I'm looking for a mixing function that given an integer from an interval <0, n) returns a random-looking integer from the same interval. The interval size n will typically be a composite non power of 2 number. I need the function to be one to one. It can only use O(1) memory, O(1) time is strongly preferred. I'm not too concerned about randomness of the output, but visually it should look random enough (see next paragraph).
I want to use this function as a pixel shuffling step in a realtime-ish renderer to select the order in which pixels are rendered (The output will be displayed after a fixed time and if it's not done yet this gives me a noisy but fast partial preview). Interval size n will be the number of pixels in the render (n = 1920*1080 = 2073600 would be a typical value). The function must be one to one so that I can be sure that every pixel is rendered exactly once when finished.
I've looked at the reversible building blocks used by hash prospector, but these are mostly specific to power of 2 ranges.
The only other method I could think of is multiply by large prime, but it doesn't give particularly nice random looking outputs.
What are some other options here?
Here is one solution based on the idea of primitive roots modulo a prime:
If a is a primitive root mod p then the function g(i) = a^i % p is a permutation of the nonzero elements which are less than p. This corresponds to the Lehmer prng. If n < p, you can get a permutation of 0, ..., n-1 as follows: Given i in that range, first add 1, then repeatedly multiply by a, taking the result mod p, until you get an element which is <= n, at which point you return the result - 1.
To fill in the details, this paper contains a table which gives a series of primes (all of which are close to various powers of 2) and corresponding primitive roots which are chosen so that they yield a generator with good statistical properties. Here is a part of that table, encoded as a Python dictionary in which the keys are the primes and the primitive roots are the values:
d = {32749: 30805,
65521: 32236,
131071: 66284,
262139: 166972,
524287: 358899,
1048573: 444362,
2097143: 1372180,
4194301: 1406151,
8388593: 5169235,
16777213: 9726917,
33554393: 32544832,
67108859: 11526618,
134217689: 70391260,
268435399: 150873839,
536870909: 219118189,
1073741789: 599290962}
Given n (in a certain range -- see the paper if you need to expand that range), you can find the smallest p which works:
def find_p_a(n):
for p in sorted(d.keys()):
if n < p:
return p, d[p]
once you know n and the matching p,a the following function is a permutation of 0 ... n-1:
def f(i,n,p,a):
x = a*(i+1) % p
while x > n:
x = a*x % p
return x-1
For a quick test:
n = 2073600
p,a = find_p_a(n) # p = 2097143, a = 1372180
nums = [f(i,n,p,a) for i in range(n)]
print(len(set(nums)) == n) #prints True
The average number of multiplications in f() is p/n, which in this case is 1.011 and will never be more than 2 (or very slightly larger since the p are not exact powers of 2). In practice this method is not fundamentally different from your "multiply by a large prime" approach, but in this case the factor is chosen more carefully, and the fact that sometimes more than 1 multiplication is required adding to the apparent randomness.
When we use A* with a non admissible heuristic we can sometimes get a non optimal path as result.
But when it is allowed to have path with zero cost, the only admissible heuristic that comes to my mind is h(x) = 0, which turns A* into a "simple" Dijkstra's algorithm.
Am I correct? Is this the only possible admissible heuristic? What is the real loss of not using an admissible heuristic? There is other path-finding algorithm that works better with zero cost paths?
An example:
Suppose the following graph (the numbers above edges shows the costs):
1 1 0 1 1
S --> V1 --> V2 --> V3 --> V4 --> G
Where:
S means start vertex
V means inner vertex
G means goal vertex
By looking the graph, we see that C(S) = 4.
What heuristic function h(x) can I use? If I use euclidian distance I got:
f(S) = g(S) + h(S)
f(S) = 0 + 5 = 5
We can see that this heuristic over-estimates the real distance, therefore for a more complex graph, it may not find the optimal solution.
Not true. The heuristic function h(x) has argument x consisting of the current search state. It returns an estimate of the distance from x to the goal. In a simple graph, x is a graph node.
