100 doors: Difference between revisions
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END Doors.</lang> |
END Doors.</lang> |
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=={{header|XPL0}}== |
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<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations |
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int Door(100); \You have 100 doors in a row |
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define Open, Closed; |
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int D, Pass, Step; |
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[for D:= 0 to 100-1 do \that are all initially closed |
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Door(D):= Closed; |
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Step:= 1; \The first time through, you visit every door |
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for Pass:= 1 to 100 do \You make 100 passes by the doors |
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[D:= Step-1; |
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repeat \if the door is closed, you open it; if it is open, you close it |
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if Door(D)=Closed then Door(D):= Open else Door(D):= Closed; |
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D:= D+Step; |
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until D>=100; |
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Step:= Step+1; \The second time you only visit every 2nd door |
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]; \The third time, every 3rd door |
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\until you only visit the 100th door |
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\What state are the doors in after the last pass? |
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Text(0, "Open: "); \Which are open? |
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for D:= 0 to 100-1 do |
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if Door(D)=Open then [IntOut(0, D+1); ChOut(0,^ )]; |
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CrLf(0); |
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Text(0, "Closed: "); \Which are closed? |
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for D:= 0 to 100-1 do |
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if Door(D)=Closed then [IntOut(0, D+1); ChOut(0,^ )]; |
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CrLf(0); |
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\Optimized: The only doors that remain open are those that are perfect squares |
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Text(0, "Open: "); |
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D:= 1; |
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repeat IntOut(0, D*D); ChOut(0,^ ); |
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D:= D+1; |
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until D*D>100; |
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CrLf(0); |
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]</lang> |
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=={{header|XSLT}}== |
=={{header|XSLT}}== |
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Revision as of 23:22, 22 April 2012
You are encouraged to solve this task according to the task description, using any language you may know.
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.
4DOS Batch
<lang 4DOS Batch> @echo off set doors=%@repeat[C,100] do step = 1 to 100
do door = %step to 100 by %step set doors=%@left[%@eval[%door-1],%doors]%@if[%@instr[%@eval[%door-1],1,%doors]==C,O,C]%@right[%@eval[100-%door],%doors] enddo
enddo </lang>
The SET line consists of three functions: <lang> %@left[n,string] ^: Return n leftmost chars in string %@right[n,string] ^: Return n rightmost chars in string %@if[condition,true-val,false-val] ^: Evaluate condition; return true-val if true, false-val if false </lang>
Here @IF is used to toggle between C and O.
6502 Assembly
optimized (Largely inspired by the optimized C implementation) <lang 6502asm>
;assume all memory is initially set to 0
inc $1 ;start out with a delta of 1
openloop: inc $200,X ;open a door at X
inc $1 ;add 2 to delta
inc $1
txa ;add delta to X
adc $1
tax
cpx #$65 ;check to see if we're at the 100th door
bmi openloop ;jump back to openloop if less than 100</lang>
8086 Assembly
ABAP
unoptimized <lang ABAP>form open_doors_unopt.
data: lv_door type i,
lv_count type i value 1.
data: lt_doors type standard table of c initial size 100.
field-symbols: <wa_door> type c.
do 100 times.
append initial line to lt_doors assigning <wa_door>.
<wa_door> = 'X'.
enddo.
while lv_count < 100.
lv_count = lv_count + 1.
lv_door = lv_count.
while lv_door < 100.
read table lt_doors index lv_door assigning <wa_door>.
if <wa_door> = ' '.
<wa_door> = 'X'.
else.
<wa_door> = ' '.
endif.
add lv_count to lv_door.
endwhile.
endwhile.
loop at lt_doors assigning <wa_door>.
if <wa_door> = 'X'.
write : / 'Door', (4) sy-tabix right-justified, 'is open' no-gap.
endif.
endloop.
endform.</lang>
optimized
Using <lang ABAP>form open_doors_opt.
data: lv_square type i value 1,
lv_inc type i value 3.
data: lt_doors type standard table of c initial size 100.
field-symbols: <wa_door> type c.
do 100 times.
append initial line to lt_doors assigning <wa_door>.
if sy-index = lv_square.
<wa_door> = 'X'.
add: lv_inc to lv_square, 2 to lv_inc.
write : / 'Door', (4) sy-index right-justified, 'is open' no-gap.
endif.
enddo.
endform.</lang>
ACL2
<lang lisp>(defun rep (n x)
(if (zp n)
nil
(cons x
(rep (- n 1) x))))
(defun toggle-every-r (n i bs)
(if (endp bs)
nil
(cons (if (zp i)
(not (first bs))
(first bs))
(toggle-every-r n (mod (1- i) n) (rest bs)))))
(defun toggle-every (n bs)
(toggle-every-r n (1- n) bs))
(defun 100-doors (i doors)
(if (zp i)
doors
(100-doors (1- i) (toggle-every i doors))))</lang>
ActionScript
unoptimized <lang actionscript>package {
import flash.display.Sprite;
public class Doors extends Sprite {
public function Doors() {
// Initialize the array
var doors:Array = new Array(100);
for (var i:Number = 0; i < 100; i++) {
doors[i] = false;
// Do the work
for (var pass:Number = 0; pass < 100; pass++) {
for (var j:Number = pass; j < 100; j += (pass+1)) {
doors[j] = !doors[j];
}
}
trace(doors);
}
}
}</lang>
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>
Aikido
<lang aikido> var doors = new int [100]
foreach pass 100 {
for (var door = pass ; door < 100 ; door += pass+1) {
doors[door] = !doors[door]
}
}
var d = 1 foreach door doors {
println ("door " + d++ + " is " + (door ? "open" : "closed"))
}
</lang>
ALGOL 68
unoptimized <lang algol68># 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$,i,(the doors[i]|"opened"|"closed")))
OD
); doors;</lang> optimized <lang algol68>PROC doors optimised = ( INT limit )VOID:
FOR i TO limit DO REAL num := sqrt(i); printf(($g" is "gl$,i,(ENTIER num = num |"opened"|"closed") )) OD
doors optimised(limit)</lang>
AmigaE
<lang amigae>PROC main()
DEF t[100]: ARRAY,
pass, door
FOR door := 0 TO 99 DO t[door] := FALSE
FOR pass := 0 TO 99
door := pass
WHILE door <= 99
t[door] := Not(t[door])
door := door + pass + 1
ENDWHILE
ENDFOR
FOR door := 0 TO 99 DO WriteF('\d is \s\n', door+1,
IF t[door] THEN 'open' ELSE 'closed')
ENDPROC</lang>
APL
unoptimized <lang APL>out←doors num ⍝ Simulates the 100 doors problem for any number of doors ⍝ Returns a boolean vector with 1 being open
out←⍳num ⍝ num steps out←⍳¨out ⍝ Count out the spacing for each step out←1=out ⍝ Make that into a boolean vector out←⌽¨out ⍝ Flip each vector around out←(num∘⍴)¨out ⍝ Copy each out to the right size out←≠/out ⍝ XOR each vector, toggling each marked door out←⊃out ⍝ Disclose the results to get a vector</lang> Sample Output:
10 10⍴doors 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 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
optimized <lang APL>out←doorsOptimized num;marks ⍝ Returns a boolean vector of the doors that would be left open
marks←⌊num*0.5 ⍝ Take the square root of the size, floored marks←(⍳marks)*2 ⍝ Get each door to be opened out←num⍴0 ⍝ Make a vector of 0s out[marks]←1 ⍝ Set the marked doors to 1</lang> Sample Output:
10 10⍴doorsOptimized 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 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
Alternate 1-line version Note that ⎕IO = 1
2|+/[1]0=D∘.|D←⍳100
The idea is that the n:th door will be flipped the same number of times as there are divisors for n. So first we make D all ints 1..100 (D←⍳100).
The next step is to find the remainders of every such int when divided by every other (D∘.|D).
This results in a 100×100 matrix which we turn into a binary one by testing if the values are equal to zero i.e. divisors.
Next: sum along axis 1, i.e. the columns. This tells us the number of divisors. Finally calculate the remainder of these when divided by 2, i.e. find which n have an odd number of divisors, i.e. will be flipped an odd number of times and thus end up open.
AppleScript
<lang AppleScript>set is_open to {} repeat 100 times
set end of is_open to false
end repeat with pass from 1 to 100
repeat with door from pass to 100 by pass set item door of is_open to not item door of is_open end
end set open_doors to {} repeat with door from 1 to 100
if item door of is_open then
set end of open_doors to door
end
end set text item delimiters to ", " display dialog "Open doors: " & open_doors</lang>
Arbre
<lang Arbre> openshut(n):
for x in [1..n] x%n==0
pass(n):
if n==100 openshut(n) else openshut(n) xor pass(n+1)
100doors():
pass(1) -> io
</lang>
Argile
<lang Argile>use std, array
close all doors for each pass from 1 to 100
for (door = pass) (door <= 100) (door += pass) toggle door
let int pass, door.
.: close all doors :. {memset doors 0 size of doors} .:toggle <int door>:. { !!(doors[door - 1]) }
let doors be an array of 100 bool
for each door from 1 to 100
printf "#%.3d %s\n" door (doors[door - 1]) ? "[ ]", "[X]"</lang>
AutoHotkey
Standard Approach
<lang autohotkey>Loop, 100
Door%A_Index% := "closed"
Loop, 100 {
x := A_Index, y := A_Index
While (x <= 100)
{
CurrentDoor := Door%x%
If CurrentDoor contains closed
{
Door%x% := "open"
x += y
}
else if CurrentDoor contains open
{
Door%x% := "closed"
x += y
}
}
}
Loop, 100 {
CurrentDoor := Door%A_Index%
If CurrentDoor contains open
Res .= "Door " A_Index " is open`n"
} MsgBox % Res</lang>
Alternative Approach
Making use of the identity:
<lang autohotkey>increment := 3, square := 1 Loop, 100
If (A_Index = square)
outstring .= "`nDoor " A_Index " is open"
,square += increment, increment += 2
MsgBox,, Succesfull, % SubStr(outstring, 2)</lang>
Optimized
<lang autohotkey>While (Door := A_Index ** 2) <= 100
Result .= "Door " Door " is open`n"
MsgBox, %Result%</lang>
Axiom
Unoptimized:<lang Axiom>(open,closed,change,open?) := (true,false,not,test); doors := bits(100,closed); for i in 1..#doors repeat
for j in i..#doors by i repeat doors.j := change doors.j
[i for i in 1..#doors | open? doors.i] </lang>Optimized:<lang Axiom>[i for i in 1..100 | perfectSquare? i] -- or [i^2 for i in 1..sqrt(100)::Integer]</lang>
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>
Batch File
unoptimized <lang dos> @echo off setlocal enableDelayedExpansion
- 0 = closed
- 1 = open
- SET /A treats undefined variable as 0
- Negation operator ! must be escaped because delayed expansion is enabled
for /l %%p in (1 1 100) do for /l %%d in (%%p %%p 100) do set /a "door%%d=^!door%%d" for /l %%d in (1 1 100) do if !door%%d!==1 (
echo door %%d is open
) else echo door %%d is closed </lang>
optimized <lang dos> @echo off setlocal enableDelayedExpansion set /a square=1, incr=3 for /l %%d in (1 1 100) do (
if %%d neq !square! (echo door %%d is closed) else ( echo door %%d is open set /a square+=incr, incr+=2 )
) </lang>
BASIC
unoptimized <lang qbasic>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</lang> optimized <lang qbasic>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</lang>
BBC BASIC
<lang bbcbasic> DIM doors%(100)
FOR pass% = 1 TO 100
FOR door% = pass% TO 100 STEP pass%
doors%(door%) EOR= TRUE
NEXT door%
NEXT pass%
FOR door% = 1 TO 100
IF doors%(door%) PRINT "Door " ; door% " is open"
NEXT door%</lang>
Befunge
<lang befunge>108p0>:18p;;>:9g!18g9p08g]
- `!0\|+relet|-1`*aap81::+]
- +1<r]!g9;>$08g1+
- 08paa[
- `#@_^._aa</lang>
BlitzMax
optimized <lang BlitzMax>Graphics 640,480 i=1 While ((i*i)<=100) a$=i*i DrawText a$,10,20*i Print i*i i=i+1 Wend Flip WaitKey </lang>
Bracmat
Bracmat is not really at home in tasks that involve addressing things by index number. Here are four solutions that each do the task, but none should win a price for cleanliness.
