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In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar
In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining.
+ Show Spoiler +Ref: [BS]: BullShit or Blizzard website, you choose which one I took this fact from
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16. + Show Spoiler +Isn't this one all about division to circuits? The prisoners must decide to enforce an order on the boxes. Therefor, they decide to randomly send each person to a different box (dubbed "his" box), and the guy opens the box, and goes to the box belonging to the name inside it, according to the what they defined as that person's box. Basically, you enforce an order of circuits here, the chances of a circuit being 51 or greater are very close to ln2 (combinatorics for defining the permutation, some calculus to calculate the limit of the series needed to explain this number). Therefor, the chance of success would be (1 - ln2)*100 percent, somewhere around 30%. Not quite sure though.
25. + Show Spoiler +Incorrect phrasing. The manager has 25 dollars in his pocket, his son has 2, and each of the guys has one (out of the 30). No dollar unaccounted for. The trick is that they payed 27, 25 are with the manager and 2 are with his son. No dollar vanished.
26. + Show Spoiler +Each domino has to be on one black square and one white square. You eliminated 2 of the same color, now there are 32 of one color and 30 of the other, therefor, it will be impossible to put the last domino, as the only two squares left are of the same color, therefor not next to eachother.
25 and 26 were instant, 16 took me a bit too long. Will get back to the rest later.
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On April 10 2011 18:45 0x64 wrote:In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining. + Show Spoiler +Ref: [BS]: BullShit or Blizzard website, you choose which one I took this fact from
Cool problem. This has a nice answer that does not depend on the order of morphing.
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Solution to the balls problem: + Show Spoiler +He pulls one out of the bag, and does something with it so that no one can retrieve it. Maybe he swallows it, and now it's in his digestive system. Maybe he throws it off the cliff before anyone gets a good look at it. The point is, everyone knows he just picked a ball, but they didn't see the color, and now they have no way of directly seeing it's color because the ball is gone. Lost. Irretrievable FOREVER!!!!!
So the only way they can determine the color of the ball is through logic. They open the bag and look at the remaining ball. It's black? Oh, then the one that he picked first MUST have been white!!!
Solution to the Templar Archon problem: + Show Spoiler + Instead of thinking about the merging and reincarnating separately, think of the two combined as one process. So what are our options? 1.) Two high templar merge, and when that archon dies, it will reincarnate as a high templar. 2.) Two dark templar merge, and when that archon dies, it will reincarnate as a high templar. 3.) One of each templar merge, and when that archon dies, it will reincarnate as a dark templar.
If we perform "Action 1" on our initial group of templar, it is equivalent to removing a high templar from the group. Our new set is the initial set, minus one high templar. Cool.
If we perform "Action 2" on our initial group of templar, then our new set is the initial one, plus one high templar, minus two dark templar. Also cool.
If we perform "Action 3" on our initial group of templar, it has the same effect as Action 1. We just lose a high templar. Our new set is the initial set, again minus one high templar. SUPER COOL.
Notice the effect of each action in terms of parity. All three of these do not change the parity of the number of dark templar, and all of them DO change the parity of the number of high templar. So...
If we have an even number of dark templar to begin with, we will have an even number of dark templar at the end. Namely, we will have zero dark templar at the end, and our last templar is a high templar.
If we have an odd number of dark templar to begin with, we will have an odd number of dark templar at the end. Namely, we will have one dark templar. He will be the last templar, just sitting their chilling. Perhaps we could write a screenplay about him, much like "The Last Samurai," but in this case he'd be "The Last Templar." Of course, he would still be played by Tom Cruise.
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The solution to the 12 coins problem:
+ Show Spoiler + Label and divide the coins in to three groups: ABCD EFGH IJKL.
