100 doors
Problem: You have 100 doors in a row that are all initially closed. You make 100 passes by the doors. The first time through, you visit every door and toggle the door (if the door is closed, you open it; if it is open, you close it). The second time you only visit every 2nd door (door #2, #4, #6, ...). The third time, every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Question: What state are the doors in after the last pass? Which are open, which are closed? [1]
Alternate: As noted in this page's discussion page, the only doors that remain open are whose numbers are perfect squares of integers. Opening only those doors is an optimization that may also be expressed.
Ada
unoptimized
<lang ada>with Ada.Text_Io; use Ada.Text_Io;
procedure Doors is
type Door_State is (Closed, Open);
type Door_List is array(Positive range 1..100) of Door_State;
The_Doors : Door_List := (others => Closed);
begin
for I in 1..100 loop
for J in The_Doors'range loop
if J mod I = 0 then
if The_Doors(J) = Closed then
The_Doors(J) := Open;
else
The_Doors(J) := Closed;
end if;
end if;
end loop;
end loop;
for I in The_Doors'range loop
Put_Line(Integer'Image(I) & " is " & Door_State'Image(The_Doors(I)));
end loop;
end Doors;</lang>
optimized
<lang ada>with Ada.Text_Io; use Ada.Text_Io;
with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions;
procedure Doors_Optimized is
Num : Float;
begin
for I in 1..100 loop
Num := Sqrt(Float(I));
Put(Integer'Image(I) & " is ");
if Float'Floor(Num) = Num then
Put_Line("Opened");
else
Put_Line("Closed");
end if;
end loop;
end Doors_Optimized;</lang>
ALGOL 68
unoptimized
# declare some constants #
INT limit = 100;
PROC doors = VOID:
(
MODE DOORSTATE = BOOL;
BOOL closed = FALSE;
BOOL open = NOT closed;
MODE DOORLIST = [limit]DOORSTATE;
DOORLIST the doors;
FOR i FROM LWB the doors TO UPB the doors DO the doors[i]:=closed OD;
FOR i FROM LWB the doors TO UPB the doors DO
FOR j FROM LWB the doors TO UPB the doors DO
IF j MOD i = 0 THEN
the doors[j] := NOT the doors[j]
FI
OD
OD;
FOR i FROM LWB the doors TO UPB the doors DO
printf(($g" is "gl$,i,(the doors[i]|"opened"|"closed")))
OD
);
doors;
optimized
PROC doors optimised = ( INT limit )VOID:
FOR i TO limit DO
REAL num := sqrt(i);
printf(($g" is "gl$,i,(ENTIER num = num |"opened"|"closed") ))
OD
;
doors optimised(limit)
AWK
unoptimized
<lang awk>BEGIN {
for(i=1; i <= 100; i++)
{
doors[i] = 0 # close the doors
}
for(i=1; i <= 100; i++)
{
for(j=i; j <= 100; j += i)
{
doors[j] = (doors[j]+1) % 2
}
}
for(i=1; i <= 100; i++)
{
print i, doors[i] ? "open" : "close"
}
}</lang>
optimized
<lang awk>BEGIN {
for(i=1; i <= 100; i++) {
doors[i] = 0 # close the doors
}
for(i=1; i <= 100; i++) {
if ( int(sqrt(i)) == sqrt(i) ) {
doors[i] = 1
}
}
for(i=1; i <= 100; i++)
{
print i, doors[i] ? "open" : "close"
}
}</lang>
BASIC
unoptimized
DIM doors(0 TO 99) FOR pass = 0 TO 99 FOR door = pass TO 99 STEP pass + 1 PRINT doors(door) PRINT NOT doors(door) doors(door) = NOT doors(door) NEXT door NEXT pass FOR i = 0 TO 99 PRINT "Door #"; i + 1; " is "; IF NOT doors(i) THEN PRINT "closed" ELSE PRINT "open" END IF NEXT i
optimized
DIM doors(0 TO 99) FOR door = 0 TO 99 IF INT(SQR(door)) = SQR(door) THEN doors(door) = -1 NEXT door FOR i = 0 TO 99 PRINT "Door #"; i + 1; " is "; IF NOT doors(i) THEN PRINT "closed" ELSE PRINT "open" END IF NEXT i
C
unoptimized
<lang c> #include <stdio.h>
int main()
{
char is_open[100] = { 0 };
int pass, door;
// do the 100 passes
for (pass = 0; pass < 100; ++pass)
for (door = pass; door < 100; door += pass+1)
is_open[door] = !is_open[door];
// output the result
for (door = 0; door < 100; ++door)
printf("door #%d is %s.\n", door+1, (is_open[door]? "open" : "closed"));
return 0;
}</lang>
optimized
This optimized version makes use of the fact that finally only the doors with square index are open, as well as the fact that .
