I am working on a watermarking project that embeds binary values (i.e 1s and 0s) in the image, for which I have to take the input from the user, and check certain conditions such as
1) no empty string
2) no other character or special character
3) no number other than 0 and 1
is entered.
The following code just checks the first condition. Is there any default function in Matlab to check whether entered string is binary
int_state = get(handles.edit1,'String'); %edit1 is the Tag of edit box
if isempty(int_state)`
fprintf('Error: Enter Text first\n');
else
%computation code
end
There is no such standard function, but the check can be easily implemented.
Use this error condition:
isempty(int_state) || any(~ismember(int_state, '01'))
It returns false (no error) if the string is non-empty and composed of '0's and '1's only.
The function ismember returns a boolean array that indicates for every character in int_state whether it is contained in the second argument, '01'. The advantage is that this can be generalized to arbitrary sets of allowed characters.
I think the 2nd and 3rd can be combined together as 1 condition: your input string can only be a combination of 0 and 1? If it is so, then a small trick with findstr can do that:
if length(findstr(input_str, '1')) + length(findstr(input_str, '0')) == length(input_str)
condition_satisfied;
end
tf = isnumeric(A) returns true if A is a numeric array and false otherwise.
A numeric array is any of the numeric types and any subclasses of those types.
isnumeric(A)
ans =
1 (when A is numeric).
I have MATLAB set to record three webcams at the same time. I want to capture and save each feed to a file and automatically increment it the file name, it will be replaced by experiment_0001.avi, followed by experiment_0002.avi, etc.
My code looks like this at the moment
set(vid1,'LoggingMode','disk');
set(vid2,'LoggingMode','disk');
avi1 = VideoWriter('X:\ABC\Data Collection\Presentations\Correct\ExperimentA_002.AVI');
avi2 = VideoWriter('X:\ABC\Data Collection\Presentations\Correct\ExperimentB_002.AVI');
set(vid1,'DiskLogger',avi1);
set(vid2,'DiskLogger',avi2);
and I am incrementing the 002 each time.
Any thoughts on how to implement this efficiently?
Thanks.
dont forget matlab has some roots to C programming language. That means things like sprintf will work
so since you are printing out an integer value zero padded to 3 spaces you would need something like this sprintf('%03d',n) then % means there is a value to print that isn't text. 0 means zero pad on the left, 3 means pad to 3 digits, d means the number itself is an integer
just use sprintf in place of a string. the s means String print formatted. so it will output a string. here is an idea of what you might do
set(vid1,'LoggingMode','disk');
set(vid2,'LoggingMode','disk');
for (n=1:2:max_num_captures)
avi1 = VideoWriter(sprintf('X:\ABC\Data Collection\Presentations\Correct\ExperimentA_%03d.AVI',n));
avi2 = VideoWriter(sprintf('X:\ABC\Data Collection\Presentations\Correct\ExperimentB_002.AVI',n));
set(vid1,'DiskLogger',avi1);
set(vid2,'DiskLogger',avi2);
end
Dim str as String
str = "30 40 50 60"
I want to count the number of substrings.
Expected Output: 4
(because there are 4 total values: 30, 40, 50, 60)
How can I accomplish this in VB6?
You could try this:
arrStr = Split(str, " ")
strCnt = UBound(arrStr) + 1
msgBox strCnt
Of course, if you've got Option Explicit set (which you should..) then declare the variables above first..
Your request doesn't make any sense. A string is a sequence of text. The fact that that sequence of text contains numbers separated by spaces is quite irrelevant. Your string looks like this:
30 40 50 60
There are not 4 separate values, there is only one value, shown above—a single string.
You could also view the string as containing 11 individual characters, so it could be argued that the "count" of the string would be 11, but this doesn't get you any further towards your goal.
In order to get the result that you expect, you need to split the string into multiple strings at each space, producing 4 separate strings, each containing a 2-digit numeric value.
Of course, the real question is why you're storing this value in a string in the first place. If they're numeric values, you should store them in an array (for example, an array of Integers). Then you can easily obtain the number of elements in the array using the LBound() and UBound() functions.
I agree with everything Cody stated.
If you really wanted to you could loop through the string character by character and count the number of times you find your delimiter. In your example, it is space delimited, so you would simply count the number of spaces and add 1, but as Cody stated, those are not separate values..
Are you trying to parse text here or what? Regardless, I think what you really need to do is store your data into an array. Make your life easier, not more difficult.
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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.