Admissibility requires that h(x) can only be an under-estimate (or equal to the goal distance). This condition is for each particular x. (You seem to be inferring the condition is for all possible x, which is far too strong. A* would be useless if this were necessary.)
The correct statement regarding the case you propose is that h(x) = 0 is necessary only when x is a state with distance zero to the goal. Any other value would be an over-estimate. However, for any other x (in the same state space) that requires transitions with total at least cost C>0 to get to the goal, we can have any h such that h(x)<=C.
Of course if x's distance to goal is zero, then x is the goal state and the search is complete. So your concern is vacuous - there are no cases where it's of interest.
Information to construct h(x) comes from your knowledge of the search space (e.g. characteristics of the graph). A bare, general graph alone doesn't provide anything useful. The best you can do is h(x) = cost of min weight outgoing edge of x for non-goal nodes and, as already discussed, h(x) = 0 for the goal. Again note this is a lower bound on distance to goal. It gives you Dijkstra!
To do better you need to know something about the graph's structure.
Edit
In your example, you are providing detailed knowledge, so making a good h is simple. You can use
/ 4 if x == S
| 3 if x == V1
h(x) = { 2 if x == V2 or V3
| 1 if x == V4
\ 0 if x == G
or you can use any other function h'(x) such that h'(x) <= h(x) for all x. For example, this would be admissible:
/ 3 if x == S
| 2 if x == V1
h'(x) = { 2 if x == V2 or V3
| 1 if x == V4
\ 0 if x == G
Addition
The OP points out that for many problems, h(x) can be hard to choose! This is precisely correct. If you can't find a good admissible heuristic, then A* is the wrong algorithm! Nonetheless, A* is very effective for problems where heuristics can be found. Examples I've tried myself:
Graphs where Euclidean distance is a good lower bound on the possible distance between any two nodes. For example, each pair of cities A and B is separated by a distance D "as the crow flies," but the road distance from A to B is at least D in length and possibly much more, i.e. its cost C is greater than or equal to D. In this case, D makes a fine heuristic because it's a low estimate.
Puzzles where "distance" to the winning state involves moving game pieces. In this case, the number of pieces currently out of position with respect to the winning state is a fine heuristic. Examples are the 8-bishop's problem from 7th Guest (number of bishops not yet in their final positions) and the Magic Square Problem (total Manhatten distance from all pieces' current positions to their correct position in the winning state).
I have polynomials of nontrivial degree (4+) and need to robustly and efficiently determine whether or not they have a root in the interval [0,T]. The precise location or number of roots don't concern me, I just need to know if there is at least one.
Right now I'm using interval arithmetic as a quick check to see if I can prove that no roots can exist. If I can't, I'm using Jenkins-Traub to solve for all of the polynomial roots. This is obviously inefficient since it's checking for all real roots and finding their exact positions, information I don't end up needing.
Is there a standard algorithm I should be using? If not, are there any other efficient checks I could do before doing a full Jenkins-Traub solve for all roots?
For example, one optimization I could do is to check if my polynomial f(t) has the same sign at 0 and T. If not, there is obviously a root in the interval. If so, I can solve for the roots of f'(t) and evaluate f at all roots of f' in the interval [0,T]. f(t) has no root in that interval if and only if all of these evaluations have the same sign as f(0) and f(T). This reduces the degree of the polynomial I have to root-find by one. Not a huge optimization, but perhaps better than nothing.
Sturm's theorem lets you calculate the number of real roots in the range (a, b). Given the number of roots, you know if there is at least one. From the bottom half of page 4 of this paper:
Let f(x) be a real polynomial. Denote it by f0(x) and its derivative f′(x) by f1(x). Proceed as in Euclid's algorithm to find
f0(x) = q1(x) · f1(x) − f2(x),
f1(x) = q2(x) · f2(x) − f3(x),
.
.
.
fk−2(x) = qk−1(x) · fk−1(x) − fk,
where fk is a constant, and for 1 ≤ i ≤ k, fi(x) is of degree lower than that of fi−1(x). The signs of the remainders are negated from those in the Euclid algorithm.
