Find the weight of five apples

Question

Problem

John bought 5 apples. You are given the weights of every group of four apples, and must then find the weights of the apples themselves.

For example, if all apples without the first one weigh 798 g, without the second - 794 g, without the third - 813 g, without the fourth - 806 g, and without the fifth - 789 g, the weights are 202, 206, 187, 194, and 211.

Rules

The solution of the problem by enumeration is allowed
Consider the number of points as follows: the number of bytes in your code. The lower the number of points, the better.

Have fun!

Welcome to Code Golf! You'll need to clarify a few parts of this, like whether it will always be five apples and what the allowed ranges of inputs will be. I also don't think fastest code is worth including here, since IIRC there's going to be a constant-time solution to this. — rydwolf, Commented Apr 6, 2022 at 16:13
Also, the solution will likely run in microseconds at most. In order to make this possible to accurately time, it would need to be run millions of times in a row. And relying on the user's CPU leads to unfairness, since people with faster computers will score better. — rydwolf, Commented Apr 6, 2022 at 16:15
I don't think this works as [fastest-code]. You can get the total weight of the apples by dividing the sum of your numbers by 4, then subtract each of your numbers to get the weight of that apple (This is the only solution). 5 additions, 5 subtractions and a single division is not something you really measure in seconds. — ovs, Commented Apr 6, 2022 at 16:23
I recommend removing rule 1 and sticking with the site defaults for IO, which avoids boilerplate code and concentrates the task down to its core. (Note that none of the answers so far implement this rule, probably because people are so used to using the site defaults.) — Jonathan Allan, Commented Apr 6, 2022 at 21:08

Kevin Cruijssen · Accepted Answer · 2022-04-07 09:59:04Z

6

MathGolf, 3 (or 5) bytes

Σ¼,

No idea why MathGolf has a single-byte a//4 builtin, but it's pretty useful here. ;)

Try it online.

With strict input of space-delimited values it would be 5 bytes instead:

ê_Σ¼,

Try it online.

Explanation:

     # Optional for stricter space-delimited input:
ê    # Push the inputs as integer-array
 _   # Duplicate it

Σ    # Sum the values together
 ¼   # Integer-divide it by 4
  ,  # Subtract each value in the list from this sum//4
     # (after which the entire stack is output implicitly)

edited Apr 7, 2022 at 9:59

answered Apr 7, 2022 at 9:33

Kevin Cruijssen

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\$\begingroup\$ Surely if the sum//4 is lower than an individual value, that means that one of the apples actually does have a negative mass (and so no need to take absolute value...)? \$\endgroup\$
– Dominic van Essen
Commented Apr 7, 2022 at 9:57
\$\begingroup\$ @DominicvanEssen I was indeed doubting a bit whether it was necessary, but you're indeed completely right! \$\endgroup\$
– Kevin Cruijssen
Commented Apr 7, 2022 at 9:58
3

\$\begingroup\$ Nice to see a MathGolf answer! To answer why there is a floor division by 4 operator, I had just implemented the ½ operator, and the next character in line was ¼. I knew it would be useful one of these days! \$\endgroup\$
– maxb
Commented Apr 8, 2022 at 13:49

Add a comment |

naffetS · Accepted Answer · 2022-04-08 14:34:18Z

5

+100

Vyxal, 9 7 6 5 bytes

∑4ḭ$-

Try it Online!

Explanation

∑4ḭ$-
          # (implicit input)
∑         # Sum
 4ḭ       # Divide the sum by four
   $-     # Swap and subtract

-2 bytes thanks to a stone arachnid

-1 byte thanks to ovs

edited Apr 8, 2022 at 14:34

answered Apr 6, 2022 at 23:06

naffetS

12.8k2 gold badges20 silver badges53 bronze badges

3

\$\begingroup\$ ∑4ḭ$- for 5 bytes \$\endgroup\$
– ovs
Commented Apr 8, 2022 at 8:45

