Code golf: find all anagrams - code-golf

A word is an anagram if the letters in that word can be re-arranged to form a different word.
Task:
The shortest source code by character count to find all sets of anagrams given a word list.
Spaces and new lines should be counted as characters
Use the code ruler
---------10--------20--------30--------40--------50--------60--------70--------80--------90--------100-------110-------120
Input:
a list of words from stdin with each word separated by a new line.
e.g.
A
A's
AOL
AOL's
Aachen
Aachen's
Aaliyah
Aaliyah's
Aaron
Aaron's
Abbas
Abbasid
Abbasid's
Output:
All sets of anagrams, with each set separated by a separate line.
Example run:
./anagram < words
marcos caroms macros
lump's plum's
dewar's wader's
postman tampons
dent tend
macho mocha
stoker's stroke's
hops posh shop
chasity scythia
...
I have a 149 char perl solution which I'll post as soon as a few more people post :)
Have fun!
EDIT: Clarifications
Assume anagrams are case insensitive (i.e. upper and lower case letters are equivalent)
Only sets with more than 1 item should be printed
Each set of anagrams should only be printed once
Each word in an anagram set should only occur once
EDIT2: More Clarifications
If two words differ only in capitalization, they should be collapsed into the same word, and it's up to you to decide which capitalization scheme to use for the collapsed word
sets of words only have to end in a new line, as long as each word is separated in some way, e.g. comma separated, or space separated is valid. I understand some languages have quick array printing methods built in so this should allow you to take advantage of that if it doesn't output space separated arrays.

Powershell, 104 97 91 86 83 chars
$k=#{};$input|%{$k["$([char[]]$_|%{$_+0}|sort)"]+=#($_)}
$k.Values|?{$_[1]}|%{"$_"}
Update for the new requirement (+8 chars):
To exclude the words that only differ in capitalization, we could just remove the duplicates (case-insensitvely) from the input list, i.e. $input|sort -u where -u stands for -unique. sort is case-insenstive by default:
$k=#{};$input|sort -u|%{$k["$([char[]]$_|%{$_+0}|sort)"]+=#($_)}
$k.Values|?{$_[1]}|%{"$_"}
Explanation of the [char[]]$_|%{$_+0}|sort -part
It's a key for the hashtable entry under which anagrams of a word are stored. My initial solution was: $_.ToLower().ToCharArray()|sort. Then I discovered I didn't need ToLower() for the key, as hashtable lookups are case-insensitive.
[char[]]$_|sort would be ideal, but sorting of the chars for the key needs to be case-insensitive (otherwise Cab and abc would be stored under different keys). Unfortunately, sort is not case-insenstive for chars (only for strings).
What we need is [string[]][char[]]$_|sort, but I found a shorter way of converting each char to string, which is to concat something else to it, in this case an integer 0, hence [char[]]$_|%{$_+0}|sort. This doesn't affect the sorting order, and the actual key ends up being something like: d0 o0 r0 w0. It's not pretty, but it does the job :)

Perl, 59 characters
chop,$_{join'',sort split//,lc}.="$_ "for<>;/ ./&&say for%_
Note that this requires Perl 5.10 (for the say function).

Haskell, 147 chars
prior sizes: 150 159 chars
import Char
import List
x=sort.map toLower
g&a=g(x a).x
main=interact$unlines.map unwords.filter((>1).length).groupBy((==)&).sortBy(compare&).lines
This version, at 165 chars satisifies the new, clarified rules:
import Char
import List
y=map toLower
x=sort.y
g&f=(.f).g.f
w[_]="";w a=show a++"\n"
main=interact$concatMap(w.nubBy((==)&y)).groupBy((==)&x).sortBy(compare&x).lines
This version handles:
Words in the input that differ only by case should only count as one word
The output needs to be one anagram set per line, but extra punctuation is acceptable

Ruby, 94 characters
h={};(h[$_.upcase.bytes.sort]||=[])<<$_ while gets&&chomp;h.each{|k,v|puts v.join' 'if v.at 1}

Python, 167 characters, includes I/O
import sys
d={}
for l in sys.stdin.readlines():
l=l[:-1]
k=''.join(sorted(l)).lower()
d[k]=d.pop(k,[])+[l]
for k in d:
if len(d[k])>1: print(' '.join(d[k]))
Without the input code (i.e. if we assume the wordlist already in a list w), it's only 134 characters:
d={}
for l in w:
l=l[:-1]
k=''.join(lower(sorted(l)))
d[k]=d.pop(k,[])+[l]
for k in d:
if len(d[k])>1: print(' '.join(d[k]))

AWK - 119
{split(toupper($1),a,"");asort(a);s="";for(i=1;a[i];)s=a[i++]s;x[s]=x[s]$1" "}
END{for(i in x)if(x[i]~/ .* /)print x[i]}
AWK does not have a join function like Python, or it could have been shorter...
It assumes uppercase and lowercase as different.

