Determine win in Tictactoe

19

2

Let's play some code golf!

Given a tic-tac-toe board state (Example:)

|x|x|o|
|x|o|x|
|o|o|x|

Determine whether a game is a win a lose or cat. Your code should output any of these options given a state. The above game should output lose

Just to be clear: a win is defined as any 3 xs in a row (diagonal, horizontal, vertical). a lose is 3 os in a row, while a cat game in none in a row.

To make things interesting, you get to determine your input structure for the state- which you must then explain. For instance xxoxoxoox is a valid state as seen above where each of the characters is read from left to right, top to bottom. [['x','x','o'],['x','o','x'],['o','o','x']] is the game in multidimensional array read in a similar way. While 0x1a9 which is hex for 110101001 might work as a suitable compression where 1 can be manipulated for xs and 0 can be manipulated for o.

But those are just some ideas, I'm sure you might have many of your own.

Ground rules:

  1. Your program must be able to accept any viable state.
  2. The form of input must be able to represent any state.
  3. "The win state must be determined from the board"
  4. Assume a complete board
  5. Win before lose for instance in the case 'xxxoooxxx'

Lowest character count wins

Dylan Madisetti

Posted 2014-06-24T02:29:13.440

Reputation: 3 381

Do we write a function? Take in an input? Do we get to choose the output format? – xnor – 2014-06-24T02:42:54.193

Output should be either win,lose,cat. For input it must be from a human (not randomly generated). Thus using def q(x):#magic code here works just as well as (/*magic code */)(prompt()) – Dylan Madisetti – 2014-06-24T02:46:16.410

1I like this input structure: (win|lose|cat) [xo]{9}, where the first word denotes whether the game is a win, lose, or cat (?) for player x. Able to represent any state. – Runer112 – 2014-06-24T02:51:33.303

Is it okay if we exit with an exception? – undergroundmonorail – 2014-06-24T02:56:22.230

@undergroundmonorail sure – Dylan Madisetti – 2014-06-24T02:58:56.163

@Runer112 added clause 3 for you. ('win')["oooxxoxxo"] would work in your proposed solution, but give the wrong answer thus not being eligible. – Dylan Madisetti – 2014-06-24T03:01:18.457

@DylanMadisetti that's why I specified that the first word must denote whether the game is a win, lose, or cat. The example you gave does not match my input structure. – Runer112 – 2014-06-24T03:13:49.920

@Runer112 You're right, and that's why it breaks rule 3. My improper state should not be able to work – Dylan Madisetti – 2014-06-24T03:15:34.000

2Can I suggest a rule like "The win state must be determined from the board" or "The input must contain no information except the board state"? – undergroundmonorail – 2014-06-24T03:16:28.687

Agreed. Checking the validity makes it more complex – Dylan Madisetti – 2014-06-24T03:18:23.103

are we to assume a complete board? otherwise a binary representatiomn wouldn't work great. – Not that Charles – 2014-06-24T03:32:02.637

Let's assume a complete board – Dylan Madisetti – 2014-06-24T03:34:00.587

3Are we assuming only legal games played? If so, certain states would be impossible, i.e. XXX OOO XXX but otherwise some full-board states include this as a fourth impossible outcome, where X wins but O also wins. – Riot – 2014-06-24T04:17:15.223

I think win before lose. If it's becoming too trivial we can throw in some of these stipulations – Dylan Madisetti – 2014-06-24T04:19:38.207

This question from the math stackexchange may be of relevance: http://math.stackexchange.com/questions/467757/determine-the-winner-of-a-tic-tac-toe-board-with-a-single-matrix-expression

– Riot – 2014-06-24T05:01:07.390

10why "cat" out of interest? – Chris – 2014-06-24T12:08:40.513

Can the representation of the input state be redundant? Meaning, can the content of a cell be written in more than one place in the input? – xnor – 2014-06-24T12:43:07.453

1@Chris cat is the terminology I learned for a tie. A cat game. No one else? – Dylan Madisetti – 2014-06-24T13:16:57.387

@xnor let's go with no, otherwise you can just check for groups of 3 in a long list of board permutations – Dylan Madisetti – 2014-06-24T13:19:39.900

7@DylanMadisetti: never heard it before and googlign for "win lose cat" came up with nothing. I'd have gone with tie or draw personally. Or in the case of this game maybe "inevitability". ;-) I don't much mind as far as the competition goes though. A string is a string. ;-) – Chris – 2014-06-24T13:22:48.703

@chris not crazy. 3rd line http://en.wikipedia.org/wiki/Tic-tac-toe

– Dylan Madisetti – 2014-06-24T14:09:02.120

@DylanMadisetti: Just because other people say it, doesn't mean you're not crazy. ;-) – Chris – 2014-06-24T14:29:30.850

Lol, "cat" :) I think "chicken" makes more sense :p (reference from a certain movie with Al Pacino) – aditsu quit because SE is EVIL – 2014-06-24T17:45:24.520

1Can I number all of the boxes? For example, with numbering 123456789 a board like {x: [1, 2, 3], o: [4, 8, 9]} would be winning for x? – Cruncher – 2014-06-24T18:58:52.267

@Cruncher Sure looks like valid input except your example isn't a complete board. Magic square ideas? – Dylan Madisetti – 2014-06-24T19:00:54.980

@DylanMadisetti it is a complete example. X has won. No need to fill in the rest. But yes, magic square ideas. What language is best at finding a subset of 3 with sum 15? lol – Cruncher – 2014-06-24T19:06:03.787

Answers

11

Ruby 2.0, 85 characters

Here's a simple bitmask-based solution in Ruby:

d=gets.hex
$><<[292,146,73,448,56,7,273,84].map{|m|d&m<1?:lose:d&m<m ?:cat: :win}.max

The board is represented as a hex number, made up of nine bits corresponding to the nine squares. 1 is an X, 0 is an O. This is just like the 0x1a9 example in the question, though the 0x is optional!

