23
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This is a problem from NCPC 2005. Roy has an apartment with only one single electrical outlet, but he has a bunch of power strips. Compute the maximum number of outlets he can have using the power strips he has. The number of outlets per power strip is given as input.

It turns out that if the number of outlets of the strips respectively are

$$p_1, p_2, \dots, p_n$$

then the number of outlets is $$1 - n + \sum_i p_i$$ ,

or

$$1 + p_1-1 + p_2-1 + \dots + p_n-1$$.

The input to the program or function is a non-empty series of positive integers.

Examples

2 3 4
> 7
2 4 6
> 10
1 1 1 1 1 1 1 1
> 1
100 1000 10000
> 11098
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  • 17
    \$\begingroup\$ And I thought you weren't supposed to chain power strips ... \$\endgroup\$ – Joey Feb 4 '16 at 20:00
  • \$\begingroup\$ As far as I can tell my Retina answer is the only answer using unary input. You might want to have a look at the comment discussion there: codegolf.stackexchange.com/questions/71047/electrical-outlet/… ... If you think that the unary solution is too much of a hack that's not in the spirit of the challenge, I'm happy for you to specify that the input should be in decimal (and will then fix my answer accordingly). \$\endgroup\$ – Martin Ender Feb 5 '16 at 13:47
  • 7
    \$\begingroup\$ Because electricity is so expensive, your code should be as short as possible as to avoid using more energy \$\endgroup\$ – cat Feb 6 '16 at 0:53
  • 1
    \$\begingroup\$ @cat Time to dig out the old hamster driven turing machine and mechanical computers. \$\endgroup\$ – Pharap Feb 6 '16 at 8:39
  • 1
    \$\begingroup\$ @immibis sure, but the output would be treated as the information contained in the byte stream not as what happens to by rendered by your terminal. \$\endgroup\$ – Martin Ender Feb 6 '16 at 9:04

39 Answers 39

24
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Jelly, 3 bytes

’S‘

Decrement (all), sum, increment. Try it here.

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29
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Retina, 3 bytes

 1

The trailing linefeed is significant.

Input is a space-separated list of unary numbers.

Try it online!

Explanation

The code simply removes all spaces as well as the 1 after them from the string. Here is why that works:

Addition in unary is simple: just concatenate the numbers which is the same as removing the delimiters. Decrementing by 1 is also simple: just remove a 1 from each number. We want 1 more than the sum of the decremented inputs though, so we simply only remove the 1s we find after spaces, thereby decrementing all but the first input.

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  • 1
    \$\begingroup\$ I'm wondering if input in unary should be allowed. \$\endgroup\$ – John Dvorak Feb 5 '16 at 11:39
  • \$\begingroup\$ @JanDvorak it is by default, unless the challenge explicitly specifies decimal input. (See the link in the answer.) Doesn't matter though, Jelly is winning anyway. \$\endgroup\$ – Martin Ender Feb 5 '16 at 12:15
  • \$\begingroup\$ @MartinBüttner There's sample data both in this question and the original assignment. Don't you think (unless otherwise stated) that it should be a necessary (though not sufficient) criterion for passing, that the code works with the verbatim sample data? \$\endgroup\$ – nitro2k01 Feb 5 '16 at 12:52
  • 1
    \$\begingroup\$ @nitro2k01 No (in that case most answers would probably be invalid). Unless the challenge explicitly specifies one particular input format we normally assume that lists can be taken in any native list format. Same goes for number formats (at least unary and taking integers as byte values are allowed by consensus unless the challenge forbids them). It's pretty much impossible to include sample data in every thinkable native input format in the challenge. \$\endgroup\$ – Martin Ender Feb 5 '16 at 12:56
  • \$\begingroup\$ @MartinBüttner Imo, that's not the problem that the recommendation is addressing. What still speaks against this is that (unless I'm mistaken) this doesn't work because unary is a supported or native number format in Retina but it happens to work when you process the string as string data. It's a hack. It's even a clever hack, but I'm still not convinced that's it's according to the rules. If space-separated unary numbers was a native format in Retina in the same way that a list of bytes is a native format in bf, I would agree the recommendation applies and I would have a different opinion. \$\endgroup\$ – nitro2k01 Feb 5 '16 at 13:33
9
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Hexagony, 18 14 bytes

.?<_(@'")>{+.!

Unfolded:

  . ? <
 _ ( @ '
" ) > { +
 . ! . .
  . . .

Try it online!

I don't think side-length 2 is possible, but there must might be a more efficient side-length 3 solution that this.

This is the usual "decrement all, sum, increment" approach, but I'll have to add diagrams later to show how exactly it works in Hexagony.