C++, 542 chars
#include <iostream>
#include <map>
#include <vector>
#include <boost/algorithm/string.hpp>
#define ci const_iterator
int main(){using namespace std;typedef string s;typedef vector<s> vs;vs l;
copy(istream_iterator<s>(cin),istream_iterator<s>(),back_inserter(l));map<s, vs> r;
for (vs::ci i=l.begin(),e=l.end();i!=e;++i){s a=boost::to_lower_copy(*i);
sort(a.begin(),a.end());r[a].push_back(*i);}for (map<s,vs>::ci i=r.begin(),e=r.end();
i!=e;++i)if(i->second.size()>1)*copy(i->second.begin(),i->second.end(),
ostream_iterator<s>(cout," "))="\n";}

Python, O(n^2)
import sys;
words=sys.stdin.readlines()
def s(x):return sorted(x.lower());
print '\n'.join([''.join([a.replace('\n',' ') for a in words if(s(a)==s(w))]) for w in words])

Related

Ruby rspec check string begins with certain values

I have a string that begins with a set start and then is filled to 252 characters by randomly generated letters. (237 random chars - so that the end length is 252 characters)
For the purpose of testing the string starts off with TESTDATAMENDNOW and the rest is random capital letters.
How can I test that the characters are all capital letters
and that the string begins with TESTDATAMENDNOW
I have tried to use regex expressions to define this but I'm not too sure how to get them to work properly, and what I have tried so far seems not to be working.
EDIT: Clarity
expect(string).to match(/\ATESTDATAMENDNOW[A-Z]{237}\z/)
237 because 252 minus the length of "TESTDATAMENDNOW" is 237.
Here's another way that does not use a regex:
str = "TESTDATAMENDNOW"
expect(string[0,str.size]).eq(str)
expect(string.delete("ABCDEFGHIJKLMNOPQRSTUVWXYZ").eq("")
expect(string.size).eq(252)

Splitting a string on variable numbers of words

The following question was posted by #ruhroe about an hour ago. I was about to post an answer when it was taken down. That's unfortunate, as I thought it was rather interesting. I'm putting it back up in case the OP sees this and also to give others an opportunity to post solutions.
The original question (which I've edited):
The problem is to split a string on some spaces in the string, based on criteria which depend in part on a number given by the user. If that number were, say, 5, each substring would contain either:
one word having 5 or more characters or
as many consecutive words (separated by spaces) as possible, provided the resulting string has at most 5 characters.
For example, if the string were:
"abcdefg fg hijkl mno pqrs tuv wx yz"
the result would be:
["abcdefg", "fg", "hijkl", "mno", "pqrs", "tuv", "wx yz"]
"abcdefg" is on a separate line because it has at least five characters.
"fg" is on a separate line because "fg" contains 5 or few characters and when combined with the following word, with a space between them, the resulting string, "fg hijkl", contains more than 5 characters.
"hijkl" is on a separate line because it satisfies both criteria.
How can I do that?
I believe this does it:
str = "abcdefg fg hijkl e mn pqrs tuv wx yz"
str.scan(/\b(?:\w{5,}|\w[\w\s]{0,3}\w|\w)\b/)
#=> ["abcdefg", "fg", "hijkl", "e mn", "pqrs", "tuv", "wx yz"]
As you iterate through the words in your collection (splitting the original string up into words should be trivial), it seems like there are three possible scenarios:
It's a blank line, and we should insert the current word into the line
It's a non-blank line, and the word can fit
It's a non-blank line, and the word can't fit and it should go into a new line
Something like this should work (note - I haven't tested this much outside of your solution. You'll definitely want to do that):
words.each do |word|
if line.blank?
# this is a new line, so start it with the current word
line << word
elsif word_can_fit_line?(line, word, length)
# the word fits, so append it to the current line
line << " #{word}"
else
# the word doesn't fit, so keep this line and start a new one with
# the current word
lines << line
line = word
end
end
# add the last line and we're done
lines << line
lines
Note that the implementation of word_can_fit_line? should be trivial - you just want to see if the current line length, plus a space, plus the word length, is less than or equal to your desired line length.

