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I want to calculate a vector of a quadratic form, extracting the submatrix from 3 by 3 by 5 arrays. However, I cannot make the quadratic form using broadcasting (i.e., macro "#."). When using “for” statement, we can calculate the vector of the quadratic form. I have no idea how to conduct matrix operations using “#.” (I am reluctant to expand the quadratic form to calculate the vector.)
By contrast, the inner product is computable using “#.”.
The example code is as follows:
using LinearAlgebra
a1=[5 7 2; 2 1 5; 6 2 3]
a2=[2 7 1; 3 7 2; 1 2 3]
a3=[8 5 9; 1 1 3; 2 2 3]
a4=[2 5 6; 3 5 1; 1 1 1]
a5=[7 8 1; 5 1 3; 1 5 2]
z=cat(a1,a2,a3,a4,a5,dims=3)
##### case of inner product
x=zeros(5,3)
wz = reshape([],0)
for k in 1:5
w = hcat(z[[1],[1],k], z[2,2,k]) * hcat(z[[1],[1],k], z[[2],[2],k])'
#println(w)
wz=vcat(wz, w)
end
#. wz=convert(Float64,wz)
wz=Matrix{Float64}(wz)
x[:,3]=wz
# [inner product] same result, the 3rd column vector [26.0, 53.0, 65.0, 29.0, 50.0]
display(x)
x=zeros(5,3)
#. x[:,3] = dot(hcat(z[1,1,:],z[2,2,:]), hcat(z[1,1,:],z[2,2,:])) # ok, working
# [inner product] same result, the 3rd column vector [26.0, 53.0, 65.0, 29.0, 50.0]
display(x)
##### case of quadratic form
x=zeros(5,3)
wy = reshape([],0)
for k in 1:5
w = hcat(z[[1],[1],k], z[[2],[2],k]) * z[[1,3],[1,3],k] * hcat(z[[1],[1],k], z[[2],[2],k])'
#println(w)
wy=vcat(wy, w)
end
#. wy=convert(Float64,wy)
wy=Matrix{Float64}(wy)
x[:,3]=wy
# [quadratic form] distinct result, the 3rd column vector [168.0, 183.0, 603.0, 103.0, 359.0]
display(x)
# generating five 2 by 2 matrices, distinct result
#. dot(hcat(z[[1],[1],:],z[[2],[2],:]), z[[1,3],[1,3],:], hcat(z[[1],[1],:],z[[2],[2],:]))
# obtaining ERROR: DimensionMismatch("arrays could not be broadcast to a common size; got a dimension with lengths 2 and 5")
#. dot(hcat(z[1,1,:],z[2,2,:]), z[[1,3],[1,3],:], hcat(z[1,1,:],z[2,2,:]))
Would you mind giving helps and suggestions how to get the calculation of 3rd column vector [168.0, 183.0, 603.0, 103.0, 359.0] (which is made from the quadratic form) in the above code using "#."?
EDIT:
Perhaps the question is about specifically how to make broadcasting work in this case. If so:
#views dot.(vcat.(z[1,1,:],z[2,2,:]),getindex.(Ref(z),Ref([1,3]),Ref([1,3]),axes(z,3)),vcat.(z[1,1,:],z[2,2,:]))
should be a possible clarification. Or with the #. macro (though it doesn't seem simpler):
#. dot(vcat(z[1,1,:],z[2,2,:]),getindex($Ref(z),$Ref([1,3]),$Ref([1,3]),$axes(z,3)),vcat(z[1,1,:],z[2,2,:]))
ORIGINAL:
One way to calculate this:
[
[z[1,1,k] z[2,2,k]]*z[[1,3],[1,3],k]*[z[1,1,k] z[2,2,k]]' |> first
for k ∈ axes(z,3)
]
giving:
5-element Vector{Int64}:
168
183
603
103
359
(the |> first turns 1x1 matrix into scalar)
Option 2:
[let t = z[[1,3],[1,3],k] ; sum(z[i,i,k]*t[i,j]*z[j,j,k] for i ∈ (1,2), j ∈ (1,2)) ; end for k ∈ 1:5]
or:
[let t = z[[1,3],[1,3],k], v = [z[1,1,k],z[2,2,k]] ; dot(v,t,v) ; end for k ∈ 1:5]
or (this is pretty cool):
map((z;t=z[[1,3],[1,3]],v=[z[1,1],z[2,2]])->dot(v,t,v), eachslice(z,dims=3))
I'm trying to tackle down a problem where the time limit is very low (1 second) and the number of cases is supposedly high.
