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to not use Jupyter because I can't figure out how to use the Joy kernel.
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<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Developing a Joy Program</title>
<link rel="stylesheet" href="/css/fonts.css">
<link rel="stylesheet" href="/css/site.css">
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<body>
<h1>Developing a Joy Program</h1>
<p>As a first attempt at writing software in Joy let's tackle
<a href="https://projecteuler.net/problem=1">Project Euler, first problem: "Multiples of 3 and 5"</a>:</p>
<blockquote>
<p>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.</p>
<p>Find the sum of all the multiples of 3 or 5 below 1000.</p>
</blockquote>
<h2>Filter</h2>
<p>Let's create a predicate that returns <code>true</code> if a number is a multiple of 3 or 5 and <code>false</code> otherwise.</p>
<h3><code>multiple-of</code></h3>
<p>First we write <code>multiple-of</code> which take two numbers and return true if the first is a multiple of the second. We use the <code>mod</code> operator, convert the remainder to a Boolean value, then invert it to get our answer:</p>
<pre><code>[multiple-of mod bool not] inscribe
</code></pre>
<h3><code>multiple-of-3-or-5</code></h3>
<p>Next we can use that with <code>app2</code> to get both Boolean values (for 3 and 5) and then use the logical OR function <code>\/</code> on them. (We use <code>popd</code> to get rid of the original number):</p>
<pre><code>[multiple-of-3-or-5 3 5 [multiple-of] app2 \/ popd] inscribe
</code></pre>
<p>Here it is in action:</p>
<pre><code>joy? [6 7 8 9 10] [multiple-of-3-or-5] map
[true false false true true]
</code></pre>
<p>Given the predicate function <code>multiple-of-3-or-5</code> a suitable program would be:</p>
<pre><code>1000 range [multiple-of-3-or-5] filter sum
</code></pre>
<p>This function generates a list of the integers from 0 to 999, filters
that list by <code>multiple-of-3-or-5</code>, and then sums the result. (I should
mention that the <code>filter</code> function has not yet been implemented.)</p>
<p>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.</p>
<p>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.</p>
<p>This seemed a little clunky, so I tried a different approach.</p>
<h2>Looking for Pattern</h2>
<p>Consider the first few terms in the series:</p>
<pre><code>3 5 6 9 10 12 15 18 20 21 ...
</code></pre>
<p>Subtract each number from the one after it (subtracting 0 from 3):</p>
<pre><code>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 ...
</code></pre>
<p>You get this lovely repeating palindromic sequence:</p>
<pre><code>3 2 1 3 1 2 3
</code></pre>
<p>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.</p>
<h3>Increment a Counter</h3>
<p>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:</p>
<pre><code>+ [+] dupdip
</code></pre>
<p>We start with a sum, the counter, and a term to add:</p>
<pre><code>joy? 0 0 3
0 0 3
</code></pre>
<p>Here is our function, quoted to let it be run with the <code>trace</code> combinator
(which is like <code>i</code> but it also prints a trace of evaluation to <code>stdout</code>.)</p>
<pre><code>joy? [+ [+] dupdip]
0 0 3 [+ [+] dupdip]
</code></pre>
<p>And here we go:</p>
<pre><code>joy? trace
0 0 3 • + [+] dupdip
0 3 • [+] dupdip
0 3 [+] • dupdip
0 3 • + 3
3 • 3
3 3 •
3 3
</code></pre>
<h3><code>PE1.1</code></h3>
<p>We can <code>inscribe</code> it for use in later definitions:</p>
<pre><code>[PE1.1 + [+] dupdip] inscribe
</code></pre>
<p>Let's try it out on our palindromic sequence:</p>
<pre><code>joy? 0 0 [3 2 1 3 1 2 3] [PE1.1] step
60 15
</code></pre>
<p>So one <code>step</code> through all seven terms brings the counter to 15 and the total to 60.</p>
<h3>How Many Times?</h3>
<p>We want all the terms less than 1000, and each pass through the palindromic sequence
will count off 15 so how many is that?</p>
<pre><code>joy? 1000 15 /
66
</code></pre>
<p>So 66 × 15 bring us to...</p>
<pre><code>66
joy? 15 *
990
</code></pre>
<p>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.</p>
<p>Start with the sum and counter:</p>
