So, I work in industrial automation, and normally program with ladder logic. So its rather odd compared to what I would consider normal programing. Anyway I needed to sort a list of numbers from smallest to biggest. I was looking through sorting algorithms trying to find one I could easily implement using ladder logic. I was having a hard time, but after some thinking I came up with something that wasn't even on the Wikipedia list of sorting algorithms. Well, It might be but I can't find it. I know this isn't very efficient sorting algorithm, but it does work. I want to know the name of it if it has one.
The basic version of this is, imagine an array of numbers. Take the first number in the list and compare it to all other numbers in the list, count the number of times that it is bigger than any of the other numbers. This accumulated value is the index number for where it goes in the output array. To place it in the array, check if there is already something written to that spot, if there is add one to the index and check again until there isn't anything in its spot. When the empty spot is found write it to the output array. Once you have done that to every number in the list you will have an output array with the same size as the input, but with it sorted smallest to biggest. I should note that this is assuming the language uses zero based indexing.
If this wasn't clear enough, I'm happy to elaborate further if needed.
I would say it's a worse version of counting sort:
It operates by counting the number of objects that possess distinct key values, and applying prefix sum on those counts to determine the positions of each key value in the output sequence
So it basically does the same thing you're doing: put each element in its final position by using counts. Counting sort uses an array to store the needed counts, you iterate the array to find them at each step for the current element.
I don't think there's a name for your exact algorithm.
Assalamualaikum (peace be upon you)
If in the list of n numbers, n-1 numbers are identical and only one number is unique,then, is there any way, other than linear search to identify the identical number as well as the unique number of the list?
You need to parse the list at least once. so the algorithm is at least of O(n), and linear search as you mentioned it does the work for you.
Given a population size P, I must generate P random, but unique objects. An object is an unordered list of X unique unordered pairs.
I am currently just using a while loop with T attempts at generating a random ordering before giving up. Currently T = some constant.
So my question is at what point should I stop attempting to generate more unique objects i.e. the reasonable value of T.
For example:
1) If I have 3 unique objects and I need just one more, I can attempt up to e.g. 4 times
2) But if I have 999 unique objects and I need just one more, I do not want to make e.g. 1000 attempts
The problem I'm dealing with doesn't absolutely require every unique ordering. The user specifies the number actually, so I want to determine at what point to say that it is not reasonable to generate any more.
I hope that makes sense
If not, a more general case:
Choosing N numbers, at what value of T does it start to get very difficult to start generating more unique random numbers from the possible N.
I'm not sure if T would be the same in both cases but maybe this second case would be sufficient for my needs. I need a relatively large threshold for small values of N and a relatively small threshold for large values of N.
Not that it matters, but this is for a basic genetic algorithm.
Are you asking for something like lottery tickets/balls selection? For that there is a well-known shuffle algorithm - Fisher–Yates-Knuth shuffle.
I've got a very large range/set of numbers, (1..1236401668096), that I would basically like to 'shuffle', i.e. randomly traverse without revisiting the same number. I will be running a Web service, and each time a request comes in it will increment a counter and pull the next 'shuffled' number from the range. The algorithm will have to accommodate for the server going offline, being able to restart traversal using the persisted value of the counter (something like how you can seed a pseudo-random number generator, and get the same pseudo-random number given the seed and which iteration you are on).
I'm wondering if such an algorithm exists or is feasible. I've seen the Fisher-Yates Shuffle, but the 1st step is to "Write down the numbers from 1 to N", which would take terabytes of storage for my entire range. Generating a pseudo-random number for each request might work for awhile, but as the database/tree gets full, collisions will become more common and could degrade performance (already a 0.08% chance of collision after 1 billion hits according to my calculation). Is there a more ideal solution for my scenario, or is this just a pipe dream?
The reason for the shuffling is that being able to correctly guess the next number in the sequence could lead to a minor DOS vulnerability in my app, but also because the presentation layer will look much nicer with a wider number distribution (I'd rather not go into details about exactly what the app does). At this point I'm considering just using a PRNG and dealing with collisions or shuffling range slices (starting with (1..10000000).to_a.shuffle, then, (10000001, 20000000).to_a.shuffle, etc. as each range's numbers start to run out).
Any mathemagicians out there have any better ideas/suggestions?
Concatenate a PRNG or LFSR sequence with /dev/random bits
There are several algorithms that can generate pseudo-random numbers with arbitrarily large and known periods. The two obvious candidates are the LCPRNG (LCG) and the LFSR, but there are more algorithms such as the Mersenne Twister.
The period of these generators can be easily constructed to fit your requirements and then you simply won't have collisions.
You could deal with the predictable behavior of PRNG's and LFSR's by adding 10, 20, or 30 bits of cryptographically hashed entropy from an interface like /dev/random. Because the deterministic part of your number is known to be unique it makes no difference if you ever repeat the actually random part of it.
Divide and conquer? Break down into manageable chunks and shuffle them. You could divide the number range e.g. by their value modulo n. The list is constructive and quite small depending on n. Once a group is exhausted, you can use the next one.
For example if you choose an n of 1000, you create 1000 different groups. Pick a random number between 1 and 1000 (let's call this x) and shuffle the numbers whose value modulo 1000 equals x. Once you have exhausted that range, you can choose a new random number between 1 and 1000 (without x obviously) to get the next subset to shuffle. It shouldn't exactly be challenging to keep track of which numbers of the 1..1000 range have already been used, so you'd just need a repeatable shuffle algorithm for the numbers in the subset (e.g. Fisher-Yates on their "indices").
