Python 2.6+ - 334 322 316 characters
397 368 366 characters uncompressed
#coding:l1
exec'xÚEPMO!½ï¯ i,P*Ýlš%ì‰=‰Ö–*†þz©‰:‡—Lò¾fÜ”bžAù,MVi™.ÐlǃwÁ„eQL&•uÏÔ‹¿1O6ǘ.€LSLÓ’¼›î”3òšL¸tŠv[ѵl»h;ÁºŽñÝ0Àë»Ç‡ÛûH.ª€¼âBNjr}¹„V5¾3Dë@¼¡•gO. ¾ô6 çÊsÃЮürÃ1&›ßVˆùZ`Ü€ÿžcx±ˆ‹sCàŽ êüRô{U¯ZÕDüE+³ŽFA÷{CjùYö„÷¦¯Î[0þøõ…(Îd®_›â»E#–Y%’›”ëýÒ·X‹d¼.ß9‡kD'.decode('zip')
The single newline is required, and I've counted it as one character.
Browser code-page mumbo jumbo might prevent a successful copy-and-paste of this code, so you can optionally generate the file from this code:
s = """
23 63 6F 64 69 6E 67 3A 6C 31 0A 65 78 65 63 27 78 DA 45 50 4D 4F 03 21
10 BD EF AF 20 69 2C 50 2A 02 DD 6C 9A 25 EC AD 07 8D 89 07 3D 89 1C D6
96 2A 86 05 02 1B AD FE 7A A9 89 3A 87 97 4C F2 BE 66 DC 94 62 9E 41 F9
2C 4D 56 15 69 99 0F 2E D0 6C C7 83 77 C1 16 84 65 51 4C 26 95 75 CF 8D
1C 15 D4 8B BF 31 4F 01 36 C7 98 81 07 2E 80 4C 53 4C 08 D3 92 BC 9B 11
EE 1B 10 94 0B 33 F2 9A 1B 4C B8 74 8A 9D 76 5B D1 B5 6C BB 13 9D 68 3B
C1 BA 8E F1 DD 30 C0 EB BB C7 87 DB FB 1B 48 8F 2E 1C AA 80 19 BC E2 42
4E 6A 72 01 7D B9 84 56 35 BE 33 44 8F 06 EB 40 BC A1 95 67 4F 08 2E 20
BE F4 36 A0 E7 CA 73 C3 D0 AE FC 72 C3 31 26 9B DF 56 88 AD F9 5A 60 DC
80 FF 9E 63 78 B1 88 8B 73 43 E0 8E A0 EA FC 52 F4 7B 55 8D AF 5A 19 D5
44 FC 45 2B B3 8E 46 9D 41 F7 7B 43 6A 12 F9 59 F6 84 F7 A6 01 1F AF CE
5B 30 FE F8 F5 85 28 CE 64 AE 5F 9B E2 BB 45 23 96 59 25 92 9B 94 EB FD
10 D2 B7 58 8B 64 BC 2E DF 39 87 6B 44 27 2E 64 65 63 6F 64 65 28 27 7A
69 70 27 29
"""
with open('golftris.py', 'wb') as f:
f.write(''.join(chr(int(i, 16)) for i in s.split()))
Testing
intetris
[ ]
[ ]
[ ]
[ ]
[ # # #]
[ ## ######]
[==========]
T2 Z6 I0 T7
Newlines must be Unix-style (linefeed only). A trailing newline on the last line is optional.
To test:
> python golftris.py < intetris
[ ]
[ ]
[ ]
[# ###]
[# ### ]
[##### ####]
[==========]
10
This code unzips the original code, and executes it with exec. This decompressed code weighs in at 366 characters and looks like this:
import sys
r=sys.stdin.readlines();s=0;p=r[:1];a='[##########]\n'
for l in r.pop().split():
n=int(l[1])+1;i=0xE826408E26246206601E>>'IOZTLSJ'.find(l[0])*12;m=min(zip(*r[:6]+[a])[n+l].index('#')-len(bin(i>>4*l&31))+3for l in(0,1,2))
for l in range(12):
if i>>l&2:c=n+l/4;o=m+l%4;r[o]=r[o][:c]+'#'+r[o][c+1:]
while a in r:s+=10;r.remove(a);r=p+r
print''.join(r),s
Newlines are required, and are one character each.
