Tutorial :Code Golf: Who has the best poker hand?


I love challenges like this, I'll hopefully submit my answer soon.

Which player has the best 7 card hand?

Given an unordered list of 9 cards (separated by a space), work out which player has the best poker hand. Here is a list of poker hand rankings. Example input:

2C 5H AS KS 2D 4D QD KH 3S  (ie: [[2C 5H] [AS KS] [2D 4D QD KH 3S]])  

First 2 cards in the array represent player 1's hand, second 2 in the array represent player 2's hand. The last 5 cards represent the community cards, cards both players share. In effect, both players have 7 cards, and you must determine which player has the best 5 card poker hand.

A card is defined as a string, with the first character representing the card value, and the second value representing the suit. Always upper-case. No card may appear twice.

The function will calculate if the hand is a draw or a win to either player. It will ouput the totals at the end of the input. The output format is defined later on in this post.


2C 5H AS KS 2D 4D QD KH 3S  (ie: [[2C 5H] [AS KS] [2D 4D QD KH 3S]])  Player 2 wins this hand.  Player 1 has a pair of 2's, player 2 has a pair of kings.    5S 6S 8H 9D 7S 8S JH TS 2H  (ie: [[5S 6S] [8H 9D] [7S 8S JH TS 2H]])  Player 1 wins this hand  Player 1 has a flush, player 2 has a straight.    2S 2H AC AS 2C AH 9H TS 2D      (ie: [[2S 2H] [AC AS] [2C AH 9H TS 2D]])  Player 1 wins this hand.  Player 1 has quads, player 2 has a full house    5S 6S 2D 4D 9S AS KD JC 9D  (ie: [[5S 6S] [2D 4D] [9S AS KD JC 9D]])  A draw.  Both players have Ace high.  

More Info

Thanks to mgroves for the following link to Project Euler which has a similar problem: http://projecteuler.net/index.php?section=problems&id=54

Test Data

We will use the Project Euler test data:


Your solution should accept that text file as input, and output a total of wins and draws.

Example Output

Output must be in this format:

1: 45  2: 32  D: 12  

Player 1 won 45 hands, player 2 won 32 hands, and there were 12 draws. (Not actual results)


  • Doesn't have to return the winning hand type, only WHO won if anyone
  • Card list input has no particular order
  • No card appears twice in the input
  • Input is always uppercase
  • Takes the Project Euler test data as an input
  • Outputs a count, of which player won the most hands and total draws in given format above


Perl, 414 398 370/458 344/416 char

Line breaks are not significant.

%M=map{$_,$Z++}0..9,T,J,Q,K,A;sub N{/.$/;$M{$`}.$&}    sub B{$s=@p=();  for$m(@_){$m-$_||($s+=2,++$p[$m])for@_}  @_=sort{$p[$b]-$p[$a]||$b-$a}@_;  $s=23 if$s<11&&($_[0]-$_[4]<5||$_[0]-$_[1]>8&&push@_,shift);  "@_"=~/.$/;$s+=14*(4<grep/$&/,@_);  $s=100*$s+$_ for@_;$s}    ++$X{B((@c=map{N}split)[0..4])<=>B(@c[5..9])}for<>;  printf"1: %d\n2: %d\nD: %d\n",@X{1,-1,0}  

This solves the "10 card" problem (10 cards are dealt, player 1 has the first 5 cards and player 2 has the second 5 cards).

The first section defines a subroutine N that can transform each card so that it has a numerical value. For non-face cards, this is a trivial mapping (5H ==> 5H) but it does transform the face cards (KC => 13C, AD => 14D).

The last section parses each line of input into cards, transforms the cards to contain numerical values, divides the cards into separate hands for the two players, and analyzes and compares those hands. Every hand increments one element of the hash %X. When all the input is parsed, %X contains the number of hands won by player 1, won by player 2, or ties.

The middle section is a subroutine that takes a set of five cards as input and produces a 12-digit number with the property that stronger poker hands will have higher-valued numbers. Here's how it works:


This is the "pair" detector. If any two cards have the same numerical value, increment a hash element for one of the cards and increase the "score" variable $s by two. Note that we will end up comparing each card to itself, so $s will be at least 10 and $p[$x] will be at least one for every card $x. If the hand contains three of a kind, then those three cards will match with the other two cards -- it will be like there are 9 matches among those three cards and the "score" will be at least 18.


