rookie1ja
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Pouring water II. - Back to the Water and Weighing Puzzles
Given three bowls: 8, 5 and 3 liters capacity, divide 8 liters in half (4 + 4 liters) with the minimum number of water transfers. Note that the 8-liter bowl is initially filled with 8 liters of water and the other two bowls are empty - that is all water you have.
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Pouring Water II. - solution
1. pour 5 litres from the 8-litre to the 5-litre bowl,
2. pour 3 litres from the 5-litre to the 3-litre bowl,
3. pour these 3 litres back to the 8-litre bowl,
4. pour the remaining 2 litres from the 5-litre to the 3-litre bowl,
5. pour 5 litres from the 8-litre to the 5-litre bowl,
6. pour the missing 1 litre from the 5-litre to the 3-litre bowl (there should be 4 litres left in the 5-litre bowl),
7. pour the 3 litres back from the 3-litre to the 8-litre bowl (and that’s it – in 8-litre bowl 4 litres).
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Pouring water III. - Back to the Water and Weighing Puzzles
Given three bowls: 7, 4 and 3 liters capacity. Only the 7-liter is full. Pouring the water the fewest number of times, make the quantities of 2, 2, and 3 liters.
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Pouring Water III. - solution
Three numerals in each number stand for litres in each bowl:
700 - 340 - 313 - 610 - 601 - 241 - 223 (overspilling 6 times)
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Pouring water IV. - Back to the Water and Weighing Puzzles
How can you measure 6 liters of water using only 4 and 9-liter bowls?
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Pouring Water IV. - solution
First fill the 9-litre bowl. Then overspill 4 litres to the 4-litre bowl (there are 5 litres in the 9-litre bowl afterwards) and pour out the water from the 4-litre bowl. And again overspill 4 litres to the 4-litre bowl and empty it. Then overspill the remaining litre to the 4-litre bowl but this time keep it there. Fill the 9-litre bowl to the top for the second time and overspill water to fill the 4-litre bowl to the top. Thus the required 6 litres stay in the 9-litre bowl.
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Pouring water V. - Back to the Water and Weighing Puzzles
Measure exactly 2 liters of water if you have:
1. 4 and 5-liter bowls
2. 4 and 3-liter bowls
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Pouring Water V. - solution
1st Fill the 5-litre bowl, overspill water from it to fill the 4-litre bowl, which you empty afterwards. Overspill the remaining 1 litre to the 4-litre bowl. Refill the 5-litre bowl and overspill water from it to fill the 4-litre bowl (where there is already 1 litre). Thus you are left with 2 litres in the 5-litre bowl.
2nd The same principle – this time from the other end. Fill the 3-litre bowl and overspill all of the water to the 4-litre bowl. Refill the 3-litre bowl and fill the 4-litre bowl to the top. And there you have 2 litres in the 3-litre bowl.
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Pouring water VI. - Back to the Water and Weighing Puzzles
Given three bowls: bowl A (8 liters capacity) filled with 5 liters of water; bowl B (5 liters capacity) filled with 3 liters of water; and bowl C (3 liters capacity) filled with 2 liters of water.
Can you measure exactly 1 liter, by transferring the water only 2 times?
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Pouring Water VI. - solution
1st Pour 1 litre from bowl A to bowl C. Thus 4 litres are left in the bowl A and bowl C is full (3 litres).
2nd Pour 2 litres from bowl C to bowl B. Doing that you have full bowl B (5 litres) and there is 1 litre left in bowl C.
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Weighing I. - Back to the Water and Weighing Puzzles
You have 10 bags with 1000 coins each. In one of the bags, all coins are forgeries. A true coin weighs 1 gram; each counterfeit coin weighs 1.1 gram.
If you have an accurate scale, which you can use only once, how can you identify the bag with the forgeries? And what if you didn't know how many bags contained counterfeit coins?