There's probably a better way to do the bitmasks then just hardcoding a big list. I'll happily take suggestions.

See it running on Ideone here.

Paul Prestidge

Posted 2014-06-24T02:29:13.440

Reputation: 2 390

1Your list contains 273 twice. And I really like the max idea! – Ventero – 2014-06-24T07:36:08.763

1Oh @Ventero, always with the obscure optimizations (thanks) – Paul Prestidge – 2014-06-24T08:19:33.190

There can be empty spaces on a board. Your input format does not account for this and therefore cannot represent any viable game state. – Stephen Ostermiller – 2014-06-24T13:57:26.010

2@StephenOstermiller rule 4:assume a complete board. You are correct that this rule is perhaps contradicted by rules 1 and 2, however if you read all the comments on the question I think this is within the spirit of the question (incomplete boards are not covered, while complete but illegal boards are.) However I think that octal would be a more user-friendly input format than hex. – Level River St – 2014-06-24T14:53:58.437

There can even be empty spaces on a complete board though. – Stephen Ostermiller – 2014-06-24T14:54:56.503

@StephenOstermiller by definition a complete board has something on every square, even though this may be illegal as you should stop playing once someone has won. Read the full discussion, especially around Riot's comment. – Level River St – 2014-06-24T15:07:49.057

1Got it, I thought complete meant something different. – Stephen Ostermiller – 2014-06-24T15:11:06.897

10

Mathematica, 84 chars

a=Input[];Which[Max@#>2,win,Min@#<1,lose,1>0,cat]&@{Tr@a,Tr@Reverse@a,Tr/@a,Total@a}

Input format: {{1, 1, 0}, {1, 0, 1}, {0, 0, 1}}

alephalpha

Posted 2014-06-24T02:29:13.440

Reputation: 23 988

What's happening here? – seequ – 2014-06-24T15:54:38.023

3@TheRare Start from the right. Tr@a is the trace of the field (sum over diagonal), Tr@Reverse@a is the trace of the flipped field (some over anti-diagonal), Tr/@a is Tr applied to each row, which gives you the sum over each row, Total@a gives you the sum over each column. So basically, you've got all 8 lines you need to check. Then the Which thing is applied to that (basically an if/elseif/else statement), where # represents that list of 8 values. if there's a 3 you win, else if there's a 0 you lose, else if 1>0 (true) cat. – Martin Ender – 2014-06-24T17:14:26.497

6

Bash: 283 262 258

Featuring a relatively friendly interface.

t(){ sed 's/X/true/g;s/O/false/g'<<<$@;}
y(){ t $(sed 's/X/Q/g;s/O/X/g;s/Q/O/g'<<<$@);}
f(){($1&&$2&&$3)||($1&&$5&&$9)||($1&&$4&&$7)||($2&&$5&&$8)||($3&&$5&&$7)||($3&&$6&&$9)||($4&&$5&&$6)||($7&&$8&&$9)}
f $(t $@)&&echo win||(f $(y $@)&&echo lose)||echo cat

To execute bash tictactoe.sh O X O X O X X O X

Note: the list of 9 positions is a standard matrix representation. It doesn't matter if the board is represented as column major or row major, read from left to right or top to bottom - games of noughts and crosses (or tic tac toe if you insist) are symmetrical, so input order should be irrelevant to the result in every correct implementation, as long as input is linear.

Edit: Thanks to h.j.k for shorter function syntax suggestion.

Riot

Posted 2014-06-24T02:29:13.440

Reputation: 4 639

Spaces are not required around <<< to save another four characters. – Michael Mior – 2018-10-02T13:10:56.850

Consider t() { ... } instead of function t? Can save some characters there. :) – h.j.k. – 2014-06-25T02:14:56.867

I'd totally forgotten about the alternative function syntax - thanks! – Riot – 2014-06-25T06:51:52.983

4

Befunge 93 - 375

Takes a binary string as input.

99>~\1-:!!|>v  
>0v>v>v   >^$>v
^+ + +    0<:p:
>#+#+#+    ^246
^+ + +    0<265
>#+#+#+    ^pp6
^+ + +    0<2++
 #+#+#+     55p
   0 0      552
  >^>^>0v   +46
v+ + +  <   ppp
>0 + + + v  444
   v!!-3:<< 246
  v_"ni"v   ppp
  0v" w"<   :+:
  \>,,,,@   266
  ->,,,@    555
  !^"cat"_^ 645
  !>:9-! ^  +:+
  >|        p:p
   >"eso"v  6p6
 @,,,,"l"<  246
            p2p
            >^ 
  v       <^  <

Reads the string. Bruteforce writes it (the right most vertical strip) as a matrix in between the

^+ + + 
>#+#+#+
^+ + + 
>#+#+#+
^+ + + 
 #+#+#+

adding lattice (idk). Determins the sum of the columns, rows, and two diagnals. Compares those values to 3 ("win") or 0 ("lose"), else if all the values equal 1 or 2 then draw ("cat").

