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I am trying to solve a problem which is described below,
Given value of f(0) and k , which are integers.
I need to find value of f( T ). where T<=1010
Recursive function is,
f(n) = 2*f(n-1) , if 4*f(n-1) <=k
k - ( 2*f(n-1) ) , if 4*f(n-1) > k
My efforts,
#include<iostream>
using namespace std;
int main(){
long k,f0,i;
cin>>k>>f0;
long operation ;
cin>>operation;
long answer=f0;
for(i=1;i<=operation;i++){
answer=(4*answer <= k )?(2*answer):(k-(2*answer));
}
cout<<answer;
return 0;
}
My code gives me right answer. But, The code will run 1010 time in worst case that gives me Time Limit Exceed. I need more efficient solution for this problem. Please help me. I don't know the correct algorithm.
If 2f(0) < k then you can compute this function in O(log n) time (using exponentiation by squaring modulo k).
r = f(0) * 2^n mod k
return 2 * r >= k ? k - r : r
You can prove this by induction. The induction hypothesis is that 0 <= f(n) < k/2, and that the above code fragment computes f(n).
Here's a Python program which checks random test cases, comparing a naive implementation (f) with an optimized one (g).
def f(n, k, z):
r = z
for _ in xrange(n):
if 4*r <= k:
r = 2 * r
else:
r = k - 2 * r
return r
def g(n, k, z):
r = (z * pow(2, n, k)) % k
if 2 * r >= k:
r = k - r
return r
import random
errs = 0
while errs < 20:
k = random.randrange(100, 10000000)
n = random.randrange(100000)
z = random.randrange(k//2)
a1 = f(n, k, z)
a2 = g(n, k, z)
if a1 != a2:
print n, k, z, a1, a2
errs += 1
print '.',
Can you use methmetical solution before progamming and compulating?
Actually,
f(n) = f0*2^(n-1) , if f(n-1)*4 <= k
k - f0*2^(n-1) , if f(n-1)*4 > k
thus, your code will write like this:
condition = f0*pow(2, operation-2)
answer = condition*4 =< k? condition*2: k - condition*2
For a simple loop, your answer looks pretty tight; one could optimise a little bit using answer<<2 instead of 4*answer, and answer<<1 for 2*answer, but quite possibly your compiler is already doing that. If you're blowing the time with this, it might be necessary to reduce the loop itself somehow.
I can't figure out a mathematical pattern that #Shannon was going for, but I'm thinking we could exploit the fact that this function will sooner or later cycle. If the cycle is short enough, then we could short the loop by just getting the answer at the same point in the cycle.
So let's get some cycle detection equipment in the form of Brent's algorithm, and see if we can cut the loop to reasonable levels.
def brent(f, x0):
# main phase: search successive powers of two
power = lam = 1
tortoise = x0
hare = f(x0) # f(x0) is the element/node next to x0.
while tortoise != hare:
if power == lam: # time to start a new power of two?
tortoise = hare
power *= 2
lam = 0
hare = f(hare)
lam += 1
# Find the position of the first repetition of length λ
mu = 0
tortoise = hare = x0
for i in range(lam):
# range(lam) produces a list with the values 0, 1, ... , lam-1
hare = f(hare)
# The distance between the hare and tortoise is now λ.
# Next, the hare and tortoise move at same speed until they agree
while tortoise != hare:
tortoise = f(tortoise)
hare = f(hare)
mu += 1
return lam, mu
f0 = 2
k = 198779
t = 10000000000
def f(x):
if 4 * x <= k:
return 2 * x
else:
return k - 2 * x
lam, mu = brent(f, f0)
t2 = t
if t >= mu + lam: # if T is past the cycle's first loop,
t2 = (t - mu) % lam + mu # find the equivalent place in the first loop
x = f0
for i in range(t2):
x = f(x)
print("Cycle start: %d; length: %d" % (mu, lam))
print("Equivalent result at index: %d" % t2)
print("Loop iterations skipped: %d" % (t - t2))
print("Result: %d" % x)
As opposed to the other proposed answers, this approach actually could use a memo array to speed up the process, since the start of the function is actually calculated multiple times (in particular, inside brent), or it may be irrelevant, depending on how big the cycle happens to be.
The algorithm you proposed already has O(n).
To come up with more efficient algorithms, there is not that much direction we can go about. Some typical options we have
1.Decease the coefficients of the linear term( but I doubt it would make a difference in this case
2.Change to O(Logn)(typically use some sort of divide and conquer technique)
3.Change to O(1)
In this case, we can do the last one.
