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# Which of the following numbers is a perfect square?

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Which of the following numbers is a perfect square?  [#permalink]

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Updated on: 22 Mar 2013, 03:21
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Which of the following numbers is a perfect square?

A. 1266
B. 1444
C. 2022
D. 4034
E. 8122

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Originally posted by emmak on 22 Mar 2013, 01:04.
Last edited by Bunuel on 22 Mar 2013, 03:21, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Which of the following numbers is a perfect square?  [#permalink]

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22 Mar 2013, 03:40
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emmak wrote:
Which of the following numbers is a perfect square?

A. 1266
B. 1444
C. 2022
D. 4034
E. 8122

A perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square.

From above we can deduce that the units digit of a perfect square cannot be 2, 3, or 7. Discard C and E.

Another property: perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: $$36=2^2*3^2$$, powers of prime factors 2 and 3 are even.

Make prime factorization of the options:

A. 1266 = 2*3*211. Discard. We could discard 1266 after we got that 1266 = 2*633 = 2*odd, so 2 in 1266 has an odd power, which means that 1266 is NOT a prefect square.

B. 1444 = 2^2*19^2 --> so, 1444 IS a perfect square.

D. 4034 = 2*2017. Discard. We could discard 4034 after we got that 4034 = 2*2017 = 2*odd, so 2 in 4034 has an odd power, which means that 4034 is NOT a prefect square.

Hope it's clear.
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Re: Which of the following numbers is a perfect square?  [#permalink]

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05 Feb 2014, 14:27
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emmak wrote:
Which of the following numbers is a perfect square?

A. 1266
B. 1444
C. 2022
D. 4034
E. 8122

Let's begin by eliminating answer choices C and E since a perfect square cannot end in 2,3,7 or 8

Now between A,B and D

Well we can see that all numbers are even for starters. This means that they also must be divisible by 4

Only B fits the bill

Hence B is the correct answer

Hope its clear
Cheers
J
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Re: Which of the following numbers is a perfect square? 1266  [#permalink]

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22 Mar 2013, 01:20
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perfect Square never ends with 2, 3, 7 and 8 so straight away you can eliminate 8122 and 2022.
38*38 = 1444
hence B.
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Re: Which of the following numbers is a perfect square?  [#permalink]

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04 Mar 2014, 02:45
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Perfect squares cannot end with 2,3,7 or 8, so options C & E are eliminated

A. 1266
B. 1444
E. 8122

Look at the above 3 options. All are divisible by 2, but only 1444 is divisible by 4.

NB - Consider last two digits of any number. If Last two digits of any number are divisible by 4, then the complete number is divisible by 4
Also, If Last digit of any number is divisible by 2, then the complete number is divisible by 3
If Last three digits of any number are divisible by 8, then the complete number is divisible by 8

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Re: Which of the following numbers is a perfect square?  [#permalink]

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25 Jun 2014, 01:48
Well we can see that all numbers are even for starters. This means that they also must be divisible by 4

Only B fits the bill

Hence B is the correct answer

Dear J,

Can you please explain the underlined portion ? It'd be really helpful
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Re: Which of the following numbers is a perfect square?  [#permalink]

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25 Jun 2014, 02:47
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Tanvr wrote:
Well we can see that all numbers are even for starters. This means that they also must be divisible by 4

Only B fits the bill

Hence B is the correct answer

Dear J,

Can you please explain the underlined portion ? It'd be really helpful

For an even number to be a prefect square it must be a multiple of 4. That's because we know that a prefect square must have an even powers of its primes, so 2 in even number must have even power to be a prefect square: 2, 4, 6, ... so in any case it must be a multiple of 4.
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Re: Which of the following numbers is a perfect square?  [#permalink]

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25 Jun 2014, 02:49
2
Tanvr wrote:
Well we can see that all numbers are even for starters. This means that they also must be divisible by 4

Only B fits the bill

Hence B is the correct answer

Dear J,

Can you please explain the underlined portion ? It'd be really helpful

Perfect squares have even powers of prime factors. What this means is that if a number is a perfect square, and it has 2 as a prime factor, the power of 2 in the number will be even i.e. it will have two 2s or four 2s or six 2s etc. Similarly, if it has 3 as a factor, it will have two 3s or four 3s or six 3s etc.
The reason for this is explained here:
http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/2010/12 ... t-squares/

Now when you see that 2 is a factor of all leftover options, you know that you will have at least two 2s i.e. the number must be divisible by 4 if the number is to be a perfect square. Check the last two digits of the numbers. If the last two digits are divisible by 4, the number will be divisible by 4. Only option (B) is divisible by 4 (because 44 is divisible by 4). Hence (B) is the correct answer.
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Re: Which of the following numbers is a perfect square?  [#permalink]

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23 Jul 2014, 08:25
Quick solution: B

the perfect square of whatever a integer number has the following possible digit 1 ; 4 ; 5 ; 6; 9
=> eliminate: C and E
A and D is divisible by 2, but not by 4 => A and D is not a perfect square

B is correct
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Re: Which of the following numbers is a perfect square?  [#permalink]

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25 Jun 2019, 23:40
All the numbers in the options are even numbers.
For an even number to be even, it should be divisible by 4. We can check divisibility by 4 by only looking at the last 2 digits of the number.
(A number id divisible by 4 if it's last 2 digits are a multiple of 4).

Option B. 1444 -> last 2 digits make 44 -> 44 is divisible by 4. Hence option B is divisible by 4 and is a possible candidate for a perfect square. None of the other even numbers are divisible by 4, so they CANNOT be perfect squares.

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Re: Which of the following numbers is a perfect square?   [#permalink] 25 Jun 2019, 23:40
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text:
MORE IN Circuits and Transmission Lines
MU Electronics and Telecom Engineering (Semester 3)
Circuits and Transmission Lines
December 2012
Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary

1 (a) For maximum power transfer find the value of ZL if
(i) ZL is impedance
(ii) ZL is pure resistance :
5 M
1 (b) find initial value of, f(t)= 20-10t-e-25t
verify using initial value theorem.
5 M
1 (c) In the network given below, initial value of IL =4A and VC=100. Find IC(0+):
5 M
1 (d) Current I1 and I2 entering at 1 and port 2 respectively of two port network are given by following equation.
I1 =0.5V1 - 0.2V2
I2 = -0.2V1 +V2
Obtain T and Π(pi) representation.
5 M
Attempt any FOUR from the following
1 (e) For network shown, find VC /V and draw pole-zero plot.
5 M

2 (a) Using mesh analysis find power supplied by the dependent source.
10 M
2 (b) Find current supplied by source.
10 M

3 (a) Write B and Q matrix for the Graph shown.
10 M
3 (b) Draw Bode plot for the function G(s). Find gain margin, phase margin and comment on stability.
$G(s)= \frac{2(s+0.25)}{s^2(s+1)(s+0.5)}$
10 M

4 (a) Switch is opened at t=0 with initial conditions as shown. Find
$v_1, \frac{dv_1}{dt}, \frac{dv_2}{dt}\ at \ time\ 0^+$
10 M
4 (b) Find Y parameter using interconnection:
10 M

5 (a) In the network key is closed at t=0. Find i1 (0+), i2(0+ and i3 (0+).
10 M
5 (b) Find i(t).
10 M

6 (a) The circuit attain steady state with switch at position (a) & is moved to position (b) at t=0. Find V(t) for t ≥ 0.
10 M
6 (b) Find Z11, Z21 and G21
10 M

7 (a) Realize following functon in Foster II form
$Z(s)=\frac{(s^2+1)(s^2+3)}{s(s^2+2)(s^2+4)}$
10 M
7 (b) Check following polynomials for Hurtwitz -
$\ \left(i\right)\ \ p\left(s\right)=S^4+4s^2+8$
$\left(ii\right)\ \ p\left(s\right)=s^4+s^3+5s^2+3s+4$
10 M