Sort the cards by (1) the number of times that card is part of a "pair" and (2) the value of the card. Thus in a hand with two 7's and two 3's, the two 7's will appear first, followed by the two 3's, followed by the kicker. In a hand with two 7's and three 3's, the three 3's will be first followed by the two 7's. The goal of this ordering is to distinguish two hands that have the same score -- a hand with a pair of 8's and a hand with a pair of 7's both have one pair, but we need to be able to tell that a pair of 8's is better.

    $s=23 if$s<11&&($_[0]-$_[4]<5||$_[0]-$_[1]>8&&push@_,shift);  

This line is the "straight" detector. A straight is worth 23 points and occurs when there are no pairs in the hand ($s<11 means only 5 "pairs" - each card matching with itself - were found) and either (1) the value of the highest card is exactly four more than the value of the lowest card ($_[0]-$_[4]==4), or (2) the highest value card is an Ace and the next highest card is a 5 ($_[0]-$_[1]==9), which means the hand has an A-2-3-4-5 straight. In the latter case, the Ace is now the least valuable card in the hand, so we manipulate @_ to reflect that (push@_,shift)


This line is the flush detector. A flush is worth 14 more points and occurs when the last character is the same for each card. The first expression ("@_"=~/.$/) has the side effect of setting $& to the last character (the suit) of the last card in the hand. The final expression (4<grep/$&/,@_) will be true if and only if all elements of @_ have the same last character.

    $s=100*$s+$_ for@_;$s}  

Creates and returns a value that begins with the hand's score and then contains the values of the cards, in order of the card's importance. Scores for the various hands will be

Hand           Score  ----------    ------  High card       10     (each card matches itself for two points)  One pair        14     (2 additional matches)  Two pair        18     (4 additional matches)  Three of a kind 22     (6 additional matches)  Straight        23     (no pair, but 23 points for straight)  Flush           24     (no pair, but 14 additional points for the flush)  Full house      26     (8 additional matches)  4 of a kind     34     (12 additional matches)  Straight flush  37     (23 + 14 points)  

which is consistent with the rules of poker. Hands with the same score can be distinguished by the values of the hand's cards, in order of importance to the hand, all the way down to the least valuable card in the hand.

The solution to the 9 card problem (two cards to player 1, two cards to player 2, the players share the next 5 cards and build their best 5 card hand) needs about 70 more strokes to choose the best 5 card hand out of the 7 cards available to each player:

%M=map{$_,$Z++}0..9,T,J,Q,K,A;sub N{/./;$M{$&}.$'}    sub A{my$I;  for$k(0..41){@d=@_;splice@d,$_,1for$k%7,$k/7;$s=@p=();  for$m(grep$_=N,@d){$m-$_||($s+=2,$p[$m]++)for@d}  @d=sort{$p[$b]-$p[$a]||$b-$a}@d;  $s=23 if$s<11&&($d[0]-$d[4]<5||$d[0]-$d[1]>8&&push@d,shift@d);  "@d"=~/.$/;$s+=14*(4<grep/$&/,@d);  $s=100*$s+$_ for@d;  $I=$s if$s>$I}$I}    ++$X{A((@c=split)[0,1,4..8])<=>A(@c[2..8])}for<>;  printf"1: %d\n2: %d\nD: %d\n",@X{1,-1,0}  


GolfScript - 151/187 chars

This program works on an input list of 10 cards per line, i.e. two 5 card hands.

n%0.@{3/5/{[zip~;.&,(!15*\[{n),*"TJQKA"+?}/]:|$),-4>=14*+1|{.2\?|@-,5\-.49?@*@+\.+@+\}/.16445=13*@+\]}%.~={0):0;;}{~>.!@+\@+\}if}/"1: "@@n"2: "@n"D: "0  

This program works on an input list of 9 cards per line, of the format described in the specifications.

n%0.@{3/.4>:D;2/2<{D+.{3/1$^.{3/1$^[zip~;.&,(!15*\[{n),*"TJQKA"+?}/]$:|),-4>=14*+1|{.2\?|@-,5\-.49?@*@+\.+@+\}/.16445=13*@+\]}%\;~}%$-1=\;}%.~={0):0;\(\}*~>.!@+\@+\}/"1: "@@n"2: "@n"D: "0  


Haskell: 793 796 806 826 864 904 901 880 863

Since the text file is inconsistent with 9 card hands, I'm just reading a line from the console and outputting who wins.