The recursion function is a piece-wise function
f(n) = 2*f(n-1) , if 4*f(n-1) <=k
k - ( 2*f(n-1) ) , if 4*f(n-1) > k
Let's tackle it by case:
case 1: if 4*f(n-1) <= k (1)(assuming the starting index is zero)
this is a obvious a geometry series
a_n = 2*a_n-1
Therefore, have the formula
Sn = 2^(n-1)f(0) ----()
Case 2: if 4*f(n-1) > k (2), we have
a_n = -2a_n-1 + k
Assuming, a_j is the element in the sequence which just satisfy condition (2)
Nestedly sub in an_1 to the formula, you will obtain the equation
an = k -2k +4k -8k... +(-2)^(n-j)* a_j
k -2k 4k -8... is another gemo series
Sn = k*(1-2^(n-j))/(1-2) ---gemo series sum formula with starting value k and ratio = -2
Therefore, we have a formula for an in the case 2
an = k * (1-2^(n-j))/(1-2) + (-2)^(n-j) * a_j ----（**）
All we left to do it to find aj which just dissatisfy condition (1) and satisfy (2)
This can be obtained in constant time again using the formula we have for case 1:
find n such that, 4*an = 4*Sn = 4*2^(n-1)*f(0)
solve for n: 4*2^(n-1)*f(0) = k, if n is not integer, take ceiling of n
In my first attempt to solve this question, I had wrong assumption that the value of the sequence is monotonically increasing but in fact the sequence might jump between case 1 and case 2. Therefore, there might not be constant algorithm to solve the problem.
However, we can use utilize the result above to skip iterative update complexity.
The overall algorithm will look something like:
start with T, K, and f(0)
compute n that make the condition switch using either (*) or (**)
update f(0) with f(n), update T - n
repeat
terminate when T-n = 0(the last iteration might over compute causing T-n<0, therefore, you need to go back a little bit if that happen)
Create a map that can store your results. Before finding f(n) check in that map, if solution is already existed or not.
If exists, use that solution.
Otherwise find it, store it for future use.
For C++:
Definition:
map<long,long>result;
Insertion:
result[key]=value
Accessing:
value=result[key];
Checking:
map<long,long>::iterator it=result.find(key);
if(it==result.end())
{
//key was not found, find the solution and insert into result
}
else
{
return result[key];
}
Use above technique for better solution.
I searched the answer for this question, i got various useful links but when i implemented the idea, i am getting wrong answer.
This is what I understood :
If m is prime, then it is very simple. Inverse modulus of any number 'a' can be calculated as:inverse_mod(a) = (a^(m-2))%m
but when m is not prime, the we have to find the prime factors of m ,
i.e. m= (p1^a1)*(p2^a2)*....*(pk^ak). Here p1,p2,....,pk are the prime factors of m and a1,a2,....,ak are their respective powers.
then we have to calculate :
m1 = a%(p1^a1),
m2 = a%(p2^a2),
.......
mk = a%(pk^ak)
then we have to combine all these remainders using Chinese Remainder Theorem (https://en.wikipedia.org/wiki/Chinese_remainder_theorem)
I implemented this idea for m=1000,000,000,but still i am getting Wrong Answer.
Here is my explanation for m=1000,000,000 which is not prime
m= (2^9)*(5^9) where 2 and 5 are m's prime factors.
let a is the number for which have to calculate inverse modulo m.
m1 = a%(2^9) = a^512
m2 = a%(5^9) = a^1953125
Our answer will be = m1*e1 + m2*e2
where e1= { 1 (mod 512)
0 (mod 1953125)
}
and e2= { 1 (mod 1953125)
0 (mod 512)
}
Now to calculate 'e1' and 'e2' , I used Extended Euclidean Algorithm.
https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm
The Code is :
void extend_euclid(lld a,lld b,lld& x,lld& y)
{
if(a%b==0)
{
x=0;
y=1;
return ;
}
extend_euclid(b,a%b,x,y);
int tmp=x;
x=y;
y=tmp-(a/b)*y;
}
Now e1= 1953125*y and e2=512*y;
So, Our final answer will be = m1*e1 + m2*e2 .
But after doing all this, I am getting wrong answer.
please explain and point out any mistakes which I have made while understanding Chinese Remainder Theorem .