<pre><code>joy? 0 0
0 0
</code></pre>
<p>We will run a program sixty-six times:</p>
<pre><code>joy? 66
0 0 66
</code></pre>
<p>This is the program we will run sixty-six times, it steps through the
palindromic sequence and sums up the terms:</p>
<pre><code>joy? [[3 2 1 3 1 2 3] [PE1.1] step]
0 0 66 [[3 2 1 3 1 2 3] [PE1.1] step]
</code></pre>
<p>Runing that brings us to the sum of the numbers less than 991:</p>
<pre><code>joy? times
229185 990
</code></pre>
<p>We need to count 9 more to reach the sum of the numbers less than 1000:</p>
<pre><code>joy? [3 2 1 3] [PE1.1] step
233168 999
</code></pre>
<p>All that remains is the counter, which we can discard:</p>
<pre><code>joy? pop
233168
</code></pre>
<p>And so we have our answer: <strong>233168</strong></p>
<p>This form uses no extra storage and produces no unused summands. It's
good but there's one more trick we can apply.</p>
<h2>A Slight Increase of Efficiency</h2>
<p>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.</p>
<p>Let's encode the term in 7 × 2 = 14 bits:</p>
<pre><code>Decimal: 3 2 1 3 1 2 3
Binary: 11 10 01 11 01 10 11
</code></pre>
<p>The number 11100111011011 in binary is 14811 in decimal notation.</p>
<h3>Recovering the Terms</h3>
<p>We can recover the terms from this number by <code>4 divmod</code>:</p>
<pre><code>joy? 14811
14811
joy? 4 divmod
3702 3
</code></pre>
<p>We want the term below the rest of the terms:</p>
<pre><code>joy? swap
3 3702
</code></pre>
<h3><code>PE1.2</code></h3>
<p>Giving us <code>4 divmod swap</code>:</p>
<pre><code>[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
</code></pre>
<p>So we want:</p>
<pre><code>joy? 0 0 14811 PE1.2
0 0 3 3702
</code></pre>
<p>And then:</p>
<pre><code>joy? [PE1.1] dip
3 3 3702
</code></pre>
<p>Continuing:</p>
<pre><code>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
</code></pre>
<h3><code>PE1.3</code></h3>
<p>Let's define:</p>
<pre><code>[PE1.3 PE1.2 [PE1.1] dip] inscribe
</code></pre>
<p>Now:</p>
<pre><code>14811 7 [PE1.3] times pop
</code></pre>
<p>Will add up one set of the palindromic sequence of terms:</p>
<pre><code>joy? 0 0
0 0
joy? 14811 7 [PE1.3] times pop
60 15
</code></pre>
<p>And we want to do that sixty-six times:</p>
<pre><code>joy? 0 0
0 0
joy? 66 [14811 7 [PE1.3] times pop] times
229185 990
</code></pre>
<p>And then four more:</p>
<pre><code>joy? 14811 4 [PE1.3] times pop
233168 999
</code></pre>
<p>And discard the counter:</p>
<pre><code>joy? pop
233168
</code></pre>
<p><em>Violà!</em></p>
<h3>Let's refactor.</h3>
<p>From these two:</p>
<pre><code>14811 7 [PE1.3] times pop
14811 4 [PE1.3] times pop
</code></pre>
<p>We can generalize the loop counter:</p>
<pre><code>14811 n [PE1.3] times pop
</code></pre>
<p>And use <code>swap</code> to put it to the left...</p>
<pre><code>n 14811 swap [PE1.3] times pop
</code></pre>
<h3><code>PE1.4</code></h3>
<p>...and so we have a new definition:</p>
<pre><code>[PE1.4 14811 swap [PE1.3] times pop] inscribe
</code></pre>
<p>Now we can simplify the program from:</p>
<pre><code>0 0 66 [14811 7 [PE1.3] times pop] times 14811 4 [PE1.3] times pop pop
</code></pre>
<p>To:</p>
<pre><code>0 0 66 [7 PE1.4] times 4 PE1.4 pop
</code></pre>
<p>Let's run it and see:</p>
<pre><code>joy? 0 0 66 [7 PE1.4] times 4 PE1.4 pop
233168
</code></pre>
<h3><code>PE1</code></h3>
<pre><code>[PE1 0 0 66 [7 PE1.4] times 4 PE1.4 pop] inscribe
joy? PE1
233168
</code></pre>
<p>Here's our joy program all in one place, as it might appear in <code>def.txt</code>:</p>
<pre><code>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
</code></pre>
<h2>Generator Version</h2>
<p>It's a little clunky iterating sixty-six times though the seven numbers then four more. In the <em>Generator Programs</em> notebook we derive a generator that can be repeatedly driven by the <code>x</code> combinator to produce a stream of the seven numbers repeating over and over again.</p>
<p>Here it is again:</p>
<pre><code>[0 swap [? [pop 14811] [] branch PE1.2] dip rest cons]
</code></pre>
<p>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 <code>x</code> combinator to
produce successive terms of our palindromic sequence. The <code>branch</code> sub-expression
