (8, 1), (4, 4), (6, 3), (0, 0), (7, 9), (3, 8), (2, 0), (1, 8), (8, 8), (4, 3), (9, 5), (5, 2)])

Permutation 获取所有排列

给定一个序列，例如上面给出得 a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9], 如果按照上面获取序列的算法，那么每次实验都作为独立得实现，序列中可以出现重复实验结果。但是再获取排列的时候则不能按照n重伯努利实验得思想进行了，获取排列不允许有重复得结果，即一个元素只能被选择一次。但是在排列中是存在元素得顺序因素得，也就是说同样得两个元素，不同得顺序为不同得排列。

非递归实现

在上面非递归算法得基础上，增加一个判断，判断该元素是否已经选择过就可以实现排列得获取。

def gen_permutations(outcomes, length):
    """
    Iterative function that enumerates the set of all sequences of
    outcomes of given length
    Permutation not allow repeat
    """

    ans = set([()])
    for dummy_idx in range(length):
        temp = set()
        for seq in ans:
            for item in outcomes:
                if item in seq:
                    continue
                new_seq = list(seq)
                new_seq.append(item)
                temp.add(tuple(new_seq))
        ans = temp
    return ans

# example for digits
def run_example1():
    """
    Example of all sequences
    """
    outcomes = set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
    #outcomes = set(["Red", "Green", "Blue"])
    #outcomes = set(["Sunday", "Mondy", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"])

    length = 2
    seq_outcomes = gen_permutations(outcomes, length)
    print "Computed", len(seq_outcomes), "sequences of", str(length), "outcomes"
    print "Sequences were", seq_outcomes

run_example1()

代码运行结果：

Computed 90 sequences of 2 outcomes
Sequences were set([(7, 3), (6, 9), (0, 7), (1, 6), (3, 7), (2, 5), (8, 5), (5, 8), (4, 0), (9, 0), (6, 7), (7, 6), (0, 4), (3, 2), (2, 6), (8, 2), (4, 5), (9, 3), (6, 0), (7, 5), (0, 1), (3, 1), (7, 8), (2, 1), (8, 9), (9, 4), (5, 1), (7, 2), (1, 5), (3, 6), (8, 6), (4, 1), (9, 7), (6, 4), (5, 4), (7, 1), (0, 5), (1, 0), (0, 8), (3, 5), (2, 7), (8, 3), (4, 6), (9, 2), (6, 1), (5, 7), (7, 4), (0, 2), (1, 3), (4, 8), (3, 0), (2, 8), (9, 8), (8, 0), (6, 2), (5, 0), (1, 4), (3, 9), (2, 3), (1, 9), (8, 7), (4, 2), (9, 6), (6, 5), (5, 3), (7, 0), (6, 8), (0, 6), (1, 7), (0, 9), (3, 4), (2, 4), (8, 4), (5, 9), (4, 7), (9, 1), (5, 6), (0, 3), (1, 2), (4, 9), (2, 9), (8, 1), (6, 3), (7, 9), (3, 8), (2, 0), (1, 8), (4, 3), (9, 5), (5, 2)])

递归实现

def gen_permutations_dfs(outcomes, length):
    """
    Iterative function that enumerates the set of all sequences of
    outcomes of given length
    Permutation not allow repeat, DFS
    """

    res = []
    seq = []
    dfs_permutation(outcomes, length, seq, res)

    return res

def dfs_permutation(outcomes, length, seq, res):
    """
    deep first search
    """

    if 0 == length:
        res.append(tuple(seq[:]))
        return

    for key in outcomes:
        if key in seq:
            continue
        seq.append(key)
        dfs_permutation(outcomes, length - 1, seq, res)
        seq.pop()

# example for digits
def run_example2():
    """
    Example of all sequences
    """
    outcomes = set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
    #outcomes = set(["Red", "Green", "Blue"])
    #outcomes = set(["Sunday", "Mondy", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"])

    length = 2
    seq_outcomes = gen_permutations_dfs(outcomes, length)
    print "Computed", len(seq_outcomes), "sequences of", str(length), "outcomes"
    print "Sequences were", seq_outcomes

run_example2()

