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Yoruichi-san

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Posts posted by Yoruichi-san

  1. Yepskerdoodles, too lazy to do a breakdown, but credit to DD, PG, and Pickett ^_^.

    Got a new puzzle in the works but it's not ready yet. I'm going on a trip until next week so I'm going to go dig up some puzzles I really liked that never got solved :( and drop some hints.

    Also, have a new TAoBk in the making and want try to revive the Chromatic Witch series, which really means a lot to me...and I want to 'publish' the big twist at the ending (which is pretty far away still) before another author comes up with it...;)

  2. Time for something short and simple ;)/

    ...

    body of water
    fall over
    or
    position
    fundamental building block
    not included
    annoyance
    bottom of your top
    something valuable
    a theme of this puzzle
    a manifestation of the id
    a different kind of building block
    something it's fun to do with spoilers
    red
    october
    aussi
    a portion of the pie
    above and beyond
    fish, pork, or steak?
    legal kind of coca
    ...
  3. I don't see a winning strategy. All the ways to get a sum of 15 from 3 digits can be represented by a magic square with the even numbers in the corners that is symmetric to the following example:

    492

    357

    816

    Making it effectively a game of tic tac toe, in which, if both players play ideally, there is no winner. However, assuming non-ideal play, the first player has 5 marks to the second player's 4 so he has more ways to win in case the second player makes a mistake, so I would play first...

    ...and get my opponent really really drunk ;P.

    • Upvote 2
  4. Hmm...well technically, I believe the term "distribute" implies a discrete division, not a probabilistic one, but I do like that variation on the game. That seems like a really fascinating extension...I'll see if I can work on it (or anyone else who wants to :))

  5. There is no information given in the question on consensus building. If I understand your assumption correctly, if 500 (captain) proposes a distribution in which no. 499 to 201 don't get anything, they will still agree to it. This implies that they fear, that killing no.500, will start an avalanche of killings. Why?
    Lets see this in more detail.
    I am assuming that it is established that if there are 202 pirates. No 202, will distribute 1 coin to no. 200, 198,.....,6,4,2, and they will all live.
    Now, if there are 203 pirates(p=203). Suppose 203 again proposes 200,198,196,...,4,2 to get 1 coin each. No one will agree, because they know they can make him walk the plank, and still get the same distribution. So, no. 203 has to propose a distribution among 202,201,199,197,195,...5.3.1 but he can pacify at most 100 and thus can never get the majority. So no matter what, in this scenario, no. 203 always dies.
    Next, assume there are 204 pirates. No matter what 204 proposes, no. 202 to 1 will have at most 100 consents (as they know, that none of them from 1-202 can die). If 203 doesnt agree, he dies next, so 204 will get 101 votes and including his own, he can survive. So, if there are 204 pirates they all live but 201,202,203,204 do not get anything.
    p=205: captain can never survive as the remaining 204 know that, when 205 dies, they will survive. (this is exactly like p=203 in the sense that, the captain dies no matter what)
    p=206 to 207: captain proposes 1 coin distribution to 2,4,6,8,...,200 and gets 100 votes. no. 201-204 will never agree. so 104 dissenters. hence the captain dies in p=206-207
    p=208: if captain dies, so do 207 to 205, so 205-208 will agree to what 208 proposes (distribute 1 coin to 2,4,6,..,200) and there will be 104 votes. so no. 208 survives
    p=209 to 215: they can propose to distribute to 1,3,5,7..,199 but will have at least 108 dissenters, so the captains will always die
    p=216: proposes to distribute 1 coin each to 2,4,6,8,...200 and survives
    Hence, the final solution:
    (1) n<=200:
    n keeps 100 - [(n-1)/2] coins for himself and gives 1 coin each to no's n-2,n-4,n-6,...
    (2) for n>200
    (2.1) Let n = 200 + 2k (k=1,3,5,7...) i.e. n=202,208,232,...
    captain proposes 1 coin each for 200,198,196,..,6,4,2 and there are exactly 100 + 2^(k-1) dissentors so the proposal is accepted
    (2.2) for n=200 + 2k (k=0,2,4,6,8..) i.e. n=204,216,264,..
    captain proposes 1 coin each for 199,197,195,..,5,3,1 and there are exactly 100 + 2^(k-1) dissentors so the proposal is accepted
    (2.3) for n = 200 + m (m is not a power of 2 and m>2)
    captain dies no matter what he proposes, next captain dies no matter what, and so on, till n reaches (2.1) or (2.2) above and follows that case respectively.