Note that the last non-vanishing remainder fk (or fk−1 when fk = 0) is a greatest common
divisor of f(x) and f′(x). The sequence f0, f1,. . ., fk (or fk−1 when fk = 0) is called a Sturm sequence for the polynomial f.
Theorem 1 (Sturm's Theorem) The number of distinct real zeros of a polynomial f(x) with
real coefficients in (a, b) is equal to the excess of the number of changes of sign in the sequence f0(a), ..., fk−1(a), fk over the number of changes of sign in the sequence f0(b), ..., fk−1(b), fk.
You could certainly do binary search on your interval arithmetic. Start with [0,T] and substitute it into your polynomial. If the result interval does not contain 0, you're done. If it does, divide the interval in 2 and recurse on each half. This scheme will find the approximate location of each root pretty quickly.
If you eventually get 4 separate intervals with a root, you know you are done. Otherwise, I think you need to get to intervals [x,y] where f'([x,y]) does not contain zero, meaning that the function is monotonically increasing or decreasing and hence contains at most one zero. Double roots might present a problem, I'd have to think more about that.
Edit: if you suspect a multiple root, find roots of f' using the same procedure.
Use Descartes rule of signs to glean some information. Just count the number of sign changes in the coefficients. This gives you an upper bound on the number of positive real roots. Consider the polynomial P.
P = 131.1 - 73.1*x + 52.425*x^2 - 62.875*x^3 - 69.225*x^4 + 11.225*x^5 + 9.45*x^6 + x^7
In fact, I've constructed P to have a simple list of roots. They are...
{-6, -4.75, -2, 1, 2.3, -i, +i}
Can we determine if there is a root in the interval [0,3]? Note that there is no sign change in the value of P at the endpoints.
P(0) = 131.1
P(3) = 4882.5
How many sign changes are there in the coefficients of P? There are 4 sign changes, so there may be as many as 4 positive roots.
But, now substitute x+3 for x into P. Thus
Q(x) = P(x+3) = ...
4882.5 + 14494.75*x + 15363.9*x^2 + 8054.675*x^3 + 2319.9*x^4 + 370.325*x^5 + 30.45*x^6 + x^7
See that Q(x) has NO sign changes in the coefficients. All of the coefficients are positive values. Therefore there can be no roots larger than 3.
So there MAY be either 2 or 4 roots in the interval [0,3].
At least this tells you whether to bother looking at all. Of course, if the function has opposite signs on each end of the interval, we know there are an odd number of roots in that interval.
It's not that efficient, but is quite reliable. You can construct the polynomial's Companion Matrix (A sparse matrix whose eigenvalues are the polynomial's roots).
There are efficient eigenvalue algorithms that can find eigenvalues in a given interval. One of them is the inverse iteration (Can find eigenvalues closest to some input value. Just give the middle point of the interval as the above value).
If the value f(0)*f(t)<=0 then you are guaranteed to have a root. Otherwise you can start splitting the domain into two parts (bisection) and check the values in the ends until you are confident there is no root in that segment.
if f(0)*f(t)>0 you either have no, two, four, .. roots. Your limit is the polynomial order. if f(0)*f(t)<0 you may have one, three, five, .. roots.
I have a homework problem for my algorithms class asking me to calculate the maximum size of a problem that can be solved in a given number of operations using an O(n log n) algorithm (ie: n log n = c). I was able to get an answer by approximating, but is there a clean way to get an exact answer?
There is no closed-form formula for this equation. Basically, you can transform the equation:
n log n = c
log(n^n) = c
n^n = exp(c)
Then, this equation has a solution of the form:
n = exp(W(c))
where W is Lambert W function (see especially "Example 2"). It was proved that W cannot be expressed using elementary operations.
However, f(n)=n*log(n) is a monotonic function. You can simply use bisection (here in python):
import math
def nlogn(c):
lower = 0.0
upper = 10e10
while True:
middle = (lower+upper)/2
if lower == middle or middle == upper:
return middle
if middle*math.log(middle, 2) > c:
upper = middle
else:
lower = middle
the O notation only gives you the biggest term in the equation. Ie the performance of your O(n log n ) algorithm could actually be better represented by c = (n log n) + n + 53.