Add a comment |

Dominic van Essen · Accepted Answer · 2022-04-07 12:36:57Z

4

BQN, 8 6 bytes

Edit: -2 bytes thanks to ovs

-+´÷⟜4

Try it at BQN online REPL

    ÷⟜4     # divide each of the input values by 4
 +´         # and then fold the 'plus' function across them
-           # using the negative of the input as a list of starting values
            # (so effectively we start the fold using each negative input value
            # in parallel as a starting value)

edited Apr 7, 2022 at 12:36

answered Apr 7, 2022 at 10:06

Dominic van Essen

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1

\$\begingroup\$ -+´÷⟜4 works for 6 bytes \$\endgroup\$
– ovs
Commented Apr 7, 2022 at 10:49
\$\begingroup\$ @ovs - Thanks! Can you explain how it works? In particular, how does +´ = fold plus work as a dyadic operator? \$\endgroup\$
– Dominic van Essen
Commented Apr 7, 2022 at 10:58
\$\begingroup\$ -+´÷⟜4 is just an ordinary fork. ´ when used dyadically does "fold with starting value". \$\endgroup\$
– Razetime
Commented Apr 7, 2022 at 11:08
\$\begingroup\$ @Razetime - But here both the left (- = negative of input) and right (÷⟜4 = input divided by 4) arguments are lists. How does that work to use them as a starting value? \$\endgroup\$
– Dominic van Essen
Commented Apr 7, 2022 at 11:11
1

\$\begingroup\$ arrays are fine to use as a starting value. adding an array and a number is okay in BQN. It distinguishes between arrays and atomic values (numbers, chars, so on) , so it doesn't fuss when given that as a starting value. \$\endgroup\$
– Razetime
Commented Apr 7, 2022 at 12:25

| Show 1 more comment

pxeger · Accepted Answer · 2022-04-06 18:52:11Z

3

Python, 32 bytes

lambda a:[sum(a)/4-x for x in a]

Attempt This Online!

answered Apr 6, 2022 at 18:52

pxeger

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naffetS · Accepted Answer · 2022-04-06 20:32:26Z

3

Haskell, 18 bytes

map=<<(-).(/4).sum

Attempt This Online!

edited Apr 6, 2022 at 20:32

answered Apr 6, 2022 at 18:57

naffetS

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Aiden Chow · Accepted Answer · 2022-04-07 00:58:57Z

3

Desmos, 16 bytes

f(l)=l.total/4-l

Try It On Desmos!

Try It On Desmos! - Prettified

answered Apr 7, 2022 at 0:58

Aiden Chow

14.2k1 gold badge17 silver badges55 bronze badges

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Shaggy · Accepted Answer · 2022-04-07 07:10:44Z

3

Japt v2.0a0 `-m`, 5 bytes

aWx÷4

Try it

answered Apr 7, 2022 at 7:10

Shaggy

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jonatjano · Accepted Answer · 2022-04-08 10:29:35Z