How to handle Combining Diacritical Marks with UnicodeUtils?

I am trying to insert spaces into a string of IPA characters, e.g. to turn ɔ̃wɔ̃tɨ into ɔ̃ w ɔ̃ t ɨ. Using split/join was my first thought:
s = ɔ̃w̃ɔtɨ
s.split('').join(' ') #=> ̃ ɔ w ̃ ɔ p t ɨ
As I discovered by examining the results, letters with diacritics are in fact encoded as two characters. After some research I found the UnicodeUtils module, and used the each_grapheme method:
UnicodeUtils.each_grapheme(s) {|g| g + ' '} #=> ɔ ̃w ̃ɔ p t ɨ
This worked fine, except for the inverted breve mark. The code changes ̑a into ̑ a. I tried normalization (UnicodeUtils.nfc, UnicodeUtils.nfd), but to no avail. I don't know why the each_grapheme method has a problem with this particular diacritic mark, but I noticed that in gedit, the breve is also treated as a separate character, as opposed to tildes, accents etc. So my question is as follows: is there a straightforward method of normalization, i.e. turning the combination of Latin Small Letter A and Combining Inverted Breve into Latin Small Letter A With Inverted Breve?
I understand your question concerns Ruby but I suppose the problem is about the same as with Python. A simple solution is to test the combining diacritical marks explicitly :
import unicodedata
liste=[]
s = u"ɔ̃w̃ɔtɨ"
comb=False
prec=u""
for char in s:
if unicodedata.combining(char):
liste.append(prec+char)
prec=""
else:
liste.append(prec)
prec=char
liste.append(prec)
print " ".join(liste)
>>>> ɔ̃ w̃ ɔ t ɨ

Regular expression to match my pattern of words, wild chars

can you help me with this:
I want a regular expression for my Ruby program to match a word with the below pattern
Pattern has
List of letters ( For example. ABCC => 1 A, 1 B, 2 C )
N Wild Card Charaters ( N can be 0 or 1 or 2)
A fixed word (for example “XY”).
Rules:
Regarding the List of letters, it should match words with
a. 0 or 1 A
b. 0 or 1 B
c. 0 or 1 or 2 C
Based on the value of N, there can be 0 or 1 or 2 wild chars
Fixed word is always in the order it is given.
The combination of all these can be in any order and should match words like below
ABWXY ( if wild char = 1)
BAXY
CXYCB
But not words with 2 A’s or 2 B’s
I am using the pattern like ^[ABCC]*.XY$
But it looks for words with more than 1 A, or 1 B or 2 C's and also looks for words which end with XY, I want all words which have XY in any place and letters and wild chars in any postion.
If it HAS to be a regex, the following could be used:
if subject =~
/^ # start of string
(?!(?:[^A]*A){2}) # assert that there are less than two As
(?!(?:[^B]*B){2}) # and less than two Bs
(?!(?:[^C]*C){3}) # and less than three Cs
(?!(?:[ABCXY]*[^ABCXY]){3}) # and less than three non-ABCXY characters
(?=.*XY) # and that XY is contained in the string.
/x
# Successful match
else
# Match attempt failed
end
This assumes that none of the characters A, B, C, X, or Y are allowed as wildcards.
I consider myself to be fairly good with regular expressions and I can't think of a way to do what you're asking. Regular expressions look for patterns and what you seem to want is quite a few different patterns. It might be more appropriate to in your case to write a function which splits the string into characters and count what you have so you can satisfy your criteria.
Just to give an example of your problem, a regex like /[abc]/ will match every single occurrence of a, b and c regardless of how many times those letters appear in the string. You can try /c{1,2}/ and it will match "c", "cc", and "ccc". It matches the last case because you have a pattern of 1 c and 2 c's in "ccc".
One thing I have found invaluable when developing and debugging regular expressions is rubular.com. Try some examples and I think you'll see what you're up against.
I don't know if this is really any help but it might help you choose a direction.
You need to break out your pattern properly. In regexp terms, [ABCC] means "any one of A, B or C" where the duplicate C is ignored. It's a set operator, not a grouping operator like () is.
What you seem to be describing is creating a regexp based on parameters. You can do this by passing a string to Regexp.new and using the result.
An example is roughly:
def match_for_options(options)
pattern = '^'
pattern << 'A' * options[:a] if (options[:a])
pattern << 'B' * options[:b] if (options[:b])
pattern << 'C' * options[:c] if (options[:c])
Regexp.new(pattern)
end
You'd use it something like this:
if (match_for_options(:a => 1, :c => 2).match('ACC'))
# ...
end
Since you want to allow these "elements" to appear in any order, you might be better off writing a bit of Ruby code that goes through the string from beginning to end and counts the number of As, Bs, and Cs, finds whether it contains your desired substring. If the number of As, Bs, and Cs, is in your desired limits, and it contains the desired substring, and its length (i.e. the number of characters) is equal to the length of the desired substring, plus # of As, plus # of Bs, plus # of Cs, plus at most N characters more than that, then the string is good, otherwise it is bad. Actually, to be careful, you should first search for your desired substring and then remove it from the original string, then count # of As, Bs, and Cs, because otherwise you may unintentionally count the As, Bs, and Cs that appear in your desired string, if there are any there.
You can do what you want with a regular expression, but it would be a long ugly regular expression. Why? Because you would need a separate "case" in the regular expression for each of the possible orders of the elements. For example, the regular expression "^ABC..XY$" will match any string beginning with "ABC" and ending with "XY" and having two wild card characters in the middle. But only in that order. If you want a regular expression for all possible orders, you'd need to list all of those orders in the regular expression, e.g. it would begin something like "^(ABC..XY|ACB..XY|BAC..XY|BCA..XY|" and go on from there, with about 5! = 120 different orders for that list of 5 elements, then you'd need more for the cases where there was no A, then more for cases where there was no B, etc. I think a regular expression is the wrong tool for the job here.

