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Given the following text description, write Python code to implement the functionality described below step by step
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Description:
 Project Euler
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.  
<center>
3
7 4
 2 4 6
 8 5 9 3
</center>
That is, 3 + 7 + 4 + 9 = 23.   
Find the maximum total from top to bottom of the triangle below
Step1:  Let's try a somewhat functional approach.
It is much easier to understand.
I like that.
Step2:  Try tuples instead of lists.
It's a little bit faster and still readable.
That's a good combination.
Step3:  Convert t4 and t15 to be tuples instead of lists.
This does not affect readability.
It is faster yet.
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Python Code:
t4 = [
          [3],
        [7,  4],
      [2,  4,  6],
    [8,  5,  9,  3],
]
t4
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)
from copy import deepcopy
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]
n = t4
%timeit foo(n)
foo(n)
n = t15
%timeit foo(n)
foo(n)
Explanation: Project Euler
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.  
<center>
3
7 4
 2 4 6
 8 5 9 3
</center>
That is, 3 + 7 + 4 + 9 = 23.   
Find the maximum total from top to bottom of the triangle below:  
<center>
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
</center>
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)
End of explanation
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)
n = t4
%timeit foo(n)
foo(n)
n = t15
%timeit foo(n)
foo(n)
Explanation: Let's try a somewhat functional approach.
It is much easier to understand.
I like that.
End of explanation
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)
n = t4
%timeit foo(n)
foo(n)
n = t15
%timeit foo(n)
foo(n)
Explanation: Try tuples instead of lists.
It's a little bit faster and still readable.
That's a good combination.
End of explanation
t4 = tuple(tuple(row) for row in t4)
t15 = tuple(tuple(row) for row in t15)
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)
n = t4
%timeit foo(n)
foo(n)
n = t15
%timeit foo(n)
foo(n)
Explanation: Convert t4 and t15 to be tuples instead of lists.
This does not affect readability.
It is faster yet.
End of explanation
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