λ-2d, 456 squares

the program can be imported in the playground with the following JSON file

{"41,13":"entry","41,18":"entry","41,23":"entry","41,28":"entry","41,32":"entry","42,13":"end_s","42,18":"end_s","42,23":"end_s","42,28":"end_s","42,32":"end_s","43,13":"end_e","43,18":"end_e","43,23":"end_e","43,28":"end_e","43,32":"end_e","39,15":"wire_nw","39,20":"wire_nw","39,25":"wire_nw","39,30":"wire_nw","39,34":"wire_nw","38,15":"wire_ne","38,20":"wire_ne","38,25":"wire_ne","38,30":"wire_ne","38,34":"wire_ne","44,13":"frame_tl","44,18":"frame_tl","44,23":"frame_tl","44,28":"frame_tl","44,32":"frame_tl","45,13":"wire_we","45,18":"wire_we","45,23":"wire_we","45,28":"wire_we","45,32":"wire_we","46,13":"wire_we","46,18":"wire_we","46,23":"wire_we","46,28":"wire_we","46,32":"wire_we","47,13":"wire_we","47,18":"wire_we","47,23":"wire_we","47,28":"wire_we","47,32":"wire_we","47,15":"wire_we","47,20":"wire_we","47,25":"wire_we","47,30":"wire_we","47,34":"wire_we","46,15":"wire_we","46,20":"wire_we","46,25":"wire_we","46,30":"wire_we","46,34":"wire_we","45,15":"wire_we","45,20":"wire_we","45,25":"wire_we","45,30":"wire_we","45,34":"wire_we","48,13":"wire_sw","48,18":"wire_sw","48,23":"wire_sw","48,28":"wire_sw","48,32":"wire_sw","48,15":"wire_nw","48,20":"wire_nw","48,25":"wire_nw","48,30":"wire_nw","48,34":"wire_nw","44,15":"wire_ne","44,20":"wire_ne","44,25":"wire_ne","44,30":"wire_ne","44,34":"wire_ne","44,14":"wire_ns","44,19":"wire_ns","44,24":"wire_ns","44,29":"wire_ns","44,33":"wire_ns","48,14":"wire_ns","48,19":"wire_ns","48,24":"wire_ns","48,29":"wire_ns","48,33":"wire_ns","42,14":"wire_nw","42,19":"wire_nw","42,24":"wire_nw","42,29":"wire_nw","42,33":"wire_nw","41,14":"wire_we","41,19":"wire_we","41,24":"wire_we","41,29":"wire_we","41,33":"wire_we","40,14":"wire_we","40,19":"wire_we","40,24":"wire_we","40,29":"wire_we","40,33":"wire_we","38,14":"wire_sw","38,19":"wire_sw","38,24":"wire_sw","38,29":"wire_sw","38,33":"wire_sw","39,14":"func_call","39,19":"func_call","39,24":"func_call","39,29":"func_call","39,33":"func_call","39,13":"joint_nsw","39,18":"joint_nsw","39,23":"joint_nsw","39,28":"joint_nsw","38,13":"wire_we","38,18":"wire_we","38,23":"wire_we","38,28":"wire_we","37,13":"wire_se","37,18":"wire_se","37,23":"wire_se","37,28":"wire_se","37,14":"wire_nswe","37,19":"wire_nswe","37,24":"wire_nswe","37,29":"wire_nswe","37,17":"wire_nw","37,22":"wire_nw","37,27":"wire_nw","37,32":"wire_nw","36,17":"wire_se","36,22":"wire_se","36,27":"wire_se","36,32":"wire_se","37,15":"wire_ns","37,20":"wire_ns","37,25":"wire_ns","37,30":"wire_ns","36,15":"wire_nw","36,20":"wire_nw","36,25":"wire_nw","36,30":"wire_nw","36,34":"wire_nw","35,15":"wire_ne","35,20":"wire_ne","35,25":"wire_ne","35,30":"wire_ne","35,34":"wire_ne","35,14":"wire_sw","35,19":"wire_sw","35,24":"wire_sw","35,29":"wire_sw","35,33":"wire_sw","36,14":"func_call","36,19":"func_call","36,24":"func_call","36,29":"func_call","36,33":"func_call","36,13":"joint_nsw","36,18":"joint_nsw","36,23":"joint_nsw","36,28":"joint_nsw","35,13":"wire_we","35,18":"wire_we","35,23":"wire_we","35,28":"wire_we","34,13":"wire_se","34,18":"wire_se","34,23":"wire_se","34,28":"wire_se","34,14":"wire_nswe","34,19":"wire_nswe","34,24":"wire_nswe","34,29":"wire_nswe","34,17":"wire_nw","34,22":"wire_nw","34,27":"wire_nw","34,32":"wire_nw","33,17":"wire_se","33,22":"wire_se","33,27":"wire_se","33,32":"wire_se","34,15":"wire_ns","34,20":"wire_ns","34,25":"wire_ns","34,30":"wire_ns","33,15":"wire_nw","33,20":"wire_nw","33,25":"wire_nw","33,30":"wire_nw","33,34":"wire_nw","32,15":"wire_ne","32,20":"wire_ne","32,25":"wire_ne","32,30":"wire_ne","32,34":"wire_ne","32,14":"wire_sw","32,19":"wire_sw","32,24":"wire_sw","32,29":"wire_sw","32,33":"wire_sw","33,14":"func_call","33,19":"func_call","33,24":"func_call","33,29":"func_call","33,33":"func_call","33,13":"joint_nsw","33,18":"joint_nsw","33,23":"joint_nsw","33,28":"joint_nsw","32,13":"wire_we","32,18":"wire_we","32,23":"wire_we","32,28":"wire_we","31,13":"wire_se","31,18":"wire_se","31,23":"wire_se","31,28":"wire_se","31,14":"wire_nswe","31,19":"wire_nswe","31,24":"wire_nswe","31,29":"wire_nswe","31,17":"wire_nw","31,22":"wire_nw","31,27":"wire_nw","31,32":"wire_nw","30,17":"wire_se","30,22":"wire_se","30,27":"wire_se","30,32":"wire_se","31,15":"wire_ns","31,20":"wire_ns","31,25":"wire_ns","31,30":"