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day12-2.p
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day12-2.p
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/*
--- Part Two ---
All this drifting around in space makes you wonder about the nature of the universe. Does history really repeat itself? You're curious whether the moons will ever return to a previous state.
Determine the number of steps that must occur before all of the moons' positions and velocities exactly match a previous point in time.
For example, the first example above takes 2772 steps before they exactly match a previous point in time; it eventually returns to the initial state:
After 0 steps:
pos=<x= -1, y= 0, z= 2>, vel=<x= 0, y= 0, z= 0>
pos=<x= 2, y=-10, z= -7>, vel=<x= 0, y= 0, z= 0>
pos=<x= 4, y= -8, z= 8>, vel=<x= 0, y= 0, z= 0>
pos=<x= 3, y= 5, z= -1>, vel=<x= 0, y= 0, z= 0>
After 2770 steps:
pos=<x= 2, y= -1, z= 1>, vel=<x= -3, y= 2, z= 2>
pos=<x= 3, y= -7, z= -4>, vel=<x= 2, y= -5, z= -6>
pos=<x= 1, y= -7, z= 5>, vel=<x= 0, y= -3, z= 6>
pos=<x= 2, y= 2, z= 0>, vel=<x= 1, y= 6, z= -2>
After 2771 steps:
pos=<x= -1, y= 0, z= 2>, vel=<x= -3, y= 1, z= 1>
pos=<x= 2, y=-10, z= -7>, vel=<x= -1, y= -3, z= -3>
pos=<x= 4, y= -8, z= 8>, vel=<x= 3, y= -1, z= 3>
pos=<x= 3, y= 5, z= -1>, vel=<x= 1, y= 3, z= -1>
After 2772 steps:
pos=<x= -1, y= 0, z= 2>, vel=<x= 0, y= 0, z= 0>
pos=<x= 2, y=-10, z= -7>, vel=<x= 0, y= 0, z= 0>
pos=<x= 4, y= -8, z= 8>, vel=<x= 0, y= 0, z= 0>
pos=<x= 3, y= 5, z= -1>, vel=<x= 0, y= 0, z= 0>
Of course, the universe might last for a very long time before repeating. Here's a copy of the second example from above:
<x=-8, y=-10, z=0>
<x=5, y=5, z=10>
<x=2, y=-7, z=3>
<x=9, y=-8, z=-3>
This set of initial positions takes 4686774924 steps before it repeats a previous state! Clearly, you might need to find a more efficient way to simulate the universe.
How many steps does it take to reach the first state that exactly matches a previous state?
*/
USING System.*.
/*
<x=-4, y=-9, z=-3>
<x=-13, y=-11, z=0 >
<x=-17, y=-7, z=15>
<x=-16, y=4, z=2 >
*/
DEFINE VARIABLE i AS INTEGER NO-UNDO.
DEFINE VARIABLE iStep AS INTEGER NO-UNDO.
DEFINE VARIABLE iVX AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iVY AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iVZ AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iX AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iXInit AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iY AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iYInit AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iZ AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE iZInit AS INTEGER EXTENT 4 NO-UNDO.
DEFINE VARIABLE j AS INTEGER NO-UNDO.
ETIME(YES).
ASSIGN
iX[1] = -4
iY[1] = -9
iZ[1] = -3
iX[2] = -13
iY[2] = -11
iZ[2] = 0
iX[3] = -17
iY[3] = -7
iZ[3] = 15
iX[4] = -16
iY[4] = 4
iZ[4] = 2
.
ASSIGN
iXInit = iX
iYInit = iY
iZInit = iZ.
FUNCTION applyGravity RETURNS LOGICAL ( i AS INTEGER, j AS INTEGER ):
IF iX[i] < iX[j] THEN ASSIGN
iVX[i] = iVX[i] + 1
iVX[j] = iVX[j] - 1.
ELSE IF iX[i] > iX[j] THEN ASSIGN
iVX[i] = iVX[i] - 1
iVX[j] = iVX[j] + 1.
IF iY[i] < iY[j] THEN ASSIGN
iVY[i] = iVY[i] + 1
iVY[j] = iVY[j] - 1.
ELSE IF iY[i] > iY[j] THEN ASSIGN
iVY[i] = iVY[i] - 1
iVY[j] = iVY[j] + 1.
IF iZ[i] < iZ[j] THEN ASSIGN
iVZ[i] = iVZ[i] + 1
iVZ[j] = iVZ[j] - 1.
ELSE IF iZ[i] > iZ[j] THEN ASSIGN
iVZ[i] = iVZ[i] - 1
iVZ[j] = iVZ[j] + 1.
END FUNCTION.
DO WHILE TRUE:
iStep = iStep + 1.
DO i = 1 TO 3:
DO j = i + 1 TO 4:
applyGravity(i, j).
END.
END.
DO i = 1 TO 4:
ASSIGN
iX[i] = iX[i] + iVX[i]
iY[i] = iY[i] + iVY[i]
iZ[i] = iZ[i] + iVZ[i].
END.
/* check each coordinate period (i.e. how many steps to get the same position and speed = 0) */
DEFINE VARIABLE iPeriodX AS INTEGER NO-UNDO.
DEFINE VARIABLE iPeriodY AS INTEGER NO-UNDO.
DEFINE VARIABLE iPeriodZ AS INTEGER NO-UNDO.
DEFINE VARIABLE lOKx AS LOGICAL NO-UNDO.
DEFINE VARIABLE lOKy AS LOGICAL NO-UNDO.
DEFINE VARIABLE lOKz AS LOGICAL NO-UNDO.
lOKx = YES.
lOKy = YES.
lOKz = YES.
DO i = 1 TO 4:
IF iPeriodX = 0 THEN
lOKx = lOKx AND iX[i] = iXInit[i] AND iVX[i] = 0.
IF iPeriodY = 0 THEN
lOKy = lOKy AND iY[i] = iYInit[i] AND iVY[i] = 0.
IF iPeriodZ = 0 THEN
lOKz = lOKz AND iZ[i] = iZInit[i] AND iVZ[i] = 0.
END.
IF iPeriodX = 0 AND lokx THEN iPeriodX = iStep.
IF iPeriodY = 0 AND loky THEN iPeriodY = iStep.
IF iPeriodZ = 0 AND lokz THEN iPeriodZ = iStep.
IF iPeriodX > 0 AND iPeriodY > 0 AND iPeriodZ > 0 THEN
LEAVE.
END.
/* The result is the LCM of the 3 periods */
function gcd returns int64 (x as int64, y as int64):
if y = 0 then return x.
return gcd(y, x mod y).
end function.
function lcm returns int64 (x as int64, y as int64):
return int64((x * y) / gcd(x, y)).
end function.
MESSAGE ETIME SKIP "LCM(" iPeriodX "," iPeriodY "," iPeriodZ ") =" lcm(iPeriodX, lcm(iPeriodY, iPeriodZ))
VIEW-AS ALERT-BOX INFO BUTTONS OK.
/*
---------------------------
Information (Press HELP to view stack trace)
---------------------------
13084
LCM( 113028 , 167624 , 231614 ) = 548525804273976
---------------------------
Aceptar Ayuda
---------------------------
*/