Problem: x1+x2....xn=C where x1,x2....xn >= 0 and is a integer. Find an algorithm that finds every point (x1,x2...xn) that solves this.
Why: I am trying to iterate a multivariable polynomial's terms. The powers of each term can be described by the points above. (You do this operation for C = 0 to C = degree of the polynomial)
I am stuck trying to make an efficient algorithm that produced only the unique solutions (non duplicates) and wanted to see if there is any existing algorithm
After some thought on this problem (and alot of paper), here is my algorithm:
It finds every combination of array of length N that sum to k and the elements are greater then or equal to 0.
This does not do trial and error to get the solution however it does involve quite alot of loops. Greater optimization can be made by creating a generating function when k and n are known beforehand.
If anyone has a better algorithm or finds a problem with this one, please post below, but for now this solves my problem.
Thank you #kcsquared and #Kermit the Frog for leading me in the right direction
""" Function that iterates a n length vector such that the combination sum is always equal to k and all elements are the natural numbers
.Returns if it was stopped or not
Invokes lambda function on every iteration
iteration_lambda (index_vector::Vector{T}, total_iteration::T)::Bool
Return true when it should end
"""
function partition(k::T, n::T, iteration_lambda::Function; max_vector = nothing, sum_vector = nothing, index_vector = nothing)::Bool where T
if n > 0
max_vector = max_vector == nothing ? zeros(T, n) : max_vector
sum_vector = sum_vector == nothing ? zeros(T, n) : sum_vector
index_vector = index_vector == nothing ? zeros(T, n) : index_vector
current_index_index::T = 1
total_iteration::T = 1
max_vector[1] = k
index_vector[1] = max(0, -(k * (n - 1)))
#label reset
if index_vector[current_index_index] <= max_vector[current_index_index]
if current_index_index != n
current_index_index += 1
sum_vector[current_index_index] = sum_vector[current_index_index - 1] + index_vector[current_index_index - 1]
index_vector[current_index_index] = max(0, -(k * (n - current_index_index - 1) + sum_vector[current_index_index]))
max_vector[current_index_index] = k - sum_vector[current_index_index]
else
if iteration_lambda(index_vector, total_iteration)
return true
end
total_iteration += 1
index_vector[end] += 1
end
#goto reset
end
if current_index_index != 1
current_index_index -= 1
index_vector[current_index_index] += 1
#goto reset
end
end
return false
end
I am learning ruby and was given the following assignment:
given two sorted arrays like the following we must merge them into one sorted array.
array_1 = [5,8,9,11]
array_2 = [4,6,7,10]
merge(array_1, array_2)
=> [4,5,6,7,8,9,10,11]
Given this brief description, implement the merge method that takes two arrays and returns
the properly sorted array containing the items from each array.
I wrote this answer:
def merge(arr1, arr2)
i = 0
k = 0
merged_arr = []
until k = arr2.count
while arr1[i] <= arr2[k]
merged_arr.push(arr1[i])
i += 1
end
merged_arr.push(arr2[k])
k += 1
end
merged_arr
end
My instructor sent out a solution, which I understand, however I don't understand why my answer does NOT work. Can someone please explain the faulty logic? Thank you!
Here is the (correct) solution:
def merge(array_1, array_2)
i = 0
k = 0
merged_array = []
while i < array_1.count
while k < array_2.count && array_1[i] > array_2[k]
merged_array << array_2[k]
k += 1
end
merged_array << array_1[i]
i += 1
end
print merged_array.inspect
end
k = arr2.count assigns the value of arr2.count to k and evaluates to k, so until k = arr2.count is never executed.
you also need to consider the unequal length of arr1 and arr2, your instructor's solution was only right if arr1.length >= arr2.length, but if arr1.length < arr2.length, then the elements from the extra length was lost in the solution.
I made a method that generates prime factors. Whatever composite number I push to it, it gives the prime factors. However, if I push a prime number into it, it wouldn't return 1 and the number itself. Instead, it would return 1 and some prime number smaller than the number pushed into the method.
I decided to shove an if statement that would cut the process short if the number pushed into turns out to be prime. Here's the code:
def get_prime_factors(number)
prime_factors = []
i = 0
primes = primes_gen(number)
if primes.include?(number)
return "Already a prime!"
end
original_number = number
while primes[i] <= original_number / 2
if number % primes[i] == 0
prime_factors << primes[i]
number = number / primes[i]
else
i = i + 1
end
if number == 1
return prime_factors
end
end
end
I fed 101 to the method and the method returned nil. This method calls the primes_gen method, which returns an array containing all primes smaller than the input value. Here it is:
def primes_gen(limit)
primes = []
i = 0
while i <= limit
primes << i if isprime?(i)
i = i + 1
end
primes.delete(0)
primes.delete(1)
return primes
end
I know there ought to be a more finessed way to fix the. If anyone wants to recommend a direction for me to explore as far as that goes, I'd be very grateful.
EDIT: Changed line 4 of the primes_gen() method to include a <= operator instead of a < operator.