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7
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Python, 24 bytes

lambda*n:1-len(n)+sum(n)

Try it online

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  • 1
    \$\begingroup\$ This assumes that the function is first assigned, then applied to the input from other variable. \$\endgroup\$ – juandesant Feb 5 '16 at 10:42
  • 1
    \$\begingroup\$ @juandesant ...which is perfectly fine. It's a function literal, which is a valid form of submission. \$\endgroup\$ – FlipTack Dec 3 '17 at 17:48
7
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Mathematica, 9 bytes

Tr[#-1]+1&
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7
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Haskell, 17 15 bytes

foldl1$(+).pred

Usage example: ( foldl1$(+).pred ) [2,4,6] -> 10.

Old version, different approach, 17 bytes: succ.sum.map pred.

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6
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J, 6 bytes

+/+1-#

Sum plus one minus length. Parenthesize and apply it, like so:

   (+/+1-#) 2 3 4
7
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6
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Labyrinth, 9 bytes

"?;)!@
+(

Try it online!

The usual primer:

  • Labyrinth is 2D and stack-based. Stacks have an infinite number of zeroes on the bottom.
  • When the instruction pointer reaches a junction, it checks the top of the stack to determine where to turn next. Negative is left, zero is forward and positive is right.

Here we start at the top left ", a no-op, heading rightward. Next is ?, which reads an int from STDIN (throwing away chars it can't parse as an integer, e.g. spaces). Now we have two cases:

If the input is positive, we turn right, performing:

(            decrement top of stack
+            add top two stack elements
             [continue loop]

If the input is zero (which occurs at EOF), we go straight ahead, performing:

;            pop zero from EOF
)            increment top of stack
!            output top of stack as number
@            halt program
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5
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Pyth, 5 bytes

hstMQ

increment(sum(map(decrement, input)))

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5
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ES6, 25 bytes

a=>a.map(n=>r+=n-1,r=1)|r
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  • 4
    \$\begingroup\$ I was going to post: "One of the rare cases when reduce wins the game" ... and it's 25 too l=>l.reduce((a,b)=>a+b-1). \$\endgroup\$ – edc65 Feb 4 '16 at 14:21
  • \$\begingroup\$ @edc65 Yeah, the (,b) is costly, but I like that version too. \$\endgroup\$ – Neil Feb 4 '16 at 15:46
5
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MATL, 3 bytes

qsQ

Try it online.

Explanation

qsQ
q      thread decrement over the input array
  s    sum
   Q   increment
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4
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05AB1E, 4 bytes

Code:

E<O>

Explanation:

E     # Evaluates input
 <    # Decrement on list
  O   # Compute the total sum
   >  # Increment on the sum
      # Implicit: output top of the stack

Takes input like an array (e.g. [3, 4, 5]).

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  • \$\begingroup\$ Very elegant for a golfing lang \$\endgroup\$ – Pharap Feb 6 '16 at 8:45
4
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Starry, 26 24 bytes

, + '`      + ** `, +'*.

Expects newline-separated integers. Try it online!

Thanks to @MartinBüttner for -2 bytes.

,           Read line as integer
 + '        Dupe and jump to label 1 if nonzero
`           Set label 0
      +     Push 1
 *          Sub
*           Add
 `          Set label 1
,           Read line as integer
 + '        Dupe and jump to label 0 if nonzero
*           Add
.           Output as integer

The loop is unrolled so that the first number is not decremented, negating the need to increment. Pushing numbers is expensive in Starry...

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  • \$\begingroup\$ I count only 20 bytes. \$\endgroup\$ – Addison Crump Feb 4 '16 at 21:11
  • 1
    \$\begingroup\$ @VoteToClose Did you count the leading spaces? (I'm assuming you're talking about the 26 byte) \$\endgroup\$ – Sp3000 Feb 4 '16 at 21:33
4
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Bash + GNU utilities, 16

If there are N power strips, then there should be N-1 separators in the comma-separated input list. All we need to do is replace the separators with - 1 + and arithmetically evaluate:

sed s/,/-1+/g|bc

Or using the same trick:

Pure Bash (no external utilities), 19

echo $[${1//,/-1+}]
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3
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APL (NARS 2000), 13 10 bytes

{1+(+/⍵)-⍴∊⍵}

Edit: Down to 10 with Lynn's (better) approach.

{1++/1-⍨⍵}

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3
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gs2, 5 bytes

(CP437-encoded.)

W&Φd'

That’s read-nums dec m1 sum inc.

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3
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CJam, 7 bytes

q~:(:+)

Test it here.

Same approach as Lynn's (decrement all, sum, increment). This also works for 8 bytes (and is maybe a bit more interesting):

q~{(+}*

This folds "decrement, add" over the list. By doing that, the decrement is only applied to all elements except the first, such that we don't need to take care of the increment separately.

