A/B Testing


Problem statement:

 Write AB testing logic in java, which can route to branches based on the percentage given.

 Example:

 A 25 
B 75
In this case, 25% of traffic has to go to "A" and 75% to "B"

Solution:

 Step 1:

 Find GCD of the percentages 25 and 75. GCD is 25

 Divide them by the GCD which results in 1 and 3 is the converted percentage.

 Step 2:

 Build request lookup map for the sum of converted percentage.
Have a counter map for each branches, for each request the selected lowest percentage branch counter will be incremented.

 Request Count            Counter Map           Converted Percentage       Lowest Percentage 

                                        A           B                    A(1)          B(3)                
        0                             0           0                    0               0                             A

         1                             1           0                    1               0                             B 

         2                             1           1                    1               0.3                          B

         3                             1           2                    1               0.6                          B

         
If we notice in the above table, out of 4 requests A was selected only once and B selected thrice.
This contributes to the respective percentage.

 Test:

 Now this request map along with sum of converted percentage 4(1+3) can be used to AB route.

 Request Count               Modulo%4              Request map lookup            Route

        1                                   1                            {0:A, 1:B, 2:B, 3:B}                      B

        2                                   2                            {0:A, 1:B, 2:B, 3:B}                      B

       3                                   3                            {0:A, 1:B, 2:B, 3:B}                      B

       4                                   0                            {0:A, 1:B, 2:B, 3:B}                      A

       5                                   1                            {0:A, 1:B, 2:B, 3:B}                      B

       6                                   2                            {0:A, 1:B, 2:B, 3:B}                      B

       7                                   3                            {0:A, 1:B, 2:B, 3:B}                      B

       8                                   0                            {0:A, 1:B, 2:B, 3:B}                      A
Once the request map is built, then it's simple modulo and lookup to select the route.

 Please find the code below, the logic can work for more than 2 routes/branches.

Comments

Post a Comment

Popular posts from this blog

How to modify buttonDataSetup for RPM Euro Gamepad to work in Fifa 18 on Windows

Image
        STEP 1      Identify the RPM Euro Gamepad/Joystick name connected to your Windows Connect your Gamepad to PC Press Windows Key and search for "Game Controllers" Select "Set up USB Game Controllers" from the search list. This will display the device name connected. Copy the name from the list. In my case it is below Update buttonDataSetup.ini Goto Documents and open Fifa folder Edit the buttonDataSetup.ini file Search for the gamepad name which obtained in the first step in the file If present modify it or else add the below to the end of the file. Controller_054 is just a unique name, you can add anything AddAlias is what important. Feel free to modify the mappings according to your needs. AddController "Controller_054" AddAlias "Controller (XBOX 360 For Windows)" AddMap XENON_LEFT_ANALOG_STICK_RIGHT VB_AI_LS_RIGHT AddMap XENON_LEFT_ANALOG_STICK_UP VB_AI_LS_UP AddMap XENON_LEFT_ANALOG_STICK_LEFT VB_AI_LS_LEFT AddMap XENON_LEFT_A...
Read more

Longest Palindrome length in sub-array with rearrange

Problem statement : Find the longest even palindrome length of the sub array from the given string of digits. The palindrome can be formed by rearranging the characters as well. Example : Given a string "1234535498" The palindrome formed can be 345543 and the length is 6. My Solution: Create a boolean array which will be true if it can form a palindrome. Find the contiguous boolean subarray with maximum number of true values. If the length is odd rerun the same with the subarray. Comments are welcome to provide a better solution or evaluate this solution. Algorithm: Iterate through the input string character by character. For each character, have a counter and increment it for each loop. Check if the character is present in the character array and get the last index of it. if the last index is not of -1, insert true in the boolean array for the last index and the counter. finally insert the character to the character array for the counter. After the...
Read more

How to compile and run using javac and java command in java11 or later

Image
In this blog we will see how to compile and run java class in command line using jdk11 . From java9 , modules are introduced in java ecosystem and with that there is a change in the command args to compile or run our class code if we intend to use modules. Please note we can also stick to the old classpath way. The sample project is available in github and the link to the same is provided at the end. It comprises of one module and an external lib folder. The project is very simple, which tells the passed text is a number or not. First we create a module and a class and then add a jar which is needed by the module. The directory structure looks like below Compile: Now to compile the Main.java, go to the project location(sampleJava folder) and run the below command. C:\Users\surat\git\repository\examples\sampleJava >  javac --module-source-path src\modules --module-path lib -d target src\modules\testModule\com\Main.java If you look into the structure we now create a modules dire...
Read more