Print the Lost numbers

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

As a big fan of the Lost TV series, I always got intrigued by the sequence of numbers that repetitively appears on the episodes. These numbers are:

$ 4, 8, 15, 16, 23, 42$ (A104101)

Using any programming language, write a code that outputs these numbers.

Scoring:

• Shortest answer wins




• The output must not contain any other numbers or letters. You may use any other character as separator, or even no separator at all.




• You cannot separate digits of the same number. $ 48_15162342 $ is a valid answer, but $481_5162342$ is not.




• You must respect the order.



• 
  

  If your code does not contain any of the numbers from the sequence, reduce your score by 30%. This rule does allow you to enter the digits separately. E.g.:

  
  
  

  abcde1fg5h
  

  
  
  

  Is a valid candidate because the answer does not contain the number $15$, only its digits. However, any $4$ or $8$ will invalidate the bonus.

  
  



• If the code does not contain any digit at all, reduce your score by 50%. Other characters like $¹$, $²$ or $³$ are still valid for this bonus.

number code-challenge integer

share|improve this question

edited 10 mins ago

Jo King

18.3k241100

asked 20 hours ago

Eduardo Hoefel

1769

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I'm tagging the win condition as code challenge, since the score is a function of code length and whether the code omits certain characters.
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we use no separator, the numbers will be run together like 4815162342. Is that OK?
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Related, closed question: codegolf.stackexchange.com/q/23808/67312
  – Giuseppe
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Vaguely related: Are you lost yet?
  – Dennis♦
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  For future reference, we have a restricted-source tag that could have been used here: although most answers are avoiding obvious solutions, I think the challenge would have been slightly more interesting if using digits were forbidden altogether.
  – Arnauld
  10 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

up vote
1
down vote

favorite

As a big fan of the Lost TV series, I always got intrigued by the sequence of numbers that repetitively appears on the episodes. These numbers are:

$ 4, 8, 15, 16, 23, 42$ (A104101)

Using any programming language, write a code that outputs these numbers.

Scoring:

• Shortest answer wins




• The output must not contain any other numbers or letters. You may use any other character as separator, or even no separator at all.




• You cannot separate digits of the same number. $ 48_15162342 $ is a valid answer, but $481_5162342$ is not.




• You must respect the order.



• 
  

  If your code does not contain any of the numbers from the sequence, reduce your score by 30%. This rule does allow you to enter the digits separately. E.g.:

  
  
  

  abcde1fg5h
  

  
  
  

  Is a valid candidate because the answer does not contain the number $15$, only its digits. However, any $4$ or $8$ will invalidate the bonus.

  
  



• If the code does not contain any digit at all, reduce your score by 50%. Other characters like $¹$, $²$ or $³$ are still valid for this bonus.

number code-challenge integer

share|improve this question

edited 10 mins ago

Jo King

18.3k241100

asked 20 hours ago

Eduardo Hoefel

1769

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I'm tagging the win condition as code challenge, since the score is a function of code length and whether the code omits certain characters.
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we use no separator, the numbers will be run together like 4815162342. Is that OK?
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Related, closed question: codegolf.stackexchange.com/q/23808/67312
  – Giuseppe
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Vaguely related: Are you lost yet?
  – Dennis♦
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  For future reference, we have a restricted-source tag that could have been used here: although most answers are avoiding obvious solutions, I think the challenge would have been slightly more interesting if using digits were forbidden altogether.
  – Arnauld
  10 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

up vote
1
down vote

favorite

up vote
1
down vote

favorite

As a big fan of the Lost TV series, I always got intrigued by the sequence of numbers that repetitively appears on the episodes. These numbers are:

$ 4, 8, 15, 16, 23, 42$ (A104101)

Using any programming language, write a code that outputs these numbers.

Scoring:

• Shortest answer wins




• The output must not contain any other numbers or letters. You may use any other character as separator, or even no separator at all.




• You cannot separate digits of the same number. $ 48_15162342 $ is a valid answer, but $481_5162342$ is not.




• You must respect the order.



• 
  

  If your code does not contain any of the numbers from the sequence, reduce your score by 30%. This rule does allow you to enter the digits separately. E.g.:

  
  
  

  abcde1fg5h
  

  
  
  

  Is a valid candidate because the answer does not contain the number $15$, only its digits. However, any $4$ or $8$ will invalidate the bonus.

  
  



• If the code does not contain any digit at all, reduce your score by 50%. Other characters like $¹$, $²$ or $³$ are still valid for this bonus.

number code-challenge integer

share|improve this question

edited 10 mins ago

Jo King

18.3k241100

asked 20 hours ago

Eduardo Hoefel

1769

As a big fan of the Lost TV series, I always got intrigued by the sequence of numbers that repetitively appears on the episodes. These numbers are:

$ 4, 8, 15, 16, 23, 42$ (A104101)

Using any programming language, write a code that outputs these numbers.

Scoring:

• Shortest answer wins




• The output must not contain any other numbers or letters. You may use any other character as separator, or even no separator at all.




• You cannot separate digits of the same number. $ 48_15162342 $ is a valid answer, but $481_5162342$ is not.




• You must respect the order.



• 
  

  If your code does not contain any of the numbers from the sequence, reduce your score by 30%. This rule does allow you to enter the digits separately. E.g.:

  
  
  

  abcde1fg5h
  

  
  
  

  Is a valid candidate because the answer does not contain the number $15$, only its digits. However, any $4$ or $8$ will invalidate the bonus.

