Last digit large number

Question

For a given list of number \$[x_1, x_2, x_3, ..., x_n]\$ find the last digit of \$x_1 ^{x_2 ^ {x_3 ^ {\dots ^ {x_n}}}}\$ Example:

[3, 4, 2] == 1
[4, 3, 2] == 4
[4, 3, 1] == 4
[5, 3, 2] == 5 

Because \3ドル ^ {(4 ^ 2)} = 3 ^ {16} = 43046721\$.

Because \4ドル ^ {(3 ^ 2)} = 4 ^ {9} = 262144\$.

Because \4ドル ^ {(3 ^ 1)} = 4 ^ {3} = 64\$.

Because \5ドル ^ {(3 ^ 2)} = 5 ^ {9} = 1953125\$.

Rules:

This is code golf, so the answer with the fewest bytes wins.

If your language has limits on integer size (ex. \2ドル^{32}-1\$) n will be small enough that the sum will fit in the integer.

Input can be any reasonable form (stdin, file, command line parameter, integer, string, etc).

Output can be any reasonable form (stdout, file, graphical user element that displays the number, etc).

Saw on code wars.

• Jonathan Frech: One question I have: In your post you only talk about numbers. Do you mean positive integers exclusively? That is I feel how it was interpreted.
• Maya: Is taking the input in reverse reasonable? Can the input be zero?
• Neil Slater: @NieDzejkob: 2**3 % 10 == 2**7 % 10 == 2**11 % 10. Other last digits have different patterns. This knowledge is necessary to complete the puzzle and cope with arrays of numbers that would otherwise be too large to process. As far as I can tell, none of the answers here actually cope with the original problem as posed on Code Wars.
• Maya: Dear OP, we need more test cases.
• rewritten: In fact the solution for [999999,213412499,34532599,4125159,53539,54256439,353259,4314319,5325329,1242149,142219,1243219,14149,1242149,124419,999999999] is straightforward, base ends with 9, so last digit is 9 when first exponent is odd, 1 when exponent is even. I would really like a codegolf that limits inputs to the available integer range, and not the full exponential.

Answer 1

JavaScript (ES7), 22 bytes

Limited to \2ドル^{53}-1\$.

a=>eval(a.join`**`)%10

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• Caleb Jay: What on earth, why does this work?!
• Shaggy: @komali_2: ** is JavaScript's exponentiation operator. The rest is pretty straightforward.
• Arnauld: @komali_2 a.join`**` is equivalent to a.join(['**']) and ['**'] is coerced to '**' by the join method.
• Neil Slater: I think that OP intends that the limit is on the sum of values, in which case this does not solve the problems as posed.
• Mayube: @AJFaraday the %10 at the end. When dividing any number by 10, the remainder (the modulus) will always be the last digit, so n % 10 will return the last digit of n

Answer 2

R, 25 bytes

Reduce(`^`,scan(),,T)%%10

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• ngm: Not sure what people liked so much about this answer.
• JayCe: I personally like to see Reduce being used.

Answer 3

Haskell, 19 bytes

(`mod`10).foldr1(^)

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Answer 4

HP 49G RPL, 36.5 bytes

Run it in APPROX mode (but enter the program in EXACT mode). Takes input on the stack with the first element deepest in the stack, as integers or reals.

WHILE DEPTH 1 - REPEAT ^ END 10 MOD

It straightforwardly exponentiates on the stack as in Sophia's solution until there is one value left, then takes it mod 10 to get the last digit.

The reason I use APPROX for computation is because 0.0^0.0 = 1 (when they are both reals), but 0^0 = ? (when they are both integers). APPROX coerces all integers to reals, so the input is fine with either. However, I use EXACT to enter the program because 10 (integer) is stored digit-by-digit, and is 6.5 bytes, but 10.0 (real) is stored as a full real number, and is 10.5 bytes. I also avoid using RPL's reduce (called STREAM) because it introduces an extra program object, which is 10 bytes of overhead. I already have one and don't want another.

Limited to the precision of an HP 49G real (12 decimal digits)

-10 bytes after empty list -> 1 requirement was removed.

-2 bytes by taking input on stack.

• JungHwan Min: Hey, could you explain how the bytecount is calculated? Just curious how that language uses nibbles.
• Jason: @JungHwanMin The HP 49G uses a 4-bit processor and BCD arithmetic, since it's a calculator. Internally most commands are transformed into 2.5 byte pointers to the routines they represent, to save space. Small numbers (0-9) are also transformed in this way.
• Arnauld: The Saturn processor is actually pretty fun to work with. A long time ago, I wrote this port of BurgerTime (in assembly) for the HP 48G(X). It was later ported to the 49G. Good memories!

Answer 5

dc, (削除) 17 (削除ここまで) 15 bytes

1[^z1<M]dsMxzA%

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Takes input from the stack, outputs to the stack. Very straightforward implementation - exponentiates until only one value is left on the stack and mod for the last digit.

Thanks to brhfl for saving two bytes!

