Simple Task Solved Thrice

Question

You should write 3 programs and/or functions in one language.

All of these programs should solve the same task but they all should give different (but valid) outputs. (I.e. for every pair of programs their should be some input which generates different (but valid) sets of output numbers.)

The task

You are given an integer n, greater than 1
You should return or output n distinct positive integers, and none of them should be divisible by n.
The order of the numbers doesn't matter and a permutation of numbers doesn't count as different outputs.

A valid triplet of programs with some input => output pairs:

program A:
    2 => 5 9
    4 => 5 6 9 10
    5 => 2 4 8 7 1

program B:
    2 => 1 11
    4 => 6 9 1 2
    5 => 4 44 444 4444 44444

program C (differs only in one number from program B):
    2 => 1 13
    4 => 6 9 1 2
    5 => 4 44 444 4444 44444

Scoring

Your score is the sum of the lengths of the 3 programs or functions.
Lower score is better.
If your programs/functions share code, the shared code should be counted into the length of every program that uses the code.

Does each program have to be able to be run from its own file without any includes, or can the programs depend on some shared module/library that is only counted once? — quintopia, Nov 26, 2015 at 19:09
@quintopia The programs/functions should not share code. If they do, the shared code should be counted into the length of all programs that use the code. — randomra, Nov 26, 2015 at 19:13

Dennis · Accepted Answer · 2015-11-27 01:19:59Z

J, 16 bytes

Function 1, 5 bytes

p:^i.

Function 2, 6 bytes

+p:^i.

Function 3, 5 bytes

>:^i.

How it works

Function 1

p:^i.     Right argument: y

   i.     Compute (0 ... y-1).
p:        Compute P, the prime at index y (zero-indexed).
  ^       Return all powers P^e, where e belongs to (0 ... y-1).

Since P is prime and P > y, y cannot divide P^e.

Function 2

+p:^i.    Right argument: y

 p:^i.    As before.
+         Add y to all results.

If y divided P^e + y, it would also divide P^e + y - y = P^e.

Function 3

>:^i.     Right argument: y

   i.     Compute (0 ... y-1).
>:        Compute y+1.
  ^       Return all powers (y+1)^e, where e belongs to (0 ... y-1).

If y divided (y+1)^e some prime factor Q of y would have to divide (y+1)^e.

But then, Q would divide both y and y+1 and, therefore, y + 1 - y = 1.

isaacg · Accepted Answer · 2015-11-26 20:07:43Z

4

Pyth, 17 16 bytes

5 bytes:

^LhQQ

Outputs:

2: [1, 3]
3: [1, 4, 16]
4: [1, 5, 25, 125]

6 bytes:

mh*QdQ

Outputs:

2: [1, 3]
3: [1, 4, 7]
4: [1, 5, 9, 13]

5 bytes:

|RhQQ

Outputs:

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

Alternate version, in increasing order: -ShQQ

edited Nov 26, 2015 at 20:07

answered Nov 26, 2015 at 19:16

isaacg

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1

\$\begingroup\$ Ooh. I like that third scheme. \$\endgroup\$
– quintopia
Nov 26, 2015 at 19:32
\$\begingroup\$ @isaacg oh, sorry \$\endgroup\$
– Maltysen
Nov 26, 2015 at 20:18

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

Dyalog APL, 16 17 bytes

1+⊢×⍳

⊢+1+⊢×⍳

1+⊢*⍳

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2015 at 19:13

lirtosiast

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Vitsy, 54 bytes

Programs:

V1V\[DV*1+N' 'O1+]

V2V\[DV*1+N' 'O1+]

V3V\[DV*1+N' 'O1+]

Outputs:

2 => 3 7
4 => 5 9 13 17
5 => 6 11 16 21 26

2 => 5 7
4 => 9 13 17 21
5 => 11 16 21 26 31

2 => 7 9
4 => 13 17 21 25 
5 => 16 21 26 31 36

How it works (using the first program as the explanation):

V1V\[DV*1+N' 'O1+]
V                  Capture the implicit input as a final global variable.
 1                 Push one to the stack for later use.
  V\[            ] Do everything in the brackets input times.
     D             Duplicate the top item of the stack.
      V            Push the global variable to the stack.
       *1+         Multiply, then add 1. This makes it non-divisible.
          N' 'O    Output the number followed by a space.
               1+  Add one to the number left in the stack.

Try it online!

