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plasmid

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Posts posted by plasmid


  1. 7 hours ago, CaptainEd said:

    Locations, as promised, for pentagon, in the columns labelled x and y.

      Reveal hidden contents

    Uploaded Images

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    Putting coordinates into a spreadsheet and calculating distances sure is a good way of checking your work; now I wish I had thought of doing that!

    I believe I have a way of doing it with five breadcrumbs aside from the five vertices. Up to you if you'd like to look at it a little longer or if I should go ahead and convert my chicken scratch to an intelligible post when I get a chance.


  2. 4 hours ago, rocdocmac said:

    Just to get my own mind at peace (square case) ...

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    One extra crumb in one of the orange regions (E or F) between center O and X or between O and Y, but not at X, Y or O,

    since BC=CD=CX=AY and BO=CO=DO (equidistances).

    Mouse moves from A to E (or F) to C, then to B (or D), and finally to D (or B).

     

    blob.png.d1e422dc9fa813b7152b1f781383579f.png

     

    I see what you're saying -- if we don't care whether the mouse goes from C to B or from C to D then that would work and you could let the mouse just flip a coin after it reaches C. The OP was phrased sort of ambiguously since I first talked about making the mouse take as many diagonals as possible (in which case that would be fine to do) and later said the specific letter sequence of ACBD (in which case that wouldn't).

    The only thing I'd say, though, is that if you drop a breadcrumb in region E, it seems like the mouse would go eat the breadcrumb there and then go to either B or D instead of C -- either of those two points would be closer to region E than C is.


  3. 3 hours ago, CaptainEd said:

    Definitely an interesting puzzle with a new principle; nice job, Plasmid!


    Partial analysis:

      Reveal hidden contents


    Let a0 be a breadcrumb on point A, a1 the first breadcrumb on the path to C, etc.
    Then come c0=C, c1..., e0=E, e1..., b0= B, b1..., d0 = D

    Working back from the last move:
    There must be a crumb on D to attract the mouse to the endpoint.
    Before that, a crumb on B, and then the mouse will see nothing but D, so no further bi are needed.
    Before that, a crumb on E. After that, e1 has to be closer to E than D is, else the mouse at E will go directly to D without going to B first. There may need to be additional ei
    Before that, a crumb on C, and afterward c1 must be closer to C than B is.
    Before that, a crumb on A, afterward a1 has to be closer to A than E is.

    Here are the endpoints and all the line segments from each endpoint to the next:
    A, Aa1 < AE, a1a2,...,axC,
    C, Cc1 <  CB, c1c2,...,cyE,
    E, Ee1 < ED , e1e2,...,ezB,
    B,
    D

    But I don’t yet see how to express the constraints on ai:i in 2...x, ci:i in 2..y, and ei:i in 2...z,  nor the values of x, y, and z.

     

    Multiple crumbs on a spot

      Reveal hidden contents

    I doubt that putting more than one crumb on a spot has any effect.  There’s no instruction in OP suggesting that the mouse is swayed by mass, only distance. If more than one crumb is at a spot, and the mouse goes there, it will eat one crumb, then the next and the next, until they’re all gone, and then will move to the nearest crumb.

     

     

    Yep, this gets harder than it looks ^_^


  4. On 5/9/2019 at 5:30 AM, jjs323 said:

    3 at c, 2 at b,1at d

    ... wut?

    16 hours ago, rocdocmac said:
      Hide contents

    For the square case ... 2 at C, 1 at B and 1 at D

    For the pentagon case, it looks like 4 at C, 3 at E, 2 at B and 1 at D, but that's probably not what is wanted!

     

    ... Wat?

    The mouse always goes and eats the nearest bit of food to its current position. So like Captain Ed says, putting multiple pieces of food at one spot wouldn't be helpful. If the mouse starts at point A of the square, then the food at either B or D will be closer than C (even if there are multiple breadcrumbs at C), so the mouse will go to the closer food at either B or D.

    14 hours ago, Dave said:

    In your 4 sided box, insert 4 small triangles against the 4 box sides, which now create 2 paths, a to c, and b to d.

    Repeat the mouse feeding exercise several times and the mouse will learn how to get to the 4 box corners using the your desired

    directions. Then remove the 4 triangles and watch what happens.

    Or, you could eat the bread yourself and feed the mouse to the neighbors cat.

    Nah, this mouse really is pretty dumb. It won't manage to learn, it'll still keep right on pursuing whatever bit of food is closest until it eats everything in sight.


  5. This gets tricky with rolls like (1, 1, 6) that look isosceles but don't actually form triangles.

    How would you handle such rolls? Would you omit rolls that don't form a triangle from analysis (equivalent to saying that if you get a roll that doesn't form a valid triangle then roll again), or count them as a failure to form an isosceles triangle? And would you consider a "straight line" throw like (3, 3, 6) to be a valid triangle albeit with zero area or nah?


  6. This sounds like it might be a homework problem, so I won’t flat out answer it but will give a hint on how to approach it.

    Spoiler

    It becomes a lot easier if you solve for the probability that you roll the dice three times and NOT get an isosceles triangle. In other words: what's the probability that you roll three times, and on each throw you roll a number that hasn't appeared yet?

