I'm learning SageMath (uses Python 3)
and playing with the Goldbach conjecture.
I wrote this function (it works!):
def Goldbach(n):
if n % 2 != 0 or n <= 2:
show("No és parell")
else:
for i in srange(n):
if is_prime(i):
for j in srange(n):
if is_prime(j) and i + j == n:
a = [i, j]
show(a)
return
Now I'm trying (no idea) to do the following plot:
Denoting by r(2k) the number of Goldbach partitions of 2k,
the conjecture affirms that r(2k) > 0 if k > 1.
I have to do a graph of points (k, r(2k)), k > 2.
How could I do it?
First of all, let us get some better implementation in Sage
of the routine counting the number r(K) (for K > 2 some even integer) of the solutions for p + q = 2k, p, q prime numbers.
We count both solutions (p, q) and (q, p) when they differ.
def r(K):
if K not in ZZ or K <= 2 or K % 2:
return None
if K == 4:
return 1
count = 0
for p in primes(3, K):
for q in primes(3, K + 1 - p):
if p + q == K:
count += 1
return count
goldbach_points = [(K, r(K)) for K in range(4, 100,2)]
show(points(goldbach_points))
This gives:
I am working in an optimization problem (A*v = b) where I would like to rank a set of alternatives X = {x1,x2,x3,x4}. However, I have the following normalization constraint: |v[i] - v[j]| <= 1, which can be in the form -1 <= v[i] - v[j] <= 1.
My code is as follows:
import cvxpy as cp
n = len(X) #set of alternatives
v = cp.Variable(n)
objective = cp.Minimize(cp.sum_squares(A*v - b))
constraints = [0 <= v]
#Normalization condition -1 <= v[i] - v[j] <= 1
for i in range(n):
for j in range(n):
constraints = [-1 <= v[i]-v[j], 1 >= v[i]-v[j]]
prob = cp.Problem(objective, constraints)
# The optimal objective value is returned by `prob.solve()`.
result = prob.solve()
# The optimal value for v is stored in `v.value`.
va2 = v.value
Which outputs:
[-0.15 0.45 -0.35 0.05]
Result, which is not close to what should be and even have negative values. I think, my code for the normalization contraint most probably is wrong.
You are not appending your constraints, instead you are overwriting them each time. Instead of this line
constraints = [-1 <= v[i]-v[j], 1 >= v[i]-v[j]]
You should have
constraints += [-1 <= v[i]-v[j], 1 >= v[i]-v[j]]
For cleanliness you may want to change this
for i in range(n):
for j in range(n):
To only consider each pair once:
for i in range(n):
for j in range(i+1, n):
link of question
http://codeforces.com/contest/615/problem/D
link of solution is
http://codeforces.com/contest/615/submission/15260890
In below code i am not able to understand why 1 is subtracted from mod
where mod=1000000007
ll d = 1;
ll ans = 1;
for (auto x : cnt) {
ll cnt = x.se;
ll p = x.fi;
ll fp = binPow(p, (cnt + 1) * cnt / 2, MOD);
ans = binPow(ans, (cnt + 1), MOD) * binPow(fp, d, MOD) % MOD;
d = d * (x.se + 1) % (MOD - 1);//why ??
}
Apart from the fact that there is the code does not make much sense as out of context as it is, there is the little theorem of Fermat:
Whenever MOD is a prime number, as 10^9+7 is, one can reduce exponents by multiples of (MOD-1) as for any a not a multiple of MOD
a ^ (MOD-1) == 1 mod MOD.
Which means that
a^b == a ^ (b mod (MOD-1)) mod MOD.
As to the code, which is efficient for its task, consider n=m*p^e where m is composed of primes smaller than p.
Then for each factor f of m there are factors 1*f, p*f, p^2*f,...,p^e*f of n. The product over all factors of n thus is the product over
p^(0+1+2+...+e) * f^(e+1) = p^( e*(e+1)/2 ) * f^(e+1)
over all factors f of m. Putting the numbers of factors as d and the product of factors of m as ans results in the combined formula
ans = ans^( e+1 ) * p^( d*e*(e+1)/2 )
d = d*(e+1)
which can now be recursively applied to the list of prime factors and their multiplicities.