<lang c> #include <stdio.h>
int main()
{
int square = 1, increment = 3, door;
for (door = 1; door <= 100; ++door)
{
printf("door #%d", door);
if (door == square)
{
printf(" is open.\n");
square += increment;
increment += 2;
}
else
printf(" is closed.\n");
}
return 0;
}</lang>
The following ultra-short optimized version demonstrates the flexibility of C loops, but isn't really considered good C style:
<lang c> #include <stdio.h>
int main()
{
int door, square, increment;
for (door = 1, square = 1, increment = 1; door <= 100; door++ == square && (square += increment += 2))
printf("door #%d is %s.\n", door, (door == square? "open" : "closed"));
return 0;
}</lang>
C++
unoptimized
<lang cpp> #include <iostream>
int main()
{
bool is_open[100] = { false };
// do the 100 passes
for (int pass = 0; pass < 100; ++pass)
for (int door = pass; door < 100; door += pass+1)
is_open[door] = !is_open[door];
// output the result
for (int door = 0; door < 100; ++door)
std::cout << "door #" << door+1 << (is_open[door]? " is open." : " is closed.") << std::endl;
return 0;
}</lang>
optimized
This optimized version makes use of the fact that finally only the doors with square index are open, as well as the fact that .
<lang cpp> #include <iostream>
int main()
{
int square = 1, increment = 3;
for (int door = 1; door <= 100; ++door)
{
std::cout << "door #" << door;
if (door == square)
{
std::cout << " is open." << std::endl;
square += increment;
increment += 2;
}
else
std::cout << " is closed." << std::endl;
}
return 0;
}</lang>
The following ultra-short optimized version demonstrates the flexibility of C++ loops, but isn't really considered good C++ style:
<lang cpp> #include <iostream>
int main()
{
for (int door = 1, square = 1, increment = 1; door <= 100; door++ == square && (square += increment += 2))
std::cout << "door #" << door << (door == square? " is open." : " is closed.") << std::endl;
return 0;
}</lang>
Common Lisp
Unoptimized / functional
This is a very unoptimized version of the problem, using recursion and quite considerable list-copying. It emphatises the functional way of solving this problem
<lang lisp>(defun visit-door (doors doornum value1 value2)
"visits a door, swapping the value1 to value2 or vice-versa"
(let ((d (copy-list doors))
(n (- doornum 1)))
(if (eq (nth n d) value1)
(setf (nth n d) value2)
(setf (nth n d) value1))
d))
(defun visit-every (doors num iter value1 value2)
"visits every 'num' door in the list"
(if (> (* iter num) (length doors))
doors
(visit-every (visit-door doors (* num iter) value1 value2)
num
(+ 1 iter)
value1
value2)))
(defun do-all-visits (doors cnt value1 value2)
"Visits all doors changing the values accordingly"
(if (< cnt 1)
doors
(do-all-visits (visit-every doors cnt 1 value1 value2)
(- cnt 1)
value1
value2)))
(defun print-doors (doors)
"Pretty prints the doors list"
(format T "~{~A ~A ~A ~A ~A ~A ~A ~A ~A ~A~%~}~%" doors))
(defun start (&optional (size 100))
"Start the program"
(let* ((open "_")
(shut "#")
(doors (make-list size :initial-element shut)))
(print-doors (do-all-visits doors size open shut))))
</lang>
Optimized
This is an optimized version of the above, using the perfect square algorithm (Note: This is non-functional as the state of the doors variable gets modified by a function call)
<lang lisp>(defun perfect-square-list (n)
"Generates a list of perfect squares from 0 up to n"
(loop for i from 1 to (sqrt n) collect (expt i 2)))
(defun open-door (doors num open)
"Sets door at num to open"
(setf (nth (- num 1) doors) open)
doors)
(defun visit-all (doors vlist open)
"Visits and opens all the doors indicated in vlist"
(if (null vlist)
doors
(visit-all (open-door doors (car vlist) open)
(cdr vlist)
open)))
(defun start2 (&optional (size 100))
"Start the program"
(print-doors (visit-all (make-list size :initial-element "#")
(perfect-square-list size) "_")))
</lang>
Optimized (2)
This version displays a much more functional solution through the use of (mapcar ...)