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Weighing I. - solution
If there is only 1 bag with forgeries, then take 1 coin from the first bag, 2 coins from the second bag ... ten coins from the tenth bag and weigh the picked coins. Find out how many grams does it weigh and compare it to the ideal state of having all original coins. The amount of grams (the difference) is the place of the bag with fake coins.
If there is more than 1 bag with forgeries, then there is lots of possible solution. I can offer you this one as an example: 1, 2, 4, 10, 20, 50, 100, 200, 500 and 1000.
Imagine you have 10 bags full of coins, in each bag 1000 coins. In one bag, there are all coins forgeries. The original coin is 1 gram light, forgery is 1.1 gram. Balancing (Edit: Weighing) just once on an accurate weighing-machine, how can you identify the bag with forgeries? And what if you didn't know how many bags contain forgeries?
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Weighing II. - Back to the Water and Weighing Puzzles
A genuine gummy drop bear has a mass of 10 grams, while an imitation gummy drop bear has a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears and the others – imitation.
Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears?
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Weighing II. - solution
Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively 0, 1, 2, 4, 7, 13, and 24 bears.
The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N.
The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs.
Now, in case you're curious, the possible weight deficits and their unique decompositions are:
3 = 0 + 1 + 2
5 = 0 + 1 + 4
6 = 0 + 2 + 4
7 = 1 + 2 + 4
8 = 0 + 1 + 7
9 = 0 + 2 + 7
10 = 1 + 2 + 7
11 = 0 + 4 + 7
12 = 1 + 4 + 7
13 = 2 + 4 + 7
14 = 0 + 1 + 13
15 = 0 + 2 + 13
16 = 1 + 2 + 13
17 = 0 + 4 + 13
18 = 1 + 4 + 13
19 = 2 + 4 + 13
20 = 0 + 7 + 13
21 = 1 + 7 + 13
22 = 2 + 7 + 13
24 = 4 + 7 + 13
25 = 0 + 1 + 24
26 = 0 + 2 + 24
27 = 1 + 2 + 24
28 = 0 + 4 + 24
29 = 1 + 4 + 24
30 = 2 + 4 + 24
31 = 0 + 7 + 24
32 = 1 + 7 + 24
33 = 2 + 7 + 24
35 = 4 + 7 + 24
37 = 0 + 13 + 24
38 = 1 + 13 + 24
39 = 2 + 13 + 24
41 = 4 + 13 + 24
44 = 7 + 13 + 24
Note that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43.
Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears?
Edit: scales (for this case) = weighing machine (like in science labs, so no pair of scales) and the plates are big enough for all possible weighings
Edit: each carton contains 50 bears
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Weighing III. - Back to the Water and Weighing Puzzles
This puzzle goes a step further from the previous one.
You have eight bags, each of them containing 48 coins. Five of those bags contain only true coins, the rest of them contain fake coins. A fake coin weighs 1 gram less than a real coin. You have an accurate scale, with the precision of up to 1 gram.
Weighing only once and using the minimum number of coins, how can you find the bags containing the fake coins?
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Weighing III. - solution
Similar to the former brain teaser.
I take out 0 (no coin from the first bag), 1 (one coin from the second bag etc.), 2, 4, 7, 13, 24, 44 coins (from the last 8th bag). Each triple is unique enabling an easy way to identify the bags with fake coins (using only 95 coins).
This puzzle is a step further than the previous one (weight of fake and non fake coins is the same as the weight of bears in the previous puzzle). You have eight bags, each of them containing 48 coins. Five of these bags contain only true coins, the rest of them contain fake coins. Fake coins weigh 1 gram less than the real coins. You do not know what bags have fake coins and what bags have real coins. You can use a scale, a dynamometer type one, with precision up to 1 gram (an accurate weighing machine). Weighing only once and using the minimum number of coins, how can you find the bags containing the fake coins?
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Weighing IV. - Back to the Water and Weighing Puzzles
One of twelve tennis balls is a bit lighter or heavier (you do not know which) than the others. How would you identify this odd ball if you could use an old two-pan balance scale only 3 times?
