As a first attempt at writing software in Joy let's tackle 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.
Let's create a predicate that returns true if a number is a multiple of 3 or 5 and false otherwise.
multiple-ofFirst we write multiple-of which take two numbers and return true if the first is a multiple of the second. We use the mod operator, convert the remainder to a Boolean value, then invert it to get our answer:
[multiple-of mod bool not] inscribe
multiple-of-3-or-5Next we can use that with app2 to get both Boolean values (for 3 and 5) and then use the logical OR function \/ on them. (We use popd to get rid of the original number):
[multiple-of-3-or-5 3 5 [multiple-of] app2 \/ popd] inscribe
Here it is in action:
joy? [6 7 8 9 10] [multiple-of-3-or-5] map
[true false false true true]
Given the predicate function multiple-of-3-or-5 a suitable program would be:
1000 range [multiple-of-3-or-5] filter sum
This function generates a list of the integers from 0 to 999, filters
that list by multiple-of-3-or-5, and then sums the result. (I should
mention that the filter function has not yet been implemented.)
Logically this is fine, but pragmatically we are doing more work than we should. 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:
+ [+] dupdip
We start with a sum, the counter, and a term to add:
joy? 0 0 3
0 0 3
Here is our function, quoted to let it be run with the trace combinator
(which is like i but it also prints a trace of evaluation to stdout.)
joy? [+ [+] dupdip]
0 0 3 [+ [+] dupdip]
And here we go:
joy? trace
0 0 3 • + [+] dupdip
0 3 • [+] dupdip
0 3 [+] • dupdip
0 3 • + 3
3 • 3
3 3 •
3 3
PE1.1We can inscribe it for use in later definitions:
[PE1.1 + [+] dupdip] inscribe
Let's try it out on our palindromic sequence:
joy? 0 0 [3 2 1 3 1 2 3] [PE1.1] step
60 15
So one step through all seven terms brings the counter to 15 and the total to 60.
We want all the terms less than 1000, and each pass through the palindromic sequence will count off 15 so how many is that?
joy? 1000 15 /
66
So 66 × 15 bring us to...
66
joy? 15 *
990
That means we want to run the full list of numbers sixty-six times to get to 990 and then, obviously, the first four terms of the palindromic sequence, 3 2 1 3, to get to 999.
Start with the sum and counter:
joy? 0 0
0 0
We will run a program sixty-six times:
joy? 66
0 0 66
This is the program we will run sixty-six times, it steps through the palindromic sequence and sums up the terms:
joy? [[3 2 1 3 1 2 3] [PE1.1] step]
0 0 66 [[3 2 1 3 1 2 3] [PE1.1] step]
Runing that brings us to the sum of the numbers less than 991:
joy? times
229185 990
We need to count 9 more to reach the sum of the numbers less than 1000:
joy? [3 2 1 3] [PE1.1] step
233168 999
All that remains is the counter, which we can discard:
joy? pop
233168
And so we have our answer: 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 for themselves and a few bytes for their list. 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.
Let's encode the term in 7 × 2 = 14 bits:
Decimal: 3 2 1 3 1 2 3
Binary: 11 10 01 11 01 10 11
The number 11100111011011 in binary is 14811 in decimal notation.
We can recover the terms from this number by 4 divmod:
joy? 14811
14811
joy? 4 divmod
3702 3
We want the term below the rest of the terms:
joy? swap
3 3702
PE1.2Giving us 4 divmod swap:
[PE1.2 4 divmod swap] inscribe
joy? 14811
14811
joy? PE1.2
3 3702
joy? PE1.2
3 2 925
joy? PE1.2
3 2 1 231
joy? PE1.2
3 2 1 3 57
joy? PE1.2
3 2 1 3 1 14
joy? PE1.2
3 2 1 3 1 2 3
joy? PE1.2
3 2 1 3 1 2 3 0
So we want:
joy? 0 0 14811 PE1.2
0 0 3 3702
And then:
joy? [PE1.1] dip
3 3 3702
Continuing:
joy? PE1.2 [PE1.1] dip
8 5 925
joy? PE1.2 [PE1.1] dip
14 6 231
joy? PE1.2 [PE1.1] dip
23 9 57
joy? PE1.2 [PE1.1] dip
33 10 14
joy? PE1.2 [PE1.1] dip
45 12 3
joy? PE1.2 [PE1.1] dip
60 15 0
PE1.3Let's define:
[PE1.3 PE1.2 [PE1.1] dip] inscribe
Now:
14811 7 [PE1.3] times pop
Will add up one set of the palindromic sequence of terms:
joy? 0 0
0 0
joy? 14811 7 [PE1.3] times pop
60 15
And we want to do that sixty-six times:
joy? 0 0
0 0
joy? 66 [14811 7 [PE1.3] times pop] times
229185 990
And then four more:
joy? 14811 4 [PE1.3] times pop
233168 999
And discard the counter:
joy? pop
233168
Violà!