Try changing primes = primes_gen(number) to primes = primes_gen(number+1) in first function and see if it works. Or try changing the i < limit condition to i <= limit in the second function.
Also, why are you deleting the 0th and 1st element in primes_gen method? Is it because of values you get for 0, 1? In which case, you can initialize with i=2.
So this code will count the total number of pairs of numbers whose difference is K. it is naive method and I need to optimize it. suggestions?
test = $stdin.readlines
input = test[0].split(" ")
numbers = test[1].split(" ")
N = input[0]
K = input[1]
count = 0
for i in numbers
current = i.to_i
numbers.shift
for j in numbers
difference = (j.to_i - current).abs
if (difference == K)
count += 1
end
end
end
puts count
Would have been nice for you to give some examples of input and output, but I think this is correct.
require 'set'
def count_diff(numbers, difference)
set = Set.new numbers
set.inject 0 do |count, num|
set.include?(num+difference) ? count+1 : count
end
end
difference = gets.split[1].to_i
numbers = gets.split.map { |num| num.to_i }
puts count_diff(numbers, difference)
Untested, hopefully actual Ruby code
Documentation for Set: http://www.ruby-doc.org/stdlib/libdoc/set/rdoc/classes/Set.html
require 'set'
numbers_set = Set.new
npairs = 0
numbers.each do |number|
if numbers_set.include?(number + K)
npairs += 1
end
if numbers_set.include?(number - K)
npairs += 1
end
numbers_set.add(number)
end
Someone deleted his post, or his post was deleted... He had the best solution, here it is :
test = $stdin.readlines
input = test[0].split(" ")
numbers = test[1].split(" ")
K = input[1]
count = 0
numbers.combination(2){|couple| couple.inject(:-).abs == K ? count++}
puts count
You don't even need N.
I do not know Ruby so I'll just give you the big idea:
Get the list
Keep a boolean array (call it arr), marking off numbers as true if the number exists in the list
Loop through the list and see if arr[num-K] and/or arr[num+K] is true where num is a number in your list
This uses up quite a bit of memory though so another method is to do the following:
Keep a hash map from an integer n to an integer count
Go through your list, adding num+K and num-K to the hash map, incrementing count accordingly
Go through your list and see if num is in the hash map. If it is, increment your counter by count
Well, I have this bit of code that is slowing down the program hugely because it is linear complexity but called a lot of times making the program quadratic complexity. If possible I would like to reduce its computational complexity but otherwise I'll just optimize it where I can. So far I have reduced down to:
def table(n):
a = 1
while 2*a <= n:
if (-a*a)%n == 1: return a
a += 1
Anyone see anything I've missed? Thanks!
EDIT: I forgot to mention: n is always a prime number.
EDIT 2: Here is my new improved program (thank's for all the contributions!):
def table(n):
if n == 2: return 1
if n%4 != 1: return
a1 = n-1
for a in range(1, n//2+1):
if (a*a)%n == a1: return a
EDIT 3: And testing it out in its real context it is much faster! Well this question appears solved but there are many useful answers. I should also say that as well as those above optimizations, I have memoized the function using Python dictionaries...
Ignoring the algorithm for a moment (yes, I know, bad idea), the running time of this can be decreased hugely just by switching from while to for.
for a in range(1, n / 2 + 1)
(Hope this doesn't have an off-by-one error. I'm prone to make these.)
Another thing that I would try is to look if the step width can be incremented.
Take a look at http://modular.fas.harvard.edu/ent/ent_py .
The function sqrtmod does the job if you set a = -1 and p = n.
You missed a small point because the running time of your improved algorithm is still in the order of the square root of n. As long you have only small primes n (let's say less than 2^64), that's ok, and you should probably prefer your implementation to a more complex one.
If the prime n becomes bigger, you might have to switch to an algorithm using a little bit of number theory. To my knowledge, your problem can be solved only with a probabilistic algorithm in time log(n)^3. If I remember correctly, assuming the Riemann hypothesis holds (which most people do), one can show that the running time of the following algorithm (in ruby - sorry, I don't know python) is log(log(n))*log(n)^3:
class Integer
# calculate b to the power of e modulo self
def power(b, e)
raise 'power only defined for integer base' unless b.is_a? Integer
raise 'power only defined for integer exponent' unless e.is_a? Integer
raise 'power is implemented only for positive exponent' if e < 0
return 1 if e.zero?
x = power(b, e>>1)
x *= x
(e & 1).zero? ? x % self : (x*b) % self
end
# Fermat test (probabilistic prime number test)
def prime?(b = 2)
raise "base must be at least 2 in prime?" if b < 2
raise "base must be an integer in prime?" unless b.is_a? Integer
power(b, self >> 1) == 1
end
# find square root of -1 modulo prime
def sqrt_of_minus_one
return 1 if self == 2
return false if (self & 3) != 1
raise 'sqrt_of_minus_one works only for primes' unless prime?