This means that without knowing the exact nature of the performance of your algorithm you wouldn't be able to calculate the exact number of operations required to process an given amount of data.
But it is possible to calculate that the maximum number of operations required to process a data set of size n is more than a certain number, or conversely that the biggest problem set that can be solved, using that algorithm and that number of operations, is smaller than a certain number.
The O notation is useful for comparing 2 algorithms, ie an O(n^2) algorithm is faster than a O(n^3) algorithm etc.
see Wikipedia for more info.
some help with logs
This question on getting random values from a finite set got me thinking...
It's fairly common for people to want to retrieve X unique values from a set of Y values. For example, I may want to deal a hand from a deck of cards. I want 5 cards, and I want them to all be unique.
Now, I can do this naively, by picking a random card 5 times, and try again each time I get a duplicate, until I get 5 cards. This isn't so great, however, for large numbers of values from large sets. If I wanted 999,999 values from a set of 1,000,000, for instance, this method gets very bad.
The question is: how bad? I'm looking for someone to explain an O() value. Getting the xth number will take y attempts...but how many? I know how to figure this out for any given value, but is there a straightforward way to generalize this for the whole series and get an O() value?
(The question is not: "how can I improve this?" because it's relatively easy to fix, and I'm sure it's been covered many times elsewhere.)
Variables
n = the total amount of items in the set
m = the amount of unique values that are to be retrieved from the set of n items
d(i) = the expected amount of tries needed to achieve a value in step i
i = denotes one specific step. i ∈ [0, n-1]
T(m,n) = expected total amount of tries for selecting m unique items from a set of n items using the naive algorithm
Reasoning
The first step, i=0, is trivial. No matter which value we choose, we get a unique one at the first attempt. Hence:
d(0) = 1
In the second step, i=1, we at least need 1 try (the try where we pick a valid unique value). On top of this, there is a chance that we choose the wrong value. This chance is (amount of previously picked items)/(total amount of items). In this case 1/n. In the case where we picked the wrong item, there is a 1/n chance we may pick the wrong item again. Multiplying this by 1/n, since that is the combined probability that we pick wrong both times, gives (1/n)2. To understand this, it is helpful to draw a decision tree. Having picked a non-unique item twice, there is a probability that we will do it again. This results in the addition of (1/n)3 to the total expected amounts of tries in step i=1. Each time we pick the wrong number, there is a chance we might pick the wrong number again. This results in:
d(1) = 1 + 1/n + (1/n)2 + (1/n)3 + (1/n)4 + ...
Similarly, in the general i:th step, the chance to pick the wrong item in one choice is i/n, resulting in:
d(i) = 1 + i/n + (i/n)2 + (i/n)3 + (i/n)4 + ... = = sum( (i/n)k ), where k ∈ [0,∞]
This is a geometric sequence and hence it is easy to compute it's sum:
d(i) = (1 - i/n)-1
The overall complexity is then computed by summing the expected amount of tries in each step:
T(m,n) = sum ( d(i) ), where i ∈ [0,m-1] = = 1 + (1 - 1/n)-1 + (1 - 2/n)-1 + (1 - 3/n)-1 + ... + (1 - (m-1)/n)-1
Extending the fractions in the series above by n, we get:
T(m,n) = n/n + n/(n-1) + n/(n-2) + n/(n-3) + ... + n/(n-m+2) + n/(n-m+1)
We can use the fact that:
n/n ≤ n/(n-1) ≤ n/(n-2) ≤ n/(n-3) ≤ ... ≤ n/(n-m+2) ≤ n/(n-m+1)
Since the series has m terms, and each term satisfies the inequality above, we get:
T(m,n) ≤ n/(n-m+1) + n/(n-m+1) + n/(n-m+1) + n/(n-m+1) + ... + n/(n-m+1) + n/(n-m+1) = = m*n/(n-m+1)
It might be(and probably is) possible to establish a slightly stricter upper bound by using some technique to evaluate the series instead of bounding by the rough method of (amount of terms) * (biggest term)
Conclusion
This would mean that the Big-O order is O(m*n/(n-m+1)). I see no possible way to simplify this expression from the way it is.
