Related
Obviously, any competent programming language has built-in subtraction. However, say you needed to implement a custom subtraction algorithm (say, to handle very large numbers). If you are using the borrowing method, you would have to first compare the inputs to determine if the result will be negative, and if so, you switch the order of the subtraction. I am curious if it is possible to write an algorithm that does not need this comparison/operand swapping?
Further, could such an algorithm be more efficient than compare-first, which in the worst case would have to compare every place value?
*I am aware that one could use complements, but this limits the size of permitted inputs (assuming finite representation size).
If you are implementing bignums, then you can combine the comparison and the subtraction so that you only look at each "limb" once, except for one. (The word "limb" comes from the Gnu MP library but it is also used elsewhere to mean the same thing.)
First, you scan from the big end of the longer number until you find a nonzero limb or you reach thr length of the smaller number. In the first case, you know that the longer number is bigger and you also know the position of its most significant limb. (If you have adopted the convention that numbers always have a nonzero most-significant limb then you can skip this step.)
Suppose now that you didn't prove the the longer number is bigger, so you know are at the same position in both numbers. Continue scanning both numbers until you find a difference. At this point, you know which number is bigger and you also know the position of the most significant limb of the larger number.
So now you know which number is bigger, and you can do the normal borrowing algorithm from the low-order ends of the numbers, stopping when you reach the high-order limb which you previously identified.
Since the two scans stop at the same place, you don't look at any limb more than once, exvept the high-order limb which stopped the first scan.
Whether it is actually worthwhile to implement this hack is something you would have to decide. The complication may not be justfiable. But it is possible.
I was trying various methods to implement a program that gives the digits of pi sequentially. I tried the Taylor series method, but it proved to converge extremely slowly (when I compared my result with the online values after some time). Anyway, I am trying better algorithms.
So, while writing the program I got stuck on a problem, as with all algorithms: How do I know that the n digits that I've calculated are accurate?
Since I'm the current world record holder for the most digits of pi, I'll add my two cents:
Unless you're actually setting a new world record, the common practice is just to verify the computed digits against the known values. So that's simple enough.
In fact, I have a webpage that lists snippets of digits for the purpose of verifying computations against them: http://www.numberworld.org/digits/Pi/
But when you get into world-record territory, there's nothing to compare against.
Historically, the standard approach for verifying that computed digits are correct is to recompute the digits using a second algorithm. So if either computation goes bad, the digits at the end won't match.
This does typically more than double the amount of time needed (since the second algorithm is usually slower). But it's the only way to verify the computed digits once you've wandered into the uncharted territory of never-before-computed digits and a new world record.
Back in the days where supercomputers were setting the records, two different AGM algorithms were commonly used:
Gauss–Legendre algorithm
Borwein's algorithm
These are both O(N log(N)^2) algorithms that were fairly easy to implement.
However, nowadays, things are a bit different. In the last three world records, instead of performing two computations, we performed only one computation using the fastest known formula (Chudnovsky Formula):
This algorithm is much harder to implement, but it is a lot faster than the AGM algorithms.
Then we verify the binary digits using the BBP formulas for digit extraction.
This formula allows you to compute arbitrary binary digits without computing all the digits before it. So it is used to verify the last few computed binary digits. Therefore it is much faster than a full computation.
The advantage of this is:
Only one expensive computation is needed.
The disadvantage is:
An implementation of the Bailey–Borwein–Plouffe (BBP) formula is needed.
An additional step is needed to verify the radix conversion from binary to decimal.
I've glossed over some details of why verifying the last few digits implies that all the digits are correct. But it is easy to see this since any computation error will propagate to the last digits.
Now this last step (verifying the conversion) is actually fairly important. One of the previous world record holders actually called us out on this because, initially, I didn't give a sufficient description of how it worked.
So I've pulled this snippet from my blog:
N = # of decimal digits desired
p = 64-bit prime number
Compute A using base 10 arithmetic and B using binary arithmetic.
If A = B, then with "extremely high probability", the conversion is correct.
For further reading, see my blog post Pi - 5 Trillion Digits.
Undoubtedly, for your purposes (which I assume is just a programming exercise), the best thing is to check your results against any of the listings of the digits of pi on the web.
And how do we know that those values are correct? Well, I could say that there are computer-science-y ways to prove that an implementation of an algorithm is correct.
More pragmatically, if different people use different algorithms, and they all agree to (pick a number) a thousand (million, whatever) decimal places, that should give you a warm fuzzy feeling that they got it right.
Historically, William Shanks published pi to 707 decimal places in 1873. Poor guy, he made a mistake starting at the 528th decimal place.
Very interestingly, in 1995 an algorithm was published that had the property that would directly calculate the nth digit (base 16) of pi without having to calculate all the previous digits!
