Odd word problem: Difference between revisions

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=={{header|Java}}==
=={{header|Java}}==
<lang java>public class OddWord {
{{incorrect|Java|Peeking ahead is not allowed}}
interface CharHandler {
This is translated from the first [[C]] version on the [http://c2.com/cgi/wiki?OddWordProblemSolutions solutions page].
CharHandler handle(char c) throws Exception;
<lang java>import java.io.BufferedReader;
}
import java.io.IOException;
final CharHandler fwd = new CharHandler() {
import java.io.StringReader;
public CharHandler handle(char c) {

System.out.print(c);
public class OddWord {
return (Character.isLetter(c) ? fwd : rev);
public static void processStream(BufferedReader in) throws IOException{
if(checkEnd(in))return;
while(true){
forward(in);
if(checkEnd(in))return;
reverse(in);
if(checkEnd(in))return;
}
}
}
};
class Reverser extends Thread implements CharHandler {
private static boolean checkEnd(BufferedReader in) throws IOException{
Reverser() {
if(peek(in) == '.'){
setDaemon(true);
System.out.println((char)in.read());
start();
return true;
}else{
System.out.print((char)in.read());
return false;
}
}
}
private Character ch; // For inter-thread comms
private static char peek(BufferedReader in) throws IOException{
private char recur() throws Exception {
in.mark(1);
notify();
char retVal = (char)in.read();
while (ch == null) wait();
char c = ch, ret = c;
in.reset();
ch = null;
return retVal;
if (Character.isLetter(c)) {
ret = recur();
System.out.print(c);
}
return ret;
}
}
public synchronized void run() {
try {
private static void forward(BufferedReader in) throws IOException{
while(Character.isLetter(peek(in))){
while (true) {
System.out.print((char)in.read());
System.out.print(recur());
notify();
}
}
} catch (Exception e) {}
}
}
public synchronized CharHandler handle(char c) throws Exception {
while (ch != null) wait();
private static void reverse(BufferedReader in) throws IOException{
ch = c;
if(Character.isLetter(peek(in))){
notify();
char character = (char)in.read();
while (ch != null) wait();
reverse(in);
return (Character.isLetter(c) ? rev : fwd);
System.out.print(character);
}
}
}
}
final CharHandler rev = new Reverser();
public static void main(String[] args) throws IOException{

processStream(new BufferedReader(new StringReader("what,is,the;meaning,of:life.")));
public void loop() throws Exception {
processStream(new BufferedReader(new StringReader("we,are;not,in,kansas;any,more.")));
CharHandler handler = fwd;
processStream(new BufferedReader(new StringReader(";what,is,the;meaning,of:life.")));
int c;
processStream(new BufferedReader(new StringReader("'we,are;not,in,kansas;any,more.")));
while ((c = System.in.read()) >= 0) {
handler = handler.handle((char) c);
}
}
}

public static void main(String...args) throws Exception {
new OddWord().loop();
}
}</lang>
}</lang>
Output is equivalent to that of the Python solution.
Output:
<pre>what,si,the;gninaem,of:efil.
we,era;not,ni,kansas;yna,more.
;what,si,the;gninaem,of:efil.
'we,era;not,ni,kansas;yna,more.</pre>


=={{header|Python}}==
=={{header|Python}}==

Revision as of 13:47, 5 November 2011

Odd word problem is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Write a program that solves the odd word problem.

Description: You are promised an input stream consisting of English letters and punctuations. It is guaranteed that

  • the words (sequence of consecutive letters) are delimited by one and only one punctuation; that
  • the stream will begin with a word; that
  • the words will be at least one letter long; and that
  • a full stop (.) appears after, and only after, the last word.

For example, what,is,the;meaning,of:life. is such a stream with six words. Your task is to reverse the letters in every other word while leaving punctuations intact, producing e.g. "what,si,the;gninaem,of:efil.", while observing the following restrictions:

  1. Only I/O allowed is reading or writing one character at a time, which means: no reading in a string, no peeking ahead, no pushing characters back into the stream, and no storing characters in a global variable for later use;
  2. You are not to explicitly save characters in a collection data structure, such as arrays, strings, hash tables, etc, for later reversal;
  3. You are allowed to use recursions, closures, continuations, threads, coroutines, etc.

Test case: work on both the "life" example given above, and the text we,are;not,in,kansas;any,more.

C

Using GCC nested function as closures. This can only be passed up the stack, not the other way around. It's also doable with makecontext, and may be possible with setjmp.

<lang c>#include <stdio.h>

  1. include <ctype.h>

int do_char(int odd, void (*f)(void)) { int c = getchar();

void write_out(void) { putchar(c); if (f) f(); }

if (!odd) putchar(c);

if (isalpha(c)) return do_char(odd, write_out);

if (odd) { if (f) f(); putchar(c); }

return c != '.'; }

int main() { int i = 1; while (do_char(i = !i, 0));

return 0; }</lang>

Factor

This example is incorrect. Please fix the code and remove this message.

