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Project Euler, first problem: "Multiples of 3 and 5"

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.
In [1]:
from notebook_preamble import J, V, define

Let's create a predicate that returns True if a number is a multiple of 3 or 5 and False otherwise.

In [2]:
define('P [3 % not] dupdip 5 % not or')
In [3]:
V('80 P')
             • 80 P
          80 • P
          80 • [3 % not] dupdip 5 % not or
80 [3 % not] • dupdip 5 % not or
          80 • 3 % not 80 5 % not or
        80 3 • % not 80 5 % not or
           2 • not 80 5 % not or
       False • 80 5 % not or
    False 80 • 5 % not or
  False 80 5 • % not or
     False 0 • not or
  False True • or
        True •

Given the predicate function P a suitable program is:

PE1 == 1000 range [P] filter sum

This function generates a list of the integers from 0 to 999, filters that list by P, and then sums the result.

Logically this is fine, but pragmatically we are doing more work than we should be; we generate one thousand integers but actually use less than half of them. A better solution would be to generate just the multiples we want to sum, and to add them as we go rather than storing them and adding summing them at the end.

At first I had the idea to use two counters and increase them by three and five, respectively. This way we only generate the terms that we actually want to sum. We have to proceed by incrementing the counter that is lower, or if they are equal, the three counter, and we have to take care not to double add numbers like 15 that are multiples of both three and five.

This seemed a little clunky, so I tried a different approach.

Consider the first few terms in the series:

3 5 6 9 10 12 15 18 20 21 ...

Subtract each number from the one after it (subtracting 0 from 3):

3 5 6 9 10 12 15 18 20 21 24 25 27 30 ...
0 3 5 6  9 10 12 15 18 20 21 24 25 27 ...
-------------------------------------------
3 2 1 3  1  2  3  3  2  1  3  1  2  3 ...

You get this lovely repeating palindromic sequence:

3 2 1 3 1 2 3

To make a counter that increments by factors of 3 and 5 you just add these differences to the counter one-by-one in a loop.

To make use of this sequence to increment a counter and sum terms as we go we need a function that will accept the sum, the counter, and the next term to add, and that adds the term to the counter and a copy of the counter to the running sum. This function will do that:

PE1.1 == + [+] dupdip
In [4]:
define('PE1.1 + [+] dupdip')
In [5]:
V('0 0 3 PE1.1')
        • 0 0 3 PE1.1
      0 • 0 3 PE1.1
    0 0 • 3 PE1.1
  0 0 3 • PE1.1
  0 0 3 • + [+] dupdip
    0 3 • [+] dupdip
0 3 [+] • dupdip
    0 3 • + 3
      3 • 3
    3 3 •
In [6]:
V('0 0 [3 2 1 3 1 2 3] [PE1.1] step')
                            • 0 0 [3 2 1 3 1 2 3] [PE1.1] step
                          0 • 0 [3 2 1 3 1 2 3] [PE1.1] step
                        0 0 • [3 2 1 3 1 2 3] [PE1.1] step
        0 0 [3 2 1 3 1 2 3] • [PE1.1] step
0 0 [3 2 1 3 1 2 3] [PE1.1] • step
              0 0 3 [PE1.1] • i [2 1 3 1 2 3] [PE1.1] step
                      0 0 3 • PE1.1 [2 1 3 1 2 3] [PE1.1] step
                      0 0 3 • + [+] dupdip [2 1 3 1 2 3] [PE1.1] step
                        0 3 • [+] dupdip [2 1 3 1 2 3] [PE1.1] step
                    0 3 [+] • dupdip [2 1 3 1 2 3] [PE1.1] step
                        0 3 • + 3 [2 1 3 1 2 3] [PE1.1] step
                          3 • 3 [2 1 3 1 2 3] [PE1.1] step
                        3 3 • [2 1 3 1 2 3] [PE1.1] step
          3 3 [2 1 3 1 2 3] • [PE1.1] step
  3 3 [2 1 3 1 2 3] [PE1.1] • step
              3 3 2 [PE1.1] • i [1 3 1 2 3] [PE1.1] step
                      3 3 2 • PE1.1 [1 3 1 2 3] [PE1.1] step
                      3 3 2 • + [+] dupdip [1 3 1 2 3] [PE1.1] step
                        3 5 • [+] dupdip [1 3 1 2 3] [PE1.1] step
                    3 5 [+] • dupdip [1 3 1 2 3] [PE1.1] step
                        3 5 • + 5 [1 3 1 2 3] [PE1.1] step
                          8 • 5 [1 3 1 2 3] [PE1.1] step
                        8 5 • [1 3 1 2 3] [PE1.1] step
            8 5 [1 3 1 2 3] • [PE1.1] step
    8 5 [1 3 1 2 3] [PE1.1] • step
              8 5 1 [PE1.1] • i [3 1 2 3] [PE1.1] step
                      8 5 1 • PE1.1 [3 1 2 3] [PE1.1] step
                      8 5 1 • + [+] dupdip [3 1 2 3] [PE1.1] step
                        8 6 • [+] dupdip [3 1 2 3] [PE1.1] step
                    8 6 [+] • dupdip [3 1 2 3] [PE1.1] step
                        8 6 • + 6 [3 1 2 3] [PE1.1] step
                         14 • 6 [3 1 2 3] [PE1.1] step
                       14 6 • [3 1 2 3] [PE1.1] step
             14 6 [3 1 2 3] • [PE1.1] step
     14 6 [3 1 2 3] [PE1.1] • step
             14 6 3 [PE1.1] • i [1 2 3] [PE1.1] step
                     14 6 3 • PE1.1 [1 2 3] [PE1.1] step
                     14 6 3 • + [+] dupdip [1 2 3] [PE1.1] step
                       14 9 • [+] dupdip [1 2 3] [PE1.1] step
                   14 9 [+] • dupdip [1 2 3] [PE1.1] step
                       14 9 • + 9 [1 2 3] [PE1.1] step
                         23 • 9 [1 2 3] [PE1.1] step
                       23 9 • [1 2 3] [PE1.1] step
               23 9 [1 2 3] • [PE1.1] step
       23 9 [1 2 3] [PE1.1] • step
             23 9 1 [PE1.1] • i [2 3] [PE1.1] step
                     23 9 1 • PE1.1 [2 3] [PE1.1] step
                     23 9 1 • + [+] dupdip [2 3] [PE1.1] step
                      23 10 • [+] dupdip [2 3] [PE1.1] step
                  23 10 [+] • dupdip [2 3] [PE1.1] step
                      23 10 • + 10 [2 3] [PE1.1] step
                         33 • 10 [2 3] [PE1.1] step
                      33 10 • [2 3] [PE1.1] step
                33 10 [2 3] • [PE1.1] step
        33 10 [2 3] [PE1.1] • step
            33 10 2 [PE1.1] • i [3] [PE1.1] step
                    33 10 2 • PE1.1 [3] [PE1.1] step
                    33 10 2 • + [+] dupdip [3] [PE1.1] step
                      33 12 • [+] dupdip [3] [PE1.1] step
                  33 12 [+] • dupdip [3] [PE1.1] step
                      33 12 • + 12 [3] [PE1.1] step
                         45 • 12 [3] [PE1.1] step
                      45 12 • [3] [PE1.1] step
                  45 12 [3] • [PE1.1] step
          45 12 [3] [PE1.1] • step
            45 12 3 [PE1.1] • i
                    45 12 3 • PE1.1
                    45 12 3 • + [+] dupdip
                      45 15 • [+] dupdip
                  45 15 [+] • dupdip
                      45 15 • + 15
                         60 • 15
                      60 15 •

So one step through all seven terms brings the counter to 15 and the total to 60.

In [7]:
1000 / 15
Out [7]:
66.66666666666667
In [8]:
66 * 15
Out [8]:
990
In [9]:
1000 - 990
Out [9]:
10
In [10]:
999 - 990
Out [10]:
9

That means we want to run the full list of numbers sixty-six times to get to 990 and then the first four numbers 3 2 1 3 to get to 999.

