Code Golf: Who has the best poker hand? - language-agnostic

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
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.
Examples
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:
http://projecteuler.net/project/poker.txt
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)
Rules
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}
++$X{B((#c=map{N}split)[0..4])<=>B(#c[5..9])}for<>;
printf"1: %d\n2: %d\nD: %d\n",#X{1,-1,0}
This solves the "10 card" problem (10 cards are dealt, player 1 has the first 5 cards and player 2 has the second 5 cards).
The first section defines a subroutine N that can transform each card so that it has a numerical value. For non-face cards, this is a trivial mapping (5H ==> 5H) but it does transform the face cards (KC => 13C, AD => 14D).
The last section parses each line of input into cards, transforms the cards to contain numerical values, divides the cards into separate hands for the two players, and analyzes and compares those hands. Every hand increments one element of the hash %X. When all the input is parsed, %X contains the number of hands won by player 1, won by player 2, or ties.
The middle section is a subroutine that takes a set of five cards as input and produces a
12-digit number with the property that stronger poker hands will have higher-valued numbers. Here's how it works:
for$m(#_){$m-$_||($s+=2,++$p[$m])for#_}
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 $x. If the hand contains three of a kind, then those three cards will match with the other two cards -- it will be like there are 9 matches among those three cards and the "score" will be at least 18.
#_=sort{$p[$b]-$p[$a]||$b-$a}#_;
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]-$_[1]==9), which means the hand has an A-2-3-4-5 straight. In the latter case, the Ace is now the least valuable card in the hand, so we manipulate #_ to reflect that (push#_,shift)
"#_"=~/.$/;$s+=14*(4<grep/$&/,#_);
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.
Bugfixes:
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!
Golfed:
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
Ungolfed:
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";

Related

Code Golf: Build Me an Arc

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
Challenge
The shortest program by character count that accepts standard input of the form X-Y R, with the following guarantees:
R is a non-negative decimal number less than or equal to 8
X and Y are non-negative angles given in decimal as multiples of 45° (0, 45, 90, 135, etc.)
X is less than Y
Y is not 360 if X is 0
And produces on standard output an ASCII "arc" from the starting angle X to the ending angle Y of radius R, where:
The vertex of the arc is represented by o
Angles of 0 and 180 are represented by -
Angles of 45 and 225 are represented by /
Angles of 90 and 270 are represented by |
Angles of 135 and 315 are represented by \
The polygonal area enclosed by the two lines is filled with a non-whitespace character.
The program is not required to produce meaningful output if given invalid input. Solutions in any language are allowed, except of course a language written specifically for this challenge, or one that makes unfair use of an external utility. Extraneous horizontal and vertical whitespace is allowed in the output provided that the format of the output remains correct.
Happy golfing!
Numerous Examples
Input:
0-45 8
Output:
/
/x
/xx
/xxx
/xxxx
/xxxxx
/xxxxxx
/xxxxxxx
o--------
Input:
0-135 4
Output:
\xxxxxxxx
\xxxxxxx
\xxxxxx
\xxxxx
o----
Input:
180-360 2
Output:
--o--
xxxxx
xxxxx
Input:
45-90 0
Output:
o
Input:
0-315 2
Output:
xxxxx
xxxxx
xxo--
xxx\
xxxx\
Perl, 235 211 225 211 207 196 179 177 175 168 160 156 146 chars
<>=~/-\d+/;for$y(#a=-$'..$'){print+(map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a),$/}
Perl using say feature, 161 149 139 chars
$ echo -n '<>=~/-\d+/;for$y(#a=-$'"'"'..$'"'"'){say map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a}' | wc -c
139
$ perl -E '<>=~/-\d+/;for$y(#a=-$'"'"'..$'"'"'){say map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a}'
Perl without trailing newline, 153 143 chars
<>=~/-\d+/;for$y(#a=-$'..$'){print$/,map$_|$y?!($t=8*($y>0)+atan2(-$y,$_)/atan2 1,1)&-$&/45==8|$t>=$`/45&$t<=-$&/45?qw(- / | \\)[$t%4]:$":o,#a}
Original version commented:
$_=<>;m/(\d+)-(\d+) (\d+)/;$e=$1/45;$f=$2/45; # parse angles and radius, angles are 0-8
for$y(-$3..$3){ # loop for each row and col
for$x(-$3..$3){
$t=atan2(-$y,$x)/atan2 1,1; # angle of this point
$t+=8if($t<0); # normalize negative angles
#w=split//,"-/|\\"x2; # array of ASCII symbols for enclosing lines
$s.=!$x&&!$y?"o":$t==$e||$t==$f?$w[$t]:$t>$e&&$t<$f?"x":$";
# if it's origin -> "o", if it's enclosing line, get symbol from array
# if it's between enclosing angles "x", otherwise space
}
$s.=$/;
}
print$s;
EDIT 1: Inlined sub, relational and equality operators return 0 or 1.
EDIT 2: Added version with comments.
EDIT 3: Fixed enclosing line at 360º. Char count increased significantly.
EDIT 4: Added a shorter version, bending the rules.
EDIT 5: Smarter fix for the 360º enclosing line. Also, use a number as fill. Both things were obvious. Meh, I should sleep more :/
EDIT 6: Removed unneeded m from match operator. Removed some semicolons.
EDIT 7: Smarter regexp. Under 200 chars!
EDIT 8: Lots of small improvements:
Inner for loop -> map (1 char)
symbol array from split string -> qw (3 chars)
inlined symbol array (6 chars, together with the previous improvement 9 chars!)
Logical or -> bitwise or (1 char)
Regexp improvement (1 char)
Use arithmethic for testing negative angles, inspired by Jacob's answer (5 chars)
EDIT 9: A little reordering in the conditional operators saves 2 chars.
EDIT 10: Use barewords for characters.
EDIT 11: Moved print inside of loop, inspired by Lowjacker's answer.
EDIT 12: Added version using say.
EDIT 13: Reuse angles characters for fill character, as Gwell's answer does. Output isn't as nice as Gwell's though, that would require 5 additional chars :) Also, .. operator doen't need parentheses.
EDIT 14: Apply regex directly to <>. Assign range operator to a variable, as per Adrian's suggestion to bta's answer. Add version without the final newline. Updated say version.
EDIT 15: More inlining. map{block}#a -> map expr,#a.
Lua, 259 characters
Slightly abuses the non-whitespace character clause to produce a dazzling display and more importantly save strokes.
m=math i=io.read():gmatch("%d+")a=i()/45 b=i()/45 r=i()for y=r,-r,-1 do for x=-r,r do c=m.atan2(y,x)/m.pi*4 c=c<0 and c+8 or c k=1+m.modf(c+.5)io.write(x==0 and y==0 and'o'or c>=a and c<=b and('-/|\\-/|\\-'):sub(k,k)or c==0 and b==8 and'-'or' ')end print()end
Input: 45-360 4
\\\|||///
\\\|||//
\\\\|//
--\\|/
----o----
--//|\\--
////|\\\\
///|||\\\
///|||\\\
Able to handle odd angles
Input: 15-75 8
|/////
|//////
|//////
|//////
///////
|//////-
////---
//-
o
MATLAB, 188 chars :)
input '';[w x r]=strread(ans,'%d-%d%d');l='-/|\-/|\-';[X Y]=meshgrid(-r:r);T=atan2(-Y,X)/pi*180;T=T+(T<=0)*360;T(T>w&T<x)=-42;T(T==w)=-l(1+w/45);T(T==x)=-l(1+x/45);T(r+1,r+1)=-'o';char(-T)
Commented code:
%%# Get the string variable (enclose in quotes, e.g. '45-315 4')
input ''
%%# Extract angles and length
[w x r]=strread(ans,'%d-%d%d');
%%# Store characters
l='-/|\-/|\-';
%%# Create the grid
[X Y]=meshgrid(-r:r);
%%# Compute the angles in degrees
T=atan2(-Y,X)/pi*180;
%%# Get all the angles
T=T+(T<=0)*360;
%# Negative numbers indicate valid characters
%%# Add the characters
T(T>w&T<x)=-42;
T(T==w)=-l(1+w/45);
T(T==x)=-l(1+x/45);
%%# Add the origin
T(r+1,r+1)=-'o';
%%# Display
char(-T)
Mathematica 100 Chars
Out of competition because graphics are too perfect :)
f[x_-y_ z_]:=Graphics#Table[
{EdgeForm#Red,Disk[{0,0},r,{x °,y °}],{r,z,1,-1}]
SetAttributes[f,HoldAll]
Invoke with
f[30-70 5]
Result
alt text http://a.imageshack.us/img80/4294/angulosgolf.png
alt text http://a.imageshack.us/img59/7892/angulos2.png
Note
The
SetAttributes[f, HoldAll];
is needed because the input
f[a-b c]
is otherwise interpreted as
f[(a-b*c)]
GNU BC, 339 chars
Gnu bc because of read(), else and logical operators.
scale=A
a=read()/45
b=read()/45
c=read()
for(y=c;y>=-c;y--){for(x=-c;x<=c;x++){if(x==0)if(y<0)t=-2else t=2else if(x>0)t=a(y/x)/a(1)else if(y<0)t=a(y/x)/a(1)-4else t=a(y/x)/a(1)+4
if(y<0)t+=8
if(x||y)if(t==a||t==b||t==b-8){scale=0;u=(t%4);scale=A;if(u==0)"-";if(u==1)"/";if(u==2)"|";if(u==3)"\"}else if(t>a&&t<b)"x"else" "else"o"};"
"}
quit
MATLAB 7.8.0 (R2009a) - 168 163 162 characters
Starting from Jacob's answer and inspired by gwell's use of any non-whitespace character to fill the arc, I managed the following solution:
[w x r]=strread(input('','s'),'%d-%d%d');
l='o -/|\-/|\-';
X=meshgrid(-r:r);
T=atan2(-X',X)*180/pi;
T=T+(T<=-~w)*360;
T(T>x|T<w)=-1;
T(r+1,r+1)=-90;
disp(l(fix(3+T/45)))
And some test output:
>> arc
0-135 4
\||||////
\|||///-
\||//--
\|/---
o----
I could reduce it further to 156 characters by removing the call to disp, but this would add an extra ans = preceding the output (which might violate the output formatting rules).