wire_ns","30,15":"wire_nw","30,20":"wire_nw","30,25":"wire_nw","30,30":"wire_nw","30,34":"wire_nw","29,15":"wire_ne","29,20":"wire_ne","29,25":"wire_ne","29,30":"wire_ne","29,34":"wire_ne","29,14":"wire_sw","29,19":"wire_sw","29,24":"wire_sw","29,29":"wire_sw","29,33":"wire_sw","30,14":"func_call","30,19":"func_call","30,24":"func_call","30,29":"func_call","30,33":"func_call","30,13":"joint_nsw","30,18":"joint_nsw","30,23":"joint_nsw","30,28":"joint_nsw","29,13":"wire_we","29,18":"wire_we","29,23":"wire_we","29,28":"wire_we","28,13":"wire_se","28,18":"wire_se","28,23":"wire_se","28,28":"wire_se","28,14":"wire_nswe","28,19":"wire_nswe","28,24":"wire_nswe","28,29":"wire_nswe","28,17":"wire_nw","28,22":"wire_nw","28,27":"wire_nw","28,32":"wire_nw","27,17":"wire_se","27,22":"wire_se","27,27":"wire_se","27,32":"wire_se","28,15":"wire_ns","28,20":"wire_ns","28,25":"wire_ns","28,30":"wire_ns","27,14":"func_call","27,19":"func_call","27,24":"func_call","27,29":"func_call","27,33":"func_call","27,13":"joint_nsw","27,18":"joint_nsw","27,23":"joint_nsw","27,28":"joint_nsw","26,13":"wire_se","26,18":"wire_se","26,23":"wire_se","26,28":"wire_se","26,14":"wire_ns","26,19":"wire_ns","26,24":"wire_ns","26,29":"wire_ns","26,15":"wire_ns","26,20":"wire_ns","26,25":"wire_ns","26,30":"wire_ns","26,16":"wire_ne","26,21":"wire_ne","26,26":"wire_ne","26,31":"wire_ne","27,16":"wire_we","27,21":"wire_we","27,26":"wire_we","27,31":"wire_we","29,16":"wire_we","29,21":"wire_we","29,26":"wire_we","29,31":"wire_we","30,16":"wire_we","30,21":"wire_we","30,26":"wire_we","30,31":"wire_we","32,16":"wire_we","32,21":"wire_we","32,26":"wire_we","32,31":"wire_we","33,16":"wire_we","33,21":"wire_we","33,26":"wire_we","33,31":"wire_we","35,16":"wire_we","35,21":"wire_we","35,26":"wire_we","35,31":"wire_we","36,16":"wire_we","36,21":"wire_we","36,26":"wire_we","36,31":"wire_we","39,16":"wire_sw","39,21":"wire_sw","39,26":"wire_sw","39,31":"wire_sw","39,17":"wire_ns","39,22":"wire_ns","39,27":"wire_ns","39,32":"wire_ns","38,16":"wire_we","38,21":"wire_we","38,26":"wire_we","38,31":"wire_we","28,16":"wire_nswe","28,21":"wire_nswe","28,26":"wire_nswe","28,31":"wire_nswe","31,16":"wire_nswe","31,21":"wire_nswe","31,26":"wire_nswe","31,31":"wire_nswe","34,16":"wire_nswe","34,21":"wire_nswe","34,26":"wire_nswe","34,31":"wire_nswe","37,16":"wire_nswe","37,21":"wire_nswe","37,26":"wire_nswe","37,31":"wire_nswe","28,33":"wire_we","31,33":"wire_we","34,33":"wire_we","37,33":"wire_we","39,12":"num_7","40,12":"num_8","41,12":"num_9","36,11":"num_8","37,11":"num_0","38,11":"num_6","33,12":"num_8","34,12":"num_1","35,12":"num_3","30,11":"num_7","31,11":"num_9","32,11":"num_4","28,12":"num_9","29,12":"num_8","27,12":"num_7","30,12":"wire_ns","36,12":"wire_ns","27,15":[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],"27,20":[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],"27,25":[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],"27,30":[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],"27,34":[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],"53,12":"func_def","55,12":"func_def","57,12":"func_def","59,12":"func_def","61,12":"func_def","54,12":"end_s","58,12":"end_s","56,12":"end_s","60,12":"end_s","62,12":"end_s","63,12":"end_e","54,11":"label","55,11":[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],"53,11":"wire_se","54,13":"joint_nse","55,16":"op_plus","57,16":"op_plus","59,16":"op_plus","61,16":"op_plus","55,15":"func_call","57,15":"func_call","59,15":"func_call","61,15":"func_call","56,13":"wire_nswe","58,13":"wire_nswe","61,13":"wire_we","60,13":"wire_nswe","62,13":"wire_nswe","59,13":"wire_we","57,13":"wire_we","55,13":"wire_we","55,14":"wire_sw","54,14":"wire_ne","56,15":"wire_nw","58,15":"wire_nw","60,15":"wire_nw","57,14":"wire_sw","59,14":"wire_sw","60,14":"func_call","61,14":"wire_sw","56,14":"func_call","58,14":"func_call","62,14":"func_call","62,15":"wire_nw","63,14":"wire_sw","63,15":"wire_ns","63,16":"func_call","63,17":"op_div","64,16":"wire_nw","64,15":"func_call","64,14":"num_4","65,15":"wire_sw","65,16":"func_call","65,17":"op_minus","66,16":"wire_nw","66,14":"func_call","67,14":"wire_nw","66,13":"wire_sw","67,12":"wire_sw","66,15":"wire_ns","67,13":"wire_ns","63,13":"wire_we","64,13":"wire_we","65,13":"wire_we","64,12":"wire_we","65,12":"wire_we","66,12":"wire_we"}