In [11]:
define('PE1 0 0 66 [[3 2 1 3 1 2 3] [PE1.1] step] times [3 2 1 3] [PE1.1] step pop')
In [12]:
J('PE1')
233168

This form uses no extra storage and produces no unused summands. It's good but there's one more trick we can apply. The list of seven terms takes up at least seven bytes. But notice that all of the terms are less than four, and so each can fit in just two bits. We could store all seven terms in just fourteen bits and use masking and shifts to pick out each term as we go. This will use less space and save time loading whole integer terms from the list.

    3  2  1  3  1  2  3
0b 11 10 01 11 01 10 11 == 14811
In [13]:
0b11100111011011
Out [13]:
14811
In [14]:
define('PE1.2 [3 & PE1.1] dupdip 2 >>')
In [15]:
V('0 0 14811 PE1.2')
                      • 0 0 14811 PE1.2
                    0 • 0 14811 PE1.2
                  0 0 • 14811 PE1.2
            0 0 14811 • PE1.2
            0 0 14811 • [3 & PE1.1] dupdip 2 >>
0 0 14811 [3 & PE1.1] • dupdip 2 >>
            0 0 14811 • 3 & PE1.1 14811 2 >>
          0 0 14811 3 • & PE1.1 14811 2 >>
                0 0 3 • PE1.1 14811 2 >>
                0 0 3 • + [+] dupdip 14811 2 >>
                  0 3 • [+] dupdip 14811 2 >>
              0 3 [+] • dupdip 14811 2 >>
                  0 3 • + 3 14811 2 >>
                    3 • 3 14811 2 >>
                  3 3 • 14811 2 >>
            3 3 14811 • 2 >>
          3 3 14811 2 • >>
             3 3 3702 •
In [16]:
V('3 3 3702 PE1.2')
                     • 3 3 3702 PE1.2
                   3 • 3 3702 PE1.2
                 3 3 • 3702 PE1.2
            3 3 3702 • PE1.2
            3 3 3702 • [3 & PE1.1] dupdip 2 >>
3 3 3702 [3 & PE1.1] • dupdip 2 >>
            3 3 3702 • 3 & PE1.1 3702 2 >>
          3 3 3702 3 • & PE1.1 3702 2 >>
               3 3 2 • PE1.1 3702 2 >>
               3 3 2 • + [+] dupdip 3702 2 >>
                 3 5 • [+] dupdip 3702 2 >>
             3 5 [+] • dupdip 3702 2 >>
                 3 5 • + 5 3702 2 >>
                   8 • 5 3702 2 >>
                 8 5 • 3702 2 >>
            8 5 3702 • 2 >>
          8 5 3702 2 • >>
             8 5 925 •
In [17]:
V('0 0 14811 7 [PE1.2] times pop')
                      • 0 0 14811 7 [PE1.2] times pop
                    0 • 0 14811 7 [PE1.2] times pop
                  0 0 • 14811 7 [PE1.2] times pop
            0 0 14811 • 7 [PE1.2] times pop
          0 0 14811 7 • [PE1.2] times pop
  0 0 14811 7 [PE1.2] • times pop
            0 0 14811 • PE1.2 6 [PE1.2] times pop
            0 0 14811 • [3 & PE1.1] dupdip 2 >> 6 [PE1.2] times pop
0 0 14811 [3 & PE1.1] • dupdip 2 >> 6 [PE1.2] times pop
            0 0 14811 • 3 & PE1.1 14811 2 >> 6 [PE1.2] times pop
          0 0 14811 3 • & PE1.1 14811 2 >> 6 [PE1.2] times pop
                0 0 3 • PE1.1 14811 2 >> 6 [PE1.2] times pop
                0 0 3 • + [+] dupdip 14811 2 >> 6 [PE1.2] times pop
