The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.
I'll leave it to somebody else to write an essay on how the limitations of inductive reasoning are handled in the philosophy of science. I'm not up for that tonight.
It's not obvious (to me) why there is only a finite set of possible functions.
We say that two possible functions f_1 and f_2 are considered to be the same if they have the same "image", i.e., f_1 and f_2 are the same iff f_1(a,b,c,d) = f_2(a,b,c,d) Ɐ a,b,c,d ∈ [e,f].
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.
I'll leave it to somebody else to write an essay on how the limitations of inductive reasoning are handled in the philosophy of science. I'm not up for that tonight.
The "puzzle" is impossible to solve because, regardless of how many trials are performed, there will always be an infinite set of possible rules that will still work.
I'll leave it to somebody else to write an essay on how the limitations of inductive reasoning are handled in the philosophy of science. I'm not up for that tonight.
I think that in the context this was given in, it's reasonable to assume that the answer was at least somewhat simple. Greatly restricting the possibilities.
Okay, here's a far more challenging and also far more interesting version of the original puzzle. I'm sure MDM will enjoy it.
As before, suppose we are considering a function F with four arguments a, b, c, and d. For a given set of values, F returns either 0 or 1. Suppose that a, b, c, and d are integers within the domain [e, f], where e<f are both integers. We say that two possible functions f_1 and f_2 are considered to be the same if they have the same "image", i.e., f_1 and f_2 are the same iff f_1(a,b,c,d) = f_2(a,b,c,d) Ɐ a,b,c,d ∈ [e,f].
Given the above premises, there are (obviously) only a finite set of possible functions F that could characterise the unknown rule. Suppose that as part of the setup, one of the possible functions F is chosen at random, with each possibility having an equal chance of being chosen.
Now here is the problem:
- What algorithm will on average minimise the number of trials required to determine F?
- How many trials does that algorithm take on average?
Here's another one:
(http://i.huffpost.com/gen/2975616/thumbs/o-MATHS-570.jpg)
Plug in the numbers 1 through 9, using each digit just once, filling each of the blank spaces. Using the identified operations and the inserted values, a correct answer would lead to a solution of 66 (those double dots stand for division).
For MDM, it is obvious there is a finite set to Cathy's question due to the fact that ABCD as well as E and F are INTEGERS. At a surface level, I agree that there is no strategy other than iterating through all of the possible steps. So the average steps would be (f-e+1)^4/2. If there is a strategy, it is not apparent to me.Agree that there is a finite set of inputs to the to-be-determined function F.
E.g., let's say F = 0 for all inputs. One black box could have =0*a*b*c*d, another could have =0*(a^2+b^2+c^2+d^2), etc., leading to an infinite set of possible functions F that could characterise the unknown rule
At a surface level, I agree that there is no strategy other than iterating through all of the possible steps. So the average steps would be (f-e+1)^4/2. If there is a strategy, it is not apparent to me.
As far as I can tell Cathy, tyir is right that you effectively have a 4 dimensional bitmap and since each voxel is randomly colored, you must sample at each voxel. There isn't a randomized algorithm for this problem. Luckily, the number of steps is finite because the range is limited to a finite number of integers.
what it means for two functions to be the same function.
what it means for two functions to be the same function.
Thus tying this thread to the marriage thread(s): "...and the two shall become one."
http://www.nytimes.com/interactive/2015/07/03/upshot/a-quick-puzzle-to-test-your-problem-solving.html?_r=0
Interesting, but the puzzle jumps to weird conclusion. People don't like hearing 'no' so they can't solve the puzzle? What?
And the many shall become one - now it's polygamy! ;)For example, if the domain and codomain of an arbitrary function Q(x) are both restricted to {0}, then there is only exactly one possible function Q(x). The notations x, x**2, x+x, or sin(x) are just different notations for the same function Q : {0} → {0}.what it means for two functions to be the same function.Thus tying this thread to the marriage thread(s): "...and the two shall become one."
What about that conclusion seems unreasonable?
Conservatives like to talk with other conservatives, and liberals like to talk with other liberals, because they all agree with each other.
If people are guessing an answer without receiving a 'no' response, then they aren't reasoning out the solution correctly.Agreed.
Maybe because it's an online puzzle with no consequences, so they don't care, or maybe they misinterpreted the question, or various other reasons. Drawing the conclusion that people would be disappointed seeing the red 'no' boxes, and would rather fail to solve the puzzle (which is also receives a negative response) than have to see them, is quite a stretch in logic.Maybe, maybe not. There seems a difference between "would rather fail to solve" vs. "don't think to test a counterexample." I've seen a preponderance in real-life of people looking to "prove their idea correct", with much fewer looking to "run a test that would prove their idea wrong."
I got a little irritated when I kept seeing 'yes' pop up. Entering the wrong sequence is required to confirm the answer.Good! Just remember to apply the same thought process in real life. Although, perhaps you do so already, in which case - keep up the good work!
I included an explicit definition of sameness for greater clarity, but it was unnecessary because a function is a relation between a domain and codomain; the notation used to express that relation is not part of the "essence" of the function.
For example, if the domain and codomain of an arbitrary function Q(x) are both restricted to {0}, then there is only exactly one possible function Q(x). The notations x, x**2, x+x, or sin(x) are just different notations for the same function Q : {0} → {0}.
Good puzzle
16 yes 10 no for the recordSpoiler: show
Also have your puzzle be wrong.
If you put in 1.9999999999, 2, and 3, then it returns "no". So their "solution" is wrong. All possible iterations of the solution I picked, will still cause it to say Yes.
Good puzzle
16 yes 10 no for the recordSpoiler: show
I can almost guarantee (without checking the Javascript) that that isn't actually rounding to 16 decimal places. Floating-point data types (https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) are rounded to a fixed [total] number of base-2 digits (or a variable number to the right of the decimal point), which means that the base-10 precision can vary.Also have your puzzle be wrong.
If you put in 1.9999999999, 2, and 3, then it returns "no". So their "solution" is wrong. All possible iterations of the solution I picked, will still cause it to say Yes.
Aside from the computer limitations I mentioned above, it's also true that 1.999999... (the repeating decimal) actually is mathematically equal to 2. See these proofs (https://en.wikipedia.org/wiki/0.999...).
Good puzzle
16 yes 10 no for the recordSpoiler: show
I can almost guarantee (without checking the Javascript) that that isn't actually rounding to 16 decimal places. Floating-point data types (https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) are rounded to a fixed [total] number of base-2 digits (or a variable number to the right of the decimal point), which means that the base-10 precision can vary.Also have your puzzle be wrong.
If you put in 1.9999999999, 2, and 3, then it returns "no". So their "solution" is wrong. All possible iterations of the solution I picked, will still cause it to say Yes.
Aside from the computer limitations I mentioned above, it's also true that 1.999999... (the repeating decimal) actually is mathematically equal to 2. See these proofs (https://en.wikipedia.org/wiki/0.999...).
I couldn't figure out how to put in an infinite number of 9's, I put in a finite number, so I put in a solution that met the "rule" but returned a "no." It would have been possible to to program it such that what I put in wouldn't have returned a "no." They didn't do that though, because they were afflicted with confirmation bias.
Will anyone here admit to guessing without getting "no" answer? If so can you tell us about it?
Confirmation bias? Their conclusion is still supported by the data-- far too many people are guessing the rule before getting a "no" result. This is true regardless of what the exact "rule" is (rising numbers vs rounded numbers rising).
Will anyone here admit to guessing without getting "no" answer? If so can you tell us about it?