  • Ace now counts lower than a 2 in an ace-low run.
  • Comparing full houses fixed (again :D).
  • Guarantees that the best version of a given hand type is chosen. For example, if a player can choose between a 2-6 run and a 3-7 run, the 3-7 run is chosen (flushes aside).
  • Now shorter than the PHP solution!


import Data.List  (%)=mod  m=map  y=foldr1  t=0<1  z=13  w=[0,1,2,3,12]  n&x|length x<n=[]|t=take n x  b?x|b=x|t=[]  n!k= \c->e(n&m(%k)c)?(n&c)  e[]=1<1  e(x:y)=all(x==)y  k q c|any null[q c,p$c\\q c]=[]|t=q c  f=5!4  s c=(sort(m(%z)c)`elem`w:[[n..n+4]|n<-[0..8]])?c  r=3!z  p=2!z  g x y|c x y<2=x|t=y  q x(_,[])=x  q _ y=y  b h=y q$m($h)$zipWith(\t f->(,)t.y g.m(f.take 5).permutations)[1..][1!1,p,k p,r,s,f,k r,4!z,s.f]  h=reverse.a.m(%z)  a v|w\\v==[]=[-1..3]|t=sort v  c x y=o(h x)$h y  o[](_:_)=2  o[]_=0  o _[]=1  o(a:b)(k:d)|a>k=1|a<k=2|t=o b d  d n(a,k)|a==[]=0|n<1=0|r>s=1|r<s=2|f/=0=f|t=d(n-length o)(a\\o,k\\u)where(r,o)=b a;(s,u)=b k;f=c o u  i x=head.findIndices(x==)  u(n:k)c@[r,s]|n%z==i r"23456789TJQKA"&&n%4==i s"HDSC"=n|t=u k c  l c=(2&c++snd(splitAt 4c),drop 2c)  main=getLine>>=print.d 5.l.m(u[0..]).words  


import Control.Exception (assert)  import Data.List (permutations, sort, intersect, findIndices, (\\))  import Data.Function (on)    (%) = mod    aceLowRun = [0,1,2,3,12]    tryTake n xs    | length xs < n = []    | otherwise = take n xs    cond ? xs    | cond = xs    | otherwise = []    eqOn n f cards = allEq (tryTake n $ map f cards) ? tryTake n cards    allEq [] = False  allEq (x:xs) = all (== x) xs    combWithPair pokerHand cards    | any null [picked1, picked2] = []    | otherwise = pokerHand cards    where      picked1 = pokerHand cards      picked2 = pair $ cards \\ picked1    straightFlush = straight . flush    quads = eqOn 4 (% 13)    fullHouse = combWithPair triples    flush = eqOn 5 (% 4)    straight cards = (sort (map (% 13) cards) `elem` runs) ? cards    where      runs = aceLowRun : [[n..n+4] | n <- [0..8]]    triples = eqOn 3 (% 13)    twoPair = combWithPair pair    pair = eqOn 2 (% 13)    single = eqOn 1 id    bestVersionOfHand [] ys = ys  bestVersionOfHand xs [] = xs  bestVersionOfHand xs ys    | compareSameRankedHands xs ys < 2 = xs    | otherwise = ys    rate rating pokerHand cards = (rating, handResult)    where      handResult = foldr1 bestVersionOfHand                          (map (pokerHand . take 5) $ permutations cards)    pokerHands = zipWith rate [1..] [      single    , pair    , twoPair    , triples    , straight    , flush    , fullHouse    , quads    , straightFlush    ]    bestHand hand = foldr1 (\xs ys -> if null (snd ys) then xs else ys)                         (map ($ hand) pokerHands)    highestVals = reverse . arrangeVals . map (% 13)    where      arrangeVals vals = if vals `intersect` aceLowRun == aceLowRun        then [-1..3]        else sort vals    compareSameRankedHands = compareSameRankedHands' `on` highestVals    compareSameRankedHands' [] [] = 0  compareSameRankedHands' (card1:cards1) (card2:cards2)    | card1 > card2 = 1    | card1 < card2 = 2    | otherwise = compareSameRankedHands' cards1 cards2    decideWinner n cards1 cards2    | null cards1 = assert (null cards2) 0    | n < 1 = 0    | rating1 > rating2 = 1    | rating1 < rating2 = 2    | cmpRes /= 0 = cmpRes    | otherwise = decideWinner                    (n - assert (length bests1 == length bests2) (length bests1))                    (cards1 \\ bests1)                    (cards2 \\ bests2)    where      (rating1, bests1) = bestHand cards1      (rating2, bests2) = bestHand cards2      cmpRes = compareSameRankedHands bests1 bests2    indexOf x = head . findIndices (x==)    toNum = toNum' [0..]    toNum' (n:ns) [rank, suit]    | n % 13 == indexOf rank "23456789TJQKA" && n % 4 == indexOf suit "HDSC" = n    | otherwise = toNum' ns [rank, suit]    cluster cards = (take 2 cards ++ snd (splitAt 4 cards), drop 2 cards)    main = getLine >>= print    . uncurry (decideWinner 5)    . cluster    . map toNum    . words  


