Cheating at Countdown | Game Dev's Revenge

Dollar store Sebastian Lague xD
You too can gladly coexist. I enjoy all the videos I can get

> raving on about how optimization is important for code that needs to deploy on a rPi
> porting the code into javascript
Yep, that's a work of a platypus :D

I'm very sorry but the best Countdown player is and always will be Maurice Moss.
But it's very impressive and fun what you did there :D

I love the Abdicates easter egg!

Love the video, and the easter eggs for the 8/10 Cats does CD people. Abdicates FTW

hes european and yet he still says "aluminum", the world has been so americanised, smh

It’s pronounced “Tree” 😛

No, you write like you always use a keybaord.
Heh.

Isn't TNETENNBA already a word? 🤔🤣

"For those of you who are under 60"

I feel offended and at the same time you're entirely correct.

I am ashamed at how quickly I would order a hoodie with your avatar embroidered on it

Nice touch with the tnetennba. Moss's 9 letter word!

knew right away this was going to be a video about tries

Please, please, please, please, please, please, i'm begging do a radial tree graph of the entire dictionary as shown in your trie explanation :')

I would try to sort every dictionary word's by the letters, then sort those "words". Sort scrambled input. search n choose k of input letters; starting from choosing 9 out of 9. max 511 combnations to lookup. and best score comes up first.

Love the subtle IT crowd reference

I wonder if Rust trie is a good idea.

Maybe you do definitely write like an 8 year old but you still write better than me!

Love the IT Crowd reference in there - Bravo!

That's a nice Tenetennba.

Hello, that's a nice tnetennba :)

"You budget Sebastian Lague", I laughed so hard.

In any case, I just wanted to mention theres a much better way to do the naive algorithm.
This certainly won't be faster than the average solve for batch data with a precomputed tree, but for single games it will be much faster.

function( letters, words, depth ):
* Remember the longest word you've seen (longest)
* For each available letter:
* Get the sublist of your available words that match the letter at "depth"
* If that list is empty, return an empty string
* Perform a recursive call to this function with the available letters except for this one, the sublist of words, and depth + 1
* If the result of the recursive call is longer than "longest" overwrite it

You french 🥖?
Damm, hope you get better 🙏😢














/J

the second you mentioned the 8oo10c version i subbed !

Really cool. But use 3 wires for the morse code? I mean, it was designed with single contact in mind, you'd need a little practice but the result would be even more discreet.

Gotta love the IT Crowd reference at the start. "Umm it actually already is a word"

What did you use to visualise the trie?

i wonder if any word culling could be performed by making a list of numbers for each word that has a bit for each character and sets whether it’s on, either way that’s where my mind went first, hadn’t thought of a trie but it seems like a great option for this.

Missed an opportunities to make the word at the end 'subscribe'

That’s amazing that you made the actual device lmaoo. I have a question - could you make this even faster if you could skip the array copy by using an integer to mark each letter in the array as used or remaining, where the nth bit corresponds to the nth letter? I don’t know the speed of an array copy vs passing an integer so idk if it actually would improve performance, but the array copy set off the 🚨 in my brain. Thanks for making great videos!

I would have just made a separate list of words for each number of letters for example a list of just 1 letter words, a list of just 2 letter words, etc then look trough them and see if you can make a 9 letter word, 8 letter word, 7 letter word, etc but your way of doing it seems much better

Vidéo très intéressante! Je n’aurais pas deviné que tu étais français… quelle condition horrible 😨

But why Javascript?

Did you sort the letters of the word before inserting them? It will help prune duplicates and should reduce the depth of the trie. Coat and taco for example would both map to the path 'acot'. You'd have to keep the words in the last leaf instead of a flag for done, though.

That’s a nice Tnetennba

You’re doing great, love your channel. Sebastian Lague is so, so fucking good that even if you were a dollar store version of him (you’re not, you’re better than that) you’d be worth watching. Keep it up!

I think this can be made a lot faster by sorting the letters in the dictionary words, as well as the clue. i.e. store "abdicates" as "aabcdeist".

This would be hilarious to try out on a friend group 😂

That’s a nice tnetennba

awesome stuff on this channel, very rewatchable too

hello its me again great vid by the way just wondering what video editor do you use im thinking of posting a ASCII video that uses compute shader to get it which should get a nice and high framerate thanks Gc edit - i mean shader graph then turn it into compute maybe in a diffrent vid

It quite annoys me that he forgets about the word car in his visualisation 😂 (not really)

Loved the "Abdicates" reference in the end!! Great vid as usual! Good luck on countdown!!!

Dentroyer. Very clever 🙄😂

Ah c'est de là que ça vient des chiffres et des lettres !

Tnetennba .. okay, that got a laugh

Couldn’t you do a very quick early out of each branch? I.e. if you had no A’s, you could ignore the A branch straight away? Worst case scenario, only 9 branches to check?

Put each branch into an high level array, and then only check each based on a quick array lookup based on each letter. If that makes sense?

Now I want to take your code and optimize it for this.. lol.. but I have work to do…