I want to understand how the modulus operator works when applied to two intervals. Adding, subtracting and multiplying two intervals is trivial to implement in code, but how do you do it for modulus?
I'd be happy if someone can show me the formula, sample code or a link which explains how it works.
Background info: You have two integers x_lo < x < x_hi and y_lo < y < y_hi. What is the the lower and upper bound for mod(x, y)?
Edit: I'm unsure if it is possible to come up with the minimal bounds in an efficient manner (without calculating the mod for all x or for all y). If so, then I'll accept an accurate but non-optimal answer for the bounds. Obviously, [-inf,+inf] is a correct answer then :) but I want a bound that is more limited in size.
It turns out, this is an interesting problem. The assumption I make is that for integer intervals, modulo is defined with respect to truncated division (round towards 0).
As a consequence, mod(-a,m) == -mod(a,m) for all a, m. Moreover, sign(mod(a,m)) == sign(a).
Definitions, before we start
Closed interval from a to b: [a,b]
Empty interval: [] := [+Inf,-Inf]
Negation: -[a,b] := [-b,-a]
Union: [a,b] u [c,d] := [min(a,c),max(b,d)]
Absolute value: |m| := max(m,-m)
Simpler Case: Fixed modulus m
It is easier to start with a fixed m. We will later generalize this to the modulo of two intervals. The definition builds up recursively. It should be no problem to implement this in your favorite programming language. Pseudocode:
def mod1([a,b], m):
// (1): empty interval
if a > b || m == 0:
return []
// (2): compute modulo with positive interval and negate
else if b < 0:
return -mod1([-b,-a], m)
// (3): split into negative and non-negative interval, compute and join
else if a < 0:
return mod1([a,-1], m) u mod1([0,b], m)
// (4): there is no k > 0 such that a < k*m <= b
else if b-a < |m| && a % m <= b % m:
return [a % m, b % m]
// (5): we can't do better than that
else
return [0,|m|-1]
Up to this point, we can't do better than that. The resulting interval in (5) might be an over-approximation, but it is the best we can get. If we were allowed to return a set of intervals, we could be more precise.
General case
The same ideas apply to the case where our modulus is an interval itself. Here we go:
def mod2([a,b], [m,n]):
// (1): empty interval
if a > b || m > n:
return []
// (2): compute modulo with positive interval and negate
else if b < 0:
return -mod2([-b,-a], [m,n])
// (3): split into negative and non-negative interval, compute, and join
else if a < 0:
return mod2([a,-1], [m,n]) u mod2([0,b], [m,n])
// (4): use the simpler function from before
else if m == n:
return mod1([a,b], m)
// (5): use only non-negative m and n
else if n <= 0:
return mod2([a,b], [-n,-m])
// (6): similar to (5), make modulus non-negative
else if m <= 0:
return mod2([a,b], [1, max(-m,n)])
// (7): compare to (4) in mod1, check b-a < |modulus|
else if b-a >= n:
return [0,n-1]
// (8): similar to (7), split interval, compute, and join
else if b-a >= m:
return [0, b-a-1] u mod2([a,b], [b-a+1,n])
// (9): modulo has no effect
else if m > b:
return [a,b]
// (10): there is some overlapping of [a,b] and [n,m]
else if n > b:
return [0,b]
// (11): either compute all possibilities and join, or be imprecise
else:
return [0,n-1] // imprecise
Have fun! :)
Let see mod(x, y) = mod.
In general 0 <= mod <= y. So it's always true: y_lo < mod < y_hi
But we can see some specific cases below:
- if: x_hi < y_lo then div(x, y) = 0, then x_low < mod < x_hi
- if: x_low > y_hi then div(x, y) > 0, then y_low < mod < y_hi
- if: x_low < y_low < y_hi < x_hi, then y_low < mod < y_hi
- if: x_low < y_low < x_hi < y_hi, then y_low < mod < x_hi
- if: y_low < x_low < y_hi < x_hi, then y_low < mod < y_hi
....
