No strings (or numbers) attached

Question

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• The Official Dyalog APL 2016 Year Game

• The monthly Dyalog APL 2017 Code Golf Challenge

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Make 12 snippets/expressions, in the same language, that result in the numbers 0 through 10, and 42 respectively, but without writing any literal numeric, string, or character data.

Build-in data, like PI() and ALPHABET(), are fine, and so are e.g. CJam's U, X, Y, Z, and A constants, and Processing's BLEND, CHORD, CENTER, BREAK, and LINES.

Every snippet must be able to stand on its own, i.e. they may not be interdependent. However, inside a single snippet, you may assign a variable and use it freely, as long as you refer to it directly by name, and not through a string containing its name.

All the snippets must be valid on the submitter’s computer at the time of submission (as reported by SE), but may not rely on unusual local conditions like number of files in a directory, the exact date or time, or specific input from the user.

Examples of valid snippets

3: INT(LOG10(YEAR(TODAY()))) because it remains true in the foreseeable future
4: CUBICROOT(LEN(CHARACTERSET())) because a 256 letter character set is very common
8: SQRT(SYSTEMTYPE()) because 64-bit systems are very common

Examples of invalid snippets

5: LEN(USERNAME()) because most people do not use “Admin” as login :-)
9: LOG10(SYSTEMMEMORY()) because it only works on systems with exactly 1 GB of memory
42: CODE("*") because it contains a string/character literal

The result of each snippet must result in an actual number (value, int, float, etc.) that can be used for further calculations using the same language as the snippet, i.e not a text string representing that number.

Only character based languages allowed.

Score is total byte count of all the 12 snippets combined. Newlines separating the snippets are not counted in.

Note that the above rules may prevent some languages from participating, even if they are Turing complete.

FAQ

Q Can the programs accept any input?
A Yes, but you may not just ask for input and enter the relevant number.

Q Are physical digits (non-data) digits allowed?
A Yes, e.g. LOG10().

Q Do Symbols in Ruby count as literals?
A Yes.

Q Does score include newlines between each snippet?
A No.

Q Is TI-BASIC "character based" enough to be valid?
A Yes.

Q Do false and true count as number literals?
A No, they are acceptable.

Q Can we use a number literal to call a function if that's the only way and the number doesn't influence the output of the function?
A Yes, if that is the normal way to write code in your language.

Q My language assumes there is a [something] at the start of each program/expression. Must I include it, or should my snippets just work if placed in the middle of a program/expression?
A They should just work in the middle of a program/expression.

Q What about regex literals?
A Forbidden, except for languages that only do regexes.

Q Is one piece of code that could print all the specified numbers acceptable?
A No, they have to be separate and mutually independent.

Q May I assume a boilerplate like int main() {}... or equivalent?
A Yes.

Q What output datatypes are allowed?
A Any numeric datatype, like int, float, etc.

Q Do I need to print the result of each snippet?
A No, making the result available for subsequent use is enough.

Q Are pre-set variables allowed?
A Yes, and they become reset (if changed) for every snippet.

Q Are π and e considered number literals?
A No, you may use them.

Q May I return 4 and 2 in different cells for 42?
A No, they must be connected as one number.

Q Bytes or characters?
A Bytes, but you may choose any desired codepage.

Q May constant functions and preset variables like J's 9:, Actually's 9, and Pretzel's 9 be used?
A Yes, if the vocabulary is finite (19 for J, 10 for Actually and Pretzel).

Answer 1

Hexagony, 13 bytes

1
2
3
4
5
6
7
8
9
10
42

Try it online!

In Hexagony, 0 through 9 are functions that multiply the current memory by 10, and then add the number represented by the function name. Therefore, the first snippet is empty as memories start off as 0.

For example, if the current memory is 65, executing the function 3 will make the current memory 653.

(To downvoters: downvote all you want; I am ready.)

Answer 2

Funciton, 1222 bytes

Apart from numeric literals, there are two ways I can produce a value (any value at all) in Funciton: stdin and lambda expressions. Stdin is a single box while a full lambda expression requires more syntax, so I’m going with stdin. However, while stdin could be anything, all of the following work regardless of what input is provided.

All of the library functions used here existed before the challenge was posted.

0 (40 bytes in UTF-16)

╔╗┌┐
║╟┤└┼┬┐
╚╝└─┘└┘

This uses the raw syntax for less-than. A value is never less than itself, so the result of this is 0.

1 (52 bytes in UTF-16)

╔╗┌─╖┌─╖
║╟┤⌑╟┤ɕ╟
╚╝╘═╝╘═╝

⌑ returns a lazy sequence containing a single element and ɕ counts the number of elements. (The lazy sequence is lazy enough that this snippet doesn’t actually evaluate stdin at all!)

2 (70 bytes in UTF-16)

╔╗┌─╖┌─╖┌─╖
║╟┤⌑╟┤ʂ╟┤ɕ╟
╚╝╘═╝╘═╝╘═╝

= 2¹. ʂ generates all subsequences of a sequence, and thus turns a sequence of n elements into one with 2ⁿ.

3 (88 bytes in UTF-16)

╔╗┌─╖┌─╖┌─╖┌─╖
║╟┤⌑╟┤ʂ╟┤ɕ╟┤♯╟
╚╝╘═╝╘═╝╘═╝╘═╝

= 2 + 1. ♯ increments a value by 1.

4 (88 bytes in UTF-16)

╔╗┌─╖┌─╖┌─╖┌─╖
║╟┤⌑╟┤ʂ╟┤ʂ╟┤ɕ╟
╚╝╘═╝╘═╝╘═╝╘═╝

= 2².

5 (106 bytes in UTF-16)

╔╗┌─╖┌─╖┌─╖┌─╖┌─╖
║╟┤⌑╟┤ʂ╟┤ʂ╟┤ɕ╟┤♯╟
╚╝╘═╝╘═╝╘═╝╘═╝╘═╝

= 4 + 1.

6 (106 bytes in UTF-16)

╔╗┌─╖┌─╖┌─╖┌─╖┌─╖
║╟┤⌑╟┤ʂ╟┤ɕ╟┤♯╟┤!╟
╚╝╘═╝╘═╝╘═╝╘═╝╘═╝

= 3 factorial.

7 (110 bytes in UTF-16)

┌───┐┌─╖┌─╖┌─╖╔╗
│┌─╖├┤ɕ╟┤ʂ╟┤⌑╟╢║
└┤A╟┘╘═╝╘═╝╘═╝╚╝
 ╘╤╝

= A(2, 2) (Ackermann function).

8 (118 bytes in UTF-16)

┌────┐┌─╖┌─╖┌─╖╔╗
│┌──╖├┤ɕ╟┤ʂ╟┤⌑╟╢║
└┤<<╟┘╘═╝╘═╝╘═╝╚╝
 ╘╤═╝

= 2 << 2 (shift-left).

9 (128 bytes in UTF-16)

┌───┐┌─╖┌─╖┌─╖┌─╖╔╗
│┌─╖├┤♯╟┤ɕ╟┤ʂ╟┤⌑╟╢║
└┤×╟┘╘═╝╘═╝╘═╝╘═╝╚╝
 ╘╤╝

= 3 × 3.

10 (146 bytes in UTF-16)

┌───┐┌─╖┌─╖┌─╖┌─╖┌─╖╔╗
│┌─╖├┤♯╟┤ɕ╟┤ʂ╟┤ʂ╟┤⌑╟╢║
└┤+╟┘╘═╝╘═╝╘═╝╘═╝╘═╝╚╝
 ╘╤╝

= 5 + 5.

42 (170 bytes in UTF-16)

┌──────┐┌─╖┌─╖┌─╖┌─╖┌─╖╔╗
│┌─╖┌─╖├┤!╟┤♯╟┤ɕ╟┤ʂ╟┤⌑╟╢║
└┤♯╟┤×╟┘╘═╝╘═╝╘═╝╘═╝╘═╝╚╝
 ╘═╝╘╤╝

= 6 × (6 + 1).

Answer 3

JavaScript, 144 141 140 138 132 125 123 bytes

With help from @edc65, @Sjoerd Job Postmus, @DocMax, @usandfriends, @Charlie Wynn and @Mwr247!

result.textContent = [

+[]                          ,// 0  (3 bytes)
-~[]                         ,// 1  (4 bytes)
-~-~[]                       ,// 2  (6 bytes)
-~-~-~[]                     ,// 3  (8 bytes)
-~Math.PI                    ,// 4  (9 bytes)
-~-~Math.PI                  ,// 5  (11 bytes)
-~-~-~Math.PI                ,// 6  (13 bytes)
Date.length                  ,// 7  (11 bytes)
(a=-~-~[])<<a                ,// 8  (13 bytes) = (2 << 2)
(a=~Math.E)*a                ,// 9  (13 bytes) = (-3 * -3)
(a=-~-~[])<<a|a              ,// 10 (15 bytes) = ((2 << 2) | 2)
(a=Date.length)*--a           // 42 (19 bytes) = (7 * 6)

];

<pre id="result"></pre>

Answer 4

Mouse-2002, 27 26 17 14 bytes

The first snippets push 0-10, and ZR+ pushes 25 then 17 and 25 17 + 42 = is 1.

