I've been trying to solve the codechef problem: http://www.codechef.com/MAY11/problems/TPRODUCT/
They have given the post-contest analysis here: http://www.codechef.com/wiki/may-2011-contest-problem-editorials
I need some help in understanding the logic discussed there:
They are talking about using logarithm in place of the function
Pi=max(Vi*PL, Vi*PR)
Math is not my strong area. [I've been trying to improve by participating in contests like this]. If someone can give a very dumbed down explanation for this problem, it would be helpful for mortals like me. Thanks.
One large problem with multiplication is that numbers get very large very fast, and there are issues with reaching the upper bounds of an int or long, and spilling over to the negatives. The logarithm allows us to keep the computations small, and then get the answer back modulo n.
In retracing the result found via dynamic programming, the naive solution is to multiply all the values together and then mod:
(x0 * x1 * x2 * ... * xk) (mod n)
this is replaced with a series of smaller computations, which avoid bound overflow:
z1 = e^(log(x0) + log(x1)) modulo n
z2 = e^(log(x2) + log(z1)) modulo n
...
zk = e^(log(xk) + log(z{k-1})) modulo n
and then zk contains the result.
Presumably, they are relying on the simple mathematical observation that if:
z = y * x
then:
log(z) = log(y) + log(x)
Thus turning multiplications into additions.
Related
I want to find out coefficients of the n degree polynomial with roots 0,1,2...n-1. Can anybody suggest a good algorithm? I tried using FFT but didn't work fast enough
The simple solution that I would use is to write a function like this:
def poly_with_root_sequence (start, end, gap):
if end < start + gap:
return Polynomial([1, -start])
else:
p1 = poly_with_root_sequence(start, end, gap*2)
p2 = poly_with_root_sequence(start+gap, end, gap*2)
return p1 * p2
answer = poly_with_root_sequence(1, n, 1)
With a naive algorithm this will take O(n^2) arithmetic operations. However some of the operations will involve very large numbers. (Note that n! has more than n digits for large n.) But we have arranged that very few of the operations will involve very large numbers.
There is still no chance of producing answers as quickly as you want unless you are using a polynomial implementation with a very fast multiplication algorithm.
https://gist.github.com/ksenobojca/dc492206f8a8c7e9c75b155b5bd7a099 advertises itself as an implementation of the FFT algorithm for multiplying polynomials in Python. I can't verify that. But it gives you a shot at going fairly fast.
As replied on Evaluating Polynomial coefficients, you can do this as simple way:
def poly(lst, x):
n, tmp = 0, 0
for a in lst:
tmp = tmp + (a * (x**n))
n += 1
return tmp
print poly([1,2,3], 2)
Imagine that I'm a bakery trying to maximize the number of pies I can produce with my limited quantities of ingredients.
Each of the following pie recipes A, B, C, and D produce exactly 1 pie:
A = i + j + k
B = t + z
C = 2z
D = 2j + 2k
*The recipes always have linear form, like above.
I have the following ingredients:
4 of i
5 of z
4 of j
2 of k
1 of t
I want an algorithm to maximize my pie production given my limited amount of ingredients.
The optimal solution of these example inputs would yield me the following quantities of pies:
2 x A
1 x B
2 x C
0 x D
= a total of 5 pies
I can solve this easily enough by taking the maximal producer of all combinations, but the number
of combos becomes prohibitive as the quantities of ingredients increases. I feel like there must
be generalizations of this type of optimization problem, I just don't know where to start.
While I can only bake whole pies, I would be still be interested in seeing a method which may produce non integer results.
You can define the linear programming problem. I'll show the usage on the example, but it can of course be generalized to any data.
Denote your pies as your variables (x1 = A, x2 = B, ...) and the LP problem will be as follows:
maximize x1 + x2 + x3 + x4
s.t. x1 <= 4 (needed i's)
x1 + 2x4 <= 4 (needed j's)
x1 + 2x4 <= 2 (needed k's)
x2 <= 1 (needed t's)
x2 + 2x3 <= 5 (needed z's)
and x1,x2,x3,x4 >= 0
The fractional solution to this problem is solveable polynomially, but the integer linear programming is NP-Complete.
The problem is indeed NP-Complete, because given an integer linear programming problem, you can reduce the problem to "maximize the number of pies" using the same approach, where each constraint is an ingredient in the pie and the variables are the number of pies.