Thank You Very Much.
The inverse of a modulo m only exists if a and m are coprime. If they are not coprime, nothing will help. For example: what is the inverse of 2 mod 4?
2*0 = 0 mod 4
2*1 = 2 mod 4
2*2 = 0 mod 4
2*3 = 2 mod 4
So no inverse.
This can indeed be computed by using the extended euclidean algorithm (although I'm not sure if you're doing it right), but the simplest way, in my opinion, is by using Euler's theorem:
a^phi(m) = 1 (mod m)
a*a^(phi(m) - 1) = 1 (mod m)
=> a^(phi(m) - 1) is the invers of a (mod m)
Where phi is the totient function:
phi(x) = x * (1 - 1/f1)(1 - 1/f2)...(1 - 1/fk)
where fi > 1 is a divisor of x (not necessarily a prime divisor)
phi(36) = 36(1 - 1/2)(1 - 1/3)(1 - 1/4)(1 - 1/6)(1 - 1/9)(1 - 1/12)(1 - 1/18)(1 - 1/36)
So it can be computed in O(sqrt n).
The exponentiation can then be computed using exponentiation by squaring.
If you want to read about how you can use the extended Euclidean algorithm to find the inverse faster, read this. I don't think the Chinese remainder theorem can help here.
I believe the following function will do what you want. Change from long to int if appropriate. It returns -1 if there is no inverse, otherwise returns a positive number in the range [0..m).
public static long inverse(long a, long m) { // mult. inverse of a mod m
long r = m;
long nr = a;
long t = 0;
long nt = 1;
long tmp;
while (nr != 0) {
long q = r/nr;
tmp = nt; nt = t - q*nt; t = tmp;
tmp = nr; nr = r - q*nr; r = tmp;
}
if (r > 1) return -1; // no inverse
if (t < 0) t += m;
return t;
}
I can't follow your algorithm to see exactly what is wrong with it, but I have a few general comments: Euler's totient function is rather slow to calculate in general, depending as it does on prime factorizations. The Chinese Remainder Theorem is useful in many contexts for combining results mod coprimes but it's not necessary here and again overcomplicates this particular issue because you end up having to factor your modulus, a very slow operation. And it's faster to implement GCD and modular inverse in a loop, rather than using recursion, though of course the two methods are equally effective.
If you're trying to compute a^(-1) mod p^k for p prime, first compute a^(-1) mod p. Given an x such that ax = 1 (mod p^(k-1)), you can "Hensel lift"---you're looking for the y between 0 and p-1 such that a(x + y p^(k-1)) = 1 (mod p^k). Doing some algebra, you find that you're looking for the y such that a y p^(k-1) = 1 - ax (mod p^k)---i.e. a y = (1 - ax)/p^(k-1) (mod p), where the division by p^(k-1) is exact. You can work this out using a modular inverse for a (mod p).
(Alternatively, simply notice that a^(p^(k-1)(p-1) - 1) = 1 (mod p^k). I mention Hensel lifting because it works in much greater generality.)
Yesterday i had this interview question, which I couldn't fully answer:
Given a function f() = 0 or 1 with a perfect 1:1 distribution, create a function f(n) = 0, 1, 2, ..., n-1 each with probability 1/n
I could come up with a solution for if n is a natural power of 2, ie use f() to generate the bits of a binary number of k=ln_2 n. But this obviously wouldn't work for, say, n=5 as this would generate f(5) = 5,6,7 which we do not want.
Does anyone know a solution?
You can build a rng for the smallest power of two greater than n as you described. Then whenever this algorithm generates a number larger than n-1, throw that number away and try again. This is called the method of rejection.
Addition
The algorithm is
Let m = 2^k >= n where k is is as small as possible.
do
Let r = random number in 0 .. m-1 generated by k coin flips
while r >= n
return r
The probability that this loop stops with at most i iterations is bounded by 1 - (1/2)^i. This goes to 1 very rapidly: The loop is still running after 30 iterations with probability less than one-billionth.
You can decrease the expected number of iterations with a slightly modified algorithm:
Choose p >= 1
Let m = 2^k >= p n where k is is as small as possible.
do
Let r = random number in 0 .. m-1 generated by k coin flips
while r >= p n
return floor(r / p)
For example if we are trying to generate 0 .. 4 (n = 5) with the simpler algorithm, we would reject 5, 6 and 7, which is 3/8 of the results. With p = 3 (for example), pn = 15, we'd have m = 16 and would reject only 15, or 1/16 of the results. The price is needing four coin flips rather than 3 and a division op. You can continue to increase p and add coin flips to decrease rejections as far as you wish.
