Welcome back with another problem to solve! Today’s problem will focus on calculating big exponentials from scratch. This problem is similar to the problem seen in the , so this one will be pretty short. We’ll fast forward a bit over the explanations and direct our attention straight to the code! Daily Coding Problem number 20 If you have problems following along, I highly recommend you check for a much deeper explanation since the concepts involved are the same. Daily Coding Problem number 20 This problem is provided by the wonderful website , an awesome platform to test your math, logic, and coding skills on more than 800 problems. Go check it out; you will not be disappointed! Project Euler We’ll be using Go for this program: headache time! :D The problem Let’s take a look at the problem: 2¹⁵ = 32768, and the sum of its digits is 3+2+7+6+8=26. What is the sum of the digits of the number 2¹⁰⁰⁰? The problem is really simple: we need to calculate 2¹⁰⁰⁰ and sum its digits to give it a result. Simple, isn’t it? Well… Exponentials To give out some simple basics, an exponential number is simply the repetition of that same number a certain number of times. For example, writing 4³ means calculating 4*4*4. What’s really cool about powers (which is also what causes us this problem) is how fast they grow: not the fastest, but still pretty fast. The function we are interested in is the green one: you can easily see how fast it grows. Even if the power function x³ grows faster initially, the exponentiation of 2 catches up pretty quickly. As in the previous article, try with your calculator: you should not be able to calculate over 2³³², which is 8.749002899x10⁹⁹. And yet, with a little bit of reasoning and a simple program, we can calculate that 2³³³ is 17498005798264095394980017816940970922825355447145699491406164851279623993595007385788105416184430592 And with the same approach, 2⁵⁰⁰ is 3273390607896141870013189696827599152216642046043064789483291368096133796404674554883270092325904157150886684127560071009217256545885393053328527589376 Cool right? Let’s see how it’s done then! The approach If there’s one thing computers are good at, . We’ll exploit this ability to repeat a simple task many, many times to iterate over all the exponents from 1 to the desired one and print out the result. The simple operation we are doing is the to increase the exponent. However, we are doing this by multiplying each digit of the starting number by 2: this way, we can manage increasingly big numbers simply by multiplying their digits. Let’s write a multiplication function. it’s repeating the same job multiple times really fast multiplication of a certain number with the number 2 https://gist.github.com/NicolaM94/46ea02bb912e31dfc8baca955b93f7bd?embedable=true#file-multiplication-go As always, let’s explain a bit what’s happening here. We create the function, which takes in two parameters: , the integer we want to multiply, and , the factor we are multiplying it for. After that, we define the variable , which we’ll explain in a moment. multiply stringedInt factor carryover Now we start looping: we loop over the length of and multiply each digit by . stringedInt factor We have some type of conversion to do, however: first off, we use the variable to save our converted digit from the string (with we convert the value to its UNICODE counterpart and then convert it to ) integer stringedInt[n]-'0' rune int then, with the function, we convert the result back to a string and add it to the front of the variable, which will be our result. fmt.Sprintf out Let’s take a moment at the math done here: the possible digits we are multiplying are 0,1,2,3,4,5,6,7,8,9, but the results somehow repeat in a certain manner. For example, . The 1 will be our carryover, which will be added to the next multiplication, If you’re not versed with modulo operations, are some resources to get an idea about it. After multiplying, we just need to add the carryover that could have been stored from the previous multiplication: . 2*2 is 4, and 2*7 is 14, so a 4 with a 1 in front of it so to keep just the unit digit we calculate integer % 5 * 2 . here integer % 5 * 2 + carryover We must store the carry though; that is done in the next block of code, which evaluates if . This is because, for values from 5 to 9, our mudulo will bring us back to the values from 0 to 4, but we need to deal with the carry. so we simply check if : if it is, we have a 1 to carryover and we store it in its variable . integer > 4 We know that the carry can only be 1 at most, integer > 4 carryover for example, if the last number multiplied was an 8, will now start with a 6, but we have a 1 stored in . We must add it to the front of to have the correct result. The final block, from lines 12 to 14, simply adds the last carry possibly left from the last multiplication: out carryover out It’s really basic math, to be honest, but describing it textually could make it worse than it is. So here’s a drawn example of the process: Wrapping up Now we just need to apply that process 1,000 times to calculate the wall of text corresponding to 2¹⁰⁰⁰! Let’s write the simple code: https://gist.github.com/NicolaM94/e13f77c1664abf3dcfa3cd0367ebcb01?embedable=true#file-wrap-go In our function, we define the variable, initially equal to . Then we loop over for 1000 times to update the result. Now, since the problem wants the sum of the digits of 2¹⁰⁰⁰, we simply create the variable and loop over summing up all the digits. main final "1" sum final just change the value in the call or the upper limit in the loop, and you’ll get whatever result you want. For example, running Also, this same algorithm can be used to calculate every exponential of every number: multiply n ...
for n := 0; n < 500; n++ {
  final = multiply(final, 5)
}
... Will calculate 5⁵⁰⁰. As in where the problem was also given by , I won’t give out the solution: you need to build this by yourself, at least, if you want to know it! besides presenting math riddles and problems, . And I love that they are so committed that publishing results and guides for their problems is actually forbidden (this is one of the first 100 problems, so I understand it is permitted). Given this, . the other article ProjectEuler.net ProjectEuler.net is really a lovely project: their main focus is to provide a tool to start learning, thinking, and reasoning about math I don’t feel it would be fair to just give out the solution to copy-paste and get the achievement Have a nice coding, guys! 😎 Time Complexity As always, let’s take a brief moment to discuss the time complexity of this algorithm. The first part of the code runs a number of times equal to the length of the , which, unfortunately, will increase at each loop. Then, we run the multiplier times to reach the nth exponent. So, technically speaking, we’ll have a complexity of , which we can generalize with some approximation to O(n²). stringedInt n O[len(stringedInt)*n] Conclusion Here it is: that’s my solution! I hope you enjoyed following along and reading about this problem. If so, please leave a clap or two: that would make my day! Also, if you are willing and can contribute to the channel, share a Ko-fi. ! Here’s my page As always, thank you so much for reading. Nicola Also published . here

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