    Correct on your analysis of parts 1 and 2, but for part 3...one thing to consider...

    Even/odd parity

    ;).

    Lol, nvm, I was reading the wrong post...(again with the whole rushing things thing :wacko: . Anyways, great job :thumbsup: .

  6. There is no information given in the question on consensus building. If I understand your assumption correctly, if 500 (captain) proposes a distribution in which no. 499 to 201 don't get anything, they will still agree to it. This implies that they fear, that killing no.500, will start an avalanche of killings. Why?
    Lets see this in more detail.
    I am assuming that it is established that if there are 202 pirates. No 202, will distribute 1 coin to no. 200, 198,.....,6,4,2, and they will all live.
    Now, if there are 203 pirates(p=203). Suppose 203 again proposes 200,198,196,...,4,2 to get 1 coin each. No one will agree, because they know they can make him walk the plank, and still get the same distribution. So, no. 203 has to propose a distribution among 202,201,199,197,195,...5.3.1 but he can pacify at most 100 and thus can never get the majority. So no matter what, in this scenario, no. 203 always dies.
    Next, assume there are 204 pirates. No matter what 204 proposes, no. 202 to 1 will have at most 100 consents (as they know, that none of them from 1-202 can die). If 203 doesnt agree, he dies next, so 204 will get 101 votes and including his own, he can survive. So, if there are 204 pirates they all live but 201,202,203,204 do not get anything.
    p=205: captain can never survive as the remaining 204 know that, when 205 dies, they will survive. (this is exactly like p=203 in the sense that, the captain dies no matter what)
    p=206 to 207: captain proposes 1 coin distribution to 2,4,6,8,...,200 and gets 100 votes. no. 201-204 will never agree. so 104 dissenters. hence the captain dies in p=206-207
    p=208: if captain dies, so do 207 to 205, so 205-208 will agree to what 208 proposes (distribute 1 coin to 2,4,6,..,200) and there will be 104 votes. so no. 208 survives
    p=209 to 215: they can propose to distribute to 1,3,5,7..,199 but will have at least 108 dissenters, so the captains will always die
    p=216: proposes to distribute 1 coin each to 2,4,6,8,...200 and survives
    Hence, the final solution:
    (1) n<=200:
    n keeps 100 - [(n-1)/2] coins for himself and gives 1 coin each to no's n-2,n-4,n-6,...
    (2) for n>200
    (2.1) Let n = 200 + 2k (k=1,3,5,7...) i.e. n=202,208,232,...
    captain proposes 1 coin each for 200,198,196,..,6,4,2 and there are exactly 100 + 2^(k-1) dissentors so the proposal is accepted
    (2.2) for n=200 + 2k (k=0,2,4,6,8..) i.e. n=204,216,264,..
    captain proposes 1 coin each for 199,197,195,..,5,3,1 and there are exactly 100 + 2^(k-1) dissentors so the proposal is accepted
    (2.3) for n = 200 + m (m is not a power of 2 and m>2)
    captain dies no matter what he proposes, next captain dies no matter what, and so on, till n reaches (2.1) or (2.2) above and follows that case respectively.

    Correct on your analysis of parts 1 and 2, but for part 3...one thing to consider...

    Even/odd parity

    ;).
  7. I feel like BD has lost a large degree of it's variety in the last few years. When I first became active here, what I term the "Golden Years", there were a lot of different types of puzzles, and puzzle solvers who were willing to try them all. Now, on this latest tour, it seems to be mostly the hard-core math puzzles that get responded to at all. I'm not sure if that's just representative of the taste of the current population of BDers or if for some reason people have just become afraid of replying and risking embarrassment, or a combination of the two. In the Golden Years, when I posted a new puzzle there was usually a reply within a few hours, and even though the first replies were usually off the mark (since I have an affinity for creative thinking/figure out what to do type puzzles), it helped get the conversation in the thread started and gave me an excuse to drop hints.

