Algorithm to find Lucky Numbers

Question

I came across this question.A number is called lucky if the sum of its digits, as well as the sum of the squares of its digits is a prime number. How many numbers between A

Answer 1

I have just solved this problem.

It's just a Dynamic Programming problem. Take DP[n](sum-square_sum) as the DP function, and DP[n](sum-square_sum) is the count of all of the numbers whose digits is less than or equal to n, with the sum and square_sum of digits of the number is respectively represented by sum and square_sum. For example:

DP[1](1-1) = 1      # only 1 satisfies the condition                        
DP[2](1-1) = 2      # both 1 and 10 satisfies the condition                        
DP[3](1-1） = 3      # 1 10 100
DP[3](2-4） = 3      # 11 110 101

Since we can easily figure out the first DP state DP[1][..][..], it is:

(0-0) => 1     (1-1) => 1    (2-4) => 1     (3-9) => 1     (4-16) => 1    
(5-25) => 1    (6-36) => 1   (7-49) => 1    (8-64) => 1    (9-81) => 1

then we can deduce DP[1] from DP[1], and then DP[3] ... DP[18] the deduce above is made by the fact that every time when n increase by 1, for example from DP[1] to DP[2], we got a new digit (0..9), and the set of (sum, square_sum) pair (i.e. DP[n]) must be updated.

Finally, we can traverse the DP[18] set and count of the numbers that are lucky.

Well, how about the time and space complexity of the algorithm above? As we know sum <= 18*9=162, square_sum <= 18*9*9 = 1458, so the set of (sum, square_sum) pair (i.e. DP[n]) is very small, less than 162*1458=236196, in fact it's much smaller than 236196; The fact is: my ruby program counting all the lucky numbers between 0 and 10^18 finishes in less than 1s.

ruby lucky_numbers.rb  0.55s user 0.00s system 99% cpu 0.556 total

and I test my program by writing a test function using brute force algorithm, and it's right for numbers less than 10^7 .