You need to tell if a number is divisible by 3, but the problem is that you don't get the direct number, you get a number k, and then need to check if the concatenation of numbers from 1 to k (123...k) is divisible by 3.
Example input:
4 // The number of cases
2
6
15
130000000
Output:
YES // Because 12 is divisible by 3
YES // Because 123456 is divisible by 3
YES // Because 123456789101112131415 is divisible by 3
NO
I've found some topics about quickly checking the divisibility, but what most time takes I think is to build the number. There are cases where the initial number is as high as 130000000 (so the final is 1234...130000000) which I thinks overflows any numeric data type.
So, what am I missing here? Is there any way to know if something is divisible by 3 without concatenating the number? Any ideas?
PD: Someone also posted the triangular numbers formula which also is a correct solution and then deleted the answer, it was:
if ((1 + num) * num / 2) % 3 == 0 ? "YES" : "NO"
Every third number is divisible by three.
Every number divisible by three has a digit sum divisible by 3.
Every third number has a digit sum divisible by 3.
In between these, every third number has a digit sum congruent to 1 and then 2 mod 3.
Take a look:
n digit sum mod 3
0 0
1 1
2 2
3 0
4 1
5 2
6 0
...
10 1
11 2
12 0
...
19 1
20 2
21 0
...
Say we have a string of digits constructed as you describe, and the number we just added was divisible mod 3. When we append the next number's digits, we are appending digits whose sum is congruent to 1 mod 3, and when added to those in our number, we will get a combined digit sum congruent to 1 mod 3, so our answer for the next one will be "no". The next one will add a number with digit sum congruent to 2 mod 3, and this causes the total to become congruent to 0 again, so the answer here is "yes". Finally, adding the next number which must be divisible by 3 keeps the digit sum congruent to 0.
The takeaway?
if n is congruent to 0 modulo 3, then the answer is "yes"
if n is congruent to 1 modulo 3, then the answer is "no"
if n is congruent to 2 modulo 3, then the answer is "yes"
In particular, your example for n=15 is wrong; the digit string obtained represents a number that should be divisible by 3, and indeed it is (try it on a big enough calculator to verify).
All that is left is to find an implementation that is fast enough and handles all the required cases. If n is guaranteed to be under ~2 billion, then you are probably safe with something like
return (n % 3) != 1;
If n can be an arbitrarily large number, never fear; you can check whether the digit sum is congruent to 0 modulo 3 by adding up the digits in linear time. If not, you can add 1 from the number by coding addition like you do it by hand on paper and then check the result of that for divisibility by 3, again in linear time. So something like:
if (digit_sum_mod_3(n) == 0) return true;
else if (digit_sum_mod_3(add_one(n)) == 0) return false;
else return true;
Then you would have something like
digit_sum_mod_3(n[1...m])
sum = 0
for k = 1 to m do
sum = sum + n[k]
// keep sum from getting too big
if sum >= 18 then
sum = sum - 18
return sum % 3
add_one(n[1...m])
// work from right to left, assume big-endian
for k = m to 1 do
if n[k] < 9 then // don't need to carry
n[k] = n[k] + 1
break
else then // need to carry
n[k] = 0
if n[1] = 0 then // carried all the way to the front
n[1] = 1
n[m+1] = 0
return n
Any three consecutive numbers sum up to 0 == a + a + 1 + a + 2 mod 3.
The answer reduces to k%3 == 0, or 2k-1 % 3 == 0. The latter is equivalent to k%3 == 2, which leaves out k%3==1 which then simplifies further to k%3 != 1.
It is a known trick in mathematics that a number is divisible by three if the sum of its individual decimal digits is divisible by three.
