using visualworks, in small talk, I'm receiving a string like '31323334' from a network connection.
I need a string that reads '1234' so I need a way of extracting two characters at a time, converting them to what they represent in ascii, and then building a string of them...
Is there a way to do so?
EDIT(7/24): for some reason many of you are assuming I will only be working with numbers and could just truncate 3s or read every other char. This is not the case, examples of strings read could include any keys on the US standard keyboard (a-z, A-Z,0-9,punctuation/annotation such as {}*&^%$...)
Following along the lines of what Max started to suggest:
x := '31323334'.
in := ReadStream on: x.
out := WriteStream on: String new.
[ in atEnd ] whileFalse: [ out nextPut: (in next digitValue * 16 + (in next digitValue)) asCharacter ].
newX := out contents.
newX will have the result '1234'. Or, if you start with:
x := '454647'
You will get a result of 'EFG'.
Note that digitValue might only recognize upper case hex digits, so an asUppercase may be needed on the string before processing.
There is usually a #fold: or #reduce: method that will let you do that. In Pharo there's also a message #allPairsDo: and #groupsOf:atATimeCollect:. Using one of these methods you could do:
| collectionOfBytes |
collectionOfBytes := '9798'
groupsOf: 2
atATimeCollect: [ :group |
(group first digitValue * 10) + (group second digitValue) ].
collectionOfBytes asByteArray asString "--> 'ab'"
The #digitValue message in Pharo simply returns the value of the digit for numerical characters.
If you're receiving the data on a stream you could replace #groupsOf:atATime: with a loop (result may be any collection that you then convert to a string like above):
...
[ stream atEnd ] whileFalse: [
result add: (stream next digitValue * 10) + (stream next digitValue) ]
...
in Smalltalk/X, there is a method called "fromHexBytes:" which the ByteArray class understands. I am not sure, but think that something similar exists in other ST dialects.
If present, you can solve this with:
(ByteArray fromHexString:'68656C6C6F31323334') asString
and the reverse would be:
'hello1234' asByteArray hexPrintString
Another possible solution is to read the string as a hex number,
fetch the digitBytes (which should give you a byte array) and then convert that to a string.
I.e.
(Integer readFrom:'68656C6C6F31323334' radix:16)
digitBytes asString
One problem with that is that I am not sure about which byte-order you will get the digitBytes (LSB or MSB), and if that is defined to be the same across architectures or converted at image loading time to use the native order. So it may be required to reverse the string at the end (to be portable, it may even be required to reverse it conditionally, depending on the endianess of the system.
I cannot test this on VisualWorks, but I assume it should work fine there, too.
I'm working with OPE IDs. One file has them with two trailing zeros, eg, [998700, 1001900]. The other file has them with one or two leading zeros for a total length of six, eg, [009987, 010019]. I want to convert every OPE ID (in both files) to an eight-digit string with exactly two leading zeros and however many zeros at the end to get it to be eight digits long.
Try this:
a = [ "00123123", "077934", "93422", "1231234", "12333" ]
a.map { |n| n.gsub(/^0*/, '00').ljust(8, '0') }
=> ["00123123", "00779340", "00934220", "001231234", "00123330"]
If you have your data parsed and stored as strings, it could be done like this, for example.
n = ["998700", "1001900", "009987", "0010019"]
puts n.map { |i|
i =~ /^0*([0-9]+?)0*$/
"00" + $1 + "0" * [0, 6 - $1.length].max
}
Output:
00998700
00100190
00998700
00100190
This example on codepad.
I'm note very sure though, that I got the description exactly right. Please check the comments and I correct in case it's not exactly what you were looking for.
With the help of the answers given by #detunized & #nimblegorilla, I came up with:
"998700"[0..-3].rjust(6, '0').to_sym
to make the first format I described (always with two trailing zeros) equal to the second.
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.
I'm outputting a set of numbered files from a Ruby script. The numbers come from incrementing a counter, but to make them sort nicely in the directory, I'd like to use leading zeros in the filenames. In other words
file_001...
instead of
file_1
Is there a simple way to add leading zeros when converting a number to a string? (I know I can do "if less than 10.... if less than 100").
Use the % operator with a string:
irb(main):001:0> "%03d" % 5
=> "005"
The left-hand-side is a printf format string, and the right-hand side can be a list of values, so you could do something like:
irb(main):002:0> filename = "%s/%s.%04d.txt" % ["dirname", "filename", 23]
=> "dirname/filename.0023.txt"
Here's a printf format cheat sheet you might find useful in forming your format string. The printf format is originally from the C function printf, but similar formating functions are available in perl, ruby, python, java, php, etc.
If the maximum number of digits in the counter is known (e.g., n = 3 for counters 1..876), you can do
str = "file_" + i.to_s.rjust(n, "0")
Can't you just use string formatting of the value before you concat the filename?
"%03d" % number
Use String#next as the counter.
>> n = "000"
>> 3.times { puts "file_#{n.next!}" }
file_001
file_002
file_003
next is relatively 'clever', meaning you can even go for
>> n = "file_000"
>> 3.times { puts n.next! }
file_001
file_002
file_003
As stated by the other answers, "%03d" % number works pretty well, but it goes against the rubocop ruby style guide:
Favor the use of sprintf and its alias format over the fairly
cryptic String#% method
We can obtain the same result in a more readable way using the following:
format('%03d', number)
filenames = '000'.upto('100').map { |index| "file_#{index}" }
Outputs
[file_000, file_001, file_002, file_003, ..., file_098, file_099, file_100]