A
B
C
D
E
F
G
H
I
J
K
ZR+

Answer 5

CJam, 27 24 bytes

U    e# 0
X    e# 1
Y    e# 2
Z    e# 3
Z)   e# 3 + 1
YZ+  e# 2 + 3
ZZ+  e# 3 + 3
AZ-  e# 10 - 3
YZ#  e# 2³
A(   e# 10 - 1
A    e# 10
EZ*  e# 14 × 3

Thanks to @MartinBüttner for -1 byte!

Try it online!

Answer 6

Perl 5, 86 75 71 66 bytes

All ^Fs are literal control characters (0x06 in ASCII), and hence a single byte.

$[          # array start index, defaults to 0                                  2
!$[         # !0 is 1                                                           3
$^F         # max sys file descriptor number, 2 on all sane systems             2
++$^F       # 2 + 1                                                             4
~-$]        # 5 - 1                                                             4
int$]       # $] is Perl version, int truncates                                 5
length$~    # 1 + 5                                                             8
~~exp$^F    # floor(e^2)                                                        7
$^F<<$^F    # 2 bitshift-right 2                                                6
-!$[+ord$/  # -1 + 10                                                          10
ord$/       # input record separator, newline by default, ord gets ASCII val    5
ord($"^$/)  # 32 + 10                                                          10

Thanks to msh210 for saving 11 bytes and Dom Hastings for 9 bytes!

Answer 7

Brainfuck, 70 bytes

+
++
+++
++++
+++++
++++++
+++++++
++++++++
+++++++++
++++++++++
--[>+<++++++]>-

Each line must be run individually.

The first 10 are self explanatory: we increment the value of the cell via each plus.

The 42 is a lot more complex. It relies on the fact the most brainfuck interpreter use 8-bit cells, meaning that all operations on it are done modulo 256. The -- sets cell #0 to 254. Then we enter a loop which runs until cell #0 is 0. Each iteration adds 1 to cell #1 and adds 6 to cell #0. This loops runs 43 times, so cell #1 is 43. Finally, we subtract 1 from cell #1 to make it 42.

I got the most efficient 42 ever found from http://esolangs.org/wiki/Brainfuck_constants

Answer 8

Darkness, 339 303 bytes

This is where Darkness really shines. Get it? :~)!

Without printing (replaced the space with \s in the first line since it won't show otherwise):

\s

█ 

██ 

███ 

████ 

█████ 

██████ 

███████ 

████████ 

█████████ 

██████████ 

██████████████████████████████████████████ 

With printing:

■ 

█■ 

██■ 

███■ 

████■ 

█████■ 

██████■ 

███████■ 

████████■ 

█████████■ 

██████████■ 

██████████████████████████████████████████■ 

Each line must be run individually in this case since the program terminates in the light (a space). However, it is possible to write this on one or several lines in the same program.

Regular darkness (█) increments a register by 1, and the ■ instruction (some sort of mini-darkness) outputs the contents of the register.

Answer 9

Prolog, 113 99 bytes

Snippets:

e-e              % 0.0
e/e              % 1.0
e/e+e/e          % 2.0
ceil(e)          % 3
ceil(pi)         % 4
ceil(e*e-e)      % 5
ceil(e+e)        % 6
floor(e*e)       % 7
ceil(e*e)        % 8
ceil(pi*e)       % 9
ceil(pi*pi)      % 10
ceil(e^e*e)      % 42

Combines the mathematical constants e and pi in different ways converted to int.

Edit: Saved 14 bytes by utilizing floats for 0-2.

Answer 10

MATL, 30 bytes

O
l
H
I
K
Kl+
HI*
KYq
HK*
IH^
K:s
IH^:sI-

H, I, and K are predefined constants for 2, 3, and 4 (like pi). O and l are functions that returns a matrix of zeros (O) or ones (l), the default size is 1x1. : makes a vector, and s sums it, so K:s makes a vector from 1 to 4 and sums it to get 10. Yq is the n-th prime function, so KYq is the 4th prime, 7.

Answer 11

C#, no usings, 234 bytes

new int()                       // 0
-~new int()                     // 1
-~-~new int()                   // 2
-~-~-~new int()                 // 3
-~-~-~-~new int()               // 4
-~-~-~-~-~new int()             // 5
-~-~-~-~-~-~new int()           // 6
-~-~-~-~-~-~-~new int()         // 7
-~-~-~-~-~-~-~-~new int()       // 8
(int)System.ConsoleKey.Tab      // 9
(int)System.TypeCode.UInt32     // 10
(int)System.ConsoleKey.Print    // 42

This is much more boring than I initially thought it was going to be. I had pretty varied ideas, such as new[]{true}.Length and true.GetHashCode() and typeof(int).Name.Length and uint.MinValue etc., but new int() beat them all.

Answer 12

PHP, 157 145 91 bytes

First time posting on Code Golf, figured I'd give it a shot. I'll get better eventually :P If you see any obvious (to you) spots where I could save characters, let me know.

EDIT: Realized I didn't need the semicolons, since these are just snippets.

EDIT2: Thanks to Blackhole for many suggestions!

LC_ALL
DNS_A
~~M_E
~~M_PI
LOCK_NB 
LC_TIME
LOG_INFO
INI_ALL
IMG_WBMP
SQL_DATE
SQL_TIME
LOG_INFO*INI_ALL

Answer 13

PowerShell, 147 bytes

These use + to implicitly cast things to integers. The later numbers use Enums from the .Net Framework unerpinnings of PowerShell which happen to have the right values.

+$a                          #0, 3 bytes (unset vars are $null, +$null == 0)
+$?                          #1, 3 bytes (bool previous result, default $true, +$true == 1)
$?+$?                        #2, 5 bytes (same as #1, twice)
$?+$?+$?                     #3, 8 bytes (beats [Int][Math]::E)
$?+$?-shl$?                  #4, 11 bytes (-shl is shift-left)
$?+$?+$?+$?+$?               #5, 14 bytes
$?+$?+$?-shl$?               #6, 14 bytes (enum value, + casts to integer)
+[TypeCode]::Int16           #7, 18 bytes
$?+$?-shl$?+$?               #8, 14 bytes
+[consolekey]::tab           #9, 18 bytes
+[TypeCode]::UInt32          #10, 19 bytes
+[consolekey]::Print         #42, 20 bytes

#Total:                      147 bytes

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• -~-~-~ used in JavaScript, C# and PHP answers would be - -bnot - -bnot - -bnot in PowerShell.

• x^y exponentiation used in Perl answers, or x**y in Python or JavaScript ES7, would be [Math]::Pow($x,$y)

• constants e and Pi are the character-heavy [Math]::E and [Math]::PI

Answer 14

Python 2, 191 159 158 157 156 149 146 bytes

My first submission ever, I hope I got everything right ! Based on the time I spent on this, I guess there's surely a better one for a few of them.

# 0 | Bytes : 5
int()

# 1 | Bytes : 5
+True

# 2 | Bytes : 6
-~True

# 3 | Bytes : 8
-~-~True

# 4 | Bytes : 10
-~-~-~True

# 5 | Bytes : 12
-~-~-~-~True

# 6 | Bytes : 14
-~-~-~-~-~True

# 7 | Bytes : 16
-~-~-~-~-~-~True

# 8 | Bytes : 15
a=True;a<<a+a+a

# 9 | Bytes : 19
a=True;(a<<a+a+a)+a

# 10 | Bytes : 20
int(`+True`+`int()`)

# 42 | Bytes : 16
~-len(`license`)
# Especially proud of this one !

Total byte count: 146

Many thanks to FryAmTheEggman !

Answer 15

DC, 35 bytes

K
KZ
IZ
Iv
EI-
FI-
IZd*
IIZ/
Ivd+
IIv-
IIZ-
IKZ-
I
EdE++

To test the snippets append a f to print the stack and pass that string to dc:

$ echo 'EdE++f' | dc
42

Answer 16

TI-BASIC, 41 bytes

0~10:

X
cosh(X
int(e
int(π
-int(-π
int(√(π³
int(π+π
int(e²
int(eπ
int(π²
Xmax

42:

int(π²/ecosh(π

In TI-BASIC, all uninitialized single-letter variables start at 0, and Xmax (the right window boundary of the graph screen) starts at 10.