Example:
2271
2+2+7+1 = 12
12 is divisible by 3, therefore so is 2271
Additionally, the sum of any three consecutive integers must be divisible by three. This is because:
((n)+(n+1)+(n+2))/3 = (3n+3)/3 = n+1 = integer
Therefore:
If k mod 3 == 0, then concatenation of 1 to k is divisible by three.
If k mod 3 == 1, then concatenation of 1 to k is not divisible by three.
If k mod 3 == 2, then it is a bit trickier. In this case, concatenation of 1 to k is divisible by three if the sum of k and the number before k (which evaluates to (k)+(k-1), which is 2k-1) is divisible by three.
Therefore, the final condition is:
(k mod 3 == 0) || ((k mod 3 == 2) && (2k-1 mod 3 == 0))
However, this can be even further simplified.
It turns out that k mod 3 can only equal 2 whenever 2k-1 mod 3 equals 0 and vice versa.
See simple graph below that shows cyclic pattern of this behavior.
Therefore, the formula can be further simplified just to:
(k mod 3 == 0) || (k mod 3 == 2)
Or, even more simply:
(k mod 3 != 1)
I realize answerer already provided this answer so I don't expect this to be the accepted answer, just giving a more thorough mathematical explanation.
A number is divisible by three if the sum of its digits is divisible by three (see here). Therefore, there is no need to "construct" your number, you need simply add the digits of the individual numbers. Thus for your 15 case, you do not need to "construct" 123456789101112131415, you just need to sum all of the digits in [1, 2, 3, 4, ... 14, 15].
This is simpler than it sounds because the problem only needs to check numbers of a very specific format: 12345789101112131415…k. You can use Gauss's method to quickly get the sum of the numbers 1 to k and then check if that sum is divisible by three using the usual methods. The code for that is:
'NO' if (k*(k+1)/2)%3 else 'YES'
If you look at the pattern that occurs as k increases (NO, YES, YES, NO, YES, YES, ...), you don't even need the multiplication or division. In short, all you need is:
'YES' if (k-1)%3 else 'NO'
Here is Python code which reads integers from a file and, if it wouldn't take too long also checks the answer the hard way so you can see that it is right. (Python numbers can be infinitely long, so you don't need to worry about overflow):
#!/usr/bin/python3
# Read integers from stdin, convert each int to a triangular number
# and output YES (or NO) if it is divisible by 3.
def sumgauss(x):
'''Return the sum from 1 to x using Gauss's shortcut'''
return (x*(x+1)/2)
def triangle(n):
'''Given an integer n, return a string with all the integers
from 1 to n concatenated. E.g., 15 -> 123456789101112131415'''
result=""
for t in range(1, k+1):
result+=str(t)
return result
import sys
for k in sys.stdin.readlines():
k=int(k)
print ( 'YES' if (k-1)%3 else 'NO', end='')
# If it wouldn't take too long, double check by trying it the hard way
if k<100000:
kstr=triangle(k)
print("\t// %s modulo 3 is %d" % (kstr, int(kstr)%3))
else:
print('\t// 123456789101112131415...%d%d%d modulo 3 is %d' %
tuple([k-2, k-1, k, sumgauss(k)%3]))
Speaking of Gauss's shortcut for summation, this problem seems a lot like a homework assignment. (Gauss invented it as a student when a teacher was trying to get the class out of his hair for a while by making them add up the numbers from 1 to 100.) If this is indeed a class assignment, please make sure the teacher knows to give the A to me and stackoverflow. Thanks!