<lang lisp>(let ((i 0))
(mapcar (lambda (x)
(if (zerop (mod (sqrt (incf i)) 1))
"_"
x))
(make-list 100 :initial-element "#")))</lang>
D
module l00door ; import std.stdio ; alias char[] DOOR ; // or alias bool[] DOOR const char OPEN = 'O' ; // then OPEN = true const char CLOSE = 'x' ; // CLOSE = false
unoptimized
DOOR flip_u(inout DOOR d) {
d[] = CLOSE ;
for(int i= 0 ; i < d.length ; i++)
for(int j = i ; j < d.length ; j += i + 1)
if (d[j] == OPEN)
d[j] = CLOSE ;
else
d[j] = OPEN ;
return d ;
}
optimized
DOOR flip_o(inout DOOR d) {
d[] = CLOSE ;
for(int i= 1 ; i*i <= d.length ; i++)
d[i*i-1] = OPEN ;
return d ;
}
test program:
void main() {
DOOR d ;
d.length = 100 ;
writefln(d.flip_u()) ;
writefln(d.flip_o()) ;
}
E
Graphical
This version animates the changes of the doors (as checkboxes).
<lang e>#!/usr/bin/env rune
var toggles := [] var gets := []
- Set up GUI (and data model)
def frame := <swing:makeJFrame>("100 doors") frame.getContentPane().setLayout(<awt:makeGridLayout>(10, 10)) for i in 1..100 {
def component := <import:javax.swing.makeJCheckBox>(E.toString(i))
toggles with= fn { component.setSelected(!component.isSelected()) }
gets with= fn { component.isSelected() }
frame.getContentPane().add(component)
}
- Set up termination condition
def done frame.addWindowListener(def _ {
to windowClosing(event) {
bind done := true
}
match _ {}
})
- Open and close doors
def loop(step, i) {
toggles[i] <- ()
def next := i + step
timer.whenPast(timer.now() + 10, fn {
if (next >= 100) {
if (step >= 100) {
# Done.
} else {
loop <- (step + 1, step)
}
} else {
loop <- (step, i + step)
}
})
} loop(1, 0)
frame.pack() frame.show() interp.waitAtTop(done)</lang>
Forth
The problem with unfactored code is it's difficult to follow. The included example works, but to comprehend the code, you're mentally juggling items on the return stack (e.g., i, j) as well as items on the data stack, plus whatever is in dictionary space. Since people can only keep seven things in his or her mind, plus or minus two, this quickly becomes an intellectual burden. This style of coding typifies Forth software development in the mid-70s and -80s, which resulted in Forth's undeserved "write-only language" reputation.
The well-factored code works much better from a readability point of view (assuming familiarity with Forth's coding conventions), but without the aid of an optimizing compiler, can suffer from slower execution performance and somewhat larger memory consumption.
Of course, this size/speed tradeoff appears in many languages, not just Forth.