Even still, I feel like there are some ways to reduce this further. ;)
Ruby, 292 276 186 chars
x,y,r=gets.scan(/\d+/).map{|z|z.to_i};s=(-r..r);s.each{|a|s.each{|b|g=Math::atan2(-a,b)/Math::PI*180/1%360;print a|b==0?'o':g==x||g==y%360?'-/|\\'[g/45%4].chr: (x..y)===g ?'*':' '};puts}
Nicer-formatted version:
x, y, r = gets.scan(/\d+/).map{|z| z.to_i}
s = (-r..r)
s.each {|a|
s.each {|b|
g = (((Math::atan2(-a,b) / Math::PI) * 180) / 1) % 360
print ((a | b) == 0) ? 'o' :
(g == x || g == (y % 360)) ? '-/|\\'[(g / 45) % 4].chr :
((x..y) === g) ? '*' : ' '
}
puts
}
I'm sure someone out there who got more sleep than I did can condense this more...
Edit 1: Switched if statements in inner loop to nested ? : operator
Edit 2: Stored range to intermediate variable (thanks Adrian), used stdin instead of CLI params (thanks for the clarification Jon), eliminated array in favor of direct output, fixed bug where an ending angle of 360 wouldn't display a line, removed some un-needed parentheses, used division for rounding instead of .round, used modulo instead of conditional add
Ruby, 168 characters
Requires Ruby 1.9 to work
s,e,r=gets.scan(/\d+/).map &:to_i;s/=45;e/=45;G=-r..r;G.map{|y|G.map{|x|a=Math.atan2(-y,x)/Math::PI*4%8;print x|y!=0?a==s||a==e%8?'-/|\\'[a%4]:a<s||a>e ?' ':8:?o};puts}
Readable version:
start, _end, radius = gets.scan(/\d+/).map &:to_i
start /= 45
_end /= 45
(-radius..radius).each {|y|
(-radius..radius).each {|x|
angle = Math.atan2(-y, x)/Math::PI * 4 % 8
print x|y != 0 ? angle==start || angle==_end%8 ? '-/|\\'[angle%4] : angle<start || angle>_end ? ' ' : 8 : ?o
}
puts
}
Perl - 388 characters
Since it wouldn't be fair to pose a challenge I couldn't solve myself, here's a solution that uses string substitution instead of trigonometric functions, and making heavy use of your friendly neighbourhood Perl's ability to treat barewords as strings. It's necessarily a little long, but perhaps interesting for the sake of uniqueness:
($x,$y,$r)=split/\D/,<>;for(0..$r-1){$t=$r-1-$_;
$a.=L x$_.D.K x$t.C.J x$t.B.I x$_."\n";
$b.=M x$t.F.N x$_.G.O x$_.H.P x$t."\n"}
$_=$a.E x$r.o.A x$r."\n".$b;$x/=45;$y/=45;$S=' ';
sub A{$v=$_[0];$x==$v||$y==$v?$_[1]:$x<$v&&$y>$v?x:$S}
sub B{$x<=$_[0]&&$y>$_[0]?x:$S}
#a=!$x||$y==8?'-':$S;
push#a,map{A$_,'\\'.qw(- / | \\)[$_%4]}1..7;
push#a,!$x?x:$S,map{B$_}1..7;
eval"y/A-P/".(join'',#a)."/";print
All newlines are optional. It's fairly straightforward:
Grab user input.
Build the top ($a) and bottom ($b) parts of the pattern.
Build the complete pattern ($_).
Define a sub A to get the fill character for an angle.
Define a sub B to get the fill character for a region.
Build an array (#a) of substitution characters using A and B.
Perform the substitution and print the results.
The generated format looks like this, for R = 4:
DKKKCJJJB
LDKKCJJBI
LLDKCJBII
LLLDCBIII
EEEEoAAAA
MMMFGHPPP
MMFNGOHPP
MFNNGOOHP
FNNNGOOOH
Where A-H denote angles and I-P denote regions.
(Admittedly, this could probably be golfed further. The operations on #a gave me incorrect output when written as one list, presumably having something to do with how map plays with $_.)
C# - 325 319 chars
using System;class P{static void Main(){var s=Console.ReadLine().Split(' ');
var d=s[0].Split('-');int l=s[1][0]-48,x,y,r,a=int.Parse(d[0]),b=int.Parse(d[1]);
for(y=l;y>=-l;y--)for(x=-l;x<=l;)Console.Write((x==0&&y==0?'o':a<=(r=((int)
(Math.Atan2(y,x)*57.3)+360)%360)&&r<b||r==b%360?
#"-/|\"[r/45%4]:' ')+(x++==l?"\n":""));}}
Newlines not significant.
Sample input/output
45-180 8
\||||||||////////
\\|||||||///////
\\\||||||//////
\\\\|||||/////
\\\\\||||////
\\\\\\|||///
\\\\\\\||//
\\\\\\\\|/
--------o
135-360 5
\
\\
\\\
\\\\
\\\\\
-----o-----
----/|\\\\\
---//||\\\\
--///|||\\\
-////||||\\
/////|||||\
Java - 304 chars
class A{public static void main(String[]a){String[]b=a[0].split("-");int e=new Integer(b[1]),r=new Integer(a[1]),g,x,y=r;for(;y>=-r;y--)for(x=-r;x<=r;)System.out.print((x==0&y==0?'o':new Integer(b[0])<=(g=((int)(Math.atan2(y,x)*57.3)+360)%360)&g<e|g==e%360?"-/|\\".charAt(g/45%4):' ')+(x++<r?"":"\n"));}}
More readable version:
class A{
public static void main(String[]a){
String[]b=a[0].split("-");
int e=new Integer(b[1]),r=new Integer(a[1]),g,x,y=r;
for(;y>=-r;y--)for(x=-r;x<=r;)System.out.print((
x==0&y==0
?'o'
:new Integer(b[0])<=(g=((int)(Math.atan2(y,x)*57.3)+360)%360)&g<e|g==e%360
?"-/|\\".charAt(g/45%4)
:' '
)+(x++<r?"":"\n"));
}
}
C (902 byte)
This doesn't use trigonometric functions (like the original perl version), so it's quite ``bloated''. Anyway, here is my first code-golf submission:
#define V(r) (4*r*r+6*r+3)
#define F for(i=0;i<r;i++)
#define C ;break;case
#define U p-=2*r+2,
#define D p+=2*r+2,
#define R *++p=
#define L *--p=
#define H *p='|';
#define E else if
#define G(a) for(j=0;j<V(r)-1;j++)if(f[j]==i+'0')f[j]=a;
#define O(i) for(i=0;i<2*r+1;i++){
main(int i,char**v){char*p,f[V(8)];
int j,m,e,s,x,y,r;p=*++v;x=atoi(p);while(*p!=45)p++;
char*h="0123";y=atoi(p+1);r=atoi(*++v);
for(p=f+2*r+1;p<f+V(r);p+=2*r+2)*p=10;
*(p-2*r-2)=0;x=x?x/45:x;y/=45;s=0;e=2*r;m=r;p=f;O(i)O(j)
if(j>e)*p=h[0];E(j>m)*p=h[1];E(j>s)*p=h[2];else*p=h[3];p++;}
if(i+1==r){h="7654";m--;e--;}E(i==r){s--;}E(i>r){s--;e++;}
else{s++;e--;}p++;}for(p=f+V(r)/2-1,i=0;i<r;i++)*++p=48;
for(i=0;i<8;i++)if(i>=x&&i<y){G(64);}else G(32);
y=y==8?0:y;q:p=f+V(r)/2-1;*p='o';switch(x){
C 0:F R 45 C 1:F U R 47 C 2:F U H C 3:F U L 92
C 4:F L 45 C 5:F D L 47 C 6:F D H C 7:F D R 92;}
if(y!=8){x=y;y=8;goto q;}puts(f);}
also, the #defines look rather ugly, but they save about 200 bytes so I kept them in, anyway. It is valid ANSI C89/C90 and compiles with very few warnings (two about atoi and puts and two about crippled form of main).

Code Golf: Four is magic

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
The puzzle
A little puzzle I heard while I was in high school went something like this...
The questioner would ask me to give him a number;
On hearing the number, the questioner would do some sort of transformation on it repeatedly (for example, he might say ten is three) until eventually arriving at the number 4 (at which point he would finish with four is magic).
Any number seems to be transformable into four eventually, no matter what.
The goal was to try to figure out the transformation function and then be able to reliably proctor this puzzle yourself.
The solution
The transformation function at any step was to
Take the number in question,
Count the number of letters in its English word representation, ignoring a hyphen or spaces or "and" (e.g., "ten" has 3 letters in it, "thirty-four" has 10 letters in it, "one hundred forty-three" has 20 letters in it).
Return that number of letters.
