Odd word problem
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:
- 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;
- You are not to explicitly save characters in a collection data structure, such as arrays, strings, hash tables, etc, for later reversal;
- 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>
- 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>
Common Lisp
Even words are straightforward. For odd words, the final punctuation is printed by a closure passed back up the caller chain. <lang lisp>(defun do-even (&aux (c (read-char)))
(if (alpha-char-p (write-char c)) (do-even) (char/= c #\.)))
(defun do-odd (&aux (c (read-char)))
(if (alpha-char-p c) (prog1 (do-odd) (write-char c)) (lambda () (write-char c) (char/= #\.))))
(defun do-char (is-even)
(if is-even (do-even) (funcall (do-odd))))
(loop for x = t then (not x)
while (do-char x))</lang>
Factor
This is a delicate program with arcane control flow. To reverse each odd word, this code uses continuations to jump-back into earlier iterations of a while loop. This trick reverses the letters by reversing the loop!
This code is difficult to follow, because it twists its control flow like spaghetti. These continuations form a singly-linked list, where each continuation contains a letter and a previous continuation. The program effectively reverses this linked list.
<lang factor>USING: continuations kernel io io.streams.string locals unicode.categories ; IN: rosetta.odd-word
<PRIVATE ! Save current continuation.
- savecc ( -- continuation/f )
[ ] callcc1 ; inline
! Jump back to continuation, where savecc will return f.
- jump-back ( continuation -- )
f swap continue-with ; inline
PRIVATE>
- read-odd-word ( -- )
f :> first-continuation! f :> last-continuation! f :> reverse! ! Read characters. Loop until end of stream. [ read1 dup ] [ dup Letter? [ ! This character is a letter. reverse [ ! Odd word: Write letters in reverse order. last-continuation savecc dup [ last-continuation! 2drop ! Drop letter and previous continuation. ] [ ! After jump: print letters in reverse. drop ! Drop f. swap write1 ! Write letter. jump-back ! Follow chain of continuations. ] if ] [ ! Even word: Write letters immediately. write1 ] if ] [ ! This character is punctuation. reverse [ ! End odd word. Fix trampoline, follow chain of continuations ! (to print letters in reverse), then bounce off trampoline. savecc dup [ first-continuation! last-continuation jump-back ] [ drop ] if write1 ! Write punctuation. f reverse! ! Begin even word. ] [ write1 ! Write punctuation. t reverse! ! Begin odd word. ! Create trampoline to bounce to (future) first-continuation. savecc dup [ last-continuation! ] [ drop first-continuation jump-back ] if ] if ] if ] while ! Drop f from read1. Then print a cosmetic newline. drop nl ;
- odd-word ( string -- )
[ read-odd-word ] with-string-reader ;</lang>
USE: rosetta.odd-word ( scratchpad ) "what,is,the;meaning,of:life." odd-word what,si,the;gninaem,of:efil. ( scratchpad ) "we,are;not,in,kansas;any,more." 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. ->
J
This task's requirement to perform buffering implicitly rather than explicitly was perplexing from a J point of view (see talk page for some of that discussion). To avoid this issue, this implementation uses a coroutine-like utility.
<lang j>instream=. 'example text'
last=. _1
getch=: 3 :0 last=: last+1 last{instream )
outstream=.
putch=: 3 :0 outstream=: outstream, y return )
isletter=: toupper ~: tolower
do_char=: 3 :0
ch=. getch if. isletter ch do. if. even do. return ch return. else. (return@[ putch bind ch) yield do_char return. end. elseif. '.' = ch do. return ch return. elseif. do. even=: -. even return ch return. end.
)
evenodd=: 3 :0
instream=: y outstream=: last=: _1 even=: 1 whilst. '.'~:char do. putch char=. do_char coroutine end. outstream
)</lang>
With this implementation:
<lang j> evenodd 'what,is,the;meaning,of:life.' what,si,the;gninaem,of:efil.
evenodd 'we,are;not,in,kansas;any,more.'
we,era;not,ni,kansas;yna,more.</lang>
That said, note that this implementation has significant overhead when compared to a more direct implementation of the algorithm.
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>
Ruby
Ruby programs use single-character strings.
Using fibers and recursion
<lang ruby>f, r = nil fwd = proc {|c|
c =~ /alpha:/ ? [(print c), fwd[Fiber.yield f]][1] : c }
rev = proc {|c|
c =~ /alpha:/ ? [rev[Fiber.yield r], (print c)][0] : c }
(f = Fiber.new { loop { print fwd[Fiber.yield r] }}).resume (r = Fiber.new { loop { print rev[Fiber.yield f] }}).resume
coro = f until $stdin.eof?
coro = coro.resume($stdin.getc)
end</lang>
Using continuations
<lang ruby>require 'continuation' unless defined? Continuation require 'stringio'
- Save current continuation.
def savecc(*data)
# With MRI 1.8 (but not 1.9), the array literal # [callcc {|cc| cc}, *data] # used the wrong return value from callcc. The workaround is to # put callcc outside the array literal. continuation = callcc {|cc| cc} [continuation, *data]
end
- Jump back to continuation, where savecc will return [nil, *data].
def jump_back(continuation)
continuation[nil]
end
def read_odd_word(input, output)
first_continuation, last_continuation = nil reverse = false # Read characters. Loop until end of stream. while c = input.getc c = c.chr # For Ruby 1.8, convert Integer to String. if c =~ /alpha:/ # This character is a letter. if reverse # Odd word: Write letters in reverse order. saving, last_continuation, c = savecc(last_continuation, c) if saving last_continuation = saving else # After jump: print letters in reverse. output.print c jump_back last_continuation end else # Even word: Write letters immediately. output.print c end else # This character is punctuation. if reverse # End odd word. Fix trampoline, follow chain of continuations # (to print letters in reverse), then bounce off trampoline. first_continuation, c = savecc(c) if first_continuation jump_back last_continuation end output.print c # Write punctuation. reverse = false # Begin even word. else output.print c # Write punctuation. reverse = true # Begin odd word. # Create trampoline to bounce to (future) first_continuation. last_continuation, = savecc unless last_continuation jump_back first_continuation end end end end output.puts # Print a cosmetic newline.
end
def odd_word(string)
read_odd_word StringIO.new(string), $stdout
end
odd_word "what,is,the;meaning,of:life." odd_word "we,are;not,in,kansas;any,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.
<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.