What's up guys, it's ya bois Christian and Cameron comin' back at ya with another Julia lesson! Today we are discussing one-liners, so hold onto your socks (because they might just get blown off!), today is a pretty fun one.
https://projecteuler.net/problem=8
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934 96983520312774506326239578318016984801869478851843 85861560789112949495459501737958331952853208805511 12540698747158523863050715693290963295227443043557 66896648950445244523161731856403098711121722383113 62229893423380308135336276614282806444486645238749 30358907296290491560440772390713810515859307960866 70172427121883998797908792274921901699720888093776 65727333001053367881220235421809751254540594752243 52584907711670556013604839586446706324415722155397 53697817977846174064955149290862569321978468622482 83972241375657056057490261407972968652414535100474 82166370484403199890008895243450658541227588666881 16427171479924442928230863465674813919123162824586 17866458359124566529476545682848912883142607690042 24219022671055626321111109370544217506941658960408 07198403850962455444362981230987879927244284909188 84580156166097919133875499200524063689912560717606 05886116467109405077541002256983155200055935729725 71636269561882670428252483600823257530420752963450Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
Hey, remember how y'all did this for the first class because I assigned it before you knew the cool Julia tricks? Well now you know the tricks, so it try again using the following:
digits(int) returns an array that is the digits of the inputed integercircshift(array, int) shifts the inputed array int times, returns a shifted arrayarray[startindex:endindex] returns an array that all the elements of the original array inside (and including) the indicesprod(array) returns the product of the inputed array (we didn't explicitly teach y'all this, but it shouldn't be too hard to pick up)push!(array, element) puts the element at the end of the input arraymaximum(array) returns the highest number in arrayNote that these are the functions I used, and it took me about 7 lines. If you think of a different way to do it, go for it!
# I pasted in the number for you, you are welcome :)
num = 7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
Below I have a pretty lengthy solution to Euler 1. Let's see what we can shorten (there are about 3 to 4 things).
https://projecteuler.net/problem=1
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
# Euler 1 extended code
total = 0
A = [total += n for n in 1:999 if n % 3 == 0 || n % 5 == 0]
pop!(A)
for n in 1:999
if n % 3 == 0 || n % 5 == 0
total += n
end
end
total
Last week we assigned Euler 35 and Euler 11 as homework, can we see some presentations? Don't worry if you struggled with it (especially Euler 11), we expected some people to not get a solution.
Below I will paste the beginnings of a solution to Euler 11 that I came up with, and as a class we can work on condensing it down.
# Euler 11 code goes here
using LinearAlgebra
A = [08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08;
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00;
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65;
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91;
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80;
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50;
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70;
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21;
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72;
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95;
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92;
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57;
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58;
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40;
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66;
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69;
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36;
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16;
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54;
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48]
products = []
# Check top left to bottom right diags
for n in 0:16, x in -8:8
push!(products, max(prod(circshift(A, n)[1:4]), prod(circshift(A', n)[1:4]), prod(circshift(diag(A, x), n)[1:4]), prod(circshift(diag(A[end:-1:1,], x), n)[1:4])))
end
maximum(max(prod(circshift(A, n)[1:4]), prod(circshift(A', n)[1:4]), prod(circshift(diag(A, x), n)[1:4]), prod(circshift(diag(A[end:-1:1,], x), n)[1:4])) for n in 0:16, x in -8:8)
maximum(products)
Here are some Euler problems, some of which we have done before. Let's see if you guys can write them in one line!
Some things to remember:
https://projecteuler.net/problem=4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Find the largest palindrome made from the product of two 3-digit numbers.
# Euler 2 code goes here
https://projecteuler.net/problem=6
The sum of the squares of the first ten natural numbers is, $$1^2 +2^2 + ... + 10^2 = 385$$ The square of the sum of the first ten natural numbers is, $$(1+2+...+10)^2 = 55^2 = 3025$$ Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025−385=2640$. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
# Euler 6 code goes here
https://projecteuler.net/problem=20
$n!$ means $n$ × ($n$ − 1) × ... × 3 × 2 × 1 For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27. Find the sum of the digits in the number 100!
Hint: use the factorial(int) function
# Euler 20 code goes here
https://projecteuler.net/problem=29
Consider all integer combinations of $a^b$ for $2 \leq a \leq 5$ and $2 \leq b \leq 5$:
$2^2=4$, $2^3=8$, $2^4=16$, $2^5=32$
$3^2=9$, $3^3=27$, $3^4=81$, $3^5=243$
$4^2=16$, $4^3=64$, $4^4=256$, $4^5=1024$
$5^2=25$, $5^3=125$, $5^4=625$, $5^5=3125$
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
$4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125$
How many distinct terms are in the sequence generated by ab for $2 \leq a \leq 100$ and $2 \leq b \leq 100$?
Hint: unique(array) returns an array of only the unique elements in the original array and length(array) returns the length of a given array. Also, use BigInts.
# Euler 29 code goes here
https://projecteuler.net/problem=28
Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25 20 7 8 9 10 19 6 1 2 11 18 5 4 3 12 17 16 15 14 13It can be verified that the sum of the numbers on the diagonals is 101. What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
Hint: what do the numbers 1, 9, and 25 have in common?
# Euler 28 code goes here
Woah! No homework this time. Next time we are going to pivot into packages and other more advanced Julia topics. If you feel so inclined, taking a look at the Primes, Random, and Test packages may prove helpful for next time!
Otherwise, I encourage anyone who is up for the challenge to finish the exercises we didn't get to and/or do any other Euler problems they find intresting. I personally think Euler 13, Euler 30, and Euler 39 are good places to start, but do whichever ones you want!