Passive event listeners – Solving Violation error in JQuery

After introducing Passive event listeners in the DOM spec, following violation occurs when using JQuery methods to capture event triggers.

[Violation] Added non-passive event listener to a scroll-blocking ‘mousewheel’ event. Consider marking event handler as ‘passive’ to make the page more responsive. See

Recently, I notice this issue in code where JQuery .on method has been used. Following is what needs to be done.

Code fragment with the issue:

$(document).on('change', "input[type='date']", function (e) {

Use { passive: true } to fix the issue:

$(document).on('change', "input[type='date']", { passive: true } , function (e) {

Refer to more info on JQuery issue tracker – Issue 2871.


Religious Discrimination: What coperates think of the God?

We are on the end of the month of March though my focus is not on women rights in this post. Because we almost everywhere hear the gender discrimination and rights of the women in the society, especially in March as the International Women’s day is widely commemorated in March 8th.

Discrimination comes in variety of forms. It can be based on the gender or the ethnicity or the religion or it can be based on any other social factor or sometimes personal factors.

In simple terms, in a society, treating a person or particular group of people differently can be identified as a discrimination, it can be favorable or in a worse way. Further classified as direct and indirect discrimination.

According to U.S. Equal Employment Opportunity Commission[1], religious discrimination is termed as;

involves treating a person (an applicant or employee) unfavorably because of his or her religious beliefs. The law protects not only people who belong to traditional, organized religions, such as Buddhism, Christianity, Hinduism, Islam, and Judaism, but also others who have sincerely held religious, ethical or moral beliefs.

Recently, there were a huge public disagreement on enacting laws and regulations targeting specific set of people in the society, especially towards women. This is a direct violation of the equal rights demanded by the constitution of Democratic Socialist Republic of Sri Lanka, article 12. Following is an extract of the article 12 (1) and (2).

12. (1) All persons are equal before the law and are entitled to the equal protection of the law.
(2) No citizen shall be discriminated against on the grounds of race, religion, language, caste, sex, political opinion, place of birth or any one of such grounds : ………………….

In such a background, we still consider factors such as gender, ethnicity, religion etc when making decisions, unfortunately for public administrative decisions, equally common to both the government and private sector. Considering the religion is highly illogical yet uncivilized in a civilized society, I believe. There are many who are atheists  (do not believe in existence of God) and agnostics (nothing is known of the existence of God). And most importantly religion is something which changes over time as it is a belief. Therefore considering religion for administrative activities cannot be accepted in any means.

Holidays in 2018 in Sri Lanka were gazetted under the gazette no 2024/55 in 24th June 2017. According to the government notification , today (30th March) is declared as a public and bank holiday in 2018. Therefore employees of government offices, banks and other related institutions are given a full day leave. Whereas, the employees in private sector are not given that luxury, as today is not a mercantile holiday. This is perfectly fine as the division is based on the employees position in the society. However, dejectedly, some corporates offering a half day leave to their employees who are catholics or christians to practice the religion purview of the good friday.

Although corporates practice such with a big heart for the betterment of the employee, isn’t this a form of discrimination? Why should we adhere to a factor like religion to arrive to an administrative decision? In such a context, why catholics or christians get a full day leave on poya day? Isn’t this a direct violation of the article 12 of the constitution? Are all treated equally? Few questions out of hundreds.

Therefore if we are heading to a civilized society, such discriminations should be eliminated. Coprates need to eliminate decisions based on employees ethnicity, religion etc… They have come a long way eliminating the gender inequality, some at least for a tag line though. The next step should be completely eliminating discrimination based on other social factors, treating equal.

Finally, as an responsible leader/ entrepreneur/ professional/ civilian/ part of the society, if you are leading an establishment which takes decisions based on the employee’s ethnicity and/or religions, step against the discrimination.

Disclaimer: The author does not want to hurt any persons’ religious views. This is mainly focusing the social impact of discrimination and expressing author’s personal views on the matter.