Solution 1. Use an indexable array. Local variables are stored in stacks. Each stack corresponds to one variable name and vice versa. Stacks can also be used as arrays, but because of how local variables are implemented, arrays cannot be declared as local variables. <lang bracmat>( 100doors-tbl = door step
. tbl$(doors.101) { Create an array. Indexing is 0-based. Add one extra for addressing element nr. 100 }
& 0:?step
& whl
' ( 1+!step:~>100:?step { ~> means 'not greater than', i.e. 'less than or equal' }
& 0:?door
& whl
' ( !step+!door:~>100:?door
& 1+-1*!(!door$doors):?doors { <number>$<variable> sets the current index, which stays the same until explicitly changed. }
)
)
& 0:?door
& whl
' ( 1+!door:~>100:?door
& out
$ ( door
!door
is
( !(!door$doors):1&open
| closed
)
)
)
& tbl$(doors.0) { clean up the array }
)</lang>
Solution 2. Use one variable for each door. In Bracmat, a variable name can be any non-empty string, even a number, so we use the numbers 1 .. 100 as variable names, but also as door numbers. When used as variable an extra level of indirection is needed. See the occurrences of ?! and !! in the following code.
<lang bracmat>( 100doors-var
= step door
. 0:?door
& whl
' ( 1+!door:~>100:?door
& closed:?!door { this creates a variable and assigns a value 'closed' to it }
)
& 0:?step
& whl
' ( 1+!step:~>100:?step
& 0:?door
& whl
' ( !step+!door:~>100:?door
& ( !!door:closed&open
| closed
)
: ?!door
)
)
& 0:?door
& whl
' ( 1+!door:~>100:?door
& out$(door !door is !!door)
)
& 0:?door
& whl
' ( 1+!door:~>100:?door
& tbl$(!door.0) { cleanup the variable }
)
)</lang>
Solution 3. Use a list and a dedicated positioning pattern to address the right door in the list. Create a new list by concatenating the skipped elements with the toggled elements. This solution is computationally unfavourable because of the many concatenations. <lang bracmat>( 100doors-list = doors door doorIndex step
. :?doors
& 0:?door
& whl
' ( 1+!door:~>100:?door
& closed !doors:?doors
)
& 0:?skip
& whl
' ( :?ndoors
& whl
' ( !doors:?skipped [!skip %?door ?doors { the [<number> pattern only succeeds when the scanning cursor is at position <number> }
& !ndoors
!skipped
( !door:open&closed
| open
)
: ?ndoors
)
& !ndoors !doors:?doors
& 1+!skip:<100:?skip
)
& out$!doors
)</lang>
Solution 4. Use a list of objects. Each object can be changed without the need to re-create the whole list. <lang bracmat>( 100doors-obj = doors door doorIndex step
. :?doors
& 0:?door
& whl
' ( 1+!door:~>100:?door
& new$(=closed) !doors:?doors
)
& 0:?skip
& whl
' ( !doors:?tododoors
& whl
' ( !tododoors:? [!skip %?door ?tododoors
& ( !(door.):open&closed
| open
)
: ?(door.)
)
& 1+!skip:<100:?skip
)
& out$!doors
)</lang>
These four functions are called in the following way: <lang bracmat>100doors-tbl$ & 100doors-var$ & 100doors-list$ & 100doors-obj$;</lang>
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>
Or really optimize it -- square of an integer is, well, computable:<lang C>#include <stdio.h>
int main() { int i; for (i = 1; i * i <= 100; i++) printf("door %d open\n", i * i);
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 only calculation that's really needed: <lang cpp>#include <iostream> //compiled with "Dev-C++" , from RaptorOne
int main() {
for(int i=1; i*i<=100; i++)
std::cout<<"Door "<<i*i<<" is open!"<<std::endl;
}</lang>
C#
Unoptimized <lang csharp>using System; class Program {
static void Main()
{
//To simplify door numbers, uses indexes 1 to 100 (rather than 0 to 99)
bool[] doors = new bool[101];
for (int pass = 1; pass <= 100; pass++)
for (int current = pass; current <= 100; current += pass)
doors[current] = !doors[current];
for (int i = 1; i <= 100; i++)
Console.WriteLine("Door #{0} " + (doors[i] ? "Open" : "Closed"), i);
}
}</lang> Optimized <lang csharp>using System; class Program {
static void Main()
{
int door = 1, inrementer = 0;
for (int current = 1; current <= 100; current++)
{
Console.Write("Door #{0} ", current);
if (current == door)
{
Console.WriteLine("Open");
inrementer++;
door += 2 * inrementer + 1;
}
else
Console.WriteLine("Closed");
}
}
}</lang> Optimized for brevity <lang csharp>using System; class Program {
static void Main()
{
double n;
for (int t = 1; t <= 100; ++t)
Console.WriteLine(t + ": " + (((n = Math.Sqrt(t)) == (int)n) ? "Open" : "Closed"));
}
}</lang>
C1R
<lang c>100_doors</lang>
CLIPS
Unoptimized
<lang clips>(deffacts initial-state
(door-count 100)
)
(deffunction toggle
(?state) (switch ?state (case "open" then "closed") (case "closed" then "open") )
)
(defrule create-doors-and-visits
(door-count ?count) => (loop-for-count (?num 1 ?count) do (assert (door ?num "closed")) (assert (visit-from ?num ?num)) ) (assert (doors initialized))
)
(defrule visit
(door-count ?max) ?visit <- (visit-from ?num ?step) ?door <- (door ?num ?state) => (retract ?visit) (retract ?door) (assert (door ?num (toggle ?state))) (if (<= (+ ?num ?step) ?max) then (assert (visit-from (+ ?num ?step) ?step)) )
)
(defrule start-printing
(doors initialized) (not (visit-from ? ?)) => (printout t "These doors are open:" crlf) (assert (print-from 1))
)
(defrule print-door
(door-count ?max) ?pf <- (print-from ?num) (door ?num ?state) => (retract ?pf) (if (= 0 (str-compare "open" ?state)) then (printout t ?num " ") ) (if (< ?num ?max) then (assert (print-from (+ ?num 1))) else (printout t crlf "All other doors are closed." crlf) )
)</lang>
Optimized
<lang clips>(deffacts initial-state
(door-count 100)
)
(deffunction is-square
(?num) (= (sqrt ?num) (integer (sqrt ?num)))
)
(defrule check-doors
(door-count ?count)
=>
(printout t "These doors are open:" crlf)
(loop-for-count (?num 1 ?count) do
(if (is-square ?num) then
(printout t ?num " ")
)
)
(printout t crlf "All other doors are closed." crlf)
)</lang>
Clojure
Unoptimized / mutable array <lang clojure>(defn doors []
(let [doors (into-array (repeat 100 false))]
(doseq [pass (range 1 101)
i (range (dec pass) 100 pass)
:while (< i 100)]
(aset doors i (not (aget doors i))))
doors))
(defn open-doors [] (for [[d n] (map vector (doors) (iterate inc 1)) :when d] n))
(defn print-open-doors []
(println "Open doors after 100 passes:" (apply str (interpose ", " (open-doors)))))</lang>
Unoptimized / functional <lang clojure>(defn doors []
(reduce (fn [doors toggle-idx] (update-in doors [toggle-idx] not))
(into [] (repeat 100 false))
(for [pass (range 1 101)
i (range (dec pass) 100 pass)
:while (< i 100)]
i)))
(defn open-doors [] (for [[d n] (map vector (doors) (iterate inc 1)) :when d] n))
(defn print-open-doors []
(println "Open doors after 100 passes:" (apply str (interpose ", " (open-doors)))))</lang>
Optimized / functional <lang clojure>(defn doors [] (reduce (fn [doors idx] (assoc doors idx true)) (into [] (repeat 100 false)) (map #(dec (* % %)) (range 1 11))))
(defn open-doors [] (for [[d n] (map vector (doors) (iterate inc 1)) :when d] n))
(defn print-open-doors []
(println "Open doors after 100 passes:" (apply str (interpose ", " (open-doors)))))</lang>
COBOL
<lang cobol> IDENTIFICATION DIVISION.
PROGRAM-ID. 100Doors.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 Current PIC 9(3) VALUE ZEROES.
01 StepSize PIC 9(3) VALUE ZEROES.
01 DoorTable.
02 Doors PIC 9(1) OCCURS 100 TIMES.
01 Idx PIC 9(3).
PROCEDURE DIVISION.
Begin.
MOVE 1 TO StepSize
PERFORM 100 TIMES
MOVE StepSize TO Current
PERFORM UNTIL Current > 100
SUBTRACT Doors(Current) FROM 1 GIVING Doors(Current)
ADD StepSize TO Current GIVING Current
END-PERFORM
ADD 1 TO StepSize GIVING StepSize
END-PERFORM
PERFORM VARYING Idx FROM 1 BY 1
UNTIL Idx > 100
IF Doors(Idx) = 0
DISPLAY Idx " is closed."
ELSE
DISPLAY Idx " is open."
END-IF
END-PERFORM
STOP RUN.</lang>
CoffeeScript
unoptimized: <lang coffeescript>doors = []
for pass in [1..100]
for i in [pass..100] by pass doors[i] = !doors[i]
console.log "Doors #{index for index, open of doors when open} are open"
- matrix output
console.log doors.map (open) -> +open </lang>
optimized:
<lang coffeescript>isInteger = (i) -> Math.floor(i) == i
console.log door for door in [1..100] when isInteger Math.sqrt door</lang>
ultra-optimized: <lang coffeescript>console.log Math.pow(i,2) for i in [1..10]</lang>
ColdFusion
Basic Solution: Returns List of 100 values: 1=open 0=closed <lang coldfusion> doorCount = 1; doorList = ""; // create all doors and set all doors to open while (doorCount LTE 100) { doorList = ListAppend(doorList,"1"); doorCount = doorCount + 1; } loopCount = 2; doorListLen = ListLen(doorList); while (loopCount LTE 100) { loopDoorListCount = 1; while (loopDoorListCount LTE 100) { testDoor = loopDoorListCount / loopCount; if (testDoor EQ Int(testDoor)) { checkOpen = ListGetAt(doorList,loopDoorListCount); if (checkOpen EQ 1) { doorList = ListSetAt(doorList,loopDoorListCount,"0"); } else { doorList = ListSetAt(doorList,loopDoorListCount,"1"); } } loopDoorListCount = loopDoorListCount + 1; } loopCount = loopCount + 1; } </lang>
Squares of Integers Solution: Returns List of 100 values: 1=open 0=closed <lang coldfusion> doorCount = 1; doorList = ""; loopCount = 1; while (loopCount LTE 100) { if (Sqr(loopCount) NEQ Int(Sqr(loopCount))) { doorList = ListAppend(doorList,0); } else { doorList = ListAppend(doorList,1); } loopCount = loopCount + 1; } </lang>
Display only <lang coldfusion>
// Display all doors <cfloop from="1" to="100" index="x"> Door #x# Open: #YesNoFormat(ListGetAt(doorList,x))#
</cfloop>
// Output only open doors
<cfloop from="1" to="100" index="x">
<cfif ListGetAt(doorList,x) EQ 1>
#x#
</cfif>
</cfloop>
</lang>
Common Lisp
Unoptimized / functional This is a very unoptimized version of the problem, using recursion and quite considerable list-copying. It emphasizes 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>
Unoptimized / imperative This is a version that closely follows the problem description and is still quite short.