Looking back at the result to check if it makes sense, we see that, if n is constant, and m gets closer and closer to n, the results will quickly increase, since the denominator gets very small. This is what we'd expect, if we for example consider the example given in the question about selecting "999,999 values from a set of 1,000,000". If we instead let m be constant and n grow really, really large, the complexity will converge towards O(m) in the limit n → ∞. This is also what we'd expect, since while chosing a constant number of items from a "close to" infinitely sized set the probability of choosing a previously chosen value is basically 0. I.e. We need m tries independently of n since there are no collisions.
If you already have chosen i values then the probability that you pick a new one from a set of y values is
(y-i)/y.
Hence the expected number of trials to get (i+1)-th element is
y/(y-i).
Thus the expected number of trials to choose x unique element is the sum
y/y + y/(y-1) + ... + y/(y-x+1)
This can be expressed using harmonic numbers as
y (Hy - Hy-x).
From the wikipedia page you get the approximation
Hx = ln(x) + gamma + O(1/x)
Hence the number of necessary trials to pick x unique elements from a set of y elements
is
y (ln(y) - ln(y-x)) + O(y/(y-x)).
If you need then you can get a more precise approximation by using a more precise approximation for Hx. In particular, when x is small it is possible to
improve the result a lot.
If you're willing to make the assumption that your random number generator will always find a unique value before cycling back to a previously seen value for a given draw, this algorithm is O(m^2), where m is the number of unique values you are drawing.
So, if you are drawing m values from a set of n values, the 1st value will require you to draw at most 1 to get a unique value. The 2nd requires at most 2 (you see the 1st value, then a unique value), the 3rd 3, ... the mth m. Hence in total you require 1 + 2 + 3 + ... + m = [m*(m+1)]/2 = (m^2 + m)/2 draws. This is O(m^2).
Without this assumption, I'm not sure how you can even guarantee the algorithm will complete. It's quite possible (especially with a pseudo-random number generator which may have a cycle), that you will keep seeing the same values over and over and never get to another unique value.
==EDIT==
For the average case:
On your first draw, you will make exactly 1 draw.
On your 2nd draw, you expect to make 1 (the successful draw) + 1/n (the "partial" draw which represents your chance of drawing a repeat)
On your 3rd draw, you expect to make 1 (the successful draw) + 2/n (the "partial" draw...)
...
On your mth draw, you expect to make 1 + (m-1)/n draws.
Thus, you will make 1 + (1 + 1/n) + (1 + 2/n) + ... + (1 + (m-1)/n) draws altogether in the average case.
This equals the sum from i=0 to (m-1) of [1 + i/n]. Let's denote that sum(1 + i/n, i, 0, m-1).
Then:
sum(1 + i/n, i, 0, m-1) = sum(1, i, 0, m-1) + sum(i/n, i, 0, m-1)
= m + sum(i/n, i, 0, m-1)
= m + (1/n) * sum(i, i, 0, m-1)
= m + (1/n)*[(m-1)*m]/2
= (m^2)/(2n) - (m)/(2n) + m
We drop the low order terms and the constants, and we get that this is O(m^2/n), where m is the number to be drawn and n is the size of the list.
There's a beautiful O(n) algorithm for this. It goes as follows. Say you have n items, from which you want to pick m items. I assume the function rand() yields a random real number between 0 and 1. Here's the algorithm:
items_left=n
items_left_to_pick=m
for j=1,...,n
if rand()<=(items_left_to_pick/items_left)
Pick item j
items_left_to_pick=items_left_to_pick-1
end
items_left=items_left-1
end
It can be proved that this algorithm does indeed pick each subset of m items with equal probability, though the proof is non-obvious. Unfortunately, I don't have a reference handy at the moment.
Edit The advantage of this algorithm is that it takes only O(m) memory (assuming the items are simply integers or can be generated on-the-fly) compared to doing a shuffle, which takes O(n) memory.