  
  



• If the code does not contain any digit at all, reduce your score by 50%. Other characters like $¹$, $²$ or $³$ are still valid for this bonus.

number code-challenge integer

number code-challenge integer

share|improve this question

edited 10 mins ago

Jo King

18.3k241100

asked 20 hours ago

Eduardo Hoefel

1769

share|improve this question

edited 10 mins ago

Jo King

18.3k241100

asked 20 hours ago

Eduardo Hoefel

1769

share|improve this question

share|improve this question

edited 10 mins ago

Jo King

18.3k241100

edited 10 mins ago

Jo King

18.3k241100

edited 10 mins ago

Jo King

18.3k241100

18.3k241100

asked 20 hours ago

Eduardo Hoefel

1769

asked 20 hours ago

Eduardo Hoefel

1769

asked 20 hours ago

Eduardo Hoefel

1769

1769

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I'm tagging the win condition as code challenge, since the score is a function of code length and whether the code omits certain characters.
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we use no separator, the numbers will be run together like 4815162342. Is that OK?
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Related, closed question: codegolf.stackexchange.com/q/23808/67312
  – Giuseppe
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Vaguely related: Are you lost yet?
  – Dennis♦
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  For future reference, we have a restricted-source tag that could have been used here: although most answers are avoiding obvious solutions, I think the challenge would have been slightly more interesting if using digits were forbidden altogether.
  – Arnauld
  10 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I'm tagging the win condition as code challenge, since the score is a function of code length and whether the code omits certain characters.
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If we use no separator, the numbers will be run together like 4815162342. Is that OK?
  – xnor
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Related, closed question: codegolf.stackexchange.com/q/23808/67312
  – Giuseppe
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Vaguely related: Are you lost yet?
  – Dennis♦
  20 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  For future reference, we have a restricted-source tag that could have been used here: although most answers are avoiding obvious solutions, I think the challenge would have been slightly more interesting if using digits were forbidden altogether.
  – Arnauld
  10 hours ago
  
  

  
  
  
  

3

3

I'm tagging the win condition as code challenge, since the score is a function of code length and whether the code omits certain characters.
– xnor
20 hours ago

I'm tagging the win condition as code challenge, since the score is a function of code length and whether the code omits certain characters.
– xnor
20 hours ago

1

1

If we use no separator, the numbers will be run together like 4815162342. Is that OK?
– xnor
20 hours ago

If we use no separator, the numbers will be run together like 4815162342. Is that OK?
– xnor
20 hours ago

2

2

Related, closed question: codegolf.stackexchange.com/q/23808/67312
– Giuseppe
20 hours ago

Related, closed question: codegolf.stackexchange.com/q/23808/67312
– Giuseppe
20 hours ago

2

2

Vaguely related: Are you lost yet?
– Dennis♦
20 hours ago

Vaguely related: Are you lost yet?
– Dennis♦
20 hours ago

3

3

For future reference, we have a restricted-source tag that could have been used here: although most answers are avoiding obvious solutions, I think the challenge would have been slightly more interesting if using digits were forbidden altogether.
– Arnauld
10 hours ago

For future reference, we have a restricted-source tag that could have been used here: although most answers are avoiding obvious solutions, I think the challenge would have been slightly more interesting if using digits were forbidden altogether.
– Arnauld
10 hours ago

 | 
show 3 more comments

20 Answers
20

active

oldest

votes

up vote
11
down vote

Lost, 29 27/2 = 13.5 bytes

%?>>>>>>>>>>
>>"*"@"

Try it online! or verify that it is deterministic

Seemed like the right language to use.

Explanation:

Lost is a 2D language where the pointer starts anywhere, going in any direction. This generally leads to a lot of double checking that the pointer hasn't entered a section early.

...>>>>>>>>>> These arrows filter all pointers that appear on the top line
............. Or going vertically


%............ This flips the flag so that the program can end
............. This stops premature termination

.?.......... Clear the stack by skipping if a value popped from the stack is positive
............. When the stack is empty, the directs the pointer down

............. The directs the pointer right
.."*".. The string literal pushes all the Lost values to the stack

............ The @ terminates the program if the % flag is switched
>>........@. Otherwise it clears the stack and repeats

............. The quote here is to prevent the pointer getting stuck
............" This happens when the pointer starts between the other quotes

share|improve this answer

edited 12 hours ago

answered 20 hours ago

Jo King

18.3k241100

add a comment | 

up vote
5
down vote

Jelly, 7/2 = 3.5 bytes

“ƲÞIȥ’Ḥ

Prints the numbers without separator, i.e., the integer $4815162342$.

Try it online!

How it works

“ƲÞIȥ’ is bijective base-250 integer literal.
Ʋ, Þ, I, and È¥ have (1-based) indices $154$, $21$, $74$, and $171$ in Jelly's code page, so they encode the integer $250^3cdot154+250^2cdot21+250cdot74+171=2407581171$.

Finally, Ḥ (unhalve) doubles the integer, yielding $2cdot2407581171=4815162342$.

Doubling is necessary, because encoding the output directly leads to “¡9)Ƙ[’, which contains a digit.

share|improve this answer

edited 20 hours ago

answered 20 hours ago

Dennis♦

183k32293726

add a comment | 

up vote
5
down vote

Python 3, 29 bytes, 14.5 points

print(ord('𩦦')*ord('湡'))

Try it online!

𩦦 （⿰馬葬） is variant character of 髒 which means "dirty". 湡 is the name of a river. And they are nothing related to this question as I known.

share|improve this answer

edited 9 hours ago

answered 14 hours ago

tsh

8,03611346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I asked about separator rules and we can use any individual separator which makes 4815 162342 valid. Thus print(*map(ord,'ዏ𧨦')) saves 1.5 points :) (print(*map(ord,'밗')) would save 2 points but has been specified as invalid).
  – Jonathan Allan
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

05AB1E, score: 10 9 7 bytes / 2 = 3.5

•‘o]Ê•·

Try it online.

Or 7 bytes alternative:

•’µ[%•R

Try it online.

Both outputting the integer 4815162342.

Explanation:

•‘o]Ê• # Compressed integer 2407581171
 · # Doubled

•’µ[%• # Compressed integer 2432615184
 R # Reversed

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •‘o]Ê• is 2407581171 and •’µ[%• is 2432615184.

---

Old 9 bytes answer outputting the list [4,8,15,16,23,42]:

•ΓƒÇ²•т;в

-1 byte (and therefore -0.5 score) thanks to @Emigna.

Longer than the other 05AB1E answer, but this outputs the list [4,8,15,16,23,42] instead of the integer 4815162342.

Try it online.

