Maximum path sum I

Problem 18

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

**3** **7** 4 2 **4** 6 8 5 **9** 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)



In [1]:

    
t4 = [
          [3],
        [7,  4],
      [2,  4,  6],
    [8,  5,  9,  3],
]
t4









    Out[1]:





[[3], [7, 4], [2, 4, 6], [8, 5, 9, 3]]



In [2]:

    
t15 = [
                                [75],
                              [95, 64],
                            [17, 47, 82],
                          [18, 35, 87, 10],
                        [20,  4, 82, 47, 65],
                      [19,  1, 23, 75,  3, 34],
                    [88,  2, 77, 73,  7, 63, 67],
                  [99, 65,  4, 28,  6, 16, 70, 92],
                [41, 41, 26, 56, 83, 40, 80, 70, 33],
              [41, 48, 72, 33, 47, 32, 37, 16, 94, 29],
            [53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14],
          [70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57],
        [91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48],
      [63, 66,  4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31],
    [ 4, 62, 98, 27, 23,  9, 70, 98, 73, 93, 38, 53, 60,  4, 23],
]
len(t15)









    Out[2]:





15



In [3]:

    
from copy import deepcopy



In [4]:

    
def foo(t):
    t = deepcopy(t)
    for i in range(len(t))[::-1]:
        r = t[i]
        try:
            nr = t[i+1]
        except IndexError:
            for j in range(len(t[i])):
                t[i][j] = (t[i][j], None)
        else:
            for j in range(len(t[i])):
                dir = (t[i+1][j+1][0] > t[i+1][j+0][0])
                t[i][j] = (t[i][j] + t[i+1][j+dir][0], dir)
    return t[0][0][0]



In [5]:

    
n = t4
%timeit foo(n)
foo(n)









    



10000 loops, best of 3: 54.8 µs per loop






    Out[5]:





23



In [6]:

    
n = t15
%timeit foo(n)
foo(n)









    



1000 loops, best of 3: 396 µs per loop






    Out[6]:





1074

Let's try a somewhat functional approach. It is much easier to understand. I like that.



In [7]:

    
def foo(t):
    old_row = []
    for row in t:
        stagger_max = map(max, zip([0] + old_row, old_row + [0]))
        old_row = list(map(sum, zip(stagger_max, row)))
   
    return max(old_row)



In [8]:

    
n = t4
%timeit foo(n)
foo(n)









    



100000 loops, best of 3: 18.2 µs per loop






    Out[8]:





23



In [9]:

    
n = t15
%timeit foo(n)
foo(n)









    



10000 loops, best of 3: 120 µs per loop






    Out[9]:





1074

Try tuples instead of lists. It's a little bit faster and still readable. That's a good combination.



In [10]:

    
def foo(t):
    old_row = tuple()
    for row in t:
        stagger_max = map(max, zip((0,) + old_row, old_row + (0,)))
        old_row = tuple(map(sum, zip(stagger_max, row)))
   
    return max(old_row)



In [11]:

    
n = t4
%timeit foo(n)
foo(n)









    



100000 loops, best of 3: 17.8 µs per loop






    Out[11]:





23



In [12]:

    
n = t15
%timeit foo(n)
foo(n)









    



10000 loops, best of 3: 115 µs per loop






    Out[12]:





1074

Convert t4 and t15 to be tuples instead of lists. This does not affect readability. It is faster yet.



In [13]:

    
t4 = tuple(tuple(row) for row in t4)
t15 = tuple(tuple(row) for row in t15)



In [14]:

    
def foo(t):
    old_row = tuple()
    for row in t:
        stagger_max = map(max, zip((0,) + old_row, old_row + (0,)))
        old_row = tuple(map(sum, zip(stagger_max, row)))
   
    return max(old_row)



In [15]:

    
n = t4
%timeit foo(n)
foo(n)









    



100000 loops, best of 3: 17.1 µs per loop






    Out[15]:





23



In [16]:

    
n = t15
%timeit foo(n)
foo(n)









    



10000 loops, best of 3: 114 µs per loop






    Out[16]:





1074

I like cell 7 the most. For me, its lists are more readable than tuples.

2017-09-02 Now I play some more.

First, Consolidate the body of the loop into a single ugly long statement. It is a little bit faster.



In [17]:

    
def foo(t):
    old_row = tuple()
    for row in t:
        old_row = tuple(map(sum, zip(
            map(max, zip((0,) + old_row, old_row + (0,))), row)))
   
    return max(old_row)



In [18]:

    
n = t4
%timeit foo(n)
foo(n)









    



100000 loops, best of 3: 16.6 µs per loop






    Out[18]:





23



In [19]:

    
n = t15
%timeit foo(n)
foo(n)









    



10000 loops, best of 3: 111 µs per loop






    Out[19]:





1074

Next, eliminate the loop by using recursion. This code is much harder for me to understand. It is slower also.

It was good exercise to play with replacing a loop with recursion. Functional programming fans should like it.



In [20]:

    
def goo(t):
    *upper_rows, bottom_row = t
    if not upper_rows:
        return tuple(bottom_row)

    maximums = goo(upper_rows)
    stagger_max = map(max, zip((0,) + maximums, maximums + (0,)))
    return tuple(map(sum, zip(stagger_max, bottom_row)))



In [21]:

    
def foo(t):
    return max(goo(t))



In [22]:

    
n = t4
%timeit foo(n)
foo(n)









    



100000 loops, best of 3: 17.2 µs per loop






    Out[22]:





23



In [23]:

    
n = t15
%timeit foo(n)
foo(n)









    



10000 loops, best of 3: 119 µs per loop






    Out[23]:





1074