AndoDaan

Posted 2014-06-24T02:29:13.440

Reputation: 2 232

4

GolfScript, 27 chars

70&.{~"win""lose"if}"cat"if

The input format is a string consisting of eight octal digits, each (redundantly) encoding three consecutive board squares:

  • The first three digits each encode a single row of the board, from top down and left to right.
  • The following three digits each encode a single column of the board, from left to right and top down.
  • The final two digits each encode one of the diagonals (first from top left to bottom right, then from bottom left to top right).

To encode a sequence (row / column / diagonal) of three squares as an octal digit, replace every x in the sequence with a 1 and every o with a 0, and interpret the resulting sequence of ones and zeros as a binary number between 0 and 7 inclusive.

This input format is quite redundant (all board positions are encoded at least twice, with the center position encoded four times), but it does unambiguously represent any possible state of a completely filled tic-tac-toe board, and does not directly encode the winner into the input.

The input may, optionally, contain spaces or other delimiters between digits. In fact, all the program really cares about is whether or not the input string contains the digits 7 or 0.

For example, the example board:

|x|x|o|
|x|o|x|
|o|o|x|

may be represented by the input:

651 643 50

For convenience, here's a GolfScript program to convert an ASCII art board layout, as shown in the challenge above, into an input string suitable for this program:

."XOxo"--[{1&!}/]:a[3/.zip"048642"{15&a=}%3/]{{2base""+}%}%" "*

This converter ignores any characters other than x and o, in either case, in its input. It produces a single digit string (complete with space delimiters as shown above) suitable for feeding into the win-determining program above, so the concatenation of these two programs can be used to determine the winner directly from the ASCII art board.

Also, here's a reverse converter, just to demonstrate that the input indeed does unambiguously represent the board:

.56,48>-- 3<{2base-3>{"ox"=}%n}%"|".@@*+);

Ps. Here's an online demo of this solution.

Ilmari Karonen

Posted 2014-06-24T02:29:13.440

Reputation: 19 513

2The input format seems like bit of a cheat as much of the work happens in producing the input. – Arkku – 2014-06-24T13:50:16.823

@Arkku: Well, yes, it is, but the question explicitly says that "you get to determine your input structure for the state - which you must then explain." It even shows a bit-packed hex string as an example of a valid input format; the only difference between that and my input format is that I reorder and duplicate some bits. – Ilmari Karonen – 2014-06-24T14:00:08.650

6It is exactly the duplication that seems like a cheat. (e.g., it does directly encode the winner as the presence of 7 or 0 in the input) – Arkku – 2014-06-24T14:03:23.547

Still that's a clever encoding, it's redundant, but makes finding the solution much more efficient than any non-redundant encoding ! – ARRG – 2014-06-24T14:29:45.123

3

Python 2 - 214 bytes

b=eval(raw_input())
s=map(sum,b)
w,l='win','lose'
e="if min(s)<1:print l;a\nif max(s)>2:print w;a"
exec e+'\ns=map(sum,zip(*b))\n'+e
m=b[1][1]
for i in 0,2:
 if m==b[0][i]==b[2][abs(i-2)]:print[l,w][m];a
print'cat'

I'm sure there are improvements to be made.

To run:

python2 tictactoe.py <<< '[[1,1,1],[1,0,1],[0,1,0]]'

which represents this board:

X|X|X
-----
X|O|X
-----
0|X|0

Exits with a NameError exception in every case except cat.

undergroundmonorail

Posted 2014-06-24T02:29:13.440

Reputation: 5 897

Whoa, I never knew about <<<! +1 just for that. – Greg Hewgill – 2014-06-24T03:22:01.293

@GregHewgill It's pretty convenient. ./whatever <<< 'blah blah blah' is the same as echo -n 'blah blah blah' | ./whatever but without having a whole separate process for echo. – undergroundmonorail – 2014-06-24T03:24:13.747

@undergroundmonorail echo in bash is actually a builtin, so doesn't fork a new process – Bob – 2014-06-24T16:43:59.110

@GregHewgill it's called a herestring – None – 2014-06-25T14:30:38.423

3

JavaScript, 420 chars

if((s&0x3F000)==0x3F000||(s&0x00FC0)==0x00FC0||(s&0x0003F)==0x0003F||(s&0x030C3)==0x030C3||(s&0x0C30C)==0x0C30C||(s&0x30C30)==0x30C30||(s&0x03330)==0x03330||(s&0x30303)==0x30303)return 'win'
if((s&0x3F000)==0x2A000||(s&0x00FC0)==0x00A80||(s&0x0003F)==0x0002A||(s&0x030C3)==0x02082||(s&0x0C30C)==0x08208||(s&0x30C30)==0x20820||(s&0x03330)==0x02220||(s&0x30303)==0x20202)return 'lose'
if((s&0x2AAAA)==0x2AAAA)return 'cat'

In this version, s contains an integer which represents the state of the game board. It is a bit array of value where two bits represents each square on the board:

  • 10 - X
  • 11 - O
  • 00 - Empty square

This solution uses bit manipulation to test each of the eight possible "three in a row" configurations (it tests them each twice, once for X and once for O).

I present this with minor minification from my Tic-Tac-Toe website where this detectWin function is in use as part of a real Tic-Tac-Toe game.

Stephen Ostermiller

Posted 2014-06-24T02:29:13.440

Reputation: 131

6Well, this could be called brute forcing it. – seequ – 2014-06-24T17:12:28.493

3

Haskell, 146 chars

To make things interesting, you get to determine your input structure for the state- which you must then explain.