Factored
100 constant /doors
create 'doors
/doors allot
'doors /doors + constant <doors
: toggle 1 over c@ xor swap c! ;
: pass dup 1- 'doors + begin dup <doors < while dup toggle over + repeat 2drop ;
: passes /doors 1 begin 2dup >= while dup pass 1+ repeat 2drop ;
: .closed over c@ if over 'doors - 1+ . then ;
: report 'doors /doors begin dup while .closed 1 /string repeat cr 2drop ;
: open 'doors /doors erase ;
: doors open passes report ;
Unfactored
: doors ( n -- )
here over erase \ open all doors
dup 0 do
here over + here i + do
i c@ 1 xor i c! \ toggle
j 1+ +loop
loop
." Closed doors: "
( n ) 0 do
here i + c@ if i 1+ . then
loop cr ;
100 doors
Fortran
unoptimized
<lang fortran> PROGRAM DOORS
INTEGER, PARAMETER :: n = 100 ! Number of doors
INTEGER :: i, j
LOGICAL :: door(n) = .TRUE. ! Initially closed
DO i = 1, n
DO j = i, n, i
door(j) = .NOT. door(j)
END DO
END DO
DO i = 1, n
WRITE(*,"(A,I3,A)", ADVANCE="NO") "Door ", i, " is "
IF (door(i)) THEN
WRITE(*,"(A)") "closed"
ELSE
WRITE(*,"(A)") "open"
END IF
END DO
END PROGRAM DOORS</lang>
optimized
<lang fortran> PROGRAM DOORS
INTEGER, PARAMETER :: n = 100 ! Number of doors
INTEGER :: i
LOGICAL :: door(n) = .TRUE. ! Initially closed
DO i = 1, SQRT(REAL(n))
door(i*i) = .FALSE.
END DO
DO i = 1, n
WRITE(*,"(A,I3,A)", ADVANCE="NO") "Door ", i, " is "
IF (door(i)) THEN
WRITE(*,"(A)") "closed"
ELSE
WRITE(*,"(A)") "open"
END IF
END DO
END PROGRAM DOORS</lang>
Groovy
unoptimized
<lang groovy>
doors = [false] * 100
(0..99).each {
it.step(100, it + 1) {
doors[it] ^= true
}
}
(0..99).each {
println("Door #${it + 1} is ${doors[it] ? 'open' : 'closed'}.")
}
</lang>
optimized a
Using square roots
<lang groovy>
(1..100).each {
println("Door #${it} is ${Math.sqrt(it).with{it==(int)it} ? 'open' : 'closed'}.")
}
</lang>
optimized b
Without using square roots
<lang groovy>
doors = ['closed'] * 100
(1..10).each { doors[it**2 - 1] = 'open' }
(0..99).each {
println("Door #${it + 1} is ${doors[it]}.")
}
</lang>
Haskell
unoptimized
<lang haskell> data Door = Open | Closed deriving Show
toggle f [] = [] toggle f (Open:xs) = Closed : f xs toggle f (Closed:xs) = Open : f xs pass k [] = [] pass k xs = ys ++ toggle (pass k) zs where (ys,zs) = splitAt k xs run n = snd $ (!! n) $ iterate (\(k,xs) -> (k+1, pass k xs)) $ (0, replicate n Closed)</lang>
optimized
(without using square roots)
<lang haskell> gate :: Eq a => [a] -> [a] -> [Door]
gate (x:xs) (y:ys) | x == y = Open : gate xs ys gate (x:xs) ys = Closed : gate xs ys gate [] _ = [] run n = gate [1..n] [k*k | k <- [1..]]</lang>
alternatively, returning a list of all open gates, it's a one-liner:
<lang haskell> run n = takeWhile (< n) [k*k | k <- [1..]]</lang>
J
unoptimized
~:/ (100 $ - {. 1:)"0 >:i.100
1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 ...
optimized
(>:i.100) e. *: i.11 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 ... 1 (<:*:i.10)} 100$0 NB. alternative 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 ...
Java
unoptimized
<lang java> public class Doors {
public static void main(String[] args) {
boolean[] doors = new boolean[100];
for(int pass = 0; pass < 100; pass++){
for(int door = pass; door < 100; door += pass + 1){
doors[door] = !doors[door];
}
}
for(int i = 0; i < 100; i++){
System.out.println("Door #" + (i + 1) + " is " + (doors[i] ?
"open." : "closed."));
}
}
}</lang>
optimized
<lang java> public class Doors {
public static void main(String[] args) {
boolean[] doors = new boolean[100];
for(int pass = 0; pass < 10; pass++){
doors[(pass + 1) * (pass + 1) - 1] = true;
}
for(int i = 0; i < 100; i++){
System.out.println("Door #" + (i + 1) + " is " + (doors[i] ?