<lang lisp>(define-modify-macro toggle () not)
(defun 100-doors ()
(let ((doors (make-array 100 :initial-element nil)))
(dotimes (i 100)
(loop for j from i below 100 by (1+ i)
do (toggle (svref doors j))))
(dotimes (i 100)
(format t "door ~a: ~:[closed~;open~]~%" (1+ i) (svref doors i)))))</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 (note however that this is imperative as it does variable mutation)
<lang lisp>(let ((i 0))
(mapcar (lambda (x)
(if (zerop (mod (sqrt (incf i)) 1))
"_" "#"))
(make-list 100)))</lang>
D
<lang d>import std.stdio;
enum DoorState { Closed, Open } alias DoorState[] Doors;
Doors flipUnoptimized(Doors doors) {
doors[] = DoorState.Closed;
foreach (i; 0 .. doors.length)
for (int j = i; j < doors.length; j += i+1)
if (doors[j] == DoorState.Open)
doors[j] = DoorState.Closed;
else
doors[j] = DoorState.Open;
return doors;
}
Doors flipOptimized(Doors doors) {
doors[] = DoorState.Closed;
for (int i = 1; i*i <= doors.length; i++)
doors[i*i - 1] = DoorState.Open;
return doors;
}
// test program void main() {
auto doors = new Doors(100);
foreach (i, door; flipUnoptimized(doors))
if (door == DoorState.Open)
write(i+1, " ");
writeln();
foreach (i, door; flipOptimized(doors))
if (door == DoorState.Open)
write(i+1, " ");
writeln();
}</lang>
- Output:
1 4 9 16 25 36 49 64 81 100 1 4 9 16 25 36 49 64 81 100
Dart
unoptimized <lang dart>main() {
for (var k = 1, x = new List(101); k <= 100; k++) {
for (int i = k; i <= 100; i += k)
x[i] = !x[i];
if (x[k]) print("$k open");
}
}</lang>
optimized version (including generating squares without multiplication) <lang dart>main() {
for(int i=1,s=3;i<=100;i+=s,s+=2)
print("door $i is open");
}</lang>
Delphi
- See Pascal
DWScript
Unoptimized <lang delphi>var doors : array [1..100] of Boolean; var i, j : Integer;
for i := 1 to 100 do
for j := i to 100 do
if (j mod i) = 0 then
doors[j] := not doors[j];
for i := 1 to 100 do
if doors[i] then
PrintLn('Door '+IntToStr(i)+' is open');</lang>
Dylan
Unoptimized <lang dylan>define method doors()
let doors = make(<array>, fill: #f, size: 100);
for (x from 0 below 100)
for (y from x below 100 by x + 1)
doors[y] := ~doors[y]
end
end;
for (x from 1 to 100)
if (doors[x - 1])
format-out("door %d open\n", x)
end
end
end</lang>
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>
Eiffel
This is my first RosettaCode submission, as well as a foray into Eiffel for myself. I've tried to adhere to the description of the problem statement, as well as showcase a few Eiffelisms shown in the documentation.
file: application.e <lang eiffel>note description: "100 Doors problem" date: "07-AUG-2011" revision: "1.0"
class APPLICATION
create make
feature {NONE} -- Initialization
doors: LINKED_LIST [DOOR] -- A set of doors once Result := create {LINKED_LIST [DOOR]}.make end
make -- Run application. local count, i: INTEGER do --initialize doors count := 100 from i := 1 until i > count loop doors.extend (create {DOOR}.make (i, false)) i := i + 1 end
-- toggle doors from i := 1 until i > count loop across doors as this loop if this.item.address \\ i = 0 then this.item.open := not this.item.open end end -- across doors i := i + 1 end -- for i
-- print results doors.do_all (agent (door: DOOR) do if door.open then io.put_string ("Door " + door.address.out + " is open.") elseif not door.open then io.put_string ("Door " + door.address.out + " is closed.") end io.put_new_line end) end -- make
end -- APPLICATION</lang>
file: door.e <lang eiffel>note description: "A door with an address and an open or closed state." date: "07-AUG-2011" revision: "1.0"
class DOOR -- Represents a door
create make
feature -- initialization
make (addr: INTEGER; status: BOOLEAN) -- create door with address and status require valid_address: addr /= '%U' valid_status: status /= '%U' do address := addr open := status ensure address_set: address = addr status_set: open = status end
feature -- access
address: INTEGER
open: BOOLEAN assign set_open
feature -- mutators
set_open (status: BOOLEAN) require valid_status: status /= '%U' do open := status ensure open_updated: open = status end
end</lang>
Ela
Standard Approach
<lang ela>let gate (x::xs) (y::ys) | x == y = Open :: gate xs ys
gate (x::xs) ys = Closed :: gate xs ys gate [] _ = []
let run n = gate [1..n] [& k*k \\ k <- [1..]]</lang>
Alternate Approach <lang ela>open Core let run n = takeWhile (<n) [& k*k \\ k <- [1..]]</lang>
Emacs Lisp
Unoptimized
<lang lisp>(defun create-doors ()
"Returns a list of closed doors
Each door only has two status: open or closed. If a door is closed it has the value 0, if it's open it has the value 1."
(let ((return_value '(0))
;; There is already a door in the return_value, so k starts at 1
;; otherwise we would need to compare k against 99 and not 100 in
;; the while loop
(k 1))
(while (< k 100)
(setq return_value (cons 0 return_value))
(setq k (+ 1 k)))
return_value))
(defun toggle-single-door (doors)
"Toggle the stat of the door at the `car' position of the DOORS list
DOORS is a list of integers with either the value 0 or 1 and it represents a row of doors.
Returns a list where the `car' of the list has it's value toggled (if open it becomes closed, if closed it becomes open)."
(if (= (car doors) 1) (cons 0 (cdr doors)) (cons 1 (cdr doors))))
(defun toggle-doors (doors step original-step)
"Step through all elements of the doors' list and toggle a door when step is 1
DOORS is a list of integers with either the value 0 or 1 and it represents a row of doors. STEP is the number of doors we still need to transverse before we arrive at a door that has to be toggled. ORIGINAL-STEP is the value of the argument step when this function is called for the first time.
Returns a list of doors"
(cond ((null doors)
'())
((= step 1)
(cons (car (toggle-single-door doors))
(toggle-doors (cdr doors) original-step original-step)))
(t
(cons (car doors)
(toggle-doors (cdr doors) (- step 1) original-step)))))
(defun main-program ()
"The main loop for the program"
(let ((doors_list (create-doors))
(k 1)
;; We need to define max-specpdl-size and max-specpdl-size to big
;; numbers otherwise the loop reaches the max recursion depth and
;; throws an error.
;; If you want more information about these variables, press Ctrl
;; and h at the same time and then press v and then type the name
;; of the variable that you want to read the documentation.
(max-specpdl-size 5000)
(max-lisp-eval-depth 2000))
(while (< k 101)
(setq doors_list (toggle-doors doors_list k k))
(setq k (+ 1 k)))
doors_list))
(defun print-doors (doors)
"This function prints the values of the doors into the current buffer.
DOORS is a list of integers with either the value 0 or 1 and it represents a row of doors. "
;; As in the main-program function, we need to set the variable
;; max-lisp-eval-depth to a big number so it doesn't reach max recursion
;; depth.
(let ((max-lisp-eval-depth 5000))
(unless (null doors)
(insert (int-to-string (car doors)))
(print-doors (cdr doors)))))
- Returns a list with the final solution
(main-program)
- Print the final solution on the buffer
(print-doors (main-program))</lang>
Erlang
optimized <lang erlang>doors() ->
F = fun(X) -> Root = math:pow(X,0.5), Root == trunc(Root) end,
Out = fun(X, true) -> io:format("Door ~p: open~n",[X]);
(X, false)-> io:format("Door ~p: close~n",[X]) end,
[Out(X,F(X)) || X <- lists:seq(1,100)].</lang>
Euphoria
unoptimised <lang Euphoria>-- doors.ex include std/console.e sequence doors doors = repeat( 0, 100 ) -- 1 to 100, initialised to false
for pass = 1 to 100 do for door = pass to 100 by pass do --printf( 1, "%d", doors[door] ) --printf( 1, "%d", not doors[door] ) doors[door] = not doors[door] end for end for
sequence oc
for i = 1 to 100 do if doors[i] then oc = "open" else oc = "closed" end if
printf( 1, "door %d is %s\n", { i, oc } )
end for </lang>
Euler Math Toolbox
<lang Euler Math Toolbox> function Doors $ doors:=zeros(1,100); $ for i=1 to 100 step 1 $ for j=i to 100 step i $ if doors[j]==0 then doors[j]:=1; $ else doors[j]:=0; $ endif; $ end; $ end; $ return doors $endfunction
A:=Doors; for i=1 to 100 step 1; if A[i]==1 then "door "|i|" is open" endif; end; </lang> Output <lang>
door 1 is open door 4 is open door 9 is open door 16 is open door 25 is open door 36 is open door 49 is open door 64 is open door 81 is open door 100 is open
</lang>
Factor
<lang Factor>USING: bit-arrays formatting fry kernel math math.ranges sequences ; IN: rosetta.doors
CONSTANT: number-of-doors 100
- multiples ( n -- range )
0 number-of-doors rot <range> ;
- toggle-multiples ( n doors -- )
[ multiples ] dip '[ _ [ not ] change-nth ] each ;
- toggle-all-multiples ( doors -- )
[ number-of-doors [1,b] ] dip '[ _ toggle-multiples ] each ;
- print-doors ( doors -- )
[
swap "open" "closed" ? "Door %d is %s\n" printf
] each-index ;
- main ( -- )
number-of-doors 1 + <bit-array> [ toggle-all-multiples ] [ print-doors ] bi ;</lang>
Falcon
Unoptimized code <lang falcon>doors = arrayBuffer( 101, false )
for pass in [ 0 : doors.len() ]
for door in [ 0 : doors.len() : pass+1 ] doors[ door ] = not doors[ door ] end
end
for door in [ 1 : doors.len() ] // Show Output
> "Door ", $door, " is: ", ( doors[ door ] ) ? "open" : "closed"
end </lang> Optimized code <lang falcon> for door in [ 1 : 101 ]: > "Door ", $door, " is: ", fract( door ** 0.5 ) ? "closed" : "open"</lang>
Fantom
Unoptimized <lang fantom>
states := (1..100).toList
100.times |i| {
states = states.map |state| { state % (i+1) == 0 ? -state : +state }
}
echo("Open doors are " + states.findAll { it < 0 }.map { -it })
</lang> Optimized <lang fantom>
echo("Open doors are " + (1..100).toList.findAll { it.toFloat.pow(0.5f).toInt.pow(2) == it})
</lang>
Forth
Unoptimized <lang forth>: toggle ( c-addr -- ) \ toggle the byte at c-addr
dup c@ 1 xor swap c! ;
100 1+ ( 1-based indexing ) constant ndoors create doors ndoors allot
- init ( -- ) doors ndoors erase ; \ close all doors
- pass ( n -- ) \ toggle every nth door
ndoors over do
doors i + toggle
dup ( n ) +loop drop ;
- run ( -- ) ndoors 1 do i pass loop ;
- display ( -- ) \ display open doors
ndoors 1 do doors i + c@ if i . then loop cr ;
init run display</lang>
Optimized <lang forth>: squared ( n -- n' ) dup * ;
- doors ( n -- )
1 begin 2dup squared >= while
dup squared .
