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A while back, you may remember that I made a thread detailing puzzles from a logic book I found. It was archived ( https://forums.somethingawful.com/showthread.php?threadid=3851344&userid=0&perpage=40&pagenumber=1 ) because I lost the book. But I have found it again, so let's pick up where we left off, shall we? The answers given in the last thread to puzzle 22 were correct. Puzzle 23: Abacus Abhorrence Progress is coming to Hyperborea in the form of the abacus. While this will mean an improved ability to count and to calculate, there is a considerable concern among the inhabitants as it will mean changes for many people. Three Hyperboreans are discussing the new technology. The three are known to be a Sororean (always tells the truth), a Nororean (always lies) and a Midrorean (Alternates between true and false, but whether it is TFTF... Or FTFT... Is unknown). Man A: 1. C still counts using his fingers and toes. 2. B needs to have his own abacus. 3. I am not the Nororean. Teen Boy B: 1. I do not need to have my own abacus. 2. I am the Midrorean. Boy C: 1. I disagree with A' first statement. 2. I am the Sororean. Who is the Sororean, Nororean, and Midrorean?
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Oct 4, 2018 00:51
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A the midorean, B nororean, C sororean? A can go F T F with B telling 2 lies and C telling 2 truths
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Oct 4, 2018 03:59
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MY INEVITABLE DEBT posted:A the midorean, B nororean, C sororean? A can go F T F with B telling 2 lies and C telling 2 truths
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Oct 4, 2018 05:34
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roomforthetuna posted:I thought A soro, B mido, C noro. This would mean A goes true true true, B goes false true, and C goes false false, which seems internally consistent to me, but your model also seems internally consistent. The "I disagree with" seems a bit ambiguous and weird both ways. i think the problem here is the "i disagree" language. i feel like that's always true no matter what it's about.
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Oct 4, 2018 08:15
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Statements A-2 and B-1 are conflicting which means they can't both be true or false. If B-1 is true then B can only be the Soro or Midi, meaning B2 must be true if Soro but thats contradictory, or False if Midro, which is also contracdictory. Therefore B1 must be false and A-2 must be true. Because A2 is true A3 MUST be true as the only way it could be false is if A was the noro which would mean A2 couldn't be true. If A2 and A3 are both true than A must be Soro. If A is the Soro than C2 must be false and C1 must be false (because A1 is true). Double false means C is the noro which makes B the midro. A sororean, B is midrorean, C is nororean.
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Oct 4, 2018 11:38
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MY INEVITABLE DEBT posted:A the midorean, B nororean, C sororean? A can go F T F with B telling 2 lies and C telling 2 truths This can't work because if A is the midorean then statement A-3 must be true, which means the only possible order is TFT, but if C is the sororean than statement A1 could not be true.
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Oct 4, 2018 11:44
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Agent, Room, you two are correct! Puzzle 24: Who Won The Chariot Race? Chariot racing is a popular sport among Hyperboreans, and is her competitive. In a highly contested race, little distance separated the first three finishers, and there was disagreement as to which one was the winner. Sororeans always tell the truth, Nororeans always lie, Midroreans alternate truth and lie. Regarding the racers, which group or groups they represent is uncertain. Based on the following, who won, and to what group or groups do A, B and C belong? A: 1. I was the winner. 2. B was second. 3. C was third. B: 1. I was the winner. 2. I was well out in front all the way. 3. C was behind A and me. C: 1. I was the winner. 2. A was far behind when I crossed the finish line. 3. B finished before A.
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Oct 5, 2018 03:14
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B1 and B2 are basically the same statement, so they are either both true or both false and B can't be the midrorean. That means B3 must be true or false along with B1 and 2. A3 is the same thing as B3 so it also is either true or false along with B1 and B2. If B is the sororean and B1/2/3 and A3 are all true, then A must be the midrorean as thats the only way A3 could be true if B is the sororean. Which means A1 would also be true, except that directly contradicts B1/2. This means B CAN'T be the sororean and therefor must be the nororean. That makes A3 false and A the midrorean. A1 must also be false but A2 is true. So ultimately A is the midrorean, B is the nororean, C is the sororean, and they placed C, B, A in order.