Another interesting solution can be derived through a Markov Chain Monte Carlo technique, the Metropolis-Hastings algorithm. This would be significantly more efficient if a large number of samples were required but it would only approach the uniform distribution in the limit.
initialize: x[0] arbitrarily
for i=1,2,...,N
if (f() == 1) x[i] = (x[i-1]++) % n
else x[i] = (x[i-1]-- + n) % n
For large N the vector x will contain uniformly distributed numbers between 0 and n. Additionally, by adding in an accept/reject step we can simulate from an arbitrary distribution, but you would need to simulate uniform random numbers on [0,1] as a sub-procedure.
def gen(a, b):
min_possible = a
max_possible = b
while True:
floor_min_possible = floor(min_possible)
floor_max_possible = floor(max_possible)
if max_possible.is_integer():
floor_max_possible -= 1
if floor_max_possible == floor_min_possible:
return floor_max_possible
mid = (min_possible + max_possible)/2
if coin_flip():
min_possible = mid
else:
max_possible = mid
My #RandomNumberGenerator #RNG
/w any f(x) that gives rand ints from 1 to x, we can get rand ints from 1 to k, for any k:
get ints p & q, so p^q is smallest possible, while p is a factor of x, & p^q >= k;
Lbl A
i=0 & s=1; while i < q {
s+= ((f(x) mod p) - 1) * p^i;
i++;
}
if s > k, goto A, else return s
//** about notation/terms:
rand = random
int = integer
mod is (from) modulo arithmetic
Lbl is a “Label”, from the Basic language, & serves as a coordinates for executing code. After the while loop, if s > k, then “goto A” means return to the point of code where it says “Lbl A”, & resume. If you return to Lbl A & process the code again, it resets the values of i to 0 & s to 1.
i is an iterator for powers of p, & s is a sum.
"s+= foo" means "let s now equal what it used to be + foo".
"i++" means "let i now equal what it used to be + 1".
f(x) returns random integers from 1 to x. **//
I figured out/invented/solved it on my own, around 2008. The method is discussed as common knowledge here. Does anyone know since when the random number generator rejection method has been common knowledge? RSVP.
Given two ranges of positive integers x: [1 ... n] and y: [1 ... m] and random real R from 0 to 1, I need to find the pair of elements (i,j) from x and y such that x_i / y_j is closest to R.
What is the most efficient way to find this pair?
Using Farey sequence
This is a simple and mathematically beautiful algorithm to solve this: run a binary search, where on each iteration the next number is given by the mediant formula (below). By the properties of the Farey sequence that number is the one with the smallest denominator within that interval. Consequently this sequence will always converge and never 'miss' a valid solution.
In pseudocode:
input: m, n, R
a_num = 0, a_denom = 1
b_num = 1, b_denom = 1
repeat:
-- interestingly c_num/c_denom is already in reduced form
c_num = a_num + b_num
c_denom = a_denom + b_denom
-- if the numbers are too big, return the closest of a and b
if c_num > n or c_denom > m then
if R - a_num/a_denom < b_num/b_denom - R then
return a_num, a_denom
else
return b_num, b_denom
-- adjust the interval:
if c_num/c_denom < R then
a_num = c_num, a_denom = c_denom
else
b_num = c_num, b_denom = c_denom
goto repeat
Even though it's fast on average (my educated guess that it's O(log max(m,n))), it can still be slow if R is close to a fraction with a small denominator. For example finding an approximation to 1/1000000 with m = n = 1000000 will take a million iterations.
The standard approach to approximating reals with rationals is computing the continued fraction series (see [1]). Put a limit on the nominator and denominator while computing parts of the series, and the last value before you break the limits is a fraction very close to your real number.
This will find a very good approximation very fast, but I'm not sure this will always find a closest approximation. It is known that
any convergent [partial value of the continued fraction expansion] is nearer to the continued fraction than any other fraction whose denominator is less than that of the convergent
but there may be approximations with larger denominator (still below your limit) that are better approximations, but are not convergents.
[1] http://en.wikipedia.org/wiki/Continued_fraction
Given that R is a real number such that 0 <= R <= 1, integers x: [1 ... n] and integers y: [1 ... m]. It is assumed that n <= m, since if n > m then x[n]/y[m] will be greater than 1, which cannot be the closest approximation to R.