I guess the best option is to use a GUID/UUID. They are made for this type of thing, and it shouldn't be hard to find an existing implementation to suit your needs.
While collisions are theoretically possible, they are extremely unlikely. To quote Wikipedia:
The probability of one duplicate would be about 50% if every person on earth owns 600 million UUIDs
I've been thinking about a math/algorithm problem and would appreciate your input on how to solve it!
If I have a number (e.g. 479), I would like to recombine its digits or combination of them to a math formula that matches the original number. All digits should be used in their original order, but may be combined to numbers (hence, 479 allows for 4, 7, 9, 47, 79) but each digit may only be used once, so you can not have something like 4x47x9 as now the number 4 was used twice.
Now an example just to demonstrate on how I think of it. The example is mathematically incorrect because I couldn't come up with a good example that actually works, but it demonstrates input and expected output.
Example Input: 29485235
Example Output: 2x9+48/523^5
As I said, my example does not add up (2x9+48/523^5 doesn't result in 29485235) but I wondered if there is an algorithm that would actually allow me to find such a formula consisting of the source number's digits in their original order which would upon calculation yield the original number.
On the type of math used, I'd say parenthesis () and Add/Sub/Mul/Div/Pow/Sqrt.
Any ideas on how to do this? My thought was on simply brute forcing it by chopping the number apart by random and doing calculations hoping for a matching result. There's gotta be a better way though?
Edit: If it's any easier in non-original order, or you have an idea to solve this while ignoring some of the 'conditions' described above, it would still help tremendously to understand how to go about solving such a problem.
For numbers up to about 6 digits or so, I'd say brute-force it according to the following scheme:
1) Split your initial value into a list (array, whatever, according to language) of numbers. Initially, these are the digits.
2) For each pair of numbers, combine them together using one of the operators. If the result is the target number, then return success (and print out all the operations performed on your way out). Otherwise if it's an integer, recurse on the new, smaller list consisting of the number you just calculated, and the numbers you didn't use. Or you might want to allow non-integer intermediate results, which will make the search space somewhat bigger. The binary operations are:
Add
subtract
multiply
divide
power
concatenate (which may only be used on numbers which are either original digits, or have been produced by concatenation).
3) Allowing square root bloats the search space to infinity, since it's a unary operator. So you will need a way to limit the number of times it can be applied, and I'm not sure what that will be (loss of precision as the answer approaches 1, maybe?). This is another reason to allow only integer intermediate values.
4) Exponentiation will rapidly cause overflows. 2^(9^(4^8)) is far too large to store all the digits directly [although in base 2 it's pretty obvious what they are ;-)]. So you'll either have to accept that you might miss solutions with large intermediate values, or else you'll have to write a bunch of code to do your arithmetic in terms of factors. These obviously don't interact very well with addition, so you might have to do some estimation. For example, just by looking at the magnitude of the number of factors we see that 2^(9^(4^8)) is nowhere near (2^35), so there's no need to calculate (2^(9^(4^8)) + 5) / (2^35). It can't possibly be 29485235, even if it were an integer (which it certainly isn't - another way to rule out this particular example). I think handling these numbers is harder than the rest of the problem put together, so perhaps you should limit yourself to single-digit powers to begin with, and perhaps to results which fit in a 64bit integer, depending what language you are using.
5) I forgot to exclude the trivial solution for any input, of just concatenating all the digits. That's pretty easy to handle, though, just maintain a parameter through the recursion which tells you whether you have performed any non-concatenation operations on the route to your current sub-problem. If you haven't, then ignore the false match.
My estimate of 6 digits is based on the fact that it's fairly easy to write a Countdown solver that runs in a fraction of a second even when there's no solution. This problem is different in that the digits have to be used in order, but there are more operations (Countdown does not permit exponentiation, square root, or concatenation, or non-integer intermediate results). Overall I think this problem is comparable, provided you resolve the square root and overflow issues. If you can solve one case in a fraction of a second, then you can brute force your way through a million candidates in reasonable time (assuming you don't mind leaving your PC on).
By 10 digits, brute force appears impossible, because you have to consider 10 billion cases, each with a significant amount of recursion required. So I guess you'll hit the limit of brute force somewhere between the two.
Note also that my simple algorithm at the top still has a lot of redundancy - it doesn't stop you doing (4,7,9,1) -> (47,9,1) -> (47,91), and then later also doing (4,7,9,1) -> (4,7,91) -> (47,91). So unless you work out where those duplicates are going to occur and avoid them, you'll attempt (47,91) twice. Obviously that's not much work when there's only 2 numbers in the list, but when there are 7 numbers in the list, you probably do not want to e.g. add 4 of them together in 6 different ways and then solve the resulting 4-number problem 6 times. Cleverness here is not required for the Countdown game, but for all I know in this problem it might make the difference between brute-forcing 8 digits, and brute-forcing 9 digits, which is quite significant.
Numbers like that, as I recall, are exceedingly rare, if extant. Some numbers can be expressed by their component digits in a different order, such as, say, 25 (5²).
Also, trying to brute-force solutions is hopeless, at best, given that the number of permutations increase extremely rapidly as the numbers grow in digits.
EDIT: Partial solution.
A partial solution solving some cases would be to factorize the number into its prime factors. If its prime factors are all the same, and the exponent and factor are both present in the digits of the number (such as is the case with 25) you have a specific solution.
Most numbers that do fall into these kinds of patterns will do so either with multiplication or pow() as their major driving force; addition simply doesn't increase it enough.
Short of building a neural network that replicates Carol Voorderman I can't see anything short of brute force working - humans are quite smart at seeing patterns in problems such as this but encoding such insight is really tough.