Explanation:

•ΓƒÇ²• # Compressed integer 1301916192
 т; # Integer 50 (100 halved)
 в # Convert the first integer to Base-50 (arbitrary): [4,8,15,16,23,42]

See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer-lists?) to understand why •ΓƒÇ²• is 1301916192, and •ΓƒÇ²•50в is [4,8,15,16,23,42].

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

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  1
  

  
  
  
  
  
  

  
  You could have •ΓƒDz•т;в for 4.5 as post-script numbers are okay for the bonus.
  – Emigna
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, nice! Thanks.
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Charcoal, 13 bytes / 2 = 6.5

ＩＥTPIHA.⁻⁸⁸℅ι

Try it online! Link is to verbose version of code. Works by subtracting the ASCII codes of the string TPIHA. from 88 and casting to string.

share|improve this answer

answered 20 hours ago

Neil

77.3k744174

add a comment | 

up vote
2
down vote

Brain-Flak, 52/2 == 26 bytes

(((((((((()()()()))))[()])())()()()))[]())

Try it online!

share|improve this answer

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

add a comment | 

up vote
2
down vote

05AB1E, 6*0.7 = 4.2 bytes

•1Z&ð“

Try it online!

Prints the number uncompressed from base-255

share|improve this answer

answered 14 hours ago

Emigna

44.2k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You beat me to it. ;)
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

JavaScript (ES7), 34/2 = 17 bytes

_=>eval(atob`NjUwNTgxMDArNDEqKjY`)

Try it online!

This decodes and evaluates the expression "65058100+41**6", which does not contain any digit once encoded in base-64.

$$65058100+41^6=65058100+4750104241=4815162341$$

---

JavaScript (ES6), 13 bytes

Boring obvious solution.

_=>4815162341

Try it online!

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Arnauld

67.3k584283

add a comment | 

up vote
2
down vote

Java 8, score: 24 bytes / 2 = 12

v->(long)''*'Ⓥ'*'䧶'

Contains an unprintable character 0x1B.

Try it online.

Explanation:

v-> // Method with empty unused parameter and long return-type
 (long) // Cast the character (and therefore the result) to a long
 '' // 27
 *'Ⓥ' // Multiplied by 9419
 *'䧶' // Multiplied by 18934

In Java, characters can be autoboxed to integers holding their unicode value. Unfortunately, the maximum supported unicode for characters is 65,535, so I can't use just two characters to multiply (since the largest two numbers that divide the expected 4,815,162,342 are 56,802 and 84,771, where the 84,771 unfortunately exceeds the maximum 65,535.

In addition, since the maximum size of an int is 32²-1 (2,147,483,647) and the result 4,815,162,342 is larger than that, an explicit cast to long, which can hold up to 64²-1 (9,223,372,036,854,775,807), is required.

---

Boring answer would have been 14 bytes without any bonuses:

v->4815162341L

Try it online.

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I like this one. Pretty short for being Java :)
  – Emigna
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Thanks. It's too bad it requires the cast to long and doesn't support very large unicode characters. If it weren't for those two restrictions mentioned, just v->'𩦦'*'湡' (15 bytes, score 7.5) would have been enough. But it's still very short indeed. :) Although Java mostly has many, many weaknesses in terms of codegolfing (duhh..), calculating with characters because we aren't allowed to use digits is one of its few strengths. Was also pretty useful in this rather similar answer of mine.
  – Kevin Cruijssen
  11 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Neim, 6 5 bytes, 3 2.5 points

Jσς§A

Explanation:

J Push 48
 σ Push 15
 ς Push 16
 § Push 23
 A Push 42
 Implicitly join the contents 
 of the stack together and print

Try it online!

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Okx

12.2k27100

add a comment | 

up vote
1
down vote

JavaScript, 143 bytes (not sure how to score)

(g=`$2*2`)=>g.repeat(6).replace(/(.)/g,(m,p,i,k='')=>
 (k=m*[g-3,g-2,g,g,+g+2,g*3-1][i]
 ,RegExp(`$g-2|$g`).test(i)?k-1:i==+g+1?k-(g/2):k))

Try it online!

Start with six 4's, multiply, add, subtract by, to, from 4 to derive output.

share|improve this answer

answered 19 hours ago

guest271314

264211

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Ok, you were trying to get the bonus, but with a code this size, it didn't compensate. Why not simply '4815162342'?
  – Eduardo Hoefel
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoHoefel Do not gather the "score" or "bonus" system, did not try to get the bonus, just tried to not use any of the numbers required at output. The code outputs the numbers without hardcoding any of the numbers. The number 4, with addition, subtraction, multiplication and the index of the number 4 within a string (or array) can be used to derive the required numbers.
  – guest271314
  19 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your score is 143*0.7=100.1
  – Jo King
  19 hours ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

PHP, 35/2=17.5

<?=zzzzzzzzzzzzzzz^NVBVKOVKLVHIVNH;

a digital approach: 40*.7=28

<?=2+2,_,5+3,_,17-2,_,17-1,_,17+6,_,7*6;

no digits, no strings: 68/2 = 34

<?=$p++,!$p++,$p+=$p,_,$p+=$p,_,~-$q=$p+$p,_,$q,_,--$p+$q,_,$p*~-$p;

---

Try them online.

share|improve this answer

answered 15 hours ago

Titus

12.7k11237

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  1
  

  
  
  
  
  
  

  
  Or just 14 bytes for <?=4815162342;
  – Jo King
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  OP hasn´t replied to wether we can omit the delimiters or not; but yeah. Why not just 10 bytes: 4815162342. Or <?=~ + 10 unprintables -> 15/2=7.5
  – Titus
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Japt, 10 bytes / 2 = 5

"óT"mc w

Test it

share|improve this answer

answered 11 hours ago

Shaggy

17.6k21663

add a comment | 

up vote
1
down vote

JavaScript (SpiderMonkey), 67 bytes / 2 = 33.5 60 bytes / 2 = 30 58 bytes / 2 = 29 48 bytes / 2 = 24

-7 bytes/3.5, -2 bytes/1 courtesy of @JoKing, -10 bytes/5 courtesy of @tsh

print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)

Try it online!