• brhfl: You can golf one byte by changing 10% to A%, and one more byte by not checking stack depth twice - just put a 1 atop the stack before executing since n^1==n: 1[^z1<M]dsMxA%
• Sophia Lechner: Good ideas! I had no idea dc would let me use A as a literal while set to decimal input. Thank you @brhfl!
• Digital Trauma: @SophiaLechner This trick works for all input bases: codegolf.stackexchange.com/a/77728/11259

Answer 6

J, 5 bytes

-3 bytes thanks to cole!

10|^/

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• cole: Does 10|^/ not work?
• Galen Ivanov: @cole Of course it works, thanks!
• Jonah: Finally, a challenge where J beats Jelly!

Answer 7

05AB1E, 4 bytes

.«mθ

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Explanation

.« # fold
 m # power
 θ # take the last digit

• Mr. Xcoder: As far as I am aware, stack-based languages can assume the input is present on the stack instead of STDIN or alternatives, so something like this should work for 4 bytes (alternatively, just place E in the header).
• Mr. Xcoder: Relevant Meta.
• Adnan: I have fixed this issue in the latest commit for future use.
• Emigna: @Mr.Xcoder: Right! I should have remembered that. So rarely need to though with implicit input. Thanks :)
• Erik the Outgolfer: @Mr.Xcoder Uh, I'm not sure if that's what the meta really means. What is a "function" in 05AB1E? I think it should simply be a string, since you can assign it to a variable, and can be eval-ed with .V. .«mθ looks more like a snippet, since, by itself, you can't assign it to a variable for later reuse. Well, Adnan now fixed the issue, but eh.

Answer 8

Pure Bash (builtins only - no external utilities), 21

echo $[${1//,/**}%10]

Input is given on the command line as a comma-separated list.

Bash integers are subject to normal signed integer limits for 64- and 32-bit versions.

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• Sophia Lechner: ^ is bitwise XOR, which is why you're getting 5 out of the test case instead of the correct 1. You'll need to add a byte to switch to **
• Digital Trauma: @SophiaLechner Yes - of course - good catch! Not sure how that ^ crept in - I had ** in previous iterations of my dev cycle.

Answer 9

Wolfram Language (Mathematica), 16 bytes

Power@##~Mod~10&

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Answer 10

Python 2 and Python 3, 30 bytes

lambda N:eval('**'.join(N))%10

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The input N is expected to be an iterable object over string representations of number literals.

Answer 11

Perl 6, 14 bytes

{([**] $_)%10}

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Uses the reduction meta square brackets with the operator **, modulo 10.

Answer 12

Julia 0.6, 30 bytes

first∘digits∘x->foldl(^,x)

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This is an anon function
∘ is the composition operator.
It is multiple bytes

Answer 13

Ruby, (削除) 41 (削除ここまで) 47 bytes

Size increase due to handling of any 0 in the input array, which needs extra consideration. Thanks to rewritten

->a{a.reverse.inject{|t,n|n<2?n:n**(t%4+4)}%10}

This is solved as I believe the original source intended, i.e. for very large exponentiations that will not fit into language native integers - the restriction is that the array will sum into 2**32-1, not that the interim calculations are also guaranteed to fit. In fact that would seem to be the point of the challenge on Code Wars. Although Ruby's native integers can get pretty big, they cannot cope with the example below processed naively with a %10 at the end

E.g.

Input: [999999,213412499,34532597,4125159,53539,54256439,353259,4314319,5325329,1242149,142219,1243219,14149,1242149,124419,999999999]

Output: 9

• rewritten: Impressive. By spending a 4 more bytes, you can also cope with much higher towers: replace n**(t%4+4) with n**((t-1)%4+1) so you get n**1 instead of n**5 etc. Kudos for the observation that at any stage 4 would be a good cycle.
• rewritten: There is an issue if the sequence has 0s
• Neil Slater: @rewritten: Good spot! I'll have to think about that. In theory the sequence should be forced to terminate 2 steps before the first zero.
• rewritten: Indeed, but that will require a lot more code, exactly 6 more bytes: n<2?n: before n**.

Answer 14

05AB1E, 8 bytes

`.gGm}T%

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• Jonathan Allan: I think you mean ...gGm}10% (or something golfier).

Answer 15

APL (Dyalog Unicode), 5 bytes

10|*⌿

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Answer 16

C# (.NET Core), 84 bytes

a=>{int i=a.Length-1,j=a[i];for(;i-->0;)j=(int)System.Math.Pow(a[i],j);return j%10;}

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• -7 bytes thanks to @raznagul

• raznagul: You can save some bytes by removing the brackers around a and by combining the loop condition with the decrement (for(var i=a.Lengt-1;i-->0;)). But using-statement must be included in he byte count.
• digEmAll: @raznagul: sorry, I'm pretty new to code-golf in C#, is it ok now ?
• raznagul: No problem. Yes this is fine now.
• raznagul: You can save 3 more bytes by using a new variable to hold the result and remove most of the index access to the array: Try it online!
• digEmAll: @raznagul: great !