the function in itself is the right structure, the left one being 5 calls to the function with the 5 arguments in differents orders

js equivalent

a=798
b=794
c=813
d=806
e=789
f=v=>w>=>x=>y=>z=>((v+w+x+y+z)/4)-v

f(a)(b)(c)(d)(e)
f(b)(c)(d)(e)(a)
f(c)(d)(e)(a)(b)
f(d)(e)(a)(b)(c)
f(e)(a)(b)(c)(d)

Language explanation can be found in the playground by loading the cheatsheet example, or in the release blog post

Engineer Toast · Accepted Answer · 2022-05-18 12:04:53Z

3

Excel, 15 bytes

=SUM(A1#)/4-A1#

Input is in cell A1 as an array. For instance, ={798;794;813;806;789}

Excel, 19 bytes

=SUM(A1:A5)/4-A1:A5

Input is in the cells A1:A5. Doesn't rely on array input. Output is wherever the formula is. It's not a very interesting solution.

edited May 18, 2022 at 12:04

answered Apr 21, 2022 at 14:18

Engineer Toast

9,5291 gold badge23 silver badges53 bronze badges

\$\begingroup\$ If you take input as an array, this simplifies to Sum(A1#)/4-A1# \$\endgroup\$
– Taylor Raine
Commented May 18, 2022 at 0:20

Add a comment |

Yousername · Accepted Answer · 2022-04-06 21:41:02Z

2

TI-Basic, 7 bytes

sum(Ans)/4-Ans

Takes input in Ans. Output is stored in Ans and displayed.

answered Apr 6, 2022 at 21:41

Yousername

4,09010 silver badges24 bronze badges

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chunes · Accepted Answer · 2022-04-06 21:44:54Z

2

Factor + `math.unicode`, 23 bytes

[ dup Σ 4 / swap n-v ]

Try it online!