share|improve this answer

edited 7 hours ago

answered 15 hours ago

guest271314

264211

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)
  – tsh
  14 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Or just print(4815162342) for 17 bytes
  – Jo King
  13 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Python 2, 16 bytes

print 4815162342

Try it online!

share|improve this answer

answered 16 hours ago

Vedant Kandoi

1166

add a comment | 

up vote
0
down vote

perl -M5.010 -Mre=eval, 32/2 == 16 bytes

This program is mostly unprintable characters -- characters which aren't even Unicode characters. Here is a hexdump of the program:

$ od -x solution
0000000 2727 7e3d 277e c0d7 8c84 869e cbd8 c7d3
0000020 ced3 d3ca c9ce cdd3 d3cc cdcb 82d8 27d6
0000040 
$

And here's the program to create the solution:

#!/opt/perl/bin/perl

use 5.026;

use strict;
use warnings;
no warnings 'syntax';

use experimental 'signatures';

my $q = ~"(?say'4,8,15,16,23,42')";
print "''=~~'$q'";

__END__

share|improve this answer

answered 8 hours ago

Abigail

39716

add a comment | 

up vote
0
down vote

Red, 50 bytes / 2 = 25

foreach c"abcdcefgaf"[prin index? find"cfgade.b"c]

Try it online!

Prints the numbers without separator

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Galen Ivanov

5,46211031

add a comment | 

up vote
0
down vote

APL(Dyalog Unicode), 18/2 = 9 bytes

×/⎕UCS'𩦦湡'

Just boring old character multiplication.

Try it online!

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Quintec

1,095517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The byte count is 18 in UTF-8. Neither 𩦦 nor 湡 exist in any APL code page, as far as I know.
  – Dennis♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis Thanks.
  – Quintec
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

Whitespace, score: 49 bytes / 2 = 24.5

[S S S T T S T T T S T T T T S S S T S N
_Push_56802][S S S T S T S S T S T T S S T S S S T T N
_Push_84771][T S S N
_Multiply][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Pseudo-code:

Integer i = 56802
Integer j = 84771
Integer k = i * j
Print k as number to STDOUT

share|improve this answer

answered 6 hours ago

Kevin Cruijssen

32.9k554176

add a comment | 

up vote
0
down vote

Z80Golf, 17 bytes * 0.5 = 8.5

00000000: 1b17 1e1a 1e19 1d1c 1b1d 2676 0a03 c5ee ..........&v....
00000010: 2f /

Try it online!

Assembly:

db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '8'	;17 rla
db 0x2F ^ '1'	;1e ld e,
db 0x2F ^ '5'	;1a 0x1a
db 0x2F ^ '1'	;1e ld e, 
db 0x2F ^ '6'	;19 0x19
db 0x2F ^ '2'	;1d dec e
db 0x2F ^ '3'	;1c inc e
db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '2'	;1d dec e
ld h, 0x76	;halt

ld a, (bc)	;load the next digit. The first char in addr in 0x0
inc bc		;get next digit
push bc		;return to the next digit which is basically a nop
xor 0x2F	;decode the digit
		;fall through into putchar. Putchar (0x8000), prints the char in register a

Assembly

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Logern

59535

add a comment | 

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20 Answers
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20

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oldest

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up vote
11
down vote

Lost, 29 27/2 = 13.5 bytes

%?>>>>>>>>>>
>>"*"@"

Try it online! or verify that it is deterministic

Seemed like the right language to use.

Explanation:

Lost is a 2D language where the pointer starts anywhere, going in any direction. This generally leads to a lot of double checking that the pointer hasn't entered a section early.

...>>>>>>>>>> These arrows filter all pointers that appear on the top line
............. Or going vertically


%............ This flips the flag so that the program can end
............. This stops premature termination

.?.......... Clear the stack by skipping if a value popped from the stack is positive
............. When the stack is empty, the directs the pointer down

............. The directs the pointer right
.."*".. The string literal pushes all the Lost values to the stack

............ The @ terminates the program if the % flag is switched
>>........@. Otherwise it clears the stack and repeats

............. The quote here is to prevent the pointer getting stuck
............" This happens when the pointer starts between the other quotes

share|improve this answer

edited 12 hours ago

answered 20 hours ago

Jo King

18.3k241100

add a comment | 

up vote
11
down vote

Lost, 29 27/2 = 13.5 bytes

%?>>>>>>>>>>
>>"*"@"

Try it online! or verify that it is deterministic

Seemed like the right language to use.

Explanation:

Lost is a 2D language where the pointer starts anywhere, going in any direction. This generally leads to a lot of double checking that the pointer hasn't entered a section early.

...>>>>>>>>>> These arrows filter all pointers that appear on the top line
............. Or going vertically


%............ This flips the flag so that the program can end
............. This stops premature termination

.?.......... Clear the stack by skipping if a value popped from the stack is positive
............. When the stack is empty, the directs the pointer down

............. The directs the pointer right
.."*".. The string literal pushes all the Lost values to the stack

............ The @ terminates the program if the % flag is switched
>>........@. Otherwise it clears the stack and repeats

............. The quote here is to prevent the pointer getting stuck
............" This happens when the pointer starts between the other quotes

share|improve this answer

edited 12 hours ago

answered 20 hours ago

Jo King

18.3k241100

add a comment | 

up vote
11
down vote

up vote
11
down vote

Lost, 29 27/2 = 13.5 bytes

%?>>>>>>>>>>
>>"*"@"

Try it online! or verify that it is deterministic

Seemed like the right language to use.

Explanation:

Lost is a 2D language where the pointer starts anywhere, going in any direction. This generally leads to a lot of double checking that the pointer hasn't entered a section early.

...>>>>>>>>>> These arrows filter all pointers that appear on the top line
............. Or going vertically


%............ This flips the flag so that the program can end
............. This stops premature termination

.?.......... Clear the stack by skipping if a value popped from the stack is positive
............. When the stack is empty, the directs the pointer down

............. The directs the pointer right
.."*".. The string literal pushes all the Lost values to the stack

............ The @ terminates the program if the % flag is switched
>>........@. Otherwise it clears the stack and repeats

............. The quote here is to prevent the pointer getting stuck
............" This happens when the pointer starts between the other quotes

share|improve this answer

edited 12 hours ago

answered 20 hours ago

Jo King

18.3k241100

Lost, 29 27/2 = 13.5 bytes

%?>>>>>>>>>>
>>"*"@"

Try it online! or verify that it is deterministic

Seemed like the right language to use.