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(Assume that each student in the group plays at least one game). Let F, H and C represent the set of students who play foot ball, hockey and cricket respectively. We first discuss cardinality for finite sets and (Hint: you can arrange $\Q^+$ in a sequence; use this to arrange $\Q$ into a sequence.) set which is a contradiction. of students who play both (foot ball & hockey) only = 12, No. To see this, note that when we add $|A|$ and $|B|$, we are counting the elements in $|A \cap B|$ twice, Because of the symmetyofthissituation,wesaythatA and B can be put into 1-1 correspondence. That is often difficult, however. If $A$ is a finite set, then $|B|\leq |A| < \infty$, Then,byPropositionsF12andF13intheFunctions section,fis invertible andf−1is a 1-1 correspondence fromBtoA. Consider sets A and B.By a transformation or a mapping from A to B we mean any subset T of the Cartesian product A×B that satisfies the following condition: . Question: Prove that N(all natural numbers) and Z(all integers) have the same cardinality. To prove that a given in nite set X … Theorem. How to prove that all maximal independent sets of a matroid have the same cardinality. We can, however, try to match up the elements of two inﬁnite sets A and B one by one. As seen, the symbol for the cardinality of a set resembles the absolute value symbol — a variable sandwiched between two vertical lines. Provided a matroid is a 2-tuple (M,J ) where M is a finite set and J is a family of some of the subsets of M satisfying the following properties: If A is subset of B and B belongs to J , then A belongs to J , =  n(F) + n(H) + n(C) - n(FnH) - n(FnC) - n(HnC) + n(FnHnC), n(FuHuC)  =  65 + 45 + 42 -20 - 25 - 15 + 8. This function is bijective. When it ... prove the corollary one only has to observe that a function with a “right inverse” is the “left inverse” of that function and vice versa. 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I presume you have sent this A2A to me following the most recent instalment of our ongoing debate regarding the ontological nature and resultant enumeration of Zero. To be precise, here is the definition. Find the total number of students in the group. A set A is countably infinite if and only if set A has the same cardinality as N (the natural numbers). The proof of this theorem is very similar to the previous theorem. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. Now that we know about functions and bijections, we can define this concept more formally and more rigorously. The idea is exactly the same as before. f:A → Bbea1-1correspondence. while the other is called uncountable. However, I am stuck in proving it since there are more than one "1", "01" = "1", same as other numbers. Fix m 2N. $$\biggl|\bigcup_{i=1}^n A_i\biggr|=\sum_{i=1}^n\left|A_i\right|-\sum_{i < j}\left|A_i\cap A_j\right|$$ but you cannot list the elements in an uncountable set. Imprint CRC Press. 1. if it is a finite set, $\mid A \mid < \infty$; or. I could not prove that cardinality is well defined, i.e. Thus according to Deﬁnition 2.3.1, the sets N and Z have the same cardinality. When A and B have the same cardinality, we write jAj= jBj. Thus, any set in this form is countable. of students who play both (hockey & cricket) only = 7, No. A set is an infinite set provided that it is not a finite set. of students who play hockey only = 18, No. The two sets A = {1,2,3} and B = {a,b,c} thus have the cardinality since we can match up the elements of the two sets in such a way that each element in each set is matched with exactly one element in the other set. For example, if $A=\{2,4,6,8,10\}$, then $|A|=5$. Math 131 Fall 2018 092118 Cardinality - Duration: 47:53. For in nite sets, this strategy doesn’t quite work. Let X m = fq 2Q j0 q 1; and mq 2Zg. For example, a consequence of this is that the set of rational numbers $\mathbb{Q}$ is countable. Since $A$ and $B$ are Cardinality of a set is a measure of the number of elements in the set. If A can be put into 1-1 correspondence with a subset of B (that is, there is a 1-1 case the set is said to be countably infinite. I can tell that two sets have the same number of elements by trying to pair the elements up. (useful to prove a set is finite) • A set is infinite when there … (b) A set S is finite if it is empty, or if there is a bijection for some integer . of students who play both foot ball &  hockey = 20, No. n(AuB)  =  Total number of elements related to any of the two events A & B. n(AuBuC)  =  Total number of elements related to any of the three events A, B & C. n(A)  =  Total number of elements related to  A. n(B)  =  Total number of elements related to  B. n(C)  =  Total number of elements related to  C. Total number of elements related to A only. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. We first discuss cardinality for finite sets and then talk about infinite sets. A useful application of cardinality is the following result. It would be a good exercise for you to try to prove this to yourself now. If S is a set, we denote its cardinality by |S|. Cantor showed that not all in・］ite sets are created equal 窶・his de・］ition allows us to distinguish betweencountable and uncountable in・］ite sets. Set S is a set consisting of all string of one or more a or b such as "a, b, ab, ba, abb, bba..." and how to prove set S is a infinity set. For finite sets, cardinalities are natural numbers: |{1, 2, 3}| = 3 |{100, 200}| = 2 For infinite sets, we introduced infinite cardinals to denote the size of sets: $$C=\bigcup_i \bigcup_j \{ a_{ij} \},$$ The cardinality of a set is denoted by $|A|$. Figure 1.13 shows one possible ordering. We have been able to create a list that contains all the elements in $\bigcup_{i} A_i$, so this correspondence with natural numbers $\mathbb{N}$. is concerned, this guideline should be sufficient for most cases. n(FnH)  =  20, n(FnC)  =  25, n(HnC)  =  15. The set whose elements are each and each and every of the subsets is the ability set. The above theorems confirm that sets such as $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ and their Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. 11 Cardinality Rules ... two sets, then the sets have the same size. Maybe this is not so surprising, because N and Z have a strong geometric resemblance as sets of points on the number line. For in nite sets, this strategy doesn’t quite work. Any set which is not finite is infinite. If $A_1, A_2,\cdots$ is a list of countable sets, then the set $\bigcup_{i} A_i=A_1 \cup A_2 \cup A_3\cdots$ Discrete Mathematics - Cardinality 17-16 More Countable Sets (cntd) The fact that you can list the elements of a countably infinite set means that the set can be put in one-to-one If $A$ and $B$ are countable, then $A \times B$ is also countable. should also be countable, so a subset of a countable set should be countable as well. We can say that set A and set B both have a cardinality of 3. set is countable. S and T have the same cardinality if there is a bijection f from S to T. Notation: means that S and T have the same cardinality. respectively. The cardinality of a set is denoted by $|A|$. Thus by applying there'll be 2^3 = 8 elements contained in the ability set. Definition of cardinality. forall s : fset_expr (A:=A), exists n, (cardinality_fset s n /\ forall s' n', eq_fset s s' -> cardinality_fset s' n' -> n' = n). 1. 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Cardinality and Bijections Definition: Set A has the same cardinality as set B, denoted |A| = |B|, if there is a bijection from A to B – For finite sets, cardinality is the number of elements – There is a bijection from n-element set A to {1, 2, 3, …, n} Following Ernie Croot's slides The cardinality of a finite set is the number of elements in the set. if you need any other stuff in math, please use our google custom search here. Any subset of a countable set is countable. If you are less interested in proofs, you may decide to skip them. Total number of elements related to C only. refer to Figure 1.16 in Problem 2 to see this pictorially). I can tell that two sets have the same number of elements by trying to pair the elements up. Click here to navigate to parent product. However, as we mentioned, intervals in $\mathbb{R}$ are uncountable. Introduction to the Cardinality of Sets and a Countability Proof - Duration: 12:14. 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text:
Remove repeated numbers from a list

So, I just managed to finish a problem that was proposed to me, and I find it very difficult since I started learning looping with for and using lists recently, I thought about sharing the problem and my solution with you guys so you could give me some advice on how I should make my code better or different ways to solve the problem, hopefully you guys can help me, thank you for your attention and sorry if I made some mistakes, I am new to programming.

Question

Write a function that receives a list with natural numbers, verify if such list has repeated numbers, if so remove them. The function should return a correspondent list with no repeated elements. The list should be returned in ascending order.

def remove_repeated_numbers(z):
x = 1
for i in z:
y = x
while y <= (len(z) - 1):
if i == z[y]:
del z[y]
y = (y - 1)              #Once z[y] is deleted, should check same spot again
y = (y + 1)
x = (x + 1)
list_finished = sorted(z[:])
return list_finished

def main():
run = True
z = []
while run:
u = int(input("Entry value: "))
if u != 0:
z.append(u)
else:
run = False
print(remove_repeated_numbers(z))

main()

• You may be interested in the SO question What is the cleanest way to do a sort plus uniq on a Python list? Sep 24 at 7:44
• Do you know that your ENTIRE script can be replaced with this ONE-liner: sorted(set(l))? This is extremely basic, set is a built-in data type that is ordered without duplicates with fast membership checking that is exactly what is designed to do this sort of thing, and then you just need to use sorted function to return a sorted (shallow) copy of the list. Type casting is by far the most efficient and readable way to do this. Sep 24 at 12:26
• @XeнεiΞэnвϵς I tried your sorted(set(l)). Didn't work. Got NameError: name 'l' is not defined. Sep 24 at 14:44
• @BolucPapuccuoglu NameError: name 'z' is not defined. Sep 24 at 15:43
• @BolucPapuccuoglu Yeah, but XeнεiΞэnвϵς said to replace ENTIRE script with that (shouting already theirs). Sep 24 at 15:48

It's a good start. Depending on what class this is for, the person (teacher?) who proposed this question to you might care about time complexity of your solution. Your solution is going to have a number of problems there, since:

• del z[y] is linear-time ("O(n)")
• That occurs within a loop over the list, within another loop over the list, making the solution worst-case cubic if I'm not mistaken.

