I have a tensor of size m x n (m rows and n columns).
For example:
[ 5 8 4 3
1 3 5 4
3 9 8 6 ]
I wish to randomly select half of the columns, and set all the values in this columns as zeros.
For our example, it will create something like this:
[ 5 0 4 0
1 0 5 0
3 0 8 0 ]
I'm aware how to set zero randomly half of all the elements,
torch.rand(x.shape) > 0.5
but done randomly without consideration in the columns, which is not helpfull for my case.
Thank you for any help,
Dave
import torch
x = torch.rand(3,4)
x
tensor([[0.0143, 0.1070, 0.9985, 0.0727],
[0.4052, 0.8716, 0.7376, 0.5495],
[0.2553, 0.2330, 0.9285, 0.6535]])
for i in [1,3] : # list has your columns which you want to make zero
x[:,i] = 0
For a work shift optimization problem, I've defined a binary variable in PuLP as follows:
pulp.LpVariable.dicts('VAR', (range(D), range(N), range(T)), 0, 1, 'Binary')
where
D = # days in each schedule we create (=28, or 4 weeks)
N = # of workers
T = types of work shift (=6)
For the 5th and 6th type of work shift (with index 4 and 5), I need to add a constraint that any worker who works these shifts must do so for seven consecutive days... and not any seven days but the seven days starting from Monday (aka a full week). I've tried defining the constraint as follows, but I'm getting an infeasible solution when I add this constraint and try to solve the problem (it worked before without it)
I know this constraint (along with the others from before) should theoretically be feasible because we manually schedule work shifts with the same set of constraints. Is there anything wrong with the way I've coded the constraint?
## looping over each worker
for j in range(N):
## looping for every Monday in the 28 days
for i in range(0,D,7):
c = None
## accessing only the 5th and 6th work shift type
for k in range(4,T):
c+=var[i][j][k]+var[i+1][j][k]+var[i+2][j][k]+var[i+3][j][k]+var[i+4][j][k]+var[i+5][j][k]+var[i+6][j][k]
problem+= c==7
If I understand correctly then your constraint requires that each worker is required to work the 4th and 5th shift in every week. This is because of c == 7, i.e. 7 of the binaries in c must be set to 1. This does not allow any worker to work in shift 0 through 3, right?
You need to change the constraint so that c == 7 is only enforced if the worker works any shift in that range. A very simple way to do that would be something like
v = list()
for k in range(4,T):
v.extend([var[i][j][k], var[i+1][j][k], var[i+2][j][k], var[i+3][j][k], var[i+4][j][k], var[i+5][j][k], var[i+6][j][k]])
c = sum(v)
problem += c <= 7 # we can pick at most 7 variables from v
for x in v:
problem += 7 * x <= c # if any variable in v is picked, then we must pick 7 of them
This is by no means the best way to model that (indicator variables would be much better), but it should give you an idea what to do.
Just to offer an alternative approach, assuming (as I read it) that for any given week a worker can either work some combination of shifts in [0:3] across the seven days, or one of the shifts [4:5] every day: we can do this by defining a new binary variable Y[w][n][t] which is 1 if in week w worker n does a restricted shift t, 0 otherwise. Then we can relate this variable to our existing variable X by adding constraints so that the values X can take depend on the values of Y.