The mathematical constant π is one byte, but e is two bytes.

Answer 17

Math++, 92 bytes total

0 (1 bytes): a

1 (2 bytes):!a

2 (3 bytes):_$e

3 (4 bytes): _$pi

4 (7 bytes): _$e+_$e

5 (8 bytes): _($e+$e)

6 (9 bytes): _$pi+_$pi

7 (8 bytes): _($e*$e)

8 (9 bytes): _($e*$pi)

9 (10 bytes): _($pi*$pi)

10 (12 bytes): _$e*_($e+$e)

42 (19 bytes): _($pi+$pi)*_($e*$e)

Answer 18

Python 2, 306 275 274 bytes

I used the fact that for any x (integer and not 0) the expression x/x equals 1 and played around with some bitwise operations.

I adjusted the snippets such that they still meet the requirements (thanks @nimi this saved me 24 bytes), but you have to manually test them. Here is the code and individual byte counts:

zero.py Bytes: 7
len({})
--------------------------
one.py  Bytes: 12
r=id(id)
r/r
--------------------------
two.py  Bytes: 17
r=id(id)
-(~r/r)
--------------------------
three.py    Bytes: 20
r=id(id)
-(~r/r)|r/r
--------------------------
four.py Bytes: 20
r=~id(id)/id(id)
r*r
--------------------------
five.py Bytes: 26
r=~id(id)/id(id)
(r*r)|r/r
--------------------------
six.py  Bytes: 25
r=~id(id)/id(id)
(r*r)|-r
--------------------------
seven.py    Bytes: 27
r=~id(id)/id(id)
-~(r*r|-r)
--------------------------
eight.py    Bytes: 24
r=-(~id(id)/id(id))
r<<r
--------------------------
nine.py Bytes: 29
r=-(~id(id)/id(id))
r-~(r<<r)
--------------------------
ten.py  Bytes: 31
r=~id(id)/id(id)
-r*((r*r)|r/r)
--------------------------
answer.py   Bytes: 37
r=-(~id(id)/id(id))
(r<<r*r)|(r<<r)|r
--------------------------
Total byte count: 274

Answer 19

Javascript (Browser Env), 155 136 130 bytes

+[]
-~[]
-~-~[]
-~-~-~[]
-~-~-~-~[]
-~-~-~-~-~[]
-~-~-~[]<<-~[]
-~-~-~-~-~-~-~[]
-~[]<<-~-~-~[]
~(~[]+[]+-[])
-~[]+[]+-[]
-~(top+top.s).length // Requires browser environment

Thanks to:
@Ismael Miguel: 155 -> 136 -> 130 bytes

Answer 20

Seriously, 39 33 bytes

Stuff in parentheses is explanations:

 (single space, pushes size of stack, which is 0 at program start)
 u (space pushes 0, u adds 1 (1))
 ⌐ (space pushes 0, ⌐ adds 2 (2))
 u⌐ (space pushes 0, u adds 1 (1), ⌐ adds 2 (3))
 ⌐⌐ (space pushes 0, ⌐⌐ adds 2 twice (4))
 ⌐P (space pushes 0, ⌐ adds 2 (2), P pushes the 2nd prime (5))
Hl▓ (H pushes "Hello, World!", l pushes length (13), ▓ pushes pi(13) (6))
QlP (Q pushes "QlP", l pushes length (3), P pushes the 3rd prime (7))
Ql╙ (Q pushes "QlP", l pushes length (3), ╙ pushes 2**3 (8))
úl▓ (ú pushes the lowercase English alphabet, l pushes length (26), ▓ pushes pi(26) (9))
 u╤ (space pushes 0, u adds 1 (1), ╤ pushes 10**1 (10))
HlPD (H pushes "Hello, World!", l pushes length (13), P pushes the 13th prime (43), D subtracts 1 (42))

Hexdumps of programs:

20
2075
20a9
2075a9
20a9a9
20a950
486cb2
516c50
516cd3
a36cb2
2075d1
486c5044

Thanks to quintopia for 6 bytes!

Answer 21

dc, 42 bytes

K
zz
OZ
zzz+
OZd*
OdZ/
zzzz*
Ozz+-
OdZ-
Oz-
O
Od+dz++

Results

0
1
2
3
4
5
6
7
8
9
10
42

There aren't many ways to generate new numbers with dc. I use O: output base, initially 10; K: precision, initially 0; z stack depth, initially 0; Z significant digits of operand. We combine these with the usual arithmetic operators.

Test program

#!/bin/bash

progs=(                                         \
    "K"                                         \
    "zz"                                        \
    "OZ"                                        \
    "zzz+"                                      \
    "OZd*"                                      \
    "OdZ/"                                      \
    "zzzz*"                                     \
    "Ozz+-"                                     \
    "OdZ-"                                      \
    "Oz-"                                       \
    "O"                                         \
    "Od+dz++"                                   \
)

a=0
results=()
for i in "${progs[@]}"
do
    results+=($(dc -e "${i}p"))
    (( a+=${#i} ))
done

echo "#dc, $a bytes"
echo
printf '    %s\n' "${progs[@]}"
echo
echo '##Results'
echo
printf '    %s\n' "${results[@]}"

Answer 22

Japt, 34 33 30 bytes

1 byte saved thanks to @ThomasKwa

T
°T
C-A
D-A
E-A
F-A
G-A
°G-A
Iq
´A
A
H+A

Here's what each of the different chars means:

T    0
A    10
B    11
C    12
D    13
E    14
F    15
G    16
H    32
I    64
q    sqrt on numbers
°    ++
´    --

Answer 23

Mathematica, 101 bytes

a-a
a/a
⌊E⌋
⌈E⌉
⌈π⌉
⌊E+E⌋
⌈E+E⌉
⌊E*E⌋
⌈E*E⌉
⌈E*π⌉
⌈π*π⌉
⌊π*π^π/E⌋

I'm pretty sure some of these are suboptimal. Those rounding brackets are really expensive.

For consistency, the first two could also be E-E and E/E of course, but I thought it's quite nifty to get 0 and 1 from a computation with undefined variables.

Answer 24

Marbelous, 98 bytes

Not terribly exciting, it relies on the ?n devices which turn any marble into a random value in the range 0..n (inclusive) a side effect of this is that ?0 turns any marble into a 0 regardless of input. I think the use of literals is permitted because the value does not affect the outcome and there is no other way to call a function once in Marbelous.

0:

00  # A hexadicemal literal: value 0
?0  # Turn any marble into a random value from the range 0..0 (inclusive)

1:

00
?0
+1  # increment by one

...

9:

00
?0
+9

10:

00
?0
+A  # increment by 10

42:

00
?0
+L  # increment by 21
+L

Answer 25

DUP, 68 bytes

[
 [
  [
[)]!
[ )]!
[  )]!
[)~_]!
[ )~_]!
[  )~_]!
[   )~_]!
[)$+]!
[ )$+~_$+]!

Try it here.

There are a LOT of ways to do this, but I'm abusing the return stack for this one.

Explanation

To fully figure this out, you need to understand DUP's behavior regarding lambdas. Instead of pushing the lambda itself to the stack, it actually pushes the current IP to the stack when the lambda is detected. That can explain the first 3 snippets, which involve lambdas.

The next snippets use the return stack. When ! is executed, the current IP is pushed to the return stack, and the top of the stack is set as the new IP to start lambda execution. ) pops a number from the return stack onto the data stack.

That's pretty much enough to explain the rest of the snippets. If you still don't get it, keep in mind that the Step button is quite handy!

Answer 26

><>, 86 bytes

• 0: ln;
• 1: lln;
• 2: llln;
• 3: lll+n;
• 4: lll:+n;
• 5: llll+n;
• 6: llll++n; or llll:+n;
• 7: lllll+n;
• 8: lllll:+n;
• 9: lllll++n; or llllll+n;
• 10: llll+:+n; or lll:l+*n;
• 42: llll*ll+*n;

Relies on stack size to get its literals.

Answer 27

MS Excel formulas, 163 151 150 143 bytes

Not exactly a programming language, but here it goes...

0:  -Z9                         (03 bytes)
1:  N(TRUE)                     (07 bytes)
2:  TYPE(T(Z9))                 (11 bytes)
3:  TRUNC(PI())                 (11 bytes)
4:  TYPE(TRUE)                  (10 bytes)
5:  ODD(PI())                   (09 bytes)
6:  FACT(PI())                  (10 bytes)
7:  ODD(PI()+PI())              (14 bytes)
8:  EVEN(PI()+PI())             (15 bytes)
9:  TRUNC(PI()*PI())            (16 bytes)
10: EVEN(PI()*PI())             (15 bytes)
42: EVEN(CODE(-PI())-PI())      (22 bytes)

PI() is used in most cases as it is the shorter way (that I am aware of) to introduce a numeric value without using a number or string literal. N converts various things (incl. booleans) to numbers, and T converts various things to text. TYPE returns 2 for a text argument and 4 for a boolean argument. TRUNC discards fractional part (i.e. rounds positive numbers down), EVEN rounds up to the next even number, and ODD rounds up to the next odd number. CODE(-PI()) is the ASCII code of the first character of the conversion to text of -π, i.e. 45 (for "-").

