Logic Puzzles and reasoning
- Galahad
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Re: Logic Puzzles and reasoning
"Bad" might not be the word for it. Just "impossible", mostly because of its subjective nature.
From a coldly logical perspective (and making assumptions of my own), any outcome will anger at least one group.
For example: I could allow the independents to declare their independence, then relocate the group which is intolerant of any disagreement (the "first" one) to that independent state (there will not be disagreement because nothing in the question suggested the independent group disagrees with any of the other groups). This would satisfy everyone but the "third" group, who will begin rioting since the relocation could count as eviction. The impossibility comes in with the following maxim:
Let Group A = Residents who want to evict any who disagree with them.
Let Group B = Residents who want Group A to be evicted, and will riot if they are not.
Let Group C = Residents who will riot if Group A is evicted (i.e do not want Group A to be evicted)
There are only two possible states: Evicted and unevicted. One cannot be both, and one cannot be neither.
So... There is a paradox. Either state will provoke at least one group.
(Oh, and one might say: but what about the two halves not talking to each other? Will that not rectify this? Sadly, no. The question said that all three groups, A to C, are scattered throughout both halves of the town. So, the only difference is that one half might not know about it.)
So...
Perhaps the real solution is to tell everyone to stop being intolerant twats.
From a coldly logical perspective (and making assumptions of my own), any outcome will anger at least one group.
For example: I could allow the independents to declare their independence, then relocate the group which is intolerant of any disagreement (the "first" one) to that independent state (there will not be disagreement because nothing in the question suggested the independent group disagrees with any of the other groups). This would satisfy everyone but the "third" group, who will begin rioting since the relocation could count as eviction. The impossibility comes in with the following maxim:
Let Group A = Residents who want to evict any who disagree with them.
Let Group B = Residents who want Group A to be evicted, and will riot if they are not.
Let Group C = Residents who will riot if Group A is evicted (i.e do not want Group A to be evicted)
There are only two possible states: Evicted and unevicted. One cannot be both, and one cannot be neither.
So... There is a paradox. Either state will provoke at least one group.
(Oh, and one might say: but what about the two halves not talking to each other? Will that not rectify this? Sadly, no. The question said that all three groups, A to C, are scattered throughout both halves of the town. So, the only difference is that one half might not know about it.)
So...
Perhaps the real solution is to tell everyone to stop being intolerant twats.
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- Light-footed
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Re: Logic Puzzles and reasoning
We should use the Systems Engineering approach.
1) It improves methods for determining the scope of needs to be met by a system
2) It improves methods for defining a system
3) It addresses the total system with all of its elements from a life-cycle perspective
4) The interactions in the overall system hierarchy will be considered
5) Organized integration
6) It Establishes a disciplined approach with appropriate review, evaluation and feedback.
1) It improves methods for determining the scope of needs to be met by a system
2) It improves methods for defining a system
3) It addresses the total system with all of its elements from a life-cycle perspective
4) The interactions in the overall system hierarchy will be considered
5) Organized integration
6) It Establishes a disciplined approach with appropriate review, evaluation and feedback.
- Raven Song
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Re: Logic Puzzles and reasoning
I change my answer on favour of Tetsudra's statistics. I go to the oub, have a drink, and order a bacon sammich
Learn the rules like a pro, so you can break them like an artist. Pablo Picasso
- VoxLupus
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Re: Logic Puzzles and reasoning
I know I am a bit late to the party, but I decided to give Galahad's puzzle a shot without looking at the solution.
I took a different approach, writing the problem as a set of linear simultaneous equations and equating the columns and rows and subtracting the resulting equations from each other, I was too lazy to do matrices I found the solution, and an almost-a-solution, only one of the diagonals was out.
I took a different approach, writing the problem as a set of linear simultaneous equations and equating the columns and rows and subtracting the resulting equations from each other, I was too lazy to do matrices I found the solution, and an almost-a-solution, only one of the diagonals was out.
- Galahad
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Re: Logic Puzzles and reasoning
A most interesting approach. Well done, Vox. That is correct.
- Galahad
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Re: Logic Puzzles and reasoning
A week has passed by, and still no puzzles.
Fear not, logicians - I am here with another!
The solution shall be revealed tomorrow evening (8 September).
Fear not, logicians - I am here with another!
In front of you is a crowd of humans, dogs and spiders. Ignoring the utter confusion (and your arachnophobia), the riddler comes to your side. You are then blindfolded.
"Humans typically have two legs; dogs, four; spiders, eight," he mutters coolly and lowly, "All in all, there are twice as many humans as dogs. The total number of LEGS is 38, and the total number of HEADS is either 8, 10 or 11. But to make it even trickier... Either two OR three of the humans are one-legged amputees.
How many humans, dogs and spiders are there, respectively?"
The solution shall be revealed tomorrow evening (8 September).
Re: Logic Puzzles and reasoning
2 spiders, 3 dogs, and 6 humans, two of which are one-legged amputees.