For all of the numbers I have ever cared to test, this converges to 4. Since "four" also has four letters in it, there would be an infinite loop here; instead it is merely referred to as magic by convention to end the sequence.
The challenge
Your challenge is to create a piece of code that will read a number from the user and then print lines showing the transformation function being repeatedly applied until "four is magic" is reached.
Specifically:
Solutions must be complete programs in and of themselves. They cannot merely be functions which take in a number-- factor in the input.
Input must be read from standard input. (Piping from "echo" or using input redirection is fine since that also goes from stdin)
The input should be in numeric form.
For every application of the transformation function, a line should be printed: a is b., where a and b are numeric forms of the numbers in the transformation.
Full stops (periods) ARE required!
The last line should naturally say, 4 is magic..
The code should produce correct output for all numbers from 0 to 99.
Examples:
> 4
4 is magic.
> 12
12 is 6.
6 is 3.
3 is 5.
5 is 4.
4 is magic.
> 42
42 is 8.
8 is 5.
5 is 4.
4 is magic.
> 0
0 is 4.
4 is magic.
> 99
99 is 10.
10 is 3.
3 is 5.
5 is 4.
4 is magic.
The winner is the shortest submission by source code character count which is also correct.
BONUS
You may also try to write a version of the code which prints out the ENGLISH NAMES for the numbers with each application of the transformation function. The original input is still numeric, but the output lines should have the word form of the number.
(Double bonus for drawing shapes with your code)
(EDIT) Some clarifications:
I do want the word to appear on both sides in all applicable cases, e.g. Nine is four. Four is magic.
I don't care about capitalization, though. And I don't care how you separate the word tokens, though they should be separated: ninety-nine is okay, ninety nine is okay, ninetynine is not okay.
I'm considering these a separate category for bonus competition with regard to the challenge, so if you go for this, don't worry about your code being longer than the numeric version.
Feel free to submit one solution for each version.
Perl, about 147 char
Loosely based on Platinum Azure's solution:
chop
($_.=
<>);#
u="433
5443554
366 887
798 866
555 766
"=~ /\d
/gx ;#4
sub r{4
-$_ ?$_
<20 ?$u
[$_ ]:(
$'? $u[
$'] :0)
+$u[18+$&]:magic}print"
$_ is ",$_=r(),'.'while
/\d
/x;
444
GolfScript - 101 96 93 92 91 90 94 86 bytes
90 → 94: Fixed output for multiples of 10.
94 → 86: Restructured code. Using base 100 to remove non-printable characters.
86 → 85: Shorter cast to string.
{n+~."+#,#6$DWOXB79Bd")base`1/10/~{~2${~1$+}%(;+~}%++=" is "\".
"1$4$4-}do;;;"magic."
Common Lisp 157 Chars
New more conforming version, now reading form standard input and ignoring spaces and hyphens:
(labels((g (x)(if(= x 4)(princ"4 is magic.")(let((n(length(remove-if(lambda(x)(find x" -"))(format nil"~r"x)))))(format t"~a is ~a.~%"x n)(g n)))))(g(read)))
In human-readable form:
(labels ((g (x)
(if (= x 4)
(princ "4 is magic.")
(let ((n (length (remove-if (lambda(x) (find x " -"))
(format nil "~r" x)))))
(format t"~a is ~a.~%" x n)
(g n)))))
(g (read)))
And some test runs:
>24
24 is 10.
10 is 3.
3 is 5.
5 is 4.
4 is magic.
>23152436
23152436 is 64.
64 is 9.
9 is 4.
4 is magic.
And the bonus version, at 165 chars:
(labels((g(x)(if(= x 4)(princ"four is magic.")(let*((f(format nil"~r"x))(n(length(remove-if(lambda(x)(find x" -"))f))))(format t"~a is ~r.~%"f n)(g n)))))(g(read)))
Giving
>24
twenty-four is ten.
ten is three.
three is five.
five is four.
four is magic.
>234235
two hundred thirty-four thousand two hundred thirty-five is forty-eight.
forty-eight is ten.
ten is three.
three is five.
five is four.
four is magic.
Python 2.x, 144 150 154 166 chars
This separates the number into tens and ones and sum them up. The undesirable property of the pseudo-ternary operator a and b or c that c is returned if b is 0 is being abused here.
n=input()
x=0x4d2d0f47815890bd2
while n-4:p=n<20and x/10**n%10or 44378/4**(n/10-2)%4+x/10**(n%10)%10+4;print n,"is %d."%p;n=p
print"4 is magic."
The previous naive version (150 chars). Just encode all lengths as an integer.
n=input()
while n-4:p=3+int('1yrof7i9b1lsi207bozyzg2m7sclycst0zsczde5oks6zt8pedmnup5omwfx56b29',36)/10**n%10;print n,"is %d."%p;n=p
print"4 is magic."
C - with number words
445 431 427 421 399 386 371 359* 356 354† 348 347 characters
That's it. I don't think I can make this any shorter.
All newlines are for readability and can be removed:
i;P(x){char*p=",one,two,three,four,five,six,sM,eight,nine,tL,elM,twelve,NP,4P,
fifP,6P,7P,8O,9P,twLQ,NQ,forQ,fifQ,6Q,7Q,8y,9Q,en,evL,thir,eL,tO,ty, is ,.\n,
4RmagicS,zero,";while(x--)if(*++p-44&&!x++)*p>95|*p<48?putchar(*p),++i:P(*p-48);
}main(c){for(scanf("%d",&c);c+(i=-4);P(34),P(c=i),P(35))P(c?c>19?P(c/10+18),
(c%=10)&&putchar(45):0,c:37);P(36);}
Below, it is somewhat unminified, but still pretty hard to read. See below for a more readable version.
i;
P(x){
char*p=",one,two,three,four,five,six,sM,eight,nine,tL,elM,twelve,NP,4P,fifP,6P,7P,8O,9P,twLQ,NQ,forQ,fifQ,6Q,7Q,8y,9Q,en,evL,thir,eL,tO,ty, is ,.\n,4RmagicS,zero,";
while(x--)
if(*++p-44&&!x++)
*p>95|*p<48?putchar(*p),++i:P(*p-48);
}
main(c){
for(scanf("%d",&c);c+(i=-4);P(34),P(c=i),P(35))
P(c?
c>19?
P(c/10+18),
(c%=10)&&
putchar(45)
:0,
c
:37);
P(36);
}
Expanded and commented:
int count; /* type int is assumed in the minified version */
void print(int index){ /* the minified version assumes a return type of int, but it's ignored */
/* see explanation of this string after code */
char *word =
/* 1 - 9 */
",one,two,three,four,five,six,sM,eight,nine,"
/* 10 - 19 */
"tL,elM,twelve,NP,4P,fifP,6P,7P,8O,9P,"
/* 20 - 90, by tens */
"twLQ,NQ,forQ,fifQ,6Q,7Q,8y,9Q,"
/* lookup table */
"en,evL,thir,eL,tO,ty, is ,.\n,4RmagicS,zero,";
while(index >= 0){
if(*word == ',')
index--;
else if(index == 0) /* we found the right word */
if(*word >= '0' && *word < 'a') /* a compression marker */
print(*word - '0'/*convert to a number*/);
else{
putchar(*word); /* write the letter to the output */
++count;
}
++word;
}
}
int main(int argc, char **argv){ /* see note about this after code */
scanf("%d", &argc); /* parse user input to an integer */
while(argc != 4){
count = 0;
if(argc == 0)
print(37/*index of "zero"*/);
else{
if(argc > 19){
print(argc / 10/*high digit*/ + 20/*offset of "twenty"*/ - 2/*20 / 10*/);
argc %= 10; /* get low digit */
if(argc != 0) /* we need a hyphen before the low digit */
putchar('-');
}
print(argc/* if 0, then nothing is printed or counted */);
}
argc = count;
print(34/*" is "*/);
print(argc); /* print count as word */
print(35/*".\n"*/);
}
print(36/*"four is magic.\n"*/);
}
About the encoded string near the beginning
The names of the numbers are compressed using a very simple scheme. Frequently used substrings are replaced with one-character indices into the name array. A "lookup table" of extra name entries is added to the end for substrings not used in their entirety in the first set. Lookups are recursive: entries can refer to other entries.
For instance, the compressed name for 11 is elM. The print() function outputs the characters e and l (lower-case 'L', not number '1') verbatim, but then it finds the M, so it calls itself with the index of the 29th entry (ASCII 'M' - ASCII '0') into the lookup table. This string is evL, so it outputs e and v, then calls itself again with the index of the 28th entry in the lookup table, which is en, and is output verbatim. This is useful because en is also used in eL for een (used after eight in eighteen), which is used in tO for teen (used for every other -teen name).
This scheme results in a fairly significant compression of the number names, while requiring only a small amount of code to decompress.
The commas at the beginning and end of the string account for the simplistic way that substrings are found within this string. Adding two characters here saves more characters later.
About the abuse of main()
argv is ignored (and therefore not declared in the compressed version), argc's value is ignored, but the storage is reused to hold the current number. This just saves me from having to declare an extra variable.
About the lack of #include
Some will complain that omitting #include <stdio.h> is cheating. It is not at all. The given is a completely legal C program that will compile correctly on any C compiler I know of (albeit with warnings). Lacking protoypes for the stdio functions, the compiler will assume that they are cdecl functions returning int, and will trust that you know what arguments to pass. The return values are ignored in this program, anyway, and they are all cdecl ("C" calling convention) functions, and we do indeed know what arguments to pass.
Output
Output is as expected:
0
zero is four.
four is magic.
1
one is three.
three is five.
five is four.
four is magic.