Finally, I hope your initial algorithm wasn't pi/4 = 1 - 1/3 + 1/5 - 1/7 + ... That may be the simplest to program, but it's also one of the slowest ways to do so. Check out the pi article on Wikipedia for faster approaches.
You could use multiple approaches and see if they converge to the same answer. Or grab some from the 'net. The Chudnovsky algorithm is usually used as a very fast method of calculating pi. http://www.craig-wood.com/nick/articles/pi-chudnovsky/
The Taylor series is one way to approximate pi. As noted it converges slowly.
The partial sums of the Taylor series can be shown to be within some multiplier of the next term away from the true value of pi.
Other means of approximating pi have similar ways to calculate the max error.
We know this because we can prove it mathematically.
You could try computing sin(pi/2) (or cos(pi/2) for that matter) using the (fairly) quickly converging power series for sin and cos. (Even better: use various doubling formulas to compute nearer x=0 for faster convergence.)
BTW, better than using series for tan(x) is, with computing say cos(x) as a black box (e.g. you could use taylor series as above) is to do root finding via Newton. There certainly are better algorithms out there, but if you don't want to verify tons of digits this should suffice (and it's not that tricky to implement, and you only need a bit of calculus to understand why it works.)
There is an algorithm for digit-wise evaluation of arctan, just to answer the question, pi = 4 arctan 1 :)
I do not know a whole lot about math, so I don't know how to begin to google what I am looking for, so I rely on the intelligence of experts to help me understand what I am after...
I am trying to find the smallest string of equations for a particular large number. For example given the number
"39402006196394479212279040100143613805079739270465446667948293404245721771497210611414266254884915640806627990306816"
The smallest equation is 64^64 (that I know of) . It contains only 5 bytes.
Basically the program would reverse the math, instead of taking an expression and finding an answer, it takes an answer and finds the most simplistic expression. Simplistic is this case means smallest string, not really simple math.
Has this already been created? If so where can I find it? I am looking to take extremely HUGE numbers (10^10000000) and break them down to hopefully expressions that will be like 100 characters in length. Is this even possible? are modern CPUs/GPUs not capable of doing such big calculations?
Edit:
Ok. So finding the smallest equation takes WAY too much time, judging on answers. Is there anyway to bruteforce this and get the smallest found thus far?
For example given a number super super large. Sometimes taking the sqaureroot of number will result in an expression smaller than the number itself.
As far as what expressions it would start off it, well it would naturally try expressions that would the expression the smallest. I am sure there is tons of math things I dont know, but one of the ways to make a number a lot smaller is powers.
Just to throw another keyword in your Google hopper, see Kolmogorov Complexity. The Kolmogorov complexity of a string is the size of the smallest Turing machine that outputs the string, given an empty input. This is one way to formalize what you seem to be after. However, calculating the Kolmogorov complexity of a given string is known to be an undecidable problem :)
Hope this helps,
TJ
There's a good program to do that here:
http://mrob.com/pub/ries/index.html
I asked the question "what's the point of doing this", as I don't know if you're looking at this question from a mathemetics point of view, or a large number factoring point of view.
As other answers have considered the factoring point of view, I'll look at the maths angle. In particular, the problem you are describing is a compressibility problem. This is where you have a number, and want to describe it in the smallest algorithm. Highly random numbers have very poor compressibility, as to describe them you either have to write out all of the digits, or describe a deterministic algorithm which is only slightly smaller than the number itself.
There is currently no general mathemetical theorem which can determine if a representation of a number is the smallest possible for that number (although a lower bound can be discovered by understanding shannon's information theory). (I said general theorem, as special cases do exist).
As you said you don't know a whole lot of math, this is perhaps not a useful answer for you...
You're doing a form of lossless compression, and lossless compression doesn't work on random data. Suppose, to the contrary, that you had a way of compressing N-bit numbers into N-1-bit numbers. In that case, you'd have 2^N values to compress into 2^N-1 designations, which is an average of 2 values per designation, so your average designation couldn't be uncompressed. Lossless compression works well on relatively structured data, where data we're likely to get is compressed small, and data we aren't going to get actually grows some.
It's a little more complicated than that, since you're compressing partly by allowing more information per character. (There are a greater number of N-character sequences involving digits and operators than digits alone.) Still, you're not going to get lossless compression that, on the average, is better than just writing the whole numbers in binary.
It looks like you're basically wanting to do factoring on an arbitrarily large number. That is such a difficult problem that it actually serves as the cornerstone of modern-day cryptography.
This really appears to be a mathematics problem, and not programming or computer science problem. You should ask this on https://math.stackexchange.com/
While your question remains unclear, perhaps integer relation finding is what you are after.
EDIT:
There is some speculation that finding a "short" form is somehow related to the factoring problem. I don't believe that is true unless your definition requires a product as the answer. Consider the following pseudo-algorithm which is just sketch and for which no optimization is attempted.