1+ repeat 2drop ;
100 doors</lang>
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>
F#
Requires #light in versions of F# prior to 2010 beta. <lang fsharp>let answerDoors =
let ToggleNth n (lst:bool array) = // Toggle every n'th door
[(n-1) .. n .. 99] // For each appropriate door
|> Seq.iter (fun i -> lst.[i] <- not lst.[i]) // toggle it
let doors = Array.create 100 false // Initialize all doors to closed
Seq.iter (fun n -> ToggleNth n doors) [1..100] // toggle the appropriate doors for each pass
doors // Initialize all doors to closed
</lang> Following is the solution using perfect squares. The coercions in PerfectSquare are, I believe, slightly different in versions prior to 2010 beta and, again, #light is required in those versions. <lang fsharp>open System let answer2 =
let PerfectSquare n =
let sqrt = int(Math.Sqrt(float n))
n = sqrt * sqrt
[| for i in 1..100 do yield PerfectSquare i |]</lang>
GAP
<lang gap>doors := function(n)
local a,j,s;
a := [ ];
for j in [1 .. n] do
a[j] := 0;
od;
for s in [1 .. n] do
j := s;
while j <= n do
a[j] := 1 - a[j];
j := j + s;
od;
od;
return Filtered([1 .. n], j -> a[j] = 1);
end;
doors(100);
- [ 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 ]</lang>
GML
<lang gml>var doors,a,i; //Sets up the array for all of the doors. for (i=1;i<=100;i+=1)
{
doors[i]=0;
}
//This first for loop goes through and passes the interval down to the next for loop. for (i=1;i<=100;i+=1)
{
//This for loop opens or closes the doors and uses the interval(if interval is 2 it only uses every other etc..)
for (a=0;a<=100;a+=i;)
{
//Opens or closes a door.
doors[a]=!doors[a];
}
}
open_doors=; //This for loop goes through the array and checks for open doors. //If the door is open it adds it to the string then displays the string. for (i=1;i<=100;i+=1)
{
if doors[i]=1
{
open_doors+="Door Number "+string(i)+" is open#";
}
}
show_message(open_doors); game_end();</lang>
Go
unoptimized <lang go>package main
import "fmt"
func main() {
doors := make([]bool, 100)
// the 100 passes called for in the task description
for pass := 1; pass <= 100; pass++ {
for door := pass-1; door < 100; door += pass {
doors[door] = !doors[door]
}
}
// one more pass to answer the question
for i, v := range doors {
if v {
fmt.Print("1")
} else {
fmt.Print("0")
}
if i%10 == 9 {
fmt.Print("\n")
} else {
fmt.Print(" ")
}
}
}</lang> Output:
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 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
optimized <lang go>package main
import "fmt"
func main() {
var door int = 1 var incrementer = 0
for current := 1; current <= 100; current++ {
fmt.Printf("Door %d ", current)
if current == door {
fmt.Printf("Open\n")
incrementer++
door += 2*incrementer + 1
} else {
fmt.Printf("Closed\n")
}
}
}</lang>
Golfscript
<lang golfscript>100:c;[{0}c*]:d; c,{.c,>\)%{.d<\.d=1^\)d>++:d;}/}/ [c,{)"door "\+" is"+}%d{{"open"}{"closed"}if}%]zip {" "*puts}/</lang>
optimized with sqrt (Original version of GolfScript has no sqrt operator, but it can be added easily; the code was tested using a work-in-progress C interpreter for a language compatible enough with Golfscript) <lang golfscript>100,{)}% {:d.sqrt 2?= {"open"}{"close"}if"door "d+" is "+\+puts}/</lang>
optimized without sqrt <lang golfscript>[{"close"}100*]:d; 10,{)2?(.d<\["open"]\)d>++:d;}/ [100,{)"door "\+" is"+}%d]zip {" "*puts}/</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 ${doors[it]}.")
}</lang>
Haskell
unoptimized <lang haskell>data Door = Open | Closed deriving Show
toggle Open = Closed toggle Closed = Open
toggleEvery :: [Door] -> Int -> [Door] toggleEvery xs k = zipWith ($) fs xs
where fs = cycle $ (replicate (k-1) id) ++ [toggle]
run n = foldl toggleEvery (replicate n Closed) [0..n]</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>
haXe
<lang haxe>class RosettaDemo {
static public function main()
{
findOpenLockers(100);
}
static function findOpenLockers(n : Int)
{
var i = 1;
while((i*i) <= n)
{
neko.Lib.print(i*i + "\n");
i++;
}
}
}</lang>
HicEst
Unoptimized <lang hicest>REAL :: n=100, open=1, door(n)
door = 1 - open ! = closed DO i = 1, n
DO j = i, n, i door(j) = open - door(j) ENDDO
ENDDO DLG(Text=door, TItle=SUM(door)//" doors open") </lang> Optimized <lang hicest>door = 1 - open ! = closed DO i = 1, n^0.5
door(i*i) = open
ENDDO DLG(Text=door, TItle=SUM(door)//" doors open") </lang>
Icon and Unicon
Icon and Unicon don't have a boolean type because most often, logic is expressed in terms of success or failure, which affects flow at run time.
Unoptimized solution. <lang icon> procedure main()
door := table(0) # default value of entries is 0
every pass := 1 to 100 do
every door[i := pass to 100 by pass] := 1 - door[i]
every write("Door ", i := 1 to 100, " is ", if door[i] = 1 then "open" else "closed")
end </lang>
Optimized solution. <lang icon> procedure main()
every write("Door ", i := 1 to 100, " is ", if integer(sqrt(i)) = sqrt(i) then "open" else "closed")
end </lang>
or
<lang icon>procedure main(args)
dMap := table("closed")
every dMap[(1 to sqrt(100))^2] := "open"
every write("Door ",i := 1 to 100," is ",dMap[i])
end</lang>
Inform 7
<lang inform7>Hallway is a room.
A toggle door is a kind of thing. A toggle door can be open or closed. It is usually closed. A toggle door has a number called the door number. Understand the door number property as referring to a toggle door. Rule for printing the name of a toggle door: say "door #[door number]".
There are 100 toggle doors.
When play begins (this is the initialize doors rule): let the next door number be 1; repeat with D running through toggle doors: now the door number of D is the next door number; increment the next door number.
To toggle (D - open toggle door): now D is closed. To toggle (D - closed toggle door): now D is open.
When play begins (this is the solve puzzle rule): let the door list be the list of toggle doors; let the door count be the number of entries in the door list; repeat with iteration running from 1 to 100: let N be the iteration; while N is less than the door count: toggle entry N in the door list; increase N by the iteration; say "Doors left open: [list of open toggle doors]."; end the story.</lang>
Informix 4GL
<lang Informix 4GL> MAIN
DEFINE
i, pass SMALLINT,
doors ARRAY[100] OF SMALLINT
FOR i = 1 TO 100
LET doors[i] = FALSE
END FOR
FOR pass = 1 TO 100
FOR i = pass TO 100 STEP pass
LET doors[i] = NOT doors[i]
END FOR
END FOR
FOR i = 1 TO 100
IF doors[i]
THEN DISPLAY i USING "Door <<& is open"
ELSE DISPLAY i USING "Door <<& is closed"
END IF
END FOR
END MAIN </lang>
Io
simple boolean list solution: <lang io> doors := List clone for(i,1,100, doors append(false)) for(i,1,100,
for(x,i,100, i, doors atPut(x - 1, doors at(x - 1) not))
) doors foreach(i, x, if(x, "Door #{i + 1} is open" interpolate println)) </lang>
Ioke
Unoptimized Object Oriented solution. <lang ioke>NDoors = Origin mimic
NDoors Toggle = Origin mimic do(
initialize = method(toggled?, @toggled? = toggled?) toggle! = method(@toggled? = !toggled?. self)
)
NDoors Doors = Origin mimic do(
initialize = method(n,
@n = n
@doors = {} addKeysAndValues(1..n, (1..n) map(_, NDoors Toggle mimic(false)))
)
numsToToggle = method(n, for(x <- (1..@n), (x % n) zero?, x))
toggleThese = method(nums, nums each(x, @doors[x] = @doors at(x) toggle))
show = method(@doors filter:dict(value toggled?) keys sort println)
)
- Test code
x = NDoors Doors mimic(100) (1..100) each(n, x toggleThese(x numsToToggle(n))) x show</lang>
J
unoptimized <lang j> ~:/ (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 ...
~:/ 0=|/~ >:i.100 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 ...</lang> optimized <lang j> (e. *:) 1+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 ...
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 ...</lang>
with formatting <lang j> 'these doors are open' ; >: I. (>:i.100) e. *: i.11 ┌────────────────────┬───────────────────────────┐ │these doors are open│1 4 9 16 25 36 49 64 81 100│ └────────────────────┴───────────────────────────┘
</lang>
Java
unoptimized <lang java> public class HundredDoors {
public static void main(String[] args) {
boolean[] doors = new boolean[101];
for (int i = 1; i <= 100; i++) {
for (int j = i; j <= 100; j++) {
if(j % i == 0) doors[j] = !doors[j];
}
}
for (int i = 1; i <= 100; i++) {
System.out.printf("Door %d: %s%n", i, doors[i] ? "open" : "closed");
}
}
}</lang>
optimized <lang java>public class Doors {
public static void main(final 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> optimized 2 <lang java>public class Doors {
public static void main(final String[] args)
{
StringBuilder sb = new StringBuilder();
for (int i = 1; i <= 10; i++)
sb.append("Door #").append(i*i).append(" is open\n");
System.out.println(sb.toString()); }
}</lang>
optimized 3 <lang java>public class Doors{
public static void main(String[] args){
int i;
for(i = 1; i < 101; i++){
double sqrt = Math.sqrt(i);
if(sqrt != (int)sqrt){
System.out.println("Door " + i + " is closed");
}else{
System.out.println("Door " + i + " is open");
}
}
}
}</lang>
JavaScript
<lang javascript>var doors = [], n = 100, i, j;
for (i = 1; i <= n; i++) { for (j = i; j <= n; j += i) { doors[j] = !doors[j]; } }
for (i = 1 ; i <= n ; i++) { if (doors[i]) console.log("Door " + i + " is open"); }</lang> outputs
door 1 is open. door 4 is open. door 9 is open. door 16 is open. door 25 is open. door 36 is open. door 49 is open. door 64 is open. door 81 is open. door 100 is open.
Using features of JavaScript 1.6, this
<lang javascript>var
n = 100, doors = [n], step, idx;
// now, start opening and closing for (step = 1; step <= n; step += 1)
for (idx = step; idx <= n; idx += step) // toggle state of door doors[idx] = !doors[idx];
// find out which doors are open var open = doors.reduce(function(open, val, idx) {
if (val) {
open.push(idx);
}
return open;
}, []); document.write("These doors are open: " + open.join(', '));</lang> outputs
these doors are open: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
K
unoptimized / converted from Q . <lang k> `closed `open ![ ; 2 ] @ #:' 1 _ = ,/ &:' 0 = t !\:/: t : ! 101</lang>
optimized / 1 origin indices <lang k> ( 1 + ! 10 ) ^ 2</lang>
/ As parameterized function : <lang k> { ( 1 + ! _ x ^ % 2 ) ^ 2 } 100 </lang>
LabVIEW
This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.
- Optimized
This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.
Liberty BASIC
<lang lb>dim doors(100) for pass = 1 to 100
for door = pass to 100 step pass
doors(door) = not(doors(door))
next door
next pass print "open doors "; for door = 1 to 100
if doors(door) then print door;" ";
next door</lang>
Logo
<lang Logo>to doors
- Problem 100 Doors
- FMSLogo
- lrcvs 2010
make "door (vector 100 1) for [p 1 100][setitem :p :door 0]
for [a 1 100 1][for [b :a 100 :a][make "x item :b :door ifelse :x = 0 [setitem :b :door 1][setitem :b :door 0] ] ]
for [c 1 100][make "y item :c :door ifelse :y = 0 [pr (list :c "Close)] [pr (list :c "Open)] ] end</lang>
Lhogho
This implementation defines 100 variables, named "1 through "100, rather than using a list. Thanks to Pavel Boytchev, the author of Lhogho, for help with the code.