OK :). My representation of a board is one of those 126 characters

ĻŃŇʼnŊœŗřŚşšŢťŦŨųŷŹźſƁƂƅƆƈƏƑƒƕƖƘƝƞƠƤƳƷƹƺƿǁǂDždžLjǏǑǒǕǖǘǝǞǠǤǯDZDzǵǶǸǽǾȀȄȍȎȐȔȜȳȷȹȺȿɁɂɅɆɈɏɑɒɕɖɘɝɞɠɤɯɱɲɵɶɸɽɾʀʄʍʎʐʔʜʯʱʲʵʶʸʽʾˀ˄ˍˎː˔˜˭ˮ˰˴˼̌

Here's the solution in 146 chars :

main=interact$(\x->case(head x)of h|elem h "ĻŃœťŦŨųŷŹƁƂƅƈƕƠƤƳƿǂdžǞǤǵǾȀȳȿɁɅɑɒɘɝɠɤɵɽʀʐʽʾː˭ˮ˰˴˼̌"->"lose";h|elem h "ƏƝƞƹǁLjǑǝȍȺɆɈɶɾʎʸ"->"cat";h->"win")

And here's how it works, as an haskell script :

import Data.List (subsequences, (\\))
import Data.Char (chr)

-- A set of indexes [0-8] describing where on the board pieces of a single color have been played
-- For example the board "OxO;Oxx;xxO" is indexes [0,2,3,8]
type Play = [Int]

-- There are 126 filled tic tac toe boards when X plays first.
--      (This is a combination of 4 OHs among 9 places : binomial(9 4) = 126)
-- perms returns a list of all such possible boards (represented by the index of their OHs).
perms = filter (\x -> 4 == length x) $ subsequences [0..8]

-- We now create an encoding for plays that brings them down to a single char.
-- The index list can be seen as an 9 bit binary word [0,2,3,8] -> '100001101'
-- This, in turn is the integer 269. The possible boards give integers between 15 and 480.
-- Let's call those PlayInts
type PlayInt = Int

permToInt [] = 0
permToInt (x:xs) = (2 ^ x) + permToInt xs 

-- Since the characters in the range 15-480 are not all printable. We offset the chars by 300, this gives the range 
-- ĻŃŇʼnŊœŗřŚşšŢťŦŨųŷŹźſƁƂƅƆƈƏƑƒƕƖƘƝƞƠƤƳƷƹƺƿǁǂDždžLjǏǑǒǕǖǘǝǞǠǤǯDZDzǵǶǸǽǾȀȄȍȎȐȔȜȳȷȹȺȿɁɂɅɆɈɏɑɒɕɖɘɝɞɠɤɯɱɲɵɶɸɽɾʀʄʍʎʐʔʜʯʱʲʵʶʸʽʾˀ˄ˍˎː˔˜˭ˮ˰˴˼̌
-- Of all distinct, printable characters
uOffset = 300

-- Transform a PlayInt to its Char representation
pIntToUnicode i = chr $ i + uOffset

-- Helper function to convert a board in a more user friendly representation to its Char
-- This accepts a representation in the form "xooxxxoxo"
convertBoard s = let play = map snd $ filter (\(c, i) -> c == 'o') $ (zip s [0..]) :: Play 
    in pIntToUnicode $ permToInt play

--
-- Now let's cook some data for our final result
--  

-- All boards as chars
allUnicode = let allInts = map permToInt perms 
    in map pIntToUnicode allInts

-- Now let's determine which boards give which outcome.

-- These are all lines, columns, and diags that give a win when filled
wins = [
        [0,1,2],[3,4,5],[6,7,8], -- lines
        [0,3,6],[1,4,7],[2,5,8], -- columns
        [0,4,8],[2,4,6] -- diagonals
    ]

isWin :: Play -> Bool   
isWin ps = let triplets = filter (\x -> 3 == length x) $ subsequences ps -- extract all triplets in the 4 or 5 moves played
    in any (\t -> t `elem` wins) triplets -- And check if any is a win line

-- These are OH wins
oWins = filter isWin perms
-- EX wins when the complement board wins
xWins = filter (isWin . complement) perms
    where complement ps = [0..9] \\ ps
-- And it's stalemate otherwise
cWins = (perms \\ oWins) \\ xWins

-- Write the cooked data to files
cookData = let toString = map (pIntToUnicode . permToInt) in do
  writeFile "all.txt" allUnicode
  writeFile "cWins.txt" $ toString cWins
  writeFile "oWins.txt" $ toString oWins
  writeFile "xWins.txt" $ toString xWins

-- Now we know that there are 48 OH-wins, 16 stalemates, and 62 EX wins (they have more because they play 5 times instead of 4).
-- Finding the solution is just checking to which set an input board belongs to (ungolfed :)
main = interact $ \x -> case (head x) of -- Only consider the first input char
    h | elem h "ĻŃœťŦŨųŷŹƁƂƅƈƕƠƤƳƿǂdžǞǤǵǾȀȳȿɁɅɑɒɘɝɠɤɵɽʀʐʽʾː˭ˮ˰˴˼̌" -> "lose" -- This string is == oWins
    h | elem h "ƏƝƞƹǁLjǑǝȍȺɆɈɶɾʎʸ" -> "cat" -- And this one == cWins
    h -> "win"

ARRG

Posted 2014-06-24T02:29:13.440

Reputation: 141

2

Ruby, 84 characters

$><<(gets.tr("01","10")[r=/0..(0|.0.)..0|000(...)*$|^..0.0.0/]?:win:~r ?:lose: :cat)

Simple, RegExp based solution. The input format is a 9-digit binary string, e.g. 110101001 for the example board given in the question.