"open." : "closed."));
}
}
}</lang>
MAXScript
unoptimized
doorsOpen = for i in 1 to 100 collect false
for pass in 1 to 100 do
(
for door in pass to 100 by pass do
(
doorsOpen[door] = not doorsOpen[door]
)
)
for i in 1 to doorsOpen.count do
(
format ("Door % is open?: %\n") i doorsOpen[i]
)
optimized
for i in 1 to 100 do
(
root = pow i 0.5
format ("Door % is open?: %\n") i (root == (root as integer))
)
Modula-3
unoptimized
<lang modula3>MODULE Doors EXPORTS Main;
IMPORT IO, Fmt;
TYPE State = {Closed, Open}; TYPE List = ARRAY [1..100] OF State;
VAR doors := List{State.Closed, ..};
BEGIN
FOR i := 1 TO 100 DO
FOR j := FIRST(doors) TO LAST(doors) DO
IF j MOD i = 0 THEN
IF doors[j] = State.Closed THEN
doors[j] := State.Open;
ELSE
doors[j] := State.Closed;
END;
END;
END;
END;
FOR i := FIRST(doors) TO LAST(doors) DO
IO.Put(Fmt.Int(i) & " is ");
IF doors[i] = State.Closed THEN
IO.Put("Closed.\n");
ELSE
IO.Put("Open.\n");
END;
END;
END Doors.</lang>
optimized
<lang modula3>MODULE DoorsOpt EXPORTS Main;
IMPORT IO, Fmt;
TYPE State = {Closed, Open}; TYPE List = ARRAY [1..100] OF State;
VAR doors := List{State.Closed, ..};
BEGIN
FOR i := 1 TO 10 DO doors[i * i] := State.Open; END;
FOR i := FIRST(doors) TO LAST(doors) DO
IO.Put(Fmt.Int(i) & " is ");
IF doors[i] = State.Closed THEN
IO.Put("Closed.\n");
ELSE
IO.Put("Open.\n");
END;
END;
END DoorsOpt.</lang>
OCaml
unoptimized
let max_doors = 100
let show_doors =
Array.iteri (fun i x -> Printf.printf "Door %d is %s\n" (i+1)
(if x then "open" else "closed"))
let flip_doors doors =
for i = 1 to max_doors do
let rec flip idx =
if idx < max_doors then begin
doors.(idx) <- not doors.(idx);
flip (idx + i)
end
in flip (i - 1)
done;
doors
let () =
show_doors (flip_doors (Array.make max_doors false))
optimized
let optimised_flip_doors doors =
for i = 1 to int_of_float (sqrt (float_of_int max_doors)) do
doors.(i*i - 1) <- true
done;
doors
let () =
show_doors (optimised_flip_doors (Array.make max_doors false))
Octave
<lang octave>doors = false(100,1); for i = 1:100
for j = i:i:100 doors(j) = !doors(j); endfor
endfor for i = 1:100
if ( doors(i) )
s = "open";
else
s = "closed";
endif
printf("%d %s\n", i, s);
endfor</lang>
Pascal
<lang pascal>Program OneHundredDoors;
var
doors : Array[1..100] of Boolean; i, j : Integer;
begin
for i := 1 to 100 do
doors[i] := False;
for i := 1 to 100 do begin
j := i;
while j <= 100 do begin
doors[j] := not doors[j]; j := j + i
end
end;
for i := 1 to 100 do begin
Write(i, ' ');
if doors[i] then
WriteLn('open')
else
WriteLn('closed');
end
end.</lang>
PHP
optimized
<lang php> <?php for ($i = 1; $i <= 100; $i++) { $root = sqrt($i); $state = ($root == ceil($root)) ? "open" : "closed"; echo "Door $i: $state"; } ?> </lang>
Perl
unoptimized
my @doors;
foreach my $pass (1 .. 100) {
foreach (1 .. 100) {
if (0 == $_ % $pass) {
if (1 == $doors[$_]) {
$doors[$_] = 0;
} else {
$doors[$_] = 1;
};
};
};
};
print "$_\t$doors[$_]\n" foreach 1 .. 100;
optimized
while( ++$i <= 100 )
{
$root = sqrt($i);
if ( int( $root ) == $root )
{
print "Door $i is open\n";
}
else
{
print "Door $i is closed\n";
}
}
Pop11
unoptimized
lvars i;
lvars doors = {% for i from 1 to 100 do false endfor %};
for i from 1 to 100 do
for j from i by i to 100 do
not(doors(j)) -> doors(j);
endfor;
endfor;
;;; Print state
for i from 1 to 100 do
printf('Door ' >< i >< ' is ' ><
if doors(i) then 'open' else 'closed' endif, '%s\n');
endfor;
optimized
for i to 100 do
lvars root = sqrt(i);
i; if root = round(root) then ' open' ><; else ' closed' ><; endif; =>
endfor;
Python
Note: both versions require Python 2.5 for the ternary syntax.