    Puzzle creators and puzzle solvers are a symbiotic relationship, although I'm not sure which started dropping off first, there isn't much hope for reviving the BD organism without somehow increasing the populations of both simultaneously :/.

  8. Wow, I'm not very good with eulogies and I didn't know SP as well as I would have liked, but I think I can sum up my view of him with one word:

    He was the kind of man who did the NY Times Crossword puzzle in pen.

    Respect. ;)

  9. It seems the harmonic function that represents my activity on BD has finally crossed the x-axis once again...;P I recently came across an extension to an old favorite that I'd like to share. The original variant has been posted here several times probably but for those who don't know it, here is a recap:

    Season 1 Recap:

    There are 5 pirates that must split 100 gold coins among themselves. They do so following the following protocol:

    1. The captain presents his proposal on how to distribute the coins.

    2. The living pirates, including the captain, take a vote on whether to pass or reject the proposal. In case of a tie, the captain has the breaking vote.

    3. If the proposal passes, the coins are divided according to it. If it is rejected, the captain walks the plank and the first mate becomes captain, the second mate (if they are called that?) becomes first mate, the third mate becomes second, etc, and return to step 1.

    Assume the pirates are all perfectly rational, do not trust each other, and prioritize in the following order:

    1. They want to survive.

    2. They want to maximize their personal loot.

    3. They want to see others walk the plank.

    What is the captain's proposal?

    Barring the cliffhanger ending, the logical next extension, then...

    Season 2 recap:

    Same as above but with n pirates, where n<100.

    And finally, the new extension...

    Season 3 spoilers:

    What if n>100?

  10. Well, I'm assuming the yoga part is referring to a mat. I don't know anything about Mugiwara though. . . All you can get by taking a letter away though is ma, at or mt.

    That is what that part is referring to :). There are enough solved that you shouldn't have to know both parts to solve each one...

    ...the order of the lines is not random

    ;)
  11. The agents floated some very creative ideas around. However, the Witch was in the details...they had to summon objects that fit into the rules of the round. For example, one agent wanted to create an earthquake, but that was not an object itself that could be created, but it was a phenomenon that could possibly be generated by creating other objects? And they had to remember that each agent could only summon one object at a time, and only objects that could be easily described in a word or two (not complex contraptions designed specifically for the task...although they could combine simple objects that different agents summoned into a complex invention if they so desired). Whatever they decided to try in the end, they had to summon it with the power of green.

  12. [spoiler=Overcame yield stress/Blue]Flow/Low

    Still not sure about Mugiwara/yoga... probably because I barely know anything about yoga.

    Correct. Hint: you don't really need to know much about yoga other than what you'd take to a beginner class ;).

  13. An ambitious Agent OmegaScales provided the final push that allowed the agents to make it to the top. Agent Flamebirde placed the key in the lock and it turned with a very satisfying click. Grinning excitedly, he pushed the hatch up and raised his head and shoulders above the opening. Closing his eyes, he allowed the warm, welcoming sunlight to drench his skin. Then he opened his eyes, and...

    "Holy ****!" he interjected, hastily ducking back into the burrow just in the nick of time, as a front of smoldering air swept over the agents. A few that managed to keep their eyes open caught a glimpse of the furious red-orange wave from whence it was emitted.

    "..." Captain Wraith slowly moved his agape jaw as realization dawned on him. "You've got to be kidding me..."

    post-7969-0-15798600-1377627562_thumb.pn

    Task: Defeat the enemy army, neutralize the dragon, save the Princess from the tower.

    Notes: The enemy army is not drawn to scale. There are twice as many enemy soldiers as agents, they are armored, armed, and highly proficient at battle. The dragon is (more or less) drawn to scale. It is circling the tower and will attack anything more than 10 m above the ground. The dragon's scales are of a substance that cannot be cut by any substance known to man (including diamond). The tower is constructed of a substance that cannot be climbed and cannot be cut. The moat has a radius of 100 m from the tower.

    Remember that agents have no specialized training (i.e. in swordplay, acrobatics, etc).

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