Details: The final "." is missing from the output.

This is delicate code with arcane control flow.

<lang factor>USING: continuations kernel io io.streams.string locals unicode.categories ; FROM: sets => in? ; IN: rosetta.odd-word

<PRIVATE ! Save current continuation.

savecc ( -- continuation/f )
   [ ] callcc1 ; inline
jump-back ( continuation -- )
   f swap continue-with ; inline

PRIVATE>

odd-word ( -- )
   f :> first-continuation!
   f :> last-continuation!
   f :> reverse!
   ! Read characters. Loop until reading "." or end of stream.
   [ read1 dup { CHAR: . f } in? ] [
       dup Letter? [
           ! This character is a letter.
           reverse [
               last-continuation savecc
               dup [
                   last-continuation!
                   2drop       ! Drop letter and previous continuation.
               ] [
                   drop
                   swap write1
                   jump-back
               ] if
           ] [ write1 ] if
       ] [
           ! This character is not a letter. Assume punctuation.
           reverse [
               savecc dup [
                   first-continuation!
                   last-continuation jump-back
               ] [ drop ] if
               write1
               f reverse!
           ] [
               write1
               t reverse!
               savecc dup [
                   last-continuation!
               ] [ first-continuation jump-back ] if
           ] if
       ] if
   ] until
   drop        ! Drop "." or f.
   reverse [
       savecc dup [
           first-continuation!
           last-continuation jump-back
       ] [ drop ] if
   ] when
   nl          ! Print a cosmetic newline.
   ;
string-odd-word ( string -- )
   [ odd-word ] with-string-reader ;</lang>
USE: rosetta.odd-word
( scratchpad ) "what,is,the;meaning,of:life." string-odd-word
what,si,the;gninaem,of:efil
( scratchpad ) "we,are;not,in,kansas;any,more." string-odd-word
we,era;not,ni,kansas;yna,more

Go

<lang go>package main

import (

   "bytes"
   "fmt"
   "io"
   "os"
   "unicode"

)

func main() {

   owp(os.Stdout, bytes.NewBufferString("what,is,the;meaning,of:life."))
   fmt.Println()
   owp(os.Stdout, bytes.NewBufferString("we,are;not,in,kansas;any,more."))
   fmt.Println()

}

func owp(dst io.Writer, src io.Reader) {

   byte_in := func () byte {
       bs := make([]byte, 1)
       src.Read(bs)
       return bs[0]
   }
   byte_out := func (b byte) { dst.Write([]byte{b}) }    
   var odd func() byte
   odd = func() byte {
       s := byte_in()
       if unicode.IsPunct(rune(s)) {
           return s
       }
       b := odd()
       byte_out(s)
       return b
   }
   for {
       for {
           b := byte_in()
           byte_out(b)
           if b == '.' {
               return
           }
           if unicode.IsPunct(rune(b)) {
               break
           }
       }
       b := odd()
       byte_out(b)
       if b == '.' {
           return
       }
   }

}</lang> Output:

what,si,the;gninaem,of:efil.
we,era;not,ni,kansas;yna,more.

A different approach, using defer: <lang go>package main

import (

   "bytes"
   "fmt"
   "io"
   "os"
   "unicode"

)

func main() {

   owp(os.Stdout, bytes.NewBufferString("what,is,the;meaning,of:life."))
   fmt.Println()
   owp(os.Stdout, bytes.NewBufferString("we,are;not,in,kansas;any,more."))
   fmt.Println()

}

func owp(dst io.Writer, src io.Reader) {

   byte_in := func () byte {
       bs := make([]byte, 1)
       src.Read(bs)
       return bs[0]
   }
   byte_out := func (b byte) { dst.Write([]byte{b}) }    
   odd := func() byte {
       for {
           b := byte_in()
           if unicode.IsPunct(int(b)) {
               return b
           }
           defer byte_out(b)
       }
       panic("impossible")
   }
   for {
       for {
           b := byte_in()
           byte_out(b)
           if b == '.' {
               return
           }
           if unicode.IsPunct(rune(b)) {
               break
           }
       }
       b := odd()
       byte_out(b)
       if b == '.' {
           return
       }
   }

}</lang>

Icon and Unicon

The following recursive version is based on the non-deferred GO version. A co-expression is used to turn the parameter to the wrapper into a character at a time stream.