Ruby, 78 characters

$><<(gets.tr("ox","xo")[r=/o...(o|.o.)...o|ooo|o_.o._o/]?:win:~r ?:lose: :cat)

Input format: xxo_xox_oox

Ventero

Posted 2014-06-24T02:29:13.440

Reputation: 9 842

1

Bash: 208 chars

y(){ tr '01' '10'<<<$@;}
f(){ x=$[($1&$2&$3)|($1&$5&$9)|($1&$4&$7)|($2&$5&$8)|($3&$5&$7)|($3&$6&$9)|($4&$5&$6)|($7&$8&$9)]; }
f $@;w=$x
f $(y $@)
([ $x -eq 1 ]&&echo lose)||([ $w -eq 1 ]&&echo win)||echo cat

To execute bash tictactoe.sh 0 1 0 1 0 1 1 0 1

Inspired by this answer.

Michael Mior

Posted 2014-06-24T02:29:13.440

Reputation: 111

1

Haskell, 169

main=interact$(\x->last$"cat":[b|(a,b)<-[("ooo","lose"),("xxx","win")],any(==a)x]).(\x->x++(foldr(zipWith(:))(repeat[])x)++map(zipWith(!!)x)[[0..],[2,1,0]]).take 3.lines

Input format: "X" is represented only by x, "O" only by o. Within each row, characters are simultaneous without spaces, etc. Rows are separated by new lines.

Generates all possible rows/columns/diagonals, then filters [("ooo","lose"),("xxx","win")] by their existence on the board, then selects the second word in the tuple, so we know which players won. We prepend "cat" so that we can take the last element of the list as our winner. If both players won, "win" will be last (list comprehensions maintain order). Since "cat" is always first, if a winner exists, it will be chosen, but otherwise a last element still exists as prepending "cat" guarantees nonemptyness.

EDIT: Shaved 3 characters by changing last list comprehension to map.

YawarRaza7349

Posted 2014-06-24T02:29:13.440

Reputation: 71

1

Dart - 119

(See dartlang.org).

Original version using RegExp: 151 chars.

main(b,{w:"cat",i,p,z}){
 for(p in["olose","xwin"])
   for(i in[0,2,3,4])
     if(b[0].contains(new RegExp('${z=p[0]}(${'.'*i}$z){2}')))
       w=p.substring(1);
  print(w);
}

Input on the command line is 11 characters, e.g., "xxx|ooo|xxx". Any non-xo character can be used as delimiter.

Leading whitespace and newlines should be omitted before counting characters, but I cut away the internal whitespace where possible. I wish there was a smaller way to make the substring.

Recusive bit-base version: 119 chars. Input must be a 9-bit number with 1s representing 'x' and 0s representing 'o'.

main(n){
  n=int.parse(n[0]);
  z(b,r)=>b>0?b&n==b&511?"win":z(b>>9,n&b==0?"lose":r):r;
  print(z(0x9224893c01c01e2254,"cat"));
}

lrn

Posted 2014-06-24T02:29:13.440

Reputation: 521

1

C,150 approx

It's midnight here and I haven't done any testing, but I'll post the concept anyway. I'll get back to it tomorrow.

User inputs two octal numbers (I wanted to use binary but as far as I know C only supports octal):

a represents the centre square, 1 for an X, 0 for an O

b is a nine-digit number representing the perimeter squares, circling round the board starting in one corner and finishing in the same corner (with repeat of that corner only), 1 for an X, 0 for an O.

There are two possible ways to win:

  1. centre square is X (a=1) and two opposite squares are also X (b&b*4096 is nonzero)

  2. three adjacent perimeter squares are X (b/8 & b & b*8 is nonzero.) This is only a valid win if the middle square is an edge square, not a corner square, therefore it is necessary to apply the mask m also, to avoid the corner square cases.

Losing is detected using the variable c, which is the inverse of b.

int a,b,c,m=010101010;
main(){
    scanf("%o%o",a,b);c=b^0111111111;
    printf("%s",(a&&b&b*4096)|(b/8&b&b*8&m)?"win":((!a&&c&c*4096)|(c/8&c&c*8)?"lose":"cat"));
}

Level River St

Posted 2014-06-24T02:29:13.440

Reputation: 22 049

You forgot to apply the mask m in the "lose" detection - c/8&c&c*8. I have re-golfed your code (without testing its operation) as follows: int a,b;t(v){return a&&v&v<<12||v/8&v&v*8&0x208208;}main(){scanf("%o%o",a,b);printf("%s",t(b)?"win":t(b^0x1249249)?"lose":"cat");} (130 chars). The repeated test was long enough to extract into a test function t(); this removes the need for c and m; the constants converted to hex to save one char each. – Toby Speight – 2015-06-05T10:59:29.797

Just spotted that the printf doesn't need a format string - just supply the result string as the format - or puts it, since the question doesn't ask for a newline after the output! (saves a further 7 chars). – Toby Speight – 2015-06-05T11:02:21.230

1

Bash, 107 103

Generates and runs a sed script.

I/O format: oxo-oox-xoo outputs lose (use a - to separate rows). Input on stdin. Requires GNU sed for the c command.