You can only balance one set of balls against another, so no reference weights and no weight measurements.
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Weighing IV. - solution
It is enough to use the pair of scales just 3 times. Let’s mark the balls using numbers from 1 to 12 and these special symbols:
x? means I know nothing about ball number x;
x< means that this ball is maybe lighter then the others;
x> means that this ball is maybe heavier then the others;
x. means this ball is “normal”.
At first, I lay on the left pan balls 1? 2? 3? 4? and on the right pan balls 5? 6? 7? 8?.
If there is equilibrium, then the wrong ball is among balls 9-12. I put 1. 2. 3. on the left and 9? 10? 11? on the right pan.
If there is equilibrium, then the wrong ball is number 12 and comparing it with another ball I find out if it is heavier or lighter.
If the left pan is heavier, I know that 12 is normal and 9< 10< 11<. I weigh 9< and 10<.
If they are the same weight, then ball 11 is lighter then all other balls.
If they are not the same weight, then the lighter ball is the one up.
If the right pan is heavier, then 9> 10> and 11> and the procedure is similar to the former text.
If the left pan is heavier, then 1> 2> 3> 4>, 5< 6< 7< 8< and 9. 10. 11. 12. Now I lay on the left pan 1> 2> 3> 5< and on the right pan 4> 9. 10. 11.
If there is equilibrium, then the suspicious balls are 6< 7< and 8<. Identifying the wrong one is similar to the former case of 9< 10< 11<
If the left pan is lighter, then the wrong ball can be 5< or 4>. I compare for instance 1. and 4>. If they weigh the same, then ball 5 is lighter the all the others. Otherwise ball 4 is heavier (is down).
If the left pan is heavier, then all balls are normal except for 1> 2> and 3>. Identifying the wrong ball among 3 balls was described earlier.
One of twelve pool balls is a bit lighter or heavier (you do not know) than the others. At least how many times do you have to use an old pair of scales to identify this ball (and be always 100% sure)?
(a pair of scales = a scale consisting of a lever resting on a fulcrum with weighing pans at each end of the lever equidistant from the fulcrum)
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Weighing V. - Back to the Water and Weighing Puzzles
On a Christmas tree there were two blue, two red, and two white balls. All seemed same. However, in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls.
Using a balance scale twice, identify the lighter balls.
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Weighing V. - solution
Lay one red and one white ball on left pan and one blue and the other white ball on the right pan. If there is equilibrium, then it is clear that there is one heavier and one lighter ball on each side. That’s why comparing white balls is enough to learn everything.
However, if at first weighing one side is heavier, then there must be a heavier white ball on that side. The next reasonable step is to compare the already weighed red ball and yet not weighed blue ball. After that, the character of each ball is clear, isn’t it?
On a Christmas tree there were two blue, two red and two white balls. All seemed the same, however in each colour pair one ball was heavier. All three lighter balls were the same weight, just like all three heavier balls. Using a pair of scales twice, identify the lighter balls.
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Weighing VI. - Back to the Water and Weighing Puzzles
There are 9 similar balls. Eight of them weigh the same and the ninth is a bit heavier.
How would you identify the heavier ball if you could use a two-pan balance scale only twice?
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Weighing VI. - solution
Divide the 9 balls into 3 groups of 3. Weigh two groups. Thus you find out which group is the heavier ball in. Choose 2 balls from this group and compare their weights. And that's it.
Having 9 balls, equally big, equally heavy. Only one of them is a bit heavier. How would you identify it if you could use a pair of scales only twice?
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Weighing VII. - Back to the Water and Weighing Puzzles
Given 27 table tennis balls, one is heavier than the others.
What is the minimum number of weighings (using a two-pan balance scale) needed to guarantee identifying the heavy one? Always. Of course, the other 26 balls weigh the same.
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Weighing VII. - solution
It is enough to use a pair of scales 3 times.