Explanation

       ! { 798 794 803 816 789 }
dup    ! { 798 794 803 816 789 } { 798 794 803 816 789 }
Σ      ! { 798 794 803 816 789 } 4000
4      ! { 798 794 803 816 789 } 4000 4
/      ! { 798 794 803 816 789 } 1000
swap   ! 1000 { 798 794 803 816 789 }
n-v    ! { 202 206 197 184 211 }

answered Apr 6, 2022 at 21:44

chunes

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naffetS · Accepted Answer · 2022-04-07 02:45:09Z

2

LOLCODE, 216 bytes

HOW IZ I f YR a
I HAS A s ITZ 0
IM IN YR l UPPIN YR i TIL BOTH SAEM i 5
s R SUM OF s a'Z SRS i
IM OUTTA YR l
IM IN YR l UPPIN YR i TIL BOTH SAEM i 5
VISIBLE DIFF OF QUOSHUNT OF s 4 a'Z SRS i
IM OUTTA YR l
IF U SAY SO

Try it online!

answered Apr 7, 2022 at 2:45

naffetS

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naffetS · Accepted Answer · 2022-04-14 16:28:44Z

2

Pip, 9 7 bytes

$+a/4-a

Attempt This Online!

-2 bytes thanks to DLosc

edited Apr 14, 2022 at 16:28

answered Apr 13, 2022 at 20:17

naffetS

12.8k2 gold badges20 silver badges53 bronze badges

\$\begingroup\$ Since arithmetic operators apply to lists itemwise, $+a/4-a works for 7 bytes. Attempt This Online! \$\endgroup\$
– DLosc
Commented Apr 14, 2022 at 16:25

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Kevin Cruijssen · Accepted Answer · 2022-04-21 13:35:44Z

2

Halfwit, 5 bytes

kJ>+<k+N+N

Try it online.

Inputs as a list of BigInts.

Explanation:

kJ          # Sum the (implicit) input-list
  >+<       # Push compressed BigInt 4n
     k+     # Integer-divide the sum by this 4
       N+N  # Subtract the values in the (implicit) input-list from this value:
       N    #  Negate the value
        +   #  Add it to each value in the (implicit) input-list
         N  #  Negate each value in the list
            # (after which the result is output implicitly)

edited Apr 21, 2022 at 13:35

answered Apr 21, 2022 at 13:22

Kevin Cruijssen

131k13 gold badges144 silver badges384 bronze badges

\$\begingroup\$ Woah, nice! I’m glad I made this lang somewhat usable. You can use the compressed encoding if you want (S instead of kJ, / instead of k+). I’ll try to fix the bigint/number bug at some point \$\endgroup\$
– emanresu A
Commented Apr 30, 2022 at 9:16
\$\begingroup\$ @emanresuA Yeah, I know about the S and / instead of kJ and k+, but I liked to write the full-out version so you know the amount of characters halved is the byte-count. :) Btw, is >+< the proper/shortest way to push 4n, or is there a better alternative? \$\endgroup\$
– Kevin Cruijssen
Commented Apr 30, 2022 at 11:40

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naffetS · Accepted Answer · 2022-05-01 18:43:01Z

2

C (gcc), 61 58 bytes

n,i;f(int*a){for(i=10;i--;)i>4?n+=a[i%5]:(a[i]=n/4-a[i]);}

Try it online!