                  0 3 • [+] dupdip 14811 2 >> 6 [PE1.2] times pop
              0 3 [+] • dupdip 14811 2 >> 6 [PE1.2] times pop
                  0 3 • + 3 14811 2 >> 6 [PE1.2] times pop
                    3 • 3 14811 2 >> 6 [PE1.2] times pop
                  3 3 • 14811 2 >> 6 [PE1.2] times pop
            3 3 14811 • 2 >> 6 [PE1.2] times pop
          3 3 14811 2 • >> 6 [PE1.2] times pop
             3 3 3702 • 6 [PE1.2] times pop
           3 3 3702 6 • [PE1.2] times pop
   3 3 3702 6 [PE1.2] • times pop
             3 3 3702 • PE1.2 5 [PE1.2] times pop
             3 3 3702 • [3 & PE1.1] dupdip 2 >> 5 [PE1.2] times pop
 3 3 3702 [3 & PE1.1] • dupdip 2 >> 5 [PE1.2] times pop
             3 3 3702 • 3 & PE1.1 3702 2 >> 5 [PE1.2] times pop
           3 3 3702 3 • & PE1.1 3702 2 >> 5 [PE1.2] times pop
                3 3 2 • PE1.1 3702 2 >> 5 [PE1.2] times pop
                3 3 2 • + [+] dupdip 3702 2 >> 5 [PE1.2] times pop
                  3 5 • [+] dupdip 3702 2 >> 5 [PE1.2] times pop
              3 5 [+] • dupdip 3702 2 >> 5 [PE1.2] times pop
                  3 5 • + 5 3702 2 >> 5 [PE1.2] times pop
                    8 • 5 3702 2 >> 5 [PE1.2] times pop
                  8 5 • 3702 2 >> 5 [PE1.2] times pop
             8 5 3702 • 2 >> 5 [PE1.2] times pop
           8 5 3702 2 • >> 5 [PE1.2] times pop
              8 5 925 • 5 [PE1.2] times pop
            8 5 925 5 • [PE1.2] times pop
    8 5 925 5 [PE1.2] • times pop
              8 5 925 • PE1.2 4 [PE1.2] times pop
              8 5 925 • [3 & PE1.1] dupdip 2 >> 4 [PE1.2] times pop
  8 5 925 [3 & PE1.1] • dupdip 2 >> 4 [PE1.2] times pop
              8 5 925 • 3 & PE1.1 925 2 >> 4 [PE1.2] times pop
            8 5 925 3 • & PE1.1 925 2 >> 4 [PE1.2] times pop
                8 5 1 • PE1.1 925 2 >> 4 [PE1.2] times pop
                8 5 1 • + [+] dupdip 925 2 >> 4 [PE1.2] times pop
                  8 6 • [+] dupdip 925 2 >> 4 [PE1.2] times pop
              8 6 [+] • dupdip 925 2 >> 4 [PE1.2] times pop
                  8 6 • + 6 925 2 >> 4 [PE1.2] times pop
                   14 • 6 925 2 >> 4 [PE1.2] times pop
                 14 6 • 925 2 >> 4 [PE1.2] times pop
             14 6 925 • 2 >> 4 [PE1.2] times pop
           14 6 925 2 • >> 4 [PE1.2] times pop
             14 6 231 • 4 [PE1.2] times pop
           14 6 231 4 • [PE1.2] times pop
   14 6 231 4 [PE1.2] • times pop
             14 6 231 • PE1.2 3 [PE1.2] times pop
             14 6 231 • [3 & PE1.1] dupdip 2 >> 3 [PE1.2] times pop
 14 6 231 [3 & PE1.1] • dupdip 2 >> 3 [PE1.2] times pop
             14 6 231 • 3 & PE1.1 231 2 >> 3 [PE1.2] times pop
           14 6 231 3 • & PE1.1 231 2 >> 3 [PE1.2] times pop
               14 6 3 • PE1.1 231 2 >> 3 [PE1.2] times pop
               14 6 3 • + [+] dupdip 231 2 >> 3 [PE1.2] times pop
                 14 9 • [+] dupdip 231 2 >> 3 [PE1.2] times pop
