Related
I understood the problem and written the below code. My code works and it prints the number of detected valid and invalid patterns when I quit the program with ctrl+z.
Here is my code:
%{
int valid = 0;
int invalid = 0;
%}
%%
([A-Z][a-zA-Z0-9]*[a-z])* {valid++;}
[a-zA-Z0-9]* {invalid++;}
%%
int yywrap(){}
int main(int argc, char **argv[])
{
printf("\n Enter inputs: \n\n");
yylex();
printf("\n\n\tNumber of VALID patterns = %d\n", valid);
printf("\tNumber of invalid patterns = %d\n\n", invalid);
return 0;
}
But I want something like this:
It should print the detected patterns, number of valid patterns and the number of invalid patterns whenever I input a new line.
There should be an EXIT command.
To achieve your goal, you should modify your code like this:
/*** Definition Section ***/
%{
int valid = 0;
int invalid = 0;
%}
/*** Rules Section ***/
%%
([A-Z][a-zA-Z0-9]*[a-z])* {printf("\n\tPattern Detected: %s ", yytext); valid++;}
[a-zA-Z0-9]* {invalid++;}
"\n" {
printf("\n\n\tNumber of VALID patterns = %d\n", valid);
printf("\tNumber of invalid patterns = %d\n\n", invalid);
valid = 0;
invalid = 0;
}
EXIT__ return 0;
%%
/*** User code section***/
int yywrap(){}
int main(int argc, char **argv[])
{
printf("\n Enter inputs: \n\n");
yylex();
return 0;
}
Here main change comes in the rule section.
Rule-1: ([A-Z][a-zA-Z0-9]*[a-z])* It detect and count valid patterns that starts with an uppercase letter, ends with a lowercase letter. In action, it prints the detected patterns and does the counting job too. Here yytext contains the text in the buffer, for this rule, it's the detected pattern.
Rule-2: [a-zA-Z0-9]* Keep a track of invalid patterns. It will help to prevent returning unmatched patterns.
Rule-3: "\n" It detects when you input a new line. In action, it prints the detected patterns, the number of valid patterns, and the number of invalid patterns whenever I input a new line. Also, reset the variables for counting to zero for the next line of input.
Rule-4: EXIT__ whenever you will input this exact command, the program will exit.
You can avoid printing the numbers of valid and invalid patterns inside the main function in the user code section.
But if you want to print the numbers of detected valid and invalid patterns at the end too, then this program will require a few modifications.
I'm trying to parse a file that (apparently) ends its lines with carriage returns, but they aren't being matched as such in Swift, despite having the same UTF8 value. I can see possible fixes for the problem, but I'm curious as to what these characters actually are.
Here's some sample code, with the output below. (CR is set using Character("\r"), although I've tried it using "\r" as well.
try f.forEach() { c in
print(c, terminator:" ") // DBG
if (c == "\r") {
print("Carriage return found!")
}
print(String(c).utf8.first!, terminator:" ")//DBG
print(String(describing:pstate)) // DBG
...
case .field:
switch c {
case CR,LF :
self.endline()
pstate = .eol
When it reaches the end of line (which shows up as such in my text editors), I get this:
. 46 field
0 48 field
13 field
I 73 field
It doesn't seem to be matching using == or in the switch statement. Is there another approach I should be using for this character?
(I'll note that the parsing works fine with files that terminate in newlines.)
I determined what the problem was. By looking at c.unicodeScalars I discovered that the end of line character was in fact "\r\n", not just "\r". As seen in my code I was only taking the first when printing it out as UTF-8. I don't know if that's something from String.forEach or in the file itself.
I know that there are tests to determine if something is a newline. Swift 5 has them directly (c.isNewline), and there is also the CharacterSet approach as noted by Bill Nattaner.
I'm happier with something that will work in my switch statement (and thus I'll define each one explicitly), but that might change if I expect to deal with a wider variety of files.
I'm a little hazy as to what the f.forEach represents, but if your variable c is of type Character then you could replace your if statement with:
if "\(c)".rangeOfCharacter( from: CharacterSet.newlines ) != nil
{
print("Carriage return found!")
}
That way you won't have to invent a list of all-possible new line characters.
I have a any challenge. I must write brainfuck-code.