EDIT: Removed equal signs from the byte count (-12!) - as pointed out by Nᴮᶻ in the comments, they are not supposed to be included.

EDIT 2: Assuming the rest of the worksheet is empty, it is possible to use a reference to an empty cell as zero (again, suggested by Nᴮᶻ) provided that we include a minus sign (or use it in other numeric expression) to resolve type ambiguity.

Answer 28

05AB1E, 40 38 24 bytes

¾
X
Y
XÌ
Y·
T;
T;>
T;Ì
TÍ
T<
T
žwT+

• Push counter_variable
• Push 1
• Push 2
• Push 1+2
• Push 2*2
• Push 10/2
• Push (10/2)+1
• Push (10/2)+2
• Push 10-2
• Push 10-1
• Push 10
• Push 32, 10, add

Answer 29

PHP, 38 37 bytes

Note: uses IBM-850 encoding.

+_          #0
+~╬         #1
+~═         #2
+~╠         #3
+~╦         #4
+~╩         #5
+~╔         #6
+~╚         #7
+~Ã         #8
+~ã         #9
+~╬¤        #10
+~╦═        #42

Explanation

Undefined constants in PHP have their name as default value. The resulting string is then negated to result in "0", "42" etc. Finally, it's converted to integer using the leading +.

Tweaks

• Saved a byte by exploiting the fact that converting any string results in 0 when converted to int, except when it matches /^[0-9.]+/

Answer 30

D1ffe7e45e, 112

0
02
020
0202
02020
020202
0202020
02020202
020202020
0202020202
02020202020
202020202020202020202020202020202020202020

Each line is a different snippet.

The numbers in the program don't count as number literals since they're only used as commands.

The final one can definitely be golfed more.

EDIT: I got the interpreter working and all snippets work. If you'd like to test if yourself, add 0f0f to the end of the snippet so the program terminates.

Answer 31

Pyth, 35 34 33 bytes

-1 byte by @Mimarik

There are a number of possibilities for some programs.

0, 1 byte

Z

1, 2 bytes

hZ
!Z

2, 3 bytes

hhZ
eCG
eCd
lyd
lyb
lyN

3, 3 bytes

l`d

4, 3 bytes

l`b
eCN

5, 4 bytes

hl`b
telG

6, 3 bytes

elG

7, 4 bytes

tttT
helG

8, 3 bytes

ttT

9, 2 bytes

tT

10, 1 byte

T

42, 4 bytes

yhyT

All of these involve either basic double (y), +1 (h) and -1 (t) commands, or l (length of a string). The Z variable is initialized to zero.

For 5, b is initialized to a newline character. Backtick gives "\n" (including the quotes, and the length of that string is 4.

Try them here!

Answer 32

Dyalog APL, 59 52 bytes

-7 bytes by @NBZ

I wasn't about to let one of @NBZ's questions go without an APL answer!

⍴⍬            0
⎕IO           1
⍴⎕SD          2
⎕WX           3
⍴⎕AI          4
⌈⍟⎕PW         5
⌊⍟⎕FR         6
⍴⎕TS          7
+⍨⍴⎕AI        8
⌊○○⎕IO        9; floor(pi^2 times ⎕IO)
⍴⎕D           10
⍎⌽⍕⌈*○≢#    42

In the last snippet, by NBZ, ≢# equals 1. ceil(e^(pi*1)) is calculated as 24, whose digits are then swapped.

Constants used:

• ⍬, is the empty numeric one-dimensional vector. Therefore, its shape ⍴⍬ is 0.
• # is a special vector of length 1.
• ⎕IO (index origin) starts at 1.
• ⎕AV, the character set, is of length 256.
• ⎕PW, the print width, is 79 characters.
• ⎕WX, window expose (whatever that is) is 3.
• ⎕FR, the float representation, is 645. I have no idea what this is either.
• ⎕D, "digits", is '0123456789'.
• ⎕TS, timestamp, has seven elements: Year, month, day, hr, min, sec, ms.
• ⎕SD, screen dimensions, has two elements: width and height.
• ⎕AI, account info, has four elements. I don't know what they are.

Answer 33

R, 123 bytes

+!pi                #  0  -   4 bytes
+TRUE               #  1  -   5 bytes
nchar(NA)           #  2  -   9 bytes
floor(pi)           #  3  -   9 bytes
cars$d[pi]          #  4  -  10 bytes
ncol(CO2)           #  5  -   9 bytes
nrow(BOD)           #  6  -   9 bytes
cars$s[pi]          #  7  -  10 bytes
sum(dim(BOD))       #  8  -  13 bytes
floor(pi*pi)        #  9  -  12 bytes
ceiling(pi*pi)      # 10  -  14 bytes
nrow(CO2)/nchar(NA) # 42  -  19 bytes
                    # Total 123 bytes

Answer 34

R, 166 165 bytes

I made a PI based answer

pi-pi                         # 0
pi/pi                         # 1
(pi+pi)/pi                    # 2 
floor(pi)                     # 3 
ceiling(pi)                   # 4 
ceiling(pi)+pi/pi             # 5 
floor(pi+pi)                  # 6 
ceiling(pi+pi)                # 7 
floor(pi*pi)-pi/pi            # 8 
floor(pi*pi)                  # 9 
ceiling(pi*pi)                # 10 
ceiling(pi^pi+pi+sqrt(pi))    # 42

Answer 35

80386 machine code, 73 bytes

Hexdump:

33 c0
33 c0 40
33 c0 40 40
33 c0 40 40 40
33 c0 40 40 40 40
33 c0 40 8d 04 80
33 c0 40 8d 04 80 40
33 c0 40 8d 04 80 40 40
33 c0 40 8d 04 c0 48
33 c0 40 8d 04 c0
33 c0 40 8d 04 c0 40
33 c0 40 8d 04 c0 8d 04 80 48 48 48

The snippets calculate the required value in the eax register, suitable as a return value from a function.

I used various combinations of the following instructions:

xor eax, eax         (33 c0)    eax = 0
inc eax              (40)       eax = eax + 1
dec eax              (48)       eax = eax - 1
lea eax, [eax+eax*4] (8d 04 80) eax = eax * 5
lea eax, [eax+eax*8] (8d 04 c0) eax = eax * 9

The numbers here are too small to use all the possibilities of the instruction set - the only noteworthy thing here is the easy multiplication by 5 and 9. For example, here is a 13-byte snippet whose result is 2016:

33 c0      xor eax, eax             eax = 0
40         inc eax                  eax = 1
8d 0c c0   lea ecx, [eax+eax*8]     ecx = 9
8d 04 89   lea eax, [ecx+ecx*4]     eax = 45
f7 e0      mul eax                  eax = 2025
2b c1      sub eax, ecx             eax = 2016

Answer 36

Visual Basic for Applications, 68 bytes

VBA has a lot of builtin constants and public enums. I opted not to use any external or MS Office-specific library - only base VBA:

-x                     '=0  (2 bytes)
x^x                    '=1  (3 bytes)
vbGet                  '=2  (5 bytes)
vbLong                 '=3  (6 bytes)
vbLet                  '=4  (5 bytes)
vbDouble               '=5  (8 bytes)
vbYes                  '=6  (5 bytes)
vbNo                   '=7  (4 bytes)
vbSet                  '=8  (5 bytes)
vbObject               '=9  (8 bytes)
vbError                '=10 (7 bytes)
vbKeyPrint             '=42 (10 bytes)

Total = 68 bytes. Notice that by default, undeclared variables (such as x above) are all set to blank Variants, that may be interpreted, depending on context, as 0 or as an empty string. However, both operators - and ^ force it to be numeric. Also, in VBA, 0^0 is evaluated to 1 (instead of generating an error as one might expect). I'll also post an answer without constants.