4
four is magic.
20
twenty is six.
six is three.
three is five.
five is four.
four is magic.
21
twenty-one is nine.
nine is four.
four is magic.
* The previous version missed the mark on two parts of the spec: it didn't handle zero, and it took input on the command line instead of stdin. Handling zeros added characters, but using stdin instead of command line args, as well as a couple of other optimzations saved the same number of characters, resulting in a wash.
† The requirements have been changed to make clear that the number word should be printed on both sides of " is ". This new version meets that requirement, and implements a couple more optimizations to (more than) account for the extra size necessary.
J, 107 112 characters
'4 is magic.',~}:('.',~":#{.,' is ',":#{:)"1]2&{.\.
(]{&(#.100 4$,#:3 u:ucp'䌵䐵吶梇禈榛ꪛ멩鮪鮺墊馊꥘誙誩墊馊ꥺ겻곋榛ꪛ멩鮪鮺'))^:a:
(Newline for readability only)
Usage and output:
'4 is magic.',~}:('.',~":#{.,' is ',":#{:)"1]2&{.\.(]{&(#.100 4$,#:3 u:ucp'䌵䐵吶梇禈榛ꪛ멩鮪鮺墊馊꥘誙誩墊馊ꥺ겻곋榛ꪛ멩鮪鮺'))^:a:12
12 is 6.
6 is 3.
3 is 5.
5 is 4.
4 is magic.
T-SQL, 413 451 499 chars
CREATE FUNCTION d(#N int) RETURNS int AS BEGIN
Declare #l char(50), #s char(50)
Select #l='0066555766',#s='03354435543668877987'
if #N<20 return 0+substring(#s,#N+1,1) return 0+substring(#l,(#N/10)+1,1) + 0+(substring(#s,#N%10+1,1))END
GO
CREATE proc M(#x int) as BEGIN
WITH r(p,n)AS(SELECT p=#x,n=dbo.d(#x) UNION ALL SELECT p=n,n=dbo.d(n) FROM r where n<>4)Select p,'is',n,'.' from r print '4 is magic.'END
(Not that I'm seriously suggesting you'd do this... really I just wanted to write a CTE)
To use:
M 95
Returns
p n
----------- ---- -----------
95 is 10.
10 is 3.
3 is 5.
5 is 4.
4 is magic.
Java (with boilerplate), 308 290 286 282 280 characters
class A{public static void main(String[]a){int i=4,j=0;for(;;)System.out.printf("%d is %s.%n",i=i==4?new java.util.Scanner(System.in).nextInt():j,i!=4?j="43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:".charAt(i)-48:"magic");}}
I'm sure Groovy would get rid of much of that.
Explanation and formatting (all comments, newlines and leading/trailing whitespace removed in count):
Reasonably straight forward, but
//boilerplate
class A{
public static void main(String[]a){
//i is current/left number, j right/next number. i=4 signals to start
//by reading input
int i=4,j=0;
for(;;)
//print in the form "<left> is <right>."
System.out.printf(
"%d is %s.%n",
i=i==4?
//<left>: if i is 4 <left> will be a new starting number
new java.util.Scanner(System.in).nextInt():
//otherwise it's the next val
j,
i!=4?
//use string to map number to its length (:;< come after 9 in ASCII)
//48 is value of '0'. store in j for next iteration
j="43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:".charAt(i)-48:
//i==4 is special case for right; print "magic"
"magic");
}
}
Edit: No longer use hex, this is less keystrokes
Windows PowerShell: 152 153 184 bytes
based on the previous solution, with more influence from other solutions
$o="03354435543668877988"
for($input|sv b;($a=$b)-4){if(!($b=$o[$a])){$b=$o[$a%10]-48+"66555766"[($a-$a%10)/10-2]}$b-=48-4*!$a
"$a is $b."}'4 is magic.'
C, 158 characters
main(n,c){char*d="03354435543668877988";for(scanf("%d",&n);n-4;n=c)printf("%d is %d.\n",n,c=n?n<19?d[n]-48:d[n%10]-"_,**+++)**"[n/10]:4);puts("4 is magic.");}
(originally based on Vlad's Python code, borrowed a trick from Tom Sirgedas' C++ solution to squeeze out a few more characters)
expanded version:
main(n, c) {
char *d = "03354435543668877988";
for (scanf("%d",&n); n-4; n = c)
printf("%d is %d.\n", n, c = n ? n<19 ? d[n]-48 : d[n%10] - "_,**+++)**"[n/10] : 4);
puts("4 is magic.");
}
Python, 129 133 137 148 chars
As a warm-up, here is my first version (improves couple of chars over previous best Python).
PS. After a few redactions now it is about twenty char's shorter:
n=input()
while n-4:p=(922148248>>n/10*3&7)+(632179416>>n%10*3&7)+(737280>>n&1)+4*(n<1);print n,'is %d.'%p;n=p
print'4 is magic.'
C#: 210 Characters.
Squished:
using C=System.Console;class B{static void Main(){int
x=0,y=int.Parse(C.ReadLine());while(x!=4)C.Write((x=y)+" is {0}.\n",x==4?"magic":""+(y=x==0?4:"03354435543668877988"[x<20?x:x%10]+"0066555766"[x/10]-96));}}
Expanded:
using C=System.Console;
class B
{
static void Main()
{
int x=0,y=int.Parse(C.ReadLine());
while(x!=4)
C.Write((x=y)+" is {0}.\n",
x==4?
"magic":
""+(y= x==0?
4:
"03354435543668877988"[x<20?x:x%10]+
"0066555766"[x/10]-96)
);
}
}
Tricks this approach uses:
Create a lookup table for number name lengths based on digits that appear in the number.
Use character array lookup on a string, and char arithmetic instead of a numeric array.
Use class name aliasing to short Console. to C.
Use the conditional (ternary) operator (?:) instead of if/else.
Use the \n with Write escape code instead of WriteLine
Use the fact that C# has a defined order of evaluation to allow assignments inside the Write function call
Use the assignment expressions to eliminate extra statements, and thus extra braces
Perl: 148 characters
(Perl: 233 181 212 206 200 199 198 185 179 149 148 characters)
Moved exceptions hash into unit array. This resulted in my being able to cut a lot of characters :-)
mobrule pointed out a nasty bug. Quick fix adds 31 characters, ouch!
Refactored for zero special case, mild golfing done as well.
Direct list access for single use rather than storing to array? Hell yes!
SO MUCH REFACTORING for just ONE bloody character. This, truly, is the life of a golfer. :-(
Oops, easy whitespace fix. 198 now.
Refactored some redundant code.
Last return keyword in r is unnecessary, shaved some more off.
Massive refactoring per comments; unfortunately I could only get it to 149 because I had to fix a bug that was present in both my earlier code and the commenters' versions.
Trying bareword "magic".
Let's get this ball rolling with a modest attempt in Perl.
#u=split'','4335443554366887798866555766';$_=<>;chop;print"$_ is ".($_=$_==4?0:$_<20?$u[$_]:($u[$_/10+18]+($_%10&&$u[$_%10]))or magic).".
"while$_
Tricks:
Too many!
JavaScript 1.8 (SpiderMonkey) - 153 Chars
l='4335443554366887798866555766'.split('')
for(b=readline();(a=+b)-4;print(a,'is '+b+'.'))b=a<20?l[a]:+l[18+a/10|0]+(a%10&&+l[a%10])
print('4 is magic.')
Usage: echo 42 | js golf.js
Output:
42 is 8.
8 is 5.
5 is 4.
4 is magic.
With bonus - 364 chars
l='zero one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty thirty fourty fifty sixty seventy eighty ninety'.split(' ')
z=function(a)a<20?l[a]:l[18+a/10|0]+(a%10?' '+l[a%10]:'')
for(b=+readline();(a=b)-4;print(z(a),'is '+z(b)+'.'))b=z(a).replace(' ','').length
print('four is magic.')
Output:
ninety nine is ten.
ten is three.
three is five.
five is four.
four is magic.
Haskell, 224 270 characters
o="43354435543668877988"
x!i=read[x!!i]
n x|x<20=o!x|0<1="0066555766"!div x 10+o!mod x 10
f x=zipWith(\a b->a++" is "++b++".")l(tail l)where l=map show(takeWhile(/=4)$iterate n x)++["4","magic"]
main=readLn>>=mapM putStrLn.f
And little more readable -
ones = [4,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8]
tens = [0,0,6,6,5,5,5,7,6,6]
n x = if x < 20 then ones !! x else (tens !! div x 10) + (ones !! mod x 10)
f x = zipWith (\a b -> a ++ " is " ++ b ++ ".") l (tail l)
where l = map show (takeWhile (/=4) (iterate n x)) ++ ["4", "magic"]
main = readLn >>= mapM putStrLn . f
C++ Stdio version, minified: 196 characters
#include <cstdio>
#define P;printf(
char*o="43354435543668877988";main(int p){scanf("%d",&p)P"%d",p);while(p!=4){p=p<20?o[p]-48:"0366555966"[p/10]-96+o[p%10]P" is %d.\n%d",p,p);}P" is magic.\n");}
C++ Iostreams version, minified: 195 characters
#include <iostream>
#define O;std::cout<<
char*o="43354435543668877988";main(int p){std::cin>>p;O p;while(p!=4){p=p<20?o[p]-48:"0366555966"[p/10]-96+o[p%10]O" is "<<p<<".\n"<<p;}O" is magic.\n";}
Original, un-minified: 344 characters
#include <cstdio>
int ones[] = { 4, 3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8 };
int tens[] = { 0, 3, 6, 6, 5, 5, 5, 9, 6, 6 };
int n(int n) {
return n<20 ? ones[n] : tens[n/10] + ones[n%10];
}
int main(int p) {
scanf("%d", &p);
while(p!=4) {
int q = n(p);
printf("%i is %i\n", p, q);
p = q;
}
printf("%i is magic\n", p);
}
Delphi: 329 characters
Single Line Version:
program P;{$APPTYPE CONSOLE}uses SysUtils;const S=65;A='EDDFEEDFFEDGGIIHHJII';B='DGGFFFJGG';function Z(X:Byte):Byte;begin if X<20 then Z:=Ord(A[X+1])-S else Z:=(Ord(B[X DIV 10])-S)+Z(X MOD 10)end;var X,Y:Byte;begin Write('> ');ReadLn(X);repeat Y:=Z(X);WriteLn(Format('%d is %d.',[X,Y]));X:=Y;until X=4;WriteLn('4 is magic.');end.