For the integers problem - there are a lot of approximation techniques in the literature for the problem if you can do with "close up to a certain bound", (for example local ratio technique or primal-dual are often used) or if you need an exact solution - exponential solution is probably your best shot. (Unless of course, P=NP)
Since all your functions are linear, it sounds like you're looking for either linear programming (if continuous values are acceptable) or integer programming (if you require your variables to be integers).
Linear programming is a standard technique, and is efficiently solvable. A traditional algorithm for doing this is the simplex method.
Integer programming is intractable in general, because adding integral constraints allows it to describe intractable combinatorial problems. There seems to be a large number of approximation techniques (for example, you might try just using regular linear programming to see what that gets you), but of course they depend on the specific nature of your problem.
I came across a question where it was asked to find the number of unique ways of reaching from point 1 to point 2 in a 2D co-ordinate plain.
Note: This can be assumed without loss of generality that x1 < x2 and y1 < y2.
Moreover the motions are constrained int he following manner. One can move only right or up. means a valid move is from (xa, ya) to (xb, yb) if xa < xb and ya < yb.
Mathematically, this can be found by ( [(x2-x1)+(y2-y1)]! ) / [(x2-x1)!] * [(y2-y1)!]. I have thought of code too.
I have approaches where I coded with dynamic programming and my approach takes around O([max(x2,y2)]^2) time and Theta( x2 * y2 ) where I can just manage with the upper or lower triangular matrix.
Can you think of some other approaches where running time is less than this? I am thinking of a recursive solution where the minimum running time is O(max(x2,y2)).
A simple efficient solution is the mathematical one.
Let x2-x1 = n and y2-y1 = m.
You need to take exactly n steps to the right, and m steps up, all is left to determine is their order.
This can be modeled as number of binary vectors with n+m elements with exactly n elements set to 1.
Thus, the total number of possibilities is chose(n,n+m) = (n+m)! / (n! * m!), which is exactly what you got.
Given that the mathematical answer is both proven and both faster to calculate - I see no reason for using a different solution with these restrictions.
If you are eager to use recursion here, the recursive formula for binomial coefficient will probably be a good fit here.
EDIT:
You might be looking for the multiplicative formula to calculate it.
To compute the answer, you can use this formula:
(n+m)!/(n!m!)=(n+1)*(n+2)/2*(n+3)/3*…*(n+m)/m
So the pseudo code is:
let foo(n,m) =
ans=1;
for i = 1 to m do
ans = ans*(n+i)/i;
done;
ans
The order of multiplications and divisions is important, if you modify it you can have an overflow even if the result is not so large.
I finally managed to write the article to describe this question in detail and complete the answer as well. Here is the link for the same. http://techieme.in/dynamic-programming-distinct-paths-between-two-points/
try this formula:
ans = (x2-x1) * (y2-y1) + 1;
I've been trying to implement a modular exponentiator recently. I'm writing the code in VHDL, but I'm looking for advice of a more algorithmic nature. The main component of the modular exponentiator is a modular multiplier which I also have to implement myself. I haven't had any problems with the multiplication algorithm- it's just adding and shifting and I've done a good job of figuring out what all of my variables mean so that I can multiply in a pretty reasonable amount of time.
The problem that I'm having is with implementing the modulus operation in the multiplier. I know that performing repeated subtractions will work, but it will also be slow. I found out that I could shift the modulus to effectively subtract large multiples of the modulus but I think there might still be better ways to do this. The algorithm that I'm using works something like this (weird pseudocode follows):
result,modulus : integer (n bits) (previously defined)
shiftcount : integer (initialized to zero)
while( (modulus<result) and (modulus(n-1) != 1) ){
modulus = modulus << 1
shiftcount++
}
for(i=shiftcount;i>=0;i--){
if(modulus<result){result = result-modulus}
if(i!=0){modulus = modulus >> 1}
}
So...is this a good algorithm, or at least a good place to start? Wikipedia doesn't really discuss algorithms for implementing the modulo operation, and whenever I try to search elsewhere I find really interesting but incredibly complicated (and often unrelated) research papers and publications. If there's an obvious way to implement this that I'm not seeing, I'd really appreciate some feedback.
I'm not sure what you're calculating there to be honest. You talk about modulo operation, but usually a modulo operation is between two numbers a and b, and its result is the remainder of dividing a by b. Where is the a and b in your pseudocode...?
Anyway, maybe this'll help: a mod b = a - floor(a / b) * b.
I don't know if this is faster or not, it depends on whether or not you can do division and multiplication faster than a lot of subtractions.