Answer 37

Jelly, 31

<blank>
‘
‘‘
‘‘‘
‘‘²
‘‘²‘
‘‘‘Ḥ
⁴H’
⁴H
‘‘‘²
⁵
⁴Ḥ⁵+

Explanation

0: Outputs 0 by default

1. Increment
2. Increment twice
3. Increment thrice
4. Increment twice and square
5. Increment twice, square, and increment
6. Increment thrice and double
7. Halve sixteen and decrement
8. Halve sixteen
9. Increment thrice and square
10. Ten

42: Double sixteen and add ten

Answer 38

Haskell, 212 210 198 196 190 187 134 bytes

Edit: After reading in the FAQ that using pi is allowed, throwing everything away and doing it all using pionly saves a lot of bytes.

pi-pi                     -- 0.0,  5 bytes
pi/pi                     -- 1.0,  5 bytes
succ$pi/pi                -- 2.0, 10 bytes
floor pi                  --   3,  8 bytes
ceiling pi                --   4, 10 bytes
succ.ceiling$pi           --   5, 15 bytes
floor$pi+pi               --   6, 11 bytes
ceiling$pi+pi             --   7, 13 bytes
pred.floor$pi*pi          --   8, 16 bytes
floor$pi*pi               --   9, 11 bytes
round$pi*pi               --  10, 11 bytes
floor$pi**pi+pi+pi        --  42, 18 bytes

However, I still like the old approach more as using a number to generate other numbers feels a bit like cheating.

sum[]                     --  0,  5 bytes
length[LT]                --  1, 10 bytes
fromEnum GT               --  2, 11 bytes
length[LT ..]             --  3, 13 bytes
length$show[()]           --  4, 15 bytes
length.show$False         --  5, 17 bytes
length$show[True]         --  6, 17 bytes
length.show$Left()        --  7, 18 bytes
length.show$Right()       --  8, 19 bytes
length$show[Just[]]       --  9, 19 bytes
length$show[LT ..]        -- 10, 18 bytes
fromEnum.succ.last$show() -- 42, 25 bytes

LT,EQ and GT are the values of the Ordering Type. As Ordering is an instance of Enum, fromEnum GT yields 2 and [LT ..] creates the list [LT,EQ,GT]. show converts all instances of the Show type class to strings.

Bonus:
42 in 42 bytes: length.show$Just$[LT ..]>>[EQ,GT]>>[GT,LT]
How long is nothing? length Nothing

Saved 3 bytes thanks to nimi and 3 bytes thanks to xnor.

Answer 39

Processing, 65 44 bytes

X
Y
Z
HSB
Z*Z
LINES
HSB*Z
TAB-Z
MITER
X+TAB
-~TAB
BOX+Y

Explanation

Using the hint from the question, I decided to use Processing constants to produce these numbers. These constants contain a numerical value. So to gather them, I used PConstants.java.

Here is the list of all the constants I used:

X      == 0
Y      == 1
Z      == 2
HSB    == 3
LINES  == 5
MITER  == 8
TAB    == 9 (char)
BOX    == 41

After this, all that was left for me to do is to choose the combinations of these constants that produce the numbers 1 to 10 and 42. Finally, I ended up with these:

println(X);//0
println(Y);//1
println(Z);//2
println(HSB);//3
println(Z*Z);//2 * 2 = 4
println(LINES);//5
println(HSB*Z);//3 * 2 = 6
println(TAB-Z);//9 - 2 = 7
println(MITER);//8
println(X+TAB);//0 + 9 = 9
println(-~TAB);//-~9 = 10
println(BOX+Y);//41 + 1 = 42

Several notes:

TAB has a char value of 9, meaning that if I print TAB, I print a literal tab. So to convert it to an int, I just add 0 to it.

And that's it!

Edit

I discovered PConstants.java. It appears to contain all of the constants that exist in Processing. All that research I did is on the Processing Reference is not needed anymore. Time to cut down on the bytes :)

Answer 40

O, 36 bytes

l
Nl
CA-
DA-
EA-
FA-
FA-)
A(((
A((
A(
A
A.+.+))

Mostly finding the difference of hex letters,

CA- = 13-11 = 2 DA- = 14-10 = 3

( and ) decrement and increment respectively,

A(( = 10-1-1 = 8

.+ duplicates the top of the stack and adds it together

A.+.+))
10.+.+))
20.+))
40))
42

The first two get the length of the stack (l), which zero, or one when a blank CodeBlock is pushed to the stack (N).

Answer 41

Pyth, 50 40 bytes

Z
hZ
hhZ
hhhZ
hhhhZ
hhhhhZ
ttttT
tttT
ttT
tT
T
yhyT

In Pyth, Z=0 and T=10. h is head, which means the next variable + 1. (e.g. hhZ == 2). t is tail, which means the next variable - 1 (e.g. tttT == 7). The last line is the fanciest: we essentially do double(1 + double(10)).

Try it out here!

Thomas Kwa has golfed a superior solution.

Answer 42

Javascript, 159 chars

No any variables

Any line between ` is a code snippet. The whole code is a test. It returns an array [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 42]. If returning string "42" instead of number 42 is ok, code can be made 3 schars less (remove |[]).

`
+[]
-~[]
!![]+!![]
~~Math.PI
-~Math.PI
-~Math.PI|!![]
Math.PI<<!![]
Math.PI<<!![]|!![]
!![]<<Math.PI
Math.PI*Math.PI^[]
(-~Math.PI|!![])<<!![]
-~Math.PI+[]+(!![]+!![])|[]
`.split('\n').filter(Boolean).map(eval)

Answer 43

k4, 41 bytes

Most of this is a cheap hack—every builtin function in k has an internal id number, and . retrieves them. Entries of the form .(x) are implemented thus; others are commented.

A question: are we supposed to generate the numbers or print them? E.g., for zero, should we return some numeric type containing zero as a value, or should we display the exact string 0? I ask because in there are several default settings that could be useful here, but they all have (32-bit) integer type, which means that in recent (3.x) versions of q/k4, they are displayed on the console with a type suffix of i appended. The following assumes 0i is a valid form of 0, but has alternatives in the comments.

\p     /  0: the default network port (none); alternatively, #(), the count of the empty list
#`     /  1: the count of a null symbol atom; alternatively, if this is considered a character literal, use .(+)
\W     /  2: the default starting day of the week (monday), counting from saturday as 0; alternatively, .(-)
.(*)   /  3
.(%)   /  4
#!`    /  5: the number of namespaces loaded by default (works in 3.2 or 3.3; in earlier versions, use this as 4 and .(&) as 5)
.(|)   /  6
\P     /  7: the default numeric precision (7 significant figures); alternatively, #.j, the count of the .j namespace; alternatively, .(^)
.(=)   /  8
.(<)   /  9
.(>)   / 10
.(var) / 42

Answer 44

C, 202 bytes

enum{a}; // 0, 8

enum{a,b}; // 1, 10

enum{a,b,c}; // 2, 12

enum{a,b,c,d}; // 3, 14

enum{a,b,c,d,e}; // 4, 16

enum{a,b,c,d,e,f}; // 5, 18

enum{a,b,c,d};c*d; // 6, 18

enum{a,b,c,d,e};d+e; // 7, 20

enum{a,b,c};c<<c; // 8, 17

enum{a,b,c};b+(c<<c); //  9, 21

enum{a,b,c};c+(c<<c); // 10, 21

enum{a,b,c,d,e,f};c*f*e+c; // 42, 26

Printed Version, 730 bytes

enum{a,b,c,d,e};char m[d]={(d<<e)+a,a};main(){printf(m);} // 0, 57

enum{a,b,c,d,e};char m[d]={(d<<e)+b,a};main(){printf(m);} // 1, 57

enum{a,b,c,d,e};char m[d]={(d<<e)+c,a};main(){printf(m);} // 2, 57

enum{a,b,c,d,e};char m[d]={(d<<e)+d,a};main(){printf(m);} // 3, 57

enum{a,b,c,d,e};char m[d]={(d<<e)+e,a};main(){printf(m);} // 4, 57

enum{a,b,c,d,e};char m[d]={(d<<e)+e+b,a};main(){printf(m);} // 5, 59

enum{a,b,c,d,e};char m[d]={(d<<e)+e+c,a};main(){printf(m);} // 6, 59

enum{a,b,c,d,e};char m[d]={(d<<e)+e+d,a};main(){printf(m);} // 7, 59

enum{a,b,c,d,e};char m[d]={(d<<e)+e+e,a};main(){printf(m);} // 8, 59

enum{a,b,c,d,e,f};char m[d]={(d<<e)+f+e,a};main(){printf(m);} // 9, 61

enum{a,b,c,d,e,f,g,h,i,j,k};char m[d]={k*d+h,k*k,a};main(){printf(m,k);} // 10, 72

enum{a,b,c,d,e,f,g,h,i,j,k};char m[d]={k*d+h,k*k,a};main(){printf(m,k*e+c);} // 42, 76

Answer 45

, 36 bytes (12 chars)

Ḁ
ḁ
Ḃ
ḃ
Ḅ
ḅ
Ḇ
ḇ
Ḉ
ḉ
Ḋ
Ḫ

Try it here (test-suite-style, Firefox only).

