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Finished up Generator_Programs.md

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@ -15,12 +15,15 @@
</code></pre>
<p>We can apply it to a quoted program consisting of some value <code>a</code> and some function <code>B</code>:</p>
<pre><code>[a B] x
[a B] dup i
[a B] [a B] i
[a B] a B
</code></pre>
<p>Let <code>B</code> function <code>swap</code> the <code>a</code> with the quote and run some function <code>C</code> on it to generate a new value <code>b</code>:</p>
<pre><code>B == swap [C] dip
[a B] a B
</code></pre>
<p>Evaluation would go like this:</p>
<pre><code>[a B] a B
[a B] a swap [C] dip
a [a B] [C] dip
a C [a B]
@ -345,45 +348,59 @@ otherwise we discard the term and the generator leaving the sum on the stack:</p
<pre><code>joy? [pop &gt;4M] [popop] [[PE2.1] dip third-term] tailrec
4613732
</code></pre>
<h2>Math</h2>
<h2>Let's Use Math</h2>
<p>Consider the Fib seq with algebraic variables:</p>
<pre><code>a b
b a+b
a+b a+b+b
a+b+b a+a+b+b+b
</code></pre>
<p>So if (a,b) and a is even then the next even term pair is (a+2b, 2a+3b)</p>
<p>So starting with <code>(a b)</code> and assuming <code>a</code> is even then the next even term pair is <code>(a+2b, 2a+3b)</code>.</p>
<p>Reconsider:</p>
<pre><code>[b a F] x
[b a F] b a F
</code></pre>
<p>From here we want to arrive at:</p>
<pre><code>(a+2b) [(2a+3b) (a+2b) F]
b a F
b a [F0] [F1] fork
</code></pre>
<p>Let's derive <code>F</code>. We have two values and we want two new values so that's a <code>clop</code>:</p>
<pre><code>b a F
b a [F0] [F1] clop
</code></pre>
<p>Where <code>F0</code> computes <code>a+2b</code>:</p>
<pre><code>F0 == over [+] ii
b a over [+] ii
---------------------
b a b [+] ii
b a + b +
b+a b +
a+2b
</code></pre>
<p>And:</p>
<pre><code> b a over [dup + +] ii
---------------------------
<p>And <code>F1</code> computes <code>2a+3b</code>:</p>
<pre><code>F1 == over [dup + +] ii
b a over [dup + +] ii
b a b [dup + +] ii
b a dup + + b dup + +
b a a + + b b + +
...
2a+3b
</code></pre>
<p>So after that we have</p>
<pre><code>[b a F] (a+2b) (2a+3b) F'
[over [dup + +] ii] [over [+] ii] clop
roll&lt; rrest [tuck] dip ccons
[b a F] b a F
[b a F] (2a+3b) (a+2b) roll&lt;
(2a+3b) (a+2b) [b a F] rrest
(2a+3b) (a+2b) [F] [tuck] dip ccons
joy? [1 0 [over [dup + +] ii] [over [+] ii] clop roll&lt; rrest [tuck] dip ccons]
[b a F] b‴ a‴ roll&lt;
b‴ a‴ [b a F] rrest
b‴ a‴ [F] [tuck] dip ccons
b‴ a‴ tuck [F] ccons
a‴ b‴ a‴ [F] ccons
a‴ [b‴ a‴ F]
</code></pre>
<p>Putting it all together (and deferring factoring) we have:</p>
<pre><code>F == [over [dup + +] ii] [over [+] ii] clop roll&lt; rrest [tuck] dip ccons
</code></pre>
<p>Let's try it out:</p>
<pre><code>joy? [1 0 [over [dup + +] ii] [over [+] ii] clop roll&lt; rrest [tuck] dip ccons]
[1 0 [over [dup + +] ii] [over [+] ii] clop roll&lt; rrest [tuck] dip ccons]
joy? x
@ -399,5 +416,7 @@ joy? x
2 8 34 144 [233 144 [over [dup + +] ii] [over [+] ii] clop roll&lt; rrest [tuck] dip ccons]
</code></pre>
<p>And so it goes...</p>
<h2>Conclusion</h2>
<p>Generator programs like these are fun and interesting.</p>
</body>
</html>

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@ -11,12 +11,16 @@ Consider the `x` combinator:
We can apply it to a quoted program consisting of some value `a` and some function `B`:
[a B] x
[a B] dup i
[a B] [a B] i
[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
Evaluation would go like this:
[a B] a B
[a B] a swap [C] dip
a [a B] [C] dip
@ -422,14 +426,16 @@ otherwise we discard the term and the generator leaving the sum on the stack:
joy? [pop >4M] [popop] [[PE2.1] dip third-term] tailrec
4613732
## Math
## Let's Use Math
Consider the Fib seq with algebraic variables:
a b
b a+b
a+b a+b+b
a+b+b a+a+b+b+b
So if (a,b) and a is even then the next even term pair is (a+2b, 2a+3b)
So starting with `(a b)` and assuming `a` is even then the next even term pair is `(a+2b, 2a+3b)`.
Reconsider:
@ -440,30 +446,48 @@ From here we want to arrive at:
(a+2b) [(2a+3b) (a+2b) F]
Let's derive `F`. We have two values and we want two new values so that's a `clop`:
b a F
b a [F0] [F1] fork
b a [F0] [F1] clop
Where `F0` computes `a+2b`:
F0 == over [+] ii
b a over [+] ii
---------------------
b a b [+] ii
b a + b +
b+a b +
a+2b
And:
And `F1` computes `2a+3b`:
F1 == over [dup + +] ii
b a over [dup + +] ii
---------------------------
b a b [dup + +] ii
b a dup + + b dup + +
b a a + + b b + +
...
2a+3b
So after that we have
[over [dup + +] ii] [over [+] ii] clop
roll< rrest [tuck] dip ccons
[b a F] (a+2b) (2a+3b) F'
[b a F] b‴ a‴ roll<
b‴ a‴ [b a F] rrest
b‴ a‴ [F] [tuck] dip ccons
b‴ a‴ tuck [F] ccons
a‴ b‴ a‴ [F] ccons
a‴ [b‴ a‴ F]
[b a F] b a F
Putting it all together (and deferring factoring) we have:
[b a F] (2a+3b) (a+2b) roll<
(2a+3b) (a+2b) [b a F] rrest
(2a+3b) (a+2b) [F] [tuck] dip ccons
F == [over [dup + +] ii] [over [+] ii] clop roll< rrest [tuck] dip ccons
Let's try it out:
joy? [1 0 [over [dup + +] ii] [over [+] ii] clop roll< rrest [tuck] dip ccons]
[1 0 [over [dup + +] ii] [over [+] ii] clop roll< rrest [tuck] dip ccons]
@ -482,3 +506,6 @@ And:
And so it goes...
## Conclusion
Generator programs like these are fun and interesting.