# Define the sets of shifts
non_restricted_shifts = [0,1,2,3]
restricted_shifts = [4,5]
# Define a binary variable Y, 1 if for week w worker n works restricted shift t
Y = LpVariable.dicts('Y', (range(round(D/7)), range(N), restricted_shifts), cat=LpBinary)
# If sum(Y[week][n][:]) = 1, the total number of non-restricted shifts for that week and n must be 0
for week in range(round(D/7)):
for n in range(N):
prob += lpSum(X[d][n][t] for d in range(week*7, week*7 + 7) for t in non_restricted_shifts) <= 1000*(1-lpSum(Y[week][n][t] for t in restricted_shifts))
# If worker n has 7 restricted shift t in week w, then Y[week][n][t] == 1, otherwise it is 0
for week in range(round(D/7)):
for n in range(N):
for t in restricted_shifts:
prob += lpSum(X[d][n][t] for d in range(week*7, week*7+7)) <= 7*(Y[week][n][t])
prob += lpSum(X[d][n][t] for d in range(week*7, week*7+7)) >= Y[week][n][t]*7
Some example output (D=14, N=2, T=6):
/ M T W T F S S / M T W T F S S / M T W T F S S / M T W T F S S
WORKER 0
Shifts: / 2 3 1 3 3 2 2 / 1 0 2 3 2 2 0 / 3 1 2 2 3 1 1 / 2 3 0 3 3 0 3
WORKER 1
Shifts: / 3 1 2 3 1 1 2 / 3 3 2 3 3 3 3 / 4 4 4 4 4 4 4 / 1 3 2 2 3 2 1
WORKER 2
Shifts: / 1 2 3 1 3 1 1 / 3 3 2 2 3 2 3 / 3 2 3 0 3 1 0 / 4 4 4 4 4 4 4
WORKER 3
Shifts: / 2 2 3 2 1 2 3 / 5 5 5 5 5 5 5 / 3 1 3 1 0 3 1 / 2 2 2 2 3 0 3
WORKER 4
Shifts: / 5 5 5 5 5 5 5 / 3 3 1 0 2 3 3 / 0 3 3 3 3 0 2 / 3 3 3 2 3 2 3
Is there any condition for writing a number N as sum of K prime numbers(prime numbers not necessarily distinct)?
Example: If N=6 and K=2 then we can write N as 6=3+3 whereas if N=11 and K=2 then we cannot represent 11 as sum of two primes.
My Approach- I deduced the condition that If K>=N then we cannot represent N as sum of K primes.Also if K=1 then by primality testing we can check whether whether N is a prime number. Also by goldbach's conjecture for even numbers(except 2) N can be represented as sum of two prime numbers.
But the main problem is that I'm not able to predict it for K>=3.
1.Well, first list out all the prime numbers less than and equal to N.
2.Brute Force Approach with backtracking method.
ex :
N = 8
k = 2.
2 2
2 3
2 5
2 7
3 3(Don't again consider 3 and 2)
3 5.
Done!
ex : 2
N = 12,
k = 4
2 2 2 2
2 2 2 3
2 2 2 5
2 2 2 7
2 2 3 3(don't again check for 2232)
2 2 3 5.
Done!
ex 3:
N = 11,
k = 3
2 2 2
2 2 3
2 2 5
2 2 7
2 2 11
2 3 3(don't check again for 232)
2 3 5
2 3 7>11(don't check for 2311)
3 3 3(don't again check the 32.. series.)
10.3 3 5
Done!
I've been trying to read a txt file containing a formatted matrix (9x9) into an int array. The txt file is selected by the user using NSOpenPanel.
An example txt file:
2 7 9 1 6 2 1 1 1
9 1 3 3 4 0 6 8 5
5 3 2 9 3 8 6 7 0
6 0 9 2 5 6 4 8 0
3 2 0 4 0 5 0 6 0
4 0 5 4 0 3 9 0 0
6 4 1 3 2 5 7 2 0
6 5 7 2 1 3 0 9 3
1 0 2 7 5 1 0 0 0
I'm really new to mac programming so any help would be greatly appreciated.
You can either do it with Objectice-C using
classes loke NSString and NSArray
reading a complete file is just
NSString *filesContent = [[NSString alloc] initWithContentsOfFile:#"file.txt"];
then you can split that array e.g at whitespace with something like
NSArray rows = [filesContent componentsSeparatedByString:#"\r\n"];
then you can split the rows at whitespace.
And or you do it the old faschioned C way.
Opening the file with fopen
reading the file line by line
splitting the line e.g. with sscanf
And filling an Array
int arr[9][[9];
Pseudo Code (not tested, just to give you an idea)
char buf [2048];
char *pc;
int arr[9][[9];
int i_rval;
int row[9];
File *fin = fopen("file_with_9_x_9_matrix.txt");
/* error handling */
int i = 0;
while ((pc = fgets(buf, sizeof(buf), fin)) != NULL) {
row = arr[i];
i_rval = sscanf("%d %d %d %d %d....", &row[0], &row[1]);
/* error handling */
i++;
}
Or you can mix something out of Objectice-C and C.
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Write code to determine if a number is divisible by 3. The input to the function is a single bit, 0 or 1, and the output should be 1 if the number received so far is the binary representation of a number divisible by 3, otherwise zero.