From these two:
14811 7 [PE1.3] times pop
14811 4 [PE1.3] times pop
We can generalize the loop counter:
14811 n [PE1.3] times pop
And use swap to put it to the left...
n 14811 swap [PE1.3] times pop
PE1.4...and so we have a new definition:
[PE1.4 14811 swap [PE1.3] times pop] inscribe
Now we can simplify the program from:
0 0 66 [14811 7 [PE1.3] times pop] times 14811 4 [PE1.3] times pop pop
To:
0 0 66 [7 PE1.4] times 4 PE1.4 pop
Let's run it and see:
joy? 0 0 66 [7 PE1.4] times 4 PE1.4 pop
233168
PE1[PE1 0 0 66 [7 PE1.4] times 4 PE1.4 pop] inscribe
joy? PE1
233168
Here's our joy program all in one place, as it might appear in def.txt:
PE1.1 + [+] dupdip
PE1.2 4 divmod swap
PE1.3 PE1.2 [PE1.1] dip
PE1.4 14811 swap [PE1.3] times pop
PE1 0 0 66 [7 PE1.4] times 4 PE1.4 pop
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.
Here it is again:
[0 swap [? [pop 14811] [] branch PE1.2] dip rest cons]
This self-referential quote contains a bit of state (the initial 0) and a "step"
function (the rest of the quote) and can be "forced" by the x combinator to
produce successive terms of our palindromic sequence. The branch sub-expression
resets the integer that encodes the terms when it reaches 0.
Let's inscribe this generator quote to keep it handy.
PE1.terms[PE1.terms [0 swap [? [pop 14811] [] branch PE1.2] dip rest cons]] inscribe
Let's try it out:
joy? PE1.terms 21 [x] times pop
3 2 1 3 1 2 3 3 2 1 3 1 2 3 3 2 1 3 1 2 3
Pretty neat, eh?
We know from above that we need sixty-six times seven then four more terms to reach up to but not over one thousand.
joy? 7 66 * 4 +
466
joy? 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
joy? [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.
joy? 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.
joy? [10 9 8 7 6 5 4 3 2 1] sum
55
Gauss famously showed that you can find the sum directly with a simple equation.
10 9 8 7 6
+ 1 2 3 4 5
---- -- -- -- --
11 11 11 11 11
11 × 5 = 55
The sum of the first N positive integers is:
(𝑛 + 1) × 𝑛 / 2
In Joy this equation could be expressed as:
dup ++ * 2 /
(Note that (𝑛 + 1) × 𝑛 will always be an even number.)
We can apply the same reasoning to the PE1 problem.
Recall that between 1 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...
joy? [3 5 6 9 10 12 15] reverse
[15 12 10 9 6 5 3]
joy? [978 980 981 984 985 987 990]
[15 12 10 9 6 5 3] [978 980 981 984 985 987 990]
joy? zip
[[978 15] [980 12] [981 10] [984 9] [985 6] [987 5] [990 3]]
joy? [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.)
...and then sum the sums...
joy? sum
6945
We arrive at 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.
6945
joy? 33 *
229185
We also have those four additional terms between 990 and 1000, they are unpaired:
993 995 996 999
I think we can just use those as is, so we can give the "sum of all the multiples of 3 or 5 below 1000" like so:
joy? 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 palindrome. 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 least common multiple 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.
Of course, having done all that, the simplest joy program for the first Project Euler problem is just:
[PE1 233168] inscribe
Fin.