This conversation is closed.

## What question would you ask to identify whether or not you were chatting with a well developed software or a person?

Imagine an experiment where you are asked to chat with one hundred people online, no sound or image, just text. Three of them are actually not real, they an extremely good automated response systems. Your task is to identify those three. You are allowed to ask only one and same question from everyone. People on the other end are specifically chosen such that none of them have similar personality. Programs are also given a unique personality. Only trick is, while you ask questions, programs observe responses of everybody else and may or may not change behavior based on that. What would your question be?

P.S. If you would like to be sure how good is 'extremely good' automated response system in the though experiment above, you may consider it to be the best of such systems you think is possible.

**Topics:**imitation intelligence personality software

## Closing Statement from Farrukh Yakubov

Now that the conversation is over I would like to leave you with more thoughts.

Imagine, this experiment took place and you asked your question, and indicated three of the participants as programs. What if this experiment was not what you thought it was, and after the experiment you were told that 100 participants were all human or all programs, or even a single person answering 100 different ways? What if the purpose of the experiment was not about the capabilities of programs, but about the people - to see how people percieve an intelligent software? Did you think about this possibility?

On the other hand, if the experiment was to test the programs, how effective do you thinki it would be to use this same question of the experiment? i.e. asking "What question would you ask to identify whether or not you were chatting with a well developed software or a person?" from each of the 100 participants.

It is up to you to chose the post experinment scenario, and you would be correct. Because, the experiment can work both ways wether you decide to look at this experiment as an attemp to test programs, or a way of understanding peoples' understanding of programs.

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## Keith W Henline 100+

## Yoka Feng 20+

## Keith W Henline 100+

## Yoka Feng 20+

## Farrukh Yakubov 50+

## Yoka Feng 20+

## Farrukh Yakubov 50+

## Keith W Henline 100+

## Farrukh Yakubov 50+

## Keith W Henline 100+

## Yoka Feng 20+

## Farrukh Yakubov 50+

Hi Yoka, I think Keith said it's wrong because just asking any one of them about the safe way out does not provide sufficient information to identify where the doors lead. You may just get lucky and ask the person that knows the truth, or it may turn out otherwise. The trick is to assume the person you are going to talk to could be both of them. If you asked any one of them which way the other one would point to as safe, they could either lie or tell the truth. But if they lie, then the other person would tell the truth, and vice versa. While having only two possible answer choices, opposite of lie is the truth, of truth is a lie.

Therefore, no matter who you ask, you either get truth about the lie, or a lie about the truth. Thus you get a lie. Now you can be sure about where the doors lead.

## Yoka Feng 20+

## Yoka Feng 20+

Why this answer can't be programmed in a smart robot?

## Yoka Feng 20+

And actually, I don't think this kind of question can help me judge a person on the internet in our real life. I'd be too lazy to answer it and pass my attention to chat with other people.

## Keith W Henline 100+

## Keith W Henline 100+

By the way I had a good laugh about your "quantum entangled particles" explanation. By the way if you have not seen Princess Bride by all means watch it some time.. 3 min. part on logic- https://www.youtube.com/watch?v=0sPVEBAtwmg

## Farrukh Yakubov 50+

If the purpose is just to provide the sorted index of a requested entry, Selection algorithm to find kth smallest item from the set has a linear complexity. But it is not ideal if random kth items are being continuously accessed.

Thus I have a solution in mind, that modifies, reuses and combines existing methods to create generic non-comparative sorting that works with a set of data (let size of the set be 'n'), where each item has arbitrary length, and does so in linear time.

Edit: I don't expect it to be same or similar to what you have in mind, its just another way of doing things.

Algorithm is explained on the next comment. This is going to be divided into few chunks due to limits of this conversation platform.

## Farrukh Yakubov 50+

It does not modify the original data set, but produces an array of pointers (referred as the map) of length n. Other memory that will be used is of size 256 integers (referred as the workspace), which is no longer required after completion of the algorithm. I'm going to start describing it from the lowest component to highest. Also, I'll use C notation to avoid wordy sentences.

First component takes advantage of pointer manipulation and underlying architecture.It is a partial Counting sort. This stage takes in only set of bytes.

1.Reset workspace to zeros.

2.for each item e in the input set, perform workspace[e]++ //offset of each entry in workspace represents a value of an item; value at the offset represents the # of items in the set that are equal to the item.

3.for i=1 to 'size of workspace', perform workspace[i]+=workspace[i-1] //value of each entry in workspace represents #of items in the set that are less than or equal to the item with value 'offset'.