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Oct 5, 2018 04:25
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Agent355 posted:snip You've proven that B has to be a nonorean, so via A3 being false, A is either also a nonorean or is a midorean. Given that, A1 also has to be false (since A is either nonorean or midorean, A1=A3). So C1 is true, by a process of elimination on A1 and B1 being false (if A didn't win and B didn't win, then C has to have won). C, therefore, is either a midorean (iff C2 false) or a sororean (iff C2 true), and so by C3 (since C1=C3), the order has to be C, B, A. B did in fact finish second, so A2 is true, and A is a midorean. But is C2 true? The puzzle states that the result was closely contested between the top 3 finishers. But C2 says that A was far behind, and while A2 did finish third, this seems to contradict with what we know. So, I propose that C won the race as a midorean, B the nonorean came second and midorean A came third.
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Oct 5, 2018 07:37
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Correct, Indeterminacy! 25. Who Will Confront The Griffin? Even in a land as idyllic as Hyperborea all is not always well. A griffin was in the land and was ravaging the flocks. Three shepherds were discussing which one of them should confront the ferocious monster, which had the body of a lion and the head and wings of an eagle. The three are known to be Sororean, who always speaks truthfully, a Nororean, who always speaks falsely and a Midrorean, who makes statements that are alternatively truthful and false or false and truthful. The three shepherds make statements as follows: A: 1. B does not even own a spear with which to confront the monster. 2. C would faint if he had to confront the Griffin. 3. I am the only one who should challenge the ferocious monster. B. 1. I do, too, own a spear. 2. I am the Sororean. 3. You cannot believe anything that C says. 4. I will be the one to challenge the monster. C. 1. I would not faint if I had to confront the Griffin. 2. I am the Midrorean. 3. B is the Sororean. 4. I agree with A's third statement. Jupiter interposes to settle the problem, and, with a wave of his hand he makes the Griffin smaller than a mouse. The three shepherds then chase it away. Which shepherd is the Sororean, which is the Midrorean, and which is the Nororean?
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Oct 6, 2018 01:57
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Suppose B is the Sororean. Then B3 says that C is the Nororean, but C3 is true, which is a contradiction. Now suppose B is the Midrorean. Then B2 is false, so B1 and B3 are true, and again C is the Nororean. That means A is the Sororean, but then A1 contradicts B1. So B is the Nororean. Then C3 is false, so C must be the Midrorean, and A is the Sororean.
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Oct 6, 2018 09:13
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This is a nice one! Junpei, once everyone has had a chance to look at everything, would you mind if I were to effortpost about B3 and C3?
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Oct 6, 2018 09:37
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Banned King is correct! Puzzle 26: Disagreeable D Four Hyperboreans area sked to which group or groups they belong. They respond below, although one of them is being disagreeable. Sororeans always tell the truth, Nororeans always lie, and Midroreans alternate truth and falsehood. A: We each belong to a different group. B: We are all in the same group. C: 1. We are not all in the same group. 2. I am in the same group as B. D: 1. I disagree with A's statement. 2. I disagree with B's statement. 3. I disagree with C's first statement. What group or groups are represented by the four Hyperboreans?
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Oct 7, 2018 19:08
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I meta-guessed that all are Midroreans, since with a single statement there is no way to know if it is a Midrorean or any other group. A: Lie B: True C.1: Lie C.2: True D.1: True D.2: Lie D.3: True
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Oct 7, 2018 19:24
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A is obviously false: there are four of them and only three groups! So D1 is true, hence D3 (= not C1 = B) must also be true. Thus, they are all Midroreans.
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Oct 7, 2018 22:49
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Correct, both of you! Puzzle 27: Two Times Four Shepherds In every group of shepherds in Hyperborea, at least one is a Sororean, who always tells the truth, and at least one is a Nororean, who always lies. A visitor approached four shepherds and asked each how many of the four were Sororeans. These answers were given: - Three of us are Sororeans. - One of us is. - Two of us are. - None of us are. The visitor approached four more shepherds on another hillside and asked how many were Nororeans. Their answers follow. -We are all Nororeans. -One of us is. -Three of us are. The fourth declined to speak. How many of the shepherds on the two hillsides were Sororeans?