I've interpreted rule 5 as "if both win and lose are possible, choose win".

Main Code

This is the actual answer.

Nothing interesting really. It defines $b as /cwin to save characters, then defines the win condition part of the script, then uses sed y/x/o/\;s$b/close/ to convert x to o and cwin to close (thereby generating the lose conditions). It then sends the two things and ccat (which will output cat if no win/lose condition is matched) to sed.

b=/cwin
v="/xxx$b
/x...x...x$b
/x..-.x.-..x$b
/x-.x.-x$b"
sed "$v
`sed y/x/o/\;s$b/close/<<<"$v"`
ccat"

Generated Code

This is the sed script generated and run by the Bash script.

In the regexes, . matches any character and after them cTEXT prints TEXT and exits if the regex is matched.

This can run as a standalone sed script. It's 125 characters long, you can count it as another solution.

/xxx/cwin
/x...x...x/cwin
/x..-.x.-..x/cwin
/x-.x.-x/cwin
/ooo/close
/o...o...o/close
/o..-.o.-..o/close
/o-.o.-o/close
ccat

user16402

Posted 2014-06-24T02:29:13.440

Reputation:

1

Python 3, 45

Input is in i, which is a list of numbers representing each row, column, and diagonal of the game board, e.g:

X X O
O X O
O O X

is represented by [6, 2, 1, 4, 6, 1, 7, 4].

Code: ('cat','lose','win')[2 if 7 in i else 0 in i]

pseudonym117

Posted 2014-06-24T02:29:13.440

Reputation: 1 053

1

CJam, 39 38 36 characters

"ᔔꉚ굌궽渒䗠脯뗠㰍㔚귇籾〳㎪䬔⹴쪳儏⃒ꈯ琉"2G#b129b:c~

This is a base converted code for

q3/_z__Wf%s4%\s4%]`:Q3'o*#"win"{Q'x3*#"lose""cat"?}?

which is 52 characters long.

The input is simply the string representation of the board starting from top left, going row by row. For example:

oxooxooox

which results in a win output. Or

oxooxoxox

which results in a cat output, etc.

The code simply does the following three things:

  • q3/_ - Split the string into parts of 3, i.e. per row
  • _z - Copy the per row array and transpose into per column array.
  • __Wf%s4% - Reverse each row and get the left to right diagonal. This is the secondary diagonal of the board.
  • \s4% - Get the main diagonal of the board
  • ]` - Wrap everything in array and stringify the array.

Now we have all possible groups of 3 from the board. We simply check for existence of "ooo" and "xxx" to determine the result.

Try it online here

Optimizer

Posted 2014-06-24T02:29:13.440

Reputation: 25 836

1

GNU sed, 25 bytes

If the input is a redundant representation of the board with separate views for columns, rows and diagonals, as used in other answers as well, then sed is very well suited for checking the end state of the game with the least bytes.

Input format: xxx ooo xxx xox xox xox xox xox (board state taken from the OP's question)

/xxx/cwin
/ooo/close
ccat

If the input format is non-redundant (xxx ooo xxx), then the sed code above works only if prepended by the line below, making the program 96 bytes long (with the needed r flag counted).

s/(.)(.)(.) (.)(.)(.) (.)(.)(.)/& \1\4\7 \2\5\8 \3\6\9 \1\5\9 \3\5\7/

seshoumara

Posted 2014-06-24T02:29:13.440

Reputation: 2 878

0

APL(NARS), 69 chars, 138 bytes

{w←3 3⍴⍵⋄x←(+/1 1⍉⊖w),(+/1 1⍉w),(+⌿w),+/w⋄3∊x:'win'⋄0∊x:'lose'⋄'cat'}

The input should be one 3x3 matrix or one linear array of 9 element that can be only 1 (for X) and 0 (for O), the result will be "cat" if nobody wins, "lose" if O wins, "win" if X wins. There is no check for one invalid board or input is one array has less than 9 element or more or check each element <2.

As a comment: it would convert the input in a 3x3 matrix, and build one array named "x" where elements are the sum each row column and diagonal.

Some test see example showed from others:

  f←{w←3 3⍴⍵⋄x←(+/1 1⍉⊖w),(+/1 1⍉w),(+⌿w),+/w⋄3∊x:'win'⋄0∊x:'lose'⋄'cat'}
  f 1 2 3
win
  f 0 0 0
lose
  f 1 0 1  1 0 1  1 0 1
win
  f 0 1 1  1 0 0  1 1 1
win
  f 0 0 1  1 0 1  1 1 0
lose
  f 1 1 0  0 1 1  1 0 0
cat
  f 1 1 0  0 1 0  0 0 1
win
  f 1 1 0  1 0 1  0 0 1
lose

RosLuP

Posted 2014-06-24T02:29:13.440

Reputation: 3 036

0

VB.net

With the example provide is encoded as the following bit-pattern

q  = &B_100101_100110_011010 ' 00 Empty, 01 = O, 10 = X

Now we can determine the result (or winner) by doing the following.