Happy 3.1415926 (π) day : 3-14

Happy 3.1415926 Day !

In every year, π day is celebrated on March 14th (3.14).

π is a super star in mathematics. It is mysterious and people keep exploring it. It is proven that this constant has no end hence it is irrational, that means it cannot be represented as a fraction. π is bigger, as big as the whole universe. In deed π is not the only irrational number we got. But π attracted many and we keep loving it.

Even though there is no known benefit of calculating the value of π people insist of doing it. For many engineering applications, having value of π for 3 or 4 decimal places would be sufficient. Other sophisticated scientific advanced might need much more accuracy but I am pretty sure, it is enough to know first 100 digits of π. But today, humans have calculated 22,459,157,718,361 digits[1] of π.

There are quite many activities are organized for the π day worldwide other than enjoying delicious pies. One of the interested things is ,contests on memorizing decimal places of π. The current world record is 70030 digits[2].

To celebrate the π day I have calculated π for 7 decimal places using Gregory and Leibniz series. It is one of the most primitive ways of calculating π though.

Happy 3.1415926 Day !



PS: Following script gives 7 decimal digits of π.

ans = 0
for k in range (1,100000000):
      temp = ((-1)**(k+1)/(2*k-1))
      ans = ans + (temp*4)
print (ans)


Social Media restriction in Sri Lanka

Currently in Sri Lanka, there is a restriction against leading social media applications such as Facebook, WhatsApp, Viber, Instagram etc. That does not mean the use of such services are illegal with in the country. There is no law to support it. This is not a ban, but just a restriction temporary imposed by the ISPs based on a request made by the government to control the tens situation arose within the country in last week.

Even though the direct access is restricted, it is possible to access these services through Proxy Servers and VPN (Virtual Private Network) applications. There are some false information, being circulated that the use of either social media services or VPN applications are illegal. There is no ground for such misleading information.

No action can be taken against anyone as long as the services are used in the right manner. Use indirect methods to access these services and use them but do not get involved in spreading racism.

At this moment, it is reported that access of some VPN services are also restricted. But there are plenty of VPN services out there which can be used for the purpose.

Do raise your voice against this restriction !!

Do use social media services responsibly !!

Euler’s Totient Function – Counting Primes

Euler came up with this function more than 200 years back. At that time this function drew less attention, indeed it had no much value as there is no direct application other than in number theory. However, today this function plays a major role in cyber security applications.

Euler’s Totient Function, also known as Phi Function, Prime Counting Function or \displaystyle \varphi(n) is used to count number of co-primes prior to a given integer. In other words, the function outputs number of co-primes which are less than the given number.

Co-primes are integers which does not share any common factors. For an example even though 8 is not a prime, it is a co-prime of 15. Because 8 = 2*2*2 and 15 = 2*5, therefore it can be seen that 8 or 15 does not share a common factor.

If a given integer is a prime, obviously all the integers less than the given number are co-primes to the given prime. Therefore, it is easy to calculate \displaystyle \varphi(n) if n is a prime. When n is a prime; in \varphi(n) = \displaystyle (n - 1)

When n is not a prime, \displaystyle \varphi(n) can be calculated as follow:

First, the number should be broken into prime factors, then after it is a matter of applying the following equation:

n = \displaystyle p_1 * p_2 * ....... p_m (prime factors of n)

\varphi(n) = \displaystyle n \prod_{i=1}^{m} ( 1 - \frac{1}{p_i} )

Multiplicative Property

This is another interesting property of \displaystyle \varphi(n) which can be used to compute \displaystyle \varphi(n).