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Oct 8, 2018 01:40
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The interesting thing is that these are both the same riddle, just with one additional statement in the first case. In the first case, the four statements are mutually exclusive, so at most one of them can be true. The premise tells us that at least one of them must be true, so exactly one statement is true. Hence, there is one Sororean, namely the one who claims that there is one Sororean in the group. The other three are clearly Nororeans. In the second case, the statements are again mutually exclusive, but now the possibility of there being two Sororeans (one of whom didn't say anything) and two Nororeans exists. In that case, none of the three statements made is true, which leaves too many Nororeans to match the count. So, again, there must be one Sororean, in this case the one who stated that there are three Nororeans, and that's the full count.
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Oct 8, 2018 19:29
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Correct, Nidoking! Puzzle 28: Mars Loses His Temper Four Hyperboreans were conversing with a visitor, who was actually Mars, the god of war, in disguise. The visito inquired as to their group or groups. Sororeans truth, Nororeans lie, Midroreans alternate. At times the answers to questions put to Hyperboreans can be very frustrating. Mars, who had a short temper, turned all four into frogs. Following are the responses that Mars received to his inquiry: A: 1. I am either a Sororean or a Nororean. B. B is a Sororean. B: 1. I am either a Sororean or a Midrorean. 2. A is a Midrorean. C: 1. I am neither a Nororean nor a Midrorean. 2. B claims falsely to be either a Sororean or a Midrorean. D: 1. I am a Sororean and A is a Nororean. 2. I disagree with C's second statement. To which groups do A, B, C and D belong?
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Oct 8, 2018 19:44
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Anybody who claims they might be a nororean can't be a nororean because nororeans always lie so they wouldn't claim to be one. Thus A is 100% either midorean or Sororean. If they're a mido the first statement is false so the second must be true, if they're soro then both statements are true. Either way statement A2 must be true and B is a sororean. B2 says A is midorean so he must be. C2 calls B a liar, which is false, and if C1 is true than C2 would have to be true as well, so both are false and C is a nororean. D1 is wrong because A is a midrorean, D2 is true because C2 is false. So A and D are midroreans, B is a sororean, and C is a nororean.
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Oct 8, 2018 20:05
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Correct! Puzzle 29: Who Is to Be Transferred? Hyperborea is beginning trade with Ethiopia, which is located in the part of the known world that is south of Mount Olympus. It is rumored that, to establish a trade base, one or more Hyperboreans will be transferred to Ethiopia. Four Hyperboreans are discussing the subject. At least one is a Sororean, at least one is a Nororean, and at least one is a Midrorean. Their statements follow: A: 1. C said that D is going to be transferred. 2. The food in Ethiopia is very good. 3. I applied for the transfer, but I am badly needed here. B: 1. The food in Ethiopia is not very good. 2. C said that D is going to be transferred. 3. A is a Sororean. C: 1. I did not say that D would be transferred. 2. D is a Nororean. D: 1. A is going to be transferred. 2. The food in Ethiopia is not very good. 3. C said that I am going to be transferred. Which group does everyone belong to?
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Oct 8, 2018 20:40
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If B3 were true, then B1 must also be true, but that would contradict A2. So B3 and B1 are false. Then, since A2 is true, A must be Midrorean. That means A1 (= B2 = D3) is false, so B is Nororean and C must be the Sororean (because D cannot be). Hence, A is Midrorean, B and D are Nororean, and C is Sororean. Also, Ethiopian food is very good (but we already knew that  ).