             14 9 [+] • dupdip 231 2 >> 3 [PE1.2] times pop
                 14 9 • + 9 231 2 >> 3 [PE1.2] times pop
                   23 • 9 231 2 >> 3 [PE1.2] times pop
                 23 9 • 231 2 >> 3 [PE1.2] times pop
             23 9 231 • 2 >> 3 [PE1.2] times pop
           23 9 231 2 • >> 3 [PE1.2] times pop
              23 9 57 • 3 [PE1.2] times pop
            23 9 57 3 • [PE1.2] times pop
    23 9 57 3 [PE1.2] • times pop
              23 9 57 • PE1.2 2 [PE1.2] times pop
              23 9 57 • [3 & PE1.1] dupdip 2 >> 2 [PE1.2] times pop
  23 9 57 [3 & PE1.1] • dupdip 2 >> 2 [PE1.2] times pop
              23 9 57 • 3 & PE1.1 57 2 >> 2 [PE1.2] times pop
            23 9 57 3 • & PE1.1 57 2 >> 2 [PE1.2] times pop
               23 9 1 • PE1.1 57 2 >> 2 [PE1.2] times pop
               23 9 1 • + [+] dupdip 57 2 >> 2 [PE1.2] times pop
                23 10 • [+] dupdip 57 2 >> 2 [PE1.2] times pop
            23 10 [+] • dupdip 57 2 >> 2 [PE1.2] times pop
                23 10 • + 10 57 2 >> 2 [PE1.2] times pop
                   33 • 10 57 2 >> 2 [PE1.2] times pop
                33 10 • 57 2 >> 2 [PE1.2] times pop
             33 10 57 • 2 >> 2 [PE1.2] times pop
           33 10 57 2 • >> 2 [PE1.2] times pop
             33 10 14 • 2 [PE1.2] times pop
           33 10 14 2 • [PE1.2] times pop
   33 10 14 2 [PE1.2] • times pop
             33 10 14 • PE1.2 1 [PE1.2] times pop
             33 10 14 • [3 & PE1.1] dupdip 2 >> 1 [PE1.2] times pop
 33 10 14 [3 & PE1.1] • dupdip 2 >> 1 [PE1.2] times pop
             33 10 14 • 3 & PE1.1 14 2 >> 1 [PE1.2] times pop
           33 10 14 3 • & PE1.1 14 2 >> 1 [PE1.2] times pop
              33 10 2 • PE1.1 14 2 >> 1 [PE1.2] times pop
              33 10 2 • + [+] dupdip 14 2 >> 1 [PE1.2] times pop
                33 12 • [+] dupdip 14 2 >> 1 [PE1.2] times pop
            33 12 [+] • dupdip 14 2 >> 1 [PE1.2] times pop
                33 12 • + 12 14 2 >> 1 [PE1.2] times pop
                   45 • 12 14 2 >> 1 [PE1.2] times pop
                45 12 • 14 2 >> 1 [PE1.2] times pop
             45 12 14 • 2 >> 1 [PE1.2] times pop
           45 12 14 2 • >> 1 [PE1.2] times pop
              45 12 3 • 1 [PE1.2] times pop
            45 12 3 1 • [PE1.2] times pop
    45 12 3 1 [PE1.2] • times pop
              45 12 3 • PE1.2 pop
              45 12 3 • [3 & PE1.1] dupdip 2 >> pop
  45 12 3 [3 & PE1.1] • dupdip 2 >> pop
              45 12 3 • 3 & PE1.1 3 2 >> pop
            45 12 3 3 • & PE1.1 3 2 >> pop
              45 12 3 • PE1.1 3 2 >> pop
              45 12 3 • + [+] dupdip 3 2 >> pop
                45 15 • [+] dupdip 3 2 >> pop
            45 15 [+] • dupdip 3 2 >> pop
                45 15 • + 15 3 2 >> pop
                   60 • 15 3 2 >> pop
                60 15 • 3 2 >> pop
              60 15 3 • 2 >> pop
            60 15 3 2 • >> pop
              60 15 0 • pop
                60 15 •