-3 bytes thanks to ceilingcat

edited May 1, 2022 at 18:43

answered Apr 30, 2022 at 18:58

naffetS

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Dominic van Essen · Accepted Answer · 2023-04-13 12:13:30Z

2

Nibbles, 4 bytes (8 nibbles)

+/+$4*~

  +         # sum of
   $        # the input
 /          # divided by
    4       # 4
+           # added to
            # (implicitly) each element of input
     *      # multiplied by
      ~     # -1

answered Apr 13, 2023 at 12:13

Dominic van Essen

36.4k2 gold badges22 silver badges60 bronze badges

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south · Accepted Answer · 2023-04-13 12:22:38Z

2

J, 7 bytes

-~4%~+/

Nothing new here.

Attempt This Online!

-~4%~+/
     +/  NB. sum
  4%~    NB. divide result by 4
-~       NB. subtract result by input, vectorized

answered Apr 13, 2023 at 12:22

south

2,1808 silver badges12 bronze badges

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Jonathan Allan · Accepted Answer · 2022-04-06 21:28:57Z

Jelly, 4? 8 bytes

With a strict interpretation of Rule 1 (can't take as a program argument due to the "single line" part) - a full program that reads from STDIN and writes to STDOUT:

ɠḲVµS÷4_

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...or with site default IO instead - a monadic Link that accepts a list of five numbers and yields a list of five numbers:

S÷4_

Try it online!

How?

ɠḲVµS÷4_ - Main Link: no arguments
ɠ        - read a line from STDIN
 Ḳ       - split that at spaces
  V      - evaluate that as Jelly code -> list of the five four-apple-weights, W
   µ     - start a new monadic chain - f(W)
    S    - sum W
      4  - four
     ÷   - divide -> sum(W)/4
       _ - subtract W (vectorises) -> [sum(W)/4-w1, sum(W)/4-w2, sum(W)/4-w3, sum(W)/4-w4, sum(W)/4-w5]
         - implicit print

Yousername · Accepted Answer · 2022-04-06 21:53:57Z

1

JavaScript (Node.js), 34 32 bytes

a=>a.map(e=>eval(a.join`+`)/4-e)

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^{-2 bytes thanks to Steffan}

edited Apr 6, 2022 at 21:53

answered Apr 6, 2022 at 21:35

Yousername

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naffetS · Accepted Answer · 2022-04-06 22:59:45Z

1

Ruby, 22 bytes

->a{a.map{a.sum/4-_1}}

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answered Apr 6, 2022 at 22:59

naffetS

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a stone arachnid · Accepted Answer · 2022-04-06 23:25:10Z

1

C (clang), 60 bytes

b,c;f(*a){for(b=c=0;b<5;c+=a[b++]);for(;b--;a[b]=c/4-a[b]);}

Try it online! Takes an array of 5 integers and modifies the array in-place.

answered Apr 6, 2022 at 23:25

a stone arachnid

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MarcMush · Accepted Answer · 2022-04-07 13:13:18Z

1

Julia 1.0, 14 bytes

!l=sum(l/4).-l

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answered Apr 7, 2022 at 13:13

MarcMush

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naffetS · Accepted Answer · 2022-04-07 14:55:04Z

1

05AB1E, 9 4 bytes

O4÷α

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-5 bytes thanks to Kevin Cruijssen

edited Apr 7, 2022 at 14:55

answered Apr 7, 2022 at 0:07

naffetS

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1

\$\begingroup\$ Input is done implicitly in 05AB1E, so you can remove the leading I. :) 05AB1E also vectorizes in a lot of cases, so you don't need the map here if you use a Duplicate of the list instead. To summarize, this could be 6 bytes: #DO4÷α, or with default I/O as lists it could be 4 bytes: O4÷α. \$\endgroup\$
– Kevin Cruijssen
Commented Apr 7, 2022 at 9:16

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Rui Barradas · Accepted Answer · 2022-04-07 19:29:45Z

1

R, 14 bytes

\(x)sum(x)/4-x

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Uses the new lambda, \, introduced in R 4.1.

answered Apr 7, 2022 at 19:29

Rui Barradas

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naffetS · Accepted Answer · 2022-04-07 19:58:38Z

1

J, 7 bytes

-~4%~+/

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answered Apr 7, 2022 at 19:58

naffetS

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naffetS · Accepted Answer · 2022-04-08 14:31:15Z