For a given number n appoint its last digit .
entrance
Input will consist of only one line in which there is only one integer n ( 1 < = n < = 2,000,000,000 ) , followed by a newline ' \ n' (ASCII 10).
exit
On the output has to find exactly one integer denoting the last digit of n .
example I
entrance: 32
exit: 2
example II:
entrance: 231231132
exit: 2
This is what I tried, but it didn't work:
+[>,]<.>++++++++++.
The last input is the newline. So you have to go two memory positions back to get the last digit of the number. And maybe you don't have to return a newline character, so the code is
,[>,]<<.
Nope sorry, real answer is
,[>,]<.
because your answer was getting one too far ;)
Depending on the interpreter, you might have to escape the return key by yourself. considering the return key is ASCII: 10, your code should look like this :
>,----- -----[+++++ +++++>,----- -----]<.
broken down :
> | //first operation (just in case your interpreter does not
support a negative pointer index)
,----- ----- | //first entry if it's a return; you don't even get in the loop
[
+++++ +++++ | //if the value was not ASCII 10; you want the original value back
>, | //every next entry
----- ----- | //check again for the the return,
you exit the loop only if the last entered value is 10
]
<. | //your current pointer is 0; you go back to the last valid entry
and you display it
Your issue is that a loop continues for forever until at the end of the loop the cell the pointer is currently on in equal to 0. Your code never prints in the loop, and never subtracts, so your loop will never end, and all that your code does is take an ASCII character as input, move one forward, take an ASCII character as input, and so on. All of your code after the end of the loop is useless, because that your loop will never end.
Please answer with the shortest possible source code for a program that converts an arbitrary plaintext to its corresponding ciphertext, following the sample input and output I have given below. Bonus points* for the least CPU time or the least amount of memory used.
Example 1:
Plaintext: The quick brown fox jumps over the lazy dog. Supercalifragilisticexpialidocious!
Ciphertext: eTh kiquc nobrw xfo smjup rvoe eth yalz .odg !uioiapeislgriarpSueclfaiitcxildcos
Example 2:
Plaintext: 123 1234 12345 123456 1234567 12345678 123456789
Ciphertext: 312 4213 53124 642135 7531246 86421357 975312468
Rules:
Punctuation is defined to be included with the word it is closest to.
The center of a word is defined to be ceiling((strlen(word)+1)/2).
Whitespace is ignored (or collapsed).
Odd words move to the right first. Even words move to the left first.
You can think of it as reading every other character backwards (starting from the end of the word), followed by the remaining characters forwards. Corporation => XoXpXrXtXoX => niaorCoprto.
Thank you to those who pointed out the inconsistency in my description. This has lead many of you down the wrong path, which I apologize for. Rule #4 should clear things up.
*Bonus points will only be awarded if Jeff Atwood decides to do so. Since I haven't checked with him, the chances are slim. Sorry.
Python, 50 characters
For input in i:
' '.join(x[::-2]+x[len(x)%2::2]for x in i.split())
Alternate version that handles its own IO:
print ' '.join(x[::-2]+x[len(x)%2::2]for x in raw_input().split())
A total of 66 characters if including whitespace. (Technically, the print could be omitted if running from a command line, since the evaluated value of the code is displayed as output by default.)
Alternate version using reduce:
' '.join(reduce(lambda x,y:y+x[::-1],x) for x in i.split())
59 characters.
Original version (both even and odd go right first) for an input in i:
' '.join(x[::2][::-1]+x[1::2]for x in i.split())
48 characters including whitespace.
Another alternate version which (while slightly longer) is slightly more efficient:
' '.join(x[len(x)%2-2::-2]+x[1::2]for x in i.split())
(53 characters)
J, 58 characters
>,&.>/({~(,~(>:#+:#i.#-#<.,+:#i.#>.)#-:)#<:##)&.><;.2,&' '
Haskell, 64 characters
unwords.map(map snd.sort.zip(zipWith(*)[0..]$cycle[-1,1])).words
Well, okay, 76 if you add in the requisite "import List".
