100 doors
Problem: You have 100 doors in a row that are all initially closed. You make 100 passes by the doors, starting with the first door every time. The first time through, you visit every door and toggle the door (if the door is closed, you open it; if it is open, you close it). The second time you only visit every 2nd door (door #2, #4, #6, ...). The third time, every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Question: What state are the doors in after the last pass? Which are open, which are closed? [1]
Alternate: As noted in this page's discussion page, the only doors that remain open are whose numbers are perfect squares of integers. Opening only those doors is an optimization that may also be expressed.
Ada
unoptimized
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;
optimized
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;
MAXScript
unoptimized
doorsOpen = for i in 1 to 100 collect false for pass in 1 to 100 do ( for door in pass to 100 by pass do ( doorsOpen[door] = not doorsOpen[door] ) ) for i in 1 to doorsOpen.count do ( format ("Door % is open?: %\n") i doorsOpen[i] )
optimized
for i in 1 to 100 do ( root = pow i 0.5 format ("Door % is open?: %\n") i (root == (root as integer)) )
Perl
unoptimized
my @doors; for my $pass (1..100) { for (1..100) { if (0 == $_ % $pass) { if (1 == $doors[$_]) { $doors[$_] = 0; } else { $doors[$_] = 1; }; }; }; }; print "$_\t$doors[$_]\n" for 1..100;
optimized
while( ++$i <= 100 ) { $root = sqrt($i); if ( int( $root ) == $root ) { print "Door $i is open\n"; } else { print "Door $i is closed\n"; } }
Pop11
unoptimized
lvars i; lvars doors = {% for i from 1 to 100 do false endfor %}; for i from 1 to 100 do for j from i by i to 100 do not(doors(j)) -> doors(j); endfor; endfor; ;;; Print state for i from 1 to 100 do printf('Door ' >< i >< ' is ' >< if doors(i) then 'open' else 'closed' endif, '%s\n'); endfor;
optimized
lvars i, ri, rri, s; for i from 1 to 100 do sqrt(i) -> ri; round(ri) -> rri; if rri = ri then 'open' else 'closed' endif -> s; printf('Door ' >< i >< ' is ' >< s, '%s\n'); endfor;
Python
Note: both versions require Python 2.5 for the ternary syntax.
unoptimized
close = 0 open = 1 doors = [close] * 100 for i in range(100): for j in range(i, 100, i+1): doors[j] = open if doors[j] is close else close for k, door in enumerate(doors): print "Door %d:" % (k+1), 'open' if door else 'close'
optimized
A version that only visits each door once:
for i in xrange(1, 101): root = i ** 0.5 print "Door %d:" % i, 'open' if root == int(root) else 'close'