If "shortest" is a well-defined concept, then in general you get "short" expressions by using small integers to large powers. If N is my integer, then I can find an integer nearby that is 0 mod 4. How close? Within +/- 2. I can find an integer within +/- 4 that is 0 mod 8. And so on. Now that's just the powers of 2. I can perform the same exercise with 3, 5, 7, etc. We can, for example, easily find the nearest integer that is simultaneously the product of powers of 2, 3, 5, 7, 11, 13, and 17, call it N_1. Now compute N-N_1, call it d_1. Maybe d_1 is "short". If so, then N_1 (expressed as power of the prime) + d_1 is the answer. If not, recurse to find a "short" expression for d_1.
We can also pick integers that are maybe farther away than our first choice; even though the difference d_1 is larger, it might have a shorter form.
The existence of an infinite number of primes means that there will always be numbers that cannot be simplified by factoring. What you're asking for is not possible, sorry.
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.
If I say to you:
"I am thinking of a number between 0 and n, and I will tell you if your guess is high or low", then you will immediately reach for binary search.
What if I remove the upper bound? i.e. I am thinking of a positive integer, and you need to guess it.
One possible method would be for you to guess 2, 4, 8, ..., until you guess 2**k for some k and I say "lower". Then you can apply binary search.
Is there a quicker method?
EDIT:
Clearly, any solution is going to take time proportional to the size of the target number. If I chuck Graham's number through the Ackermann function, we'll be waiting a while whatever strategy you pursue.
I could offer this algorithm too: Guess each integer in turn, starting from 1.
It's guaranteed to finish in a finite amount of time, but yet it's clearly much worse than my "powers of 2" strategy. If I can find a worse algorithm (and know that it is worse), then maybe I could find a better one?
For example, instead of powers of 2, maybe I can use powers of 10. Then I find the upper bound in log_10(n) steps, instead of log_2(n) steps. But I have to then search a bigger space. Say k = ceil(log_10(n)). Then I need log_2(10**k - 10**(k-1)) steps for my binary search, which I guess is about 10+log_2(k). For powers of 2, I have roughly log_2(log_2(n)) steps for my search phase. Which wins?
What if I search upwards using n**n? Or some other sequence? Does the prize go to whoever can find the sequence that grows the fastest? Is this a problem with an answer?
Thank you for your thoughts. And my apologies to those of you suggesting I start at MAX_INT or 2**32-1, since I'm clearly drifting away from the bounds of practicality here.
FINAL EDIT:
Hi all,
Thank you for your responses. I accepted the answer by Norman Ramsey (and commenter onebyone) for what I understood to be the following argument: for a target number n, any strategy must be capable of distinguishing between (at least) the numbers from 0..n, which means you need (at least) O(log(n)) comparisons.
However seveal of you also pointed out that the problem is not well-defined in the first place, because it's not possible to pick a "random positive integer" under the uniform probability distribution (or, rather, a uniform probability distribution cannot exist over an infinite set). And once I give you a nonuniform distribution, you can split it in half and apply binary search as normal.
This is a problem that I've often pondered as I walk around, so I'm pleased to have two conclusive answers for it.
If there truly is no upper bound, and all numbers all the way to infinity are equally likely, then there is no optimum way to do this. For any finite guess G, the probability that the number is lower than G is zero and the probability that it is higher is 1 - so there is no finite guess that has an expectation of being higher than the number.
RESPONSE TO JOHN'S EDIT:
By the same reasoning that powers of 10 are expected to be better than powers of 2 (there's only a finite number of possible Ns for which powers of 2 are better, and an infinite number where powers of 10 are better), powers of 20 can be shown to be better than powers of 10.
So basically, yes, the prize goes to fastest-growing sequence (and for the same sequence, the highest starting point) - for any given sequence, it can be shown that a faster growing sequence will win in infinitely more cases. And since for any sequence you name, I can name one that grows faster, and for any integer you name, I can name one higher, there's no answer that can't be bettered. (And every algorithm that will eventually give the correct answer has an expected number of guesses that is infinite, anyway).
People (who have never studied probability) tend to think that "pick a number from 1 to N" means "with equal probability of each", and they act according to their intuitive understanding of probability.
Then when you say "pick any positive integer", they still think it means "with equal probability of each".
This is of course impossible - there exists no discrete probability distribution with domain the positive integers, where p(n) == p(m) for all n, m.
So, the person picking the number must have used some other probability distribution. If you know anything at all about that distribution, then you must base your guessing scheme on that knowledge in order to have the "fastest" solution.