Doorknob · Accepted Answer · 2015-11-26 19:38:15Z

2

Perl, 79

One char added to each program because this requires the -n flag.

for$a(0..$_-1){say$_*$a+1}
for$a(1..$_){say$_*$a+1}
for$a(2..$_+1){say$_*$a+1}

Fairly straightforward.

answered Nov 26, 2015 at 19:38

Doorknob

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LegionMammal978 · Accepted Answer · 2015-11-26 22:24:48Z

2

Mathematica, 12 + 12 + 12 = 36 bytes

# Range@#-1&
# Range@#+1&
#^Range@#+1&

Tests:

# Range@#-1&[10]
(* -> {9, 19, 29, 39, 49, 59, 69, 79, 89, 99} *)
# Range@#+1&[10]
(* -> {11, 21, 31, 41, 51, 61, 71, 81, 91, 101} *)
#^Range@#+1&[10]
(* -> {11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001} *)

edited Nov 26, 2015 at 22:24

answered Nov 26, 2015 at 20:41

LegionMammal978

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\$\begingroup\$ Could you add some example outputs? \$\endgroup\$
– Paŭlo Ebermann
Nov 26, 2015 at 22:00

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Martin Ender · Accepted Answer · 2015-11-26 22:47:15Z

2

CJam, 8 + 8 + 8 = 24 bytes

{,:)))+}
{_,f*:)}
{)_(,f#}

These are three unnamed functions which expect n to be on the stack and leave a list of integers in its place. I'm not sure this is optimal, but I'll have to hunting for a shorter solution later.

Test suite.

Results:

{,:)))+}
2 => [1 3]
3 => [1 2 4]
4 => [1 2 3 5]
5 => [1 2 3 4 6]

{_,f*:)}
2 => [1 3]
3 => [1 4 7]
4 => [1 5 9 13]
5 => [1 6 11 16 21]

{)_(,f#}
2 => [1 3]
3 => [1 4 16]
4 => [1 5 25 125]
5 => [1 6 36 216 1296]

The first one also works as

{_),:)^}

or

{_(,+:)}

edited Nov 26, 2015 at 22:47

answered Nov 26, 2015 at 19:30

Martin Ender

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\$\begingroup\$ your Results show all three giving the same output when n=2 \$\endgroup\$
– Sparr
Nov 26, 2015 at 20:50
\$\begingroup\$ @Sparr That's allowed. Check the examples in the challenge. The only requirement is that they compute different functions, not different results on every input. \$\endgroup\$
– Martin Ender
Nov 26, 2015 at 20:51
\$\begingroup\$ Ahh, they have to have different results on some input. That's what I misread. \$\endgroup\$
– Sparr
Nov 26, 2015 at 20:51

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2

Python 2, 79 bytes

lambda n:range(1,n*n,n)
lambda n:range(1,2*n*n,2*n)
lambda n:range(1,3*n*n,3*n)

Three anonymous function that start at 1 and count up by each of n, 2*n, 3*n for n terms.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2015 at 22:49

xnor

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Seriously, 20 bytes

,;r*1+

,;R*1+

,;R1+*1+

Yeah, this isn't optimal...

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2015 at 19:37

lirtosiast

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Ypnypn · Accepted Answer · 2015-11-26 19:57:30Z

1

Par, 16 bytes

The custom encoding, described here, uses only one byte per character.

✶″{*↑                   ## 3 => (0 1 2) => (0 3 6)  => (1 4 7)
✶″U{ⁿ↑                  ## 3 => (1 2 3) => (3 9 27) => (4 10 28)
✶U¡↑◄                   ## 3 => (1 2 3) =>             (1 2 4)

Outputs

2 => (1 3)
3 => (1 4 7)
4 => (1 5 9 13)
5 => (1 6 11 16 21)

2 => (3 5)
3 => (4 10 28)
4 => (5 17 65 257)
5 => (6 26 126 626 3126)

2 => (1 3)
3 => (1 2 4)
4 => (1 2 3 5)
5 => (1 2 3 4 6)

edited Nov 26, 2015 at 19:57

answered Nov 26, 2015 at 19:46

Ypnypn

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Haskell, 54 bytes

f n=n+1:[1..n-1]
g n=5*n+1:[1..n-1]
h n=9*n+1:[1..n-1]

These three functions are pretty straightforward so…

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 27, 2015 at 2:18

arjanen

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alephalpha · Accepted Answer · 2015-11-27 03:49:37Z

1

Octave, 11 + 13 + 13 = 37 bytes

@(a)1:a:a^2
@(a)a-1:a:a^2
@(a)(1:a)*a+1

answered Nov 27, 2015 at 3:49

alephalpha

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quintopia · Accepted Answer · 2015-11-27 12:53:39Z

1

Python 2, 125 bytes

N=input();print[i*N+1for i in range(N)]
N=input();print[i*N+1for i in range(1,N+1)]
N=input();print[i*N+1for i in range(2,N+2)]

Each line here is a complete program. The most obvious solution in my mind.