<lang Logo>to doors ;Problem 100 Doors ;Lhogho
for "p [1 100] [ make :p "false ]
for "a [1 100 1] [ for "b [:a 100 :a] [ if :b < 101 [ make :b not thing :b ] ] ]
for "c [1 100] [ if thing :c [ (print "door :c "is "open) ] ] end
doors</lang>
Lua
<lang lua>is_open = {}
for door = 1,100 do is_open[door] = false end
for pass = 1,100 do
for door = pass,100,pass do
is_open[door] = not is_open[door]
end
end
for i,v in next,is_open do
if v then
print ('Door '..i..':','open')
else
print ('Door '..i..':', 'close')
end
end</lang>
M4
<lang m4>define(`_set', `define(`$1[$2]', `$3')')dnl define(`_get', `defn(`$1[$2]')')dnl define(`for',`ifelse($#,0,``$0,`ifelse(eval($2<=$3),1, `pushdef(`$1',$2)$5`'popdef(`$1')$0(`$1',eval($2+$4),$3,$4,`$5')')')')dnl define(`opposite',`_set(`door',$1,ifelse(_get(`door',$1),`closed',`open',`closed'))')dnl define(`upper',`100')dnl for(`x',`1',upper,`1',`_set(`door',x,`closed')')dnl for(`x',`1',upper,`1',`for(`y',x,upper,x,`opposite(y)')')dnl for(`x',`1',upper,`1',`door x is _get(`door',x) ')dnl</lang>
Mathematica
unoptimized 1 <lang mathematica>n=100; tmp=ConstantArray[-1,n]; Do[tmpi;;;;i*=-1;,{i,n}]; Do[Print["door ",i," is ",If[tmpi==-1,"closed","open"]],{i,1,Length[tmp]}]</lang>
unoptimized 2 <lang mathematica>f[n_] = "Closed"; Do[Do[If[f[n] == "Closed", f[n] = "Open", f[n] = "Closed"], {n, k, 100, k}], {k, 1, 100}]; Table[f[n], {n, 1, 100}]</lang>
optimized 1 <lang mathematica>Do[Print["door ",i," is ",If[IntegerQ[Sqrt[i]],"open","closed"]],{i,100}]</lang>
optimized 2 <lang mathematica>n=100; a=Range[1,Sqrt[n]]^2 Do[Print["door ",i," is ",If[MemberQ[a,i],"open","closed"]],{i,100}]</lang>
optimized 3 <lang mathematica>n=100 nn=1 a=0 For[i=1,i<=n,i++,
If[i==nn, Print["door ",i," is open"]; a++; nn+=2a+1; , Print["door ",i," is closed"]; ];
]</lang>
These will only give the indices for the open doors: unoptimized 2 <lang mathematica>Pick[Range[100], Xor@@@Array[Divisible[#1,#2]&, {100,100}]]</lang>
optimized 4 <lang mathematica>Range[Sqrt[100]]^2</lang>
MATLAB
Iterative Method
Unoptimized <lang MATLAB> a=zeros(1,100); for b=1:100; for i=b:b:100;
if a(i)==1
a(i)=0;
else
a(i)=1;
end
end end a </lang> Optimized <lang MATLAB> for x=1:100;
if sqrt(x) == floor(sqrt(x)) a(i)=1; end
end a </lang> More Optimized <lang MATLAB> a = zeros(100,1); for counter = 1:sqrt(100);
a(counter^2) = 1;
end a </lang>
Vectorized Method
<lang MATLAB>function [doors,opened,closed] = hundredDoors()
%Initialize the doors, make them booleans for easy vectorization
doors = logical( (1:1:100) );
%Go through the flipping process, ignore the 1 case because the doors
%array is already initialized to all open
for initialPosition = (2:100)
doors(initialPosition:initialPosition:100) = not( doors(initialPosition:initialPosition:100) );
end
opened = find(doors); %Stores the numbers of the open doors
closed = find( not(doors) ); %Stores the numbers of the closed doors
end</lang>
Known-Result Method
<lang MATLAB> doors((1:10).^2) = 1;
doors </lang>
MAXScript
unoptimized <lang maxscript>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]
)</lang> optimized <lang maxscript>for i in 1 to 100 do (
root = pow i 0.5
format ("Door % is open?: %\n") i (root == (root as integer))
)</lang>
Mercury
<lang Mercury>:- module doors.
- - interface.
- - import_module array, io, int.
- - type door ---> open ; closed.
- - type doors == array(door).
- - func toggle(door) = door.
- - pred walk(int::in, doors::in, doors::out) is semidet.
- - pred walks(int::in, int::in, doors::in, doors::out) is det.
- - pred main(io::di, io::uo) is det.
- - implementation.
toggle(open) = closed. toggle(closed) = open.
walk(N, !D) :- walk(N, N, !D).
- - pred walk(int::in, int::in, doors::in, doors::out) is semidet.
walk(At, By, !D) :-
semidet_lookup(!.D, At - 1, Door),
slow_set(!.D, At - 1, toggle(Door), !:D),
( walk(At + By, By, !D) -> true ; true ).
walks(N, End, !D) :-
( N =< End, walk(N, !D) -> walks(N + 1, End, !D) ; true ).
main(!IO) :-
io.write(Doors1, !IO), io.nl(!IO),
array.init(100, closed, Doors0),
walks(1, 100, Doors0, Doors1).</lang>
Metafont
<lang metafont>boolean doors[]; for i = 1 upto 100: doors[i] := false; endfor for i = 1 upto 100:
for j = 1 step i until 100: doors[j] := not doors[j]; endfor
endfor for i = 1 upto 100:
message decimal(i) & " " & if doors[i]: "open" else: "close" fi;
endfor end</lang>
Mirah
<lang Mirah>import java.util.ArrayList
class Door :state
def initialize @state=false end
def closed?; !@state; end def open?; @state; end
def close; @state=false; end def open; @state=true; end
def toggle if closed? open else close end end
def toString; Boolean.toString(@state); end end
doors=ArrayList.new 1.upto(100) do
doors.add(Door.new)
end
1.upto(100) do |multiplier|
index = 0
doors.each do |door|
Door(door).toggle if (index+1)%multiplier == 0
index += 1
end
end
i = 0 doors.each do |door|
puts "Door #{i+1} is #{door}."
i+=1
end </lang>
MIPS Assembly
<lang mips>.data
doors: .space 100 num_str: .asciiz "Number " comma_gap: .asciiz " is " newline: .asciiz "\n"
.text main:
- Clear all the cells to zero
li $t1, 100 la $t2, doors
clear_loop:
sb $0, ($t2) add $t2, $t2, 1 sub $t1, $t1, 1 bnez $t1, clear_loop
- Now start the loops
li $t0, 1 # This will the the step size li $t4, 1 # just an arbitrary 1
loop1:
move $t1, $t0 # Counter la $t2, doors # Current pointer add $t2, $t2, $t0 addi $t2, $t2, -1
loop2:
lb $t3, ($t2) sub $t3, $t4, $t3 sb $t3, ($t2) add $t1, $t1, $t0 add $t2, $t2, $t0 ble $t1, 100, loop2
addi $t0, $t0, 1 ble $t0, 100, loop1
# Now display everything la $t0, doors li $t1, 1
loop3:
li $v0, 4 la $a0, num_str syscall li $v0, 1 move $a0, $t1 syscall
li $v0, 4 la $a0, comma_gap syscall
li $v0, 1 lb $a0, ($t0) syscall
li $v0, 4, la $a0, newline syscall
addi $t0, $t0, 1 addi $t1, $t1, 1 bne $t1, 101 loop3
</lang>
ML/I
<lang ML/I>MCSKIP "WITH" NL "" 100 doors MCINS %. MCSKIP MT,<> "" Doors represented by P1-P100, 0 is closed MCPVAR 100 "" Set P variables to 0 MCDEF ZEROPS WITHS NL AS <MCSET T1=1 %L1.MCSET PT1=0 MCSET T1=T1+1 MCGO L1 UNLESS T1 EN 101 > ZEROPS "" Generate door state MCDEF STATE WITHS () AS <MCSET T1=%A1. MCGO L1 UNLESS T1 EN 0 closed<>MCGO L0 %L1.open> "" Main macro - no arguments "" T1 is pass number "" T2 is door number MCDEF DOORS WITHS NL AS <MCSET T1=1 "" pass loop %L1.MCGO L4 IF T1 GR 100 "" door loop MCSET T2=T1 %L2.MCGO L3 IF T2 GR 100 MCSET PT2=1-PT2 MCSET T2=T2+T1 MCGO L2 %L3.MCSET T1=T1+1 MCGO L1 %L4."" now output the result MCSET T1=1 %L5.door %T1. is STATE(%PT1.) MCSET T1=T1+1 MCGO L5 UNLESS T1 GR 100 > "" Do it DOORS</lang>
MMIX
See 100 doors/MMIX
Modula-2
unoptimized <lang modula2>MODULE Doors; IMPORT InOut;
TYPE State = (Closed, Open); TYPE List = ARRAY [1 .. 100] OF State;
VAR
Doors: List; I, J: CARDINAL;
BEGIN
FOR I := 1 TO 100 DO
FOR J := 1 TO 100 DO
IF J MOD I = 0 THEN
IF Doors[J] = Closed THEN
Doors[J] := Open
ELSE
Doors[J] := Closed
END
END
END
END;
FOR I := 1 TO 100 DO
InOut.WriteCard(I, 3);
InOut.WriteString(' is ');
IF Doors[I] = Closed THEN
InOut.WriteString('Closed.')
ELSE
InOut.WriteString('Open.')
END;
InOut.WriteLn END
END Doors.</lang>
optimized <lang modula2>MODULE DoorsOpt; IMPORT InOut;
TYPE State = (Closed, Open); TYPE List = ARRAY [1 .. 100] OF State;
VAR
Doors: List; I: CARDINAL;
BEGIN
FOR I := 1 TO 10 DO Doors[I*I] := Open END;
FOR I := 1 TO 100 DO
InOut.WriteCard(I, 3);
InOut.WriteString(' is ');
IF Doors[I] = Closed THEN
InOut.WriteString('Closed.')
ELSE
InOut.WriteString('Open.')
END;
InOut.WriteLn
END
END DoorsOpt.</lang>
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>
MOO
<lang moo>is_open = make(100); for pass in [1..100]
for door in [pass..100]
if (door % pass)
continue;
endif
is_open[door] = !is_open[door];
endfor
endfor
"output the result"; for door in [1..100]
player:tell("door #", door, " is ", (is_open[door] ? "open" : "closed"), ".");
endfor</lang>
MUMPS
<lang MUMPS>doors new door,pass For door=1:1:100 Set door(door)=0 For pass=1:1:100 For door=pass:pass:100 Set door(door)='door(door) For door=1:1:100 If door(door) Write !,"Door",$j(door,4)," is open" Write !,"All other doors are closed." Quit Do doors Door 1 is open Door 4 is open Door 9 is open Door 16 is open Door 25 is open Door 36 is open Door 49 is open Door 64 is open Door 81 is open Door 100 is open All other doors are closed.</lang>
NetRexx
unoptimized <lang netrexx>/* NetRexx */ options replace format comments java crossref savelog symbols binary
True = Rexx(1 == 1) False = Rexx(\True)
doors = False
loop i_ = 1 to 100
loop j_ = 1 to 100 if 0 = (j_ // i_) then doors[j_] = \doors[j_] end j_ end i_
loop d_ = 1 to 100
if doors[d_] then state = 'open' else state = 'closed'
say 'Door Nr.' Rexx(d_).right(4) 'is' state end d_
</lang>
optimized (Based on the Java 'optimized' version)
<lang netrexx>/* NetRexx */ options replace format comments java crossref savelog symbols binary
True = (1 == 1) False = \True
doors = boolean[100]
loop i_ = 0 to 9
doors[(i_ + 1) * (i_ + 1) - 1] = True; end i_
loop i_ = 0 to 99
if doors[i_] then state = 'open' else state = 'closed'
say 'Door Nr.' Rexx(i_ + 1).right(4) 'is' state end i_
</lang>
optimized 2 (Based on the Java 'optimized 2' version)
<lang netrexx>/* NetRexx */ options replace format comments java crossref savelog symbols binary
resultstring =
loop i_ = 1 to 10
resultstring = resultstring || 'Door Nr.' Rexx(i_ * i_).right(4) 'is open\n' end i_
say resultstring </lang>
optimized 3 <lang netrexx>/* NetRexx */
loop i = 1 to 10
say 'Door Nr.' i * i 'is open.' end i
</lang>
Objeck
optimized <lang objeck> bundle Default {
class Doors {
function : Main(args : String[]) ~ Nil {
doors := Bool->New[100];
for(pass := 0; pass < 10; pass += 1;) {
doors[(pass + 1) * (pass + 1) - 1] := true;
};
for(i := 0; i < 100; i += 1;) {
IO.Console->GetInstance()->Print("Door #")->Print(i + 1)->Print(" is ");
if(doors[i]) {
"open."->PrintLine();
}
else {
"closed."->PrintLine();
};
};
}
}
} </lang>
OCaml
unoptimized <lang ocaml>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))</lang>
optimized <lang ocaml>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))</lang>
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>
OpenEdge/Progress
<lang Progress (OpenEdge ABL)>DEFINE VARIABLE lopen AS LOGICAL NO-UNDO EXTENT 100. DEFINE VARIABLE idoor AS INTEGER NO-UNDO. DEFINE VARIABLE ipass AS INTEGER NO-UNDO. DEFINE VARIABLE cresult AS CHARACTER NO-UNDO.
DO ipass = 1 TO 100:
idoor = 0.
DO WHILE idoor <= 100:
idoor = idoor + ipass.
IF idoor <= 100 THEN
lopen[ idoor ] = NOT lopen[ idoor ].
END.
END.
DO idoor = 1 TO 100:
cresult = cresult + STRING( lopen[ idoor ], "1 /0 " ).
