Comments on: DX. Dumb Robots

By: Eric

Eric — Mon, 19 May 2008 05:18:02 +0000

For an added challenge, it turns out you can solve the puzzle using only one direction of travel.

By: jpincott

jpincott — Mon, 19 May 2008 04:34:37 +0000

For the robot puzzle, if we label the commands thus:

L = Move left one step
R = Move right one step
Gn = Goto instruction n
Cn = Conditional goto - Gn iff currently standing on a parachute

then the following program will suffice:

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By: nklein

nklein — Mon, 19 May 2008 04:12:12 +0000

DUMB ROBOTS SPOILER

The shortest algorithm I am sure of has 7 instructions.
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I had assumed instruction time was negligible compared to moving time in arriving at the algorithm. If every instruction takes equal time, we can eliminate 3 and 4. Both robots move leftward one space every three ticks until the rightmost robot finds the other parachute. Then, the rightmost robot accelerates to one square leftward every two ticks.

By: cstarbi

cstarbi — Mon, 19 May 2008 02:07:10 +0000

Something that I found interesting: Find out if this code is the fastest way to cause a collision, and, if not, what code is? (I assumed in my solution that travel time is much much longer than the time it takes to detect a parachute or perform a goto)

By: strauss

strauss — Fri, 16 May 2008 11:12:42 +0000

(* Man, 27 by hand is quite a treat! Here’s the same result in Mathematica; I love the compactness of this language! *)

Collatz[1] = 1;

Collatz[n_] := If[EvenQ[n], n/2, 3 n + 1];

DoCollatz[n_] := (Print[Length[#] - 1, ” steps: “, #, “\n”]) &@ FixedPointList[Collatz, n]

(* so typing in DoCollatz[27] spits out the sequence starting at 27;

DoCollatz /@ Range[1,200] spits out results for 1 through 200 *)

By: cybersekkin

cybersekkin — Fri, 16 May 2008 05:05:21 +0000

#I (being a dumb chemical robot) made an error going by hand
#so put this together as a simple test for a single number (27))

#!/usr/bin/perl
#
#
my $num = 27;
my $count = 0;
my $test3 = 0;

until( $test3 == 1 )
{
print “$count number is: $num\n”;

if($num eq 1)
{
$test3 = 1;
}
#get last digit if even divide by 2
if(!($num % 2))
{
$num = $num / 2;
}
#if not multiply by 3 and add 1
else
{
$num = (($num * 3) + 1)
}
$count++;
}

By: intchanter

intchanter — Wed, 14 May 2008 22:16:24 +0000

I whipped up some Python code that any of you can hack on to try out the Collatz function. By posting it here, I am releasing it into the public domain, so have fun with it.

————-
#!/usr/bin/env python

def collatz(start):
    '''
    Run through the collatz series for a given number and return the number of
    iterations and the largest number found.
    '''
    current = start
    iter = 0
    max = start
    while current != 1:
        iter += 1
        if current % 2: # odd
            current = current * 3 + 1
        else: # even
            current = current / 2
        if current > max:
            max = current
    return (iter, max)

if __name__ == "__main__":
    target = 1000
    max_tuple = (0, 0, 0)
    iter_tuple = (0, 0, 0)
    for start in xrange(1, target + 1):
        (iter, max) = collatz(start)
        if max > max_tuple[2]:
            max_tuple = (start, iter, max)
        if iter > iter_tuple[1]:
            iter_tuple = (start, iter, max)
    print 'Largest integer hit:', max_tuple
    print 'Longest series hit:', iter_tuple
————-