unoptimized
close = 0
open = 1
doors = [close] * 100
for i in range(100):
for j in range(i, 100, i+1):
doors[j] = open if doors[j] is close else close
print "Door %d:" % (i+1), 'open' if doors[i] else 'close'
for k, door in enumerate(doors):
print "Door %d:" % (k+1), 'open' if door else 'close'
optimized
A version that only visits each door once:
for i in xrange(1, 101):
root = i ** 0.5
print "Door %d:" % i, 'open' if root == int(root) else 'close'
R
unoptimized
doors_puzzle <- function(ndoors=100,passes=100) {
doors <- rep(FALSE,ndoors)
for (ii in seq(1,passes)) {
mask <- seq(0,ndoors,ii)
doors[mask] <- !doors[mask]
}
return (which(doors == TRUE))
}
doors_puzzle()
optimized
## optimized version... we only have to to up to the square root of 100 seq(1,sqrt(100))**2
Ruby
unoptimized
<lang ruby> n = 100 Open = "open" Closed = "closed" def Open.f
Closed
end def Closed.f
Open
end doors = [Closed] * (n+1) for mul in 1..n
for x in 1..n
doors[mul*x] = (doors[mul*x] || break).f
end
end doors.each_with_index {
|b, i|
puts "Door #{i} is #{b}" if i>0
} </lang>
optimized
<lang ruby> n = 100 doors = ["closed"] * (n+1) for x in 1..n
doors[x**2] = (doors[x**2] || break) && "open"
end doors.each_with_index {
|b, i|
puts "Door #{i} is #{b}" if i>0
} </lang>
Scheme
unoptimized
<lang scheme> (define *max-doors* 100)
(define (show-doors doors)
(let door ((i 0)
(l (vector-length doors)))
(cond ((= i l)
(newline))
(else
(printf "~nDoor ~a is ~a"
(+ i 1)
(if (vector-ref doors i) "open" "closed"))
(door (+ i 1) l)))))
(define (flip-doors doors)
(define (flip-all i)
(cond ((> i *max-doors*) doors)
(else
(let flip ((idx (- i 1)))
(cond ((>= idx *max-doors*)
(flip-all (+ i 1)))
(else
(vector-set! doors idx (not (vector-ref doors idx)))
(flip (+ idx i))))))))
(flip-all 1))
(show-doors (flip-doors (make-vector *max-doors* #f))) </lang>
optimized
<lang scheme> (define (optimised-flip-doors doors)
(define (flip-all i)
(cond ((> i (floor (sqrt *max-doors*))) doors)
(else
(vector-set! doors (- (* i i) 1) #t)
(flip-all (+ i 1)))))
(flip-all 1))
(show-doors (optimised-flip-doors (make-vector *max-doors* #f))) </lang>
Smalltalk
Unoptimized
<lang smalltalk>|a| a := Array new: 100 . 1 to: 100 do: [ :i | a at: i put: false ].
1 to: 100 do: [ :pass |
pass to: 100 by: pass do: [ :door | a at: door put: (a at: door) not . ]
].
"output" 1 to: 100 do: [ :door |
( 'door #%1 is %2' %
{ door . (a at: door) ifTrue: [ 'open' ] ifFalse: [ 'closed' ] } ) displayNl
]</lang>
Optimized
<lang smalltalk>|a| a := (1 to: 100) collect: [ :x | false ]. 1 to: 10 do: [ :i | a at: (i squared) put: true ]. 1 to: 100 do: [ :i |
( 'door #%1 is %2' % { i .
