ThunderCloud
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Quoting, emphasis added:
A sees B=334 C=334, he thinks:
- I can have 334 or 335
- whether I pass or give up, B will know I have not seen B=335 C=334 -> B will announce 334
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Actually A cannot make the second deduction. Hypothetically, had A in fact seen B=335 and C=334, he would have known only that his number was either A=334 or A=333. So he would still have to pass (or give up) the first time he is asked, regardless. His response is therefore not very immediately helpful to B...
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Let's start with what would happen if giving up hadn't been allowed. I will call the logicians A, B, and C. A goes first, then B, then C.
1) Distribution 334,334,334: All pass 333 times, then A passes, B wins, C wins.
2) Distribution 335,334,334: All pass 333 times, then A passes, B wins, C wins.
3) Distribution 335,333,334: All pass 333 times, then A wins, B wins, C wins.
4) Distribution 334,335,334: All pass 333 times, then A wins, B wins.
5) Distribution 334,335,333: All pass 333 times, then A wins, B wins.
6) Distribution 334,334,335: All pass 333 times, then A wins, B wins, C wins.
7) Distribution 334,333,335: All pass 333 times, then A wins, B wins, C wins.
Now, assume that A has 334 and B has 335. C can't distinguish between cases 4 and 5, and thus he will never be able to announce his number. So if giving up is allowed, then C would give up immediately if he saw that A had 334 and B had 335.
So here's what happens in the actual scenario: A passes, B passes, C passes, A passes. Now B announces 334, because C would have given up rather than passing if B had had 335. Now C announces 334, because if he had had 335 then B would have had no way of already having distinguished between cases 6 and 7. Finally A announces 334, as B wouldn't already have been able to distinguish between cases 2 and 3 if A had had 335.
Summary:
pass
pass
pass
pass
win
win
win
I think you are on the right track, and very close.
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Could they communicate beforehand so that they can deduce a plan?
No. The logicians cannot agree upon a strategy in advance. However, you may assume each of them to be "perfect" in that they will deduce all that they logically can. You may further assume that each logician will assume the others to behave this way as well (i.e., that it is generally known that all three logicians are "perfect").
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This did not happen:
1a) Number One sees 334 and 335 and announces 334
1b) Number Two and Three announce 1003-(sum of numbers they see)
1c) Number One announces 334
So this must have happened:
2a) Number One sees 334 and 334, gives up and commits suicide
2b) Number Two and Three announce 334
I know a better and harder version, if I find it, I'll post it. Does it matter if it is in French?
I am not sure I follow your reasoning here -- care to elaborate further?
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Still unanswered...
Can you find the complete sequence of responses? It is considerably shorter than one might initially think...
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well, not entirely sure i agree. for example, how would you denote 1/13?
you need a consistent way of denoting any rational number, not just ones that repeat the last digit.
personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part.
so
_ 0.9 = 1
would be more accurate (although more cumbersome.)
I might have missed the gist of this question the first time around...
Another way to express 1/13 is 0.076923076923076923...
It is clumsy compared with the bar method in this example (unlike .999...), but still understandable.
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well, not entirely sure i agree. for example, how would you denote 1/13?
you need a consistent way of denoting any rational number, not just ones that repeat the last digit.
personally, i like the bar method, where the repeated values have a bar over the top over the length of the repeating part.
so
_ 0.9 = 1
would be more accurate (although more cumbersome.)
______
1/13 = 0.076923
The bar covers the part that repeats.
here's my challenge for you.
accurately add 1/13 to 45/89 in decimal representation.
This can be done, but would be rather tedious (particularly as the decimal expansion for 45/89 has a relatively long repeating digit string). The sum will also have a repeating pattern, of some size less than or equal to the LCM of the repeating digit string lengths of the two operands.
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Both examples B and C have more than two vertices with odd degree (odd number of lines intersecting). Therefore, these cannot be drawn without retracing a line or lifting one's pencil. However, examples A and D have only vertices of even degree (4, to be exact), and can therefore be drawn without any retracing.
Or rather: A and D have only two vertices with odd degree. So both can be drawn without lifting the pencil or retracing a line (in both cases, you must start at one of the odd vertices and end at the other).
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Both examples B and C have more than two vertices with odd degree (odd number of lines intersecting). Therefore, these cannot be drawn without retracing a line or lifting one's pencil. However, examples A and D have only vertices of even degree (4, to be exact), and can therefore be drawn without any retracing.
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I ask U whether V has a crush on them. Since I have not asked U's sweetheart a question yet, U must answer spitefully (untruthfully). If U says no, then in fact U is V's sweetheart (and V is W's and W is U's) and V will answer my next question (What is U's name?) honestly, from which I can deduce everyone's identity. On the other hand, if U says yes, then U is not V's crush; therefore W is V's crush, V is U's crush, and (more importantly) U is W's crush. So in that case W will answer my next question honestly (What is U's name?) from which I can deduce everyone's identity.
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The PO, I think, would only choose the distribution of red/blue, not the individuals that receive a certain color.The puzzle states: "There is a spy who informs the PO of the strategy. So in placing the hats, the PO does his best to foil it." I interpreted this to mean the PO placed the hats himself, as diabolically as he could.
I follow you. And the OP, again, to be fair to you, does not preclude your comment.
The difference I drew was between knowing the strategy in general terms and being privy to the details of its implementation. And that is a finer distinction than the OP makes.
BTW, This may distinguish between Phil's and Rainman's (post 11) approaches. I don't think your PO tactic can defeat Rainman's solution. Or can it?
The Evil PO would be out of luck with Rainman's approach, so far as I can tell.