Sample output:
$ cat data
2
6
15
130000000
130000001
$ ./k3.py < data
YES // 12 modulo 3 is 0
YES // 123456 modulo 3 is 0
YES // 123456789101112131415 modulo 3 is 0
NO // 123456789101112131415...129999998129999999130000000 modulo 3 is 1
YES // 123456789101112131415...129999999130000000130000001 modulo 3 is 0
The first 32 triangular numbers:
$ seq 32 | ./k3.py
NO // 1 modulo 3 is 1
YES // 12 modulo 3 is 0
YES // 123 modulo 3 is 0
NO // 1234 modulo 3 is 1
YES // 12345 modulo 3 is 0
YES // 123456 modulo 3 is 0
NO // 1234567 modulo 3 is 1
YES // 12345678 modulo 3 is 0
YES // 123456789 modulo 3 is 0
NO // 12345678910 modulo 3 is 1
YES // 1234567891011 modulo 3 is 0
YES // 123456789101112 modulo 3 is 0
NO // 12345678910111213 modulo 3 is 1
YES // 1234567891011121314 modulo 3 is 0
YES // 123456789101112131415 modulo 3 is 0
NO // 12345678910111213141516 modulo 3 is 1
YES // 1234567891011121314151617 modulo 3 is 0
YES // 123456789101112131415161718 modulo 3 is 0
NO // 12345678910111213141516171819 modulo 3 is 1
YES // 1234567891011121314151617181920 modulo 3 is 0
YES // 123456789101112131415161718192021 modulo 3 is 0
NO // 12345678910111213141516171819202122 modulo 3 is 1
YES // 1234567891011121314151617181920212223 modulo 3 is 0
YES // 123456789101112131415161718192021222324 modulo 3 is 0
NO // 12345678910111213141516171819202122232425 modulo 3 is 1
YES // 1234567891011121314151617181920212223242526 modulo 3 is 0
YES // 123456789101112131415161718192021222324252627 modulo 3 is 0
NO // 12345678910111213141516171819202122232425262728 modulo 3 is 1
YES // 1234567891011121314151617181920212223242526272829 modulo 3 is 0
YES // 123456789101112131415161718192021222324252627282930 modulo 3 is 0
NO // 12345678910111213141516171819202122232425262728293031 modulo 3 is 1
YES // 1234567891011121314151617181920212223242526272829303132 modulo 3 is 0
Actually the answer is pretty straight forward, if the sum of the digits divisible by three then the number is also divisible by 3.
string ans=(((1 + num) * num) / 2) % 3 == 0 ? "YES" : "NO";
according to the problem sum of digit can be considered as sum of numbers from 1 to n, sum=(n*(n+1))/2
*Make sure you divide the whole thing by 2
Another approach:
string ans=n % 3 !=1 ? "YES" : "NO";
You can prove that if n or n-2 is divisible by 3, then the sum up to n is divisible by 3 (e.g., in your case sum(1...8), sum(1..9), sum(1..11), etc.).
I have a m x n tensor (Tensor 1) and another k x 2 tensor (Tensor 2) and I wish to extract all the values of Tensor 1 using indices based on Tensor 2. For example;
Tensor1
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
[torch.DoubleTensor of size 4x5]
Tensor2
2 1
3 5
1 1
4 3
[torch.DoubleTensor of size 4x2]
And the function would yield;
6
15
1
18
The first solution that comes into mind is to simply loop through indexes and pick the correspoding values:
function get_elems_simple(tensor, indices)
local res = torch.Tensor(indices:size(1)):typeAs(tensor)
local i = 0
res:apply(
function ()
i = i + 1
return tensor[indices[i]:clone():storage()]
end)
return res
end
Here tensor[indices[i]:clone():storage()] is just a generic way to pick an element from a multi-dimensional tensor. In k-dimensional case this is exactly analogous to tensor[{indices[i][1], ... , indices[i][k]}].
This method works fine if you don't have to extract lots of values (the bottleneck is :apply method which is not able to use many optimization techniques and SIMD instructions because the function it executes is a black box). The job can be done way more efficiently: the method :index does exactly what you need... with a one-dimensional tensor. Multi-dimensional target/index tensors need to be flattened:
function flatten_indices(sp_indices, shape)
sp_indices = sp_indices - 1
local n_elem, n_dim = sp_indices:size(1), sp_indices:size(2)
local flat_ind = torch.LongTensor(n_elem):fill(1)
local mult = 1
for d = n_dim, 1, -1 do
flat_ind:add(sp_indices[{{}, d}] * mult)
mult = mult * shape[d]
end
return flat_ind
end
function get_elems_efficient(tensor, sp_indices)
local flat_indices = flatten_indices(sp_indices, tensor:size())
local flat_tensor = tensor:view(-1)
return flat_tensor:index(1, flat_indices)
end
The difference is drastic:
n = 500000
k = 100
a = torch.rand(n, k)
ind = torch.LongTensor(n, 2)
ind[{{}, 1}]:random(1, n)
ind[{{}, 2}]:random(1, k)
elems1 = get_elems_simple(a, ind) # 4.53 sec
elems2 = get_elems_efficient(a, ind) # 0.05 sec
print(torch.all(elems1:eq(elems2))) # true
What is the Equation that on increasing the integer x returns an alternate of 0 and 1
example
x = 22
result 1
x = 23
result 0
x = 24
result 1
Based on the example data, it would be modulo 2. Assuming x is an int (and C/C++/C#):
(x + 1) % 2;
In C or C++ this would be
int y = (x+1)%2;
mathematically,
y = (x+1) modulo 2
It's called modulo. You can use mod by 2 after adding one in the value.