Examples:
input "0": (0) output 1
inputs "1,0,0": (4) output 0
inputs "1,1,0,0": (6) output 1
This is based on an interview question. I ask for a drawing of logic gates but since this is stackoverflow I'll accept any coding language. Bonus points for a hardware implementation (verilog etc).
Part a (easy): First input is the MSB.
Part b (a little harder): First input is the LSB.
Part c (difficult): Which one is faster and smaller, (a) or (b)? (Not theoretically in the Big-O sense, but practically faster/smaller.) Now take the slower/bigger one and make it as fast/small as the faster/smaller one.
There's a fairly well-known trick for determining whether a number is a multiple of 11, by alternately adding and subtracting its decimal digits. If the number you get at the end is a multiple of 11, then the number you started out with is also a multiple of 11:
47278 4 - 7 + 2 - 7 + 8 = 0, multiple of 11 (47278 = 11 * 4298)
52214 5 - 2 + 2 - 1 + 4 = 8, not multiple of 11 (52214 = 11 * 4746 + 8)
We can apply the same trick to binary numbers. A binary number is a multiple of 3 if and only if the alternating sum of its bits is also a multiple of 3:
4 = 100 1 - 0 + 0 = 1, not multiple of 3
6 = 110 1 - 1 + 0 = 0, multiple of 3
78 = 1001110 1 - 0 + 0 - 1 + 1 - 1 + 0 = 0, multiple of 3
109 = 1101101 1 - 1 + 0 - 1 + 1 - 0 + 1 = 1, not multiple of 3
It makes no difference whether you start with the MSB or the LSB, so the following Python function works equally well in both cases. It takes an iterator that returns the bits one at a time. multiplier alternates between 1 and 2 instead of 1 and -1 to avoid taking the modulo of a negative number.
def divisibleBy3(iterator):
multiplier = 1
accumulator = 0
for bit in iterator:
accumulator = (accumulator + bit * multiplier) % 3
multiplier = 3 - multiplier
return accumulator == 0
Here... something new... how to check if a binary number of any length (even thousands of digits) is divisible by 3.
-->((0))<---1--->()<---0--->(1) ASCII representation of graph
From the picture.
You start in the double circle.
When you get a one or a zero, if the digit is inside the circle, then you stay in that circle. However if the digit is on a line, then you travel across the line.
Repeat step two until all digits are comsumed.
If you finally end up in the double circle then the binary number is divisible by 3.
You can also use this for generating numbers divisible by 3. And I wouldn't image it would be hard to convert this into a circuit.
1 example using the graph...
11000000000001011111111111101 is divisible by 3 (ends up in the double circle again)
Try it for yourself.
You can also do similar tricks for performing MOD 10, for when converting binary numbers into base 10 numbers. (10 circles, each doubled circled and represent the values 0 to 9 resulting from the modulo)
EDIT: This is for digits running left to right, it's not hard to modify the finite state machine to accept the reverse language though.
NOTE: In the ASCII representation of the graph () denotes a single circle and (()) denotes a double circle. In finite state machines these are called states, and the double circle is the accept state (the state that means its eventually divisible by 3)
Heh
State table for LSB:
S I S' O
0 0 0 1
0 1 1 0
1 0 2 0
1 1 0 1
2 0 1 0
2 1 2 0
Explanation: 0 is divisible by three. 0 << 1 + 0 = 0. Repeat using S = (S << 1 + I) % 3 and O = 1 if S == 0.
State table for MSB:
S I S' O
0 0 0 1
0 1 2 0
1 0 1 0
1 1 0 1
2 0 2 0
2 1 1 0
Explanation: 0 is divisible by three. 0 >> 1 + 0 = 0. Repeat using S = (S >> 1 + I) % 3 and O = 1 if S == 0.
S' is different from above, but O works the same, since S' is 0 for the same cases (00 and 11). Since O is the same in both cases, O_LSB = O_MSB, so to make MSB as short as LSB, or vice-versa, just use the shortest of both.
Here is an simple way to do it by hand.
Since 1 = 22 mod 3, we get 1 = 22n mod 3 for every positive integer.
Furthermore 2 = 22n+1 mod 3. Hence one can determine if an integer is divisible by 3 by counting the 1 bits at odd bit positions, multiply this number by 2, add the number of 1-bits at even bit posistions add them to the result and check if the result is divisible by 3.