Explanation:

Lost is a 2D language where the pointer starts anywhere, going in any direction. This generally leads to a lot of double checking that the pointer hasn't entered a section early.

...>>>>>>>>>> These arrows filter all pointers that appear on the top line
............. Or going vertically


%............ This flips the flag so that the program can end
............. This stops premature termination

.?.......... Clear the stack by skipping if a value popped from the stack is positive
............. When the stack is empty, the directs the pointer down

............. The directs the pointer right
.."*".. The string literal pushes all the Lost values to the stack

............ The @ terminates the program if the % flag is switched
>>........@. Otherwise it clears the stack and repeats

............. The quote here is to prevent the pointer getting stuck
............" This happens when the pointer starts between the other quotes

share|improve this answer

edited 12 hours ago

answered 20 hours ago

Jo King

18.3k241100

share|improve this answer

share|improve this answer

edited 12 hours ago

edited 12 hours ago

edited 12 hours ago

answered 20 hours ago

Jo King

18.3k241100

answered 20 hours ago

Jo King

18.3k241100

answered 20 hours ago

Jo King

18.3k241100

18.3k241100

add a comment | 

add a comment | 

up vote
5
down vote

Jelly, 7/2 = 3.5 bytes

“ƲÞIȥ’Ḥ

Prints the numbers without separator, i.e., the integer $4815162342$.

Try it online!

How it works

“ƲÞIȥ’ is bijective base-250 integer literal.
Ʋ, Þ, I, and È¥ have (1-based) indices $154$, $21$, $74$, and $171$ in Jelly's code page, so they encode the integer $250^3cdot154+250^2cdot21+250cdot74+171=2407581171$.

Finally, Ḥ (unhalve) doubles the integer, yielding $2cdot2407581171=4815162342$.

Doubling is necessary, because encoding the output directly leads to “¡9)Ƙ[’, which contains a digit.

share|improve this answer

edited 20 hours ago

answered 20 hours ago

Dennis♦

183k32293726

add a comment | 

up vote
5
down vote

Jelly, 7/2 = 3.5 bytes

“ƲÞIȥ’Ḥ

Prints the numbers without separator, i.e., the integer $4815162342$.

Try it online!

How it works

“ƲÞIȥ’ is bijective base-250 integer literal.
Ʋ, Þ, I, and È¥ have (1-based) indices $154$, $21$, $74$, and $171$ in Jelly's code page, so they encode the integer $250^3cdot154+250^2cdot21+250cdot74+171=2407581171$.

Finally, Ḥ (unhalve) doubles the integer, yielding $2cdot2407581171=4815162342$.

Doubling is necessary, because encoding the output directly leads to “¡9)Ƙ[’, which contains a digit.

share|improve this answer

edited 20 hours ago

answered 20 hours ago

Dennis♦

183k32293726

add a comment | 

up vote
5
down vote

up vote
5
down vote

Jelly, 7/2 = 3.5 bytes

“ƲÞIȥ’Ḥ

Prints the numbers without separator, i.e., the integer $4815162342$.

Try it online!

How it works

“ƲÞIȥ’ is bijective base-250 integer literal.
Ʋ, Þ, I, and È¥ have (1-based) indices $154$, $21$, $74$, and $171$ in Jelly's code page, so they encode the integer $250^3cdot154+250^2cdot21+250cdot74+171=2407581171$.

Finally, Ḥ (unhalve) doubles the integer, yielding $2cdot2407581171=4815162342$.

Doubling is necessary, because encoding the output directly leads to “¡9)Ƙ[’, which contains a digit.

share|improve this answer

edited 20 hours ago

answered 20 hours ago

Dennis♦

183k32293726

Jelly, 7/2 = 3.5 bytes

“ƲÞIȥ’Ḥ

Prints the numbers without separator, i.e., the integer $4815162342$.

Try it online!

How it works

“ƲÞIȥ’ is bijective base-250 integer literal.
Ʋ, Þ, I, and È¥ have (1-based) indices $154$, $21$, $74$, and $171$ in Jelly's code page, so they encode the integer $250^3cdot154+250^2cdot21+250cdot74+171=2407581171$.

Finally, Ḥ (unhalve) doubles the integer, yielding $2cdot2407581171=4815162342$.

Doubling is necessary, because encoding the output directly leads to “¡9)Ƙ[’, which contains a digit.

share|improve this answer

edited 20 hours ago

answered 20 hours ago

Dennis♦

183k32293726

share|improve this answer

share|improve this answer

edited 20 hours ago

edited 20 hours ago

edited 20 hours ago

answered 20 hours ago

Dennis♦

183k32293726

answered 20 hours ago

Dennis♦

183k32293726

answered 20 hours ago

Dennis♦

183k32293726

183k32293726

add a comment | 

add a comment | 

up vote
5
down vote

Python 3, 29 bytes, 14.5 points

print(ord('𩦦')*ord('湡'))

Try it online!

𩦦 （⿰馬葬） is variant character of 髒 which means "dirty". 湡 is the name of a river. And they are nothing related to this question as I known.

share|improve this answer

edited 9 hours ago

answered 14 hours ago

tsh

8,03611346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I asked about separator rules and we can use any individual separator which makes 4815 162342 valid. Thus print(*map(ord,'ዏ𧨦')) saves 1.5 points :) (print(*map(ord,'밗')) would save 2 points but has been specified as invalid).
  – Jonathan Allan
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
5
down vote

Python 3, 29 bytes, 14.5 points

print(ord('𩦦')*ord('湡'))

Try it online!

𩦦 （⿰馬葬） is variant character of 髒 which means "dirty". 湡 is the name of a river. And they are nothing related to this question as I known.

share|improve this answer

edited 9 hours ago

answered 14 hours ago

tsh

8,03611346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I asked about separator rules and we can use any individual separator which makes 4815 162342 valid. Thus print(*map(ord,'ዏ𧨦')) saves 1.5 points :) (print(*map(ord,'밗')) would save 2 points but has been specified as invalid).
  – Jonathan Allan
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

Python 3, 29 bytes, 14.5 points

print(ord('𩦦')*ord('湡'))

Try it online!