The trivial and common solution is to scrap the whole routine and replace it with calls to two built-ins:

from typing import Iterable, List

def remove_repeated_numbers(numbers: Iterable[int]) -> List[int]:
return sorted(set(numbers))

I will leave it as an exercise to you to read the documentation and determine the complexity of this implementation.

This will construct a set from the numbers, which guarantees uniqueness but not order; and then constructs a sorted list from that. Interpreted literally, the question doesn't call for the existence of a main but I don't know how picky the problem is.

Other things to note:

• Don't use single-letter variables
• Add type hints for greater clarity

Are we sure that the function works? Will it still work when we make changes (perhaps to improve its performance)?

Python comes with some unit-testing facilities to help us ensure that we don't break the function. One that's very easy to use is doctest, which lets us put the tests right by the code itself.

We could start by testing an empty list:

def remove_repeated_numbers(z):
"""
Return a sorted copy of the input list, with duplicates removed.

>>> remove_repeated_numbers([])
[]
"""
⋮
⋮

We write each test as a line beginning with >>> and an expression to evaluate, with the expected result on the following line.

We can run the tests with some easy boilerplate (which we don't need to understand right now):

if __name__ == '__main__':
import doctest
doctest.testmod()

To be sure that our tests work properly, let's add a test that we know will fail:

>>> remove_repeated_numbers([])
None

That gives us a useful diagnostic:

**********************************************************************
File "268315.py", line 7, in __main__.remove_repeated_numbers
Failed example:
remove_repeated_numbers([])
Expected:
None
Got:
[]
**********************************************************************

Let's remove that, and add some more tests:

    >>> remove_repeated_numbers([0])
[0]
>>> remove_repeated_numbers([0, 0])
[0]
>>> remove_repeated_numbers([0, 1])
[0, 1]
>>> remove_repeated_numbers([1, 0])
[0, 1]
>>> remove_repeated_numbers([0, 0, 1, 1, 1])
[0, 1]
>>> remove_repeated_numbers([0, 1, 1, 1, 0])
[0, 1]
"""

That's good, and we now have more confidence in what we're doing.

We can now change the implementation to use a set, as suggested in a different answer, but I'll propose a smaller change, and that is to sort the input first:

for i in sorted(z):

That initially seems like it's making more work (after all, the input could be much longer than the output), but it will reduce the later operations, because now all the duplicates are adjacent, so we don't need to examine the whole list to find them.

Now, instead of the expensive del z[y] (which needs to shuffle all the subsequent elements in the list), we can copy to a new list to return to the user. Note that this changes the behaviour - I believe for the better. The code presented modifies the actual array passed to the function (though the modified array is only deduplicated, not sorted). It's generally less surprising to return a copy so that the original data is unchanged.

Just to demonstrate, try this code with the function as posted:

a = [ 2, 1, 1, 2]
print(remove_repeated_numbers(a))
print(a)

With this code, the function returns [1, 2] and a is changed to [2, 1].

So let's create our result array:

result = []
⋮
return result

Now our loop is very simple: for each element we see, if it's the same as the one we most recently added to result, then ignore it, else add it to result:

for i in sorted(z):
if i != result[-1]:
result.append(i)

That's not quite right, because result starts off empty. We need to always include the first element, so we must change the test to also succeed when result is empty - i.e. not result:

for i in sorted(z):
if not result or i != result[-1]:
result.append(i)

I think that's easier to understand than code that deletes items from the input list.

Full modified code

For reference, here's the full program from this approach:

def remove_repeated_elements(input_list):
"""
Return a sorted copy of the input list, with duplicates removed.

>>> remove_repeated_elements([])
[]
>>> remove_repeated_elements([0])
[0]
>>> remove_repeated_elements([0, 0])
[0]
>>> remove_repeated_elements([0, 1])
[0, 1]
>>> remove_repeated_elements([1, 0])
[0, 1]
>>> remove_repeated_elements([0, 0, 1, 1, 1])
[0, 1]
>>> remove_repeated_elements([0, 1, 1, 1, 0])
[0, 1]
>>> remove_repeated_elements(['review', 'code', 'code', 'review'])
['code', 'review']
"""
result = []
for element in sorted(input_list):
if not result or result[-1] < element:
result.append(element)
return result

if __name__ == '__main__':
import doctest
doctest.testmod()

Note that I changed the name slightly, as this will work with any type that can be ordered (such as strings - see the last test), not just numbers. The answer that recommends using set also works with more than just numbers (but requires a type that can be hashed, as well as ordered).

I also changed the test to use <, so it will work for any (admittedly theoretical, and highly improbable) type that can be sorted but not compared for inequality.

• I really like this answer, and I suspect indeed the eliminating the duplicates in a sorted list is the core of the exercise -- otherwise it could just as well require the elements to be returned in the order they were in the initial list. I do find not result slightly difficult to read, and would favor len(result) == 0 instead as I find it more "immediate" in meaning; is not result considered more idiomatic? Sep 24 at 9:40
• There's some debate on what's preferred - some Python people prefer the length test and others the truth test. My opinion on that isn't very strong. Sep 24 at 10:36
• Since sorted only require comparison via __lt__, you could write result[-1] < element for the rare cases __ne__ is not defined. Also, what's the rationale of using result += [element] rather than result.append(element)? Sep 24 at 14:20
• @301: Both good points - I'll modify accordingly. Sep 24 at 14:27

The first thing to know about Python is that most basic operations on lists have helper built-ins. Your code could use set() as the holding structure which will ensure uniqueness for you without any work.

With that out of the way, I think your assignment is more about showing your work so let's take that view on it:

• Your run variable is not needed. You can use break keyword to get out of the while loop.
• You should try to do the early escape first in conditions to avoid nesting. In your case, you can say "if input is 0, break" then everything below assumes that value is not 0 so you don't need an else.
• Double loop is overkill. if <numer> not in <array>: then add the number to array. It's slow but easy. You can do better with a set or a dictionary but that's excessive work for what you are doing.
• If you're working in Python avoid index-based iteration. You can/should use iterables: for value in entered_data:.
• Finally, avoid short names on variables. They create unmaintainable code and keep mental overhead high.


text:
# How to implement combining 2 ingredients to create a product?

In my game, you can combine a variety of ingredients to make a product. For example, you could combine water and dirt to make mud.

What I am currently doing to achieve this is by listing each combination possibility that a particular ingredient has. The one thing I managed to do to make it a little more efficient is that I managed to remove duplicates (So for example, if I have coded ingredientA + IngredientB = Cheese, there isn't another entry for IngredientB + Ingredient A = Cheese).

The problem I have with my method is that it is very annoying adding new ingredients. If I do so, I have to go back to every previous ingredient I have coded, and add it's interaction with the ingredient(s) I want to add.

The amount of ingredients I have means this is becoming increasingly time consuming.

Is there a better way to do this?

• Maybe it would help you if you just build a better editor for your recipes. Unity editor scripting is a very powerful tool. You can use it to create whole new UIs for creating and maintaining custom assets inside the Unity editor. May 15, 2017 at 15:02
• I was considering this also, I already have a rudimentary custom interface setup already, but the problem still remains that every new ingredient takes longer to implement. I'm not very experience with the editor scripting, so my implementation clearly could be better. Thanks for your help though! May 15, 2017 at 15:27

Each product that exists in your world could be listed in an array or table in a database.

Each product can then have a collection of Recipes.

Instead of each Recipe actually containing a Product, the can simply reference the Products necessary to produce another Product.

For example, let's say Cheese is made from Milk and Chemicals. :) All these things will exist on the same level wherever you decide to store them. Cheese will have a Recipe. This Recipe will reference Milk and Chemicals. Assuming Cheese has another Recipe (say, Milk and Fart), this other Recipe will also reference the necessary products.

I believe this way, there's no need to worry about backward compatibility.

• Thanks Alot! I'm fairly sure you've solved my problem. One small thing I'm struggling to wrap my head around in what you outlined above, is how I should code the recognition of the ingredients. Currently, I take the itemID of ingredient A, and then search an ingredientA array with ingredientB's itemID. This yields me the itemID of the product. It's not immediately obvious to me how I would search for the correct recipe in your example. May 15, 2017 at 15:07
• I stopped being lazy and wrote out some pseudo code and figured it out (it was obvious). Thanks alot for your help. On the off chance someone was similarly confused: I just take both the ID's in an if statement. if (ID1 == 5 && ID2 == 23) { spawn product X } May 15, 2017 at 15:23

Following what Didia said, what you want sounds like a Composite pattern (https://en.wikipedia.org/wiki/Composite_pattern).