# now just try all numbers (each succeeds with probability 1/2)
2.upto(self) do |b|
e = self >> 1
e >>= 1 while (e & 1).zero?
x = power(b, e)
next if [1, self-1].include? x
loop do
y = (x*x) % self
return x if y == self-1
raise 'sqrt_of_minus_one works only for primes' if y == 1
x = y
end
end
end
end
# find a prime
p = loop do
x = rand(1<<512)
next if (x & 3) != 1
break x if x.prime?
end
puts "%x" % p
puts "%x" % p.sqrt_of_minus_one
The slow part is now finding the prime (which takes approx. log(n)^4 integer operation); finding the square root of -1 takes for 512-bit primes still less than a second.
Consider pre-computing the results and storing them in a file. Nowadays many platforms have a huge disk capacity. Then, obtaining the result will be an O(1) operation.
(Building on Adam's answer.)
Look at the Wikipedia page on quadratic reciprocity:
x^2 ≡ −1 (mod p) is solvable if and only if p ≡ 1 (mod 4).
Then you can avoid the search of a root precisely for those odd prime n's that are not congruent with 1 modulo 4:
def table(n):
if n == 2: return 1
if n%4 != 1: return None # or raise exception
...
Based off OP's second edit:
def table(n):
if n == 2: return 1
if n%4 != 1: return
mod = 0
a1 = n - 1
for a in xrange(1, a1, 2):
mod += a
while mod >= n: mod -= n
if mod == a1: return a//2 + 1
It looks like you're trying to find the square root of -1 modulo n. Unfortunately, this is not an easy problem, depending on what values of n are input into your function. Depending on n, there might not even be a solution. See Wikipedia for more information on this problem.
Edit 2: Surprisingly, strength-reducing the squaring reduces the time a lot, at least on my Python2.5 installation. (I'm surprised because I thought interpreter overhead was taking most of the time, and this doesn't reduce the count of operations in the inner loop.) Reduces the time from 0.572s to 0.146s for table(1234577).
def table(n):
n1 = n - 1
square = 0
for delta in xrange(1, n, 2):
square += delta
if n <= square: square -= n
if square == n1: return delta // 2 + 1
strager posted the same idea but I think less tightly coded. Again, jug's answer is best.
Original answer: Another trivial coding tweak on top of Konrad Rudolph's:
def table(n):
n1 = n - 1
for a in xrange(1, n // 2 + 1):
if (a*a) % n == n1: return a
Speeds it up measurably on my laptop. (About 25% for table(1234577).)
Edit: I didn't notice the python3.0 tag; but the main change was hoisting part of the calculation out of the loop, not the use of xrange. (Academic since there's a better algorithm.)
Is it possible for you to cache the results?
When you calculate a large n you are given the results for the lower n's almost for free.
One thing that you are doing is repeating the calculation -a*a over and over again.
Create a table of the values once and then do look up in the main loop.
Also although this probably doesn't apply to you because your function name is table but if you call a function that takes time to calculate you should cache the result in a table and just do a table look up if you call it again with the same value. This save you the time of calculating all of the values when you first run but you don't waste time repeating the calculation more than once.
I went through and fixed the Harvard version to make it work with python 3.
http://modular.fas.harvard.edu/ent/ent_py
I made some slight changes to make the results exactly the same as the OP's function. There are two possible answers and I forced it to return the smaller answer.
import timeit
def table(n):
if n == 2: return 1
if n%4 != 1: return
a1=n-1
def inversemod(a, p):
x, y = xgcd(a, p)
return x%p
def xgcd(a, b):
x_sign = 1
if a < 0: a = -a; x_sign = -1
x = 1; y = 0; r = 0; s = 1
while b != 0:
(c, q) = (a%b, a//b)
(a, b, r, s, x, y) = (b, c, x-q*r, y-q*s, r, s)
return (x*x_sign, y)
def mul(x, y):
return ((x[0]*y[0]+a1*y[1]*x[1])%n,(x[0]*y[1]+x[1]*y[0])%n)
def pow(x, nn):
ans = (1,0)
xpow = x
while nn != 0:
if nn%2 != 0:
ans = mul(ans, xpow)
xpow = mul(xpow, xpow)
nn >>= 1
return ans
for z in range(2,n) :
u, v = pow((1,z), a1//2)
if v != 0:
vinv = inversemod(v, n)
if (vinv*vinv)%n == a1:
vinv %= n
if vinv <= n//2:
return vinv
else:
return n-vinv
tt=0
pri = [ 5,13,17,29,37,41,53,61,73,89,97,1234577,5915587277,3267000013,3628273133,2860486313,5463458053,3367900313 ]
for x in pri:
t=timeit.Timer('q=table('+str(x)+')','from __main__ import table')
tt +=t.timeit(number=100)
print("table(",x,")=",table(x))
print('total time=',tt/100)
This version takes about 3ms to run through the test cases above.
For comparison using the prime number 1234577
OP Edit2 745ms
The accepted answer 522ms
The above function 0.2ms