When, p and q are prime numbers:

\varphi(n) = \displaystyle \varphi(p) * \varphi(q)

as a fact, if p and q are prime numbers then, \varphi(p) = \displaystyle (p - 1) * (q - 1)


Calculating \displaystyle \varphi(24) :

24 = \displaystyle 2^3 * 3

Using the above equation:

\varphi(n) = \displaystyle n \prod_{i=1}^{m} ( 1 - \frac{1}{p_i} )

n = 24, m = 2 and primes are 2, 3:

\varphi(24) = \displaystyle 24 * ( 1 - \frac{1}{2} ) * ( 1 - \frac{1}{3} )

\varphi(24) = \displaystyle 24 * ( \frac{1}{2} ) * ( \frac{2}{3} )

\varphi(24) = \displaystyle 24 * ( \frac{1}{3} )

\varphi(24) = \displaystyle 8

Will check if the answer is correct, using the besic definition:

Following are all the integers less than 24, starting from 1

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23

Remove co-prime of 24:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23

Remaining number count is 8, therefore both answers match.

Python Implementation

It is possible to use the above equation to calculate \varphi(n) \displaystyle in Python. However, it is easier to use the definition of prime counting rather than using the equation.

The algorithm would be as follow:

  1. \varphi(n) \displaystyle needs to be calculated.
  2. Start from k = 2, counter is 0.
  3. Check if the selected integer (k) is a co-prime to n.
  4. If it is a co-prime increase the counter.

To check if k is a co-prime to n, GCD operation can be used.

def calculate_phi(n):
    phi_n = 1
    count = 0
    i = 2
    while (i<n):
        r = gcd_euc(n, i)
        i = i + 1
            phi_n = phi_n + 1
    return phi_n


Above mentioned methods can be used to calculate \displaystyle \varphi(n) of a given integer.  Results of the phi function is often applied in cryptography.

Square and Multiply algorithm

Square and Multiply algorithm is an interesting algorithm which is also known as binary exponentiation algorithm as well. The algorithm is very useful against calculation of large integer powers of a number.

For an example, like calculating; \displaystyle 3^4 (which is multiplying 3, four times), this is pretty straight forward:

3 * 3 = \displaystyle 3^2
3^2 * 3  = \displaystyle 3^3
3^3 * 3  = \displaystyle 3^4

In order to get the answer 3 rounds of calculations are required.

In that same manner, to calculate \displaystyle 5^{13} , 12 rounds of multiplication rounds are required. Nevertheless it is possible to use the following method and reduce number of required rounds.

5^1 * 5^0 =\displaystyle 5^1
5^1 * 5^1 = \displaystyle 5^2
5^2 * 5^2  = \displaystyle 5^4
5^4 * 5^4  = \displaystyle 5^8
5^8 * 5^4  = \displaystyle 5^{12}
5^{12} * 5^1  = \displaystyle 5^{13}

Above is a short cut method to obtain the answer, however things might go weird when the exponent is really large, for example something like \displaystyle 3^{105} or \displaystyle 2^{846} .

Even using the shortcut method seems quite cumbersome but better than the first method.

However, in such situations Square and Multiply method comes handy. Even in the second method, we have applied Square and Multiply operation to get the answer. Using this algorithm make it easier to apply the logic behind, especially when it needs to be implemented as a computer program.

Taking back the earlier example, check the operations carried out:

5^1 * 5^0 =\displaystyle 5^1
(Square) 5^1 * 5^1 = \displaystyle 5^2
(Square) 5^2 * 5^2  = \displaystyle 5^4
(Square) 5^4 * 5^4  = \displaystyle 5^8
(Multiply) 5^8 * 5^4  = \displaystyle 5^{12}
(Multiply) 5^{12} * 5^1  = \displaystyle 5^{13}

what if the above is arranged as follow:

5^1 * 5^0 =\displaystyle 5^1
(Square) 5^1 * 5^1 = \displaystyle 5^2
(Multiply) 5^2 * 5^1  = \displaystyle 5^3
(Square) 5^3 * 5^3  = \displaystyle 5^6
(Square) 5^6 * 5^6  = \displaystyle 5^{12}
(Multiply) 5^{12} * 5^1  = \displaystyle 5^{13}

Even though that there is no much improvement in computational rounds, the second approach aligns with the Square and Multiple method. So as you might already have understood, the Square and Multiple method determines the combination of operations that needs to be carried out to reach to the final answer.