EDIT @Sherlock9 saved two bytes.

edited Nov 27, 2015 at 12:53

answered Nov 26, 2015 at 19:21

quintopia

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Haskell, 50

f n=n+1:[1..n-1]
f n=1:[n+1..2*n-1]
f n=[1,n+1..n^2]

Examples:

 f1 5=[6,1,2,3,4]
 f2 5=[1,6,7,8,9]
 f3 5=[1,6,11,16,21]

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 27, 2015 at 17:28

proud haskeller

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Golfscript, 50 51 57 bytes

A Golfscript version of what used to be quintopia's Python code. Each function takes n off the stack.

{.,{1$*)}%\;}:f;    i*n+1 for i in range(n)
{.,{)1$*)}%\;}:g;   i*n+1 for i in range(1,n+1)
{.,{1$)\?}%\;}:h;   (n+1)**i for i in range(n)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Nov 26, 2015 at 20:13

Sherlock9

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pizzapants184 · Accepted Answer · 2017-05-15 22:02:41Z

TI-Basic (TI-84 Plus CE), 55 40 bytes total

PRGM:C 12 bytes
    seq(AnsX+1,X,1,Ans

PRGM:B 14 bytes
    seq(AnsX+1,X,2,Ans+1

PRGM:C 14 bytes
    seq(AnsX+1,X,3,Ans+2

Simple, similar to many other answers here, each displays a list of the numbers (X+A)N+1 for X in range(N) and with A being which program (1, 2, or 3).

Old solution (55 bytes):

PRGM:C 17 bytes
    Prompt N
    For(X,1,N
    Disp XN+1
    End

PRGM:B 19 bytes
    Prompt N
    For(X,2,N+1
    Disp XN+1
    End

PRGM:C 19 bytes
    Prompt N
    For(X,3,N+2
    Disp XN+1
    End

Simple, similar to many other answers here, each displays the numbers (X+A)N+1 for X in range(N) and with A being which program (1, 2, or 3).

Stack Exchange Network

Simple Task Solved Thrice

The task

Scoring

16 Answers 16

J, 16 bytes

Function 1, 5 bytes

Function 2, 6 bytes

Function 3, 5 bytes

How it works

Function 1

Function 2

Function 3

Pyth, 17 16 bytes

5 bytes:

6 bytes:

5 bytes:

Dyalog APL, 16 17 bytes

Vitsy, 54 bytes

Programs:

Outputs:

How it works (using the first program as the explanation):

Perl, 79

Mathematica, 12 + 12 + 12 = 36 bytes

CJam, 8 + 8 + 8 = 24 bytes

Python 2, 79 bytes

Seriously, 20 bytes

Par, 16 bytes

Outputs

Haskell, 54 bytes

Octave, 11 + 13 + 13 = 37 bytes

Python 2, 125 bytes

Haskell, 50

Golfscript, 50 51 57 bytes

TI-Basic (TI-84 Plus CE), 55 40 bytes total

Your Answer

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number
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Hot Network Questions

Simple Task Solved Thrice

The task

Scoring

16 Answers 16

J, 16 bytes

Function 1, 5 bytes

Function 2, 6 bytes

Function 3, 5 bytes

How it works

Function 1

Function 2

Function 3

Pyth, 17 16 bytes

5 bytes:

6 bytes:

5 bytes:

Dyalog APL, 16 17 bytes

Vitsy, 54 bytes

Programs:

Outputs:

How it works (using the first program as the explanation):

Perl, 79

Mathematica, 12 + 12 + 12 = 36 bytes

CJam, 8 + 8 + 8 = 24 bytes

Python 2, 79 bytes

Seriously, 20 bytes

Par, 16 bytes

Outputs

Haskell, 54 bytes

Octave, 11 + 13 + 13 = 37 bytes

Python 2, 125 bytes

Haskell, 50

Golfscript, 50 51 57 bytes

TI-Basic (TI-84 Plus CE), 55 40 bytes total

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