yeah more graphomata episodes! love this game. -if you look at the leaderboards you may come to realize that this is an understatement- the "score" is based on whether you've made an optimal tradeoff between cost and cycles. "cost" is the number of commands times the number of _condition_ commands, so if you used A Join A as the redundant command in C10 you would get full points. note that the optimal tradeoff between cost and cycles is based on other solutions that the server knows about, so you can get a score that's temporarily _better_ than the maximum if you come up with a new solution. the scores are renormalized regularly, though, so it won't last for long.
I almost gave up on the 'delayed' achievement, but when I jump to the end and see you achieve it I decide to try it again, and to my surprise I make it. Thank you for encouraging me to do it. Can't wait for the special area episode, where a new mechanic is introduced, and when you try to not use the new mechanic :)
fun facts: there is currently no way to know the names of f and g without testing them your first video of this game is on the official website if you win a level while using root, your score turns imaginary
"If you win a level while using root, your score turns imaginary." ...I would call that a dad joke, but it's a bit more involved than most of such. It's a math professor joke. The game really never specified what "root" was a root of, did it?
@@jagpreetsingh6711 Plenty of levels are able to be made more optimal if you use root, which can give a higher score. Solves that pass normal cases but not optional ones also count as root, so if you have some jank solution that doesn't scale you can get plenty of points above max that way too.
Honestly, fitting a polynomial to the data isn't even that hard since you can easily just start with an N⁴ polynomial and maybe some of the coefficients are zero, but what I really want to know is if there are some later levels which have exponential or even factorial complexity but also still try and fit the numbers. Probably not, since you'd only be able to have a handful of tests, but it's funny to imagine.
I think I'm following this fairly well, but it would certainly help me if you went through the different inputs to show how each one works; though I understand you probably don't want to. However, that "build a Y star" one, I haven't even the faintest clue how it's supposed to know when to stop building a chain based on how long they're supposed to be and how many there are. I can think of a solution where it just increments the last chain built, builds a new chain, then builds it up while it counts one of the previous ones; but this would be extremely worker-heavy beyond a few chains.
You can use a worker on a chain to store a counter (when it walks off the end of the chain, the counter is done counting). With two workers, you can store two numbers - for instance, one to store how much longer you have to build in the current chain, and one to store how many chains you need. There are later levels that require much more fancy tricks, of course, but that one's supposed to be about learning the general concept.
You can make B join B (or anything join itself) to "waste" a block for looping purposes without counting it as an conditional
yeah more graphomata episodes! love this game. -if you look at the leaderboards you may come to realize that this is an understatement-
the "score" is based on whether you've made an optimal tradeoff between cost and cycles. "cost" is the number of commands times the number of _condition_ commands, so if you used A Join A as the redundant command in C10 you would get full points.
note that the optimal tradeoff between cost and cycles is based on other solutions that the server knows about, so you can get a score that's temporarily _better_ than the maximum if you come up with a new solution. the scores are renormalized regularly, though, so it won't last for long.
Hello internet and welcome to graph theory
o7
amazing
Carol bob alice
outro contains a very good usage of "cba"
I almost gave up on the 'delayed' achievement, but when I jump to the end and see you achieve it I decide to try it again, and to my surprise I make it. Thank you for encouraging me to do it.
Can't wait for the special area episode, where a new mechanic is introduced, and when you try to not use the new mechanic :)
fun facts:
there is currently no way to know the names of f and g without testing them
your first video of this game is on the official website
if you win a level while using root, your score turns imaginary
But how do you get complex?
(Like a+ib because there is an achievement for that)
@jagpreetsingh6711 if you solve normally, but get a better solution with root. It’s possible on a few reasonably early levels like A11
Great, i really love complex scoring system
"If you win a level while using root, your score turns imaginary."
...I would call that a dad joke, but it's a bit more involved than most of such. It's a math professor joke. The game really never specified what "root" was a root of, did it?
@@jagpreetsingh6711 Plenty of levels are able to be made more optimal if you use root, which can give a higher score. Solves that pass normal cases but not optional ones also count as root, so if you have some jank solution that doesn't scale you can get plenty of points above max that way too.
It's less puzzle and more programming. Still fun though, wish I had more time for it
I started playing Graphomata because of you :)
Last Battle Advanced
Oh great what’s next? Counting meta puzzle games?
WHERE'S MY PORTAL 1.5!!!
flp;kfdkga;opikgbrs
I honestly don't know what to do with those episodes, hard to find a natural between-episodes-breaking-point
It feels more like linked list data structure instead of graph theory
graph is just linked list with more link
I wish there were more videos about this game in general, it's quite good, but unpopular, and i have been hardstuck on a few levels
Honestly, fitting a polynomial to the data isn't even that hard since you can easily just start with an N⁴ polynomial and maybe some of the coefficients are zero, but what I really want to know is if there are some later levels which have exponential or even factorial complexity but also still try and fit the numbers.
Probably not, since you'd only be able to have a handful of tests, but it's funny to imagine.
There are levels where exponential complexity is natural and they do fit it properly.
I think I'm following this fairly well, but it would certainly help me if you went through the different inputs to show how each one works; though I understand you probably don't want to.
However, that "build a Y star" one, I haven't even the faintest clue how it's supposed to know when to stop building a chain based on how long they're supposed to be and how many there are.
I can think of a solution where it just increments the last chain built, builds a new chain, then builds it up while it counts one of the previous ones; but this would be extremely worker-heavy beyond a few chains.
You can use a worker on a chain to store a counter (when it walks off the end of the chain, the counter is done counting). With two workers, you can store two numbers - for instance, one to store how much longer you have to build in the current chain, and one to store how many chains you need.
There are later levels that require much more fancy tricks, of course, but that one's supposed to be about learning the general concept.
please do more of this!
programming in visual programming
Graph spaghetti advanced
i am currently in egypt
edit so icely heart disapears hehe
Haskell when? Also, carol ❤
spaghetti js code
pasta python code
12:20 what happened at row 50 column 1?
f(x)=x^2