Python - 69 chars
(including whitespace and linebreaks)
This handles all I/O.
for w in raw_input().split():
o=""
for c in w:o=c+o[::-1]
print o,
Perl, 78 characters
For input in $_. If that's not acceptable, add six characters for either $_=<>; or $_=$s; at the beginning. The newline is for readability only.
for(split){$i=length;print substr$_,$i--,1,''while$i-->0;
print"$_ ";}print $/
C, 140 characters
Nicely formatted:
main(c, v)
char **v;
{
for( ; *++v; )
{
char *e = *v + strlen(*v), *x;
for(x = e-1; x >= *v; x -= 2)
putchar(*x);
for(x = *v + (x < *v-1); x < e; x += 2)
putchar(*x);
putchar(' ');
}
}
Compressed:
main(c,v)char**v;{for(;*++v;){char*e=*v+strlen(*v),*x;for(x=e-1;x>=*v;x-=2)putchar(*x);for(x=*v+(x<*v-1);x<e;x+=2)putchar(*x);putchar(32);}}
Lua
130 char function, 147 char functioning program
Lua doesn't get enough love in code golf -- maybe because it's hard to write a short program when you have long keywords like function/end, if/then/end, etc.
First I write the function in a verbose manner with explanations, then I rewrite it as a compressed, standalone function, then I call that function on the single argument specified at the command line.
I had to format the code with <pre></pre> tags because Markdown does a horrible job of formatting Lua.
Technically you could get a smaller running program by inlining the function, but it's more modular this way :)
t = "The quick brown fox jumps over the lazy dog. Supercalifragilisticexpialidocious!"
T = t:gsub("%S+", -- for each word in t...
function(w) -- argument: current word in t
W = "" -- initialize new Word
for i = 1,#w do -- iterate over each character in word
c = w:sub(i,i) -- extract current character
-- determine whether letter goes on right or left end
W = (#w % 2 ~= i % 2) and W .. c or c .. W
end
return W -- swap word in t with inverted Word
end)
-- code-golf unit test
assert(T == "eTh kiquc nobrw xfo smjup rvoe eth yalz .odg !uioiapeislgriarpSueclfaiitcxildcos")
-- need to assign to a variable and return it,
-- because gsub returns a pair and we only want the first element
f=function(s)c=s:gsub("%S+",function(w)W=""for i=1,#w do c=w:sub(i,i)W=(#w%2~=i%2)and W ..c or c ..W end return W end)return c end
-- 1 2 3 4 5 6 7 8 9 10 11 12 13
--34567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890
-- 130 chars, compressed and written as a proper function
print(f(arg[1]))
--34567890123456
-- 16 (+1 whitespace needed) chars to make it a functioning Lua program,
-- operating on command line argument
Output:
$ lua insideout.lua 'The quick brown fox jumps over the lazy dog. Supercalifragilisticexpialidocious!'
eTh kiquc nobrw xfo smjup rvoe eth yalz .odg !uioiapeislgriarpSueclfaiitcxildcos
I'm still pretty new at Lua so I'd like to see a shorter solution if there is one.
For a minimal cipher on all args to stdin, we can do 111 chars:
for _,w in ipairs(arg)do W=""for i=1,#w do c=w:sub(i,i)W=(#w%2~=i%2)and W ..c or c ..W end io.write(W ..' ')end
But this approach does output a trailing space like some of the other solutions.
For an input in s:
f=lambda t,r="":t and f(t[1:],len(t)&1and t[0]+r or r+t[0])or r
" ".join(map(f,s.split()))
Python, 90 characters including whitespace.
TCL
125 characters
set s set f foreach l {}
$f w [gets stdin] {$s r {}
$f c [split $w {}] {$s r $c[string reverse $r]}
$s l "$l $r"}
puts $l
Bash - 133, assuming input is in $w variable
Pretty
for x in $w; do
z="";
for l in `echo $x|sed 's/\(.\)/ \1/g'`; do
if ((${#z}%2)); then
z=$z$l;
else
z=$l$z;
fi;
done;
echo -n "$z ";
done;
echo
Compressed
for x in $w;do z="";for l in `echo $x|sed 's/\(.\)/ \1/g'`;do if ((${#z}%2));then z=$z$l;else z=$l$z;fi;done;echo -n "$z ";done;echo
Ok, so it outputs a trailing space.