<lang Icon>procedure main() every OddWord(!["what,is,the;meaning,of:life.",

               "we,are;not,in,kansas;any,more."])

end

procedure OddWord(stream) #: wrapper for demonstration

  write("Input stream: ",stream)
  writes("Output stream: ") & eWord(create !stream,'.,;:') & write()

end

procedure eWord(stream,marks) #: handle even words

  repeat {                               
     repeat 
        writes(@stream) ? if ="." then return else if any(marks) then break
     if writes(oWord(stream,marks)) == '.' then return
     }

end

procedure oWord(stream,marks) #: handle odd words (reverse)

  if any(marks,s := @stream) then return s
  return 1(oWord(stream,marks), writes(s))  

end</lang>

Output:

Input stream: what,is,the;meaning,of:life.
Output stream: what,si,the;gninaem,of:efil.
Input stream: we,are;not,in,kansas;any,more.
Output stream: we,era;not,ni,kansas;yna,more.

A slightly different solution which uses real I/O from stdin is: <lang Unicon>procedure main(A)

   repeat (while writes((any(&letters, c := reads(&input,1)),c))) |
          (if "." ~== writes(c) then "." ~== writes(rWord()))     | break
   write()

end

procedure rWord(c)

   c1 := rWord((any(&letters, c1 := reads(&input,1)),c1))
   writes(\c)
   return c1

end</lang> And some sample runs:

->rw
what,is,the;meaning,of:life.
what,si,the;gninaem,of:efil.
->rw
we,are;not,in,kansas;any,more.
we,era;not,ni,kansas;yna,more.
->

Java

<lang java>public class OddWord {

   interface CharHandler {

CharHandler handle(char c) throws Exception;

   }
   final CharHandler fwd = new CharHandler() {

public CharHandler handle(char c) { System.out.print(c); return (Character.isLetter(c) ? fwd : rev); }

   };
   class Reverser extends Thread implements CharHandler {

Reverser() { setDaemon(true); start(); } private Character ch; // For inter-thread comms private char recur() throws Exception { notify(); while (ch == null) wait(); char c = ch, ret = c; ch = null; if (Character.isLetter(c)) { ret = recur(); System.out.print(c); } return ret; } public synchronized void run() { try { while (true) { System.out.print(recur()); notify(); } } catch (Exception e) {} } public synchronized CharHandler handle(char c) throws Exception { while (ch != null) wait(); ch = c; notify(); while (ch != null) wait(); return (Character.isLetter(c) ? rev : fwd); }

   }
   final CharHandler rev = new Reverser();
   public void loop() throws Exception {

CharHandler handler = fwd; int c; while ((c = System.in.read()) >= 0) { handler = handler.handle((char) c); }

   }
   public static void main(String...args) throws Exception {

new OddWord().loop();

   }

}</lang> Output is equivalent to that of the Python solution.

Python

<lang python>from sys import stdin, stdout

def char_in(): return stdin.read(1) def char_out(c): stdout.write(c)

def odd(prev = lambda: None): a = char_in() if not a.isalpha(): prev() char_out(a) return a != '.'

# delay action until later, in the shape of a closure def clos(): char_out(a) prev()

return odd(clos)

def even(): while True: c = char_in() char_out(c) if not c.isalpha(): return c != '.'

e = False while odd() if e else even(): e = not e</lang> Running:<lang>$ echo "what,is,the;meaning,of:life." | python odd.py what,si,the;gninaem,of:efil. $ echo "we,are;not,in,kansas;any,more." | python odd.py we,era;not,ni,kansas;yna,more.</lang>

Scheme

Output is identical to python. <lang lisp>(define (odd)

 (let ((c (read-char)))
   (if (char-alphabetic? c)
     (let ((r (odd)))

(write-char c) r)

     (lambda () (write-char c) c))))

(define (even)

 (let ((c (read-char)))
   (write-char c)
   (if (char-alphabetic? c)
     (even)
     c)))

(let loop ((i #f))

 (let ((c (if i ((odd)) (even))))
   (if (char=? c #\.)
     (exit)
     (loop (not i)))))</lang>

Tcl

Although the input is handled as strings, they're all as single-character strings.

Works with: Tcl version 8.6

<lang tcl>package require Tcl 8.6

proc fwd c {

   expr {[string is alpha $c] ? "[fwd [yield f][puts -nonewline $c]]" : $c}

} proc rev c {

   expr {[string is alpha $c] ? "[rev [yield r]][puts -nonewline $c]" : $c}

} coroutine f while 1 {puts -nonewline [fwd [yield r]]} coroutine r while 1 {puts -nonewline [rev [yield f]]} for {set coro f} {![eof stdin]} {} {

   set coro [$coro [read stdin 1]]

}</lang> Output is identical to Python and Scheme versions.

The only difference between the two coroutines (apart from the different names used when flipping back and forth) is the timing of the write of the character with respect to the recursive call.