IF idoor MODULO 10 = 0 THEN
cresult = cresult + "~r":U.
END.
MESSAGE cresult VIEW-AS ALERT-BOX. </lang>
Oz
<lang oz>declare
NumDoors = 100 NumPasses = 100
fun {NewDoor} closed end
fun {Toggle Door}
case Door of closed then open
[] open then closed
end
end
fun {Pass Doors I}
{List.mapInd Doors
fun {$ Index Door}
if Index mod I == 0 then {Toggle Door}
else Door
end
end}
end
Doors0 = {MakeList NumDoors}
{ForAll Doors0 NewDoor}
DoorsN = {FoldL {List.number 1 NumPasses 1} Pass Doors0}
in
%% print open doors
{List.forAllInd DoorsN
proc {$ Index Door}
if Door == open then
{System.showInfo "Door "#Index#" is open."}
end end }</lang>
Output:
Door 1 is open. Door 4 is open. Door 9 is open. Door 16 is open. Door 25 is open. Door 36 is open. Door 49 is open. Door 64 is open. Door 81 is open. Door 100 is open.
PARI/GP
Unoptimized version. <lang parigp> v=vector(d=100);/*set 100 closed doors*/ for(i=1,d,forstep(j=i,d,i,v[j]=1-v[j])); for(i=1,d,if(v[i],print("Door ",i," is open."))) </lang> Optimized version. <lang parigp>for(n=1,10,print("Door ",n^2," is open."))</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>
Optimized version.
<lang pascal>program OneHundredDoors;
{$APPTYPE CONSOLE}
uses
math, sysutils;
var
AOpendoors : String; ACloseDoors : String; i : Integer;
begin
for i := 1 to 100 do
begin
if (sqrt(i) = floor(sqrt(i))) then
AOpenDoors := AOpenDoors + IntToStr(i) + ';'
else
ACloseDoors := ACloseDoors + IntToStr(i) +';';
end;
WriteLn('Open doors: ' + AOpenDoors);
WriteLn('Close doors: ' + ACloseDoors);
end. </lang>
PHP
optimized <lang php><?php for ($i = 1; $i <= 100; $i++) { $root = sqrt($i); $state = ($root == ceil($root)) ? 'open' : 'closed'; echo "Door {$i}: {$state}\n"; } ?></lang>
unoptimized <lang php><?php $toggleState = array('open' => 'closed', 'closed' => 'open'); $doors = array_fill(1, 100, 'closed'); for ($pass = 1; $pass <= 100; ++$pass) { for ($nr = 1; $nr <= 100; ++$nr) { if ($nr % $pass == 0) { $doors[$nr] = $toggleState[$doors[$nr]]; } } } for ($nr = 1; $nr <= 100; ++$nr) printf("Door %d is %s\n", $nr, $doors[$nr]); ?></lang>
Perl
unoptimized
<lang perl>my @doors; for my $pass (1 .. 100) {
for (1 .. 100) {
if (0 == $_ % $pass) {
$doors[$_] = not $doors[$_];
};
};
};
print "Door $_ is ", $doors[$_] ? "open" : "closed", "\n" for 1 .. 100;</lang>
optimized
<lang perl>print "Door $_ is open\n" for map { $_**2 } 1 .. 10;</lang> <lang perl>print "Door $_ is ", qw"closed open"[int sqrt == sqrt], "\n" for 1..100;</lang> <lang perl>while( ++$i <= 100 ) {
$root = sqrt($i);
if ( int( $root ) == $root )
{
print "Door $i is open\n";
}
else
{
print "Door $i is closed\n";
}
}</lang>
Perl 6
unoptimized
<lang perl6>my @doors = False xx 101;
($_ = !$_ for @doors[0, * + $_ ...^ * > 100]) for 1..100;
say "Door $_ is ", <closed open>[ @doors[$_] ] for 1..100;</lang>
optimized
<lang perl6>say "Door $_ is open" for map {$^n ** 2}, 1..10;</lang>
Here's a version using the cross meta-operator instead of a map:
<lang perl6> say "Door $_ is open" for 1..10 X** 2;</lang>
This one prints both opened and closed doors:
<lang perl6>say "Door $_ is ", <closed open>[.sqrt == .sqrt.floor] for 1..100;</lang>
PicoLisp
unoptimized <lang PicoLisp>(let Doors (need 100)
(for I 100
(for (D (nth Doors I) D (cdr (nth D I)))
(set D (not (car D))) ) )
(println Doors) )</lang>
optimized <lang PicoLisp>(let Doors (need 100)
(for I (sqrt 100)
(set (nth Doors (* I I)) T) )
(println Doors) )</lang>
Output in both cases:
(T NIL NIL T NIL NIL NIL NIL T NIL NIL NIL NIL NIL NIL T NIL NIL NIL NIL NIL NIL NIL NIL T NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL T NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL T NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL T NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL T NIL NIL NIL N IL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL NIL T)
With formatting: <lang PicoLisp>(let Doors (need 100)
(for I (sqrt 100)
(set (nth Doors (* I I)) T) )
(make
(for (N . D) Doors
(when D (link N)) ) ) )</lang>
Output:
(1 4 9 16 25 36 49 64 81 100)
Piet
Pike
<lang pike>array onehundreddoors() {
array doors = allocate(100);
foreach(doors; int i;)
for(int j=i; j<100; j+=i+1)
doors[j] = !doors[j];
return doors;
}</lang> optimized version: <lang pike>array doors = map(enumerate(100,1,1), lambda(int x)
{
return sqrt((float)x)%1 == 0.0;
});</lang>
<lang pike>write("%{%d %d %d %d %d %d %d %d %d %d\n%}\n", doors/10)</lang> output:
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 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
PL/I
<lang PL/I> declare door(100) bit (1) aligned; declare closed bit (1) static initial ('0'b),
open bit (1) static initial ('1'b);
declare (i, inc) fixed binary;
door = closed; inc = 1; do until (inc >= 100);
do i = inc to 100 by inc;
door(i) = ^door(i); /* close door if open; open it if closed. */
end;
inc = inc+1;
end;
do i = 1 to 100;
put skip edit ('Door ', trim(i), ' is ') (a);
if door(i) then put edit (' open.') (a);
else put edit (' closed.') (a);
end; </lang>
Pop11
unoptimized <lang pop11>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;</lang>
optimized <lang pop11>for i to 100 do
lvars root = sqrt(i); i; if root = round(root) then ' open' ><; else ' closed' ><; endif; =>
endfor;</lang>
PostScript
Bruteforce:<lang PostScript>/doors [ 100 { false } repeat ] def
1 1 100 { dup 1 sub exch 99 {
dup doors exch get not doors 3 1 roll put
} for } for doors pstack</lang>Shows: <lang>[true false false true false false false false true false ...<90 doors later>... true]</lang>
PowerShell
unoptimized
<lang powershell>$doors = @(0..99) for($i=0; $i -lt 100; $i++) {
$doors[$i] = 0 # start with all doors closed
} for($i=0; $i -lt 100; $i++) {
$step = $i + 1
for($j=$i; $j -lt 100; $j = $j + $step) {
$doors[$j] = $doors[$j] -bxor 1
}
} foreach($doornum in 1..100) {
if($doors[($doornum-1)] -eq $true) {"$doornum open"}
else {"$doornum closed"}
}</lang>
unoptimized Pipeline
<lang powershell>$doors = 1..100 | ForEach-Object {0} 1..100 | ForEach-Object { $a=$_;1..100 | Where-Object { -not ( $_ % $a ) } | ForEach-Object { $doors[$_-1] = $doors[$_-1] -bxor 1 }; if ( $doors[$a-1] ) { "door opened" } else { "door closed" } } </lang>
unoptimized Pipeline 2
<lang powershell>$doors = 1..100 | ForEach-Object {0} $visited = 1..100 1..100 | ForEach-Object { $a=$_;$visited[0..([math]::floor(100/$a)-1)] | Where-Object { -not ( $_ % $a ) } | ForEach-Object { $doors[$_-1] = $doors[$_-1] -bxor 1;$visited[$_/$a-1]+=($_/$a) }; if ( $doors[$a-1] ) { "door opened" } else { "door closed" } } </lang>
ProDOS
Uses math module. <lang ProDOS>enableextensions enabledelayedexpansion editvar /newvar /value=0 /title=closed editvar /newvar /value=1 /title=open editvar /newvar /range=1-100 /increment=1 /from=2 editvar /newvar /value=2 /title=next
- doors
for /alloccurrences (!next!-!102!) do editvar /modify /value=-open- editvar /modify /value=-next-=+1 if -next- /hasvalue=100 goto :cont else goto :doors
- cont
printline !1!-!102! stoptask</lang>
Prolog
unoptimized
<lang Prolog>doors_unoptimized(N) :- length(L, N), maplist(init, L), doors(N, N, L, L1), affiche(N, L1).
init(close).
doors(Max, 1, L, L1) :- !,
inverse(1, 1, Max, L, L1).
doors(Max, N, L, L1) :- N1 is N - 1, doors(Max, N1, L, L2), inverse(N, 1, Max, L2, L1).
inverse(N, Max, Max, [V], [V1]) :-
!,
0 =:= Max mod N -> inverse(V, V1); V1 = V.
inverse(N, M, Max, [V|T], [V1|T1]) :- M1 is M+1, inverse(N, M1, Max, T, T1), ( 0 =:= M mod N -> inverse(V, V1); V1 = V).
inverse(open, close).
inverse(close, open).
affiche(N, L) :- forall(between(1, N, I), ( nth1(I, L, open) -> format('Door ~w is open.~n', [I]); true)). </lang>
optimized
<lang Prolog>doors_optimized(N) :- Max is floor(sqrt(N)), forall(between(1, Max, I), ( J is I*I,format('Door ~w is open.~n',[J]))).
</lang>
PureBasic
unoptimized <lang purebasic>Dim doors.i(100)
For x = 1 To 100
y = x While y <= 100 doors(y) = 1 - doors(y) y + x Wend
Next
OpenConsole() PrintN("Following Doors are open:") For x = 1 To 100
If doors(x) Print(Str(x) + ", ") EndIf
Next Input()</lang>
optimized <lang PureBasic>OpenConsole() PrintN("Following Doors are open:") For i = 1 To 100
root.f = Sqr(i) If root = Int(root) Print (Str(i) + ", ") EndIf
Next Input()</lang>
Output:
Following Doors are open: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100,
Python
unoptimized <lang python>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'
</lang>
optimized
A version that only visits each door once:
<lang python>for i in xrange(1, 101):
root = i ** 0.5 print "Door %d:" % i, 'open' if root == int(root) else 'close'</lang>
One liner using a list comprehension, item lookup, and is_integer
<lang python>print '\n'.join(['Door %s is %s' % (i, ('closed', 'open')[(i**0.5).is_integer()]) for i in xrange(1, 101)])</lang>
One liner using a generator expression, ternary operator, and modulo
<lang python>print '\n'.join('Door %s is %s' % (i, 'closed' if i**0.5 % 1 else 'open') for i in range(1, 101))</lang>
<lang python> for i in list(range(1, 101)):
if i**0.5 % 1: state='open'
else: state='close'
print ("Door {}:{}".format(i, state))
</lang>
Q
unoptimized <lang q>`closed`open mod[;2]count each 1 _ group raze where each 0=t mod\:/:t:til 101</lang>
optimized <lang q>`closed`open (1+til 100) in `int$xexp[;2] 1+til 10</lang>
R
unoptimized <lang r>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()</lang>
optimized <lang r>## optimized version... we only have to to up to the square root of 100 seq(1,sqrt(100))**2</lang>
optimized <lang r>x <- rep(1, 100) for (i in 1:100-1) {
x <- xor(x, rep(c(rep(0,i),1), length.out=100))
} which(!x)</lang>
REALbasic
<lang realbasic>//True=Open; False=Closed
Dim doors(100) As Boolean //Booleans default to false
For j As Integer = 1 To 100
For i As Integer = 1 to 100
If i Mod j = 0 Then
doors(i) = Not doors(i)
End If
Next
Next</lang>
REBOL
Unoptimized
<lang rebol>doors: array/initial 100 'closed repeat i 100 [
door: at doors i forskip door i [change door either 'open = first door ['closed] ['open]]
]</lang>
Optimized
<lang rebol>doors: array/initial 100 'closed repeat i 10 [doors/(i * i): 'open] </lang>
Retro
<lang Retro>: squared ( n-n ) dup * ;
- doors ( n- ) [ 1 repeat 2over squared > 0; drop dup squared putn space 1+ again ] do 2drop ;
100 doors</lang>
REXX
<lang rexx>door. = 0 do inc = 1 to 100
do d = inc to 100 by inc door.d = \door.d end
end say "The open doors after 100 passes:" do i = 1 to 100
if door.i = 1 then say i
end </lang> Here is another version, solving it the hard way.