A resulting product is the same thing as it's parts. In your case, a cheese would be a product just as much as milk and rennet. Each product could have an array of "product required" to create it, but in the end they're all the same conceptually. So when you add a new product, you just need to specify it's required products to create it:

                          milk
cheese <
/          rennet
mac n' cheese <
\         flour
pasta <
egg

In this case, you'd have

List<Product> list_products = new List<Product>();

To add mac n' cheese, simply push a new value in the list

list_products.add(new Product("mac n' cheese",new Tuple<string,string>("cheese","pasta")));

I used Tuple wince you said that 2 product create another, but Lists or even arrays could do the job here.

Of course the code is probably not working (or working awfully with string comparison, problems with localization, etc.), so see it as pseudocode, but you should get the gist of it. The cool thing is that you could (and should IMO) put your products in a nice JSON or XML format that is easily parsable and scalable with this kind of data. Also, a hashtable of ("ID",Product) would probably be easier to manage.


text:
## Asian Journal of Mathematics

### Dirac Lie Groups

#### Abstract

A classical theorem of Drinfel'd states that the category of simply connected Poisson Lie groups $H$ is isomorphic to the category of Manin triples $(\mathfrak{d, g, h})$, where $\mathfrak{h}$ is the Lie algebra of $H$. In this paper, we consider Dirac Lie groups, that is, Lie groups $H$ endowed with a multiplicative Courant algebroid $A$ and a Dirac structure $E \subseteq \mathbb{A}$ for which the multiplication is a Dirac morphism. It turns out that the simply connected Dirac Lie groups are classified by so-called Dirac Manin triples. We give an explicit construction of the Dirac Lie group structure defined by a Dirac Manin triple, and develop its basic properties.

#### Article information

Source
Asian J. Math. Volume 18, Number 5 (2014), 779-816.

Dates
First available in Project Euclid: 2 December 2014


text:
# Protein preparation with CAMPARI¶

With the latest implemented features it is now straightforward to use CAMPARI to directly generate a receptor mol2 file for SEED. As CAMPARI’s pdb parser is fairly robust, in the vast majority of the cases it will be able to read a raw pdb without the need of any preprocessing.

The input files required by CAMPARI to describe the receptor are a sequence file listing all the residues in the system and a pdb or another structural file to read the atom coordinates (the atoms for which no coordinates are provided will be rebuilt by CAMPARI). Additionally CAMPARI only requires a key file with a list of keywords specifying the calculation to run. We provide some key file templates to perform a standard protein preparation protocol for docking. For details please refer to CAMPARI v4 documentation.

The sequence file can be extracted from the SEQRES of the pdb using the script convert_SEQRES_toseq.sh provided in the CAMPARI tool directoty.

Before starting the preparation with CAMPARI be sure to be running version 4. Once the key-file is specified, CAMPARI can be run as easy as:

campari -k keyfile.key >& log

By default CAMPARI will write as structural output both a pdb and a mol2 file with the ALT_TYPE_SET specification as required by SEED.

We recommend using CHARMM parameters for docking with SEED. Using AMBER/GAFF force field is still experimental, but can be done (see Gaff preparation (WORK IN PROGRESS)). CHARMM parameters can be specified in the key-file with the following keyword:

PARAMETERS  <CAMPARI_PATH>/params/abs4.2_charmm36.prm

The output mol2 files will contain CHARMM partial charges (in the charge column) and the CAMPARI biotypes (in the ALT_TYPE_SET line), which needs to be mapped to the corresponding CHARMM atom types by running:

bash convert_CAMP_BIO_to_CGENFF.sh CAMP_BIO_to_CGENFF.map MOL2FILE > OUTPUTFILE

A protein receptor can be prepared with CAMPARI using the two keyfiles tmd_build.template.key and cons_mini_abs_internal.template.key (make sure to correct filenames and paths).

tmd_build.template.key rebuilds the system starting from the provided pdb. Missing residues are added and missing sidechains are build in random conformations. After that a specific relaxation protocol (refer to keyword TMD_RELAX in CAMPARI documentation) is carried out to remove any major steric clashes.

The mol2 output of the run (suffix _RELAXED.mol2 or _END.mol2) is almost ready to be used in SEED (only the remapping of CHARMM atom types is needed). Note that the RELAXED files contains the system coordinates after the relaxation, whereas the END files are the final snapshots of the simulation, in case for example additional molecular dynamics or Monte Carlo is run after relaxation.

We recommend to additionally run a short minimization on the relaxed output (pdb with suffix _RELAXED.pdb). Minimization can be performed in internal coordinates with cons_mini_abs_internal.template.key. We recommend the use of a freeze file (keyword FRZFILE, using mode A) to allow only side-chain terminal dihedrals (those ending with a hydrogen) to move during minimization. This can be accomplished for example by extracting the indices of the non-terminal dihedrals (the ones we would like to freeze) from a CAMPARI log file (keyword TMDREPORT enabled), with a command like the following:

sed -n '/Summary of Rotation/,/End of Summary of Rotation/p' log | sed '/Mol.  #/,/Atom/d' | tail -n +2 | head -n -1 | awk '$3 > 10 {print$1}'

If the system is made up by multiple chains or molecules, make sure to include also constraints for rigid body translations and rotations in the FRZFILE.

For the equivalent target preparation protocol for AMBER/GAFF, refer to Writing a SEED-compatible receptor with AMBER parameters.


text:
# 13.4: ML and Bayesian Tests for State-Dependent Diversification

Now that we can calculate the likelihood for state-dependent diversification models, formulating ML and Bayesian tests follows the same pattern we have encountered before. For ML, some comparisons are nested and so you can use likelihood ratio tests. For example, we can compare the full BiSSe model (Maddison et al. 2007), with parameters q01, q10, λ0, λ1, μ0, μ1 with a restricted model with parameters q01, q10, λall, μall. Since the restricted model is a special case of the full model where λ0 = λ1 = λall and μ0 = μ1 = μall, we can compare the two using a likelihood ratio test, as described earlier in the book. Alternatively, we can compare a series of BiSSe-type models by comparing their AICc scores.

For example, I will apply this approach to the example of self-incompitability. I will use data from Goldberg and Igic (2012), who provide a phylogenetic tree and data for 356 species of Solanaceae. All species were classified as having any form of self incompatibility, even if the state is variable among populations. The data, along with a stochastic character map of state changes, are shown in Figure 13.4. Applying the BiSSe models to these data and assuming that q01 ≠ q10, we obtain the following results:

Model Number of parameters Parameter estimates lnL AIC
Character-independent model 4 λ = 0.65, μ = 0.57 -945.96 1899.9
q01 = 0.16, q10 = 0.09
Speciation rate depends on character 5 μ = 0.57 -945.57 1901.1
λ0 = 0.69, λ1 = 0.63
q01 = 0.17, q10 = 0.08
Extinction rate depends on character 5 λ = 0.65 -943.93 1897.9
μ0 = 0.45, μ1 = 0.67
q01 = 0.22, q10 = 0.06
Full character-dependent model 6 λ0 = 0.49, λ1 = 0.79 -941.94 1895.9
μ0 = 0.20, μ1 = 0.84
q01 = 0.29, q10 = 0.05

From this, we conclude that models where the character influences diversification fit best, with the full model receiving the most support. We can't discount the possibility that the character only influences extinction and not speciation, since that model is within 2 AIC units of the best model.

Alternatively, we can carry out a Bayesian test for state-dependent diversification.

Bayesian test for state-dependent diversification

Like other models in the book, this requires setting up an MCMC algorithm that samples the posterior distributions of our model parameters (FitzJohn 2012). In this case:

1. Sample a set of starting parameter values, q01, q10, λ0, λ1, μ0, μ1, from their prior distributions. For example, one could set prior distribution for all parameters as exponential with a mean and variance of λpriori (note that, as usual, the choice for this parameter should depend on the units of tree branch lengths you are using). We then select starting values for all parameters from the prior.
2. Given the current parameter values, select new proposed parameter values using the proposal density Q(p′|p). For all parameter values, we can use a uniform proposal density with width wp, so that Q(p′|p) U(p − wp/2, p + wp/2). We can either choose all parameter values simultaneously, or one at a time (the latter is typically more effective).
3. Calculate three ratios:
• The prior odds ratio. This is the ratio of the probability of drawing the parameter values p and p′ from the prior. Since we have exponential priors for all parameters, we can calculate this ratio as: $$R_{prior} = \frac{\lambda_{prior_i} e^{-\lambda_{prior_i} p'}}{\lambda_{prior_i} e^{-\lambda_{prior_i} p}}=e^{\lambda_{prior_i} (p-p')} \label{13.11}$$
• The proposal density ratio. This is the ratio of probability of proposals going from p to p′ and the reverse. We have already declared a symmetrical proposal density, so that Q(p′|p)=Q(p|p′) and Rproposal = 1.
• The likelihood ratio. This is the ratio of probabilities of the data given the two different parameter values. We can calculate these probabilities from the approach described in the previous section.
4. Find Raccept as the product of the prior odds, proposal density ratio, and the likelihood ratio. In this case, the proposal density ratio is 1, so (eq. 13.12):

Raccept = Rprior ⋅ Rlikelihood


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# Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2

Students can Download Maths Chapter 2 Measurements Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2

Question 1.
Find the area of rhombus PQRS shown in the following figures.