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The PO, I think, would only choose the distribution of red/blue, not the individuals that receive a certain color.The puzzle states: "There is a spy who informs the PO of the strategy. So in placing the hats, the PO does his best to foil it." I interpreted this to mean the PO placed the hats himself, as diabolically as he could.
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I am not sure if this satisfies the requirements about not impacting the eco-system.
suppose you create a "collapsible fish bowl" made of a water proof material [that will not leak]. leave the sides down.
Note: This could be "very large"
When champ swims into the area of the cage:
- close the sides and
- lift it such that the tops of the sides are just above [inches perhaps] the water level. and
- weigh the enclosure.
- lower the enclosure
- drop the sides
As soon as champ safely exits the area of the cage:
- close the sides and
- lift it such that the tops of the sides are just above [inches perhaps] the water level. and
- weigh the enclosure.
- lower the enclosure
- drop the sides
The difference in weight will give you the difference between Champ's weight and the weight of the water volume displaced by Champ.
On reflection, this may be a non-trivial calculation, but the concept may work.
I think the environmental officials would be alright with that approach. Well done!
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I place one coin on one side of each scale (or, if both sides must be engaged, I place 7 of the coins on one side and one coin on the other side). Obviously, the heavier side of each of the scales is the one with the coin on it. Any scale that answers correctly is not the reverse scale; any scale that answers incorrectly is not the normal scale. So, at this point, with one weighing on each scale we can positively identify one of them, either the normal or reversed, whichever one the random scale did not imitate. However, I can't see a method of bounding the number of additional weighings necessary to definitely isolate the random scale. If it randomly imitates the normal scale for 1000 trials in a row, it would take at least as many weighings to isolate, it seems...
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Upon seeing a 49/50 split amongst their peers, half the prisoners will assume a 50/50 split and guess the minority color, while the other half will assume a 51/49 split and guess the majority color. Upon seeing a 48 / 51 split, each prisoner assumes a 49/51 split and guesses the minority color. Upon seeing any other distribution, each prisoner guesses his hat to be the majority color.
Actually, that conniving PO could foil this plan too, saving only 48...
Half of the prisoners will assume a 50/50 distribution upon seeing a 49/50 split, and will assume a 48/52 distribution upon seeing a 48/51 split. The other half will assume a 49/51 distribution in both cases. In every other case, all prisoners guess the majority color.
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Upon seeing a 49/50 split amongst their peers, half the prisoners will assume a 50/50 split and guess the minority color, while the other half will assume a 51/49 split and guess the majority color. Upon seeing a 48 / 51 split, each prisoner assumes a 49/51 split and guesses the minority color. Upon seeing any other distribution, each prisoner guesses his hat to be the majority color.
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have half the prisioners think of blue as 1 and red as 0, and guess their own hat color as the total of all 1's %2
and the other half have red as 1 and blue as 0. trying to come up with a better strat.
I would give each prisoner that counts blue as 1 a red hat, one prisoner who counts red as 1 a red hat, and the other 49 prisoners (who all count red as 1) blue hats. 51/49 split, deviously distributed. Everyone who is counting blue as 1 (and is wearing a red hat) sees 49 blue hats and therefore guesses blue and is hauled back off to jail. Everyone who counts red as 1 and is wearing a blue hat sees 51 red hats and therefore guesses red. The lone prisoner who counts red hats as 1 and is wearing a red hat sees 50 red hats and therefore guesses blue. No parolees leaving prison this year...
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Each prisoner should guess that his hat color is the same as the majority of the other hats he sees UNLESS he sees 49 of one color and 50 of the other, or he sees 48 of one and 51 of the other, in which case he should guess his hat is the minority color.
I think the worst the PO could do in that case would be to give a 49/51 distribution, in which case the 49 would be freed. -
With a tip of the hat to wolfgang...
Lake Champlain has long been surrounded by rumors of a mysterious creature akin to the Loch Ness Monster. The locals call him "Champ," and he is officially a protected species in both Vermont and New York State.
And after so many decades of stories and speculation, Champ has been spotted!! Scientists, eager to learn as much as possible about the potentially prehistoric lake monster before he disappears again, have quickly assembled all of the maritime and scientific equipment that they might possibly need. However, owing to Champ's thoroughly protected status, officials have mandated that no experiment shall be conducted which might pose an immediate risk to Champ's life or health; in particular, they stipulate that Champ can not be removed from the lake at any time, even momentarily (nor can the ecosystem of the lake be impacted such as by draining the lake).
Can you find a method to determine Champ's inherent buoyancy?
Can you find a method to determine how much Champ would weigh if he were allowed to be hoisted out of the water and placed on a scale?
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I may be missing something obvious, but why is the expected time to see "willywilly" not 26^10?
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So are we saying only one case is true?
Yes. Both sides of the relations specify real numbers, which are well-ordered. So, exactly one of the relations must be true.
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Sum_{i = 1 to infinity](0.1^i) = 1/9 as a geometric series (also easily proven algebraically). 0.999... = 9 * (0.111...) = 9 * Sum_{i = 1 to infinity](0.1^i) = 9 * 1/9 = 1. The equality is therefore true. By extension, both of the inequalities are not true.
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52 weeks in a year
64 squares on a chess board
12 months in a year
2 G on a R...?
60 seconds in a minute-
1
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By construction, the diameter of the infinite sequence of circles must add up to the height of the original triangle, which is 12 by Pythagorus. So, the sum of the circumferences is 12*Pi.
Numbered Foreheads Concluded
in New Logic/Math Puzzles
Posted