Code Golf: Validate Sudoku Grid

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.
Introduction
A valid Sudoku grid is filled with numbers 1 to 9, with no number occurring more than once in each sub-block of 9, row or column. Read this article for further details if you're unfamiliar with this popular puzzle.
Challenge
The challenge is to write the shortest program that validates a Sudoku grid that might not be full.
Input will be a string of 9 lines of 9 characters each, representing the grid. An empty cell will be represented by a .. Your output should be Valid if the grid is valid, otherwise output Invalid.
Example
Input
123...789
...456...
456...123
789...456
...123...
564...897
...231...
897...564
...564...
Output
Valid
Input
123456789
987654321
123456789
123456789
987654321
123456789
123456789
987654321
123456789
Output
Invalid
Code Golf Rules
Please post your shortest code in any language that solves this problem. Input and output may be handled via stdin and stdout or by other files of your choice.
Winner will be the shortest solution (by byte count) in a language with an implementation existing prior to the posting of this question. So while you are free to use a language you've just made up in order to submit a 0-byte solution, it won't count, and you'll probably get downvotes.
Golfscript: 56
n%{zip''+9/.{'.'-..&=}%$0=\}:|2*{3/}%|;**"InvV"3/="alid"
C: 165 162 161 160 159
int v[1566],x,y=9,c,b;main(){while(y--)for(x=9;x--+1;)if((c
=getchar()*27)>1242)b|=v[x+c]++|v[y+9+c]++|v[x-x%3+y/3+18+c]
++;puts(b?"Invalid":"Valid");return 0;}
The two newlines are not needed. One char saved by josefx :-) ...
Haskell: 207 230 218 195 172
import List
t=take 3
h=[t,t.drop 3,drop 6]
v[]="V"
v _="Inv"
f s=v[1|v<-[s,transpose s,[g=<<f s|f<-h,g<-h]],g<-map(filter(/='.'))v,g/=nub g]++"alid\n"
main=interact$f.lines
Perl: 168 128
$_=join'',<>;#a=/.../g;print+(/(\d)([^\n]{0,8}|(.{10})*.{9})\1/s
+map"#a[$_,$_+3,$_+6]"=~/(\d).*\1/,0..2,9..11,18..20)?Inv:V,alid
The first regex checks for duplicates that are in the same row and column; the second regex handles duplicates in the "same box".
Further improvement is possible by replacing the \n in the first regex with a literal newline (1 char), or with >= Perl 5.12, replacing [^\n] with \N (3 char)
Earlier, 168 char solution:
Input is from stdin, output is to stderr because it makes things so easy. Linebreaks are optional and not counted.
$_=join'',<>;$m=alid.$/;$n=Inv.$m;/(\d)(\N{0,8}|(.{10})*.{9})\1/s&&
die$n;#a=/.../g;for$i(0,8,17){for$j($i..$i+2){
$_=$a[$j].$a[$j+3].$a[$j+6];/(\d).*\1/&&die$n}}die"V$m"
Python: 230 221 200 185
First the readable version at len=199:
import sys
r=range(9)
g=[raw_input()for _ in r]
s=[[]for _ in r*3]
for i in r:
for j in r:
n=g[i][j]
for x in i,9+j,18+i/3*3+j/3:
<T>if n in s[x]:sys.exit('Invalid')
<T>if n>'.':s[x]+=n
print'Valid'
Since SO doesn't display tab characters, I've used <T> to represent a single tab character.
PS. the same approach minEvilized down to 185 chars:
r=range(9)
g=[raw_input()for _ in r]
s=['']*27
for i in r:
for j in r:
for x in i,9+j,18+i/3*3+j/3:n=g[i][j];s[x]+=n[:n>'.']
print['V','Inv'][any(len(e)>len(set(e))for e in s)]+'alid'
Perl, 153 char
#B contains the 81 elements of the board.
&E tests whether a subset of #B contains any duplicate digits
main loop validates each column, "block", and row of the puzzle
sub E{$V+="#B[#_]"=~/(\d).*\1/}
#B=map/\S/g,<>;
for$d(#b=0..80){
E grep$d==$_%9,#b;
E grep$d==int(($_%9)/3)+3*int$_/27,#b;
E$d*9..$d*9+8}
print$V?Inv:V,alid,$/
Python: 159 158
v=[0]*244
for y in range(9):
for x,c in enumerate(raw_input()):
if c>".":
<T>for k in x,y+9,x-x%3+y//3+18:v[k*9+int(c)]+=1
print["Inv","V"][max(v)<2]+"alid"
<T> is a single tab character
Common Lisp: 266 252
(princ(let((v(make-hash-table))(r "Valid"))(dotimes(y 9)(dotimes(x
10)(let((c(read-char)))(when(>(char-code c)46)(dolist(k(list x(+ 9
y)(+ 18(floor(/ y 3))(- x(mod x 3)))))(when(>(incf(gethash(+(* k
9)(char-code c)-49)v 0))1)(setf r "Invalid")))))))r))
Perl: 186
Input is from stdin, output to stdout, linebreaks in input optional.
#y=map/\S/g,<>;
sub c{(join'',map$y[$_],#$h)=~/(\d).*\1/|c(#_)if$h=pop}
print(('V','Inv')[c map{$x=$_;[$_*9..$_*9+8],[grep$_%9==$x,0..80],[map$_+3*$b[$x],#b=grep$_%9<3,0..20]}0..8],'alid')
(Linebreaks added for "clarity".)
c() is a function that checks the input in #y against a list of lists of position numbers passed as an argument. It returns 0 if all position lists are valid (contain no number more than once) and 1 otherwise, using recursion to check each list. The bottom line builds this list of lists, passes it to c() and uses the result to select the right prefix to output.
One thing that I quite like is that this solution takes advantage of "self-similarity" in the "block" position list in #b (which is redundantly rebuilt many times to avoid having #b=... in a separate statement): the top-left position of the ith block within the entire puzzle can be found by multiplying the ith element in #b by 3.
More spread out:
# Grab input into an array of individual characters, discarding whitespace
#y = map /\S/g, <>;
# Takes a list of position lists.
# Returns 0 if all position lists are valid, 1 otherwise.
sub c {
# Pop the last list into $h, extract the characters at these positions with
# map, and check the result for multiple occurences of
# any digit using a regex. Note | behaves like || here but is shorter ;)
# If the match fails, try again with the remaining list of position lists.
# Because Perl returns the last expression evaluated, if we are at the
# end of the list, the pop will return undef, and this will be passed back
# which is what we want as it evaluates to false.
(join '', map $y[$_], #$h) =~ /(\d).*\1/ | c(#_) if $h = pop
}
# Make a list of position lists with map and pass it to c().
print(('V','Inv')[c map {
$x=$_; # Save the outer "loop" variable
[$_*9..$_*9+8], # Columns
[grep$_%9==$x,0..80], # Rows
[map$_+3*$b[$x],#b=grep$_%9<3,0..20] # Blocks