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3
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C, 60 59 55 bytes

x;main(s){while(~scanf("%i",&x))s+=x-1;printf("%i",s);}
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3
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Perl 6, 14 bytes

{1+[+] --«@_}

usage

my &f = {1+[+] --«@_}

say f([2,3,4]) # 7
say f([2,4,6]) # 10
say f([1,1,1,1,1,1,1,1]) # 1
say f([100,1000,10000]) # 11098
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  • \$\begingroup\$ I fully intended to edit my answer to the very same thing ( see the html comment) \$\endgroup\$ – Brad Gilbert b2gills Feb 5 '16 at 4:36
  • \$\begingroup\$ 11 bytes: {.sum-$_+1} \$\endgroup\$ – nwellnhof Nov 22 '18 at 23:58
3
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Seriously, 7 bytes

,;l@Σ-u

Try it online!

Explanation:

,;l@Σ-u
,        push input
 ;       dupe
  l@     push length (n), swap
    Σ-u  push sum, subtract n, add one
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2
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Perl 6, 20 bytes

put 1+sum --«@*ARGS

( You can use << instead of « )

Usage:

$ perl6 -e 'put 1+sum --«@*ARGS' 100 1000 10000
11098
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  • \$\begingroup\$ « is a Perl operator? \$\endgroup\$ – user253751 Feb 4 '16 at 19:56
  • \$\begingroup\$ @immibis Actually it's part of several Perl 6 operators @arraya »+« @arrayb ++«@array @array».method @array»++ « a 'space separated' list of words » Several of those are what is known as Meta operators, in that they combine with other operators. ( Perl 5 doesn't have those operators currently. ) \$\endgroup\$ – Brad Gilbert b2gills Feb 4 '16 at 21:37
2
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Perl 5 23+2=25 or 19+2=21

Requires -ap flags:

map{$.+=($_-1)}@F;$_=$.

Saved in a file and run as

perl -ap file.pl

EDIT: Another answer, smaller (19+2) but basically copied from dev-null answer:

$.+=$_-1for@F;$_=$.
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2
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F#, 25 bytes

Seq.fold(fun s n->s+n-1)1

This is a function that takes in an array/list/sequence of integers and returns the required result.

How it works:

Seq.fold allows you to apply a function to every element of a sequence while carrying some state around while it does so. The result of the function as applied to the first element will give the state that will be put into the function for the second element, and so forth. For example, to sum up the list [1; 3; 4; 10], you'd write it like this:

Seq.fold (fun sum element -> sum + element) 0 [1; 3; 4; 10]
         (       function to apply        ) ^ (sequence to process)
                                     ( initial state )

Which would be applied like so:

// First, initial state  + first element
0 + 1  = 1
// Then, previous state + next element until the end of the sequence
1 + 3  = 4
4 + 4  = 8
8 + 10 = 18

With the last state being the return value of Seq.fold.

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2
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𝔼𝕊𝕄𝕚𝕟, 5 chars / 7 bytes

ï⒭+‡_

Try it here (Firefox only).

Uses a custom encoding with 10-bit chars (thx @Dennis!). Run encode('ï⒭+‡_') in the JS console to get encoded form, and decode(/*ENCODED TEXT HERE*/) to decode encoded form.

Explanation

Translates to Javascript ES6 as:

i=>i.reduce(($,_)=>$+--_)
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  • \$\begingroup\$ Interesting encoding. \$\endgroup\$ – lirtosiast Feb 4 '16 at 22:56
  • \$\begingroup\$ It works quite nicely, too. \$\endgroup\$ – Mama Fun Roll Feb 5 '16 at 2:19
2
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Mornington Crescent, 1909 1873 1839 bytes

Take Northern Line to Stockwell
Take Victoria Line to Seven Sisters
Take Victoria Line to Victoria
Take Circle Line to Victoria
Take Circle Line to Bank
Take Circle Line to Hammersmith
Take Circle Line to Cannon Street
Take Circle Line to Hammersmith
Take Circle Line to Cannon Street
Take Circle Line to Bank
Take Circle Line to Hammersmith
Take District Line to Upminster
Take District Line to Hammersmith
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Turnham Green
Take District Line to Hammersmith
Take District Line to Turnham Green
Take District Line to Notting Hill Gate
Take Circle Line to Notting Hill Gate
Take Circle Line to Bank
Take Circle Line to Embankment
Take Northern Line to Stockwell
Take Northern Line to Embankment
Take Circle Line to Temple
Take Circle Line to Hammersmith
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Blackfriars
Take Circle Line to Embankment
Take District Line to Parsons Green
Take District Line to Bank
Take Circle Line to Hammersmith
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Upminster
Take District Line to Becontree
Take District Line to Parsons Green
Take District Line to Embankment
Take Circle Line to Blackfriars
Take Circle Line to Bank
Take Northern Line to Angel
Take Northern Line to Bank
Take Circle Line to Bank
Take District Line to Upminster
Take District Line to Bank
Take Circle Line to Bank
Take Northern Line to Mornington Crescent

Try it online!