There are variables for numbers from 0-256. (One of the benefits of choosing from UTF-8 is that you can dedicate a whole charset of 256 characters just to this kind of stuff. (Of course, byte count is a side effect.))

Answer 46

J, 46 43 42 bytes

Thanks to Nᴮᶻ for the help and exploration!

%_         NB. Inverse of infinity
_.         NB. Indeterminate value, evaluates to 1
2:_
3:_
4:_
5:_
6:_
7:_
8:_
9:_
+~5:_      NB. verb magic
(7:*6:)_   NB. 7 * 6

Here are some other programs:

(number): (representation)
1:  *_             NB. signum of infinity                                          same bytes
1:  #a:            NB. The tally of the Ace/Empty box                              +1 byte
1:  1:_            NB. The constant function with infinity;                        +1 byte
2:  #_`_           NB. The tally of infinity gerund-ed with itself                 +1 byte
2:  #_,_           NB. (same)                                                      +1 byte
2:  #_;_           NB. (same)                                                      +1 byte
42: (*:+])6:_      NB. verb magic / 6 squared plus 6                               +1 byte
42: +:<.o.7:_      NB. double floor pi times 7                                     +1 byte
42: <.(#a.)%6:_    NB. Tally of alphabet (256) divided by 6 (6:_) floored (<.);    +3 bytes
42: a+*~(a=:6:_)   NB. 6 squared plus 6.                                           +4 bytes
42: (a=:6:_)+:*a   NB. 6 squared plus 6.                                           +4 bytes
42: #C.>.%:^@^2:_  NB. I'm not entirely sure how this works                        +5 bytes

Answer 47

Perl 6, 84 81 bytes

+$           #  0 
++$          #  1
-+^++$       #  2
-+^e         #  3
-+^π         #  4
1+-+^π       #  5
floor τ      #  6
-+^τ         #  7
floor e*π    #  8
floor π*π    #  9
-+^(τ+e)     # 10
-+^(e**e*e)  # 42

Constants used

> e
2.71828182845905
> π
3.14159265358979
> τ
6.28318530717959

An attempt at some explanation

#0: $ is an anonymous state variable, the + coerces it to a number, and by default its value is 0
#1: ++$ increases the state var by 1, giving us 1
#2: +^ as a prefix is the Integer bitwise negation operator, use that with ++$ (1) and we get -2, use - to negate it and give 2
#3: Use +^ with e, which is coerced to Int, giving -3 which then gets negated with -
#4: Same as #3 but with π
#5: Same as #4 but add 1
I think the rest are easy to follow with the above bits explained.

Answer 48

Visual Basic for Applications, no named constants, 134 bytes

That's tougher:

-x                     '=0  (2 bytes)
x^x                    '=1  (3 bytes)
x^x+x^x                '=2  (7 bytes)
x^x+x^x+x^x            '=3  (11 bytes)
Len(Year(x))           '=4  (12 bytes)
Len(Year(x)&-x)        '=5  (15 bytes)
Day(x)Mod Month(x)     '=6  (18 bytes)
Len(Time())-x^x        '=7  (15 bytes)
Len(Time())            '=8  (11 bytes)
Len(Time()&-x)         '=9  (14 bytes)
Len(Date())            '=10 (11 bytes)
Day(x)+Month(x)        '=42 (15 bytes)

Total = 134 bytes. Again, - and ^ force undeclared variables (Variants) to be treated as numeric, and in VBA 0^0 evals to 1.

The concatenation operator force numbers to string form, so -x is 0 and &-x is & "0".

For Day, Month and Year functions, zero corresponds to december 31, 1899.

I also used the fact that (at least for most locales), VBA date strings have 10 characters and time strings have eight...

Thanks! This was a funny challenge.

Answer 49

Hoon, 89 bytes

`@`&
`@`|
+(|)
+:!=(+)
+:!=(-<)
+:!=(->)
+:!=(+<)
+:!=(+>)
+:!=(-<-)
+:!=(-<+)
+:!=(->-)
+:!=(->->-)

0 is just "yes" (0) casted to an atom

1 is just "no" (1) casted to an atom

2 is incremented "no" - .+ automatically casts to an atom, so no need for @

3 - 42 are the fun ones: It's using the != rune, which compiles an arbitrary Hoon expression to Nock. In this case, it's compiling axis navigation syntax into the Nock expression [0 axis], which I then take the tail of with +:expr.

In Hoon, all variable accesses are actually just indexing into a binary tree, and it allows you to index manually as well. The syntax is alternating sets of -/+ and </>, each one being head/tails for the current node. I also use +axis, which is a bit shorter for 8, 9, and 42.

Answer 50

Haskell, 295 bytes

0: length[]
1: length[()]
2: length[(),()]
3: length[(),(),()]
4: length[(),(),(),()]
5: length[(),(),(),(),()]
6: length[(),(),(),(),(),()]
7: length[(),(),(),(),(),(),()]
8: length[(),(),(),(),(),(),(),()]
9: length[(),(),()]*length[(),(),()]
10: length[(),()]*length[(),(),(),(),()]
42: length[(),(),(),(),(),()]*length[(),(),(),(),(),(),()]

Answer 51

R, 100 bytes

Copied the solutions for 6 and 7 from @MiloChen here.

+F                   # | 0     | 2 bytes
+T                   # | 1     | 2 bytes
T+T                  # | 2     | 3 bytes
T+T+T                # | 3     | 5 bytes
T+T+T+T              # | 4     | 7 bytes
T+T+T+T+T            # | 5     | 9 bytes
nrow(BOD)            # | 6     | 9 bytes
cars$s[pi]           # | 7     | 10 bytes
cars$s[pi]+T         # | 8     | 12 bytes
(pi*pi)%/%T          # | 9     | 11 bytes
(pi*pi)%/%T+T        # | 10    | 13 bytes
(pi^pi+pi+pi)%/%T    # | 42    | 17 bytes
                     # | total | 100 bytes

This saves some bytes compared to the 123 and 165 solutions.

Answer 52

Jolf, 26 bytes

<empty>
lέ
lq
mfπ
Uά
½t
mfτ
mΓCt
wwt
wt
t
mQ

Okay, so I have a 42-builtin. I'm obsessed with that number. Also for forty-two, lm,t: the length of the integer partitions of ten.

Explanations

The empty program is a truth machine, and, since given no input, just output's a zero.

lέ is the length of a tab, i.e., 1.

lq is the length of the source code, i.e., 2.

mfπ is the floor of pi, i.e., 3.

Uά is the square root of 16, i.e., 4.

½t is half ten, i.e., 5.

mfτ is the floor of tau (i.e. twice pi), i.e., 6

mΓCt is the length of the collatz sequence beginning with ten, i.e. 7.

wwt is 10-1-1, i.e., 8.

wt is 10-1, i.e., 9.

t is 10.

mQ is the builtin for 42.

---

Alternate solution

Except for 0 and 10, a solution with @<char with charcode value of N> would be golfier solution for N, however, even though it doesn't use any literals (it actually reads the source code's next character), it isn't in the spirit of the challenge. This would tally 20 bytes.

Answer 53

Ruby, 194 bytes

[]<=>[]
[[]].size
[[],[]].size
Dir.name.size
Hash.name.size
Float.name.size
Method.name.size
Integer.name.size
NilClass.name.size
TrueClass.name.size
FalseClass.name.size
Integer.name.size*Method.name.size

Answer 54

Perl6 (117 112) no state

e-e
e/e
e.floor
e.round
pi.ceiling
floor e+e
tau.floor
tau.ceiling
floor pi*e
floor pi*pi
round pi*pi
tau.floor*tau.ceiling

Answer 55

BASH, 318 bytes

0 to 3 are exit code, other ones are stored in A

bytes count is split as follow :2-5-5-26-23-32-26-38-26-33-32-70

#0
ls
#1
cd \\
#2
ls \\
#3
cd \\
A=$?
A=$(($A+$A+$A))
#4
ls \\
A=$?
A=$(($A*$A))
#5
cd \\
A=$?
A=$(($A+$A+$A+$A+$A))
#6
ls \\
A=$?
A=$(($A+$A+$A))
#7
cd \\
A=$?
A=$(($A+$A+$A+$A+$A+$A+$A))
#8
ls \\
A=$?
A=$(($A*$A*$A))
#9
A=$?
A=$(($A+$A+$A))
#10
ls \\
A=$?
A=$(($A+$A+$A+$A+$A))
#42
@
A=$?
cd \\
B=$?
cd \\
C=$?
C=$(($C+$C+$C))
B=$(($A-$B))
A=$(($B/$C))