Another way to speed up the subtraction approach is to use binary search. If you want a mod b, you need to subtract b from a until a is smaller than b. So basically you need to find k such that:
a - k*b < b, k is min
One way to find this k is a linear search:
k = 0;
while ( a - k*b >= b )
++k;
return a - k*b;
But you can also binary search it (only ran a few tests but it worked on all of them):
k = 0;
left = 0, right = a
while ( left < right )
{
m = (left + right) / 2;
if ( a - m*b >= b )
left = m + 1;
else
right = m;
}
return a - left*b;
I'm guessing the binary search solution will be the fastest when dealing with big numbers.
If you want to calculate a mod b and only a is a big number (you can store b on a primitive data type), you can do it even faster:
for each digit p of a do
mod = (mod * 10 + p) % b
return mod
This works because we can write a as a_n*10^n + a_(n-1)*10^(n-1) + ... + a_1*10^0 = (((a_n * 10 + a_(n-1)) * 10 + a_(n-2)) * 10 + ...
I think the binary search is what you're looking for though.
There are many ways to do it in O(log n) time for n bits; you can do it with multiplication and you don't have to iterate 1 bit at a time. For example,
a mod b = a - floor((a * r)/2^n) * b
where
r = 2^n / b
is precomputed because typically you're using the same b many times. If not, use the standard superconverging polynomial iteration method for reciprocal (iterate 2x - bx^2 in fixed point).
Choose n according to the range you need the result (for many algorithms like modulo exponentiation it doesn't have to be 0..b).
(Many decades ago I thought I saw a trick to avoid 2 multiplications in a row... Update: I think it's Montgomery Multiplication (see REDC algorithm). I take it back, REDC does the same work as the simpler algorithm above. Not sure why REDC was ever invented... Maybe slightly lower latency due to using the low-order result into the chained multiplication, instead of the higher-order result?)
Of course if you have a lot of memory, you can just precompute all the 2^n mod b partial sums for n = log2(b)..log2(a). Many software implementations do this.
If you're using shift-and-add for the multiplication (which is by no means the fastest way) you can do the modulo operation after each addition step. If the sum is greater than the modulus you then subtract the modulus. If you can predict the overflow, you can do the addition and subtraction at the same time. Doing the modulo at each step will also reduce the overall size of your multiplier (same length as input rather than double).
The shifting of the modulus you're doing is getting you most of the way towards a full division algorithm (modulo is just taking the remainder).
EDIT Here is my implementation in Python:
def mod_mul(a,b,m):
result = 0
a = a % m
b = b % m
while (b>0):
if (b&1)!=0:
result += a
if result >= m: result -= m
a = a << 1
if a>=m: a-= m
b = b>>1
return result
This is just modular multiplication (result = a*b mod m). The modulo operations at the top are not needed, but serve as a reminder that the algorithm assumes a and b are less than m.
Of course for modular exponentiation you'll have an outer loop that does this entire operation at each step doing either squaring or multiplication. But I think you knew that.
For modulo itself, I'm not sure. For modulo as part of the larger modular exponential operation, did you look up Montgomery multiplication as mentioned in the wikipedia page on modular exponentiation? It's been a while since I've looked into this type of algorithm, but from what I recall, it's commonly used in fast modular exponentiation.
edit: for what it's worth, your modulo algorithm seems ok at first glance. You're basically doing division which is a repeated subtraction algorithm.
That test (modulus(n-1) != 1) //a bit test?
-seems redundant combined with (modulus<result).
Designing for hardware implementation i would be conscious of the smaller/greater than tests implying more logic (subtraction) than bitwise operations and branching on zero.
If we can do bitwise tests easily, this could be quick:
m=msb_of(modulus)
while( result>0 )
{
r=msb_of(result) //countdown from prev msb onto result
shift=r-m //countdown from r onto modulus or
//unroll the small subtraction
takeoff=(modulus<<(shift)) //or integrate this into count of shift
result=result-takeoff; //necessary subtraction
if(shift!=0 && result<0)
{ result=result+(takeoff>>1); }
} //endwhile
if(result==0) { return result }
else { return result+takeoff }
(code untested may contain gotchas)
result is repetively decremented by modulus shifted to match at most significant bits.
After each subtraction: result has a ~50/50 chance of loosing more than 1 msb. It also has ~50/50 chance of going negative,
addition of half what was subtracted will always put it into positive again. > it should be put back in positive if shift was not=0
The working loop exits when result is underrun and 'shift' was 0.
