CaptainEd

Gavin, mature reflection suggests to me that my assumptions ( 1 ) and ( 2 ) above are good assumptions, that make an interesting, challenging puzzle, but that ( 3 ) makes it impossible. ( 1 ) makes it possible to keep some information about each flag; ( 2 ) if you had infinite precision arithmetic, then one number could act as all of memory, large enough to contain a map of the entire maze, so restricting to finite sized numbers makes the puzzle interesting; ( 3 ) If the robot could leave breadcrumbs (Tremaux's algorithm, noted by DejMar), then you could imagine a strategy in which the robot uses the maze itself as the unbounded memory, tallying flags that have no breadcrumbs on them, and figuring out a way to determine that it has seen all of the boundary and all of the insides. So, did you somehow accidentally leave out one of the rules (I doubt it, as you a careful guy, but I can hope...), and the robot IS permitted to leave more than just a flag in a maze cell?

More of the A cases are ( ) ( ) than are ( ( ) )

Here are the rest of the cases

A related thought (perhaps the same thought :-) ). However, I don't think my thought panned out...

Bonanova, thanks for the clarification for how to use the explicit bin-centered velocities. But even more, I'm most excited by the observations that (a) only 4 of the 24 cases have mixed results ( either A or E ), and the notion of complementary cases (3124, 1243) and (2134, 1342). I'm less convinced of the assignment of I/(n+1) for velocities (I don't have clear enough vision) But it seems to me that the notion of complementary cases renders those assignments unnecessary. As you showed, 20 of the cases give unequivocal results without using explicit velocities, and the other four can be easily proven to be exact complements, so that those four add up to exactly two cases, also without explicit velocities. From this, I think we should be able to automatically analyze most of the 6! Cases, and list the remainder. Perhaps it will be obvious that we can automatically find complementary pairs again. As you said, the rationalization for the next coefficient is still missing... Oh well!

I'm still slow. How do those numbers help us know the annihilation order? Suppose there are 4 bullets, with velocities 2/5, 4/5, 3/5, 1/5 Don't I still need to compute collision times to discover whether it's zilch or escape? I gather from your note that these numbers are sufficient to recognize the pattern.

Bubbled, no I don't. You've done what I was suggesting. And the results match Bonanova's formula! Amazing! I sure wish I understood what must be a VERY SIMPLE model of this that allows such a simple formula. i'm not seeing it, given that rank ordering of velocities doesn't even fully determine (zilch vs. escape). This is an exciting puzzle. Thanks for writing this program.

Rank order of velocities does not always determine outcome, which is why I can't compute the probabilities in a closed form

I agree with bubbled about Jasen's statement: Bubbled, I agree that you have to recognize that bullets disappear, but I disagree that you have to simulate each second; merely calculate collision timestamps

Error! I left out a case: corrected cases are: 20000 samples of this yields 0.371. This is close enough to bonanova's statement of 3/8 being spot on, that I'm looking forward EAGERLY to the simple analytical explanation :-)

I've run 20000 samples, with p(zilch-4) of 0.319

My Monte Carlo for zilch-4 yields 0.357.

I'm surprised that the choice of annihilation is so simple as these programs suggest. Perhaps there's a constraint I haven't comprehended yet. Suppose the bullets fired on one second intervals are b0, b1, b2, and b3, And suppose their velocities are v0, v1, v2, v3. Even the 4-bullet case seems to have several outcomes, one of which is that b2 catches b1 before b1 catches b0. Nowhere in these code bodies is there a three-bullet comparison. In fact, even that comparison isn't quite enough, as it's possible that b3 could catch b2 before b1 catches b0. Am I missing something in the code? or am i missing something in the problem?

I assume that ( 1 ) as the number of flags is bounded (between 100 and 1000), we can require enough memory to store a record for each found flag, and can declare how long a record is ( 2 ) a memory cell has finite precision, and that the dimensions of the maze could be arbitrarily greater than the largest value of a cell. So, although we could store some information about each found flag, we could NOT store coordinates. In fact we cannot even store coordinates of one spot (after all, we may have walked a googolplex number of steps in a straight line). ( 3 ) we are NOT permitted to leave anything in a square except a flag, and are not permitted to distinguish the orientation of the flag. Are these all true assumptions about the rules?

Bonanova showed that right triangles could be partitioned into similar but non-congruent triangles. All I showed is that no other triangles can be partitioned into similar triangles.

Neither. It doesn't prove that the condition can be met for right triangles, only that IF the condition can be met at all, it can only be for right triangles.

Here's my attempt at a proof that only right triangles are solutions to the OP. oops, sorry, don't know how to attach the files INSIDE the spoiler... .

I no longer remember whether I ever knew this, but... So, I've caught up with Rob

If we interpret the clause as "...d has to decrease by two to keep the same median of (a,b,c,d)", I get an unequivocal answer (edited from a moment ago)

BMAD, I'm also having difficulty interpreting the words "...d has to decrease by two to keep the same median". The only median mentioned so far is the median of a and b. As Rob and Araver have said, the value of d has no effect on the median of a and b. So, should we interpret the fragment above as: "...d has to decrease by two to keep the same median of (a,b,c,d)"?

Araver, I'm assuming that the shortest path is what counts, and that a stakes assignment with two or three equal numbers counts as zero. I think the coin-flipping is window dressing...Of course, Bonanova's opinion is more reliable :-) Here's my answer at 8 steps:

ok, here's my submission for N=5:

i fear I still don't understand