Eg if input string is helloworld I want the output to be like:
do
he
we
low
hell
hold
roll
well
word
hello
lower
world
...
all the way up to the longest word that is an anagram of a substring of helloworld. Like in Scrabble for example.
The input string can be any length, but rarely more than 16 chars.
I've done a search and come up with structures like a trie, but I am still unsure of how to actually do this.
The structure used to hold your dictionary of valid entries will have a huge impact on efficiency. Organize it as a tree, root being the singular zero letter "word", the empty string. Each child of root is a single first letter of a possible word, children of those being the second letter of a possible word, etc., with each node marked as to whether it actually forms a word or not.
Your tester function will be recursive. It starts with zero letters, finds from the tree of valid entries that "" isn't a word but it does have children, so you call your tester recursively with your start word (of no letters) appended with each available remaining letter from your input string (which is all of them at that point). Check each one-letter entry in tree, if valid make note; if children, re-call tester function appending each of remaining available letters, and so on.
So for example, if your input string is "helloworld", you're going to first call your recursive tester function with "", passing the remaining available letters "helloworld" as a 2nd parameter. Function sees that "" isn't a word, but child "h" does exist. So it calls itself with "h", and "elloworld". Function sees that "h" isn't a word, but child "e" exists. So it calls itself with "he" and "lloworld". Function sees that "e" is marked, so "he" is a word, take note. Further, child "l" exists, so next call is "hel" with "loworld". It will next find "hell", then "hello", then will have to back out and probably next find "hollow", before backing all the way out to the empty string again and then starting with "e" words next.
I couldn't resist my own implementation. It creates a dictionary by sorting all the letters alphabetically, and mapping them to the words that can be created from them. This is an O(n) start-up operation that eliminates the need to find all permutations. You could implement the dictionary as a trie in another language to attain faster speedups.
The "getAnagrams" command is also an O(n) operation which searches each word in the dictionary to see if it is a subset of the search. Doing getAnagrams("radiotelegraphically")" (a 20 letter word) took approximately 1 second on my laptop, and returned 1496 anagrams.
# Using the 38617 word dictionary at
# http://www.cs.umd.edu/class/fall2008/cmsc433/p5/Usr.Dict.Words.txt
# Usage: getAnagrams("helloworld")
def containsLetters(subword, word):
wordlen = len(word)
subwordlen = len(subword)
if subwordlen > wordlen:
return False
word = list(word)
for c in subword:
try:
index = word.index(c)
except ValueError:
return False
word.pop(index)
return True
def getAnagrams(word):
output = []
for key in mydict.iterkeys():
if containsLetters(key, word):
output.extend(mydict[key])
output.sort(key=len)
return output
f = open("dict.txt")
wordlist = f.readlines()
f.close()
mydict = {}
for word in wordlist:
word = word.rstrip()
temp = list(word)
temp.sort()
letters = ''.join(temp)
if letters in mydict:
mydict[letters].append(word)
else:
mydict[letters] = [word]
An example run:
>>> getAnagrams("helloworld")
>>> ['do', 'he', 'we', 're', 'oh', 'or', 'row', 'hew', 'her', 'hoe', 'woo', 'red', 'dew', 'led', 'doe', 'ode', 'low', 'owl', 'rod', 'old', 'how', 'who', 'rho', 'ore', 'roe', 'owe', 'woe', 'hero', 'wood', 'door', 'odor', 'hold', 'well', 'owed', 'dell', 'dole', 'lewd', 'weld', 'doer', 'redo', 'rode', 'howl', 'hole', 'hell', 'drew', 'word', 'roll', 'wore', 'wool','herd', 'held', 'lore', 'role', 'lord', 'doll', 'hood', 'whore', 'rowed', 'wooed', 'whorl', 'world', 'older', 'dowel', 'horde', 'droll', 'drool', 'dwell', 'holed', 'lower', 'hello', 'wooer', 'rodeo', 'whole', 'hollow', 'howler', 'rolled', 'howled', 'holder', 'hollowed']
The data structure you want is called a Directed Acyclic Word Graph (dawg), and it is described by Andrew Appel and Guy Jacobsen in their paper "The World's Fastest Scrabble Program" which unfortunately they have chosen not to make available free online. An ACM membership or a university library will get it for you.
I have implemented this data structure in at least two languages---it is simple, easy to implement, and very, very fast.