Dim g = {21, 1344, 86016, 66576, 16644, 4161, 65379, 4368}
Dim w = If(g.Any(Function(p)(q And p)=p),"Lose",If(g.Any(Function(p)(q And p*2)=p*2),"Win","Cat"))

Adam Speight

Posted 2014-06-24T02:29:13.440

Reputation: 1 234

0

J - 97 bytes

Well, the simplest approach available. The input is taken as 111222333, where the numbers represent rows. Read left-to-right. Player is x and enemy is o. Empty squares can be anything except x or o.

f=:(cat`lose>@{~'ooo'&c)`('win'"_)@.('xxx'&c=:+./@(r,(r|:),((r=:-:"1)(0 4 8&{,:2 4 6&{)@,))3 3&$)

Examples: (NB. is a comment)

   f 'xoxxoxxox' NB. Victory from first and last column.
win
   f 'oxxxooxxx' NB. Victory from last row.
win
   f 'ooxxoxxxo' NB. The example case, lost to a diagonal.
lose
   f 'xxooxxxoo' NB. Nobody won.
cat
   f 'xoo xx ox' NB. Victory from diagonal.
win

Ungolfed code an explanation

row   =: -:"1                        Checks if victory can be achieved from any row.
col   =: -:"1 |:                     Checks if victory can be achieved from any column.
diag  =: -:"1 (0 4 8&{ ,: 2 4 6&{)@, Checks if victory can be achieved from diagonals.
check =: +./@(row,col,diag) 3 3&$    Checks all of the above and OR's them.

f     =: (cat`lose >@{~ 'ooo'&check)`('win'"_)@.('xxx'&check)
Check if you have won ........................@.('xxx'&check)
 If yes, return 'win' .............. ('win'"_)
 If not                   (cat`lose >@{~ 'ooo'&check)
  Check if enemy won ................... 'ooo'&check
   If yes, return 'lose'   ---`lose >@{~
   If not, return 'cat'    cat`---- >@{~

seequ

Posted 2014-06-24T02:29:13.440

Reputation: 1 714

0

Python 2, 120 bytes

b=0b101001110
l=[448,56,7,292,146,73,273,84]
print(['Win'for w in l if w&b==w]+['Lose'for w in l if w&~b==w]+['Cat'])[0]

Or Python, 115 bytes from the Python shell (2 or 3):

b=0b101001110;l=[448,56,7,292,146,73,273,84];(['Win'for w in l if w&b==w]+['Lose'for w in l if w&~b==w]+['Cat'])[0]

The board variable is set to the binary format described in the question: 1 for X, 0 for O, left-to-right, top-to-bottom. In this case, 101001110 represents

XOX
OOX
XXO

Which leads to output: Cat

Cees Timmerman

Posted 2014-06-24T02:29:13.440

Reputation: 625

What is the input format? – seequ – 2014-06-24T17:32:17.583

0

Python (73 62 chars)

The input is four lower-case strings representing four distinct views of the same board, all concatenated into a single string: by row, by column, right-diagonal, left-diagonal.

UPDATE

Thanks to theRare for pointing this out with a good counter-example! Each view of the board, along with each segment (row or column) within a board has to be separated by a character that is neither an "x" or an "o" so that the structure of the board is preserved even after concatenation. The borders around each view of the board will be square brackets ("[" and "]"), and the separator between rows/columns will be a pipe character "|".

This makes the algorithm simple--just look for "xxx" or "ooo" for a win or loss, respectively. Otherwise it's a tie (cat).

E.g. the board (reading left-to-right, top-to-bottom)...

X|X|X X|O|X O|X|O

...is represented as "[xxx|xox|oxo]" (by rows) + "[xxo|xox|xxo]" (by columns) + "[xoo]" (right diag) + [xoo]" (left diag) = "[xxx|xox|oxo][xxo|xox|xxo][xoo][xoo]".

This Python statement prints the game result given the variable s as input:

print 'win' if 'xxx' in s else 'lose' if 'ooo' in s else 'cat'

bob

Posted 2014-06-24T02:29:13.440

Reputation: 109

Does this work for board OXX XOO XOX (it should be cat)? – seequ – 2014-06-24T18:50:14.567

No...no it doesn't. Good catch! I guess my solution was a little too simple... Oops! – bob – 2014-06-24T19:41:08.203

I can't say this type of solution didn't cross my mind. :) – seequ – 2014-06-24T19:47:12.543

0

Javascript 1.6, 71 chars

I'm assuming input as an array game which contains each row, each column and each diag as a 3 char string. Similar to bob's answer, but it comes in an array, not as a concatenated string.

alert(game.indexOf("xxx")>=0?"win":game.indexOf("ooo")>=0?"lose":"cat")

EDIT @nyuszika7h's comment (67 chars)

alert(~game.indexOf("xxx")?"win":~game.indexOf("ooo")?"lose":"cat")

JNF

Posted 2014-06-24T02:29:13.440

Reputation: 131

You can use ~game.indexOf("xxx") instead of game.indexOf("xxx")>=0, same for the other one. – nyuszika7h – 2014-07-05T12:54:40.473

0

JavaScript, 133, 114 characters

r = '/(1){3}|(1.{3}){2}1|(1.{4}){2}1|(1\|.1.\|1)/';alert(i.match(r)?'WIN':i.match(r.replace(/1/g,0))?'LOSS':'CAT')

The input i is a simple string with delimiters for the rows, i.e. 100|001|100

Edit: updated my method to replace the 1s in the regex with zeroes to check for the loss case.

thomaux

Posted 2014-06-24T02:29:13.440

Reputation: 219

You can remove the spaces around = and the quotes around the regex literal. Also, 1... is one character shorter than 1.{3}. – nyuszika7h – 2014-07-05T12:56:30.470

1r.test(i) is also one character shorter than i.match(r). – nyuszika7h – 2014-07-05T12:58:27.253

0

Haskell (69 chars)

i x=take 4$(x>>=(\y->case y of{'7'->"win";'0'->"lose";_->""}))++"cat"

This takes the same input as described by this answer. More specifically, the input is 8 octal values, describing the binary value of each row, column, and diagonal. The code makes every instance of 7 "win", every instance of 0 "lose", and removes everything else. Then it adds "cat" to the end and takes the first 4 chars from the result.