Solution:
(i) Given the diagonals d1 = 16 cm ; d2 = 8 cm
Area of the rhombus = $$\frac{1}{2}$$(d1 × d2) sq. units
= $$\frac{1}{2}$$ × 16 × 8 cm2 = 64 cm2
Area of the rhombus = 64 cm2

(ii) Given base b = 15 cm ; Height h = 11 cm
Area of the rhombus = (base × height) sq. units
= 15 × 11 cm2 = 165 cm2
Area of the rhombus = 165 cm2

Question 2.
Find the area of a rhombus whose base is 14 cm and height is 9 cm.
Solution:

Given base b = 14 cm ; Height h = 9 cm
Area of the rhombus = b × h sq. units
= 14 × 9 cm2 = 126 cm2

Question 3.
Find the missing value.

Solution:
(i) Given diagonal d1 = 19 cm ; d2 = 16 cm
Area of the rhombus = $$\frac{1}{2}$$(d1 × d2) sq. units = $$\frac{1}{2}$$ × 19 × 16
= 152 cm2

(ii) Given diagonal d1 = 26 m ; Area of the rhombus = 468 sq. m
= $$\frac{1}{2}$$(d1 × d2) = 468 ; (26 × d2) = 468 × 2
d2 = $$\frac{468 \times 2}{26}$$ = d2 = 36 m

(iii) Given diagonal d2 = 12 mm; Area of the rhombus = 180 sq. m
$$\frac{1}{2}$$(d1 × d2) = 180
$$\frac{1}{2}$$(d1 × 12) = 180
d1 × 12 = 180 × 2
d1 = $$\frac{180 \times 2}{12}$$
d1 = 30 mm
Diagonal d1 = 30 mm
Tabulating the results we have

Question 4.
The area of a rhombus is 100 sq. cm and length of one of its diagonals is 8 cm. Find the length of the other diagonal.
Solution:
Given the length of one diagonal d1 = 8 cm ; Area of the rhombus = 100 sq. cm
$$\frac{1}{2}$$(d1 × d2) = 100
$$\frac{1}{2}$$ × 8 × d2 = 100
8 × d2 = 100 × 2
d2 = $$\frac{100 \times 2}{8}$$ = 25 cm
Length of the other diagonal d2 = 25 cm

Question 5.
A sweet is in the shape of rhombus whose diagonals are given as 4 cm and 5 cm. The surface of the sweet should be covered by an aluminum foil. Find the cost of aluminum foil used for 400 such sweets at the rate of ₹ 7 per 100 sq. cm.
Solution:
Diagonals d1 = 4 cm and d2 = 5 cm
Area of one rhombus shaped sweet = $$\frac{1}{2}$$(d1 × d2) sq. units = $$\frac{1}{2}$$ × 4× 5 cm2 = 10 cm2
Aluminum foil used to cover 1 sweet = 10 cm2
∴ Aluminum foil used to cover 400 sweets = 400 × 10 = 4000 cm2
Cost of aluminum foil for 100 cm2 = ₹ 7
∴ Cost of aluminum foil for 4000 cm2 = $$\frac{4000}{100}$$ × 7 = ₹ 280
∴ Cost of aluminum foil used = ₹ 280.

Objective Type Questions

Question 6.
The area of the rhombus with side 4 cm and height 3 cm is
(i) 7 sq. cm
(ii) 24 sq. cm
(iii) 12 sq. cm
(iv) 10 sq. cm
Solution:
(iii) 12 sq. cm
Hint:
Area = Base × Height = 4 × 3 = 12 cm2

Question 7.
The area of the rhombus when both diagonals measuring 8 cm is
(i) 64 sq. cm
(ii) 32 sq. cm
(iii) 30 sq. cm
(iv) 16 sq. cm
Solution:
(ii) 32 sq. cm
Hint:
Area = $$\frac{1}{2}$$(d1 × d2) = $$\frac{1}{2}$$ × 8 × 8 = 32

Question 8.
The area of the rhombus is 128 sq. cm. and the length of one diagonal is 32 cm. The length of the other diagonal is
(i) 12 cm
(ii) 8 cm
(iii) 4 cm
(iv) 20 cm
Solution:
(ii) 8 cm
Hint:
$$\frac{1}{2}$$ × d1 × d2 = 128 ⇒ d2 = $$\frac{128 \times 2}{32}$$ = 8cm

Question 9.
The height of the rhombus whose area 96 sq. m and side 24 m is
(i) 8 m
(ii) 10 m
(iii) 2 m
(iv) 4 m
Solution:
(iv) 4 m
Hint:
Area = Base × height = 96 ⇒ height = $$\frac{96}{24}$$ = 4

Question 10.
The angle between the diagonals of a rhombus is
(i) 120°
(ii) 180°
(iii) 90°
(iv) 100°
Solution:
(iii) 90°
Hint:
Angles of a rhombus bisect at right angles.


text:
# How can Tensorflow be used with the flower dataset to compile and fit the model?

PythonServer Side ProgrammingProgrammingTensorflow

The flower dataset can be compiled and fit to the model using the ‘compile’ and ‘fit’ methods respectively. To the ‘fit’ method, the training dataset as well as the validation dataset are passed as parameters. The number of epochs is also defined in the ‘fit’ method.

We will be using the flowers dataset, which contains images of several thousands of flowers. It contains 5 sub-directories, and there is one sub-directory for every class.

We are using the Google Colaboratory to run the below code. Google Colab or Colaboratory helps run Python code over the browser and requires zero configuration and free access to GPUs (Graphical Processing Units). Colaboratory has been built on top of Jupyter Notebook.

print("The model is being compiled")
model.compile(
optimizer='adam',
loss=tf.losses.SparseCategoricalCrossentropy(from_logits=True),
metrics=['accuracy'])
print("The model is being fit to the data")
model.fit(
train_ds,
validation_data=val_ds,
epochs=3
)

Code credit: https://www.tensorflow.org/tutorials/load_data/images

## Output

The model is being compiled
The model is being fit to the data
Epoch 1/3
92/92 [==============================] - 107s 1s/step - loss: 1.3570 - accuracy: 0.4183 - val_loss: 1.0730 - val_accuracy: 0.5913
Epoch 2/3
92/92 [==============================] - 101s 1s/step - loss: 1.0185 - accuracy: 0.5927 - val_loss: 1.0041 - val_accuracy: 0.6199
Epoch 3/3
92/92 [==============================] - 95s 1s/step - loss: 0.8691 - accuracy: 0.6529 - val_loss: 0.9985 - val_accuracy: 0.6281
<tensorflow.python.keras.callbacks.History at 0x7f2cdcbbba90>

## Explanation

• Once the layers are created and the data is trained, the next step is to compile the model that has been built.
• Once the compiling is done, the model is fit to the input dataset.
• The validation accuracy is a low value in comparison to the training accuracy value.
• This means that our model overfits.
Published on 11-Feb-2021 06:35:17

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text:
Congenital toxoplasmosis (CT) is an infectious disease caused by placental transfer of Toxoplasma gondii protozoa to the fetus and can result in neurological and neurocognitive deficits or death.1,2 Most cases are asymptomatic in immunocompetent individuals, even though some patients may experience self-limiting symptoms such as febrile conditions, lymphadenopathy, hepatosplenomegaly, and eventual skin rash.3,4 The pathogenesis of toxoplasmosis varies by strain and host susceptibility based on an individual’s genetic traits,5 causing significant damage and even death in immunocompetent adults.6 In pregnant women, placental transference of the protozoan is related to severe toxoplasmosis in neonates, which may be related to the genetic characteristics of T. gondii isolates prevalent in animals and humans.7

T. gondii infection is predominantly widespread in tropical climates, especially at low altitudes, as the oocysts survive better in these environments.8,9 It is estimated that approximately 51.4% of pregnant women worldwide are infected with the parasite.10 In Brazil, T. gondii infection ranges from 50–80% in women of reproductive age.11

Maternal serological screening during prenatal care enables the identification of infected and vulnerable pregnant women for follow-up as a means to reduce the risk of fetal infection.12 The severity of congenital toxoplasmosis is inversely related to the gestational period at the time of infection. At the beginning of pregnancy, transmission is low, but the pathology can be severe. At the end of pregnancy, high transmission occurs, but with a lower risk of miscarriage.13 Treatment during pregnancy is related to a lower risk of infection and decreased severity in babies.14,15 Maternal health issues in Brazil have been discussed in 1980, with the implementation of different maternal health policies and programs,16 in addition to greater access to prenatal care in health units. In Italy, Slovenia, France, and Austria, the prenatal program is mandatory, whereas in the United States of America, lack of screening has contributed to greater disease severity, which is only observed at birth.17–20

Epidemiological research in urban and rural environments shows geographical territory together with the perception of the disease spread. Thus, detection of risk factors associated with T. gondii infection can lead to improvements in preventive health strategies.21,22 In this context, the present study aimed to assess the prevalence of T. gondii infection and the factors associated with congenital toxoplasmosis in pregnant women with assistance from the Public Health Service at Ouro Preto, Brazil.