Divide the 27 balls to 3 groups, 9 balls in each. Compare 2 groups – the heavier one contains the ball. If there is equilibrium, then the ball is in the third group. Thus we know the 9 suspicious balls.
Divide the 9 balls to 3 groups of 3. Compare 2 groups, and as mentioned above, identify the group of 3 suspicious balls.
Compare 2 balls (of the 3 possibly heavier ones) and you know everything.
So we used a pair of scales 3 times to identify the heavier ball.
Having 27 table tennis balls, one is heavier than the others. How many times (minimum) do you need to use a pair of scales to identify it.
Edit: What is the minimum number of weighings needed to guarantee a confirmation (always!) of which ball is the heavy one? Of course, the other 26 balls weigh the same.
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Weighing VIII. - Back to the Water and Weighing Puzzles
Suppose that the objects to be weighed may range from 1 to 121 pounds at 1-pound intervals: 1, 2, 3,..., 119, 120 and 121. After placing one such weight on either of two weighing pans of a pair of scales, one or more precalibrated weights are then placed in either or both pans until a balance is achieved, thus determining the weight of the object.
If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects?
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Weighing VIII. - solution
There are necessary at least 5 weights to bring into balance any of the 121 possible objects. And they weigh as follows: 1, 3, 9, 27, 81g.
Suppose that the objects to be weighed may range from 1 to 121 pounds at 1-pound intervals: 1, 2, 3,..., 119, 120, 121. After placing one such weight on either of two weighing pans of a pair of scales, one or more precalibrated weights are then placed in either or both pans until a balance is achieved, thus determining the weight of the object. If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such weights that would be sufficient to bring into balance any of the 121 possible objects?
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Hourglass I. - Back to the Water and Weighing Puzzles
Having 2 sand-glasses: one 7-minute and the second one 4-minute, how can you correctly time 9 minutes?
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Sand-Glass I. - solution
Turn both sand-glasses. After 4 minutes turn upside down the 4-min sand-glass. When the 7-min sand-glass spills the last grain, turn the 7-min upside down. Then you have 1 minute in the 4-min sand-glass left and after spilling everything, in the 7-min sand-glass there will be 1 minute of sand down (already spilt). Turn the 7-min sand-glass upside down and let the 1 minute go back. And that's it. 4 + 3 + 1 + 1 = 9
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Hourglass II. - Back to the Water and Weighing Puzzles
A teacher of mathematics used an unconventional method to measure a 15-minute time limit for a test. He just used 7 and 11-minutehourglasses. During the whole time he turned sandglasses only 3 times (turning both hourglasses at once count as one flip).
Explain how the teacher measured 15 minutes.
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Sand-Glass II. - solution
When the test began, the teacher turned both 7min and 11min sand-glasses. After the 7min one spilt its last grain, he turned it upside down (the 11min one is still to spill sand for another 4 minutes). When the 11min sand-glass was spilt, he turned the 7min one upside down for the last time. And that’s it.
A teacher of mathematics used an unconventional method to measure time for a test lasting 15 minutes. He used just a sand-glass, which spills in 7 minutes and a second sand-glass, which spills in 11 minutes. During the whole time he turned sand-glasses only 3 times. Explain how the teacher measured 15 minutes.
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Burning Fuses - Back to the Water and Weighing Puzzles
Your job is to measure 45 minutes, if you have only two cords and matches to light the cords.
1. The two cords are twisted from various materials and so their different segments can burn at different rates.
2. Each cord burns from end to end in exactly one hour.
Describe your way of measuring the 45 minutes.
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Igniter Cords - solution
Start fire on both ends of one igniter cord and on one end of the second igniter cord. The very moment the first cord (where both ends burn) stops burning (that is after 30 minutes), start fire on the other end of the second cord (otherwise it would burn another 30 minutes). Thus the second igniter cord burns just 15 minutes from then. And that is all together 45 minutes.