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Oct 9, 2018 00:40
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Correct! Puzzle 30: An Outlier The land of Hyperborea has several well-established conventions to which all Hyperboreans should adhere. Certainly, their unusual standards of veracity are important to the land's traditions. There are the few odd inhabitants, though, who do not accept the value of conventions. The following statements are made by four inhabitants who are engaged in discussion. One is a Sororean, one is a Nororean, and one is a Midrorean. However, the last one does not follow the customary rules of veracity and must be considered an Outlier, who neither always tells the truth, nor always lies. They tell both, but not alternatively. Which is the Sororean, which is the Nororean, which is the Midrorean, and which is the Outlier? A: 1. My statements are not all truthful. 2. We are overworked. 3. We are all lucky to be here. 4. We Hyperboreans are favored by the gods. B: 1. I agree with A's third statement. 2. Every time I see a visitor, I think maybe it is one of the gods, in disguise. 3. I am doing more than my share of the work. 4. My statements are all truthful. C: 1. My statements are all truthful. 2. D's second statement is false. 3. The gods do not visit us in disguise. 4. We are all overworked. D: 1. C's first statement is truthful. 2. B's third statement is truthful. 3. My statements are all truthful. 4. The gods frequently visit us in disguise.
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Oct 10, 2018 04:40
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A1 must be true. If it were a lie, then it would have to be true, which is impossible. So A is not Nonorean However, since A1 confirms other statements are false, A cannot be Sororean. D cannot be Sororean, as D1 confirming C1 in that case would mean both D and C are Sororean, when there is only one. As C3 is false, given the information from previous problems, C is not Sororean either. This means B must be the Sororean. Since at least one of C's statements is false, D1 is also false. Since B is the Sororean, D2 must be true, meaning C2 is false. Since C1, C2, and C3 are all false, C must either be Nororean or the Outlier. Since D's statements follow a False-True-False-True pattern, he is either Midorean or the Outlier. If C was the Outlier, then D must be Midorean and A must be Nonorean. However, A is not the Nonorean, as shown earlier, therefore, C is Nonorean. As given from the beginning of this puzzle section, the Hyperboreans are favored by the gods, meaning A4 is true. Since both A1 and A4 are true, this means that A is the Outlier and D is the Midorean.
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Oct 10, 2018 06:52
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Correct! Puzzle 31: Five Hyperborean Heroes. Hyperboreans recognized heroes as those warriors who had single-handedly vanquished monsters. Five Hyperboreans who had accomplished this feat were Actaeon, Ceyx, Minos, Nisus, and Pyramus. They had defeated one sea serpent, two griffins and two chimeras. Of the heroes, little is known of their veracity, other than the fact that one-and ONLY one-is an Outlier. Based on their statements below, can you determine what group each belongs to, and who defeated what monster type? Actaeon: 1. Everything Minos states is false. 2. I did not defeat a griffin. 3. Ceyx defeated a griffin. 4. I agree with Nisus's first statement. Ceyx: 1. I defeated a sea serpent. 2. Minos did not defeat a chimaera. 3. Nisus defeated a chimaera. 4. Actaeon defeated a griffin. Minos: 1. I agree with Actaeon's first statement. 2. Ceyx did not defeat a chimaera. 3. Actaeon is not a Sororean. 4. I defeated a sea serpent. Nisus: 1. Minos's third statement is false. 2. I defeated a chimaera. 3. Actaeon did not defeat a griffin. 4. Pyramus defeated a chimaera. Pyramus: 1. Nisus defeated a sea serpent. 2. Minos is a Midrorean. 3. Actaeon is not a Nororean. 4. Ceyx defeated a chimaera. Bonus: Some of the names of these heroes are references. What are they referencing?