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  • \$\begingroup\$ "90% of all instructions involve takeing the District Line." It's because District is where all the arithmetic stations are. However, on TIO, it doesn't seem to work for any of the examples. \$\endgroup\$ – NieDzejkob Mar 9 '18 at 14:17
  • \$\begingroup\$ 1873 bytes by using shorter line names where possible \$\endgroup\$ – NieDzejkob Mar 15 '18 at 17:04
  • \$\begingroup\$ TIO's interpreter has a bug and doesn't implement Turnham Green \$\endgroup\$ – pppery Jul 29 '18 at 2:04
  • \$\begingroup\$ Nice catch. I've sent a PR that fixes it upstream. \$\endgroup\$ – NieDzejkob Jul 29 '18 at 8:33
1
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Python 3, 79 bytes

import sys
print(sum(map(lambda x: int(x)-1, sys.stdin.readline().split()))+1)
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  • \$\begingroup\$ Looks like you're counting a newline as two bytes. Maybe replace it with a semi-colon to save a byte. A few spaces can be removed too. \$\endgroup\$ – Daffy Feb 6 '16 at 3:06
1
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Ruby, 30 bytes

$*.inject(1){|s,v|s+=v.to_i-1}

Simple enough - starting from 1, add up the supplied numbers, each -1 (command line args are in $*). Shame inject is such a long word.

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1
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PowerShell, 19 bytes

$args-join'-1+'|iex

Note that 1 + p1-1 + p2-1 + ... + pn-1 is equivalent to p1-1 + p2-1 + ... + pn.

Takes input as separate command-line arguments with $args. We -join those together with a -1+ delimiter to create a string, such as 2-1+3-1+4. The string is then piped to Invoke-Expression (similar to eval), and outputs the result.

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1
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Perl, 21 + 2 = 23 bytes

$a+=$_-1for@F;say++$a

Requires -a and -E:

$ perl -aE'$a+=$_-1for@F;say++$a'<<<'2 3 4'
7
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  • \$\begingroup\$ You can use the -a flag to get @F variable with already split elements, and replace -n with -p so you dont'need say, reducing it to 21+2: $a+=$_-1for@F;$_=++$a \$\endgroup\$ – ChatterOne Feb 4 '16 at 16:02
  • \$\begingroup\$ Using -p instead of say is the same because I need to use $_= anyway. \$\endgroup\$ – andlrc Feb 4 '16 at 17:10
  • \$\begingroup\$ @ChatterOne -a is a good idea! \$\endgroup\$ – andlrc Feb 4 '16 at 17:17
1
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Brainfuck, 15 bytes

Assumption: The , operator returns 0 once all input has been exhausted, and there are no extension cords with 0 plugs. Also, the IO needs to be in byte values instead of ASCII character codes.

+>,[-[-<+>],]<.

Explanation: This uses 2 registers. A "Value" accumulator register, representing the number of devices that can be plugged in, and a "current cord" register that keeps track of the value of the current cord. It starts off by incrementing the value by 1, for the existing outlet. Then, for each extension cord, it subtracts one from the value since a plug is being taken up, then increments the value by the number of plugs.

Most online interpreters don't operate in raw byte input mode. To test it online, use this code:

+>,[->-[>+<-----]>---[-<+>]<[-<->]<[-<+>],]<.
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  • \$\begingroup\$ Can I test the program somewhere? \$\endgroup\$ – Pål GD Feb 4 '16 at 18:12
  • \$\begingroup\$ Thanks, corrected those mistakes. I'm not aware of any online interpreters that operate in byte mode. I can throw together an implementation that subtracts '0' from the inputs that will run on any online interpreter. \$\endgroup\$ – Ethan Feb 4 '16 at 18:21
  • \$\begingroup\$ If you want to test out the code, run it here: copy.sh/brainfuck Don't put spaces in between the numerical values. Unfortunately, since it's operating in ASCII mode, the demo code will only work on single digit values. However, the 15 byte version will work properly on an value <= 255. After you run it, view the memory dump to view the final value. \$\endgroup\$ – Ethan Feb 4 '16 at 18:29
  • \$\begingroup\$ One day BF will have proper standards for expected IO and we'll just be able to say 'using standard 3' instead of e.g. 'input and output are all asciii terminated by null char'. \$\endgroup\$ – Pharap Feb 6 '16 at 8:48

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