GolfScript 258 241 247/341 217/299 char

Solution for the 10 card problem. Only the last couple of newlines are significant:

10:T):J):Q):K):A;0:a;0:b;0:d;"\r\n"%{' '/5/{.{)\;}/4*-+++!:f;{);~}%{$0:z(%{.z-  !99*+:z}%}5*.{+}*99/:P!{..)\(@4+-!2*\;\.2<~9+-!\;+}and:s;[s f*6P=4P=f s P 6$]\;}  %.~={;;d):d;}{~>{a):a;}{b):b;}if}if}/  '1: 'a'  2: 'b'  D: 'd n  

The 9 card problem currently needs about 80 more characters.

10:T):J):Q):K):A;0:a;0:b;0:d;"\r\n"%{' '/);{('Z'%+}2*[0$2>\7<]  {:H;7,{H=:x;H{x=!},:I;6,{I=:x;I{x=!},}/}%{.{)\;}/4*-+++!:f;  {);~}%{$0:z(%{.z-!99*+:z}%}5*.{+}*99/:P!{..)\(@4+-!2*\;\.2<~9+-!\;+}and:s;[  s f*6P=4P=f s P 6$]\;}%{[\].~>{~;}{~\;}if}*}%.~={;;d):d;}{~>{a):a;}{b):b;}if}if}/  '1: 'a'  2: 'b'  D: 'd n  

Less golfed version of 10 card problem.

10:T;11:J;12:Q;13:K;14:A;              # map for face cards  0:a;0:b;0:d;                           # other initialization    "\r\n"%                                # split input on \n  {                                      # on each line of input      ' '/                               #  divide line into ten cards      5/                                 #  split into five card hands        {.                                 #  on each of the two hands             {)\;}%                        #   chop last character of each card           .(5*\;\{+}*=                  #   check sum of elem == 5*1st elem           :f;                           #   this is the flush flag             {);~}%$                       #   reduce cards to numerical values             0:z;{.z- 20%{}                               {;z 20+}if:z}%{-1*}$      #   detect pairs             .(:h;;                        #   extract value of highest card               20h>{..)\(@4+-!2*\;\          # detect straight               .2<~9+-!\;+}and:s;        # s=2 for regular straight, s=1 for A-5 straight                                           # result of this mapping - 6 elem array           [ 0$                          #   #6 - cards in the hand             .{20/}%{+}*:P               #   #5 - number of pairs             s                           #   #4 - is this a straight?             f                           #   #3 - is this a flush?             4P=                         #   #2b - is this a full house?             h 59>                       #   #2 - is this 4 of a kind?             s f *                       #   #1 - is this a straight flush?           ]-1%         \;      }/        \.@.@                             # put [hand1 hand2 hand1 hand2] on stack        =                                 # check hand1==hand2        {;;d):d;}                         # if equal, increment d (draw)         {>{a):a;}                      # if >, increment a (player 1 wins)         {b):b;}if                      # if <, increment b (player 2 wins)       }if  }/                                          # output results      '1: 'a'  2: 'b'  D: 'd n  


C, 665+379 chars

Here's my answer in 2 parts.

The first is a complete 7 card evaluator, including the "AddCard" macro A. It returns a 32-bit number ranking the hand. The high nibble is the type, bits 13..25 indicate the high card(s) and bits 0..12 indicate the kicker(s). When comparing the results, the better hand will always have the larger value.