𩦦 （⿰馬葬） is variant character of 髒 which means "dirty". 湡 is the name of a river. And they are nothing related to this question as I known.

share|improve this answer

edited 9 hours ago

answered 14 hours ago

tsh

8,03611346

Python 3, 29 bytes, 14.5 points

print(ord('𩦦')*ord('湡'))

Try it online!

𩦦 （⿰馬葬） is variant character of 髒 which means "dirty". 湡 is the name of a river. And they are nothing related to this question as I known.

share|improve this answer

edited 9 hours ago

answered 14 hours ago

tsh

8,03611346

share|improve this answer

share|improve this answer

edited 9 hours ago

edited 9 hours ago

edited 9 hours ago

answered 14 hours ago

tsh

8,03611346

answered 14 hours ago

tsh

8,03611346

answered 14 hours ago

tsh

8,03611346

8,03611346

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I asked about separator rules and we can use any individual separator which makes 4815 162342 valid. Thus print(*map(ord,'ዏ𧨦')) saves 1.5 points :) (print(*map(ord,'밗')) would save 2 points but has been specified as invalid).
  – Jonathan Allan
  2 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I asked about separator rules and we can use any individual separator which makes 4815 162342 valid. Thus print(*map(ord,'ዏ𧨦')) saves 1.5 points :) (print(*map(ord,'밗')) would save 2 points but has been specified as invalid).
  – Jonathan Allan
  2 hours ago
  

  
  
  
  

I asked about separator rules and we can use any individual separator which makes 4815 162342 valid. Thus print(*map(ord,'ዏ𧨦')) saves 1.5 points :) (print(*map(ord,'밗')) would save 2 points but has been specified as invalid).
– Jonathan Allan
2 hours ago

I asked about separator rules and we can use any individual separator which makes 4815 162342 valid. Thus print(*map(ord,'ዏ𧨦')) saves 1.5 points :) (print(*map(ord,'밗')) would save 2 points but has been specified as invalid).
– Jonathan Allan
2 hours ago

add a comment | 

up vote
3
down vote

05AB1E, score: 10 9 7 bytes / 2 = 3.5

•‘o]Ê•·

Try it online.

Or 7 bytes alternative:

•’µ[%•R

Try it online.

Both outputting the integer 4815162342.

Explanation:

•‘o]Ê• # Compressed integer 2407581171
 · # Doubled

•’µ[%• # Compressed integer 2432615184
 R # Reversed

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •‘o]Ê• is 2407581171 and •’µ[%• is 2432615184.

---

Old 9 bytes answer outputting the list [4,8,15,16,23,42]:

•ΓƒÇ²•т;в

-1 byte (and therefore -0.5 score) thanks to @Emigna.

Longer than the other 05AB1E answer, but this outputs the list [4,8,15,16,23,42] instead of the integer 4815162342.

Try it online.

Explanation:

•ΓƒÇ²• # Compressed integer 1301916192
 т; # Integer 50 (100 halved)
 в # Convert the first integer to Base-50 (arbitrary): [4,8,15,16,23,42]

See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer-lists?) to understand why •ΓƒÇ²• is 1301916192, and •ΓƒÇ²•50в is [4,8,15,16,23,42].

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could have •ΓƒDz•т;в for 4.5 as post-script numbers are okay for the bonus.
  – Emigna
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, nice! Thanks.
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

05AB1E, score: 10 9 7 bytes / 2 = 3.5

•‘o]Ê•·

Try it online.

Or 7 bytes alternative:

•’µ[%•R

Try it online.

Both outputting the integer 4815162342.

Explanation:

•‘o]Ê• # Compressed integer 2407581171
 · # Doubled

•’µ[%• # Compressed integer 2432615184
 R # Reversed

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •‘o]Ê• is 2407581171 and •’µ[%• is 2432615184.

---

Old 9 bytes answer outputting the list [4,8,15,16,23,42]:

•ΓƒÇ²•т;в

-1 byte (and therefore -0.5 score) thanks to @Emigna.

Longer than the other 05AB1E answer, but this outputs the list [4,8,15,16,23,42] instead of the integer 4815162342.

Try it online.

Explanation:

•ΓƒÇ²• # Compressed integer 1301916192
 т; # Integer 50 (100 halved)
 в # Convert the first integer to Base-50 (arbitrary): [4,8,15,16,23,42]

See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer-lists?) to understand why •ΓƒÇ²• is 1301916192, and •ΓƒÇ²•50в is [4,8,15,16,23,42].

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could have •ΓƒDz•т;в for 4.5 as post-script numbers are okay for the bonus.
  – Emigna
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, nice! Thanks.
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

05AB1E, score: 10 9 7 bytes / 2 = 3.5

•‘o]Ê•·

Try it online.

Or 7 bytes alternative:

•’µ[%•R

Try it online.

Both outputting the integer 4815162342.

Explanation:

•‘o]Ê• # Compressed integer 2407581171
 · # Doubled

•’µ[%• # Compressed integer 2432615184
 R # Reversed

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •‘o]Ê• is 2407581171 and •’µ[%• is 2432615184.

---

Old 9 bytes answer outputting the list [4,8,15,16,23,42]:

•ΓƒÇ²•т;в

-1 byte (and therefore -0.5 score) thanks to @Emigna.

Longer than the other 05AB1E answer, but this outputs the list [4,8,15,16,23,42] instead of the integer 4815162342.

Try it online.

Explanation:

•ΓƒÇ²• # Compressed integer 1301916192
 т; # Integer 50 (100 halved)
 в # Convert the first integer to Base-50 (arbitrary): [4,8,15,16,23,42]

See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer-lists?) to understand why •ΓƒÇ²• is 1301916192, and •ΓƒÇ²•50в is [4,8,15,16,23,42].

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

05AB1E, score: 10 9 7 bytes / 2 = 3.5

•‘o]Ê•·

Try it online.