(a at: i) ifTrue: [ 'open' ]
ifFalse: [ 'closed' ] }
) displayNl
]</lang>
SETL
Unoptimized
<lang SETL>program hundred_doors;
const toggle := {['open', 'closed'], ['closed', 'open']};
doorStates := ['closed'] * 100;
(for interval in [1..100])
doorStates := [if i mod interval = 0 then
toggle(prevState) else
prevState end:
prevState = doorStates(i)];
end;
(for finalState = doorStates(i))
print('door', i, 'is', finalState);
end;
end program;</lang> If 'open' weren't a reserved word, we could omit the single quotes around it.
Optimized
Exploits the fact that squares are separated by successive odd numbers. Use array replication to insert the correct number of closed doors in between the open ones. <lang SETL>program hundred_doors;
doorStates := (+/ [['closed'] * oddNum with 'open': oddNum in [1,3..17]]);
(for finalState = doorStates(i))
print('door', i, 'is', finalState);
end;
end program; </lang>
Tcl
unoptimized
<lang tcl>package require Tcl 8.5 set n 100 set doors [concat - [lrepeat $n 0]] for {set i 1} {$i <= $n} {incr i} {
for {set step 1} {$step <= $n} {incr step} {
if {$i % $step == 0} {
lset doors $i [expr { ! [lindex $doors $i]}]
}
}
} for {set i 1} {$i <= $n} {incr i} {
puts [format "door %d is %s" $i [expr {[lindex $doors $i] ? "open" : "closed"}]]
}</lang>
optimized
<lang tcl>package require Tcl 8.5 for {set i 1} {$i <= $n} {incr i} {
set root [::tcl::mathfunc::sqrt $i]
puts [format "door %d is %s" $i [expr {int($root) == $root ? "open" : "closed"}]]
}</lang>
UNIX Shell
<lang bash>#! /bin/bash
declare -a doors for((i=1; i <= 100; i++)); do
doors[$i]=0
done
for((i=1; i <= 100; i++)); do
for((j=i; j <= 100; j += i)); do
echo $i $j doors[$j]=$(( doors[j] ^ 1 ))
done
done
for((i=1; i <= 100; i++)); do
if [[ ${doors[$i]} -eq 0 ]]; then
op="closed"
else
op="open"
fi echo $i $op
done</lang>
Vedit macro language
Unoptimized
This implementation uses a free edit buffer as data array and for displaying the results.
A closed door is represented by a character '-' and an open door by character 'O'.
<lang vedit>
Buf_Switch(Buf_Free)
Ins_Char('-', COUNT, 100) // All doors closed
for (#1 = 1; #1 <= 100; #1++) {
for (#2 = #1; #2 <= 100; #2 += #1) {
Goto_Col(#2)
Ins_Char((Cur_Char^0x62), OVERWRITE) // Toggle between '-' and 'O'
}
} </lang>
Optimized <lang vedit> Buf_Switch(Buf_Free) Ins_Char('-', COUNT, 100) for (#1=1; #1 <= 10; #1++) {
Goto_Col(#1*#1)
Ins_Char('O', OVERWRITE)
} </lang>
Output:
O--O----O------O--------O----------O------------O--------------O----------------O------------------O
Visual Basic .NET
Platform: .NET
Language Version: 9.0+
unoptimized
Module Module1
Sub Main()
Dim doors(100) As Boolean 'Door 1 is at index 0
For pass = 1 To 100
For door = pass - 1 To 99 Step pass
doors(door) = Not doors(door)
Next
Next
For door = 0 To 99
Console.WriteLine("Door # " & (door + 1) & " is " & If(doors(door), "Open", "Closed"))
Next
Console.ReadLine()
End Sub
End Module
optimized
Module Module1
Sub Main()
Dim doors(100) As Boolean 'Door 1 is at index 0
For i = 1 To 10
doors(i ^ 2 - 1) = True
Next
For door = 0 To 99
Console.WriteLine("Door # " & (door + 1) & " is " & If(doors(door), "Open", "Closed"))
Next
Console.ReadLine()
End Sub
End Module