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Oct 14, 2018 17:23
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Answer: Actaeon is a Nororean and defeated a griffin, Ceyx is a Sororean and defeated a sea serpent, Minos is the Outlier (FTTF) and defeated a griffin, Nisus is a Midrorean (FTFT) and defeated a chimaera, and Pyramus is a Nororean and defeated a chimaera. My Logic: Actaeon's third statement claims Ceyx defeated a griffin while Minos states that Ceyx did not defeat a chimaera, these line up with each other, but not with Actaeon's first statement that Minos always lies, so either statement 1 is false, statement 3 is false, or both are false. These wouldn't fit a Midrorean's pattern so Actaeon is either a Nororean or the Outlier. As he definitely can't be a Sororean, statement 4 must also be false, which makes Minos's third statement true and Nisus's first statement false as well. Skipping to Minos, his third statement has been proven true, but his first statement that he "agrees with Actaeon's first statement" (that everything Minos himself says is a lie) must be false. This makes Minos the Outlier because he has a definite pattern of F?T? which a Midrorean cannot have. As there can only be one Outlier and Actaeon has at least 2 false statements in a non-Midrorean order, Actaeon must be a Nororean. Since Actaeon is a Nororean, his remaining statement of "I did not defeat a griffin" must be false (he defeated a griffin) and we can also confirm from his third statement that Ceyx did not defeat a griffin. Ceyx's statement 4 is true, therefore he's either a FTFT Midrorean or a Sororean, that's all we can tell for sure at the moment. Statements 1 and 3 from Nisus are definitely false, so he's either a FTFT Midrorean or another Nororean and we'll leave him at that for now. Pyramus claims in his second statement that Minos is a Midrorean and in his third that Actaeon is not a Nororean. We know Minos is the Outlier and Actaeon is a Nororean, making this two consecutive false statements so he must be a Nororean. As this is the case, we also now know that Ceyx did not defeat a chimaera and Nisus did not defeat a sea serpent. Going back to Ceyx, it is impossible for both him and Nisus to be FTFT Midrorean's because his third statement is identical to Nisus's second. Ceyx has been ruled out as a Nororean and Nisus has been ruled out as a Sororean, so Ceyx must be a Sororean. This means Ceyx is telling the truth about killing a sea serpent and Nisus is telling the truth about killing a chimaera, making Nisus a FTFT Midrorean. This leads to Nisus's fourth statement of "Pyramus defeated a chimaera" being true. As the only thing that remains is a griffin, Minos must have killed the remaining griffin.
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Oct 14, 2018 20:40
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Correct! And that is the final puzzle on The Land of Hyperborean! Section 4: Sir Hector Revisited These puzzles deal with placing items (and persons) just so. The approach and thinking processes in accomplishing this different from the puzzles in the previous sections. No diagrams, no assumptions, no considerations of veracity. Your challenge for each puzzle in this part of simply take the items listed and place them in the order described. Puzzle 32: Sir Hector's Piscatoral Prowess In Sir Hector's time, part of the training of young men for Knighthood involve instruction in recreational activities such as hunting, falconry, fishing, dancing, and playing the harp. Sir Hector particularly enjoyed fishing during his spare time and was proud of his skill at the sport. During one fishing season he caught at least one of each eight varieties of fish in five fishing excursions. Given the following information, what was the order in which you copy a varieties of fish? The catfish was caught before the northern pike, but after the walleyed pike, which was caught up before the muskellunge. The rainbow trout preceded the perch. The largemouth bass came after the muskellunge and the walleyed pike which was caught during the second fishing excursion. The perch and the northern pike were caught after the muskellunge, which was caught before the smallmouth bass, which was caught before the perch and the catfish but after the rainbow trout which followed the largemouth bass. No fish were caught during the first fishing excursion of the season, and a total of only two fish were caught during the first three excursions. Fishing was especially good during the fourth excursion, in which the perch, the largemouth bass and the rainbow trout were among the fish caught. Two varieties of fish were not caught until the final excursion.