Start off by balancing ABCD and EFGH Three possibilities, and I will address the simplest ones first: 1) ABCD = EFGH, find the fake coin in IJKL in two tries, you should know how to solve this, weigh ABC against IJK, and then I against J if it doesn't balance. 2) ABCD < EFGH, this is the same as the next one, just switch labels
3) ABCD > EFGH Weigh ABCEF against DIJKL, remember IJKL are now real coins and we have three possiblities:
1. ABCEF=DIJKL, we have now cleared A B C E F D by balancing the scale, we know one between G and H is fake and it weighs less than a real coin, balance G against H and find the lighter one.
2. ABCEF>DIJKL, D is clear because it was in the heavier group in step two, E and F are clear because they could only be lighter if they are fake. One of the A B C is fake and the fake coin is heavier, balance A against B and find the fake coin.
3. ABCEF<DIJKL, A B C are clear since they could only be heavier if they were fake. D E F are suspects here, either D is too heavy or E / F is too light. Balance E against F and find the fake coin.
Edit: just found out that everhate's solution from last page is much less convoluted, i somehow managed to axe that idea when brainstorming and came up with this ugly beast instead >_<.
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On April 10 2011 10:33 Darkren wrote:Show nested quote +On April 10 2011 08:46 Husnan wrote:On April 10 2011 05:46 cmpcmp wrote:Solution to the 3 princesses / most eligible bachelor puzzle Yes, there is a logical, reasonable, and not stupid solution if that's what ur thinking. Solution + Show Spoiler +You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger. Explanation + Show Spoiler +There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her. Wow, beautiful! That logic doesnt work because the younger sister could lie and tell u the middle one is the youngest. U could also ask the older sister if the middle one is younger and than her and she will answer u yes. + Show Spoiler +U could ask do u think lying to someone is ok? If u pick the younger one she will tell u no(lie) If u pick the older one she will tell u no If u pick the middle one she will answer yes because she lies sometimes and tell the truth other times
U then know if she answers yes that she is the middle sister and if she answers no u marrie her + Show Spoiler +No, yours doesn't work and the one you quoted works fine. The middle sister's answer is essentially random so you cannot ask a question that results in any possibility of marrying the person you asked. As a result, you have to ask a yes/no answer where yes = marrying one of the other two and no = marrying the other; if you get the middle sister the answer conveniently doesn't matter as both are eligible and thus the randomness is a nonfactor.
So, all you have to do is find a question which produces one answer for [NOT middle] and one answer for [middle] regardless of whether it's asked to the eldest or the youngest. The age question is such a question, because
-From the perspective of the eldest sister, she is the eldest and the liar is the youngest (which is the truth), therefore the middle sister is older than the liar.
-From the perspective of the youngest sister, she is the eldest and the truthteller is the youngest (the opposite of reality, since the youngest sister always lies), therefore she will also say that the middle sister is older than the liar. Remember that the youngest is compelled to lie and cannot choose not to, so her answer will be consistent.
So you ask sister A "is sister B older than sister C?".
-If sister A is the middle sister, either B or C are fine so the answer doesn't matter. -If sister A is the eldest and B is the middle sister, she will answer "yes" so you should pick C (the younger = liar) -If sister A is the eldest and B is the youngest, she will answer "no" so you should pick B (the younger = liar) -If sister A is the youngest and B is the middle, she will answer "yes" so you should pick C (the older = truthteller) -If sister A is the youngest and B is the eldest, she will answer "no" so you should pick B (the older = truthteller)
So we end up with
-If sister A answers "yes", pick C. Otherwise, pick B. This works for all permutations of A,B,C.
Edit: Didn't see a solution for 18 in the thread, so in case anyone wants it (it's incredibly simple algebra):
+ Show Spoiler +n lifeboats * n men per lifeboat = n^2 men
n-10 lifeboats after 10 sink * n+10 men per lifeboat = (n-10)(n+10) remaining men
drowned men = n^2 - (n-10)(n+10)
= n^2 - (n^2 - 10n + 10n - 100)
= 100
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13) + Show Spoiler +Uhm, isn't that why there is a binary expression for every number?