7 is more than probably golfable a bit more, as is 42

Answer 56

Racket 205 bytes

(- pi pi)
(/ pi pi)
(/(+ pi pi)pi)
(floor pi)
(ceiling pi)
(sub1(floor(+ pi pi)))
(floor(+ pi pi))
(ceiling(+ pi pi))
(floor(sub1(* pi pi)))
(floor(* pi pi))
(ceiling(* pi pi))
(floor(+ pi pi(expt pi pi)))

Output:

0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
42.0

Answer 57

Haskell, 133 bytes

Pi-based, three floating point answers with the rest being integers:

 0: (5)  pi-pi
 1: (5)  pi/pi
 2: (10) (pi+pi)/pi
 3: (8)  round pi
 4: (10) ceiling pi
 5: (15) succ.ceiling$pi
 6: (11) floor$pi+pi
 7: (13) ceiling$pi+pi
 8: (16) floor$cosh pi-pi
 9: (11) floor$pi*pi
10: (11) round$pi*pi
42: (18) floor$pi**pi+pi+pi

Answer 58

SmileBASIC, 73 71 bytes

#NO 'unknown use (maybe dialog box results?)
#UP 'd-pad up
#CHKZ 'used in SPCHK and BGCHK to get the z coordinate
#TRED 'text color red
#LEFT 'd-pad left
#BQLSF 'biquad low shelf filter
#BQHSF 'biquad high shelf filter
#BQPEQ 'biquad peaking equalizer
#CHKI 'used in SPCHK to check definition number
#TBLUE 'text color blue
#TPURPLE 'text color purple
#BQHSF*#BQPEQ '42

Answer 59

tcl, 376

$argc
incr i
llength [list {} {}]
llength [list {} {} {}]
llength [list {} {} {} {}]
llength [list {} {} {} {} {}]
llength [list {} {} {} {} {} {}]
llength [list {} {} {} {} {} {} {}]
llength [list {} {} {} {} {} {} {} {}]
llength [list {} {} {} {} {} {} {} {} {}]
llength [list {} {} {} {} {} {} {} {} {} {}]
expr [llength [list {} {} {} {} {} {} {}]]*[llength [list {} {} {} {} {} {}]]

To try it, on the first line just prepend puts and in the others wrap them by puts [ before and ] after.

demo

Answer 60

Sinclair ZX81/Timex TX1000/1500 BASIC

After some discussion with Adám below, you have the following on the ZX81** (0 to 10 inclusive and 42), where explanations may be necessary, see the relevant note in the square brackets:

[1] .
[2] NOT .
(PI+PI)/PI
INT PI
INT PI+NOT .
INT (PI+PI)-NOT .
INT (PI+PI)
INT (PI+PI)+NOT .
INT (PI*PI+COS PI)
LEN STR$ PI
INT (PI*PI)+NOT .
[3] INT (PI**PI+PI+PI)

Notes:

1. This is a short-hand of 0.0
2. Not zero is 1, this is two keystrokes (so two bytes), which is shorter then PI/PI (three keystrokes)
3. ** is to the power of, so it's the integer value of PI to the value of PI plus PI+PI

To see this in action, you can type it in as follows using Direct or Online mode in ZX BASIC:

PRINT .,NOT .,(PI+PI)/PI,INT PI,INT PI+NOT .,INT (PI+PI)-NOT .,INT (PI+PI),INT (PI+PI)+NOT .,INT (PI*PI+ COS PI),LEN STR$ PI,INT (PI*PI)+NOT .,INT (PI**PI+PI+PI)

Here is a more evil listing that I came up with that prints 1 to 10 inclusive (try typing it into a ZX81 emulator - actually don't):

1 PRINT LEN STR$ USR VAL (STR$ INT (EXP PI+PI)+STR$ (PI-PI)+STR$ (PI/PI+(PI/PI)))
2 PRINT (PI+PI)/PI
3 PRINT INT PI
4 PRINT INT (SGN PI+PI)
5 PRINT INT PI+(PI/PI)
6 PRINT INT (PI+PI)
7 PRINT INT ((PI+PI)+((PI/PI)))
8 PRINT INT (PI+PI+PI-SGN (PI))
9 PRINT LEN STR$ PI
10 PRINT INT (PI*PI)+PI/PI

Answer 61

Lua 5.3, 146 bytes

Uses a lot of length operators. Version probably doesn't matter but it is dependent on the value of _VERSION so I might as well include it for clarity.

0. #{}
1. #{_G}
2. #{_G}+#{_G}
3. #type(x)
4. #type{}-#{_G}
5. #type{}
6. #type{}+#{_G}
7. #_VERSION
8. #type(load)
9. #type(load)+#{_G}
10. #_VERSION+#type(x)
42. - -(#type{}-#{_G}..#{_G}+#{_G})

Annotated with explainations:

0. #{}                 -- {} is an empty table, # is the length operator, length of empty table is 0.
1. #{_G}               -- {_G} is a table containing the global environment (itself a table), so there is one item.
2. #{_G}+#{_G}         -- 1+1
3. #type(x)            -- x is not set and so defaults to nil, type(nil) returns the string "nil", length of "nil" is 3.
4. #type{}-#{_G}       -- 5-1
5. #type{}             -- type of table is "table", length of "table" is 5, brackets not required when only argument is a table literal.
6. #type{}+#{_G}       -- 5+1
7. #_VERSION           -- _VERSION contains the current version as a string ("Lua 5.3"), of length 7.
8. #type(load)         -- type of 'load' is "function", length of "function" is 8.
9. #type(load)+#{_G}   -- 8+1
10. #_VERSION+#type(x) -- 7+3
42. - -(#type{}-#{_G}..#{_G}+#{_G}) -- concatenation operator implicitly converts numerical operands to strings, resulting in "4".."2"=="42". Double negation converts it to a number.

Answer 62

Japt, 22 bytes

0:  T       T == 0
1:  v       Implicit input 0; Number.v gives 1 if it's even, 0 otherwise
2:  y       Number.y(other) gives GCD of two numbers; `other` defaults to 2
3:  D-A     A == 10, D == 13; 13-10
4:  Gq      G == 16; 16.sqrt()
5:  Az      Number.z(other) gives this/other rounded towards 0; `other` defaults to 2
6:  Cz      C == 12
7:  Ez      E == 14
8:  Iq      I == 64
9:  AÉ      É is a shortcut for `-1`
10: A       Obviously.
42: H+A     H == 32; 32+10

Try it online: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 42

Answer 63

MathGolf, 20 bytes

p   # 0
î   # 1
⌠   # 2
d±  # 3
☻√  # 4
♂½  # 5
B½  # 6
D½  # 7
☻½  # 8
d²  # 9
♂   # 10
J∞  # 42

Try it online.

Explanation:

p   # Print the top of the stack (or 0 if there is no input, nor anything on the stack)
î   # 1-indexed loop index (1 by default)
⌠   # + 2 (which becomes 0+2 by default)
d±  # Push -3, and then pop and push its absolute value
☻√  # Push 16, square-rooted
♂½  # Push 10, halved
B½  # Push 12, halved
D½  # Push 14, halved
F½  # Push 16, halved
d²  # Push -3, squared
♂   # Push 10
J∞  # Push 21, doubled

Of course there any many alternatives for some.. i.e. 6 can alternatively be X√; τi; ╟x; M¼; C½; N¼; O¼; P¼; etc. (Try it online).

Answer 64

dc, 44

Score does not include newlines between each snippet.

If CJam's U, X, Y, Z, and A constants are allowed, then I assume dc's representation of 10-15 as A-F is also OK.

z
zZ
CI-
DI-
EI-
FI-
II+E-
II+D-
II+C-
II+B-
A
IB+CI-*

@yeti's answer is better.

Answer 65

beeswax, 50 46 bytes

(62 57 bytes, if IPs are not initialized)

Explanation of instructions:

P    increment top lstack value
M    decrement top lstack value
z    all lstack values = 0
F    all lstack values = top
G    get value from absolute(row,column) = lstack(2nd,3rd) value.
@    flip 1st and 3rd lstack values
.    top=top*2nd value
+    top=top+2nd value
!    bitwise NOT
(bel) ASCII control character 7, not printable

Instructions for number creation without using numbers, with multiple examples, if the byte lengths are the same. Prepend z if the bee is not in an initialized state:

 initialized  to 0                         not initialized to 0
                                                  prepend z
no instruction              [0,0,0]•       z
P                           [0,0,1]•       zP
PP                          [0,0,2]•       zPP
PPP                         [0,0,3]•       zPPP
PPF. or PPPP or PPF+        [0,0,4]•       zPPF.
PPF.P or PPF+P or PPPPP     [0,0,5]•       zPPF.P
PPPF+ or PPF++ or PPFP.     [0,0,6]•       zPPPF+
PPPF+P or PPF..M            [0,0,7]•       zPPF..M
(bel)*1FG                   [0,0,7]•  ← this would make ‘7’ one byte shorter
PPF..                       [0,0,8]•       zPPF..
PPPF.                       [0,0,9]•       zPPPF.
PPPF.P                      [0,0,10]•      zPPPF.P
*PFG                        [0,0,42]•      *PFG (in this configuration the IP is always initialized)

You can clone my beeswax GitHub repository here.

