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Convert syntax highlighter spec.

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This commit is contained in:
Simon Forman
2021-11-19 13:57:36 -08:00
parent 31c26cd235
commit 3f40e30c6f
12 changed files with 361 additions and 361 deletions
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docs/sphinx_docs/notebooks/Developing.rst
+35 -35
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@@ -12,7 +12,7 @@ As an example of developing a program in Joy let's take the first problem from t
Find the sum of all the multiples of 3 or 5 below 1000.
.. code:: ipython2
.. code:: python
from notebook_preamble import J, V, define
@@ -22,11 +22,11 @@ Sum a range filtered by a predicate
Let's create a predicate that returns ``True`` if a number is a multiple
of 3 or 5 and ``False`` otherwise.
.. code:: ipython2
.. code:: python
define('P == [3 % not] dupdip 5 % not or')
.. code:: ipython2
.. code:: python
V('80 P')
@@ -108,11 +108,11 @@ the counter to the running sum. This function will do that:
PE1.1 == + [+] dupdip
.. code:: ipython2
.. code:: python
define('PE1.1 == + [+] dupdip')
.. code:: ipython2
.. code:: python
V('0 0 3 PE1.1')
@@ -131,7 +131,7 @@ the counter to the running sum. This function will do that:
3 3 .
.. code:: ipython2
.. code:: python
V('0 0 [3 2 1 3 1 2 3] [PE1.1] step')
@@ -219,7 +219,7 @@ total to 60.
How many multiples to sum?
^^^^^^^^^^^^^^^^^^^^^^^^^^
.. code:: ipython2
.. code:: python
1000 / 15
@@ -232,7 +232,7 @@ How many multiples to sum?
.. code:: ipython2
.. code:: python
66 * 15
@@ -245,7 +245,7 @@ How many multiples to sum?
.. code:: ipython2
.. code:: python
1000 - 990
@@ -260,7 +260,7 @@ How many multiples to sum?
We only want the terms *less than* 1000.
.. code:: ipython2
.. code:: python
999 - 990
@@ -276,11 +276,11 @@ We only want the terms *less than* 1000.
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.
.. code:: ipython2
.. code:: python
define('PE1 == 0 0 66 [[3 2 1 3 1 2 3] [PE1.1] step] times [3 2 1 3] [PE1.1] step pop')
.. code:: ipython2
.. code:: python
J('PE1')
@@ -305,7 +305,7 @@ integer terms from the list.
3 2 1 3 1 2 3
0b 11 10 01 11 01 10 11 == 14811
.. code:: ipython2
.. code:: python
0b11100111011011
@@ -318,11 +318,11 @@ integer terms from the list.
.. code:: ipython2
.. code:: python
define('PE1.2 == [3 & PE1.1] dupdip 2 >>')
.. code:: ipython2
.. code:: python
V('0 0 14811 PE1.2')
@@ -349,7 +349,7 @@ integer terms from the list.
3 3 3702 .
.. code:: ipython2
.. code:: python
V('3 3 3702 PE1.2')
@@ -376,7 +376,7 @@ integer terms from the list.
8 5 925 .
.. code:: ipython2
.. code:: python
V('0 0 14811 7 [PE1.2] times pop')
@@ -518,11 +518,11 @@ integer terms from the list.
And so we have at last:
.. code:: ipython2
.. code:: python
define('PE1 == 0 0 66 [14811 7 [PE1.2] times pop] times 14811 4 [PE1.2] times popop')
.. code:: ipython2
.. code:: python
J('PE1')
@@ -542,17 +542,17 @@ Let's refactor
14811 n [PE1.2] times pop
n 14811 swap [PE1.2] times pop
.. code:: ipython2
.. code:: python
define('PE1.3 == 14811 swap [PE1.2] times pop')
Now we can simplify the definition above:
.. code:: ipython2
.. code:: python
define('PE1 == 0 0 66 [7 PE1.3] times 4 PE1.3 pop')
.. code:: ipython2
.. code:: python
J('PE1')
@@ -581,11 +581,11 @@ 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.
.. code:: ipython2
.. code:: python
define('PE1.terms == [0 swap [dup [pop 14811] [] branch [3 &] dupdip 2 >>] dip rest cons]')
.. code:: ipython2
.. code:: python
J('PE1.terms 21 [x] times')
@@ -598,7 +598,7 @@ produce a stream of the seven numbers repeating over and over again.
We know from above that we need sixty-six times seven then four more
terms to reach up to but not over one thousand.
.. code:: ipython2
.. code:: python
J('7 66 * 4 +')
@@ -611,7 +611,7 @@ terms to reach up to but not over one thousand.
Here they are...
~~~~~~~~~~~~~~~~
.. code:: ipython2
.. code:: python
J('PE1.terms 466 [x] times pop')
@@ -624,7 +624,7 @@ Here they are...
...and they do sum to 999.
~~~~~~~~~~~~~~~~~~~~~~~~~~
.. code:: ipython2
.. code:: python
J('[PE1.terms 466 [x] times pop] run sum')
@@ -638,7 +638,7 @@ 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.
.. code:: ipython2
.. code:: python
J('0 0 PE1.terms 466 [x [PE1.1] dip] times popop')
@@ -654,7 +654,7 @@ A little further analysis renders iteration unnecessary.
Consider finding the sum of the positive integers less than or equal to
ten.
.. code:: ipython2
.. code:: python
J('[10 9 8 7 6 5 4 3 2 1] sum')
@@ -686,11 +686,11 @@ positive integers is:
(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.)
.. code:: ipython2
.. code:: python
define('F == dup ++ * 2 floordiv')
.. code:: ipython2
.. code:: python
V('10 F')
@@ -727,7 +727,7 @@ And ending with:
If we reverse one of these two blocks and sum pairs...
.. code:: ipython2
.. code:: python
J('[3 5 6 9 10 12 15] reverse [978 980 981 984 985 987 990] zip')
@@ -737,7 +737,7 @@ If we reverse one of these two blocks and sum pairs...
[[978 15] [980 12] [981 10] [984 9] [985 6] [987 5] [990 3]]
.. code:: ipython2
.. code:: python
J('[3 5 6 9 10 12 15] reverse [978 980 981 984 985 987 990] zip [sum] map')
@@ -750,7 +750,7 @@ If we reverse one of these two blocks and sum pairs...
(Interesting that the sequence of seven numbers appears again in the
rightmost digit of each term.)
.. code:: ipython2
.. code:: python
J('[ 3 5 6 9 10 12 15] reverse [978 980 981 984 985 987 990] zip [sum] map sum')
@@ -771,7 +771,7 @@ additional unpaired terms between 990 and 1000:
So we can give the "sum of all the multiples of 3 or 5 below 1000" like
so:
.. code:: ipython2
.. code:: python
J('6945 33 * [993 995 996 999] cons sum')