Following steps needs to be carried out :

  1. Get the binary representation of the exponent.
  2. Bits are read from left to right (MSB first) and it should start with a 1.
  3. Starting value\displaystyle n^0
  4. Start scanning bits from the left.
  5. As mentioned above the first bit must be 1.
  6. If scanned bit is 1 then, square the value and then multiply by \displaystyle n
  7. If scanneed bit is 0 then, square the value.
  8. Repeat this for all the bits.

As an example, \displaystyle 5^{13} can be taken. Binary representation of 13 is: 13_{ten}  = \displaystyle 1101_{two}

Since the first bit is 1, initially the value is squared and multiplied. Then next bit is taken, it is 1, therefore the value is square and multiplied again.  Then the third bit is 0, therefore the value is only squared. Final bit is 1, the value needs to be squared and multiplied. Pretty easy right !

Time to check this method for a larger exponent. Lets take following : \displaystyle 3^{105}

\displaystyle 105_{ten} = 1101001_{two}

(Square  ) 3^0 * 3^0 =\displaystyle 3^0
(Multiply) 3^0 * 3^1 =\displaystyle 3^1
(Square ) 3^1 * 3^1 =\displaystyle 3^2
(Multiply) 3^2 * 3^1 =\displaystyle 3^3
(Square ) 3^3 * 3^3 =\displaystyle 3^6
(Square ) 3^6 * 3^6 =\displaystyle 3^{12}
(Multiply) 3^{12} * 3^1 =\displaystyle 3^{13}
(Square ) 3^{13} * 3^{13} =\displaystyle 3^{26}
(Square ) 3^{26} * 3^{26} =\displaystyle 3^{52}
(Square ) 3^{52} * 3^{52} =\displaystyle 3^{104}
(Multiply) 3^{104} * 3^1 =\displaystyle 3^{105}

Python Implementation

Following is the python code:

string bin(int num)

The above function takes an integer as the argument and then return the binary equivalent as a string. It starts with 0b prefix. Therefore it is required to consider [2:] of the string. As the first bit is always 1, the final output of the first step is always equal to x. That is the reason for starting the for loop from 3.

def exp_func(x, y):
    exp = bin(y)
    value = x

    for i in range(3, len(exp)):
        value = value * value
            value = value*x
    return value


Square and Multiply algorithm is a very useful algorithm which can be used to calculate values of integers having really large exponents. The number of calculation rounds is relatively less compared to the brute force method.

Deduce remainder of large numbers [Manual Method]

Often articles are published on Python algorithms in this blog. But today’s article is all about calculating the remainder of two numbers by hand, manually.

When two numbers are divided, it produces two things. One is termed as the Quotient and the other is termed as the Remainder. Based on the application, more focus could be given either to the Quotient or to the Remainder. Today, will focus on the Remainder and find how to deduce the Remainder when a large number is divide by a comparatively small number without doing extensive calculations.

According to the Euclidean division, this can be written as follow:

let x – numerator, y – denominator, k – quotient and r – remainder

x = \displaystyle k*y + r

Using the above equation, will try to find the remainder of 27 / 12.

27 = \displaystyle k*12 + r

It is possible to try applying k = 1 ; k = 2 and then calculate r, at k = 2 we get the answer which is r = 3 . This algorithm is promising, however when the numerator (x) is relatively a large number, more computation is required.

For an example, try to find the remainder of \displaystyle (25^{243}/12)

Assuming this calculation is done by hand, as the numerator is a large number it is not possible to compute the remainder using the above algorithm, of course it is possible but it might take lot of computational steps and time. Therefore another approach is required.