Your actual question is actually a lot more interesting than what I answered (and harder). I've never been any good at statistitcs (and it's been a while since I did any), but intuitively, I'd say that the run-time complexity of that algorithm would probably something like an exponential. As long as the number of elements picked is small enough compared to the size of the array the collision-rate will be so small that it will be close to linear time, but at some point the number of collisions will probably grow fast and the run-time will go down the drain.
If you want to prove this, I think you'd have to do something moderately clever with the expected number of collisions in function of the wanted number of elements. It might be possible do to by induction as well, but I think going by that route would require more cleverness than the first alternative.
EDIT: After giving it some thought, here's my attempt:
Given an array of m elements, and looking for n random and different elements. It is then easy to see that when we want to pick the ith element, the odds of picking an element we've already visited are (i-1)/m. This is then the expected number of collisions for that particular pick. For picking n elements, the expected number of collisions will be the sum of the number of expected collisions for each pick. We plug this into Wolfram Alpha (sum (i-1)/m, i=1 to n) and we get the answer (n**2 - n)/2m. The average number of picks for our naive algorithm is then n + (n**2 - n)/2m.
Unless my memory fails me completely (which entirely possible, actually), this gives an average-case run-time O(n**2).
The worst case for this algorithm is clearly when you're choosing the full set of N items. This is equivalent to asking: On average, how many times must I roll an N-sided die before each side has come up at least once?
Answer: N * HN, where HN is the Nth harmonic number,
a value famously approximated by log(N).
This means the algorithm in question is N log N.
As a fun example, if you roll an ordinary 6-sided die until you see one of each number, it will take on average 6 H6 = 14.7 rolls.
Before being able to answer this question in details, lets define the framework. Suppose you have a collection {a1, a2, ..., an} of n distinct objects, and want to pick m distinct objects from this set, such that the probability of a given object aj appearing in the result is equal for all objects.
If you have already picked k items, and radomly pick an item from the full set {a1, a2, ..., an}, the probability that the item has not been picked before is (n-k)/n. This means that the number of samples you have to take before you get a new object is (assuming independence of random sampling) geometric with parameter (n-k)/n. Thus the expected number of samples to obtain one extra item is n/(n-k), which is close to 1 if k is small compared to n.
Concluding, if you need m unique objects, randomly selected, this algorithm gives you
n/n + n/(n-1) + n/(n-2) + n/(n-3) + .... + n/(n-(m-1))
which, as Alderath showed, can be estimated by
m*n / (n-m+1).
You can see a little bit more from this formula:
* The expected number of samples to obtain a new unique element increases as the number of already chosen objects increases (which sounds logical).
* You can expect really long computation times when m is close to n, especially if n is large.
In order to obtain m unique members from the set, use a variant of David Knuth's algorithm for obtaining a random permutation. Here, I'll assume that the n objects are stored in an array.
for i = 1..m
k = randInt(i, n)
exchange(i, k)
end
here, randInt samples an integer from {i, i+1, ... n}, and exchange flips two members of the array. You only need to shuffle m times, so the computation time is O(m), whereas the memory is O(n) (although you can adapt it to only save the entries such that a[i] <> i, which would give you O(m) on both time and memory, but with higher constants).
Most people forget that looking up, if the number has already run, also takes a while.
The number of tries nessesary can, as descriped earlier, be evaluated from:
T(n,m) = n(H(n)-H(n-m)) ⪅ n(ln(n)-ln(n-m))
which goes to n*ln(n) for interesting values of m
However, for each of these 'tries' you will have to do a lookup. This might be a simple O(n) runthrough, or something like a binary tree. This will give you a total performance of n^2*ln(n) or n*ln(n)^2.
For smaller values of m (m < n/2), you can do a very good approximation for T(n,m) using the HA-inequation, yielding the formula:
2*m*n/(2*n-m+1)
As m goes to n, this gives a lower bound of O(n) tries and performance O(n^2) or O(n*ln(n)).
All the results are however far better, that I would ever have expected, which shows that the algorithm might actually be just fine in many non critical cases, where you can accept occasional longer running times (when you are unlucky).