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ball justice.+ Show Spoiler + He could pick a black one, but then get them to check what was left in the bag. Since a black ball is left they would "know" he drew a white one and he would go free. And if things didn't go to plan he would at least expose that he was being cheated and might get a "fair" shot at avoiding justice. Not sure what any of this would have to do with where he is. I guess he could "accidentally" drop his black ball off the cliff to get them to check the bag.
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27-Templar problem : + Show Spoiler +If we say that the dark templar is 1 and the high templar is 0, we have a xor operation (see http://en.wikipedia.org/wiki/Exclusive_or )As we know, the XOR operation is commutative (the order doesn't matter) so we can merge in the order we want. Let's say we have M dark templar and N high templar : Let's merge all the dark templars together : if M is even, we get one high templar, otherwise we get one dark templar. Let's merge all the high templars together : we get an high templar. (ie 0 XOR 0 XOR 0 XOR 0... =0) Result : If M is even, we get an high templar. If M is odd, we get an dark templar.
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On April 10 2011 13:03 stepover12 wrote:26. suppose you have a chess board with 2 opposite corners cut out as in picture
there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not? Your chessboard is sideways. In proper chess diagrams, The board is shown as if the white and black pieces were on the bottom and top of the board, respectively. An empty board would look like this: Just a nitpick. + Show Spoiler [solution] +Answer would be no. each domino piece would, by necessity, cover 1 black square and 1 white square. Because there are 32 white squares and 30 black squares, once you've placed 30 dominos, the 31st would have to cover 2 white squares, which is not possible.
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edit: goddamnit double post
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A puzzle of mine:
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
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On April 11 2011 01:05 Tintti wrote: A puzzle of mine:
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
+ Show Spoiler +The exact number of finite things there is to know?
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On April 11 2011 01:14 mishimaBeef wrote:Show nested quote +On April 11 2011 01:05 Tintti wrote: A puzzle of mine:
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
+ Show Spoiler +The exact number of finite things there is to know?
+ Show Spoiler +Didn't actually think of that myself. There is another one.
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On April 11 2011 01:31 Tintti wrote:Show nested quote +On April 11 2011 01:14 mishimaBeef wrote:On April 11 2011 01:05 Tintti wrote: A puzzle of mine:
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
+ Show Spoiler +The exact number of finite things there is to know? + Show Spoiler +Didn't actually think of that myself. There is another one.
+ Show Spoiler +The knowledge that he knows all there is to know?
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Let's imagine a universe where there is a finite number of things people may know. Is that a limit on the number of things an individual may know, or a limit on the number of things that are knowable?
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On April 11 2011 03:22 gyth wrote:Show nested quote +Let's imagine a universe where there is a finite number of things people may know. Is that a limit on the number of things an individual may know, or a limit on the number of things that are knowable?
The latter, although I'm not really sure how they actually differ :D
The wording I used in this riddle in my opinion suggests the correct answer more easily than the original one, which was something like "What's the last thing one may ever know?" I reckoned that as the riddlers here would anyway ask for clarifications on this rather obscure question I'd might as well post that in a more precise form.
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although I'm not really sure how they actually differ The difference between you being able to only hold 7 facts in your mind versus there only being 7 facts in the whole universe. If it was a limit on his mind + Show Spoiler +then the last thing he'd know is that he couldn't know everything.
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One that's similar to the liters of water problem.
Here's what you have:
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
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Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between). Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?
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In Madadia, a rather strange and misguided assassin, from his hidden position, uses a high-powered rifle to shoot someone in the foot from 50 feet away. The bullet travels at 1300 feet per second. Both the person being shot at and the assassin are at sea level. What will be the first evidence to the person of the attack? (As in how will he know he as been shot.)
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Fenrax
United States5018 Posts
On April 11 2011 06:34 MusicalPulse wrote: One that's similar to the liters of water problem.
Here's what you have:
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
+ Show Spoiler + - sell 3-liter jug - empty one 8-liter jug into pool 1 - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them - fill pools 3+4 with 4 liters of water each - fill one 8-liter jug with the 8 liter water from pool 1 - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them - fill pools 1+2 with exactly 4 liters - attack Terran with 4-pool
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