Answer 66

Staq, 60 bytes

&              0
&i             1
&ii            2
&iii           3
&iiq           4
&iiqi          5
&iiqii         6
&iiiqdd        7
&iiiqd         8
&iiiq          9
&iiiqi        10
&iiqii&+qX+   42

Instructions:

& push 0 on top of stack
i increment top stack value
d decrement top stack value
q square top stack value
+ add top and 2nd stack values, put sum on top of stack
X delete 2nd stack value

Answer 67

Cardinal, 50 bytes

Instructions, with alternatives if they have the same lengths:

0:  IPs are initialized to 0, otherwise, use instruction 0
1:  +
2:  ++
3:  +++
4:  ++++     or ++=*    or ++=t
5:  +++++    or ++=t+   or ++=*+
6:  +++=*    or ++=**
7:  +++=*+   or ++=tt-
8:  ++=tt
9:  +++=t
10: ++=tt*   or +++=t+
42: +++=*=t*

Explanation:

IPs in Cardinal carry length 2 stacks around. The topmost value is called active, the bottom value is called passive. Both can switch roles.

+ increment active value
- decrement active value
* addition : active=active+passive
t multiplication: active=active*passive
= set passive = active value

Answer 68

Actually, 32 bytes (non-competing)

╜
╒K
 ⌐
╦L
╦K
╦L⌐
╦Lτ
╠²L
╦L╙
╦L²
kY╤
╒º╦+L

All full programs. Try it online!

Explanation

• ╒ is ln(2)
• ╦ is pi
• ╠ is e
• L is floor
• K is ceil
• τ is double
• ² is square

For the last one:

╒º╦+L
╒      ln(2)
 º     convert from radian to degree (39.71440802747729)
  ╦+   + pi (42.85600068106709)
    L  floor

Answer 69

Jelly, 26 bytes (non-competing)

00: 
01: ‘
02: ‘‘
03: ‘‘‘
04: ‘‘Ḥ
05: ⁵H
06: ⁵H‘
07: ⁴H’
08: ⁴H
09: ⁵’
10: ⁵
42: ⁴Ḥ+⁵

Try it online! (paste each one of the snippets in the 'Code' field.)

Answer 70

Brain-Flak, 172 143 bytes (non-competing)

(<>)
(())
(()())
(()()())
(()()()())
(()()()()())
((()()()){})
((()()()){}())
((()()()()){})
(((()()())){}{})
((()()()()()){})
(((((()()()){}()){})){}{})

Try it online! (copy-paste yourself)

Answer 71

QBIC, 14 bytes (non-competing)

Thank you @Adám, for clarifying part of the rules. Brought the byecount down to 14.

This one's easy in QBIC, but marked non-competing because (this part of) QBIC was made after the challenge was posted.

a
q
r
s
t
u
v
w
x
y
z
v*w

All lower-case letters are numerical variables in QBIC, which default to 0. Exceptions are the letters q-z, which get initialised to 1-10. 42 is simply 6*7.

Answer 72

Ruby, 149 147 bytes

$.
-~$.
-~-~$.
-~-~-~$.
-~-~-~-~$.
-~-~-~-~-~$.
-~-~-~$.>>~$.
-~-~-~-~-~-~-~$.
-~$.>>~-~-~$.
-~$.>>~-~-~$.|-~$.
-~-~-~-~-~$.>>~$.
-~-~-~-~-~$.>>~-~-~$.|-~-~$.

Explanation:

$.is the number of the last line read from the current input file. Since we are not reading any file, this is initialized with 0.

1 through 5 are just the previous number +1, then I can use shifts for multiplication (negative right shift saves 1 byte) and or for addition (can't use +/- because of priority).

So, for example, 42 is 5>>-3|2

Answer 73

Excel VBA, 59 bytes

Enums from the Excel object available to VBA. n is said to be an uninitialized variable and thus holds the default value of 0 as a Variant.

These snippets may be called to the VBE immediate window by prepending with ?. e.g. ?xlOr.

Further information on each response is commented to the right

00: -n          ''  2 bytes;  Uninitialized var `n` defaults to `0`
01: n^n         ''  3 bytes;  0^0=1
02: xlOr        ''  4 bytes;  Excel.XlAutoFilterOperator.xlOr
03: xlSet       ''  5 bytes;  Excel.XlCubeFieldType.xlSet
04: xlWKS       ''  5 bytes;  Excel.XlFileFormat.xlWKS
05: xlDXF       ''  5 bytes;  Excel.XlPictureConvertorType.xlDXF
06: xlCSV       ''  5 bytes;  Excel.XlFileFormat.xlCSV
07: xlCGM       ''  5 bytes;  Excel.XlPictureConvertorType.xlCGM
08: xlEPS       ''  5 bytes;  Excel.XlPictureConvertorType.xlEPS
09: xlTIF       ''  5 bytes;  Excel.XlPictureConvertorType.xlTIF
10: xlPCX       ''  5 bytes;  Excel.XlPictureConvertorType.xlPCX
42: xlWJ3+xlOr  ''  10 bytes; Excel.XlFileFormat.xlWJ3+Excel.XlAutoFilterOperator.xlOr

Answer 74

Husk, 42 bytes

With string and character literals forbidden as well as numbers, all Husk has to work with are sequence builtins: N (all natural numbers), ø (empty list), and the İ family. (And functions which never get evaluated, on account of being wrapped, but that's a bit far from useful.)

Lø        Length of the empty list.
¬ø        Negation of the empty list.
←İp       First prime. I wanted to use \. here, but even though there are no digits . is a (deliberate) corner case of numeric literal syntax and not a special constant.
←İπ       First digit of pi.
□←tN      Second natural number, squared.
←İ5       First power of 5. (Power sequences only exist for 2, 3, 5, and 7; the digits are arbitrary labels.)
D←İ3      First power of 3, times 2.
←İ7       First power of 7.
½LΘİ€     Length of the list of Euro denominations with a zero tacked on to the beginning, over 2.
□←İπ      First digit of pi, squared.
←İ⁰       First power of 10.
D→D←İ⁰    First power of 10, times 2, plus 1, times 2.

Answer 75

Java 8, 300 bytes

My only goal was to get every value using only java.lang.Math and integer casting. Because of that, it's probably not the best answer in terms of golf. I have fun making it, however, and that's what counts :)

If I had more time, I would have gone further and only used the constants Math.PI and Math.E.

(int)(Math.E-Math.E)              // 0
(int)(Math.E/Math.E)              // 1
(int)Math.E                       // 2
(int)Math.PI                      // 3
(int)(Math.E/Math.E+Math.PI)      // 4
(int)(Math.E+Math.E)              // 5
(int)(Math.PI+Math.PI)            // 6
(int)(Math.E+Math.E+(int)Math.E)  // 7
(int)(Math.E+Math.E+Math.E)       // 8
(int)(Math.PI*Math.PI)            // 9
(int)Math.ceil(Math.PI*Math.PI)   // 10
(int)Math.ceil(Math.pow(Math.PI,Math.PI)+Math.E+Math.E) //42

TIO code to evaluate all of these at once

Answer 76

C, 78 bytes

Quite a handful of undefined behaviours.

The implementation is gcc (GCC) 5.3.1 20151207 (Red Hat 5.3.1-2), using this online C sandbox.

a-a                //1-1
a                  //1
a+a                //1+1
a+a+a              //1+1+1
++a*a              //2*2
++a+a++            //3+2
++a*a++            //3*2
a+++a+a+a          //1+2+2+2
++a<<a             //2<<2
++a*++a            //3*3
a+++a*++a          //1+3*3
(a=++a*a++)*a+a    //(a=3*2)*a+a

a defaults to 1 being the first argument of main.

Boilerplate example for the last one:

#include <stdio.h>
int main(int a)
{
    printf("%d\n",(a=++a*a++)*a+a);
    return 0;
}

---

without undefined behaviors, 88 bytes

a-a              //1-1
a                //1
a+a              //1+1
a+a+a            //1+1+1
a<<a+a           //1<<2
a<<a+a|a         //(1<<2)|1
a+a+a<<a         //3<<1
a+a+a<<a|a       //(3<<1)|1
a+a<<a+a         //2<<2
a+a<<a+a|a       //(2<<2)|1
a++;a<<a|a       //(2<<2)|2
a=a+a+a<<a;a*a+a //a=3<<1;a*a+a