resets the integer that encodes the terms when it reaches 0.</p>
<p>Let's <code>inscribe</code> this generator quote to keep it handy.</p>
<h3><code>PE1.terms</code></h3>
<pre><code>[PE1.terms [0 swap [? [pop 14811] [] branch PE1.2] dip rest cons]] inscribe
</code></pre>
<p>Let's try it out:</p>
<pre><code>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
</code></pre>
<p>Pretty neat, eh?</p>
<h3>How Many Terms?</h3>
<p>We know from above that we need sixty-six times seven then four more terms to reach up to but not over one thousand.</p>
<pre><code>joy? 7 66 * 4 +
466
</code></pre>
<h3>Here they are...</h3>
<pre><code>joy? PE1.terms 466 [x] times pop
</code></pre>
<blockquote>
<p>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</p>
</blockquote>
<h3>...and they do sum to 999.</h3>
<pre><code>joy? [PE1.terms 466 [x] times pop] run sum
999
</code></pre>
<p>Now we can use <code>PE1.1</code> to accumulate the terms as we go, and then <code>pop</code> the
generator and the counter from the stack when we're done, leaving just the sum.</p>
<pre><code>joy? 0 0 PE1.terms 466 [x [PE1.1] dip] times popop
233168
</code></pre>
<h2>A Little Further Analysis...</h2>
<p>A little further analysis renders iteration unnecessary.
Consider finding the sum of the positive integers less than or equal to ten.</p>
<pre><code>joy? [10 9 8 7 6 5 4 3 2 1] sum
55
</code></pre>
<p>Gauss famously showed that you can find the sum directly with a simple equation.</p>
<p><a href="https://en.wikipedia.org/wiki/File:Animated_proof_for_the_formula_giving_the_sum_of_the_first_integers_1%2B2%2B...%2Bn.gif">Observe</a>:</p>
<pre><code> 10 9 8 7 6
+ 1 2 3 4 5
---- -- -- -- --
11 11 11 11 11
11 × 5 = 55
</code></pre>
<p>The sum of the first N positive integers is:</p>
<pre><code>(𝑛 + 1) × 𝑛 / 2
</code></pre>
<p>In Joy this equation could be expressed as:</p>
<pre><code>dup ++ * 2 /
</code></pre>
<p>(Note that <code>(𝑛 + 1) × 𝑛</code> will always be an even number.)</p>
<h3>Generalizing to Blocks of Terms</h3>
<p>We can apply the same reasoning to the <code>PE1</code> problem.</p>
<p>Recall that between 1 and 990 inclusive there are sixty-six "blocks" of seven terms each, starting with:</p>
<pre><code>3 5 6 9 10 12 15
</code></pre>
<p>And ending with:</p>
<pre><code>978 980 981 984 985 987 990
</code></pre>
<p>If we reverse one of these two blocks and sum pairs...</p>
<pre><code>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]
</code></pre>
<p>(Interesting that the sequence of seven numbers appears again in the rightmost digit of each term.)</p>
<p>...and then sum the sums...</p>
<pre><code>joy? sum
6945
</code></pre>
<p>We arrive at 6945.</p>
<h3>Pair Up the Blocks</h3>
<p>Since there are sixty-six blocks and we are pairing them up, there must be thirty-three pairs, each of which sums to 6945.</p>
<pre><code>6945
joy? 33 *
229185
</code></pre>
<p>We also have those four additional terms between 990 and 1000, they are unpaired:</p>
<pre><code>993 995 996 999
</code></pre>
<h3>A Simple Solution</h3>
<p>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:</p>
<pre><code>joy? 6945 33 * [993 995 996 999] cons sum
233168
</code></pre>
<h3>Generalizing</h3>
<p>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.</p>
<p>The sequence of differences will always be a palindrome. Consider an interval spanning the least common multiple of 𝑛 and 𝑚:</p>
<pre><code>| | | | | | | |
| | | | |
</code></pre>
<p>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.</p>
<p>Geometrically, the actual values of 𝑛 and 𝑚 and their <em>least common multiple</em> 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.</p>
<h1>The Simplest Program</h1>
<p>Of course, having done all that, the simplest joy program for the first Project Euler problem is just:</p>
<pre><code>[PE1 233168] inscribe
</code></pre>
<p>Fin.</p>
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# Developing a Joy Program
As a first attempt at writing software in Joy let's tackle
[Project Euler, first problem: "Multiples of 3 and 5"](https://projecteuler.net/problem=1):
> 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.