Formated:
program P;
{$APPTYPE CONSOLE}
uses
SysUtils;
const
S = 65;
A = 'EDDFEEDFFEDGGIIHHJII';
B = 'DGGFFFJGG';
function Z(X:Byte):Byte;
begin
if X<20
then Z := Ord(A[X+1])-S
else Z := (Ord(B[X DIV 10])-S) + Z(X MOD 10);
end;
var
X,Y: Byte;
begin
Write('> ');
ReadLn(X);
repeat
Y:=Z(X);
WriteLn(Format('%d is %d.' , [X,Y]));
X:=Y;
until X=4;
WriteLn('4 is magic.');
end.
Probably room for some more squeezing... :-P
C# 314 286 283 274 289 273 252 chars.
Squished:
252
Normal:
using C = System.Console;
class P
{
static void Main()
{
var x = "4335443554366877798866555766";
int m, o, v = int.Parse(C.ReadLine());
do {
C.Write("{0} is {1}.\n", o = v, v == 4 ? (object)"magic" : v = v < 20 ? x[v] - 48 : x[17 + v / 10] - 96 + ((m = v % 10) > 0 ? x[m] : 48));
} while (o != 4);
C.ReadLine();
}
}
Edit Dykam: Did quite some carefull insertions and changes:
Changed the l.ToString() into a cast to object of the string "magic".
Created a temporary variable o, so I could move the break outside the for loop, that is, resulting in a do-while.
Inlined the o assignment, aswell the v assignment, continueing in inserting the calculation of l in the function arguments altogether, removing the need for l. Also inlined the assignment of m.
Removed a space in int[] x, int[]x is legit too.
Tried to transform the array into a string transformation, but the using System.Linq was too much to make this an improvement.
Edit 2 Dykam
Changed the int array to a char array/string, added proper arithmics to correct this.
Lua, 176 Characters
o={[0]=4,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8}t={3,6,6,5,5,5,7,6,6}n=0+io.read()while n~=4 do a=o[n]or o[n%10]+t[(n-n%10)/10]print(n.." is "..a..".")n=a end print"4 is magic."
or
o={[0]=4,3,3,5,4,4
,3,5,5,4,3,6,6,8,8
,7,7,9,8,8}t={3,6,
6,5,5,5,7,6,6}n=
0+io.read()while
n ~= 4 do a= o[n
]or o[n%10]+t[(n
-n%10)/10]print(
n.." is "..a.."." )n=a
end print"4 is magic."
C - without number words
180 175* 172 167 characters
All newlines are for readability and can be removed:
i;V(x){return"\3#,#6$:WOXB79B"[x/2]/(x%2?1:10)%10;}main(c){for(scanf("%d",&c);
c-4;)i=c,printf("%d is %d.\n",i,c=c?c>19?V(c/10+19)+V(c%10):V(c):4);puts(
"4 is magic.");}
Slightly unminified:
i;
V(x){return"\3#,#6$:WOXB79B"[x/2]/(x%2?1:10)%10;}
main(c){
for(scanf("%d",&c);c-4;)
i=c,
printf("%d is %d.\n",i,c=c?c>19?V(c/10+19)+V(c%10):V(c):4);
puts("4 is magic.");
}
* The previous version missed the mark on two parts of the spec: it didn't handle zero, and it took input on the command line instead of stdin. Handling zero added characters, but using stdin instead of command line args saved even more, resulting in a net savings.
perl, 123 122 characters
Just realized that there is no requirement to output to STDOUT, so output to STDERR instead and knock off another character.
#u='0335443554366887798866555766'=~/./g;$_+=<>;warn"$_ is ",$_=$_-4?$_<20?$u[$_]||4:$u[chop]+$u[$_+18]:magic,".\n"until/g/
And, a version that returns spelled out numbers:
279 278 276 280 characters
#p=(Thir,Four,Fif,Six,Seven,Eigh,Nine);#n=("",One,Two,Three,Four,Five,#p[3..6],Ten,Eleven,Twelve,map$_.teen,#p);s/u//for#m=map$_.ty,Twen,#p;$n[8].=t;sub n{$n=shift;$n?$n<20?$n[$n]:"$m[$n/10-2] $n[$n%10]":Zero}$p+=<>;warnt$m=n($p)," is ",$_=$p-4?n$p=()=$m=~/\w/g:magic,".\n"until/c/
While that meets the spec, it is not 100% well formatted. It returns an extra space after numbers ending in zero. The spec does say:
"I don't care how you separate the word tokens, though they should be separated"
That's kind of weaselly though.
A more correct version at
282 281 279 283 characters
#p=(Thir,Four,Fif,Six,Seven,Eigh,Nine);#n=("\x8",One,Two,Three,Four,Five,#p[3..6],Ten,Eleven,Twelve,map$_.teen,#p);s/u//for#m=map$_.ty,Twen,#p;$n[8].=t;sub n{$n=shift;$n?$n<20?$n[$n]:"$m[$n/10-2]-$n[$n%10]":Zero}$p+=<>;warn$m=n($p)," is ",$_=$p-4?n$p=()=$m=~/\w/g:magic,".\n"until/c/
Python:
#!/usr/bin/env python
# Number of letters in each part, we don't count spaces
Decades = ( 0, 3, 6, 6, 6, 5, 5, 7, 6, 6, 0 )
Smalls = ( 0, 3, 3, 5, 4, 4, 3, 5, 5, 4 )
Teens = ( 6, 6, 8, 8, 7, 7, 9, 8, 8 )
def Count(n):
if n > 10 and n < 20: return Teens[n-11]
return Smalls[n % 10 ] + Decades [ n / 10 ]
N = input()
while N-4:
Cnt = Count(N)
print "%d is %d" % ( N, Cnt)
N = Cnt
print "4 is magic"
C++, 171 characters (#include omitted)
void main(){char x,y,*a="03354435543668877988";scanf("%d",&x);for(;x-4;x=y)y=x?x<19?a[x]-48:"_466555766"[x/10]+a[x%10]-96:4,printf("%d is %d.\n",x,y);puts("4 is magic.");}
Ruby, 164 characters
n=gets.to_i;s="03354435543668877987";if n==0;puts"0 is 4.";else;puts"#{n} is #{n=(n<20)?s[n]-48:"0066555766"[n/10]-48+s[n%10]-48}." until n==4;end;puts"4 is magic."
decoded:
n = gets.to_i
s = "03354435543668877987"
if n == 0
puts "0 is 4."
else
puts "#{n} is #{n = (n < 20) ? s[n] - 48 : "0066555766"[n / 10] - 48 + s[n % 10] - 48}." until n == 4
end
puts "4 is magic."
Lua 185 190 199
added periods, added io.read, removed ()'s on last print
n=io.read();while(n~=4)do m=('43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:'):sub(n+1):byte()-48;print(n,' is ',m,'.')n=m;end print'4 is magic.'
with line breaks
n=io.read()
while (n~=4) do
m=('43354435543668877988699;::9;;:699;::9;;:588:998::9588:998::9588:998::97::<;;:<<;699;::9;;:699;::9;;:'):sub(n+1):byte()-48;
print(n,' is ',m,'.')
n=m;
end
print'4 is magic.'