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Oct 16, 2018 05:51
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Let "X < Y" be "Fish X was caught before Fish Y". From the second paragraph, we can deduce the following: 1. Walleyed Pike < Catfish < Northern Pike 2. Walleyed Pike < Muskellunge 3. Rainbow Trout < Perch 4. Muskellunge < Largemouth Bass 5. Walleyed Pike < Largemouth Bass 6. Muskellunge < Perch 7. Muskellunge < Northern Pike 8. Muskellunge < Smallmouth Bass 9. Smallmouth Bass < Perch 10. Smallmouth Bass < Catfish 11. Largemouth Bass < Rainbow Trout < Smallmouth Bass From paragraph 3, Expedition A: - Expedition B: Walleyed Pike, ? Expedition C: ? Expedition D: Perch, Largemouth Bass, Rainbow Trout, ? Expedition E: ? From statement 3, we can place the Rainbow Trout before the Perch. From statements 9 and 11, we can place the Smallmouth Bass between the Rainbow Trout and the Perch, and the Largemouth Bass before the Rainbow Trout. From statement 4 and statement 2, we can place the Muskellunge as the second fish caught in the first three expeditions, after the Walleyed Pike and before the Largemouth Bass. This also satisfies statements 6 and 8. The remaining two, the Catfish and the Northern Pike, must be the two unique varieties caught on the last expedition, which satisfies statements 7 and 10. From statement 1, we can place the Catfish before the Northern Pike. Final Order: 1 - Walleyed Pike (Expedition B) 2 - Muskellunge (either after Walleyed Pike on Expedition B, or on Expedition C) 3 - Largemouth Bass (Expedition D) 4 - Rainbow Trout (Expedition D) 5 - Smallmouth Bass (Expedition D) 6 - Perch (Expedition D) 7 - Catfish (Expedition E) 8 - Northern Pike (Expedition E)
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Oct 16, 2018 09:51
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Correct! Puzzle 33: The First Tournament A tournament was scheduled in the local province and seven knights, including Sir Hector, registered for the competition. It was agreed that every participant would joust with each of the others. Every knight's number of wins would be totaled to determine the winner and the rankings of the other six knights. What was the order in which the knights finished in the tournament, and how many jousts did each win? -Sir Intrepid finished ahead of Sir Gallant and behind Sir Victor. -Sir Able's ranking was an odd number, as was Sir Victor's. -Sir Bold and Sir Able finished behind Sir Intrepid. -Sir Staunch won four jousts. -Sir Hector finished two places behind Sir Bold. -Sir Gallant finished two places ahead of Sir Able.
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Oct 25, 2018 01:56
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Victor, Intrepid, Staunch, Bold, Gallant, Hector, Able, I think? Only iffy part there is Intrepid vs. Staunch. It works if they won 6, 5, 4, 3, 2, 1, 0 jousts in that order. Round robin is more complicated than that but either we don't have enough information or I'm too dumb to parse out the possibilities there. The key is that Intrepid finished in front of a bunch of people: Gallant, Able, and Bold, and also Hector since he finished behind Bold. Including Intrepid himself that's 5 people Victor has to be in front of, and since his rank is also odd he can only be first. I didn't really keep track of my logic after that. Princey fucked around with this message at 05:08 on Oct 25, 2018 |
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Oct 25, 2018 05:02
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Correct! Puzzle 34: The Second Tournament Tournaments were a regular part of the activities of Sir Hector and his fellow knights. In one tournament, seven knights participated. Every knight jousted with each of the others and each joust counted one point for the winner. In one joust, the two knights fought to a standoff, so each was awared one-half point. From the statements below, determine the rankings of the combatants in the tournament (There were no ties) and the number of points each one earned. 1. The one who finished in first place fought to a standoff with one of the others. 2. Sir Bold did not finish behind Sir Gallant. 3. Sir Gallant finished one point ahead of Sir Able, who finished two places in ranking ahead of Sir Staunch. 4. Sir Victor would have moved ahead of both Sir Hector and Sir Staunch if he had won his joust with Sir Hector. 5. Sir Intrepid finished behind Sir Staunch in number of points earned.