## Filter
Let's create a predicate that returns `true` if a number is a multiple of 3 or 5 and `false` otherwise.
### `multiple-of`
First 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-5`
Next 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.
## Looking for Pattern
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.
### Increment a Counter
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
### `PE1.1`
We 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.
### How Many Times?
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.
## A Slight Increase of Efficiency
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.
### Recovering the Terms
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
### `PE1.2`
Giving 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
### `PE1.3`
Let'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!*
### Let's refactor.
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
## 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.
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?
### How Many Terms?
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
### Here they are...
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
### ...and they do sum to 999.
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...
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.
[Observe](https://en.wikipedia.org/wiki/File:Animated_proof_for_the_formula_giving_the_sum_of_the_first_integers_1%2B2%2B...%2Bn.gif):
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.)
### Generalizing to Blocks of Terms
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.
### Pair Up the Blocks
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
### A Simple Solution
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
### Generalizing
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.
# The Simplest Program
Of course, having done all that, the simplest joy program for the first Project Euler problem is just:
[PE1 233168] inscribe
Fin.

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# Generator Programs
## Using `x` to Generate Values
Cf. [Self-reproducing and reproducing programs](https://www.kevinalbrecht.com/code/joy-mirror/jp-reprod.html) by Manfred von Thun
Consider the `x` combinator:
x == dup i
We can apply it to a quoted program consisting of some value `a` and some function `B`:
[a B] x
[a B] a B
Let `B` function `swap` the `a` with the quote and run some function `C` on it to generate a new value `b`:
B == swap [C] dip
[a B] a B
[a B] a swap [C] dip
a [a B] [C] dip
a C [a B]
b [a B]
Now discard the quoted `a` with `rest` then `cons` `b`:
b [a B] rest cons
b [B] cons
[b B]
Altogether, this is the definition of `B`:
B == swap [C] dip rest cons
## An Example
We can make a generator for the Natural numbers (0, 1, 2, ...) by using
`0` for the initial state `a` and `[dup ++]` for `[C]`.
We need the `dup` to leave the old state value behind on the stack.
Putting it together:
[0 swap [dup ++] dip rest cons]
Let's try it:
joy? [0 swap [dup ++] dip rest cons]
[0 swap [dup ++] dip rest cons]
joy? [x]
[0 swap [dup ++] dip rest cons] [x]
joy? trace
[0 swap [dup ++] dip rest cons] • x
[0 swap [dup ++] dip rest cons] • 0 swap [dup ++] dip rest cons
[0 swap [dup ++] dip rest cons] 0 • swap [dup ++] dip rest cons
0 [0 swap [dup ++] dip rest cons] • [dup ++] dip rest cons
0 [0 swap [dup ++] dip rest cons] [dup ++] • dip rest cons
0 • dup ++ [0 swap [dup ++] dip rest cons] rest cons
0 0 • ++ [0 swap [dup ++] dip rest cons] rest cons
0 0 • 1 + [0 swap [dup ++] dip rest cons] rest cons
0 0 1 • + [0 swap [dup ++] dip rest cons] rest cons
0 1 • [0 swap [dup ++] dip rest cons] rest cons
0 1 [0 swap [dup ++] dip rest cons] • rest cons
0 1 [swap [dup ++] dip rest cons] • cons
0 [1 swap [dup ++] dip rest cons] •
After one application of `x` the quoted program contains 1 and 0 is below it on the stack.