Your job is to measure 45 minutes, if you have only two igniter cords and matches to light the cords. The two igniter cords have the following features:
1. They are twisted from various materials and also different parts can burn at different speed (e. g. after ten minutes they will not burn at the same point).
2. Every cord burns from ignition to the end exactly one hour.
Describe your way of measuring the 45 minutes.
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Honestants and Swindlecants I. - Back to the Logic Problems
These are typical logic problems which can be solved by using classic logic operations.
There are two kinds of people on a mysterious island. There are so-called Honestants who speak always the truth, and the others are Swindlecants who always lie.
Three fellows (A, B and C) are having a quarrel at the market. A gringo goes by and asks the A fellow: "Are you an Honestant or a Swindlecant?" The answer is incomprehensible so the gringo asks B: "What did A say?" B answers: "A said that he is a Swindlecant." And to that says the fellow C: "Do not believe B, he is lying!" Who is B and C?
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Honestants and Swindlecants I. - solution
It is impossible that any inhabitant of such an island says: „I am a liar.“ An honestant would thus be lying and a swindlecant would be speaking truth. So B must have been lying that A said "I (A) am swindlecant" and therefore B is a swindlecant. And that means that C is telling the truth saying B is lying – so C is an honestant. However, it is not clear what is A.
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Honestants and Swindlecants II. - Back to the Logic Problems
Afterwards he meets another two aborigines. One says: "I am a Swindlecant or the other one is an Honestant." Who are they?
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Honestants and Swindlecants II. - solution
Logical disjunction is a statement "P or Q". Such a disjunction is false if both P and Q are false. In all other cases it is true. Note that in everyday language, use of the word "or" can sometimes mean "either, but not both" (e.g., "would you like tea or coffee?"). In logic, this is called an "exclusive disjunction" or "exclusive or" (xor).
So if A was a swindlecant, then his statement would be false (thus A would have to be an honestant and B would have to be a swindlecant). However, that would cause a conflict which implicates that A must be an honestant. In that case at least one part of his statement is true and as it can’t be the first one, B must be an honestant, too.
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Honestants and Swindlecants III. - Back to the Logic Problems
Our gringo displeased the sovereign with his intrusive questions and was condemned to death. But there was also a chance to save himself by solving the following logic problem. The gringo was shown two doors - one leading to a scaffold and the second one to freedom (both doors were the same) and only the door guards knew what was behind the doors. The sovereign let the gringo put one question to one guard. And because the sovereign was an honest man he warned that exactly one guard is a Swindlecant.
What question can save the gringo's life?
Classic wording:
Two Doors
You are travelling down a country lane to a distant village. You reach a fork in the road and find a pair of identical twin sisters standing there. One of the sisters always tells the truth and the other always lies.
If you are allowed to ask only one question to one of the sisters to find the correct road to the village, what is your question?
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Honestants and Swindlecants III. - solution
There are a few types of questions:
- Indirect question: „Hey you, what would the other guard say, if I asked him where this door leads?“ The answer is always negated.
- Tricky question: „Hey you, does an honestant stand at the door to freedom?“ The answer will be YES, if I am asking an honestant who is standing at the door to freedom, or if I am asking a swindlecant standing again at the same door. So I can walk through the door. A similar deduction can be made for negative answer.
- Complicated question: „Hey you, what would you say, if I asked you ...?“ An honestant is clear, but a swindlecant should lie. However, he is forced by the question to lie two times and thus speak the truth.
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Honestants and Swindlecants IV. - Back to the Logic Problems
Our gringo was lucky and survived. On his way to the pub he met three aborigines. One made this statement: "We are all Swindlecants." The second one concluded: "Just one of us is an honest man." Who are they?
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Honestants and Swindlecants IV. - solution
The first one must be a swindlecant (otherwise he would bring himself into a liar paradox), and so (knowing that the first one is lying) there must be at least one honestant among them. If the second one is lying, then (as the first one stated) the third one is an honestant, but that would make the second one speak the truth. So the second one is an honestant and C is a swindlecant.