A simple-minded approach is to generate all the "substrings" and, for each of them, check whether it's an element of the set of acceptable words. E.g., in Python 2.6:
import itertools
import urllib
def words():
f = urllib.urlopen(
'http://www.cs.umd.edu/class/fall2008/cmsc433/p5/Usr.Dict.Words.txt')
allwords = set(w[:-1] for w in f)
f.close()
return allwords
def substrings(s):
for i in range(2, len(s)+1):
for p in itertools.permutations(s, i):
yield ''.join(p)
def main():
w = words()
print '%d words' % len(w)
ss = set(substrings('weep'))
print '%d substrings' % len(ss)
good = ss & w
print '%d good ones' % len(good)
sgood = sorted(good, key=lambda w:(len(w), w))
for aword in sgood:
print aword
main()
will emit:
38617 words
31 substrings
5 good ones
we
ewe
pew
wee
weep
Of course, as other responses pointed out, organizing your data purposefully can greatly speed-up your runtime -- although the best data organization for a fast anagram finder could well be different... but that will largely depend on the nature of your dictionary of allowed words (a few tens of thousands, like here -- or millions?). Hash-maps and "signatures" (based on sorting the letters in each word) should be considered, as well as tries &c.
What you want is an implementation of a power set.
Also look at Eric Lipparts blog, he blogged about this very thing a little while back
EDIT:
Here is an implementation I wrote of getting the powerset from a given string...
private IEnumerable<string> GetPowerSet(string letters)
{
char[] letterArray = letters.ToCharArray();
for (int i = 0; i < Math.Pow(2.0, letterArray.Length); i++)
{
StringBuilder sb = new StringBuilder();
for (int j = 0; j < letterArray.Length; j++)
{
int pos = Convert.ToInt32(Math.Pow(2.0, j));
if ((pos & i) == pos)
{
sb.Append(letterArray[j]);
}
}
yield return new string(sb.ToString().ToCharArray().OrderBy(c => c).ToArray());
}
}
This function gives me the powersets of chars that make up the passed in string, I then can use these as keys into a dictionary of anagrams...
Dictionary<string,IEnumerable<string>>
I created my dictionary of anagrams like so... (there are probably more efficient ways, but this was simple and plenty quick enough with the scrabble tournament word list)
wordlist = (from s in fileText.Split(new string[] { Environment.NewLine }, StringSplitOptions.RemoveEmptyEntries)
let k = new string(s.ToCharArray().OrderBy(c => c).ToArray())
group s by k).ToDictionary(o => o.Key, sl => sl.Select(a => a));
Like Tim J, Eric Lippert's blog posts where the first thing to come to my mind. I wanted to add that he wrote a follow-up about ways to improve the performance of his first attempt.
A nasality talisman for the sultana analyst
Santalic tailfans, part two
I believe the Ruby code in the answers to this question will also solve your problem.
I've been playing a lot of Wordfeud on my phone recently and was curious if I could come up with some code to give me a list of possible words. The following code takes your availble source letters (* for a wildcards) and an array with a master list of allowable words (TWL, SOWPODS, etc) and generates a list of matches. It does this by trying to build each word in the master list from your source letters.
I found this topic after writing my code, and it's definitely not as efficient as John Pirie's method or the DAWG algorithm, but it's still pretty quick.
public IList<string> Matches(string sourceLetters, string [] wordList)
{
sourceLetters = sourceLetters.ToUpper();
IList<string> matches = new List<string>();
foreach (string word in wordList)
{
if (WordCanBeBuiltFromSourceLetters(word, sourceLetters))
matches.Add(word);
}
return matches;
}
public bool WordCanBeBuiltFromSourceLetters(string targetWord, string sourceLetters)
{
string builtWord = "";
foreach (char letter in targetWord)
{
int pos = sourceLetters.IndexOf(letter);
if (pos >= 0)
{
builtWord += letter;
sourceLetters = sourceLetters.Remove(pos, 1);
continue;
}
// check for wildcard
pos = sourceLetters.IndexOf("*");
if (pos >= 0)
{
builtWord += letter;
sourceLetters = sourceLetters.Remove(pos, 1);
}
}
return string.Equals(builtWord, targetWord);
}