0 [1 swap [dup ++] dip rest cons]
We can use `x` as many times as we like to get as many terms as we like:
joy? x x x x x pop
0 1 2 3 4 5
### `direco`
Let's define a helper function:
[direco dip rest cons] inscribe
That makes our generator quote into:
[0 swap [dup ++] direco]
## Making Generators
We want to define a function that accepts `a` and `[C]` and builds our quoted program:
a [C] G
-------------------------
[a swap [C] direco]
Working in reverse:
[a swap [C] direco] cons
a [swap [C] direco] concat
a [swap] [[C] direco] swap
a [[C] direco] [swap]
a [C] [direco] cons [swap]
Reading from the bottom up:
[direco] cons [swap] swap concat cons
Or:
[direco] cons [swap] swoncat cons
### make-generator
[make-generator [direco] cons [swap] swoncat cons] inscribe
Let's try it out:
joy? 0 [dup ++] make-generator
[0 swap [dup ++] direco]
And generate some values:
joy? x x x pop
0 1 2
### Powers of Two
Let's generate powers of two:
joy? 1 [dup 1 <<] make-generator
[1 swap [dup 1 <<] direco]
We can drive it using `times` with the `x` combinator.
joy? 10 [x] times pop
1 2 4 8 16 32 64 128 256 512
## Generating Multiples of Three and Five
Look at the treatment of the Project Euler Problem One in the
[Developing a Program](/notebooks/Developing_a_Program.html)
notebook and you'll see that we might be interested in generating an endless cycle of:
3 2 1 3 1 2 3
To do this we want to encode the numbers as pairs of bits in a single integer:
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 using `4 divmod`.
joy? 14811 [4 divmod swap] make-generator
[14811 swap [4 divmod swap] direco]
joy? x
3 [3702 swap [4 divmod swap] direco]
joy? x
3 2 [925 swap [4 divmod swap] direco]
joy? x
3 2 1 [231 swap [4 divmod swap] direco]
joy? x
3 2 1 3 [57 swap [4 divmod swap] direco]
joy? x
3 2 1 3 1 [14 swap [4 divmod swap] direco]
joy? x
3 2 1 3 1 2 [3 swap [4 divmod swap] direco]
joy? x
3 2 1 3 1 2 3 [0 swap [4 divmod swap] direco]
joy? x
3 2 1 3 1 2 3 0 [0 swap [4 divmod swap] direco]
joy? x
3 2 1 3 1 2 3 0 0 [0 swap [4 divmod swap] direco]
joy? x
3 2 1 3 1 2 3 0 0 0 [0 swap [4 divmod swap] direco]
...we get a generator that works for seven cycles before it reaches zero.
### Reset at Zero
We need a function that checks if the int has reached zero and resets it if so.
That's easy enough to write:
? [pop 14811] [] branch
I don't like that we're checking every time even though we know we
only need to reset the integer every seventh time, but this way we
can include this function in the generator (rather than wrapping the
generator in something to do it only every seventh iteration.) So
the "forcing" function is just `x`.
### `PE1.1.check`
[PE1.1.check ? [pop 14811] [] branch] inscribe
### `PE1.1`
[PE1.1 4 divmod swap] inscribe
Now we can `make-generator`:
joy? 14811 [PE1.1.check PE1.1] make-generator
[14811 swap [PE1.1.check PE1.1] direco]
We can then "force" the generator with `x` to get as many terms as we like:
joy? 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
### Run 466 times
In the PE1 problem [PE1 problem](/notebooks/Developing_a_Program.html)
we are asked to sum all the multiples of three and five
less than 1000. It's worked out that we need to use our cycle of seven numbers
sixty-six times and then four more.
joy? 7 66 * 4 +
466
If we drive our generator 466 times and sum the stack we get 999:
joy? 14811 [PE1.1.check PE1.1] make-generator
[14811 swap [PE1.1.check PE1.1] direco]
joy? 466 [x] times pop enstacken sum
999
If you want to see how this is used read the
[Developing a Program](/notebooks/Developing_a_Program.html) notebook.