Therefore, the best approximation of R with the denominator d will be either floor(R*d) / d or ceil(R*d) / d.
The problem can be solved in O(m) time and O(1) space (in Python):
from __future__ import division
from random import random
from math import floor
def fractionize(R, n, d):
error = abs(n/d - R)
return (n, d, error) # (numerator, denominator, absolute difference to R)
def better(a, b):
return a if a[2] < b[2] else b
def approximate(R, n, m):
best = (0, 1, R)
for d in xrange(1, m+1):
n1 = min(n, int(floor(R * d)))
n2 = min(n, n1 + 1) # ceil(R*d)
best = better(best, fractionize(R, n1, d))
best = better(best, fractionize(R, n2, d))
return best
if __name__ == '__main__':
def main():
R = random()
n = 30
m = 100
print R, approximate(R, n, m)
main()
Prolly get flamed, but a lookup might be best where we compute all of the fractional values for each of the possible values.. So a simply indexing a 2d array indexed via the fractional parts with the array element containing the real equivalent. I guess we have discrete X and Y parts so this is finite, it wouldnt be the other way around.... Ahh yeah, the actual searching part....erm reet....
Rather than a completely brute force search, do a linear search over the shortest of your lists, using round to find the best match for each element. Maybe something like this:
best_x,best_y=(1,1)
for x in 1...n:
y=max(1,min(m,round(x/R)))
#optional optimization (if you have a fast gcd)
if gcd(x,y)>1:
continue
if abs(R-x/y)<abs(R-bestx/besty):
best_x,best_y=(x,y)
return (best_x,best_y)
Not at all sure whether the gcd "optimization" will ever be faster...
The Solution:
You can do this O(1) space and O(m log(n)) time:
there is no need to create any list to search,
The pseudo code may be is buggy but the idea is this:
r: input number to search.
n,m: the ranges.
for (int i=1;i<=m;i++)
{
minVal = min(Search(i,1,n,r), minVal);
}
//x and y are start and end of array:
decimal Search(i,x,y,r)
{
if (i/x > r)
return i/x - r;
decimal middle1 = i/Cill((x+y)/2);
decimal middle2 = i/Roof((x+y)/2);
decimal dist = min(middle1,middle2)
decimal searchResult = 100000;
if( middle > r)
searchResult = Search (i, x, cill((x+y)/2),r)
else
searchResult = Search(i, roof((x+y)/2), y,r)
if (searchResult < dist)
dist = searchResult;
return dist;
}
finding the index as home work to reader.
Description: I think you can understand what's the idea by code, but let trace one of a for loop:
when i=1:
you should search within bellow numbers:
1,1/2,1/3,1/4,....,1/n
you check the number with (1,1/cill(n/2)) and (1/floor(n/2), 1/n) and doing similar binary search on it to find the smallest one.
Should do this for loop for all items, so it will be done m time. and in each time it takes O(log(n)). this function can improve by some mathematical rules, but It will be complicated, I skip it.
If the denominator of R is larger than m then use the Farey method (which the Fraction.limit_denominator method implements) with a limit of m to get a fraction a/b where b is smaller than m else let a/b = R. With b <= m, either a <= n and you are done or else let M = math.ceil(n/R) and re-run the Farey method.
def approx2(a, b, n, m):
from math import ceil
from fractions import Fraction
R = Fraction(a, b)
if R < Fraction(1, m):
return 1, m
r = R.limit_denominator(m)
if r.numerator > n:
M = ceil(n/R)
r = R.limit_denominator(M)
return r.numerator, r.denominator
>>> approx2(113, 205, 50, 200)
(43, 78)
It might be possible to just run the Farey method once using a limiting denominator of min(ceil(n/R), m) but I am not sure about that:
def approx(a, b, n, m):
from math import ceil
from fractions import Fraction
R = Fraction(a, b)
if R < Fraction(1, m):
return 1, m
r = R.limit_denominator(min(ceil(n/R), m))
return r.numerator, r.denominator
Is there any known algorithm that can generate a shuffled range [0..n) in linear time and constant space (when output produced iteratively), given an arbitrary seed value?
Assume n may be large, e.g. in the many millions, so a requirement to potentially produce every possible permutation is not required, not least because it's infeasible (the seed value space would need to be huge). This is also the reason for a requirement of constant space. (So, I'm specifically not looking for an array-shuffling algorithm, as that requires that the range is stored in an array of length n, and so would use linear space.)
