Apologies to Eugene Nonko - I'd overlooked your remarks about the great
similarity between the
14 segments. Plagiarism rules!
Mike
On 13/01/2022 23:34, Eugene Nonko wrote:
I did solve it, but I didn't use J for that one.
I wrote a translator in Perl that converted their assembly into C. C,
because it's easier to read, plus I thought that some brute forcing might
be required and compiled C will run faster than interpreting their assembly.
While studying that generated C code I discovered that there are 14 very
similar segments, each segment belonging to one of two types. Segments of
one type increased the value of z and segments of another type decreased
the value of z given some condition. That allowed me to reduce the size of
search space from 9^14 (too large) to 9^7 (very manageable).
-- Eugene
On Thu, Jan 13, 2022 at 2:17 PM Raul Miller <rauldmil...@gmail.com> wrote:
https://adventofcode.com/2021/day/24
I have not completed the day 24 puzzle.
The day 24 puzzle has a sequence of instructions representing a
calculation to verify a model number (and conceptually enable features
based on that model number -- though part A of the puzzle does not
provide any details about that).
The processing unit performs integer calculations, and has four
registers: W, X, Y and Z.
There are six instructions, one which inputs a digit of the model
number, an add instruction, a multiply instruction, an integer
division instruction, a modulo instruction and an equals instruction.
The input instruction always inputs to register W (and is the only
instruction used to update W). The div instruction always divides by 1
or 26. The modulo instruction is always used to find a remainder
modulo 26 (and always only operates on positive values, or 0 for the
numerator). The multiply instruction seems to always multiply by 0 or
powers of 26. (To load a value into a register, the register is first
multiplied by 0 and then has another value added to it.)
A certain amount of simplifications are possible using math identities
and range constraints. But I do not have much more that's useful to
say until I've found a way of solving the puzzle.
Here's a snapshot of where I'm at, trying to work through these issues
(looks sloppy with a proportionally spaced font):
...
digit13 NB. W234=: W216 inp 13 [ 1 thru 9
0 NB. X235=: X223 mul 0 [ 0
add Z228 Y232 NB. X236=: X235 add Z233 [ 9 thru 5520918021
mod 26|Z233 26 NB. X237=: X236 mod 26 [ 0 thru 25
mul <.Z210%26 X223 NB. Z238=: Z233 div 26 [ 0 thru 212343000
add X237 _4 NB. X239=: X237 add -4 [ _4 thru 21
eql X239 digit13 NB. X240=: X239 eql W234 [ 0 thru 1
eql X240 0 NB. X241=: X240 eql 0 [ 0 thru 1
0 NB. Y242=: Y232 mul 0 [ 0
25 NB. Y243=: Y242 add 25 [ 25
mul 25 X241 NB. Y244=: Y243 mul X241 [ 0 thru 25
add Y244 1 NB. Y245=: Y244 add 1 [ 1 thru 26
mul <.Z228%26 Y245 NB. Z246=: Z238 mul Y245 [ 0 thru 5520918000
0 NB. Y247=: Y245 mul 0 [ 0
digit13 NB. Y248=: Y247 add W234 [ 1 thru 9
add digit13 7 NB. Y249=: Y248 add 7 [ 8 thru 16
mul Y249 X241 NB. Y250=: Y249 mul X241 [ 0 thru 16
add Z246 Y250 NB. Z251=: Z246 add Y250 [ 0 thru 5520918016
I am currently working on some tree unification mechanisms (maximum
common subtree elimination to let me better inspect the partially
resolved calculations).
I believe Eugene Nonko has solved this one, though I do not know if he
found J useful in his approach.
FYI,
--
Raul
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