I am stuck on this problem for quite a while, it is basically reverse engineering bulls and cow game.
Read more here: http://rosettacode.org/wiki/Bulls_and_cows
I am not able to develop a logic for the problem given below, if you can think of a solving approach please comment the same.
Problem Statement:
Given few clue words(of form ABCD/DBCA etc) and the number of cows and bulls for each word,program
should be able to work out the actual word by evaluating the given clue words and generate the output secret word.
TEST CASES:
Input:
4
DBCC 0 2
CDAB 2 1
CAAD 1 2
CDDA 2 0
Output:
BDAA
The idea is to reduce the space of possible solutions. Before you start, all 4^4 combinations are possible. After you read the first clue [DBCC 0 2 ], you can eliminate a number of possible solutions, in this particular example you can eliminate all states which have a D in the first place, all which have a B in the second place and so on. Just eliminate all possible solutions which do not "fit" the current clue.
Do this with each clue, until only one solution is left. Another interesting problem of course is how to generate good clue patterns.
The way I did it is:
1. Generate all possible words, put them in a list (array)
2. Randomly select one of them (first question)and ask for clues
3. Take the answer (let's say it is 2,1)
4. Start comparing that question with
first, second, ..., to the last word from the list
5. if they give the same clue: count them, plac
Related
I believe these questions should yield the same answer:
(The question is "what are the odds of throwing at least two 1's with 3 dice")
https://www.wolframalpha.com/input/?i=1%2C1+on+3+6-sided+dice
(solution: 15/216)
and
https://www.wolframalpha.com/input/?i=at+least+two+1+on+3+6-sided+dice
(solution: 16/216)
It seems the first does not seem take into account 1,1,1 as solution. Perhaps it interprets the question as exactly two ones. But when showing the example roll on the bottom and pressing reroll, it does show up as a solution. This seems like a bug.
Can anyone shed some light on this?
---- UPDATE ----
So the bug boils down to:
When asking WolframAlpha the question:
"What are the odds of rolling exactly two 1's (and no more) with 5 dice?" (https://www.wolframalpha.com/input/?i=1%2C1+on+3+6-sided+dice)
The solution: 15/216 is correct, but the example box below shows 1,1,1 as possible result (press reroll a couple of times). This is especially unclear when you throw 5 dice and seeing 1,1,1,4,1 as valid result. It's very unclear if it interprets as exactly 1,1 or at least 1,1.
As #paxdiablo pointed out, Wolfram Alpha answers https://www.wolframalpha.com/input/?i=exactly+two+1+on+3+6-sided+dice with the same probability, but without the bug (1,1,1 does not show up as valid answer).
I would suggest that it is interpreting 1,1 on 3 6-sided dice as exactly two 1's, not at least two. That discounts the 1,1,1 solution which explains the disparity:
1 1 2-5 (5)
1 2-5 1 (5)
2-5 1 1 (5)
----
(15)
1 1 1 (1)
----
(16)
That gels with the precise probability given of 5/72 (15/216).
As to why it's putting forward 1,1,1 when you throw the dice, I would say that is a a bug if it's meant to agree with the input. That certainly appears to be the case as it always has the 1,1 at the start.
Interestingly enough, the two 1 on 3 6-sided dice query generates the same probability but, after trying 50 dice rolls, the 1,1,1 combo never appeared. Perhaps the natural language processing is different between the two different areas.
As an aside, I have submitted feedback to Wolfram Alpha to see what they say about it. If they ever get back to me, I'll post the response here.
Update 1: Standard form letter response has been received, so hopefully a further response will come shortly:
We appreciate your feedback regarding Wolfram|Alpha. The issue you reported has been passed along to our development team for review. Thank you for helping us improve Wolfram|Alpha.
This is my first post on Stack Overflow, so please excuse my mistakes if I'm doing something wrong.
Ok so I'm trying to find an algorithm/function/something that can calculate how many times I have to do the same type of shuffle of 52 playing cards to get back to where I started.
The specific shuffle I'm using goes like this:
-You will have two piles.
-You have the deck with the back facing up. (Lets call this pile 1)
-You will now alternate between putting a card in the back of pile 1 Example: Let's say you have 4 cards in a pile, back facing up, going from 4 closest to the ground and 1 closest to the sky (Their order is 4,3,2,1. You take card 1 and put it beneath card 4 mening card 1 is now closest to the ground and card 4 is second closest, order is now 1,4,3,2. and putting one in pile 2. -Pile 2 will "stack downwards" meaning you will always put the new card at the bottom of that pile. (Back always facing up)
-The first card will always get put at the back of pile 1.
-Repeat this process until all cards are in pile 2.
-Now take pile 2 and do the exact same thing you just did.