PhP Code
function get_num_name($num){
switch($num){
case 1:return 'one';
case 2:return 'two';
case 3:return 'three';
case 4:return 'four';
case 5:return 'five';
case 6:return 'six';
case 7:return 'seven';
case 8:return 'eight';
case 9:return 'nine';
}
}
function num_to_words($number, $real_name, $decimal_digit, $decimal_name){
$res = '';
$real = 0;
$decimal = 0;
if($number == 0)
return 'Zero'.(($real_name == '')?'':' '.$real_name);
if($number >= 0){
$real = floor($number);
$decimal = number_format($number - $real, $decimal_digit, '.', ',');
}else{
$real = ceil($number) * (-1);
$number = abs($number);
$decimal = number_format($number - $real, $decimal_digit, '.', ',');
}
$decimal = substr($decimal, strpos($decimal, '.') +1);
$unit_name[1] = 'thousand';
$unit_name[2] = 'million';
$unit_name[3] = 'billion';
$unit_name[4] = 'trillion';
$packet = array();
$number = strrev($real);
$packet = str_split($number,3);
for($i=0;$i<count($packet);$i++){
$tmp = strrev($packet[$i]);
$unit = $unit_name[$i];
if((int)$tmp == 0)
continue;
$tmp_res = '';
if(strlen($tmp) >= 2){
$tmp_proc = substr($tmp,-2);
switch($tmp_proc){
case '10':
$tmp_res = 'ten';
break;
case '11':
$tmp_res = 'eleven';
break;
case '12':
$tmp_res = 'twelve';
break;
case '13':
$tmp_res = 'thirteen';
break;
case '15':
$tmp_res = 'fifteen';
break;
case '20':
$tmp_res = 'twenty';
break;
case '30':
$tmp_res = 'thirty';
break;
case '40':
$tmp_res = 'forty';
break;
case '50':
$tmp_res = 'fifty';
break;
case '70':
$tmp_res = 'seventy';
break;
case '80':
$tmp_res = 'eighty';
break;
default:
$tmp_begin = substr($tmp_proc,0,1);
$tmp_end = substr($tmp_proc,1,1);
if($tmp_begin == '1')
$tmp_res = get_num_name($tmp_end).'teen';
elseif($tmp_begin == '0')
$tmp_res = get_num_name($tmp_end);
elseif($tmp_end == '0')
$tmp_res = get_num_name($tmp_begin).'ty';
else{
if($tmp_begin == '2')
$tmp_res = 'twenty';
elseif($tmp_begin == '3')
$tmp_res = 'thirty';
elseif($tmp_begin == '4')
$tmp_res = 'forty';
elseif($tmp_begin == '5')
$tmp_res = 'fifty';
elseif($tmp_begin == '6')
$tmp_res = 'sixty';
elseif($tmp_begin == '7')
$tmp_res = 'seventy';
elseif($tmp_begin == '8')
$tmp_res = 'eighty';
elseif($tmp_begin == '9')
$tmp_res = 'ninety';
$tmp_res = $tmp_res.' '.get_num_name($tmp_end);
}
break;
}
if(strlen($tmp) == 3){
$tmp_begin = substr($tmp,0,1);
$space = '';
if(substr($tmp_res,0,1) != ' ' && $tmp_res != '')
$space = ' ';
if($tmp_begin != 0){
if($tmp_begin != '0'){
if($tmp_res != '')
$tmp_res = 'and'.$space.$tmp_res;
}
$tmp_res = get_num_name($tmp_begin).' hundred'.$space.$tmp_res;
}
}
}else
$tmp_res = get_num_name($tmp);
$space = '';
if(substr($res,0,1) != ' ' && $res != '')
$space = ' ';
$res = $tmp_res.' '.$unit.$space.$res;
}
$space = '';
if(substr($res,-1) != ' ' && $res != '')
$space = ' ';
if($res)
$res .= $space.$real_name.(($real > 1 && $real_name != '')?'s':'');
if($decimal > 0)
$res .= ' '.num_to_words($decimal, '', 0, '').' '.$decimal_name.(($decimal > 1 && $decimal_name != '')?'s':'');
return ucfirst($res);
}
//////////// testing ////////////////
$str2num = 12;
while($str2num!=4){
$str = num_to_words($str2num, '', 0, '');
$str2num = strlen($str)-1;
echo $str . '=' . $str2num .'<br/>';
if ($str2num == 4)
echo 'four is magic';
}
////// Results /////////
Twelve =6
Six =3
Three =5
Five =4
four is magic
Perl - 130 chars
5.12.1 (130 chars) 121 123 132 136 140
# 1 2 3 4 5 6 7 8 9 100 11 12 13 14
#23456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123
#u='4335443554366887798866555766'=~/./g;$_=pop;say"$_ is ",$_=$_-4?$_<20?$u[$_]:$u[$_/10+18]+(($_%=10)&&$u[$_]):magic,"."until/\D/
5.10.1 (134 chars) 125 127 136 140 144
# 1 2 3 4 5 6 7 8 9 100 11 12 13 14
#23456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 123456789 1234
#u='4335443554366887798866555766'=~/./g;$_=pop;print"$_ is ",$_=$_-4?$_<20?$u[$_]:$u[$_/10+18]+(($_%=10)&&$u[$_]):magic,".\n"until/\D/
Change History:
20100714:2223 - reverted change at the attention of mobrule, but ($_%10&&$u[$_%10]) → (($_%=10)&&$u[$_]), which is the same # of chars, but I did it in case someone might see a way to improve it
20100714:0041 - split//,'...' → '...'=~/./g
20100714:0025 - ($_%10&&$u[$_%10]) → $u[$_%10]
20100713:2340 - while$_ → until/\D/ + removed unnecessary parentheses
20100713:xxxx - $=<>;chop; → $_=pop; - courtesy to mobrule
Note: I was tired of improving others' answers in comments, so now I'm being greedy and can just add my changes here :) This is a split off from Platinum Azure's answer - credit in part to Hobbs, mobrule, and Platinum Azure.
Shameless Perl with Number Words (329 characters)
Adapted fairly directly from P Daddy's C code, with some tweaks to p() to make it do the same thing using Perl primitives instead of C ones, and a mostly-rewritten mainloop. See his for an explanation. Newlines are all optional.
#t=(qw(zero one two three four five six sM eight nine
tL elM twelve NP 4P fifP 6P 7P 8O 9P twLQ NQ forQ fifQ
6Q 7Q 8y 9Q en evL thir eL tO ty 4SmagicT)," is ",".\n");
sub p{local$_=$t[pop];1while s/[0-Z]/$t[-48+ord$&]/e;
print;length}$_=<>;chop;while($_-4){
$_=($_>19?(p($_/10+18),$_&&print("-"),$_%=10)[0]:0)+p$_;
p 35;p$_;p 36}p 34
Side note: it's too bad that perl print just returns true/false; if it returned a count it would save me 7 strokes.
Ruby, 141 chars:
n=gets.to_i;m="4335443554366887798866555766";loop{s=n;n=n>20?m[18+n/10]+m[n%10]-96: m[n]-48;puts"#{s} is #{n==s ? 'magic': n}.";n==s &&break}
while(true)
{
string a;
ReadLine(a)
WriteLine(4);
}

Code Golf: Rotating Maze

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
Code Golf: Rotating Maze
Make a program that takes in a file consisting of a maze. The maze has walls given by #. The maze must include a single ball, given by a o and any number of holes given by a #. The maze file can either be entered via command line or read in as a line through standard input. Please specify which in your solution.
Your program then does the following:
1: If the ball is not directly above a wall, drop it down to the nearest wall.
2: If the ball passes through a hole during step 1, remove the ball.
3: Display the maze in the standard output (followed by a newline).
Extraneous whitespace should not be displayed.
Extraneous whitespace is defined to be whitespace outside of a rectangle
that fits snugly around the maze.
4: If there is no ball in the maze, exit.
5: Read a line from the standard input.
Given a 1, rotate the maze counterclockwise.
Given a 2, rotate the maze clockwise.
Rotations are done by 90 degrees.
It is up to you to decide if extraneous whitespace is allowed.
If the user enters other inputs, repeat this step.
6: Goto step 1.
You may assume all input mazes are closed. Note: a hole effectively acts as a wall in this regard.
You may assume all input mazes have no extraneous whitespace.
The shortest source code by character count wins.
Example written in javascript:
http://trinithis.awardspace.com/rotatingMaze/maze.html
Example mazes:
######
#o ##
######
###########
#o #
# ####### #
#### #
#########
###########################
# #
# # # #
# # # ##
# # ####o####
# # #
# #
# #########
# #
######################
Perl, 143 (128) char
172 152 146 144 143 chars,
sub L{my$o;$o.=$/while s/.$/$o.=$&,""/meg;$_=$o}$_.=<>until/
/;{L;1while s/o / o/;s/o#/ #/;L;L;L;print;if(/o/){1-($z=<>)||L;$z-2||L&L&L;redo}}
Newlines are significant.
Uses standard input and expects input to contain the maze, followed by a blank line, followed by the instructions (1 or 2), one instruction per line.
Explanation:
sub L{my$o;$o.="\n"while s/.$/$o.=$&,""/meg;$_=$o}
L is a function that uses regular expressions to rotate the multi-line expression $_ counterclockwise by 90 degrees. The regular expression was used famously by hobbs in my favorite code golf solution of all time.
$_.=<>until/\n\n/;
Slurps the input up to the first pair of consecutive newlines (that is, the maze) into $_.
L;1 while s/o / o/;s/o#/ */;
L;L;L;print
To drop the ball, we need to move the o character down one line is there is a space under it. This is kind of hard to do with a single scalar expression, so what we'll do instead is rotate the maze counterclockwise, move the ball to the "right". If a hole ever appears to the "right" of the ball, then the ball is going to fall in the hole (it's not in the spec, but we can change the # to an * to show which hole the ball fell into). Then before we print, we need to rotate the board clockwise 90 degrees (or counterclockwise 3 times) so that down is "down" again.
if(/o/) { ... }
Continue if there is still a ball in the maze. Otherwise the block will end and the program will exit.
1-($z=<>)||L;$z-2||L+L+L;redo
Read an instruction into $z. Rotate the board counterclockwise once for instruction "1" and three times for instruction "2".
If we used 3 more characters and said +s/o[#*]/ */ instead of ;s/o#/ */, then we could support multiple balls.
A simpler version of this program, where the instructions are "2" for rotating the maze clockwise and any other instruction for rotating counterclockwise, can be done in 128 chars.
sub L{my$o;$o.=$/while s/.$/$o.=$&,""/meg;$_=$o}$_.=<>until/
/;L;{1while s/o / o/+s/o#/ #/;L,L,L;print;if(/o/){2-<>&&L,L;redo}}
GolfScript - 97 chars
n/['']/~{;(#"zip-1%":|3*~{{." o"/"o "*"#o"/"# "*.#>}do}%|~.n*."o"/,(}{;\~(2*)|*~\}/\[n*]+n.+*])\;
This isn't done as well as I hoped (maybe later).