# Methods

A cross-sectional study was conducted with pregnant women between April and December 2020, with assistance from the Public Health Service at Ouro Preto, Brazil.

After signing the assent form (for children’s parents or responsible girls < 18 years old) or consent form (for women ≥ 18 years old), the non-probabilistic sample, by convenience, included 131 pregnant women, regardless of age. After being recruited, specific IgM and IgG antibody levels against T. gondii were determined. The women also responded to a structured questionnaire that was used to obtain information regarding the socioeconomic (age, education, occupation, income, and place of residence), environmental (access to water, sewage network, urban waste collection, contact with domestic animals, animal disposals at home, gardening, use of personal protective equipment (PPE) in gardening, handwashing after using the bathroom), gestational (gestational age, number of pregnancies, neuroocular complications of the firstborn, abortion, eclampsia, body mass index (BMI) pre-gestational and gestational weight gain), clinical (stool parasitological examination, type of protozoan in stool, blood pressure, kidney disease, urine culture, and types of bacteria), and food (consumption of unpasteurized milk, raw vegetables and meat, fruit/vegetable washing process, cross contamination in meal preparation, handwashing before food preparation) conditions. Their weight (kg) and height (cm) were recorded using a stadiometer, which was attached to the Welmy® electronic scale.

The income variable was classified based on the Brazilian minimum wage in REAIS (R$), R$ 1,045.00 (US\$188). The cutoff point for the education variable was 12 years of study for categorizing women who had studied until high school. The pre-gestational body mass index variable was classified as “not overweight” for BMI < 24.9 kg/m2 and as “overweight” for BMI > 25.0 kg/m2. The gestational weight gain variable was classified as shown in Table 1.

Table 1. Recommended weight gain according to pre-pregnancy maternal body mass index (BMI)
Nutritional status before pregnancy BMI (kg/m2) Weight gain during pregnancy (Kg) Weight gain per week in the 2nd and 3rd trimester (Kg)
Underweight <18.5 12.5 – 18 0.5
Eutrophic 18.5-24.9 11 – 16 0.4
Overweight 25.0-29.9 7 – 11.5 0.3
Obesity ≥30.0 5 - 9 0.2

Source: Institute of Medicine (IOM-2009)

Stool samples were collected and sent to the Inflammation Immunobiology Laboratory (LABIIN) for conservation by the addition of mercury, iodine, and formaldehyde (MIF). Subsequently, the samples were sent to the Clinical Analysis Laboratory of the same institution for analysis using the Hoffman, Pons, and Janer method for parasite identification.23

Microsoft Excel 2013 was used for data storage. Data analysis was performed using Stata (Stata Corp V.13, Texas, USA).24 Descriptive analysis included the distribution of frequencies and the absolute numbers of categorical variables. The prevalence of exposure and non-exposure to T. gondii associated with the 95% confidence interval were estimated for the included population. To verify the relationship between T. gondii exposure and the explanatory variables, univariate logistic regression was used, expressed by the odds ratio adjusted for age. The explanatory variables that presented a p-value < 0.20 in the univariate analysis, were selected for multivariate analysis. The forward data entry method was chosen wherein all variables selected in the univariate analysis were inserted simultaneously in the model and were then removed individually based on the probability of a likelihood-ratio statistic based on conditional parameter estimates, starting from the least significant. The final model included all variables that were statistically associated with outcomes, showing a p-value ≤ 0.05. The Hosmer-Lemeshow test was used to analyze or adjust the model.

# Results

Of the 131 pregnant women evaluated, 59.54% reported being in their second pregnancy; however, the prevalence of T. gondii infection was 45.8% (n = 60) and in this group, multiparas seemed to be more likely to be infected with T. gondii (P < 0.009). The average age of this group of women was 27.3 ± 6.6 years. The participants’ sociodemographic, gestational, clinical, and nutritional data, as well as BMI distribution are shown in Table 2.

Table 2. Univariate analysis with the characterization of pregnant women evaluated in 2020, Ouro Preto - MG, Brazil
Variables N % Pregnant women
Uninfected (n = 71) Infected by T. gondii (n=60) P-value
Socioeconomic variables
Age
Adolescents (<18 years) 10 7.63 7 3 0.343***
Adults (>18 years) 121 92.37 64 57
Education
< Complete high school 55 41.98 28 27 0.52*
> Complete high school 76 58.02 43 33
Employment
Unemployed 1 0.76 0 1 0.532**
Employment bond 61 43.56 29 32
Student 15 11.45 10 5 0.252**
Housewife 54 41.22 32 22 0.262**
Income
Up to 1 minimum wage 51 38.93 26 25
1-2 minimum wage 0 0 0 0
2-3 minimum wage 69 52.67 38 31 0.657**
>4 minimum wage 11 8.4 7 4 0.519**
Habitation
City 79 60.31 47 32 0.134*
District 52 39.69 24 28
Environmental variables
No 18 13.74 8 10 0.371*
Yes 113 86.26 63 50
No 14 10.69 7 7 0.739*
Yes 117 89.31 64 53
Contact with pets
Yes 66 50.38 31 35
Pet in the house (but without direct contact) 22 16.79 12 10 0.538**
No 43 32.82 28 15 0.407**
Animal waste disposed of in the house
No 64 48.85 39 25 0.130*
Yes 67 51.15 32 35
Gestational variables
Gestational age
First trimester 37 28.24 23 14
Second trimester 49 37.40 24 25 0.224**
Third trimester 45 34.35 24 21 0.421**
Pregnancy
Primipartity 53 40.46 36 17 0.009*
Multiparity 78 59.54 35 43
Neuro-ocular complications in the firstborn
Yes 8 6.11 2 6 0.142***
No 123 93.89 69 54
Pre-gestational BMI
Not Overweight 55 41.98 34 21 0.136*
Overwight 76 58.02 37 39
Gestational weight gain
Eutrophic 43 35.11 24 22
Low 44 33.59 27 17 0.379**
Excessive 41 31.30 20 21 0.752**
Clinical variables
Parasitological examination of stools (N=78)
Negative 67 85.90 40 27 0.056***
Positive 11 14.10 3 8
Blood pressure
Normotensive 115 89.84 61 54
Hypertensive 5 3.91 1 4 0.148**
Hypotensive 8 6.25 8 0
Dietary variables
Handwashing before food preparation
No 7 5.34 6 1 0.124***
Yes 124 94.66 65 59

Footnote: *Pearson’s Chi-square test **Chi-square partition test ***Fisher’s exact test. Note: IPE - Individual Protection Equipment

Most of the pregnant women reported that they had completed high school education (58.02%), had a job (43.56%), had an income between two and three minimum wages (52.67%), lived in the urban area of the city (60.31%), and had access to basic sanitation such as running water (86.26%), a sewage network (89.31%), and urban waste collection (97.71%) at their residence. Half of the participants reported having contact with domestic animals (50.38%). Among the pregnant women, 25.19% had gardening habits, and of these, 84.85% did not wear gloves. Further, 51.91% of the pregnant women reported that they did not wash their hands after bathroom use.

Regarding clinical conditions, most of the women evaluated were normotensive and did not have kidney disease (95.42%). Parasitological examination of feces showed 14.10% of positivity, in which 72.73% presented the Entamoeba coli protozoa. Among the pregnant women who underwent urine culture (n = 128), Escherichia coli was present in 36.36% of the samples.

