尋找蒙赫陣列(monge array)的每行的最小值

原創

2020-02-20 18:39

monge array是一個矩陣，滿足第j行的最小值的列號C(j)大於等於第i行的最小值的列號C(i)，當j>=i的時候，有如下特性

－－對於任何行列，正對角線的和<=反對角線的和

#!/usr/bin/python

'''file name: monge.py
   --P111
   --input one m*n monge array, find leftmost minimum value, O(m + nlgm)
   --author: zevolo, 2012.05.16

proof:

first to guees the result:
T(m,n) = T(m/2, n) + theta(m + n)
    use the recurrence tree:
      T(m, n) = c(m + n) + c(m/2 + n) + c(m/4 + n) + ... + c(m/(2^(lgm-1)) + n)
              = cm (1+1/2+1/4+...+1/(2^(lgm-1))) + cnlgm
              = cm 2(1 - 1/(2^lgm)) + cnlgm
then
assume T(m, n) <= am + bnlgm    (a, b are constant)
    T(m,n) = T(m/2, n) + c(m+n)
          <= (am/2 + bnlg(m/2)) + c(m+n)
          =  am + bnlgm - am/2 - bn + cm + cn
          =  am + bnlgm + (c - a/2)m + (c - b)n

'''

def find_odd_min(l, m, n, r):
    start = 0
    end = 0
    ret = []
    r.append(n - 1) #sentinel
    for i in range(0, m):
        k = i / 2
        if i % 2 == 0:
            ret.append(r[k])
        else:
            end = r[k+1]
            pos = start
            val = l[i][pos]
            for j in range(start + 1, end + 1):
                if l[i][j] < val:
                    val = l[i][j]
                    pos = j
            start = end

            ret.append(pos)
    return ret


def find_min(l, m, n):
    if m == 1:
        pos = 0
        val = l[0][0]
        for i in range(1, n):
            if val > l[0][i]:
                pos = i
                val = l[0][i]
        r = [pos]
        return r

    sub = []
    for i in range(0, m, 2):
        sub.append(l[i])

    r = find_min(sub, (m+1)/2, n)
    ret = find_odd_min(l, m, n, r)
    return ret

if __name__ == '__main__':
    l = [ [10, 17, 13, 28, 23],
          [17, 22, 16, 29, 23],
          [24, 28, 22, 34, 24],
          [11, 13, 6,  17, 7],
          [45, 44, 32, 37, 23],
          [36, 33, 19, 21, 6],
          [75, 66, 51, 54, 34]
        ]
    r = find_min(l, 6, 5)

    print l
    print r