## A generator for the Fibonacci Sequence.
Consider:
[b a F] x
[b a F] b a F
The obvious first thing to do is just add `b` and `a`:
[b a F] b a +
[b a F] b+a
From here we want to arrive at:
b [b+a b F]
Let's start with `swons`:
[b a F] b+a swons
[b+a b a F]
Considering this quote as a stack:
F a b b+a
We want to get it to:
F b b+a b
So:
F a b b+a popdd over
F b b+a b
And therefore:
[b+a b a F] [popdd over] infra
[b b+a b F]
But we can just use `cons` to carry `b+a` into the quote:
[b a F] b+a [popdd over] cons infra
[b a F] [b+a popdd over] infra
[b b+a b F]
Lastly:
[b b+a b F] uncons
b [b+a b F]
Putting it all together:
F == + [popdd over] cons infra uncons
fib_gen == [1 1 F]
Let's call `F` `fib_gen`:
[fib_gen + [popdd over] cons infra uncons] inscribe
We can just write the initial quote and then "force" it with `x`:
joy? [1 1 fib_gen] 10 [x] times
1 2 3 5 8 13 21 34 55 89 [144 89 fib_gen]
It skips the first term (1) but if that bothers you you can just prepend it to the program:
1 [1 1 fib_gen] 10 [x] times
## Project Euler Problem Two
> By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
Now that we have a generator for the Fibonacci sequence, we need a function that adds a term in the sequence to a sum if it is even, and `pop`s it otherwise.
```python
define('PE2.1 == dup 2 % [+] [pop] branch')
```
And a predicate function that detects when the terms in the series "exceed four million".
```python
define('>4M == 4000000 >')
```
Now it's straightforward to define `PE2` as a recursive function that generates terms in the Fibonacci sequence until they exceed four million and sums the even ones.
```python
define('PE2 == 0 fib_gen x [pop >4M] [popop] [[PE2.1] dip x] primrec')
```
```python
J('PE2')
```
4613732
Here's the collected program definitions:
fib == + swons [popdd over] infra uncons
fib_gen == [1 1 fib]
even == dup 2 %
>4M == 4000000 >
PE2.1 == even [+] [pop] branch
PE2 == 0 fib_gen x [pop >4M] [popop] [[PE2.1] dip x] primrec
### Even-valued Fibonacci Terms
Using `o` for odd and `e` for even:
o + o = e
e + e = e
o + e = o
So the Fibonacci sequence considered in terms of just parity would be:
o o e o o e o o e o o e o o e o o e
1 1 2 3 5 8 . . .
Every third term is even.
```python
J('[1 0 fib] x x x') # To start the sequence with 1 1 2 3 instead of 1 2 3.
```
1 1 2 [3 2 fib]
Drive the generator three times and `popop` the two odd terms.
```python
J('[1 0 fib] x x x [popop] dipd')
```
2 [3 2 fib]
```python
define('PE2.2 == x x x [popop] dipd')
```
```python
J('[1 0 fib] 10 [PE2.2] times')
```
2 8 34 144 610 2584 10946 46368 196418 832040 [1346269 832040 fib]
Replace `x` with our new driver function `PE2.2` and start our `fib` generator at `1 0`.
```python
J('0 [1 0 fib] PE2.2 [pop >4M] [popop] [[PE2.1] dip PE2.2] primrec')
```
4613732
## How to compile these?
You would probably start with a special version of `G`, and perhaps modifications to the default `x`?
## An Interesting Variation
```python
define('codireco == cons dip rest cons')
```
```python
V('[0 [dup ++] codireco] x')
```
. [0 [dup ++] codireco] x
[0 [dup ++] codireco] . x
[0 [dup ++] codireco] . 0 [dup ++] codireco
[0 [dup ++] codireco] 0 . [dup ++] codireco
[0 [dup ++] codireco] 0 [dup ++] . codireco
[0 [dup ++] codireco] 0 [dup ++] . cons dip rest cons
[0 [dup ++] codireco] [0 dup ++] . dip rest cons
. 0 dup ++ [0 [dup ++] codireco] rest cons
0 . dup ++ [0 [dup ++] codireco] rest cons
0 0 . ++ [0 [dup ++] codireco] rest cons
0 1 . [0 [dup ++] codireco] rest cons
0 1 [0 [dup ++] codireco] . rest cons
0 1 [[dup ++] codireco] . cons
0 [1 [dup ++] codireco] .
```python
define('G == [codireco] cons cons')
```
```python
J('230 [dup ++] G 5 [x] times pop')
```
230 231 232 233 234