4.First component does not proceed with constructing sorted array, but instead provides a way find index of each item as if the set was sorted. Index of 'someItem' from the input set in a sorted set would be workspace[someItem]. Higer level component will obtain index for each item exacly once.

Second component is a radix sort, but bytes will be used for grouping instead of bits. The map is initialized such that map[i] contains adress of set[i]. At each iteration, the first component is used to divide each subsequent set up to 256 groups, until they no longer need sorting, i.e. is of length 1. Also, actual items in the set will not be moved around, instead only the pointers in the map are modified such that map[i] is the index of ith item in a "sorted" set.

Complexity is explained on the next comment.

## Farrukh Yakubov 50+

Counting sort(first subcomponent) has complexity O(n+k), k is maximum possible value of each integer item (256 in this case), n is length of the current subset. This is a stable non-comparative sort.

Radix sort using stable non-comparative sort has execution time of Θ(d(n+k)), d is length of items in the set. n is size of the set. For arbitrary length items, upper bound should be O(p(n+k)). p is an avarage length of items in set. p.s. items of length less than p, will no longer be in subsets of size larger than 1 after p iterations.

I may not use the same method if the nature of the input is known beforehand.

Final comment, in the process I discovered this platform does not include anything after 'less than' symbol.

## Danger Lampost 20+

If each input element were allowed to be a random 64-bit integer, the size of your work space would be 16 quintillion bytes, which would be an issue.

Am I missing something?

## Timo X

## Keith W Henline 100+

Another way to phrase it is: Which door will the other guy tell me go through? and then go through the other one.

Good work out Farrukh, Yoka and Timo... remember it is the journey that is most important and all of you took the same journey. Because you got different answers should in no way spoil your journey because there "is" no destination, the destination is an illusion. Buddha put it this way:

"Nirvana is this moment seen directly. There is no where else than here. The only gate is now. The only doorway is your own body and mind. There’s nowhere to go. There’s nothing else to be. There’s no destination. It’s not something to aim for in the afterlife. It’s simply the quality of this moment."

## natasha nikulina 50+

If i choose life i'll go to the door that the lier will show me, no matter what i asked.

edited

Freedom is a lie, the lier will show me the door to freedom, if i ask for it. If i ask the door to sudden death, he will show me the same door, to freedom , because he is a lier.

If ask the person who always tells truth, where is the door to freedom , he will show me the door to sudden death, because , it's a real freedom. If i ask him where is the door to sudden death , he will show me the door to sudden death, because he always tells truth.

## Keith W Henline 100+

## natasha nikulina 50+

It's me,not Farrukh.

I've experienced the beauty of logic on the way to..., but now i see the flaws.

Frankly, i don't see any version of explanation that can eliminate the uncertainty.

In case, there is such and you know it, please share !

## Farrukh Yakubov 50+

## Farrukh Yakubov 50+

This must be one of the hardest questions to answer, due to many problems and not all of them can be compared. Perhaps this would be a good question of choice for the above thought experiment. :)

Science is key to move everything forward, and computer science seems to be the beating heart of the current era. I am not sure what I would want to tackle first, but I would let my interests lead the way.

## Farrukh Yakubov 50+

-this is a response to your response of Keith's response to you main comment. :)

## natasha nikulina 50+

OK then, enjoy it !

## Keith W Henline 100+

## Keith W Henline 100+

## natasha nikulina 50+

I mean, my ego is thin enough :)

Your riddle and that episode from " The Princes Bride " gave me an aha moment and i am grateful for that. Actually, those two are in perfect congruence. Probably i was a bit upset that there seemed to be nobody who was interested, but on the other hand, it's not easy to language what i've got, so it's OK anyway.

Thank you !

## Danger Lampost 20+

Am I missing something?

## Keith W Henline 100+

## Yoka Feng 20+

## Lawren Jones 30+

If the person says 'Yes' and is lying, then the other person would truthfully say 'No,' so the right-hand door leads to freedom.

If the person says 'Yes' and is truthful, then the other person would deceitfully say 'Yes,' so the right-hand door leads to freedom.

This is a logic question which would be much easier for an advanced computer to figure out than a person.

If the person says 'No' and is truthful, then the other person would deceitfully say 'No' and the left-hand door leads to freedom.

If the person says 'No' and is lying, then the other person would truthfully say 'Yes' and the left-hand door leads to freedom.

So, if the person says 'Yes,' then the right-hand door leads to freedom. And if the person says 'No,' then the left-hand door leads to freedom.

## Keith W Henline 100+

## Keith W Henline 100+

## Bryan Maloney 30+

## Keith W Henline 100+