And so we have at last:

In [18]:
define('PE1 0 0 66 [14811 7 [PE1.2] times pop] times 14811 4 [PE1.2] times popop')
In [19]:
J('PE1')
233168

Let's refactor.

  14811 7 [PE1.2] times pop
  14811 4 [PE1.2] times pop
  14811 n [PE1.2] times pop
n 14811 swap [PE1.2] times pop
In [20]:
define('PE1.3 14811 swap [PE1.2] times pop')

Now we can simplify the definition above:

In [21]:
define('PE1 0 0 66 [7 PE1.3] times 4 PE1.3 pop')
In [22]:
J('PE1')
233168

Here's our joy program all in one place. It doesn't make so much sense, but if you have read through the above description of how it was derived I hope it's clear.

PE1.1 == + [+] dupdip
PE1.2 == [3 & PE1.1] dupdip 2 >>
PE1.3 == 14811 swap [PE1.2] times pop
PE1 == 0 0 66 [7 PE1.3] times 4 PE1.3 pop

Generator Version

It's a little clunky iterating sixty-six times though the seven numbers then four more. In the Generator Programs notebook we derive a generator that can be repeatedly driven by the x combinator to produce a stream of the seven numbers repeating over and over again.

In [23]:
define('PE1.terms [0 swap [dup [pop 14811] [] branch [3 &] dupdip 2 >>] dip rest cons]')
In [24]:
J('PE1.terms 21 [x] times')
3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 [0 swap [dup [pop 14811] [] branch [3 &] dupdip 2 >>] dip rest cons]

We know from above that we need sixty-six times seven then four more terms to reach up to but not over one thousand.

In [25]:
J('7 66 * 4 +')
466

Here they are...

In [26]:
J('PE1.terms 466 [x] times pop')
3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3

...and they do sum to 999.

In [27]:
J('[PE1.terms 466 [x] times pop] run sum')
999

Now we can use PE1.1 to accumulate the terms as we go, and then pop the generator and the counter from the stack when we're done, leaving just the sum.

In [28]:
J('0 0 PE1.terms 466 [x [PE1.1] dip] times popop')
233168

A little further analysis renders iteration unnecessary.

Consider finding the sum of the positive integers less than or equal to ten.

In [29]:
J('[10 9 8 7 6 5 4 3 2 1] sum')
55

Instead of summing them, observe:

  10  9  8  7  6
+  1  2  3  4  5
---- -- -- -- --
  11 11 11 11 11
  
  11 * 5 = 55

From the above example we can deduce that the sum of the first N positive integers is:

(N + 1) * N / 2

(The formula also works for odd values of N, I'll leave that to you if you want to work it out or you can take my word for it.)

In [30]:
define('F dup ++ * 2 floordiv')
In [31]:
V('10 F')
      • 10 F
   10 • F
   10 • dup ++ * 2 floordiv
10 10 • ++ * 2 floordiv
10 11 • * 2 floordiv
  110 • 2 floordiv
110 2 • floordiv
   55 •

Generalizing to Blocks of Terms

We can apply the same reasoning to the PE1 problem.

Between 0 and 990 inclusive there are sixty-six "blocks" of seven terms each, starting with:

[3 5 6 9 10 12 15]

And ending with:

[978 980 981 984 985 987 990]

If we reverse one of these two blocks and sum pairs...

In [32]:
J('[3 5 6 9 10 12 15] reverse [978 980 981 984 985 987 990] zip')
[[978 15] [980 12] [981 10] [984 9] [985 6] [987 5] [990 3]]
In [33]:
J('[3 5 6 9 10 12 15] reverse [978 980 981 984 985 987 990] zip [sum] map')
[993 992 991 993 991 992 993]

(Interesting that the sequence of seven numbers appears again in the rightmost digit of each term.)

In [34]:
J('[ 3 5 6 9 10 12 15] reverse [978 980 981 984 985 987 990] zip [sum] map sum')
6945

Since there are sixty-six blocks and we are pairing them up, there must be thirty-three pairs, each of which sums to 6945. We also have these additional unpaired terms between 990 and 1000:

993 995 996 999

So we can give the "sum of all the multiples of 3 or 5 below 1000" like so:

In [35]:
J('6945 33 * [993 995 996 999] cons sum')
233168

It's worth noting, I think, that this same reasoning holds for any two numbers n and m the multiples of which we hope to sum. The multiples would have a cycle of differences of length k and so we could compute the sum of Nk multiples as above.

The sequence of differences will always be a palidrome. Consider an interval spanning the least common multiple of n and m:

|   |   |   |   |   |   |   |
|      |      |      |      |

Here we have 4 and 7, and you can read off the sequence of differences directly from the diagram: 4 3 1 4 2 2 4 1 3 4.

Geometrically, the actual values of n and m and their lcm don't matter, the pattern they make will always be symmetrical around its midpoint. The same reasoning holds for multiples of more than two numbers.

The Simplest Program

Of course, the simplest joy program for the first Project Euler problem is just:

PE1 == 233168

Fin.

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