bonanova

6. Everyone who is not here, please indicate by raising your hand.

Depends on what is meant by "have." [1] If have means "own," then yes. I can give someone something that I do not own. e.g. if I stole it. To give something, one only needs the ability to determine who controls it. If I control it, I can pass its control to someone else. So ... [2] If have means "possess the control of" then no. As stated, the paradox arises from the different antecedents of "with sorrow." Sorrow is the consequence of giving, not a possession before the act. But the language permits that interpretation by its form. Cute.

When a denial implicitly asserts a truth you have a paradox. Kind of an empty one, tho -- as was pointed out, nothing was predicted!

Putting it another way, the more certain the man is [on Friday night, should he live that long] that he cannot be hanged on Saturday [the last possible day for the hanging] the more surprised he is when it happens. It's possible because of the prisoner's unwavering belief in the truth of the judge's statements. As with other semantic paradoxes, the assumption that what is being asserted is true comes into play. If the prisoner didn't believe what the judge said was true, he might [expectantly] dread a Saturday hanging, thereby precluding it! The moral: Never believe a judge. It might cost your head.

It's the fact that one statement can be a contradiction. [1] "I am lying." Spreading that over two statements does not change the nature of the paradox: [2] "I am telling the truth." [3] "The previous statement is a lie." Here, one can simply eliminate statement [2], which carries no information, and change [3] into [4] "This statement is a lie." which is equivalent to statement [1]. To my mind the paradox arises from an explicit assertion of something's falseness using a vehicle [declarative sentence] which implicitly asserts its truth.

Destiny and free will become paradoxical if and only if [1] they both determine the outcome of events. [2] they apply to the same event at the same time. Clearly, two independent forces cannot each have their way in the same matter [e.g. going to the doctor] at the same time [today, at 10:00, the time of my appointment.] Either [if Destiny has its way] I become simply an observer, capable only of telling someone after the fact that I did [or didn't] keep my appointment or I choose to keep my appointment and am forced [lamely] into the supposition that [my then unknown] Destiny must have been that I do so. Destiny and free will cannot both be the deciding factor and be independent of each other. What does that mean? One conclusion is that one or the other simply does not exist. [At least] one of them is simply an illusion: one that people discuss, but that owns no real affective power. Usually at this point we cling to our free will and throw Destiny out the window. But experience suggests that particular outcomes seem to happen despite the choices of others to the contrary: one might assert that Hitler was destined to fail, despite protracted and horrendous [free] choices made, by many, to obtain his success. This is the seemingly favorite way to invoke Destiny - something that happens over an extended period of time, contrary to many free choices [or other inanimate obstacles] to the contrary. This is the point of [2] at the beginning of my post. Free choices and Destiny can coexist if they to not apply to the same matter at the same time. It might be my Destiny, born out of my underlying desire for health, to recover from an illness even though I decide not to keep today's appointment with the doctor. Eventually I will seek and get the help I need. In this sense, free will applies to the microscopic decisions and strategies I employ moment by moment, and can logically co-exist with a Destiny which sees eventual outcomes -- outcomes that are outside my ability directly to create. Coexistence is non paradoxical -- both can determine certain outcomes -- if they each operate in their respective, disjoint arenas. Finally, if one believes in an overpowering Destiny that applies to every event in every arena, and if, in the face of that prospect, one simply gives up an active role in life, becoming only its spectator, it can be accurately said that one has so chosen. A paradox of a different type.

What underlies paradoxes of this type is the syntactical rule that a declarative sentence is by its nature an assertion of some particular truth. To use a presumed assertion of truth to deny that same truth is paradoxical: One cannot convey usable knowledge by asserting a denial. Nor can one meaningfully deny a truth: the coin has two paradoxical sides: [1] "I am asserting a falsehood." or "I am lying." [2] "I am not asserting something that is true." or "I am not telling the truth." Putting it another way, it's physically possible to speak the words, "I am lying." But when one undertakes a linear analysis of what has happened when the words are spoken, one is drawn into the syntactical analogy of a Moebius Strip: a piece of paper having a physical connection of its two sides. The circular reasoning forced on the mind by a linear analysis of such statements creates a pleasantly frustrating tease, and the desire for consistency and meaning leaves one in a disturbingly uncomfortable state. Long live paradoxes...