     


  7. Spoiler

    Call the original number N and the new number M. Define N1 as the units digit of N, N2 as the tens digit of N, etc. If you use the same numbering system for M then the fact that M = 2N means that M1 = 2N1 mod 10, M2 = 2N2 mod 10 + int(N1/5), M3 = 2N3 mod 10 + int(N2/5), etc. And the fact that you can move the units digit of N to the front to also generate M means that M1 = N2, M2 = N3, … Mmax = N1.

    Combining the facts that M1 = 2N1 mod 10 and M1 = N2 means that N2 = 2N1 mod 10

    M2 = 2N2 mod 10 + int(N1/5) and M2 = N3 means N3 = 2N2 mod 10 + int(N1/5)

    M3 = 2N3 mod 10 + int(N2/5)  …

    Well, I could keep writing all that stuff out, but you get the idea. If you’re given the ones digit, you can solve for the rest of the digits. So we can just go through all the possible values for the ones digit and find out if we can generate a number that fits the description of the problem. Note that we can skip the cases where N1 = 0 or N1 = 1 because in those cases it would be impossible for M (where N1 is now the lead digit) to be 2N. I wrote a spreadsheet to do those calculations for the various potential starting values N1 and will paste the results.

    image.png.2fdebf772e1f49623a73749c438a5e94.png

    Now we’re looking for a number where the original N1 digit (which becomes the Mmax digit) would be 2Nmax + int(Nmax-1/5) which is what you know must be true if M = 2N. In English: if N1 is even we’re looking for a number with an Nmax of N1/2 and Nmax-1 less than 5, and if N1 is odd then we’re looking for an Nmax of N1/2 rounded down and Nmax-1 of 5 or greater. So taking the case of N1 = 2, we can look down the list of subsequent digits until we find a potential Nmax where Nmax = 1 and Nmax-1 < 5, which occurs at digit 18. That gives the number 105263157894736842. And manually checking, we see that in fact moving the ones digit to the front gives the same value as multiplying the original number by 2, which is 210526315789473684. A similar thing can be done for the other potential starting units digits to find more numbers that would be a solution to the problem.

     


  8. 1 hour ago, CaptainEd said:

    Silly question: are we required to make the paths as straight lines: AC, CB, BD?

    That's definitely not a silly question. In the question I had in mind, the paths don't need to be straight lines. But down the line when we solve the pentagon case and start thinking about generalizations, we could talk about solving for cases with straight line paths.


  9. On 4/23/2019 at 9:26 AM, BMAD said:

     I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth?

    You could be telling the truth because such a number does exist. Of course I can't say with certainty that you're telling the truth because I don't know whether you actually have such a number in mind right now. Back in a bit with a description of how to handle this problem.


  10. In case this helps clarity: the warm up problem is to take this setup with the mouse at A (I told you it was cute!) and breadcrumbs at B, C, D, and your job is to place more breadcrumbs to ensure the mouse goes first to C, then B, then D. Ideally, using as few breadcrumbs as possible.

    image.png.c1ce68bc91cc88818f746af38fba2138.png

    The real problem is to take this setup with breadcrumbs at points B, C, D, E and place more breadcrumbs to make the mouse go first to C, then E, then B, then D. Again, using as few breadcrumbs as possible.

    image.png.eb74b8f6f74acb02aa52292354f295b1.png

     

     


  11. Warning: This is a problem I have not yet found an optimal (or even very good) solution to. But it seems awfully non-trivial, and not in a genre that I've seen before, so I'm throwing it out there for the Den. ( @bonanova that means you should take a look.)

    You have a pet mouse. It's an awfully cute mouse. Kind of like those mice on Pinky and the Brain. And you'd like to make it even cuter by teaching it some tricks. The only problem is that your mouse is, well... let's just say it has the brain of a mouse so it's kind of hard to teach it any tricks. But it is very good at eating food. In fact, it will always manage to find the closest morsel of food and go eat it, then find the closest from its new position and go eat that, etc. until it eats everything in sight. So, you'd like to "teach" your mouse to do some tricks given that behavior.

    The first exercise is to place morsels of food at the corners of a square and place breadcrumbs so the mouse runs to diagonal corners as much as possible. In other words, suppose you have a square with points A, B, C, D in clockwise order with the mouse starting at point A and you'd like to make the mouse go to point C (diagonal from point A), then point B (one of the two remaining), then point D (diagonal from point B). There are already some breadcrumbs at points B, C, and D (since there need to be breadcrumbs there if you want the mouse to stop at those points) and your goal is to place breadcrumbs to make the mouse go from A to C to B to D while placing as few breadcrumbs as possible.

    That first exercise might be a challenge for your kid brother, but for a BrainDen level challenge (the one I haven't convinced myself I've found an optimal solution for yet): Instead of a square, suppose you have a regular pentagon with points A, B, C, D, and E in clockwise order and want to get the mouse to go from A to C to E to B to D?


  12. Push your feet upon on my pedals
    While the rubber meets the road
    Then as urge for fun unsettles
    Lose my grip, so off we go

    My engine wrapped around a tree
    Your limbs are but a price to pay
    All for this pursuit of glee
    On single foot now make your way


  13. On 12/13/2018 at 2:16 PM, CaptainEd said:

    Plasmid, In what sense is this not a total solution?

    Well, I've shown a lot of ways where Arcsin could win, but I don't actually prove that Arctan can win in cases that aren't covered in the above ramblings. An ideal solution would be along the lines of: "If the composition of the Praesidium meets these criteria then Arcsin has a strategy that's guaranteed to win, and if it doesn't fit those criteria then Arctan has a strategy that's guaranteed to win."

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