1

jq, 20 9 bytes

add/4-.[]

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-11 bytes thanks to @ovs

edited Apr 8, 2022 at 14:31

answered Apr 7, 2022 at 19:34

naffetS

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\$\begingroup\$ [add/4-.[]] works for 11. And I'd say you can omit the outer pair of brackets for 9 \$\endgroup\$
– ovs
Commented Apr 8, 2022 at 8:41

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DeathIncarnate · Accepted Answer · 2022-04-13 09:36:59Z

1

Burlesque, 9 bytes

J++4./j?-

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J   # Duplicate
++  # Sum
4./ # Divide by 4
j   # Swap
?-  # Subtract from each

answered Apr 13, 2022 at 9:36

DeathIncarnate

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user36046user36046 · Accepted Answer · 2022-05-05 19:41:48Z

x86 32-bit machine code, 32 bytes

00000000: 87ca 31c0 6a05 5903 448a fce2 fac1 e802  ..1.j.Y.D.......
00000010: 6a05 5950 2b44 8afc 8944 8afc 58e2 f4c3  j.YP+D...D..X...

Assembly

section .text
	global func
func:					;void func(int *ecx);
	; int *edx=ecx; int eax=0;
	xchg ecx, edx
	xor eax,eax

	; for(ecx=5;ecx>0;ecx--)eax+=edx[ecx-1]
	push 0x5
	pop ecx
	add:
	add eax, [edx + 4*ecx-4]
	loop add

	; eax = eax / 4
	shr eax, 2

	; for(ecx=5;ecx>0;ecx--)edx[ecx-1]=eax-edx[ecx-1];
	push 0x5
	pop ecx
	sub:
	push eax
	sub eax, [edx + 4*ecx-4]
	mov [edx + 4*ecx-4], eax
	pop eax
	loop sub

	; return
	ret

Takes a pointer to an array of 5 integers in ECX (fastcall convention), and modifies the array in place with the results.

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Neil · Accepted Answer · 2022-04-06 18:38:33Z

0

Charcoal, 7 bytes

Ｉ⁻÷Σθ⁴θ

Try it online! Link is to verbose version of code. Explanation:

    θ   Input list
   Σ    Take the sum
  ÷     Integer divided by
     ⁴  Literal integer `4`
 ⁻      Vectorised subtract
      θ Input list
Ｉ       Cast to string
        Implicitly print

9 bytes for a version that works with two or more apples, not just five:

Ｉ⁻÷Σθ⊖Ｌθθ

Try it online! Link is to verbose version of code.

answered Apr 6, 2022 at 18:38

Neil

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chunes · Accepted Answer · 2022-04-06 22:14:51Z

0

Red, 28 bytes

func[v][v -((sum v)/ 4)* -1]

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answered Apr 6, 2022 at 22:14

chunes

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Find the weight of five apples

Problem

Rules

40 Answers 40

MathGolf, 3 (or 5) bytes

Vyxal, 9 7 6 5 bytes

Explanation

BQN, 8 6 bytes

Python, 32 bytes

Haskell, 18 bytes

Desmos, 16 bytes

Japt v2.0a0 -m, 5 bytes

λ-2d, 456 squares

js equivalent

Excel, 15 bytes

Excel, 19 bytes

TI-Basic, 7 bytes

Factor + math.unicode, 23 bytes

Explanation

LOLCODE, 216 bytes

Pip, 9 7 bytes

Halfwit, 5 bytes

C (gcc), 61 58 bytes

Nibbles, 4 bytes (8 nibbles)

J, 7 bytes

Jelly, 4? 8 bytes

How?

JavaScript (Node.js), 34 32 bytes

Ruby, 22 bytes

C (clang), 60 bytes

Julia 1.0, 14 bytes

05AB1E, 9 4 bytes

R, 14 bytes

J, 7 bytes

jq, 20 9 bytes

Burlesque, 9 bytes

x86 32-bit machine code, 32 bytes

Assembly

Charcoal, 7 bytes

Red, 28 bytes

Your Answer

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