I'm aware of question 162606, but it doesn't present an answer to this particular question - the mappings from permutation indexes to permutations given in that question would require a huge seed value space.
Ideally, it would act like a LCG with a period and range of n, but the art of selecting a and c for an LCG is subtle. Simply satisfying the constraints for a and c in a full period LCG may satisfy my requirements, but I am wondering if there are any better ideas out there.
Based on Jason's answer, I've made a simple straightforward implementation in C#. Find the next largest power of two greater than N. This makes it trivial to generate a and c, since c needs to be relatively prime (meaning it can't be divisible by 2, aka odd), and (a-1) needs to be divisible by 2, and (a-1) needs to be divisible by 4. Statistically, it should take 1-2 congruences to generate the next number (since 2N >= M >= N).
class Program
{
IEnumerable<int> GenerateSequence(int N)
{
Random r = new Random();
int M = NextLargestPowerOfTwo(N);
int c = r.Next(M / 2) * 2 + 1; // make c any odd number between 0 and M
int a = r.Next(M / 4) * 4 + 1; // M = 2^m, so make (a-1) divisible by all prime factors, and 4
int start = r.Next(M);
int x = start;
do
{
x = (a * x + c) % M;
if (x < N)
yield return x;
} while (x != start);
}
int NextLargestPowerOfTwo(int n)
{
n |= (n >> 1);
n |= (n >> 2);
n |= (n >> 4);
n |= (n >> 8);
n |= (n >> 16);
return (n + 1);
}
static void Main(string[] args)
{
Program p = new Program();
foreach (int n in p.GenerateSequence(1000))
{
Console.WriteLine(n);
}
Console.ReadKey();
}
}
Here is a Python implementation of the Linear Congruential Generator from FryGuy's answer. Because I needed to write it anyway and thought it might be useful for others.
import random
import math
def lcg(start, stop):
N = stop - start
# M is the next largest power of 2
M = int(math.pow(2, math.ceil(math.log(N+1, 2))))
# c is any odd number between 0 and M
c = random.randint(0, M/2 - 1) * 2 + 1
# M=2^m, so make (a-1) divisible by all prime factors and 4
a = random.randint(0, M/4 - 1) * 4 + 1
first = random.randint(0, M - 1)
x = first
while True:
x = (a * x + c) % M
if x < N:
yield start + x
if x == first:
break
if __name__ == "__main__":
for x in lcg(100, 200):
print x,
Sounds like you want an algorithm which is guaranteed to produce a cycle from 0 to n-1 without any repeats. There are almost certainly a whole bunch of these depending on your requirements; group theory would be the most helpful branch of mathematics if you want to delve into the theory behind it.
If you want fast and don't care about predictability/security/statistical patterns, an LCG is probably the simplest approach. The wikipedia page you linked to contains this (fairly simple) set of requirements:
The period of a general LCG is at most
m, and for some choices of a much less
than that. The LCG will have a full
period if and only if:
c and m are relatively prime,
a - 1 is divisible by all prime factors of m
a - 1 is a multiple of 4 if m is a multiple of 4
Alternatively, you could choose a period N >= n, where N is the smallest value that has convenient numerical properties, and just discard any values produced between n and N-1. For example, the lowest N = 2k - 1 >= n would let you use linear feedback shift registers (LFSR). Or find your favorite cryptographic algorithm (RSA, AES, DES, whatever) and given a particular key, figure out the space N of numbers it permutes, and for each step apply encryption once.
If n is small but you want the security to be high, that's probably the trickiest case, as any sequence S is likely to have a period N much higher than n, but is also nontrivial to derive a nonrepeating sequence of numbers with a shorter period than N. (e.g. if you could take the output of S mod n and guarantee nonrepeating sequence of numbers, that would give information about S that an attacker might use)
See my article on secure permutations with block ciphers for one way to do it.
Look into Linear Feedback Shift Registers, they can be used for exactly this.
The short way of explaining them is that you start with a seed and then iterate using the formula
x = (x << 1) | f(x)
where f(x) can only return 0 or 1.
If you choose a good function f, x will cycle through all values between 1 and 2^n-1 (where n is some number), in a good, pseudo-random way.
Example functions can be found here, e.g. for 63 values you can use
f(x) = ((x >> 6) & 1) ^ ((x >> 5) & 1)