My question is: How many times do I have to repeat this process until I get back where I started?
Side notes:
- If this is a common way of shuffling cards and there already is a solution, please let me know.
- I'm still new to math and coding so if writing up an equation/algorithm/code for this is really easy then don't laugh at me pls ;<.
- Sorry if I'm asking this at the wrong place, I don't know how all this works.
- English isn't my main language and I'm not a native speaker either so please excuse any bad grammar and/or other grammatical errors.
I do however have a code that does all of this (Link here) but I'm unsure if it's the most effective way to do it, and it hasn't given a result yet so I don't even know if it works. If you wan't to give tips or suggestions on how to change it then please do, I would really appreciate it. It's done in scratch however because I can't write in any other languages... sorry...
Thanks in advance.
Any fixed shuffle is equivalent to a permutation; what you want to know is the order of that permutation. This can be computed by decomposing the permutation into cycles and then computing the least common multiple of the cycle lengths.
I'm not able to properly understand your algorithm, but here's an example of shuffling 8 elements and then finding the number of times that shuffle needs to be repeated to get back to an unshuffled state.
Suppose the sequence starts as 1,2,3,4,5,6,7,8 and after one shuffle, it's 3,1,4,5,2,8,7,6.
The number 1 goes to position 2, then 2 goes to position 5, then 5 goes to position 4, then 4 goes to position 3, then 3 goes to position 1. So the first cycle is (1 2 5 4 3).
The number 6 goes to position 8, then 8 goes to position 6. So the next cycle is (6 8).
The number 7 stays in position 7, so this is a trivial cycle (7).
The lengths of the cycles are 5, 2 and 1, so the least common multiple is 10. This shuffle takes 10 iterations to get back to the intitial state.
If you don't mind sitting down with pen and paper for a while, you should be able to follow this procedure for your own shuffling algorithm.
I am doing a few thing on Information Retrieval and have an exam coming up and I am absolutely clueless. First of, could anyone recommend me the shortest and best description possible for what PageRank actually is in Information Retrieval? Maybe even a good short video or your own description. I know Google use to, or did use it.
I know there are a lot of questions here but I could use as MUCH help as possible in a short length of time.
So my first question (taken from past papers, and making my own examples):
I am wanting to take a table such as:
A B C
A 0 1 0
B 1 0 1
C 0 0 0
And create a graph. I believe this is correct but unsure (I could use a "yes that is correct" or a "no":
And if I was given a graph such as:
The table would be:
A B C
A 0 1 0
B 0 0 1
C 0 0 0
Is that correct? If not, could I please get help and get it described somewhere? The lecture I am reading is not great at explaining and my lecturer isn't great at helping either.
Next I will probably be asked to use Teleportation Probability on the first table. This I desperately need help in. If the probability(the special a symbol)=1/2, does this mean multiply everything, including the 0's in the table such as 0x1/2? also 1x1/2? This is for the matrix of transition probabilities.
Next would be, how can I calculate PageRank from the above matrix. Using matrix multiplication. In words or in Pseudocode.
Another question I want to know is, will a user's page rank on twitter increase if they follow another user? I was assuming this would be a no because they are not following the user back?
Does a user's pagerank depend on how frequently you find said user if you start at a random user and click on another random persona and such till you find them? I assume this one is definitely not true. Because they might not be following said user.
I know this is a lot to ask. Does anyone have tutorials I can follow for either that are not complicated and I can look at and get it mastered today?
Thanks I really appreciate all your help. I know not one person can answer them all but can help provide assistance for some.
here's my stab at answering your questions:
good learning resource:
http://en.wikipedia.org/wiki/PageRank#Simplified_algorithm (no doubt you've see it already, but it's a pretty good one). Start there, understand the algorithm first, then do the implementation.
this might be a good simple method to implement?
http://pr.efactory.de/e-pagerank-algorithm.shtml
or this:
http://www.cs.princeton.edu/~chazelle/courses/BIB/pagerank.htm
I'm guessing you can program in Python (common school language), in that case you might be interested in a package for handling graphs which has pagerank calculations: http://networkx.lanl.gov/reference/generated/networkx.algorithms.link_analysis.pagerank_alg.pagerank.html. If you have to write your own pagerank algorithm (very doable), you could use that to check the results.
For the matrix -> graph conversion question: your professor needs to specify how directionality is encoded in the matrix. Does a 1 at B,C specify a link from B to C or from C to B? My guess would be B to C. If that's true, your first graph is wrong there, but the second graph is ok. Directionality is very important in PageRank.
I believe the Teleportation probability is the probability that a random walker executing a new step will jump to a random node in the graph. It's in the wikipedia page under "damping factor". I don't know how it ties into multiplying numbers in your matrix.