Don't try to read this code. The variable names are literally chosen at random in search of the highest compression (with different variable names, I saw as much as 342 characters after compression). A more understandable version follows:
import sys
board = sys.stdin.readlines()
score = 0
blank = board[:1] # notice that I rely on the first line being blank
full = '[##########]\n'
for piece in board.pop().split():
column = int(piece[1]) + 1 # "+ 1" to skip the '[' at the start of the line
# explanation of these three lines after the code
bits = 0xE826408E26246206601E >> 'IOZTLSJ'.find(piece[0]) * 12
drop = min(zip(*board[:6]+[full])[column + x].index('#') -
len(bin(bits >> 4 * x & 31)) + 3 for x in (0, 1, 2))
for i in range(12):
if bits >> i & 2: # if the current cell should be a '#'
x = column + i / 4
y = drop + i % 4
board[y] = board[y][:x] + '#' + board[y][x + 1:]
while full in board: # if there is a full line,
score += 10 # score it,
board.remove(full) # remove it,
board = blank + board # and replace it with a blank line at top
print ''.join(board), score
The crux is in the three cryptic lines I said I'd explain.
The shape of the tetrominoes is encoded in the hexadecimal number there. Each tetronimo is considered to occupy a 3x4 grid of cells, where each cell is either blank (a space) or full (a number sign). Each piece is then encoded with 3 hexadecimal digits, each digit describing one 4-cell column. The least significant digits describe the left-most columns, and the least significant bit in each digit describes the top-most cell in each column. If a bit is 0, then that cell is blank, otherwise it's a '#'. For example, the I tetronimo is encoded as 00F, with the four bits of the least-significant digit set on to encode the four number signs in the left-most column, and the T is 131, with the top bit set on the left and the right, and the top two bits set in the middle.
The entire hexadecimal number is then shift one bit to the left (multiplied by two). This will allow us to ignore the bottom-most bit. I'll explain why in a minute.
So given the current piece from the input, we find the index into this hexadecimal number where the 12 bits describing it's shape begin, then shift that down so that bits 1–12 (skipping bit 0) of the bits variable describe the current piece.
The assignment to drop determines how many rows from the top of the grid the piece will fall before landing on other piece fragments. The first line finds how many empty cells there are at the top of each column of the playing field, while the second finds the lowest occupied cell in each column of the piece. The zip function returns a list of tuples, where each tuple consists of the nth cell from each item in the input list. So, using the sample input board, zip(board[:6] + [full]) will return:
[
('[', '[', '[', '[', '[', '[', '['),
(' ', ' ', ' ', ' ', ' ', ' ', '#'),
(' ', ' ', ' ', ' ', '#', '#', '#'),
(' ', ' ', ' ', ' ', ' ', '#', '#'),
(' ', ' ', ' ', ' ', ' ', ' ', '#'),
(' ', ' ', ' ', ' ', ' ', '#', '#'),
(' ', ' ', ' ', ' ', ' ', '#', '#'),
(' ', ' ', ' ', ' ', '#', '#', '#'),
(' ', ' ', ' ', ' ', ' ', '#', '#'),
(' ', ' ', ' ', ' ', ' ', '#', '#'),
(' ', ' ', ' ', ' ', '#', '#', '#'),
(']', ']', ']', ']', ']', ']', ']')
]
We select the tuple from this list corresponding to the appropriate column, and find the index of the first '#' in the column. This is why we appended a "full" row before calling zip, so that index will have a sensible return (instead of throwing an exception) when the column is otherwise blank.
Then to find the lowest '#' in each column of the piece, we shift and mask the four bits that describe that column, then use the bin function to turn that into a string of ones and zeros. The bin function only returns significant bits, so we need only calculate the length of this string to find the lowest occupied cell (most significant set bit). The bin function also prepends '0b', so we have to subtract that. We also ignore the least significant bit. This is why the hexadecimal number is shift one bit to the left. This is to account for empty columns, whose string representations would have the same length as a column with only the top cell full (such as the T piece).
For example, the columns of the I tetromino, as mentioned earlier, are F, 0, and 0. bin(0xF) is '0b1111'. After ignoring the '0b', we have a length of 4, which is correct. But bin(0x0) is 0b0. After ignoring the '0b', we still have a length of' 1, which is incorrect. To account for this, we've added an additional bit to the end, so that we can ignore this insignificant bit. Hence, the +3 in the code is there to account for the extra length taken up by the '0b' at the beginning, and the insignificant bit at the end.
All of this occurs within a generator expression for three columns ((0,1,2)), and we take the min result to find the maximum number of rows the piece can drop before it touches in any of the three columns.
The rest should be pretty easy to understand by reading the code, but the for loop following these assignments adds the piece to the board. After this, the while loop removes full rows, replacing them with blank rows at the top, and tallies the score. At the end, the board and score are printed to the output.
T2 ...and then the start position. – Kayleen+500bonus though ;-) – Colenecoleopterancode-golfin your profile). You shant be swift dispatching justice in areas you are a "casual observer". What makes you think i "derided" others work? What i said is that I am puzzled that usually solutions are notably shorter. For each <problem, language> pair there is a kind oflimit inferiorto which solutions tend to converge, eventually. In regards to explanations, pls just ask what is unclear (on condition that you know the language already, since "those margins are too narrow to contain" a full tutorial). – Volgograd