I have written myPercolation.ml.
open MyUnionFind
module type MyPercolationSig = sig
type percolation
val create_percolation : int -> percolation
val open_site : percolation -> int -> int -> unit
val is_open : percolation -> int -> int -> bool
val is_full : percolation -> int -> int -> bool
val can_percolates : percolation -> bool
end
module MyPercolation : MyPercolationSig = struct
exception IndexOutOfBounds;;
type percolation =
{n : int;
sites: bool array;
union : MyUnionFind.union_find};;
let create_percolation n =
{n = n; sites = Array.make (n*n) false; union = MyUnionFind.create_union (n*n)};;
let open_site p i j =
let {n;_;union} = p
in
if not (is_open p i j) then
begin
sites.(index_of n i j) <- true;
if i - 1 >= 1 && i - 1 <= n && is_open n (i-1) j then
MyUnionFind.union union (index_of n i j) (index_of n (i-1) j)
else if i + 1 >= 1 && i + 1 <= n && is_open n (i+1) j then
MyUnionFind.union union (index_of n i j) (index_of n (i+1) j)
else if j - 1 >= 1 && j - 1 <= n && is_open n i (j-1) then
MyUnionFind.union union (index_of n i j) (index_of n i (j-1))
else if j + 1 >= 1 && j + 1 <= n && is_open n i (j+1) then
MyUnionFind.union union (index_of n i j) (index_of n i (j+1))
end;;
let index_of n i j = n * (i - 1) + j;;
let is_open {n;sites;_} i j =
if i < 1 || i > n || j < 1 || j > n then
raise IndexOutOfBounds
else
sites.(index_of n i j);;
let is_full {n;_;union} i j =
let rec is_connected_top j' =
if j = 0 then false
else
if MyUnionFind.is_connected union (index_of n i j) (index_of n 0 j') then true
else is_connected_top (j'-1)
in is_connected_top n;;
let can_percolates p =
let {n;_;_} = p
in
let rec is_full_bottom j =
if j = 0 then false
else
if is_full p n j then true
else is_full_bottom (j-1)
end
Please ignore the package MyUnionFind package. It is just a homemade implementation for union-find algorithm.
when I try to compile the myPercolation.ml, I got such an error:
$ ocamlc -c myPercolation.ml
File "myPercolation.ml", line 25, characters 11-12:
Error: Syntax error: '}' expected
File "myPercolation.ml", line 25, characters 8-9:
Error: This '{' might be unmatched
I think the error is talking about let {n;_;union} = p in function of let open_site p i j.
I have read through that line and all code many times, but I still don't see any mismatched {} in that line.
can anyone help please?
Another possible error: {n;_;_} should be {n;_} Only 1 underscore is necessary. Think of it like the _ wildcard in a match statement.
The expression let {n; _; union} = p is not well formed OCaml. I think what you want is let {n; union} = p. The way to handle fields you don't care about in a record pattern is not to mention them.
Update:
As rgrinberg points out, a much better way to describe the problem is that the _ has to appear as the last field. That's why the compiler is expecting to see } afterward. It might be good style to include the _ as an indicator that you're purposely matching only a subset of the fields of the record. You can, in fact, turn on a compiler option that checks for this.
Update 2:
The warning for incomplete record patterns is warning number 9, and also is associated with the letter R. Here's how to use R:
$ ocaml -w +R
OCaml version 4.00.0
# type r = { a: int; b: char };;
type r = { a : int; b : char; }
# let {a} = {a=3; b='x'} in a;;
Warning 9: the following labels are not bound in this record pattern:
b
Either bind these labels explicitly or add '; _' to the pattern.
- : int = 3
The command-line syntax is the same for the compiler.