x = 22
result = (x+1) modulo 2
In programming languages, it's often called %:
x = 22
result = (x+1) % 2 //<< result 1
x = 23
result = (x+1) % 2 //<< result 0
and so on..
I have a small question about vector and matrix.
Suppose a vector V = {v1, v2, ..., vn}. I generate a n-by-n distance matrix M defined as:
M_ij = | v_i - v_j | such that i,j belong to [1, n].
That is, each element M_ij in the square matrix is the absolute distance of two elements in V.
For example, I have a vector V = {1, 3, 3, 5}, the distance matrix will be
M=[
0 2 2 4;
2 0 0 2;
2 0 0 2;
4 2 2 0; ]
It seems pretty simple. Now comes to the question. Given such a matrix M, how to obtain the initial V?
Thank you.
Based on some answer for this question, it seems that the answer is not unique. So, now suppose that all the initial vector has been normalized to 0 mean and 1 variance. The question is: Given such a symmetric distance matrix M, how to decide the initial normalized vector?
You can't. To give you an idea of why, consider these two cases:
V1 = {1,2,3}
M1 = [ 0 1 2 ; 1 0 1 ; 2 1 0 ]
V2 = {3,4,5}
M2 = [ 0 1 2 ; 1 0 1 ; 2 1 0 ]
As you can see, a single M could be the result of more than one V. Therefore, you can't map backwards.
There is no way to determine the answer uniquely, since the distance matrix is invariant to adding a constant to all elements and to multiplying all the values by -1. Assuming that element 1 is equal to 0, and that the first nonzero element is positive, however, you can find an answer. Here is the pseudocode:
# Assume v[1] is 0
v[1] = 0
# e is value of first non-zero vector element
e = 0
# ei is index of first non-zero vector element
ei = 0
for i = 2...n:
# if all vector elements have been 0 so far
if e == 0:
# get the current distance from element 1 and its index
# this new element may still be 0
e = d[1,i]
ei = i
v[i] = e
elseif d[1,i] == d[ei,i] + v[ei]: # v[i] <= v[1]
# v[i] is to the left of v[1] (assuming v[ei] > v[1])
v[i] = -d[1,i]
else:
# some other case; v[i] is to the right of v[1]
v[i] = d[1,i]
I don't think it is possible to find the original vector, but you can find a translation of the vector by taking the first row of the matrix.
If you let M_ij = | v_i - v_j | and you translate all v_k for k\in [1,n] you will get
M_ij = | v-i + 1 - v_j + 1 |
= | v_i - v_j |
Hence, just take the first row as the vector and find one initial point to translate the vector to.
Correction:
Let v_1 = 0, and let l_k = | v_k | for k\in [2,n] and p_k the parity of v_k
Let p_1 = 1
for(int i = 2; i < n; i++)
if( | l_i - l_(i+1) | != M_i(i+1) )
p_(i+1) = - p_i
else
p_(i+1) = p_i
doing this for all v_k for k\in [2,n] in order will show the parity of each v_k in respect to the others
Then you can find a translation of the original vector with the same or opposite direction
Update (For Normalized vector):
Let d = Sqrt(v_1^2 + v_2^2 + ... + v_n^2)
Vector = {0, v_1 / d, v_2 / d, ... , v_n / d}
or
{0, -v_1 / d, -v_2 / d, ... , -v_n / d}