Observability and APM are way, way older than depicted in this simplistic post!
Readit Newsantsou commented on Observability's past, present, and future blog.sherwoodcallaway.com... · Posted by u/shcallaway
antsou commented on Portugal just ran on 100% renewables for six days in a row canarymedia.com/articles/... · Posted by u/Anon84
It really doesn't tell what really happened:
-This was only due to hydro production...
-The amount of rain that fell was consistent along those days, which is definitely not usual.
-The energy produced was mainly exported to Spain, but at ZERO or even NEGATIVE euros.
Does this help? Doesn't seem to me! And besides having to pay full price to the renewables producers (while exporting the same energy at ZERO), which amounts to a significant energy deficit, they are talking about even more H2 or whatever schemes??? This is the reason why Portugal has had one of the biggest rises of electricity prices in Europe over the last 15 years...
I played with this in the past (https://news.ycombinator.com/item?id=34461113#34462521), but am willing to take another stab at it.
Store one 64 bit bitboard - a set bit means that a piece is present at that place. An unset bit means that no piece is after that position. After the bitboard, store a list of 32 4 bit integers, where the order of the pieces in the list corresponds to the order of the bits set. If there are less than 16 bits set in the bitboard, ignore the last items in the list.
0000 - 0x0 - black pawn
0001 - 0x1 - black pawn (can be en-passant'd)
0010 - 0x2 - black knight
0011 - 0x3 - black bishop
0100 - 0x4 - black rook (castling unavailable)
0101 - 0x5 - black rook (castling available)
0110 - 0x6 - black king
0111 - 0x7 - black queen
1000 - 0x8 - white pawn
1001 - 0x9 - white pawn (can be en-passant'd)
1010 - 0xA - white knight
1011 - 0xB - white bishop
1100 - 0xC - white rook (castling unavailable)
1101 - 0xD - white rook (castling available)
1110 - 0xE - white king
1111 - 0xF - white queen
The starting position would be represented with a bitboard of 0xFFFF00000000FFFF, with a list of [0xD, 0xA, 0xB, 0xF, 0xE, 0xB, 0xA, 0xD, 0x8, 0x8, 0x8, 0x8, 0x8, 0x8, 0x8, 0x8, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0x5, 0x2, 0x3, 0x7, 0x6, 0x3, 0x2, 0x5], using the same position-to-number scheme in the blog post
Edit: 32 pieces, not 16. Thanks to the peanut gallery for catching it quickly
Edit2: To store which player is next: do nothing for white. For black, if there are 32 pieces, flip the bitboard upside down. (To check if it's black's turn, verify that black pawns are "below" white pawns, which is illegal before captures are made.) If there are less than 32 pieces and it's black's turn, invert the bitboard. (To check, count the number of set bits.) This is entirely taken from https://news.ycombinator.com/item?id=37526484
I believe you can also do it in a slightly different way. For each square on the board:
0 - encodes no piece in the square; at least 32 of them, so 32 bits (4 bytes)
1xxxx - xxxx being your encoding: encodes remaining pieces; at most 32 of them, so max 32 x 5 bits (20 bytes)