# 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

In [8]:

    66 * 15

Out[8]:

    990

In [9]:

    1000 - 990

Out[9]:

    10

We only want the terms less than 1000.

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] . i 6 [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] . i 5 [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] . i 4 [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] . i 3 [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] . i 2 [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] . i 1 [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] . i 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.