<lang rexx> /*REXX program to solve the 100 door puzzle, the hard-way version. */
parse arg doors . /*get the first argument (# of doors.) */ if doors== then doors=100 /*not specified? Then assume 100 doors*/
/* 0 = closed. */
/* 1 = open. */
door.=0 /*assume all that all doors are closed.*/
do j=1 for doors /*process a pass-through for all doors.*/ do k=j by j to doors /* ... every Jth door from this point. */ door.k=\door.k /*toggle the "openness" of the door. */ end end
say say 'After' doors "passes, the following doors are open:" say
do n=1 for doors if door.n then say right(n,20) end
say </lang> Output:
After 100 passes, the following doors are open:
1
4
9
16
25
36
49
64
81
100
Here is another version, solving it the easy way (version 1). <lang rexx> /*REXX program to solve the 100 door puzzle, the easy-way version. */
parse arg doors . /*get the first argument (# of doors.) */ if doors== then doors=100 /*not specified? Then assume 100 doors*/
/* 0 = closed. */
/* 1 = open. */
door.=0 /*assume all that all doors are closed.*/ say say 'For the' doors "doors problem, the following doors are open:" say
do j=1 for doors /*process an easy pass-through. */
p=j*j /*square the door number. */
/*An alternative: P=J**2 */
if p>doors then leave /*if too large, we're done. */
say right(p,20)
end
say </lang> Output:
For the 100 doors problem, the following doors are open:
1
4
9
16
25
36
49
64
81
100
Here is another easy-way solution (version 2), but for 1,000 doors. <lang rexx> /*REXX program to solve the 100 door puzzle, the easy-way version 2.*/
parse arg doors . /*get the first argument (# of doors.) */ if doors== then doors=100 /*not specified? Then assume 100 doors*/
doors=1000
/* 0 = closed. */
/* 1 = open. */
door.=0 /*assume all that all doors are closed.*/ say say 'For the' doors "doors problem, the open doors are:" say
do j=1 for doors while j*j<=doors /*limit the pass-throughs. */ say right(j**2,20) end
say </lang> Output:
For the 1000 doors problem, the open doors are:
1
4
9
16
25
36
49
64
81
100
121
144
169
196
225
256
289
324
361
400
441
484
529
576
625
676
729
784
841
900
961
Ruby
unoptimized; Ruby-way
(I tried to show as much of Ruby syntax and conventions as possible. Of course, instead of a class you could just use generic true/false, but that's not the point, is it?)
<lang ruby>class Door
attr_reader :state
def initialize
@state=:closed
end
def close; @state=:closed; end def open; @state=:open; end
def closed?; @state==:closed; end def open?; @state==:open; end
def toggle if closed? open else close end end
def to_s; @state.to_s; end end
doors=Array.new(100){Door.new} 1.upto(100) do |multiplier| doors.each_with_index do |door, i| door.toggle if (i+1)%multiplier==0 end end
doors.each_with_index{|door, i| puts "Door #{i+1} is #{door}."}</lang>
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 (1..n).each do |i|
puts "Door #{i} is #{i**0.5 == (i**0.5).round ? "open" : "closed"}"
end </lang>
generic true/false, with another way of handling the inner loop demonstrating Range#step <lang ruby> doors = [false] * 100 100.times do |i|
(i .. doors.length).step(i+1) do |j| doors[j] = !doors[j] end
end puts doors.inspect
</lang>
Run BASIC
<lang Runbasic>dim doors(100) print "Open doors "; for i = 1 to 100
for door = i to 100 step i
doors(door) = (doors(door) <> 1)
if i = door and doors(door) = 1 then print i;" ";
next door
next i</lang>Output:
Open doors 1 4 9 16 25 36 49 64 81 100
S-lang
<lang s-lang>variable door,
isOpen = Char_Type [101], pass;
for (door = 1; door <= 100; door++) {
isOpen[door] = 0;
}
for (pass = 1; pass <= 100; pass++) {
for (door = pass; door <= 100; door += pass) {
isOpen[door] = not isOpen[door];
}
}
for (door = 1; door <= 100; door++) {
if (isOpen[door]) {
print("Door " + string(door) + ":open");
} else {
print("Door " + string(door) + ":close");
}
}</lang>
Salmon
Here's an unoptimized version: <lang Salmon>variable open := <<(* --> false)>>; for (pass; 1; pass <= 100)
for (door_num; pass; door_num <= 100; pass)
open[door_num] := !(open[door_num]);;;
iterate (door_num; [1...100])
print("Door ", door_num, " is ",
(open[door_num] ? "open.\n" : "closed.\n"));;</lang>
And here's an optimized one-line version:
<lang Salmon>iterate (x; [1...10]) { iterate (y; [(x-1)*(x-1)+1...x*x-1]) { print("Door ", y, " is closed.\n"); }; print("Door ", x*x, " is open.\n"); };</lang>
And a shorter optimized one-line version:
<lang Salmon>variable y:=1;for(x;1;x<101)"Door "~sprint(x)~" is "~(x==y*y?{++y;return"open";}:"closed")!;</lang>
Sather
<lang sather>class MAIN is
main is
pass, door :INT;
doors :ARRAY{BOOL} := #(100);
loop
doors[0.upto!(99)] := false;
end;
pass := 0;
loop while!(pass < 100);
door := pass;
loop while! (door < 100);
doors[door] := ~doors[door];
door := door + pass + 1
end;
pass := pass + 1;
end;
loop
door := 0.upto!(99);
#OUT + (door+1) + " " + doors[door] + "\n";
end;
end;
end;</lang>
SAS
<lang sas>data _null_;
open=1;
close=0;
array Door{100};
do Pass = 1 to 100;
do Current = Pass to 100 by Pass;
if Door{Current} ne open
then Door{Current} = open;
else Door{Current} = close;
end;
end;
NumberOfOpenDoors = sum(of Door{*});
put "Number of Open Doors: " NumberOfOpenDoors;
run;</lang>
Scala
<lang scala>for { i <- 1 to 100
r = 1 to 100 map (i % _ == 0) reduceLeft (_^_) } println (i +" "+ (if (r) "open" else "closed"))</lang>
The map operation maps each door (i) to a boolean sequence of toggles, one for each pass: true toggles, false leaves the same.
The reduceLeft method combines all the toggles sequentially, using the XOR operator.
And then we just need to output the result.
"Optimized" version: <lang scala>val o = 1 to 10 map (i => i * i) println("open: " + o) println("closed: " + (1 to 100 filterNot o.contains))</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>
(moved from the Racket language entry, may be redundant)
<lang scheme>#lang racket
- Like "map", but the proc must take an index as well as the element.
(define (map-index proc seq)
(for/list ([(elt i) (in-indexed seq)]) (proc elt i)))
- Applies PROC to every STEPth element of SEQ, leaving the others
- unchanged.
(define (map-step proc step seq)
(map-index
(lambda (elt i)
((if (zero? (remainder i step) )
proc
values) elt))
seq))
(define (toggle-nth n seq)
(map-step not n seq))
(define (solve seq)
(for/fold ([result seq])
([(_ pass) (in-indexed seq)])
(toggle-nth (add1 pass) result)))
(for ([(door index) (in-indexed (solve (make-vector 100 #f)))])
(when door (printf "~a is open~%" index)))</lang>
optimized:
<lang scheme>#lang racket
(for-each (lambda (x) (printf "~a is open\n" x))
(filter (lambda (x)
(exact-integer? (sqrt x)))
(sequence->list (in-range 1 101))))</lang>
Seed7
unoptimized <lang seed7>$ include "seed7_05.s7i";
const proc: main is func
local
var array boolean: doorOpen is 100 times FALSE;
var integer: pass is 0;
var integer: index is 0;
var array[boolean] string: closedOrOpen is [boolean] ("closed", "open");
begin
for pass range 1 to 100 do
for key index range doorOpen do
if index rem pass = 0 then
doorOpen[index] := not doorOpen[index];
end if;
end for;
end for;
for key index range doorOpen do
write(index lpad 3 <& " is " <& closedOrOpen[doorOpen[index]] rpad 7);
if index rem 5 = 0 then
writeln;
end if;
end for;
end func;</lang>
optimized <lang seed7>$ include "seed7_05.s7i";
const proc: main is func
local
var integer: index is 0;
var integer: number is 0;
var array[boolean] string: closedOrOpen is [boolean] ("closed", "open");
begin
for index range 1 to 100 do
number := sqrt(index);
write(index lpad 3 <& " is " <& closedOrOpen[number**2 = index] rpad 7);
if index rem 5 = 0 then
writeln;
end if;
end for;
end func;</lang>
Output of both programs:
1 is open 2 is closed 3 is closed 4 is open 5 is closed 6 is closed 7 is closed 8 is closed 9 is open 10 is closed 11 is closed 12 is closed 13 is closed 14 is closed 15 is closed 16 is open 17 is closed 18 is closed 19 is closed 20 is closed 21 is closed 22 is closed 23 is closed 24 is closed 25 is open 26 is closed 27 is closed 28 is closed 29 is closed 30 is closed 31 is closed 32 is closed 33 is closed 34 is closed 35 is closed 36 is open 37 is closed 38 is closed 39 is closed 40 is closed 41 is closed 42 is closed 43 is closed 44 is closed 45 is closed 46 is closed 47 is closed 48 is closed 49 is open 50 is closed 51 is closed 52 is closed 53 is closed 54 is closed 55 is closed 56 is closed 57 is closed 58 is closed 59 is closed 60 is closed 61 is closed 62 is closed 63 is closed 64 is open 65 is closed 66 is closed 67 is closed 68 is closed 69 is closed 70 is closed 71 is closed 72 is closed 73 is closed 74 is closed 75 is closed 76 is closed 77 is closed 78 is closed 79 is closed 80 is closed 81 is open 82 is closed 83 is closed 84 is closed 85 is closed 86 is closed 87 is closed 88 is closed 89 is closed 90 is closed 91 is closed 92 is closed 93 is closed 94 is closed 95 is closed 96 is closed 97 is closed 98 is closed 99 is closed 100 is open
Slate
Unoptimized <lang slate>define: #a -> (Array newSize: 100). a infect: [| :_ | False].
a keysDo: [| :pass |
pass to: a indexLast by: pass do: [| :door | a at: door infect: #not `er]].
a keysAndValuesDo: [| :door :isOpen |
inform: 'door #' ; door ; ' is ' ; (isOpen ifTrue: ['open'] ifFalse: ['closed'])].</lang>
Optimized <lang slate>define: #a -> (Array newSize: 100). a infect: [| :_ | False].
0 below: 10 do: [| :door | a at: door squared put: True]. a keysAndValuesDo: [| :door :isOpen |
inform: 'door #' ; door ; ' is ' ; (isOpen ifTrue: ['open'] ifFalse: ['closed'])].</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>
Unoptimized, using Morphs <lang smalltalk> | m w h smh smw delay closedDoor border subMorphList |
closedDoor := Color black. border := Color veryLightGray. delay := Delay forMilliseconds: 50. w := World bounds corner x. h := (World bounds corner y) / 2. smw := w/100. smh := h/2.
m := BorderedMorph new position: 0@h. m height: smh; width: w; borderColor: border. m color: Color veryLightGray.