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Honestants and Swindlecants V. - Back to the Logic Problems
In the pub the gringo met a funny guy who said: "If my wife is an Honestant, then I am Swindlecant." Who is this couple?
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Honestants and Swindlecants V. - solution
It is important to explore the statement as a whole. Truth table of any implication is as follows:
truth truth truth
truth lie lie
lie truth truth
lie lie truthP Q P=>Q
In this logical conditional („if-then“ statement) p is a hypothesis (or antecedent) and q is a conclusion (or consequent).
It is obvious, that the husband is not a Swindlecant, because in that case one part of the statement (Q) „ ... then I am Swindlecant.“ would have to be a lie, which is a conflict. And since A is an Honestant, the whole statement is true.
If his wife was an Honestant too, then the second part of statement (Q) „ ... then I am Swindlecant.“ would have to be true, which is a conflict again. Therefore the man is an Honestant and his wife is a Swindlecant. Or is it a paradox?
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Honestants and Swindlecants VI. - Back to the Logic Problems
When the gringo wanted to pay and leave the pub, the bartender told him how much his drink costed. It was quite expensive, so he asked the bartender if he spoke the truth. But the gringo did not hear the whispered answer so he asked a man sitting next to him about it. And the man said: "The bartender said yes, but he is a big liar." Who are they?
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Honestants and Swindlecants VI. - solution
This one seems not clear to me. However, the bartender and the man sitting next to the gringo must be one honestant and one swindlecant (not knowing who is who).
1. the bartender must have said: "Yes, I speak the truth" (no matter who he is)
2. the man sitting next to gringo said: "The bartender said yes, but he is a big liar.", which is true only if BOTH parts of the sentence are true (for logical conjuction see http://en.wikipedia....al_conjunction)
o if it's true - the man is an honestant and the bartender a swindlecant,
o if it's false = "he is a big liar" is false - bartender is an honestant and the man is a swindlecant.
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Honestants and Swindlecants VII. - Back to the Logic Problems
Going out of the pub, the gringo heard about a fantastic buried treasure. He wanted to be sure so he asked another man who replied:
"On this island is a treasure, only if I am an honest man."
So shall he go and find the treasure?
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Honestants and Swindlecants VII. - solution
It is important to explore the statement as a whole. Truth table of any logical equivalence is as follows:
truth truth truth
truth lie lie
lie truth lie
lie lie truthP Q P<=>Q
If the man is an Honestant, then the whole statement must be true. One part of it, where he said that he is an honest man is true then and so the other part (about the treasure) must be true, too. However, if he is a Swindlecant, the whole statement is a lie. The part mentioning that he is an honest man is in that case of course a lie. Thus the other part must be truth. So there must be a treasure on the island, no matter what kind of man said the sentence.
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Honestants and Swindlecants VIII. - Back to the Logic Problems
Thinking about the treasure, the gringo forgot what day it was, so he asked four aborigines and got these answers:
A: Yesterday was Wednesday.
B: Tomorrow will be Sunday.
C: Today is Friday.
D: The day before yesterday was Thursday.
Because everything you need to know is how many people lied, I will not tell. What day of the week was it?
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Honestants and Swindlecants VIII. - solution
The important thing was what we did not need to know. So if we knew how many people lied we would know the answer. And one more thing – B and D said the same.
If all of them lied, there would be 4 possible days to choose from (which one is not clear).
If only one of them spoke the truth, it could be A or C, so 2 possible days (not clear again).
If two of them were honest, it would have to be B and D saying that it was Saturday.
Neither 3 nor all 4 could have been honest because of an obvious conflict.
So it was Saturday.
Pouring water I.
in Water and Weighing Puzzles
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Pouring water I. - Back to the Water and Weighing Puzzles
If you had a 5-liter bowl and a 3-liter bowl, and an unlimited access to water, how would you measure exactly 4 liters?
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