Adventofcode.Com 2017 is live :)

If anybody wants to join this advent, we can use our own private leaderboard,  with my code 130669-ab1f69bf.

I'm in. Here's my solution.

Thought about using $zstrip, but while it can remove duplicates, it can't leave only duplicates:

w $zstrip("11223","=E")
123

Also, for second task it says:

1212 produces 6: the list contains 4 items, and all four digits match the digit 2 items ahead.
1221 produces 0, because every comparison is between a 1 and a 2.

Shouldn't 1221 produce 3 as first 2 and second 1 match.

Currently failed on second task with 10938.

Bert, thanks for sharing!

Here is my solution for Day1.

Darn it Bert, these things are addictive!

There's no way I'm getting up at 5am to even possibly get on the main leader board, but its fun to see how we all approach it.

I'll post mine up here....

https://gist.github.com/SeanConnelly/065cb51fc6572c6bf3dac4a9973fb54f

And maybe a few JavaScript ones as well if I get time...

https://gist.github.com/SeanConnelly/e0623e13241fd94fbf8df96b09755f95

Just for fun, one liners get an extra point - everyone rolls eyes :)

Day three part 2...

s (x,y)=0,d=7,g(0,0)=1,a="+1  -1" f i=1:1 {s e=$s(d=7:1,1:d+2),d=$s($g(g(x+$e(a,e,e+2),y+$e(a,e-2,e)))="":e,1:d),x=x+$e(a,d,d+2),y=y+$e(a,d-2,d),g(x,y)=$g(g(x,y+1))+$g(g(x,y-1))+$g(g(x+1,y))+$g(g(x+1,y+1))+$g(g(x+1,y-1))+$g(g(x-1,y))+$g(g(x-1,y+1))+$g(g(x-1,y-1)) i g(x,y)>t ret g(x,y)}

I wonder, guys, how did you solve that the word is Anagram in Day4:

A valid passphrase must contain no two words that are anagrams of each other - that is, a passphrase is invalid if any word's letters can be rearranged to form any other word in the passphrase.

For example:

• abcde fghij is a valid passphrase.
• abcde xyz ecdab is not valid - the letters from the third word can be rearranged to form the first word.
• a ab abc abd abf abj is a valid passphrase, because all letters need to be used when forming another word.
• iiii oiii ooii oooi oooo is valid.
• oiii ioii iioi iiio is not valid - any of these words can be rearranged to form any other word.

 

My solution is:

ClassMethod IsAnagram(word,test) as %Boolean {

if $L(word)'=$L(test) return 0

f i=1:1:$L(test) if $seq(test($E(test,i))),$seq(word($E(word,i)))

set a=$O(word(""))

while a'="" {

if word(a)'=$g(test(a)) return 0

set a=$O(word(a))

}

return 1

}

The full solution is here.

Today's task was pretty difficult, it takes few hours to solve it. I made a mistake in first part even if I got the right answer, but after when I rewrote my first part, the second part was solved quite fast. But I lost so much time to figure it out.

We decided to promote and support the contest, and launched a challenge on Global Masters:

We would track the private leaderboard with code: 130669-ab1f69bf. Join this private leaderboard, solve tasks with Caché ObjectScript be the leader in the end and get the endless fame plus 10,000 GM points! 

2nd and 3rd places would get 5,000 and 3,000 points too.

Yes, there is some handicap for current members:

But we wouldn't change anything: we believe you have enough time and talent to beat them!

Let's have fun and let the COS power be with you! ;)