1 to: 100 do: [ :pos || sm | sm := BorderedMorph new height: smh ; width: smw ; borderColor: border; color: closedDoor; position: (smw*pos)@h. m addMorph: sm asElementNumber: pos].
m openInWorld. delay wait. subMorphList := m submorphs. "display every step" [1 to: 100 do: [ :step | step to: 100 by: step do: [ :pos | | subMorph | subMorph := subMorphList at: pos. subMorph color: subMorph color negated. delay wait]]] fork. </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>
SNOBOL4
unoptimized <lang snobol4> DEFINE('PASS(A,I),O') :(PASS.END) PASS O = 0 PASS.LOOP O = O + I EQ(A<O>,1) :S(PASS.1)F(PASS.0) PASS.0 A<O> = 1 :S(PASS.LOOP)F(RETURN) PASS.1 A<O> = 0 :S(PASS.LOOP)F(RETURN) PASS.END
MAIN D = ARRAY(100,0) I = 0
MAIN.LOOP I = LE(I,100) I + 1 :F(OUTPUT) PASS(D,I) :(MAIN.LOOP)
OUTPUT I = 1 ; OPEN = 'Opened doors are: ' OUTPUT.LOOP OPEN = OPEN EQ(D,1) " " I I = LE(I,100) I + 1 :S(OUTPUT.LOOP)F(OUTPUT.WRITE) OUTPUT.WRITE OUTPUT = OPEN
END </lang>
A run of this using CSNOBOL4 looks like this:
$ snobol4 100doors.sno
The Macro Implementation of SNOBOL4 in C (CSNOBOL4) Version 1.3+
by Philip L. Budne, January 23, 2011
SNOBOL4 (Version 3.11, May 19, 1975)
Bell Telephone Laboratories, Incorporated
No errors detected in source program
Opened doors are: 1 4 9 16 25 36 49 64 81 100
Normal termination at level 0
100doors.sno:18: Last statement executed was 19
(There are command flags to remove the header and the summary, but these have been left in to keep the original SNOBOL4 experience intact.)
optimized <lang snobol4> MAIN D = ARRAY(100,0) I = 1
MAIN.LOOP LE(I, 10) :F(OUTPUT) D = 1 I = I + 1 :(MAIN.LOOP)
OUTPUT I = 1 ; O = 'Opened doors are: ' OUTPUT.LOOP O = O EQ(D,1) " " I I = LE(I,100) I + 1 :S(OUTPUT.LOOP)F(OUTPUT.WRITE) OUTPUT.WRITE OUTPUT = O END </lang>
The output of this version is almost identical to the above.
Tcl
unoptimized
<lang tcl>package require Tcl 8.5 set n 100 set doors [concat - [lrepeat $n 0]] for {set step 1} {$step <= $n} {incr step} {
for {set i $step} {$i <= $n} {incr i $step} {
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 $i] ? "open" : "closed"}]]
}</lang>
optimized
<lang tcl>package require Tcl 8.5 set doors [lrepeat [expr {$n + 1}] closed] for {set i 1} {$i <= sqrt($n)} {incr i} {
lset doors [expr {$i ** 2}] open
} for {set i 1} {$i <= $n} {incr i} {
puts [format "door %d is %s" $i [lindex $doors $i]]
}</lang>
graphical
Inspired by the E solution, here's a visual representation <lang tcl>package require Tcl 8.5 package require Tk
array set door_status {}
- create the gui
set doors [list x] for {set i 0} {$i < 10} {incr i} {
for {set j 0} {$j < 10} {incr j} {
set k [expr {1 + $j + 10*$i}]
lappend doors [radiobutton .d_$k -text $k -variable door_status($k) \
-indicatoron no -offrelief flat -width 3 -value open]
grid [lindex $doors $k] -column $j -row $i
}
}
- create the controls
button .start -command go -text Start label .i_label -text " door:" entry .i -textvariable i -width 4 label .step_label -text " step:" entry .step -textvariable step -width 4 grid .start - .i_label - .i - .step_label - .step - -row $i grid configure .start -sticky ew grid configure .i_label .step_label -sticky e grid configure .i .step -sticky w
proc go {} {
global doors door_status i step
# initialize the door_status (all closed)
for {set d 1} {$d <= 100} {incr d} {
set door_status($d) closed
}
# now, begin opening and closing
for {set step 1} {$step <= 100} {incr step} {
for {set i 1} {$i <= 100} {incr i} {
if {$i % $step == 0} {
[lindex $doors $i] [expr {$door_status($i) eq "open" ? "deselect" : "select"}]
update
after 50
}
}
}
}</lang>
TI-83 BASIC
Unoptimized
<lang ti83b>PROGRAM:DOORS100
- ClrHome
- Disp "SETTING UP LIST"
- Disp "PLEASE WAIT..."
- For(I,1,100,1)
- 0→L1(I)
- End
- ClrHome
- Disp "Pass"
- For(I,1,100,1)
- For(J,I,100,I)
- Output(2,1,I)
- not(L1(J))→L1(J)
- End
- End
- ClrHome
</lang>
Optimized
<lang ti83b>PROGRAM:DOORSOPT
- For(I,1,100,1)
- not(fPart(√(I)))→L1(I)
- End
</lang>
TI-89 BASIC
<lang ti89b>Define doors(fast) = Func
Local doors,i,j
seq(false,x,1,100) → doors
If fast Then
For i,1,10,1
true → doors[i^2]
EndFor
Else
For i,1,100,1
For j,i,100,i
not doors[j] → doors[j]
EndFor
EndFor
EndIf
Return doors
EndFunc</lang>
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT DICT doors create COMPILE LOOP door=1,100
LOOP pass=1,100
SET go=MOD (door,pass)
DICT doors lookup door,num,cnt,status
IF (num==0) THEN
SET status="open"
DICT doors add door,num,cnt,status
ELSE
IF (go==0) THEN
IF (status=="closed") THEN
SET status="open"
ELSE
SET status="closed"
ENDIF
DICT doors update door,num,cnt,status
ENDIF
ENDIF
ENDLOOP
ENDLOOP ENDCOMPILE DICT doors unload door,num,cnt,status </lang> Output (variable status):
status = *
1 = open
2 = closed
3 = closed
4 = open
5 = closed
6 = closed
7 = closed
8 = closed
9 = open
10 = closed
11 = closed
12 = closed
13 = closed
14 = closed
15 = closed
16 = open
17 = closed
18 = closed
19 = closed
20 = closed
21 = closed
22 = closed
23 = closed
24 = closed
25 = open
26 = closed
27 = closed
28 = closed
29 = closed
30 = closed
31 = closed
32 = closed
33 = closed
34 = closed
35 = closed
36 = open
37 = closed
38 = closed
39 = closed
40 = closed
41 = closed
42 = closed
43 = closed
44 = closed
45 = closed
46 = closed
47 = closed
48 = closed
49 = open
50 = closed
51 = closed
52 = closed
53 = closed
54 = closed
55 = closed
56 = closed
57 = closed
58 = closed
59 = closed
60 = closed
61 = closed
62 = closed
63 = closed
64 = open
65 = closed
66 = closed
67 = closed
68 = closed
69 = closed
70 = closed
71 = closed
72 = closed
73 = closed
74 = closed
75 = closed
76 = closed
77 = closed
78 = closed
79 = closed
80 = closed
81 = open
82 = closed
83 = closed
84 = closed
85 = closed
86 = closed
87 = closed
88 = closed
89 = closed
90 = closed
91 = closed
92 = closed
93 = closed
94 = closed
95 = closed
96 = closed
97 = closed
98 = closed
99 = closed
100 = open
TXR
<lang txr>@(do (defun 100-doors ()
(let ((doors (vector 100)))
(each ((i (range 0 99)))
(for ((j i)) ((< j 100)) ((inc j (+ i 1)))
(flip [doors j])))
doors))
(each ((counter (range 1))
(door (list-vector (100-doors))))
(format t "door ~a is ~a\n" counter (if door "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 $op
done</lang>
Optimised version <lang bash>#!/bin/bash
for i in {1..100}; do
door[$i*$i]=1
[ -z ${door[$i]} ] && echo "$i closed" || echo "$i open"
done</lang>
Ursala
The doors are represented as a list of 100 booleans initialized to false. The pass function takes a number and a door list to a door list with doors toggled at indices that are multiples of the number. The main program folds the pass function (to the right) over the list of pass numbers from 100 down to 1, numbers the result, and filters out the numbers of the open doors. <lang Ursala>#import std
- import nat
doors = 0!* iota 100
pass("n","d") = remainder\"n"?l(~&r,not ~&r)* num "d"
- cast %nL
main = ~&rFlS num pass=>doors nrange(100,1)</lang> optimized version: <lang Ursala>#import nat
- cast %nL
main = product*tiiXS iota10</lang> output:
<1,4,9,16,25,36,49,64,81>
VBScript
Unoptimized <lang VBScript>Dim doorIsOpen(100), pass, currentDoor, text
For currentDoor = 0 To 99 doorIsOpen(currentDoor) = False Next
For pass = 0 To 99 For currentDoor = pass To 99 Step pass + 1 doorIsOpen(currentDoor) = Not doorIsOpen(currentDoor) Next Next
For currentDoor = 0 To 99 text = "Door #" & currentDoor + 1 & " is " If doorIsOpen(currentDoor) Then text = text & "open." Else text = text & "closed." End If WScript.Echo(text) Next</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
VHDL
unoptimized <lang vhdl>library IEEE; use IEEE.STD_LOGIC_1164.ALL;
entity DOORS is port (CLK: in std_logic; OUTPUT: out std_logic_vector(1 to 100)); end DOORS;
architecture Behavioral of DOORS is begin process (CLK) variable TEMP: std_logic_vector(1 to 100); begin --setup closed doors TEMP := (others => '0');
--looping through for i in 1 to TEMP'length loop for j in i to TEMP'length loop if (j mod i) = 0 then TEMP(j) := not TEMP(j); end if; end loop; end loop;
--assign output OUTPUT <= TEMP; end process; end Behavioral; </lang>
Visual Basic .NET
unoptimized <lang vbnet>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</lang> optimized <lang vbnet>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</lang>
Wrapl
Unoptimized <lang wrapl>MOD Doors;
IMP Agg.Table; IMP Std.String; IMP IO.Terminal USE Out;
VAR door <- {}; EVERY door[1:to(100), "closed"];
DEF toggle(num) door[num] <- door[num] = "open" => "closed" // "open";
EVERY WITH pass <- 1:to(100), num <- pass:to(100, pass) DO toggle(num);
Out:write('Doors {door @ String.T}.');
END Doors.</lang> Optimized <lang wrapl>MOD Doors;
IMP IO.Terminal USE Out;
DEF open <- ALL 1:to(100) ^ 2 \ $ <= 100; DEF closed <- ALL 1:to(100) \ NOT $ IN open;
Out:write('Doors {open} are open.\n'); Out:write('Doors {closed} are closed.\n');
END Doors.</lang>
XPL0
<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations int Door(100); \You have 100 doors in a row define Open, Closed; int D, Pass, Step;
[for D:= 0 to 100-1 do \that are all initially closed
Door(D):= Closed;
Step:= 1; \The first time through, you visit every door for Pass:= 1 to 100 do \You make 100 passes by the doors
[D:= Step-1;
repeat \if the door is closed, you open it; if it is open, you close it
if Door(D)=Closed then Door(D):= Open else Door(D):= Closed;
D:= D+Step;
until D>=100;
Step:= Step+1; \The second time you only visit every 2nd door
]; \The third time, every 3rd door
\until you only visit the 100th door
\What state are the doors in after the last pass? Text(0, "Open: "); \Which are open? for D:= 0 to 100-1 do
if Door(D)=Open then [IntOut(0, D+1); ChOut(0,^ )];
CrLf(0);
Text(0, "Closed: "); \Which are closed? for D:= 0 to 100-1 do
if Door(D)=Closed then [IntOut(0, D+1); ChOut(0,^ )];
CrLf(0);
\Optimized: The only doors that remain open are those that are perfect squares Text(0, "Open: "); D:= 1; repeat IntOut(0, D*D); ChOut(0,^ );
D:= D+1;
until D*D>100; CrLf(0); ]</lang>
XSLT
See: 100 doors/XSLT
Yorick
Unoptimized, iterative <lang yorick>doors = array(0, 100); for(i = 1; i <= 100; i++)
for(j = i; j <= 100; j += i)
doors(j) ~= 1;
print, where(doors);</lang>
Unoptimized, vectorized <lang yorick>doors = array(0, 100); for(i = 1; i <= 100; i++)
doors(i::i) ~= 1;
print, where(doors);</lang>
Optimized <lang yorick>print, indgen(1:long(sqrt(100)))^2</lang>
All of the above output:
[1,4,9,16,25,36,49,64,81,100]
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