To understand easily, will consider following generalize form; \displaystyle (x^{p}/y)


  1. Break the numerator (x) into two numbers where y needs to be a factor of one of the numbers. Let numbers are a and b, then a = k * y and x = a + b
  2. When there are some combinations available for a and b try to obtain the minimum value for b which satisfies the requirement mentioned in step 1. Obtaining b = 1 is the ideal, if possible.
  3. Once above steps are concluded, the equation reduces to \displaystyle ((a+b)^{p}/y) , further it can be deduce to \displaystyle ((k*y + b)^{p}/y)
  4. Once it is reduced, the remainder can be easily obtained by solving \displaystyle ((b)^{p}/y)
  5. In a scenario where b = 1, then the answer is 1 as \displaystyle ((1)^{p}/y) is always 1.

Getting back to the example mentioned above, \displaystyle (25^{243}/12)

  1. 25 can be broken into 1 and 24, a = 24 and b = 1, 12 become a factor of a therefore the combination satisfies the requirement.
  2. Then solve \displaystyle (1^{243}/12) , which is equal to 1.

Results can be verified with the following python code as well:

<pre>print('Remainder - ',(25**243)%12)</pre>

In such scenarios, where b is not possible to take as 1, it is required to solve the equation numerically. But even though the numerator is reduced, still it requires more computational effort.

Algorithm – part 2

When b is not equal to 1, following method can be used. It follows patterns recognition and then deduce the answer.

To understand the concept behind, will take a slightly different version of the above example, assume that it is required to calculate the remainder of \displaystyle (26^{243}/12)

At this time, as the numerator is 26, it is not possible to break into two numbers in which b = 1. When a = 12, b = 2. Therefore following algorithm needs to be used to deduce the remainder from the reduced equation.

The reduced equation for \displaystyle (29^{243}/12) is \displaystyle (2^{243}/12)

Then take the reduced equation as \displaystyle (2^{p}/12) where p = 243

The next part is identify the pattern by applying p = 1, p = 2, p = 3 ……… This needs to be done until a pattern can be recognized.

at p = 1 \displaystyle (2/12) remainder is 2
at p = 2 \displaystyle (4/12) remainder is 4
at p = 3 \displaystyle (8/12) remainder is 8
at p = 4 \displaystyle (16/12) remainder is 4
at p = 5 \displaystyle (32/12) remainder is 8
at p = 6 \displaystyle (64/12) remainder is 4

A pattern, now can be recognized. It seems when p is odd remainder is 8 and when p is even the remainder is 4 for all p when p>1. As we are looking for the remainder when p = 243 which is an odd number, as a result the remainder should be 8.

Verify the answer with the following python script.

print('Remainder - ',(26**243)%12)

Recognizing the pattern is quite cumbersome but it is always easier than working out the whole equation.

Finally, below is the mathematical proof for the above algorithm:

Mathematical Proof

Let \displaystyle (x^{p}/y) : where y ≠ 0 and x and y are positive real integers.

In Euclidean division, x = k * y + r ———————(1)

applying (1) , \displaystyle ((k*y+r)^{p}/y)

Term (k*y+r), can be treated as binomial expansion as a = k*y and b = r

(a+b)^{p} = \displaystyle c_{1}*a^{p} + c_{2}*a^{p-1}*b + c_{3}*a^{p-2}*b^{2}..... + c_{p-2}*a^{2}*b^{p-2} + c_{p-1}*a*b^{p-1} + b^{p} ———–(2)

In equation no. 2, it can be noticed that each term has a factor of a except the last term, which is \displaystyle b^{p} , that means each term is divisible with no reminder except the last term. Therefore that is the only term which generates a remainder. Calculating the remainder for the last term gives the remainder of the whole equation, that is why it is possible to reduce the original equation to:

\displaystyle ((b)^{p}/y)


When calculating prime numbers, it is often required to check if a given number has factors. One of the efficient ways is to divide given numbers and check if there is a remainder. For such operations this algorithm can be used specially when the numerator and the denominator are relatively large (in exponent format). Furthermore, if it is possible to break numbers into a and b (step 1 in Algorithm) such a way that b = 1, at a glance the remainder can be obtained. Therefore this is a faster technique that can be mastered to calculate the remainder manually and also can be implemented in computer aided calculations as well.

This article is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported (CC BY-SA 3.0) license.