Among those pregnant women in their second pregnancy; 96.95% had no history of eclampsia, 19.08% reported having a prior abortion, 6.11% reported neurological complications in their firstborn child, and 58.02% were “overweight” before pregnancy. Further, 37.40% were in the second trimester of the current pregnancy, and the weight inadequacy was 33.59% of the evaluated sample.

Regarding the volunteers’ eating habits (dietary variables), most of the sample consumed pasteurized milk (89.31%), raw vegetables (99.24%), and raw/undercooked meat (67.94%). Overall, 50.38% reported the habit of cleaning/washing fruits and vegetables under running water. When investigating cross-contamination, we observed that 75.57% washed their knives before reusing them. The volunteers reported that they routinely washed their hands before preparing meals (94.66%).

The Table 3 demonstrates the odds ratios for women infected with T. gondii. Among all factors, multiparous were 2.6 times more likely to become infected with T. gondii than primiparous (OR: 2.60; 95% CI: 1.25-5.39). This result was observed even after adjusting the odds ratio for age.

Table 3. Multivariate analysis: Risk or protection factors associated with T. gondii infection in pregnant women evaluated in 2020 at Ouro Preto, MG, Brazil.
Variables T. gondii-infected pregnant women variables
Gross OR (IC 95%) p Adjusted OR (IC 95%) P-value
Social variables
Habitation
City 1 0.704 1 0.685
District 0.78 (0.23-2.70) 0.77 (0.22-2.70)
Environmental variables
Animal waste disposed of in the house
No 1 0.552 1 0.566
Yes 0.72 (0.24-2.12) 0.73 (0.25-2.15)
Gestational variables
Pregnancy
Primiparity 1 0.010 1 0.018
Multiparity 2.60(1.25-5.39) 2.50 (1.17-5.36)
Children complications
No 1 0.664 1 0.664
Yes 1.74 (0.14-21.53) 1.74 (0.14-21.53)
Pre-gestational BMI
Without overweight 1 1 0.397
With overweight 0.62 (0.20-1.92) 0.408 0.60 (0.18-1.95)
Clinical variables
Parasitic examination in stools (N=78)
Negative 1 0.370 1 0.366
Positive 2.45(0.34-17.38) 2.47(0.35-17.6)
Blood pressure
No 1 0.183 1 0.203
Yes 4.51 (0.49-41.67) 4.23 (0.45-39.17)
Dietary variables
Handwashing before food preparation
No 0.74 (0.05-11.25) 0.827 0.75(0.05-11.54) 0.838
Yes 1 1

Footnote: ORa - age-adjusted odds ratio. Logistic Regression

The variables of access to garbage collection services, gardening habits, use of PPE in gardening, hand washing after using the bathroom, abortion, eclampsia, type of protozoan in the stool, kidney disease, urine culture and presence of bacteria, consumption of unpasteurized milk, consumption of raw vegetables, consumption of raw meat, fruit and vegetable washing process, and cross-contamination during meal preparation did not show any statistically significant differences.

# Discussion

This study was carried out in the Ouro Preto microregion, which is one of the most important historic cities in the central region of Minas Gerais state in Brazil. The economic activities revolve around mineral extraction, metallurgical industry, tourism, and agriculture. Despite increasing tourist and financial activities, this city has precarious water treatment with 75.6% of the households having adequate sanitation but lack a water treatment system.

Regarding the factors associated with T. gondii infection, our results showed that multiparas presented as an important risk factor as they showed a 2.6 times greater risk of being infected compared to primiparas. In contrast, a recent study conducted in Turkey reported that primipara women had a significantly higher risk (P < 0.02) of infection with the parasite compared to multipara women.25 However, Onduru & Aboud, (2021)26 reported that in the Temeke district, Dar es Salaam, Tanzania, the probability of a pregnant woman giving birth to a child with a toxoplasmosis was 28.5 times higher in multiparas than in primiparas. After adjusting for the effect of confusion and association in the multivariate model, the robustness of the multipara variable increased to 40.89 times more likely for pathology in the child. There is great cultural diversity among studies from different countries and Brazil, as the culture seems to show overprotection vis-à-vis the firstborn, and through the observation and personal reports of pregnant women, the subsequent pregnancies seem to be more “peaceful,” with less stress and more experience. Such facts may contribute to less food care and greater exposure to risk factors such as contact with infected soil and untreated water. As biological explanations, we present the following points in our results: greater vulnerability of uninfected multiparas to the parasite probably occurs due to continuous changes in the immunological mechanisms inherent to pregnancy, resulting from a partial immune response suppression (due to the need for tolerance to the fetus) and/or due to hormonal imbalances characteristic of the gestational condition. Further, unfavorable living conditions of the population in underdeveloped countries with less financial acquisition for larger families can result in a greater chance of infection in women, especially during multiparity, when they are more vulnerable to parasitic infection.

Further, in the study by Berger et al. (2009),27 a predisposition to T. gondii infection was observed in multigravida French pregnant women from 1995 to 2003, thus consolidating the variable as a risk factor for the signs and symptoms of toxoplasmosis.

Effective screening programs are essential for implementing public health policies. In our study, we showed 45.8% positivity for anti-T. gondii IgG, and other studies carried out in different regions of Brazil showed 45% positivity in the central-west region of Minas Gerais, 50.2% positivity in Mato Grosso do Sul, and 59.8% positivity in Porto Alegre.28,29

In contrast, European countries, have demonstrated a tendency towards reduced seroprevalence in pregnant women.30 The relevant risk factors include lack of basic sanitation, inadequate eating and hygiene habits, contact with contaminated animals and soil, territoriality, climatic characteristics,7,27,29,31,32 and the presence of more virulent strains of T. gondii.33 The greatest concern during parasitic infection during pregnancy is the damage it can cause to the fetus. Thus, atypical strains of T. gondii circulating in Brazil,34 may also be associated with severe cases of ocular toxoplasmosis 35 from congenital infection.

High rates of toxoplasmosis seroprevalence are expected in rural areas such as farms, due to the presence of domestic cats and the common presence of intermediate hosts, such as commensal rodents.36 Despite the possibility of the infectious form spreading in these areas, the present study found no interference from the geographical location in the disease spread. The municipality where our study was conducted had surface and underground water collection points and water treatment. For historical and bureaucratic reasons, there was no treatment of water meters, making it impossible to control water consumption in this region.37 Although this study did not present statistical significance in relation to basic sanitation, consumption of untreated water is considered a risk factor for the transmission of oocysts, mainly in developing countries.

One of the determining factors in the epidemiology of pathologies is the socioeconomic level.38 According to Hotez (2008),39 seroprevalence in the upper, middle, and lower socioeconomic strata in Brazil was 23%, 62%, and 84%, respectively. By combining sanitation and socioeconomic status, the prevalence rate in the adult population can reach 90%.40 In the present study, more than a third of the pregnant women survived on up to a minimum wage; however, there was no significant difference in the serotype of T. gondii.

Consumption of undercooked/roasted meat is common in more than half of the group studied, whereas in the United Kingdom, pregnant women are warned about the consumption of unpasteurized milk and undercooked meat, and risk assessment for T. gondii infection is carried out individually during prenatal care.41 In the study by Chandrasena et al. (2016),42 with pregnant women in Sri Lanka, almost all women had the habit of washing their hands after handling raw meat. Although meat consumption is high, the meat is well cooked and prepared with traditional sauces, reducing the risk of parasite infection. In a study conducted in Western Iran, the prevalence of pregnant women infected with T. gondii was high among those who had consumed undercooked eggs and meat as well as unwashed vegetables.42

In general, the diverse prevalence (20–90%) of T. gondii infection can be understood by distinction between the populations studied, diagnostic methods, and specific risk factors that facilitate disease spread, including nutritional insecurity and/or contact with vector animals for T. gondii infection.43–45 Finally, this study demonstrated that the high prevalence of T. gondii seropositivity in pregnant women may be related to multiparity, which should be considered a risk factor for toxoplasmosis.

# Conclusions

Based on our data, we propose rigorous monitoring of pregnant women during prenatal care, in both primigravida and multigravida women, using screening tests, the provision of guidance to pregnant women on the risk factors for T. gondii infection. To target the relevant risk factors, it is essential to develop educational programs in communities together with preventive measures in the public health service that must be maintained, in addition to reinforcing the application of public health policies.

## Acknowledgements

AT (Process # 305634/2017-8) and VFAN (Process # 306036/2019-3) are grateful to the National Council for Scientific and Technological Development (CNPq) for the research productivity fellowship. PVS, DNMT, BAAM and NSG thank Coordination of Improvement of Higher Education Personnel (CAPES) for their research scholarship. The authors also thank to Minas Gerais Research Funding Foundation (FAPEMIG), UFOP, and UFRN for financial and infrastructure support during this research.