Given positive integers b, c, m where (b < m) is True it is to find a positive integer e such that
(b**e % m == c) is True
where ** is exponentiation (e.g. in Ruby, Python or ^ in some other languages) and % is modulo operation. What is the most effective algorithm (with the lowest big-O complexity) to solve it?
Example:
Given b=5; c=8; m=13 this algorithm must find e=7 because 5**7%13 = 8
From the % operator I'm assuming that you are working with integers.
You are trying to solve the Discrete Logarithm problem. A reasonable algorithm is Baby step, giant step, although there are many others, none of which are particularly fast.
The difficulty of finding a fast solution to the discrete logarithm problem is a fundamental part of some popular cryptographic algorithms, so if you find a better solution than any of those on Wikipedia please let me know!
This isn't a simple problem at all. It is called calculating the discrete logarithm and it is the inverse operation to a modular exponentation.
There is no efficient algorithm known. That is, if N denotes the number of bits in m, all known algorithms run in O(2^(N^C)) where C>0.
Python 3 Solution:
Thankfully, SymPy has implemented this for you!
SymPy is a Python library for symbolic mathematics. It aims to become a full-featured computer algebra system (CAS) while keeping the code as simple as possible in order to be comprehensible and easily extensible. SymPy is written entirely in Python.
This is the documentation on the discrete_log function. Use this to import it:
from sympy.ntheory import discrete_log
Their example computes \log_7(15) (mod 41):
>>> discrete_log(41, 15, 7)
3
Because of the (state-of-the-art, mind you) algorithms it employs to solve it, you'll get O(\sqrt{n}) on most inputs you try. It's considerably faster when your prime modulus has the property where p - 1 factors into a lot of small primes.
Consider a prime on the order of 100 bits: (~ 2^{100}). With \sqrt{n} complexity, that's still 2^{50} iterations. That being said, don't reinvent the wheel. This does a pretty good job. I might also add that it was almost 4x times more memory efficient than Mathematica's MultiplicativeOrder function when I ran with large-ish inputs (44 MiB vs. 173 MiB).
Since a duplicate of this question was asked under the Python tag, here is a Python implementation of baby step, giant step, which, as #MarkBeyers points out, is a reasonable approach (as long as the modulus isn't too large):
def baby_steps_giant_steps(a,b,p,N = None):
if not N: N = 1 + int(math.sqrt(p))
#initialize baby_steps table
baby_steps = {}
baby_step = 1
for r in range(N+1):
baby_steps[baby_step] = r
baby_step = baby_step * a % p
#now take the giant steps
giant_stride = pow(a,(p-2)*N,p)
giant_step = b
for q in range(N+1):
if giant_step in baby_steps:
return q*N + baby_steps[giant_step]
else:
giant_step = giant_step * giant_stride % p
return "No Match"
In the above implementation, an explicit N can be passed to fish for a small exponent even if p is cryptographically large. It will find the exponent as long as the exponent is smaller than N**2. When N is omitted, the exponent will always be found, but not necessarily in your lifetime or with your machine's memory if p is too large.
For example, if
p = 70606432933607
a = 100001
b = 54696545758787
then 'pow(a,b,p)' evaluates to 67385023448517
and
>>> baby_steps_giant_steps(a,67385023448517,p)
54696545758787
This took about 5 seconds on my machine. For the exponent and the modulus of those sizes, I estimate (based on timing experiments) that brute force would have taken several months.
Discrete logarithm is a hard problem
Computing discrete logarithms is believed to be difficult. No
efficient general method for computing discrete logarithms on
conventional computers is known.
I will add here a simple bruteforce algorithm which tries every possible value from 1 to m and outputs a solution if it was found. Note that there may be more than one solution to the problem or zero solutions at all. This algorithm will return you the smallest possible value or -1 if it does not exist.
def bruteLog(b, c, m):
s = 1
for i in xrange(m):
s = (s * b) % m
if s == c:
return i + 1
return -1
print bruteLog(5, 8, 13)
and here you can see that 3 is in fact the solution:
print 5**3 % 13
There is a better algorithm, but because it is often asked to be implemented in programming competitions, I will just give you a link to explanation.
as said the general problem is hard. however a prcatical way to find e if and only if you know e is going to be small (like in your example) would be just to try each e from 1.
btw e==3 is the first solution to your example, and you can obviously find that in 3 steps, compare to solving the non discrete version, and naively looking for integer solutions i.e.
e = log(c + n*m)/log(b) where n is a non-negative integer
which finds e==3 in 9 steps