#define U unsigned  #define c(a)X=a;i=C=0;while(X){C|=(X&1)<<i++;X/=4;}  #define A(h,c)h[c&7]+=c,h[3]|=c  U C,i,X;  U E(U h[]){  U a=h[0]+h[1]+h[2]+h[4]-(h[3]&-16),t,v,k,e=a&0x55555540,o=a&0xAAAAAA80;  if(v=e&o/2){t=7;k=h[3]^v;i=0;while(k/=4)i++;k=1<<2*i;}  else if(v=o&o-1){t=6;v/=2;k=o/2^v;}  else if(e>1&o>1){t=6;v=o/2;k=(i=e&e-1)?i:e;}  else{a=h[3];  if(t=h[i=1]-(a&1)&4||h[i=2]-(a&2)&8||h[i=4]-(a&4)&16||h[i=0]-(a&8)&32)a=h[i];  a&=-64;v=a|a>>26&16;t*=5;  if(v=v&v<<2&v<<4&v<<6&v<<8){t+=4;a=v&=~(v/2);}  else if(t)for(i=(h[i]&63)/(i?i:8),v=a;i-->5;)a&=a-1;  else if(v=o/2)t=3;  else if (e){o=e&e-1;v=(i=o&o-1)?o:e;t=1+(o>0);}  k=a^v;k&=k-1;k&=k-(i==0);}  c(v);v=C/8;c(k);  return t<<28|v<<13|C/8;}  

The second is the input processor. It parses the project Euler file as 2+2+5 cards (ignoring the 10th card). It uses the Parse macro, P to create 32-bit values representing each card. The representation is 0A0K0Q0J0T090807060504030200shdc. A hand is stored as an array of 5 ints.

char*gets(char*);char*strchr(char*,char);  #define P(c)X=strchr(R,*c++)-R;C=1<<strchr(S,*c++)-S|64<<X*2;c++;  #define L(n)for(i=0;i<n;i++)  U g[5],h[5];  char*c,b[32];  char*S="CDHS";  char*R="23456789TJQKA";  int d,r[3]={0};  main(q){while(c=gets(b)){  L(2){P(c)A(g,C);}  L(2){P(c)A(h,C);}  L(5){P(c)A(g,C);A(h,C);}  d=E(g)-E(h);  r[d>0?0:d<0?1:2]++;  L(7)g[i]=h[i]=0;  }L(3)printf("%c:%d\n","12D"[i],r[i]);}  

I'm sure there are a few more characters to be trimmed off. I'll add an explanation soon.

The evaluator runs @17.6 Million hands/second on my 3Ghz Core2 Duo. That's only 3.5x slower than the PokerSource evaluator, which uses at least 56K of lookup tables.


PHP, 799 chars

Line breaks are not significant. This takes input from the linked url, which is different from the example input (doesn't deal with community cards). Processing is similar to mobrule's perl answer, with a different scoring method.

<?php  function s($i){$o=array_map('intval',$i);$f=(count(array_unique(str_replace($o,'',$i)))==1);  sort($o);$v=array_count_values($o);arsort($v);$u=array_keys($v);$h=max($u);$m=$u[0];$c=reset($v);  $p=count($v);$e=$c==1&&$o[4]==14&&$o[3]==5;$r=$o==range($o[0],$o[0]+4)||$e;$q=$e?5:$h;  $s=($f&&$r&&($h==12)?2<<11:($f&&$r?(2<<10)+$q:0))+($c==4?(2<<9)+$m:0)+($c==3&&$p==2?(2<<8)+$m:0)+($f?(2<<7)+$h:0)+  ($r?(2<<6)+$q:0)+($c==3?(2<<5)+$m:0)+($c==2&&$p==3?(2<<4)+$m:0)+($p==4?(2<<3)+$m:0);$s+=!$s?$h:0;return array($s,$u);}    foreach(file($argv[1]) as $d){  list($y,$z)=array_chunk(explode(' ',trim(strtr($d,array('T'=>10,'J'=>11,'Q'=>12,'K'=>13,'A'=>14)))),5);  $y=s($y);$z=s($z);$w=$y[0]-$z[0];$x=1;while(!$w&&$x<5){$w=$y[1][$x]-$z[1][$x++];}if(!$w)@$t++;elseif($w<0)@$l++;else @$k++;}  @print "1: $k\n2: $l\nD: $t";  

Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com
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