This study was approved by the Research Ethics Committee of the Universidade Federal de Ouro Preto under the protocol CAAE: 23467219.7.0000.5150. All women were informed about the research, as well as read and assigned the consent to their participation.

None.

## Authorship contributions

PVS: conceptualization, resource, writing-original draft preparation, DNMT: data curation, resource, writing-original draft preparation, BAAM: data curation, writing-original draft preparation, ZMS: data curation, resource, VFAN: data curation, resource, NSG: conceptualization, resource, writing-original draft preparation, AT: Supervision, conceptualization, writing-reviewing and editing.

## Competing interests

The authors declare that the research was conducted without conflict of interest.

## Correspondence to:

André Talvani; [email protected]

Departamento de Ciências Biológicas/ICEB, UFOP

&

Nathalia Sernizon Guimarães; [email protected]

Escola de Nutrição, ENUT, UFOP

Campus Morro do Cruzeiro, 35400-000 Ouro Preto, MG, Brazil


text:
### Theory:

Manu's favourite dinner is 'Nan with Mushroom gravy'.

There are three different varieties of Nan and two different varieties of mushroom gravy available in the restaurant.

In how many different combinations can he choose his dinner?

At a time he can get one of the Nan's variety and one of varieties in mushroom gravy.

Let us denote the three Nan varieties as $$A$$, $$B$$ and $$C$$, and the two different gravy varieties as $$M$$ and $$N$$.

Now he can eat one of these: $$A-M$$, $$A-N$$, $$B-M$$, $$B-N$$, $$C-M$$, and $$C-N$$.

Thus he can have six different combinations and we know that we have counted all the possibilities. This form of listing the counts is called as 'systematic listing'.
Let us take another situation where we use the systematic listing:

You have three different colour balloons. In how many different ways you can arrange balloons in a line?

The three colours are: Red, Yellow and Green.

Now we will arrange the balloons in different orders by fixing one colour in the first place.

Let us first fix Red colour in first place.

 First Second Third Red Yellow Green Red Green Yellow

Let us fix Yellow colour in the first place.

 First Second Third Yellow Red Green Yellow Green Red

Let us fix Green colour in the first place.

 First Second Third Green Red Yellow Green Yellow Red


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# Vertex (graph theory)

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A graph with 6 vertices and 7 edges where the vertex number 6 on the far-left is a leaf vertex or a pendant vertex

In graph theory, a vertex (plural vertices) or node is the fundamental unit of which graphs are formed: an undirected graph consists of a set of vertices and a set of edges (unordered pairs of vertices), while a directed graph consists of a set of vertices and a set of arcs (ordered pairs of vertices). In a diagram of a graph, a vertex is usually represented by a circle with a label, and an edge is represented by a line or arrow extending from one vertex to another.

From the point of view of graph theory, vertices are treated as featureless and indivisible objects, although they may have additional structure depending on the application from which the graph arises; for instance, a semantic network is a graph in which the vertices represent concepts or classes of objects.

The two vertices forming an edge are said to be the endpoints of this, and the edge is said to be incident to the vertices. A vertex w is said to be adjacent to another vertex v if the graph contains an edge (v,w). The neighborhood of a vertex v is an induced subgraph of the graph, formed by all vertices adjacent to v.

## Types of vertices

The degree of a vertex in a graph is the number of edges incident to it. An isolated vertex is a vertex with degree zero; that is, a vertex that is not an endpoint of any edge. A leaf vertex (also pendant vertex) is a vertex with degree one. In a directed graph, one can distinguish the outdegree (number of outgoing edges) from the indegree (number of incoming edges); a source vertex is a vertex with indegree zero, while a sink vertex is a vertex with outdegree zero.

A cut vertex is a vertex the removal of which would disconnect the remaining graph; a vertex separator is a collection of vertices the removal of which would disconnect the remaining graph into small pieces. A k-vertex-connected graph is a graph in which removing fewer than k vertices always leaves the remaining graph connected. An independent set is a set of vertices no two of which are adjacent, and a vertex cover is a set of vertices that includes the endpoint of each edge in the graph. The vertex space of a graph is a vector space having a set of basis vectors corresponding with the graph's vertices.

A graph is vertex-transitive if it has symmetries that map any vertex to any other vertex. In the context of graph enumeration and graph isomorphism it is important to distinguish between labeled vertices and unlabeled vertices. A labeled vertex is a vertex that is associated with extra information that enables it to be distinguished from other labeled vertices; two graphs can be considered isomorphic only if the correspondence between their vertices pairs up vertices with equal labels. An unlabeled vertex is one that can be substituted for any other vertex based only on its adjacencies in the graph and not based on any additional information.

Vertices in graphs are analogous to, but not the same as, vertices of polyhedra: the skeleton of a polyhedron forms a graph, the vertices of which are the vertices of the polyhedron, but polyhedron vertices have additional structure (their geometric location) that is not assumed to be present in graph theory. The vertex figure of a vertex in a polyhedron is analogous to the neighborhood of a vertex in a graph.

In a directed graph, the forward star of a vertex ${\displaystyle u}$ is defined as its outgoing edges. In a Graph ${\displaystyle G}$ with the set of vertices ${\displaystyle V}$ and the set of edges ${\displaystyle E}$, the forward star of ${\displaystyle u}$ can be described as

${\displaystyle \{(u,v)\in E\}.\ }$[1]

## References

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• Berge, Claude, Théorie des graphes et ses applications. Collection Universitaire de Mathématiques, II Dunod, Paris 1958, viii+277 pp. (English edition, Wiley 1961; Methuen & Co, New York 1962; Russian, Moscow 1961; Spanish, Mexico 1962; Roumanian, Bucharest 1969; Chinese, Shanghai 1963; Second printing of the 1962 first English edition. Dover, New York 2001)
• {{#invoke:citation/CS1|citation

|CitationClass=book }}

• {{#invoke:citation/CS1|citation

|CitationClass=book }}

• {{#invoke:citation/CS1|citation

|CitationClass=book }}

• {{#invoke:citation/CS1|citation

|CitationClass=book }}


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# What is the net force acting on the mass (enter first) and what is the effective spring constant?

## Question:

The mass m = 11.3 kg shown in the figure is displaced a distance x = 0.149 m to the right from its equilibrium position. k₁ = 117 N/m and k₂ = 205 N/m.

What is the net force acting on the mass (enter first) and what is the effective spring constant?

## Combinations of Springs:

Suppose we have two coil springs, and first, we arrange them in series connection, then we will notice that even if we employ a small force, the system undergoes substantial deformation. This suggests that in a series arrangement, the effective stiffness of the system becomes smaller than the stiffness of the individual springs incorporated in the system. If we arrange the same springs in a parallel connection, the reverse happens, that is, the system becomes stiffer than the individual springs.

Given data:

• {eq}m=\rm 11.3 \ kg {/eq} is the mass attached to the springs.
• {eq}k_1=\rm 117 \ N/m {/eq} is the spring constant of the left spring.
• {eq}k_2=\rm 205 \ N/m {/eq} is the spring constant of the right spring.
• {eq}x=\rm 0.149 \ m {/eq} is the deformation of the springs.

The spring force exerted by the left spring is:

{eq}\begin{align} F_1&=k_1x \\[0.3cm] &=\rm 117 \ \frac{N}{m}\times 0.149 \ m \\[0.3cm] &=\rm 17.43 \ N \end{align} {/eq}

The spring force exerted by the right spring is:

{eq}\begin{align} F_2&=k_2x \\[0.3cm] &=\rm 205 \ \frac{N}{m}\times 0.149 \ m \\[0.3cm] &=\rm 30.55 \ N \end{align} {/eq}

The net force acting on the mass is:

{eq}\begin{align} F_{net}&=F_1+F_2 \\[0.3cm] &=\rm 17.43 \ N+30.55 \ N \\[0.3cm] &\simeq \color{blue}{\boxed { \rm 48.0 \ N}} \end{align} {/eq}

Therefore, the net force acting on the mass is {eq}\color{blue}{\boxed { \rm 48.0 \ N}} {/eq}.

The distortion is the same in both springs, so it is nothing but a parallel arrangement of springs. The equivalent or effective spring constant is given by:

{eq}\begin{align} k_{eq}&=k_1+k_2 \\[0.3cm] &=\rm 117 \ \frac{N}{m}+205 \ \frac{N}{m} \\[0.3cm] &= \color{blue}{\boxed { \rm 322 \ N/m}} \end{align} {/eq}

Therefore, the effective spring constant is {eq}\color{blue}{\boxed { \rm 322 \ N/m}} {/eq}.

Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
201K

After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.