Answer 17

C (gcc), 56

• Saved 4 bytes thanks to @JonathanFrech

Recursive function r() called from macro f - normal stack limits apply.

R;r(int*n){R=pow(*n,n[1]?r(n+1):1);}
#define f(n)r(n)%10

Input given as a zero-terminated int array. This is under the assumption that none of the x_n are zero.

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• Jonathan Frech: ) r( -> )r(.
• Jonathan Frech: Also, if you want to use UB, you could golf r(int*n){return pow to R;r(int*n){R=pow.

Answer 18

Ruby, (削除) 24 (削除ここまで) 20 bytes

->a{eval(a*'**')%10}

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Answer 19

PHP, (削除) 50 (削除ここまで) 44 bytes

-6 bytes from @Shaggy

<?=eval("return ".join('**',$argv).";")%10?>

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• Ethan: btw, you can use join() to shave off 3 bytes because it's an alias of implode(), but I don't agree with how you're passing in input, you're just setting an array before your script is called. I think you should parse $argv or $argn.
• Maya: You're supposed to return the last digit, not the whole thing.
• Shaggy: 44 bytes
• Night2: Well, if this kind of input is valid, then it can be even shorter: 40 bytes

Answer 20

Japt -h, 7 bytes

OvUqp)ì

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Explanation:

OvUqp)ì
Ov ) // Japt eval:
 q // Join
 U // Input with
 p // Power method
 ì // Split into an array of numbers
-h // Return the last number

• Shaggy: Weird, that wouldn't work for me.
• Shaggy: 6 bytes
• Oliver: @Shaggy :P
• Shaggy: Ah, for Jaysis' sake! :\ That's happening more & more often!

Answer 21

Japt -h, (削除) 7 (削除ここまで) 6 bytes

If input can be taken in reverse order then the first character can be removed.

Limited to 2**53-1.

Ôr!p ì

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---

Explanation

Ô :Reverse the array
 r :Reduce by
 !p : Raising the current element to the power of the current total, initially the first element
 ì :Split to an array of digits
 :Implicitly output the last element

• Etheryte: I got the exact same answer without the flag so this looks like the optimal way for now.
• Shaggy: @Nit: until it's confirmed we can take input in reverse :)
• Etheryte: @Oliver Yeah, but you're still using a flag. Personally I think the byte count without flags is the most accurate scoring result.
• LegionMammal978: @Nit Shouldn't a flag add 3 bytes by meta consensus?
• Shaggy: @LegionMammal978, not any more.

Answer 22

Jelly, 6 bytes

U*@/DṪ

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Answer 23

Python 2, (削除) 45 (削除ここまで) 43 bytes

lambda x:reduce(lambda b,a:a**b,x[::-1])%10

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Answer 24

Excel VBA, 60 bytes

An anonymous VBE immediate window function that takes input from range [A1:XFD1]

s=1:For i=-[Count(1:1)]To-1:s=Cells(1,-i)^s:Next:?Right(s,1)

Answer 25

Stax, 7 bytes

üT(Çó╖⌐

Run and debug it

Algorithm like in my Python answer

Answer 26

CJam, 14 bytes

q~_,({~#]}*~A%

Should work for any input, since CJam isn't limited to 64 bit integers

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Answer 27

Python 3, 55 bytes

p=lambda l,i=-1:not l or f'{l[0]**int(p(l[1:],0))}'[i:] 

Older versions (削除)

p=lambda l,i=-1:len(l)and f'{l[0]**int(p(l[1:],0))}'[i:]or 1 (60 bytes)

(削除ここまで)

• mypetlion: Wouldn't that have to be p=lambda...? Python can't handle recursive anonymous lambdas, so if you need your function to be named, it needs to be part of your solution, and the naming counts towards your byte count for code-golf challenges.

Answer 28

Python 3, 47 bytes

f=lambda x,y=1:f(x[:-1],x[-1]**y)if x else y%10

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Answer 29

Brain-Flak, 161 bytes

Includes +1 for -r

([][()]){({}[()]<({}<(({}))>[()]){({}<(({})<({}<>)({<({}[()])><>({})<>}{}<><{}>)>)>[()])}{}{}>)}{}({}((()()()()()){})(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{})

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The example [3, 4, 2] takes longer than 60 seconds, so the TIO link uses [4, 3, 2].

The -r can be removed if input can be taken in reverse order for a byte count of 160.

# Push stack size -1
([][()])
# While there are 2 numbers on the stack
{({}[()]<
 # Duplicate the second number on the stack (we're multiplying this number by itself)
 ({}<(({}))>[()])
 # For 0 .. TOS
 {({}<
 # Copy TOS
 (({})<
 # Multiple Top 2 numbers
 ({}<>)({<({}[()])><>({})<>}{}<><{}>)
 # Paste the old TOS
 >)
 # End for (and clean up a little)
 >[()])}{}{}
# End While (and clean up)
>)}{}
# Mod 10
({}((()()()()()){})(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{})

Answer 30

Pari/GP, 32 bytes

a->fold((x,y)->y^x,Vecrev(a))%10

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