There will be 4 possible answers: "lose", "cat", "win" followed by an 'l', and "win" followed by a 'c', which the rules don't prohibit :)

Example usage:

i "65153806" --outputs "lose"

TheBrownMotie

Posted 2014-06-24T02:29:13.440

Reputation: 1

0

J : 83

(;:'lose cat win'){::~>:*(-&(+/@:(*./"1)@;@(;((<0 1)&|:&.>@(;|.)(,<)|:)))-.)3 3$'x'=

Usage: just append a string of x's and o's and watch the magic work. eg. 'xxxoooxxx'.

The inner verb (+/@:(*./"1)@;@(;((<0 1)&|:&.>@(;|.)(,<)|:))) basically boxes together the original binary matrix, with the transpose boxed together with the 2 diagonals. These results are razed together ; row sums are taken to determine wins, and then summed. further I'll call this verb Inner.

For finding the winner, the difference of the scores between the normal and inversed binary matrices is taken by the hook (-&Inner -.).

The rest of the code simply makes the outputs, and selects the right one.

jpjacobs

Posted 2014-06-24T02:29:13.440

Reputation: 3 440

0

J - 56 (26?) char

Input is given a 3x3 matrix of nine characters, because J can support that as a datatype, LOL.

(win`lose`cat{::~xxx`ooo<./@i.<"1,<"1@|:,2 7{</.,</.@|.)

Examples:

   NB. 4 equivalent ways to input the example board
   (3 3 $ 'xxoxoxoox') ; (_3 ]\ 'xxoxoxoox') ; ('xxo','xox',:'oox') ; (];._1 '|xxo|xox|oox')
+---+---+---+---+
|xxo|xxo|xxo|xxo|
|xox|xox|xox|xox|
|oox|oox|oox|oox|
+---+---+---+---+
   (win`lose`cat{::~xxx`ooo<./@i.<"1,<"1@|:,2 7{</.,</.@|.) 3 3 $ 'xxoxoxoox'
lose
   wlc =: (win`lose`cat{::~xxx`ooo<./@i.<"1,<"1@|:,2 7{</.,</.@|.)
   wlc (3 3 $ 'xoxoxooxo')
cat
   wlc (3 3 $ 'xxxoooxxx')
win

If we are allowed the Golfscriptish encoding of octal digits redundantly representing the state of each row, column, and diagonal, then it's just 26 characters:

   win`lose`cat{::~7 0<./@i.] 6 5 1 6 4 3 5 0
lose
   f=:win`lose`cat{::~7 0<./@i.]
   f  7 0 7 5 5 5 5 5
win

algorithmshark

Posted 2014-06-24T02:29:13.440

Reputation: 8 144

0

T-SQL (2012),110

select max(iif(@&m=0,'lose',iif(@&m=m,'win','cat')))from(VALUES(292),(146),(73),(448),(56),(7),(273),(84))z(m)

Input is a hex number. This is pretty much a translation of the ruby solution into T-SQL pretty nice and neat.

Michael B

Posted 2014-06-24T02:29:13.440

Reputation: 1 551

0

Java 7, 260 bytes

String c(int[]s){int a[]=new int[8],x=0,y;for(;x<3;x++){for(y=0;y<3;a[x]+=s[x*3+y++]);for(y=0;y<3;a[x+3]+=s[y++%3]);}for(x=0;x<9;y=s[x],a[6]+=x%4<1?y:0;a[7]+=x%2<1&x>0&x++<8?y:0);x=0;for(int i:a)if(i>2)return"win";for(int i:a)if(i<1)return"loose";return"cat";}

Ungolfed & test cases:

Try it here.

class M{
  static String c(int[] s){
    int a[] = new int[8],
        x = 0,
        y;
    for(; x < 3; x++){
      for(y = 0; y < 3; a[x] += s[x * 3 + y++]);
      for (y = 0; y < 3; a[x + 3] += s[y++ % 3]);
    }
    for(x = 0; x < 9; y = s[x],
                      a[6] += x % 4 < 1
                               ? y
                               : 0,
                      a[7] += x % 2 < 1 & x > 0 & x++ < 8
                               ? y
                               : 0);
    x = 0;
    for(int i : a){
      if(i > 2){
        return "win";
      }
    }
    for(int i : a){
      if(i < 1){
        return "loose";
      }
    }
    return "cat";
  }

  public static void main(String[] a){
    /*  xxo
        xox
        oox  */
    System.out.println(c(new int[]{ 1, 1, 0, 1, 0, 1, 0, 0, 1 }));
    /*  xxx
        ooo
        xxx  */
    System.out.println(c(new int[]{ 1, 1, 1, 0, 0, 0, 1, 1, 1 }));
    /*  xxo
        oox
        xox  */
    System.out.println(c(new int[]{ 1, 1, 0, 0, 0, 1, 1, 0, 1 }));
  }
}

Output:

loose
win
cat

Kevin Cruijssen

Posted 2014-06-24T02:29:13.440

Reputation: 67 575