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Oct 26, 2018 03:14
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1st: Bold: 5.5 2nd: Gallant: 5 3rd: Able: 4 4th: Hector: 3 5th: Staunch: 2 6th: Victor: 1.5 7th: Intrepid: 0 I had a bunch of things written up here but they were eaten by the forums, RIP Aesculus fucked around with this message at 04:39 on Oct 26, 2018 |
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Oct 26, 2018 04:32
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Okay, here's what I've got so far: Bold - 5.5 Gallant - 5 Able - 4 Hector 3 Staunch - 2.5 Victor - 2 Intrepid - 0 I'm pretty sure I'm right about the order -- Bold is ahead of Gallant is ahead of Able, is two ahead of Staunch who is ahead of two people, so that's four people placed right there. Hector has to be ahead of Staunch for Victor to be able to pass Staunch by beating Hector. Beating Hector is only a swing of one point, so Victor has to be right below Staunch to be able to manage moving two places (and Staunch has to have the half point). However, I don't know what to do about the points, because they can't be right. With seven people all jousting against every other person, there should be 6 + 5 + 4 + 3 + 2 + 1 = 21 jousts. Each joust is worth one point. So the point total should be 21. But this solution adds up to 22. That said, if the rankings are right there's not much that could be adjusted to fix it. If you lower Victor's point total by 1 it ruins his ability to pass Staunch and Hector by flipping the outcome of his fight with Hector. If you lower Hector, Staunch, AND Victor by one point, now you're down too far, to 19 points total. That can't be compensated for by increasing the other scores, either, since we know first place fought to a draw once, making 5.5 the highest possible score achievable. It says there was one draw joust, which I took to mean there was ONLY one draw joust, so throwing half points around also doesn't work. Even if we open up the possibility of more ties, that leads to a scenario where there are multiple possible solutions rather than a single one. And it explicitly says there were no point ties, so that option's out too. So yeah. Edited for clarity. Edit again: Wow I'm dumb, well done Aesculus. Princey fucked around with this message at 04:45 on Oct 26, 2018 |
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Oct 26, 2018 04:38
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Correct! Sorry for the delay. Puzzle 35: The Third Tournament Knightly jousts and tilts have become quiet popular in the province, and sixteen combatants participated in the most recent tournament. Tournament rules provided that each knight would joust with one other participant. The winners would advance to the second round and each would joust with one of the other seven remaining knights. The third round would consist of the four undefeated knights, each jousting with one of the others. The tournament winner would be decided between the two combatants reaching the fourth round. Based on the information below, list the participants who jousted in the second round, the third round, and the fourth round, and identify the tournament winner. Sir Resourceful won his joust with Sir Virtue, but lost to Sir Steadfast. Sir Gallant won his joust with Sir Victor, but lost his joust with Sir Resolute, who one against Sir Bold. Sir Hector won against Sir Stanch, but lost his joust with sir Valorous, who won against Sir True. Sir Intrepid lost his joust with Sir Resolute, as did Sir Valorous. Sir Able won against Sir Admirable, but lost to Sir Bold, who won his jousts with Sir Loyal and Sir Steadfast, who won his joust with Sir Faithful.
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Nov 4, 2018 22:59
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This puzzle is all about counting how many jousts each knight fought. Resolute defeated Intrepid, Gallant, Valorous, and Bold, so he was the winner of the tournament, and Sir Bold defeated Loyal, Steadfast, and Able, so he was the other combatant in the fourth round. The rest of the knights can be identified with similar logic - Steadfast and Valorous were the other knights who defeated two opponents, so they're the remaining knights in the third round, and so on. I filled in a bracket as I went:
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Nov 5, 2018 00:50
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Bingo! Puzzle 36: Sir Hector's Knightly Accouterments As a youth, Hector was quite poor, and when knighthood was conferred upon him he could ill afford the weapons and armor of a knight. Therefore, he arranged to assist and perform routine services for the armorer until the value of the needed paraphernalia was earned. In this way, Sir Hector acquired his knightly accouterments, piecemeal. Can you list the order in which he acquired his equipment, based on the information below? The mail gauntlets were acquired before the battle ax, the sword and the lance, but after the boots and the dagger. The padded doublet, the shield, and the breastplate were acquired before the visor, the mail hose, and the shin guards, but after the sword sheathe, which was acquired before the sword and the mail hood. The mail body armor was acquired before the dagger, the mail hose, the mail gauntlets and the battle ax, which was acquired before the sword sheath and the headpiece, which was acquired after the lance. The sword was acquired before the mail hose, the breastplate, and the visor, but after the padded doublet, the boots, and the dagger. The padded doublet was acquired after the battle ax and the mail body armor, but before the shield, which was acquired after the breastplate. The mail hood was acquired before the lance and the shin guards, which were acquired before the lance. The boots were acquired before the sword sheath and the mail gauntlets, but after the dagger. The mail hose was acquired after the shield and the sword heath, but before the visor, which was acquired before the headpiece and the mail hood.
