Chris Jester-Young

If the book hasn't talked about let yet, has it talked about internal define? :-) If so, that's a way to get around the lack of let.

I believe I even mentioned this in item 1 of my review comments.

@morbidCode My comment about correctness is conditioned upon being able to separate out the next and prev like using my let (or something similar in idea). Without being able to separate out the two, your test isn't quite correct.

Sass can sometimes be a pain to work with (as shown in the first diff linked above) but I still haven't found anything better, so. ;-)

@morbidCode I put in two links, the game itself as well as the code repo. The actual changes I worked on at the time are here: 1 2

@morbidCode Cool, I'm glad to be of help! Especially with regard to the redundancy. I'm surprised about not being able to use let (it's a very basic Scheme construct!) but it's what it is. :-)

@morbidCode It doesn't look obviously bogus. I tested 65535 for fun, which enabled me to find a bug in my version. :-P

All right, I'll go to my next appointment now. I'll check back later tonight and see how else I can be of help. :-D

As for the code I was working on, if you're curious, I pushed some changes to EmojiSweeper (code repo) this morning. :-)

@morbidCode I hope the above ^ answered your questions some! Please feel free to ask further.

and then you can do the tests you need to against prev and next.

(let* ((prev (modexp base (/ exp 2) m))
       (next (remainder (square prev) m)))
  ...)

In my code, I had (something like):

As for ^, you can check for that by checking the value (remainder (square (modexp base (/ exp 2) m)) m) against 1. You may need to stash the value using a let for later use.

2. (remainder (expmod ... m) m) is completely redundant since the return value of the (expmod ... m) is already supposed to be clipped to mod m (if not, then your expmod has a bug), so the remainder does nothing useful.

1. It allowed the base value of the recursive calls to get very, very big, since (square base) values are not clipped at each stage.

Your original approach of doing (remainder (expmod (square base) (/ exp 2) m) m) has two problems:

Okay, I have 30 minutes to answer some of your questions. Let's see how far I get.

@morbidCode Sorry, I was working on other code. Um. I'll try to answer your questions tomorrow after I check in my code.

@morbidCode (expmod base (/ exp 2) m) is the recursive case. You then square the result to get what you need, then you mod m so you can compare the result to 1.

@morbidCode Clamping x to mod m means (modulo x m), so that the result is always between 0 and m-1. The return value of expmod should, by contract, always be clamped to mod m, so the remainder call in your (remainder (expmod (square base) (/ exp 2) m) m) is redundant.

@morbidCode "You should, however, clamp the result of square to mod m" means use (expmod (remainder (square base) m) (/ exp 2) m) instead. Or, as in my version, (remainder (square (expmod base (/ exp 2) m)) m). Notice how, in both cases, the shape of the expression is (remainder (square ...)), not (remainder (expmod ...)).

@quartata Just writing to let you know I haven't forgotten your program! I have some comments to write on it, but I'm busy preparing for a lightning talk I'm giving tomorrow, so I'll try to do it after this weekend. :-)

@orlp ;-)

@orlp Yep, come join us on #chez on freenode. :-D

@poi830 I'm not a mod any more so I can't ding you for this, but that's seriously not cool.

@AlexA. :-D

@EasterlyIrk It's too much effort for me to log into Steam at the moment, sorry.

@AlexA. cky944

:shrug:

@El'endiaStarman That explains it, really: it sounds like it stops reading after 100 characters.

@Doorknob I read that as fiancée instead of fleece, lol.

@NathanMerrill Sorry; see you in Round 1B or 1C. ;-)

@EasterlyIrk Aye.

@EasterlyIrk steamcommunity.com/id/cky944

/me goes and eats tasty dinner.

So I'm going to go for 1B or 1C, if I can be bothered. ;-)

I believe I'm not going to pass Round 1A. The 1000th place is already at score 55, and I'm not going to bother to solve problem C.

@poi830 Correct.

@MarsUltor Yeah, in that case unless you can solve B and C super-fast, you may have to wait it out till Round 1B or 1C.

You still have to solve B and C faster than most contestants.

@MarsUltor It is. In that you have to get on the top 1000.

@MarsUltor Unsurprisingly, you mean.

@AlexA. I wish.

Round 3, probably not.

@AlexA. I've gotten to Round 2 before.

@AlexA. If you are among the top 1000 contestants, you get to go to Round 2.