Or 7 bytes alternative:

•’µ[%•R

Try it online.

Both outputting the integer 4815162342.

Explanation:

•‘o]Ê• # Compressed integer 2407581171
 · # Doubled

•’µ[%• # Compressed integer 2432615184
 R # Reversed

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •‘o]Ê• is 2407581171 and •’µ[%• is 2432615184.

---

Old 9 bytes answer outputting the list [4,8,15,16,23,42]:

•ΓƒÇ²•т;в

-1 byte (and therefore -0.5 score) thanks to @Emigna.

Longer than the other 05AB1E answer, but this outputs the list [4,8,15,16,23,42] instead of the integer 4815162342.

Try it online.

Explanation:

•ΓƒÇ²• # Compressed integer 1301916192
 т; # Integer 50 (100 halved)
 в # Convert the first integer to Base-50 (arbitrary): [4,8,15,16,23,42]

See this 05AB1E tip of mine (sections How to compress large integers? and How to compress integer-lists?) to understand why •ΓƒÇ²• is 1301916192, and •ΓƒÇ²•50в is [4,8,15,16,23,42].

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

share|improve this answer

share|improve this answer

edited 11 hours ago

edited 11 hours ago

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

answered 12 hours ago

Kevin Cruijssen

32.9k554176

answered 12 hours ago

Kevin Cruijssen

32.9k554176

32.9k554176

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could have •ΓƒDz•т;в for 4.5 as post-script numbers are okay for the bonus.
  – Emigna
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, nice! Thanks.
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could have •ΓƒDz•т;в for 4.5 as post-script numbers are okay for the bonus.
  – Emigna
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, nice! Thanks.
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

1

1

You could have •ΓƒDz•т;в for 4.5 as post-script numbers are okay for the bonus.
– Emigna
12 hours ago

You could have •ΓƒDz•т;в for 4.5 as post-script numbers are okay for the bonus.
– Emigna
12 hours ago

@Emigna Ah, nice! Thanks.
– Kevin Cruijssen
12 hours ago

@Emigna Ah, nice! Thanks.
– Kevin Cruijssen
12 hours ago

add a comment | 

up vote
2
down vote

Charcoal, 13 bytes / 2 = 6.5

ＩＥTPIHA.⁻⁸⁸℅ι

Try it online! Link is to verbose version of code. Works by subtracting the ASCII codes of the string TPIHA. from 88 and casting to string.

share|improve this answer

answered 20 hours ago

Neil

77.3k744174

add a comment | 

up vote
2
down vote

Charcoal, 13 bytes / 2 = 6.5

ＩＥTPIHA.⁻⁸⁸℅ι

Try it online! Link is to verbose version of code. Works by subtracting the ASCII codes of the string TPIHA. from 88 and casting to string.

share|improve this answer

answered 20 hours ago

Neil

77.3k744174

add a comment | 

up vote
2
down vote

up vote
2
down vote

Charcoal, 13 bytes / 2 = 6.5

ＩＥTPIHA.⁻⁸⁸℅ι

Try it online! Link is to verbose version of code. Works by subtracting the ASCII codes of the string TPIHA. from 88 and casting to string.

share|improve this answer

answered 20 hours ago

Neil

77.3k744174

Charcoal, 13 bytes / 2 = 6.5

ＩＥTPIHA.⁻⁸⁸℅ι

Try it online! Link is to verbose version of code. Works by subtracting the ASCII codes of the string TPIHA. from 88 and casting to string.

share|improve this answer

answered 20 hours ago

Neil

77.3k744174

share|improve this answer

share|improve this answer

answered 20 hours ago

Neil

77.3k744174

answered 20 hours ago

Neil

77.3k744174

answered 20 hours ago

Neil

77.3k744174

77.3k744174

add a comment | 

add a comment | 

up vote
2
down vote

Brain-Flak, 52/2 == 26 bytes

(((((((((()()()()))))[()])())()()()))[]())

Try it online!

share|improve this answer

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

add a comment | 

up vote
2
down vote

Brain-Flak, 52/2 == 26 bytes

(((((((((()()()()))))[()])())()()()))[]())

Try it online!

share|improve this answer

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

add a comment | 

up vote
2
down vote

up vote
2
down vote

Brain-Flak, 52/2 == 26 bytes

(((((((((()()()()))))[()])())()()()))[]())

Try it online!

share|improve this answer

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

Brain-Flak, 52/2 == 26 bytes

(((((((((()()()()))))[()])())()()()))[]())

Try it online!

share|improve this answer

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

share|improve this answer

share|improve this answer

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

answered 16 hours ago

DJMcMayhem♦

40.5k11143307

40.5k11143307

add a comment | 

add a comment | 

up vote
2
down vote

05AB1E, 6*0.7 = 4.2 bytes

•1Z&ð“

Try it online!

Prints the number uncompressed from base-255

share|improve this answer

answered 14 hours ago

Emigna

44.2k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You beat me to it. ;)
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

05AB1E, 6*0.7 = 4.2 bytes

•1Z&ð“

Try it online!

Prints the number uncompressed from base-255

share|improve this answer

answered 14 hours ago

Emigna

44.2k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You beat me to it. ;)
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

05AB1E, 6*0.7 = 4.2 bytes

•1Z&ð“

Try it online!

Prints the number uncompressed from base-255

share|improve this answer

answered 14 hours ago

Emigna

44.2k431133

05AB1E, 6*0.7 = 4.2 bytes

•1Z&ð“

Try it online!

Prints the number uncompressed from base-255

share|improve this answer

answered 14 hours ago

Emigna

44.2k431133

share|improve this answer

share|improve this answer

answered 14 hours ago

Emigna

44.2k431133

answered 14 hours ago

Emigna

44.2k431133

answered 14 hours ago

Emigna

44.2k431133

44.2k431133

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You beat me to it. ;)
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You beat me to it. ;)
  – Kevin Cruijssen
  12 hours ago
  

  
  
  
  

You beat me to it. ;)
– Kevin Cruijssen
12 hours ago

You beat me to it. ;)
– Kevin Cruijssen
12 hours ago

add a comment | 

up vote
2
down vote

JavaScript (ES7), 34/2 = 17 bytes

_=>eval(atob`NjUwNTgxMDArNDEqKjY`)

Try it online!

