Codebreaker

Problem #20 is to find the sum of the digits in the number 100!

This can be easily solved with Python as the math package comes with factorial function.

from math import factorial
reduce(lambda x, y: int(x)+int(y), str(factorial(100)))

A friend of mine, Leon, told me about the Project Euler. One of the problems is to find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Here’s my one-line Python solution. l is a string that contains all 1000 digits above.

max(map(lambda a,b,c,d,e: int(a)*int(b)*int(c)*int(d)*int(e), l[:-4], l[1:-3], l[2:-2], l[3:-1], l[4:]))

Python for the win!

I’ve been using a tuple as a substitution of a ternary operator in Python. Even though it is not exactly a ternary operator and has some drawbacks, but it works fine in most cases.

w = (u, v)[x < y]

When x < y is true, the whole expression of the tuple, (u, v)[x < y], will be evaluated as v because true means 1 here. Likewise, false will cause the tuple expression to be evaluated as u.

I had another trick to use with lambda functions.

lambda w: x < y and v or u

This might be a little less obvious to see what’s going on. When x < y is false, the following part, and v, will not be evaluated due to short-circuit evaluation, and it will jump to the next part, or u. Thus, this whole expression is to be evaluated as u. Similarly, when x < y is true, and v part has to be evaluated so it the whole expression will be evaluated as v.

However, as acute readers might already noticed, this trick has a major drawback. When v is a falsy value such as 0, False or None, the whole expression will be evaluated as u regardless of the result of x < y. So, one should refrain from using this trick unless it is impossible for v to be a falsy value.

Last Saturday, Cody told me that there is a ternary operator equivalent in Python.

w = v if x < y else u

Strictly speaking, it is not a ternary operator, rather an expression. But, in my opinion, it is more intuitive thus it makes it easier to maintain the code.