运行结果：

Computed 90 sequences of 2 outcomes
Sequences were [(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 0), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 0), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 0), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 0), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (5, 8), (5, 9), (6, 0), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 7), (6, 8), (6, 9), (7, 0), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 8), (7, 9), (8, 0), (8, 1), (8, 2), (8, 3), (8, 4), (8, 5), (8, 6), (8, 7), (8, 9), (9, 0), (9, 1), (9, 2), (9, 3), (9, 4), (9, 5), (9, 6), (9, 7), (9, 8)]

组合

组合与排列最大得区别是组合不关心顺序，所以组合得数量要比排列少。
，组合可以表示为

非递归实现

def gen_combination(outcomes, length):
    """
    Iterative function that enumerates the set of all sequences of
    outcomes of given length
    Permutation not allow repeat
    """

    ans = set([()])
    for dummy_idx in range(length):
        temp = set()
        for seq in ans:
            for item in outcomes:
                if item in seq:
                    continue
                new_seq = list(seq)
                new_seq.append(item)
                temp.add(tuple(sorted(new_seq)))
        ans = temp
    return ans


def run_example():
    """
    Examples of sorted sequences of outcomes
    """
    # example for digits
    outcomes = set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
    #outcomes = set(["Red", "Green", "Blue"])
    #outcomes = set(["Sunday", "Mondy", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"])

    length = 2
    seq_outcomes = gen_combination(outcomes, length)
    print "Computed", len(seq_outcomes), "sorted sequences of", str(length) ,"outcomes"
    print "Sequences were", seq_outcomes

run_example()

运行结果：

Computed 45 sorted sequences of 2 outcomes
Sequences were set([(5, 9), (6, 9), (1, 3), (4, 8), (5, 6), (2, 8), (4, 7), (0, 7), (4, 6), (8, 9), (1, 6), (3, 7), (2, 5), (0, 3), (5, 8), (1, 2), (6, 7), (2, 9), (1, 5), (3, 6), (0, 4), (3, 5), (2, 7), (2, 6), (4, 5), (1, 4), (3, 9), (2, 3), (1, 9), (4, 9), (0, 8), (7, 9), (0, 1), (6, 8), (3, 4), (5, 7), (2, 4), (3, 8), (0, 6), (1, 8), (1, 7), (0, 9), (0, 5), (7, 8), (0, 2)])

组合的递归实现

def gen_combination_dfs(outcomes, length):
    """
    Iterative function that enumerates the set of all sequences of
    outcomes of given length
    Permutation not allow repeat, DFS
    """

    res = []
    seq = []
    idx = 0
    dfs_combination(outcomes, length, idx, seq, res)

    return res

def dfs_combination(outcomes, length, idx, seq, res):
    """
    deep first search
    """
    if idx + length > len(outcomes):
        return

    if 0 == length:
        res.append(tuple(seq[:]))
        return

    for i in range(idx, len(outcomes) - length + 1):
        seq.append(outcomes[i])
        dfs_combination(outcomes, length - 1, i + 1, seq, res)
        seq.pop()

def run_example2():
    """
    Examples of sorted sequences of outcomes
    """
    # example for digits
    print "dfs combination"
    outcomes = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
    #outcomes = set(["Red", "Green", "Blue"])
    #outcomes = set(["Sunday", "Mondy", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"])

    length = 2
    seq_outcomes = gen_combination_dfs(outcomes, length)
    print "Computed", len(seq_outcomes), "sorted sequences of", str(length) ,"outcomes"
    print "Sequences were", seq_outcomes

run_example2()

运行结果：

Computed 45 sorted sequences of 2 outcomes
Sequences were [(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9), (7, 8), (7, 9), (8, 9)]