For the Twitter question - yes, I think you have it right. Linking to (or presumably following) a second person does nothing directly to the the first person's pagerank, but it likely increases the second person's pagerank. In practice, there could be secondary effects, like the second person noticing that the first person is interesting and following them back.
second to last question - yes, one formulation of the pagerank algorithm is as a random walk along links with the frequency of encountering a node (page) going into the pagerank.
good luck!
This question already has an answer here:
Closed 10 years ago.
Possible Duplicate:
Need help in building efficient exhaustive search algorithm
Imagine that you must open a locked door by inputting the correct 4-digit code on a keypad. After every keypress the lock evaluates the sequence of the last 4 digits inputted, i.e. by entering 123456 you have evaluated 3 codes: 1234, 2345 and 3456.
What is the shortest sequence of keypresses to evaluate all 10^4 different combinations?
Is there a method for traversing the entire space easy enough for a human to follow?
I have pondered this from time to time since a friend of mine had to brute force such a lock, to not having to spend the night outdoors in wintertime.
My feeble attempts at wrapping my head around it
With a code of length L=4 digits and an "alphabet" of digits of size D=10 the length of the optimal sequence cannot be shorter than D^L + L - 1. In simulations of smaller size than [L,D] = [4,10] I have obtained optimal results by semi-randomly searching the space. However I do not know if a solution exists for an arbitrary [L,D] pair and would not be able to remember the solution if I ever had to use it.
Lessons learned so far
When planning to spend the night at a friends house in another town, be sure to not arrive at 1 am if that person is going out to party and won't hear her cell phone.
I think you want a http://en.wikipedia.org/wiki/De_Bruijn_sequence - "a cyclic sequence of a given alphabet A with size k for which every possible subsequence of length n in A appears as a sequence of consecutive characters exactly once."
The link Evgeny provided should answer both of your quests. This answer is a bit offtopic, but you ask for a solution for humans.
In the real world you should probably rely more on Social engineering or heuristics, and after that on mathematics. I give a case on real life:
I went to visit an apartment and I found out that my cellphone was dead. Now way of contacting the person doing the visit. I was about to go back when I saw that the door used a keypad 0 - 9 and A B. I made several assumptions:
The code is 5 digits long. The length is pretty standard depending on the region you are in. I based this assumption on buildings I had access before (legally :D).
The code starts with numbers, then either A or B (based on my own building).
The keypad was not brand new. Conclusion, the numbers used in the code were a bit damaged. I knew with certainty which numbers were not in the code, and three of the four number in the code (given my previous assumptions)
By the amount of keys damaged I assumed the code didn't contain repeated keys (7 were damaged, it was clear A was used, B not used )
At the end I had 3 numbers which were in the code for sure, 2 candidates for the last number and I was sure A was at the end. On key was just slightly damaged compared to the others.
I just had to enumerate permutations starting with the candidate which seemed the more damaged, which give me 4! + 4! = 48 tries. Believe me, at the 5th try the door was opened. If I can give my 2 cents, the old put a key and open the door is still the most reliable method to restrict access to a building.
i am programming a card game and i need to sort a stack of cards by their rank. so that they form a gapless sequence.
in this special game the card with value 2 could be used as a wild card, so for example the cards
2 3 5
should be sorted like this
3 2 5
because the 2 replaces the 4, otherwise it would not be a valid sequence.
however the cards
2 3 4
should stay like they are.
restriction: there an be only one '2' used as a wildcard.
2 2 3 4
would also stay like it is, because the first 2 would replace the ACE (or 1, whatever you call it).
the following would not be a valid input sequence, since one of the 2s must be use as a wildcard and one not. it is not possible to make up a gapless sequence then.
2 4 2 6
now i have a difficulty to figure out if a 2 is used as a wildcard or not. once i got that, i think i can do the rest of the sorting
thanks for any algorithmic help on this problem!
EDIT in response to your clarification to your new requirement:
You're implying that you'll never get data for which a gapless sequence cannot be formed. (If only I could have such guarantees in the real world.) So:
Do you have a 2?
No: your sequence must already be gapless.
Yes: You need to figure out where to put it.
Sort your input. Do you see a gap? Since you can only use one 2 as a wildcard, there can be at most one gap.
No: treat the 2 as a legitimate number two.
Yes: move the 2 to the gap to fill it in.
EDIT in response to your new requirement:
In this case, just look for the highest single gap, and plug it with a 2 if you have a 2 available.
Original answer:
Since your sequence must be gapless, you could count the number of 2s you have and the sizes of all the gaps that are present. Then just fill in the highest gap for which you have a sufficient number of 2s.