The only way to calculate how "fast" a given guessing scheme is, is to calculate its expected number of guesses to find the answer. You can only do this by assuming a probability distribution for the target number. For example, if they have picked n with probability (1/2) ^ n, then I think your best guessing scheme is "1", "2", "3",... (average 2 guesses). I haven't proved it, though, maybe it's some other sequence of guesses. Certainly the guesses should start small and grow slowly. If they have picked 4 with probability 1 and all other numbers with probability 0, then your best guessing scheme is "4" (average 1 guess). If they have picked a number from 1 to a trillion with uniform distribution, then you should binary search (average about 40 guesses).
I say the only way to define "fast" - you could look at worst case. You have to assume a bound on the target, to prevent all schemes having the exact same speed, namely "no bound on the worst case". But you don't have to assume a distribution, and the answer for the "fastest" algorithm under this definition is obvious - binary search starting at the bound you selected. So I'm not sure this definition is terribly interesting...
In practice, you don't know the distribution, but can make a few educated guesses based on the fact that the picker is a human being, and what numbers humans are capable of conceiving. As someone says, if the number they picked is the Ackermann function for Graham's number, then you're probably in trouble. But if you know that they are capable of representing their chosen number in digits, then that actually puts an upper limit on the number they could have chosen. But it still depends what techniques they might have used to generate and record the number, and hence what your best knowledge is of the probability of the number being of each particular magnitude.
Worst case, you can find it in time logarithmic in the size of the answer using exactly the methods you describe. You might use Ackermann's function to find an upper bound faster than logarithmic time, but then the binary search between the number guessed and the previous guess will require time logarithmic in the size of the interval, which (if guesses grow very quickly) is close to logarithmic in the size of the answer.
It would be interesting to try to prove that there is no faster algorithm (e.g., O(log log n)), but I have no idea how to do it.
Mathematically speaking:
You cannot ever correctly find this integer. In fact, strictly speaking, the statement "pick any positive integer" is meaningless as it cannot be done: although you as a person may believe you can do it, you are actually picking from a bounded set - you are merely unconscious of the bounds.
Computationally speaking:
Computationally, we never deal with infinites, as we would have no way of storing or checking against any number larger than, say, the theoretical maximum number of electrons in the universe. As such, if you can estimate a maximum based on the number of bits used in a register on the device in question, you can carry out a binary search.
Binary search can be generalized: each time set of possible choices should be divided into to subsets of probability 0.5. In this case it's still applicable to infinite sets, but still requires knowledge about distribution (for finite sets this requirement is forgotten quite often)...
My main refinement is that I'd start with a higher first guess instead of 2, around the average of what I'd expect them to choose. Starting with 64 would save 5 guesses vs starting with 2 when the number's over 64, at the cost of 1-5 more when it's less. 2 makes sense if you expect the answer to be around 1 or 2 half the time. You could even keep a memory of past answers to decide the best first guess. Another improvement could be to try negatives when they say "lower" on 0.
If this is guessing the upper bound of a number being generated by a computer, I'd start with 2**[number of bits/2], then scale up or down by powers of two. This, at least, gets you the closest to the possible values in the least number of jumps.
However, if this is a purely mathematical number, you can start with any value, since you have an infinite range of values, so your approach would be fine.
Since you do not specify any probability distribution of the numbers (as others have correctly mentioned, there is no uniform distribution over all the positive integers), the No Free Lunch Theorem give the answer: any method (that does not repeat the same number twice) is as good as any other.
Once you start making assumptions about the distribution (f.x. it is a human being or binary computer etc. that chooses the number) this of course changes, but as the problem is stated any algorithm is as good as any other when averaged over all possible distributions.
Use binary search starting with MAX_INT/2, where MAX_INT is the biggest number your platform can handle.
No point in pretending we can actually have infinite possibilities.
UPDATE: Given that you insist on entering the realms of infinity, I'll just vote to close your question as not programming related :-)
The standard default assumption of a uniform distribution for all positive integers doesn't lead to a solution, so you should start by defining the probability distribution of the numbers to guess.
I'd probably start my guessing with Graham's Number.
The practical answer within a computing context would be to start with whatever is the highest number that can (realistically) be represented by the type you are using. In case of some BigInt type you'd probably want to make a judgement call about what is realistic... obviously ultimately the bound in that case is the available memory... but performance-wise something smaller may be more realistic.
Your starting point should be the largest number you can think of plus 1.
There is no 'efficient search' for a number in an infinite range.
EDIT: Just to clarify, for any number you can think of there are still infinitely more numbers that are 'greater' than your number, compared to a finite collection of numbers that are 'less' than your number. Therefore, assuming the chosen number is randomly selected from all positive numbers, you have zero | (approaching zero) chance of being 'above' the chosen number.
I gave an answer to a similar question "Optimal algorithm to guess any random integer without limits?"
Actually, provided there algorithm not just searches for the conceived number, but it estimates a median of the distribution of the number that you may re-conceive at each step! And also the number could be even from the real domain ;)