} 0..8], # Generates 1 column, row and block each time
'alid')
Perl: 202
I'm reading Modern Perl and felt like coding something... (quite a cool book by the way:)
while(<>){$i++;$j=0;for$s(split//){$j++;$l{$i}{$s}++;$c{$j}{$s}++;
$q{(int(($i+2)/3)-1)*3+int(($j+2)/3)}{$s}++}}
$e=V;for$i(1..9){for(1..9){$e=Inv if$l{$i}{$_}>1or$c{$i}{$_}>1or$q{$i}{$_}>1}}
print $e.alid
Count is excluding unnecessary newlines.
This may require Perl 5.12.2.
A bit more readable:
#use feature qw(say);
#use JSON;
#$json = JSON->new->allow_nonref;
while(<>)
{
$i++;
$j=0;
for $s (split //)
{
$j++;
$l{$i}{$s}++;
$c{$j}{$s}++;
$q{(int(($i+2)/3)-1)*3+int(($j+2)/3)}{$s}++;
}
}
#say "lines: ", $json->pretty->encode( \%l );
#say "columns: ", $json->pretty->encode( \%c );
#say "squares: ", $json->pretty->encode( \%q );
$e = V;
for $i (1..9)
{
for (1..9)
{
#say "checking {$i}{$_}: " . $l{$i}{$_} . " / " . $c{$i}{$_} . " / " . $q{$i}{$_};
$e = Inv if $l{$i}{$_} > 1 or $c{$i}{$_} > 1 or $q{$i}{$_} > 1;
}
}
print $e.alid;
Ruby — 176
f=->x{x.any?{|i|(i-[?.]).uniq!}}
a=[*$<].map{|i|i.scan /./}
puts f[a]||f[a.transpose]||f[a.each_slice(3).flat_map{|b|b.transpose.each_slice(3).map &:flatten}]?'Invalid':'Valid'
Lua, 341 bytes
Although I know that Lua isn't the best golfing language, however, considering it's size, I think it's worth posting it ;).
Non-golfed, commented and error-printing version, for extra fun :)
i=io.read("*a"):gsub("\n","") -- Get input, and strip newlines
a={{},{},{}} -- checking array, 1=row, 2=columns, 3=squares
for k=1,3 do for l=1,9 do a[k][l]={0,0,0,0,0,0,0,0,0}end end -- fillup array with 0's (just to have non-nils)
for k=1,81 do -- loop over all numbers
n=tonumber(i:sub(k,k):match'%d') -- get current character, check if it's a digit, and convert to a number
if n then
r={math.floor((k-1)/9)+1,(k-1)%9+1} -- Get row and column number
r[3]=math.floor((r[1]-1)/3)+3*math.floor((r[2]-1)/3)+1 -- Get square number
for l=1,3 do v=a[l][r[l]] -- 1 = row, 2 = column, 3 = square
if v[n] then -- not yet eliminated in this row/column/square
v[n]=nil
else
print("Double "..n.." in "..({"row","column","square"}) [l].." "..r[l]) --error reporting, just for the extra credit :)
q=1 -- Flag indicating invalidity
end
end
end
end
io.write(q and"In"or"","Valid\n")
Golfed version, 341 bytes
f=math.floor p=io.write i=io.read("*a"):gsub("\n","")a={{},{},{}}for k=1,3 do for l=1,9 do a[k][l]={0,0,0,0,0,0,0,0,0}end end for k=1,81 do n=tonumber(i:sub(k,k):match'%d')if n then r={f((k-1)/9)+1,(k-1)%9+1}r[3]=f((r[1]-1)/3)+1+3*f((r[2]-1)/3)for l=1,3 do v=a[l][r[l]]if v[n]then v[n]=nil else q=1 end end end end p(q and"In"or"","Valid\n")
Python: 140
v=[(k,c) for y in range(9) for x,c in enumerate(raw_input()) for k in x,y+9,(x/3,y/3) if c>'.']
print["V","Inv"][len(v)>len(set(v))]+"alid"
ASL: 108
args1["\n"x2I3*x;{;{:=T(T'{:i~{^0}?})}}
{;{;{{,0:e}:;{0:^},u eq}}/`/=}:-C
dc C#;{:|}C&{"Valid"}{"Invalid"}?P
ASL is a Golfscript inspired scripting language I made.

Resources