(These are my notes and not an explanation)
n/['']/~ #[M I]
{
;(# #[I c M]
"zip-1%":|3*~ #rotate
{{." o"/"o "*"#o"/"# "*.#>}do}% #drop
|~ #rotate back
.n* #"display" -> [I c M d]
."o"/,( #any ball? -> [I c M d ?]
}{ #d is collected into an array -> [I c M]
;\~(2*)|*~ #rotate
\ #stack order
}/
\[n*]+n.+*])\; #output
Rebmu: 298 Characters
I'm tinkering with with my own experiment in Code Golf language design! I haven't thrown matrix tricks into the standard bag yet, and copying GolfScript's ideas will probably help. But right now I'm working on refining the basic gimmick.
Anyway, here's my first try. The four internal spaces are required in the code as it is, but the line breaks are not necessary:
.fFS.sSC L{#o#}W|[l?fM]H|[l?m]Z|[Tre[wH]iOD?j[rvT]t]
Ca|[st[xY]a KrePC[[yBKx][ntSBhXbkY][ntSBhYsbWx][xSBwY]]ntJskPCmFkSk]
Ga|[rtYsZ[rtXfZ[TaRE[xY]iTbr]iTbr]t]B|[gA|[ieSlFcA[rnA]]]
MeFI?a[rlA]aFV[NbIbl?n[ut[++n/2 TfCnIEfLtBRchCbSPieTHlTbrCHcNsLe?sNsZ]]
gA|[TfCaEEfZfA[prT][pnT]nn]ulBbr JmoADjPC[3 1]rK4]
It may look like a cat was on my keyboard. But once you get past a little space-saving trick (literally saving spaces) called "mushing" it's not so bad. The idea is that Rebmu is not case sensitive, so alternation of capitalization runs is used to compress the symbols. Instead of doing FooBazBar => foo baz bar I apply distinct meanings. FOObazBAR => foo: baz bar (where the first token is an assignment target) vs fooBAZbar => foo baz bar (all ordinary tokens).
When the unmush is run, you get something more readable, but expanded to 488 characters:
. f fs . s sc l: {#o#} w: | [l? f m] h: | [l? m] z: | [t: re [w h] i od?
j [rv t] t] c: a| [st [x y] a k: re pc [[y bk x] [nt sb h x bk y] [nt sb
h y sb w x] [x sb w y]] nt j sk pc m f k s k] g: a| [rt y s z [rt x f z
[t: a re [x y] i t br] i t br] rn t] b: | [g a| [ie s l f c a [rn a]]]
m: e fi? a [rl a] a fv [n: b i bl? n [ut [++ n/2 t: f c n ie f l t br
ch c b sp ie th l t br ch c n s l e? s n s z]] g a| [t: f c a ee f z f
a [pr t] [pn t] nn] ul b br j: mo ad j pc [3 1] r k 4]
Rebmu can run it expanded also. There are also verbose keywords as well (first instead of fs) and you can mix and match. Here's the function definitions with some comments:
; shortcuts f and s extracting the first and second series elements
. f fs
. s sc
; character constants are like #"a", this way we can do fL for #"#" etc
L: {#o#}
; width and height of the input data
W: | [l? f m]
H: | [l? m]
; dimensions adjusted for rotation (we don't rotate the array)
Z: | [t: re [w h] i od? j [rv t] t]
; cell extractor, gives series position (like an iterator) for coordinate
C: a| [
st [x y] a
k: re pc [[y bk x] [nt sb h x bk y] [nt sb h y sb w x] [x sb w y]] nt j
sk pc m f k s k
]
; grid enumerator, pass in function to run on each cell
G: a| [rt y s z [rt x f z [t: a re [x y] i t br] i t br] t]
; ball position function
B: | [g a| [ie sc l f c a [rn a]]]
W is the width function and H is the height of the original array data. The data is never rotated...but there is a variable j which tells us how many 90 degree right turns we should apply.
A function Z gives us the adjusted size for when rotation is taken into account, and a function C takes a coordinate pair parameter and returns a series position (kind of like a pointer or iterator) into the data for that coordinate pair.
There's an array iterator G which you pass a function to and it will call that function for each cell in the grid. If the function you supply ever returns a value it will stop the iteration and the iteration function will return that value. The function B scans the maze for a ball and returns coordinates if found, or none.
Here's the main loop with some commenting:
; if the command line argument is a filename, load it, otherwise use string
m: e fi? a [rl a] a
; forever (until break, anyway...)
fv [
; save ball position in n
n: B
; if n is a block type then enter a loop
i bl? n [
; until (i.e. repeat until)
ut [
; increment second element of n (the y coordinate)
++ n/2
; t = first(C(n))
t: f C n
; if-equal(first(L), t) then break
ie f l t br
; change(C(B), space)
ch C B sp
; if-equal(third(L),t) then break
ie th L t br
; change(C(n), second(L))
ch C n s L
; terminate loop if "equals(second(n), second(z))"
e? s n s z
]
]
; iterate over array and print each line
g a| [t: f c a ee f z f a [pr t] [pn t] nn]
; unless the ball is not none, we'll be breaking the loop here...
ul b br
; rotate according to input
j: mo ad j pc [3 1] r k 4
]
There's not all that much particularly clever about this program. Which is part of my idea, which is to see what kind of compression one could get on simple, boring approaches that don't rely on any tricks. I think it demonstrates some of Rebmu's novel potential.
It will be interesting to see how a better standard library could affect the brevity of solutions!
Latest up-to-date commented source available on GitHub: rotating-maze.rebmu
Ruby 1.9.1 p243
355 353 characters
I'm pretty new to Ruby, so I'm sure this could be a lot shorter - theres probably some nuances i'm missing.
When executed, the path to the map file is the first line it reads. I tried to make it part of the execution arguments (would have saved 3 characters), but had issues :)
The short version:
def b m;m.each_index{|r|i=m[r].index(?o);return r,i if i}end;def d m;x,y=b m;z=x;
while z=z+1;c=m[z][y];return if c==?#;m[z-1][y]=" "; return 1 if c==?#;m[z][y]=?o;end;end;
def r m;m.transpose.reverse;end;m=File.readlines(gets.chomp).map{|x|x.chomp.split(//)};
while a=0;w=d m;puts m.map(&:join);break if w;a=gets.to_i until 0<a&&a<3;
m=r a==1?m:r(r(m));end
The verbose version:
(I've changed a bit in the compressed version, but you get the idea)
def display_maze m
puts m.map(&:join)
end
def ball_pos m
m.each_index{ |r|
i = m[r].index("o")
return [r,i] if i
}
end
def drop_ball m
x,y = ball_pos m
z=x
while z=z+1 do
c=m[z][y]
return if c=="#"
m[z-1][y]=" "
return 1 if c=="#"
m[z][y]="o"
end
end
def rot m
m.transpose.reverse
end
maze = File.readlines(gets.chomp).map{|x|x.chomp.split(//)}
while a=0
win = drop_ball maze
display_maze maze
break if win
a=gets.to_i until (0 < a && a < 3)
maze=rot maze
maze=rot rot maze if a==1
end
Possible improvement areas:
Reading the maze into a clean 2D array (currently 55 chars)
Finding and returning (x,y) co-ordinates of the ball (currently 61 chars)
Any suggestions to improve are welcome.
Haskell: 577 509 527 244 230 228 chars
Massive new approach: Keep the maze as a single string!
import Data.List
d('o':' ':x)=' ':(d$'o':x)
d('o':'#':x)=" *"++x
d(a:x)=a:d x
d e=e
l=unlines.reverse.transpose.lines
z%1=z;z%2=l.l$z
t=putStr.l.l.l
a z|elem 'o' z=t z>>readLn>>=a.d.l.(z%)|0<1=t z
main=getLine>>=readFile>>=a.d.l
Nods to #mobrule's Perl solution for the idea of dropping the ball sideways!
Python 2.6: ~ 284 ~ characters
There is possibly still room for improvement (although I already got it down a lot since the first revisions).
All comments or suggestions more then welcome!
Supply the map file on the command line as the first argument:
python rotating_maze.py input.txt
import sys
t=[list(r)[:-1]for r in open(sys.argv[1])]
while t:
x=['o'in e for e in t].index(1);y=t[x].index('o')
while t[x+1][y]!="#":t[x][y],t[x+1][y]=" "+"o#"[t[x+1][y]>" "];x+=1
for l in t:print''.join(l)
t=t[x][y]=='o'and map(list,(t,zip(*t[::-1]),zip(*t)[::-1])[input()])or 0
C# 3.0 - 650 638 characters
(not sure how newlines being counted)
(leading whitespace for reading, not counted)
using System.Linq;
using S=System.String;
using C=System.Console;
namespace R
{
class R
{
static void Main(S[]a)
{
S m=S.Join("\n",a);
bool u;
do
{
m=L(m);
int b=m.IndexOf('o');
int h=m.IndexOf('#',b);
b=m.IndexOf('#',b);
m=m.Replace('o',' ');
u=(b!=-1&b<h|h==-1);
if (u)
m=m.Insert(b-1,"o").Remove(b,1);
m=L(L(L(m)));
C.WriteLine(m);
if (!u) return;
do
{
int.TryParse(C.ReadLine(),out b);
u=b==1|b==2;
m=b==1?L(L(L(m))):u?L(m):m;
}while(!u);
}while(u);
}
static S L(S s)
{
return S.Join("\n",
s.Split('\n')
.SelectMany(z => z.Select((c,i)=>new{c,i}))
.GroupBy(x =>x.i,x=>x.c)
.Select(g => new S(g.Reverse().ToArray()))
.ToArray());
}
}
}
Reads from commandline, here's the test line I used:
"###########" "#o #" "# ####### #" "#### #" " #########"
Relied heavily on mobrule's Perl answer for algorithm.