B starts with 3 liters. Filling C [takes 1 liter] leaves 2 [not 4] liters in B.

Each weighing has three outcomes: Left side is [lighter than] [equal to] [heavier than] Right side. Three weighings can thus discern among [3]x[3]x[3]= 27 cases. We have only 24 cases: one of 12 balls is heavier or lighter than the rest. So we can solve the problem, so long as ... [1] The first weighing reduces the cases to no more than 9. [2] The second weighing reduces the cases to no more than 3. [3] The third weighing then distinguishes among 3 or fewer cases. First weighing: Set aside four balls. Why? Because, if the first weighing balances, we have 8 [fewer than 9] cases: One of the 4 excluded balls is heavier or lighter. [1] Weigh 1 2 3 4 5 6 7 8 Outcome 1[a] 1 2 3 4 balances 5 6 7 8. Since only one ball is odd, 1 2 3 4 5 6 7 8 must all be normal. We have 8 cases: 9 10 11 or 12 is H or L [shorthand: H=heavier; L=lighter] [2] Weigh 1 2 3 9 10 11 [we know 1 2 3 are normal] If this balances, 9 10 11 are also normal, and 12 is H or L. [3] Weigh 12 [any other ball] If 12 rises, then 12L; if 12 falls, then 12H. If [2] 9 10 11 rises, we have 3 cases: 9L, 10L or 11L [3] Weigh 9 10. If [3] balances, then 11L; If 10 rises, then 10L; If 9 rises, then 9L If [2] 9 10 11 falls, we have 3 cases: 9H, 10H or 11H [3] Weigh 9 10. If [3] balances, then 11H; If 10 falls, then 10H; If 9 falls, then 9H end of Outcome 1[a]: balance. Outcome 1: 1 2 3 4 falls, and 5 6 7 8 rises. This gives 8 cases: 1H, 2H, 3H, 4H, 5L, 6L, 7L, or 8L The second weighing in this case becomes tricky. Remember each of its three outcomes can lead to no more than 3 cases for the third weighing to resolve. Again, we exclude a number of balls and involve the others. We exclude any three of the balls. Why? Because if the other [included] balls balance, we have exactly 3 cases. Without loss of generality we exclude balls 1 2 3. Since that leaves an odd number of balls, 4 5 6 7 8, we need to use one of the normal balls. Finally we choose which three to weigh against the others. And here's the only hard part of this problem. We must mix some of the possibly light balls with some of the possibly heavy balls. Otherwise, one of the outcomes of the second weighing will leave us with more than 3 cases, and the third weighing will not resolve this. [2] weigh 4[H] 5[L] 6[L] 1[normal] 7[L] 8[L] in parentheses I've indicated the POSSIBLE cases that we have determined: 4[H] means 4 is heavier if it's not normal. Outcome 2[a]: 4 5 6 balances 1 7 8 These balls are all normal. We have 3 cases: 1H 2H or 3H. [3] weigh 1 2 If [3] balances, then 1 and 2 are normal, and 3H if 2 falls, then 2H if 1 falls, then 1H Outcome 2b: 4 5 6 falls, and 1 7 8 rises. We have 3 cases: 4H 7L or 8L [3] weigh 7 8 if [3] balances, then 7 and 8 are normal, and 4H if 7 rises, then 7L if 8 rises, then 8L Outcome 2[c]: 4 5 6 rises, and 1 7 8 falls. We have 2 cases: 5L or 6L. [3] weigh 1[normal] 5[L] If 5 rises, then 5L If balance, then 5 is normal, and 6L Now we can go back and take the remaining case Outcome 1[c]: 1 2 3 4 rises, and 5 6 7 8 falls. to distinguish among the remaining 8 cases: 1L 2L 3L 4L 5H 6H 7H and 8H exactly as we analyzed Outcome 1b. Simply substitute H for L and v.v. Problem solved.

The puzzle asks: Who am I to Barbara - NOT who is Barbara to me. Hence the correct answer is ... Daughter. Unless of course I am a male. Then the correct answer is not in the choices listed, it's Son-in-Law.