- Trace
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Re: Logic Puzzles and reasoning
I will second Lee's answer. 6 humans, 3 dogs, 2 spiders. 2 amputees, 11 heads.
You can start by determining the number of amputees. As the final number is even, there must be two.
From there, you know that you need at least one dog. At this point, you have 6 legs and 2 heads.
After some guesswork and trial and error, you can find the rest in time. The key is maintaining the dog-human ratio.
You can start by determining the number of amputees. As the final number is even, there must be two.
From there, you know that you need at least one dog. At this point, you have 6 legs and 2 heads.
After some guesswork and trial and error, you can find the rest in time. The key is maintaining the dog-human ratio.
"I change shapes just to hide in this place, but I'm still, I'm still an animal" -Miike Snow, 'Animal'
"Where there's life, there's hope."-Terence
"Where there's life, there's hope."-Terence
- Galahad
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Re: Logic Puzzles and reasoning
To those who have submitted answers: Thank you for the participation and interest. May I please remind you, however, that the rules of the game ask that you submit only your answer itself, not your explanation. This allows others the opportunity to answer independently.
- Rakuen Growlithe
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Re: Logic Puzzles and reasoning
That's just the sort of thing they taught you algebra for.
"If all mankind minus one, were of one opinion, and only one person were of the contrary opinion, mankind would be no more justified in silencing that one person, than he, if he had the power, would be justified in silencing mankind."
~John Stuart Mill~
“Give me the liberty to know, to utter, and to argue freely according to conscience, above all liberties.”
~John Milton~
~John Stuart Mill~
“Give me the liberty to know, to utter, and to argue freely according to conscience, above all liberties.”
~John Milton~
- Trace
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Re: Logic Puzzles and reasoning
Whoops. My apologies.Galahad wrote:To those who have submitted answers: Thank you for the participation and interest. May I please remind you, however, that the rules of the game ask that you submit only your answer itself, not your explanation. This allows others the opportunity to answer independently.
"I change shapes just to hide in this place, but I'm still, I'm still an animal" -Miike Snow, 'Animal'
"Where there's life, there's hope."-Terence
"Where there's life, there's hope."-Terence
- Galahad
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Re: Logic Puzzles and reasoning
Leeward and Trace are correct! Well done.
The answer is indeed 6 humans (with 2 amputees among them), 3 dogs and 2 cats.
To find the solution, begin by determining how many amputees there are.
If you let x = number of dogs and y = number of spiders, the equation for the number of legs becomes:
4x + (2)(2)(x) + 8y - (2 or 3) = 38.
(That is, four legs for each dog, two legs for each human (which will be twice the amount of dogs), eight legs for each spider, and subtract the missing legs due to the amputees.)
When simplified:
8x + 8y = 40 or 41.
Now, note that we should be able to divide by 8. It makes more sense for 40 to be used, as is not only divisible wholly by 8 but if we divided 41 by 8 we would get a fraction, which makes no sense - how can we have half spiders, or fractions of dogs? We can only have whole creatures.
Therefore, logic tells us then that there must be 2 amputees, so that the answer is a neat, whole 40. (We add 2 to both sides instead of 3).
x + y = 5 is what we are left with.
Now, we can figure out the answer intuitively. Keep in mind that there are either 8, 10 or 11 heads. So, let us think up some numbers according to the rules above, and see which one works:
Dogs / Humans / Spiders / Heads
1 / 2 / 4 / 7
2/ 4 / 3 / 9
3 / 6 / 2 / 11
The third option is the only one that meets the last rule - the one about the number of heads, and it is thus our answer.
The answer is indeed 6 humans (with 2 amputees among them), 3 dogs and 2 cats.
To find the solution, begin by determining how many amputees there are.
If you let x = number of dogs and y = number of spiders, the equation for the number of legs becomes:
4x + (2)(2)(x) + 8y - (2 or 3) = 38.
(That is, four legs for each dog, two legs for each human (which will be twice the amount of dogs), eight legs for each spider, and subtract the missing legs due to the amputees.)
When simplified:
8x + 8y = 40 or 41.
Now, note that we should be able to divide by 8. It makes more sense for 40 to be used, as is not only divisible wholly by 8 but if we divided 41 by 8 we would get a fraction, which makes no sense - how can we have half spiders, or fractions of dogs? We can only have whole creatures.
Therefore, logic tells us then that there must be 2 amputees, so that the answer is a neat, whole 40. (We add 2 to both sides instead of 3).
x + y = 5 is what we are left with.
Now, we can figure out the answer intuitively. Keep in mind that there are either 8, 10 or 11 heads. So, let us think up some numbers according to the rules above, and see which one works:
Dogs / Humans / Spiders / Heads
1 / 2 / 4 / 7
2/ 4 / 3 / 9
3 / 6 / 2 / 11
The third option is the only one that meets the last rule - the one about the number of heads, and it is thus our answer.