My Rotation method (L) can probably be improved.
Handles wall-less case.

Has anyone seen a programming puzzle similar to this?

"Suppose you want to build a solid panel out of rows of 4×1 and 6×1 Lego blocks. For structural strength, the spaces between the blocks must never line up in adjacent rows. As an example, the 18×3 panel shown below is not acceptable, because the spaces between the blocks in the top two rows line up.
There are 2 ways to build a 10×1 panel, 2 ways to build a 10×2 panel, 8 ways to build an 18×3 panel, and 7958 ways to build a 36×5 panel.
How many different ways are there to build a 64×10 panel? The answer will fit in a 64-bit signed integer. Write a program to calculate the answer. Your program should run very quickly – certainly, it should not take longer than one minute, even on an older machine. Let us know the value your program computes, how long it took your program to calculate that value, and on what kind of machine you ran it. Include the program’s source code as an attachment.
"
I was recently given a programming puzzle and have been racking my brains trying to solve it. I wrote some code using c++ and I know the number is huge...my program ran for a few hours before I decided just to stop it because the requirement was 1 minute of run time even on a slow computer. Has anyone seen a puzzle similar to this? It has been a few weeks and I can't hand this in anymore, but this has really been bugging me that I couldn't solve it correctly. Any suggestions on algorithms to use? Or maybe possible ways to solve it that are "outside the box". What i resorted to was making a program that built each possible "layer" of 4x1 and 6x1 blocks to make a 64x1 layer. That turned out to be about 3300 different layers. Then I had my program run through and stack them into all possible 10 layer high walls that have no cracks that line up...as you can see this solution would take a long, long, long time. So obviously brute force does not seem to be effective in solving this within the time constraint. Any suggestions/insight would be greatly appreciated.
The main insight is this: when determining what's in row 3, you don't care about what's in row 1, just what's in row 2.
So let's call how to build a 64x1 layer a "row scenario". You say that there are about 3300 row scenarios. That's not so bad.
Let's compute a function:
f(s, r) = the number of ways to put row scenario number "s" into row "r", and legally fill all the rows above "r".
(I'm counting with row "1" at the top, and row "10" at the bottom)
STOP READING NOW IF YOU WANT TO AVOID SPOILERS.
Now clearly (numbering our rows from 1 to 10):
f(s, 1) = 1
for all values of "s".
Also, and this is where the insight comes in, (Using Mathematica-ish notation)
f(s, r) = Sum[ f(i, r-1) * fits(s, i) , {i, 1, 3328} ]
where "fits" is a function that takes two scenario numbers and returns "1" if you can legally stack those two rows on top of each other and "0" if you can't. This uses the insight because the number of legal ways to place scenario depends only on the number of ways to place scenarios above it that are compatible according to "fits".
Now, fits can be precomputed and stored in a 3328 by 3328 array of bytes. That's only about 10 Meg of memory. (Less if you get fancy and store it as a bit array)
The answer then is obviously just
Sum[ f(i, 10) , {i, 1, 3328} ]
Here is my answer. It's Haskell, among other things, you get bignums for free.
EDIT: It now actually solves the problem in a reasonable amount of time.
MORE EDITS: With a sparse matrix it takes a half a second on my computer.
You compute each possible way to tile a row. Let's say there are N ways to tile a row. Make an NxN matrix. Element i,j is 1 if row i can appear next to row j, 0 otherwise. Start with a vector containing N 1s. Multiply the matrix by the vector a number of times equal to the height of the wall minus 1, then sum the resulting vector.
module Main where
import Data.Array.Unboxed
import Data.List
import System.Environment
import Text.Printf
import qualified Data.Foldable as F
import Data.Word
import Data.Bits
-- This records the index of the holes in a bit field
type Row = Word64
-- This generates the possible rows for given block sizes and row length
genRows :: [Int] -> Int -> [Row]
genRows xs n = map (permToRow 0 1) $ concatMap comboPerms $ combos xs n
where
combos [] 0 = return []
combos [] _ = [] -- failure
combos (x:xs) n =
do c <- [0..(n `div` x)]
rest <- combos xs (n - x*c)
return (if c > 0 then (x, c):rest else rest)
comboPerms [] = return []
comboPerms bs =
do (b, brest) <- choose bs
rest <- comboPerms brest
return (b:rest)
choose bs = map (\(x, _) -> (x, remove x bs)) bs
remove x (bc#(y, c):bs) =
if x == y
then if c > 1
then (x, c - 1):bs
else bs
else bc:(remove x bs)
remove _ [] = error "no item to remove"
permToRow a _ [] = a
permToRow a _ [_] = a
permToRow a n (c:cs) =
permToRow (a .|. m) m cs where m = n `shiftL` c
-- Test if two rows of blocks are compatible
-- i.e. they do not have a hole in common
rowCompat :: Row -> Row -> Bool
rowCompat x y = x .&. y == 0
-- It's a sparse matrix with boolean entries
type Matrix = Array Int [Int]
type Vector = UArray Int Word64
-- Creates a matrix of row compatibilities
compatMatrix :: [Row] -> Matrix
compatMatrix rows = listArray (1, n) $ map elts [1..n] where
elts :: Int -> [Int]
elts i = [j | j <- [1..n], rowCompat (arows ! i) (arows ! j)]
arows = listArray (1, n) rows :: UArray Int Row
n = length rows
-- Multiply matrix by vector, O(N^2)
mulMatVec :: Matrix -> Vector -> Vector
mulMatVec m v = array (bounds v)
[(i, sum [v ! j | j <- m ! i]) | i <- [1..n]]
where n = snd $ bounds v
initVec :: Int -> Vector
initVec n = array (1, n) $ zip [1..n] (repeat 1)
main = do
args <- getArgs
if length args < 3
then putStrLn "usage: blocks WIDTH HEIGHT [BLOCKSIZE...]"
else do
let (width:height:sizes) = map read args :: [Int]
printf "Width: %i\nHeight %i\nBlock lengths: %s\n" width height
$ intercalate ", " $ map show sizes
let rows = genRows sizes width
let rowc = length rows
printf "Row tilings: %i\n" rowc
if null rows
then return ()
else do
let m = compatMatrix rows
printf "Matrix density: %i/%i\n"
(sum (map length (elems m))) (rowc^2)
printf "Wall tilings: %i\n" $ sum $ elems
$ iterate (mulMatVec m) (initVec (length rows))
!! (height - 1)
And the results...
$ time ./a.out 64 10 4 6
Width: 64
Height 10
Block lengths: 4, 6
Row tilings: 3329
Matrix density: 37120/11082241
Wall tilings: 806844323190414
real 0m0.451s
user 0m0.423s
sys 0m0.012s
Okay, 500 ms, I can live with that.
I solved a similar problem for a programming contest tiling a long hallway with tiles of various shapes. I used dynamic programming: given any panel, there is a way to construct it by laying down one row at a time. Each row can have finitely many shapes at its end. So for each number of rows, for each shape, I compute how many ways there are to make that row. (For the bottom row, there is exactly one way to make each shape.) Then the shape of each row determines the number of shapes that the next row can take (i.e. never line up the spaces). This number is finite for each row and in fact because you have only two sizes of bricks, it is going to be small. So you wind up spending constant time per row and the program finishes quickly.
To represent a shape I would just make a list of 4's and 6's, then use this list as a key in a table to store the number of ways to make that shape in row i, for each i.

Find a pattern of binary numbers using shift-right and bitwise-AND?

I'm attempting to write a function in assembly that will detect if a longer binary number contains a smaller binary pattern.
Example:
Does 100111 contain 1001?
When I read this problem I figured that I would do a bitwise-AND with the large number and its smaller pattern while shifting right (logical) each time in a loop.
So, in my head I thought it would do:
100111 AND 1001 = 0
Shift-right 1
010011 AND 1001 = 0
Shift-right 1
001001 AND 1001 = 1 // Pattern FOUND!
and repeat this until either the number was shifted until it was zero or the AND returned 1.
However, I think I must have something confused because this is returning 1 for most things I put in, on the first run of the loop. Am I confused on my usage of AND?
The problem is that "partial matches" also return a non-zero value for your AND check:
100111 AND 001001 = 000001
So this tests if any of the bits match, but you want to make sure all bits are the same. The result of the AND needs to be equal to the pattern you are searching:
x = 100111
if (x AND 1001 == 1001)
print "found"
Bitwise AND does not work the way you expect (judging from the samples and ignoring the notation which seems to suggest you are using bitwise AND as the logical AND of bits). AND only takes the bits that are set to 1 "into account". E.g 1111 AND 1001 == 1001.
You need to use XOR and compare against 0 for match (remember the mask the bits you are not comparing from the result). In your example a match is found when (N ^ 1001) & 1111 == 0000
In order to make sure that both the 0 and 1 bits match your search pattern, you need to do something like this:
if ((InputPattern AND SearchMask) == SearchPattern)
{
// then match
}
The SearchMask should be all 1 bits, of a length equal to your SearchPattern. For example, you could have SearchMask == 1111, SearchPattern == 1001.
You should AND and then test against the search pattern:
if ((TestPattern & SearchPattern) == SearchPattern)
{
// then match
}
(where & represents bitwise AND)