Example: 5710=1110012.
There are 2 bits at odd positions, and 2 bits at even positions. 2*2 + 2 = 6 is divisible by 3. Hence 57 is divisible by 3.
Here is also a thought towards solving question c). If one inverts the bit order of a binary integer then all the bits remain at even/odd positions or all bits change. Hence inverting the order of the bits of an integer n results is an integer that is divisible by 3 if and only if n is divisible by 3. Hence any solution for question a) works without changes for question b) and vice versa. Hmm, maybe this could help to figure out which approach is faster...
You need to do all calculations using arithmetic modulo 3. This is the way
MSB:
number=0
while(!eof)
n=input()
number=(number *2 + n) mod 3
if(number == 0)
print divisible
LSB:
number = 0;
multiplier = 1;
while(!eof)
n=input()
number = (number + multiplier * n) mod 3
multiplier = (multiplier * 2) mod 3
if(number == 0)
print divisible
This is general idea...
Now, your part is to understand why this is correct.
And yes, do homework yourself ;)
The idea is that the number can grow arbitrarily long, which means you can't use mod 3 here, since your number will grow beyond the capacity of your integer class.
The idea is to notice what happens to the number. If you're adding bits to the right, what you're actually doing is shifting left one bit and adding the new bit.
Shift-left is the same as multiplying by 2, and adding the new bit is either adding 0 or 1. Assuming we started from 0, we can do this recursively based on the modulo-3 of the last number.
last | input || next | example
------------------------------------
0 | 0 || 0 | 0 * 2 + 0 = 0
0 | 1 || 1 | 0 * 2 + 1 = 1
1 | 0 || 2 | 1 * 2 + 0 = 2
1 | 1 || 0 | 1 * 2 + 1 = 0 (= 3 mod 3)
2 | 0 || 1 | 2 * 2 + 0 = 1 (= 4 mod 3)
2 | 1 || 2 | 2 * 2 + 1 = 2 (= 5 mod 3)
Now let's see what happens when you add a bit to the left. First, notice that:
22n mod 3 = 1
and
22n+1 mod 3 = 2
So now we have to either add 1 or 2 to the mod based on if the current iteration is odd or even.
last | n is even? | input || next | example
-------------------------------------------
d/c | don't care | 0 || last | last + 0*2^n = last
0 | yes | 1 || 0 | 0 + 1*2^n = 1 (= 2^n mod 3)
0 | no | 1 || 0 | 0 + 1*2^n = 2 (= 2^n mod 3)
1 | yes | 1 || 0 | 1 + 1*2^n = 2
1 | no | 1 || 0 | 1 + 1*2^n = 0 (= 3 mod 3)
1 | yes | 1 || 0 | 2 + 1*2^n = 0
1 | no | 1 || 0 | 2 + 1*2^n = 1
input "0": (0) output 1
inputs "1,0,0": (4) output 0
inputs "1,1,0,0": (6) output 1
shouldn't this last input be 12, or am i misunderstanding the question?
Actually the LSB method would actually make this easier. In C:
MSB method:
/*
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_msb(char *input) {
unsigned value = 0;
char *p = input;
if (*p == '1') {
value &= 1;
}
p++;
while (*p) {
if (*p != ',') {
value <<= 1;
if (*p == '1') {
ret &= 1;
}
}
p++;
}
return (value % 3 == 0) ? 1 : 0;
}
LSB method:
/*
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_lsb(char *input) {
unsigned value = 0;
unsigned mask = 1;
char *p = input;
while (*p) {
if (*p != ',') {
if (*p == '1') {
value &= mask;
}
mask <<= 1;
}
p++;
}
return (value % 3 == 0) ? 1 : 0;
}
Personally I have a hard time believing one of these will be significantly different to the other.
I think Nathan Fellman is on the right track for part a and b (except b needs an extra piece of state: you need to keep track of if your digit position is odd or even).
I think the trick for part C is negating the last value at each step. I.e. 0 goes to 0, 1 goes to 2 and 2 goes to 1.
A number is divisible by 3 if the sum of it's digits is divisible by 3.
So you can add the digits and get the sum:
if the sum is greater or equal to 10 use the same method
if it's 3, 6, 9 then it is divisible
if the sum is different than 3, 6, 9 then it's not divisible