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Nov 6, 2018 04:03
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Mail body armor, dagger, boots, gauntlets, battle axe, sheath, doublet, sword, breastplate, shield, mail hose, visor, mail hood, shin guards, lance, headpiece is what I came up with. I diagrammed it out, mostly chipping away from the ends. First pass was just a general numbers thing (if A was before three different things, then A can't be any of the last three spots), and then ran through the list checking specific items against each other until it was narrowed down. I was gonna post the diagram but it'd have to be in pre or code tags to make sense and they don't play nice with spoilers, sadly.
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Nov 6, 2018 20:52
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You are correct! Puzzle 37: Seasonings In Sir Hector's time, effective ways to preserve food were not available. There was considerable reliance on seasonings to enhance flavor and prolong freshness. A traveling spice trader was set upon by thieves, who scattered his display of seasoning. Sir Hector happened on the scene and quickly routed the thieves. He offered to assist the trader in re-positioning the seasonings on his display rack. The trader gladly accepted, and explained the order in which the seasonings should be placed in three rows on the rack. Can you help by listing the 21 seasonings in their correct arrangement in the three rows? ("To the right or left of" means in the same row. "Above" or "Below" means in the same column.) The mace should be below the oregano and to the left of the cardamom, which should be above the thyme, which should be to the left of the basil and the rosemary. The cloves should be above the dillweed, to the right of the paprika, the oregano, and the nutmeg, and to the left of the cinnamon, which should be above the rosemary. The tarragon should be below the cayenne and to the left of the thyme, the fennel, and the garlic, which should be below the mace. The cumin should be above the cayenne, to the right of the paprika and to the left of the oregano and the cinnamon. The rosemary should be to the right of the thyme and the fennel, which should b below the dillweed, which should be to the right of the mace, the poppy seed and the saffron. The ginger should be above the cardamom, to the right of the oregano and the paprika, which should be above the saffron. The curry should be below the cinnamon, and to the right of the cardamom and the dillweed. The poppy seed should be above the basil, which should be to the right of the thyme, the garlic, and the chili powder, which should be below the saffron.
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Nov 7, 2018 02:04
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 The trick here is to first find the trios of spices that have to be in the same column, and then use the right/left clues to see in what order they have to be. Edit: One additional step actually, while you can find all the trios of spices going by just above/below clues, the exact order from top to bottom of a few of the groups of three needs some context from the left/right clues as well, but I think I only had to move like 2-3 spices up or down a slot for it all to work out. KennyMan666 fucked around with this message at 13:55 on Nov 7, 2018 |
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Nov 7, 2018 02:32
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Correct! And that is the last puzzle in Sir Hector Revisited! Section 5: Matching Wits Not all matches are made in heaven-especially not the ones in the mythical kingdom which is the setting for this chapter. These matches involve seven-league boots, magic weapons, and royal dinnerware, and they're made by a motley crew of ogres, wizards and impoverished queens. In these puzzles you'll be converting words into mathematical symbols and formulas. You'll find them useful tools, particularly in the later puzzles when the numbers multiply! Puzzle 38: In the Dark Planning to roam the countryside and prey upon its defenseless people, the ogre reached into his dark closet. There he had stored four six-league boots and eight seven-league boots. How many boots did he have to pull out of the close to make sure he had a pair that matched?
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Nov 7, 2018 23:29
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| # ? May 4, 2024 09:05 |
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Three, since he could have one of each after pulling two, so the third has to make a match (since the puzzle didn't specify anything about right and left boots). KennyMan666 fucked around with this message at 10:48 on Nov 8, 2018 |
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Nov 8, 2018 00:26
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