This decodes and evaluates the expression "65058100+41**6", which does not contain any digit once encoded in base-64.

$$65058100+41^6=65058100+4750104241=4815162341$$

---

JavaScript (ES6), 13 bytes

Boring obvious solution.

_=>4815162341

Try it online!

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Arnauld

67.3k584283

add a comment | 

up vote
2
down vote

JavaScript (ES7), 34/2 = 17 bytes

_=>eval(atob`NjUwNTgxMDArNDEqKjY`)

Try it online!

This decodes and evaluates the expression "65058100+41**6", which does not contain any digit once encoded in base-64.

$$65058100+41^6=65058100+4750104241=4815162341$$

---

JavaScript (ES6), 13 bytes

Boring obvious solution.

_=>4815162341

Try it online!

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Arnauld

67.3k584283

add a comment | 

up vote
2
down vote

up vote
2
down vote

JavaScript (ES7), 34/2 = 17 bytes

_=>eval(atob`NjUwNTgxMDArNDEqKjY`)

Try it online!

This decodes and evaluates the expression "65058100+41**6", which does not contain any digit once encoded in base-64.

$$65058100+41^6=65058100+4750104241=4815162341$$

---

JavaScript (ES6), 13 bytes

Boring obvious solution.

_=>4815162341

Try it online!

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Arnauld

67.3k584283

JavaScript (ES7), 34/2 = 17 bytes

_=>eval(atob`NjUwNTgxMDArNDEqKjY`)

Try it online!

This decodes and evaluates the expression "65058100+41**6", which does not contain any digit once encoded in base-64.

$$65058100+41^6=65058100+4750104241=4815162341$$

---

JavaScript (ES6), 13 bytes

Boring obvious solution.

_=>4815162341

Try it online!

share|improve this answer

edited 11 hours ago

answered 11 hours ago

Arnauld

67.3k584283

share|improve this answer

share|improve this answer

edited 11 hours ago

edited 11 hours ago

edited 11 hours ago

answered 11 hours ago

Arnauld

67.3k584283

answered 11 hours ago

Arnauld

67.3k584283

answered 11 hours ago

Arnauld

67.3k584283

67.3k584283

add a comment | 

add a comment | 

up vote
2
down vote

Java 8, score: 24 bytes / 2 = 12

v->(long)''*'Ⓥ'*'䧶'

Contains an unprintable character 0x1B.

Try it online.

Explanation:

v-> // Method with empty unused parameter and long return-type
 (long) // Cast the character (and therefore the result) to a long
 '' // 27
 *'Ⓥ' // Multiplied by 9419
 *'䧶' // Multiplied by 18934

In Java, characters can be autoboxed to integers holding their unicode value. Unfortunately, the maximum supported unicode for characters is 65,535, so I can't use just two characters to multiply (since the largest two numbers that divide the expected 4,815,162,342 are 56,802 and 84,771, where the 84,771 unfortunately exceeds the maximum 65,535.

In addition, since the maximum size of an int is 32²-1 (2,147,483,647) and the result 4,815,162,342 is larger than that, an explicit cast to long, which can hold up to 64²-1 (9,223,372,036,854,775,807), is required.

---

Boring answer would have been 14 bytes without any bonuses:

v->4815162341L

Try it online.

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I like this one. Pretty short for being Java :)
  – Emigna
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Thanks. It's too bad it requires the cast to long and doesn't support very large unicode characters. If it weren't for those two restrictions mentioned, just v->'𩦦'*'湡' (15 bytes, score 7.5) would have been enough. But it's still very short indeed. :) Although Java mostly has many, many weaknesses in terms of codegolfing (duhh..), calculating with characters because we aren't allowed to use digits is one of its few strengths. Was also pretty useful in this rather similar answer of mine.
  – Kevin Cruijssen
  11 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

Java 8, score: 24 bytes / 2 = 12

v->(long)''*'Ⓥ'*'䧶'

Contains an unprintable character 0x1B.

Try it online.

Explanation:

v-> // Method with empty unused parameter and long return-type
 (long) // Cast the character (and therefore the result) to a long
 '' // 27
 *'Ⓥ' // Multiplied by 9419
 *'䧶' // Multiplied by 18934

In Java, characters can be autoboxed to integers holding their unicode value. Unfortunately, the maximum supported unicode for characters is 65,535, so I can't use just two characters to multiply (since the largest two numbers that divide the expected 4,815,162,342 are 56,802 and 84,771, where the 84,771 unfortunately exceeds the maximum 65,535.

In addition, since the maximum size of an int is 32²-1 (2,147,483,647) and the result 4,815,162,342 is larger than that, an explicit cast to long, which can hold up to 64²-1 (9,223,372,036,854,775,807), is required.

---

Boring answer would have been 14 bytes without any bonuses:

v->4815162341L

Try it online.

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I like this one. Pretty short for being Java :)
  – Emigna
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Thanks. It's too bad it requires the cast to long and doesn't support very large unicode characters. If it weren't for those two restrictions mentioned, just v->'𩦦'*'湡' (15 bytes, score 7.5) would have been enough. But it's still very short indeed. :) Although Java mostly has many, many weaknesses in terms of codegolfing (duhh..), calculating with characters because we aren't allowed to use digits is one of its few strengths. Was also pretty useful in this rather similar answer of mine.
  – Kevin Cruijssen
  11 hours ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Java 8, score: 24 bytes / 2 = 12

v->(long)''*'Ⓥ'*'䧶'

Contains an unprintable character 0x1B.

Try it online.

Explanation:

v-> // Method with empty unused parameter and long return-type
 (long) // Cast the character (and therefore the result) to a long
 '' // 27
 *'Ⓥ' // Multiplied by 9419
 *'䧶' // Multiplied by 18934

In Java, characters can be autoboxed to integers holding their unicode value. Unfortunately, the maximum supported unicode for characters is 65,535, so I can't use just two characters to multiply (since the largest two numbers that divide the expected 4,815,162,342 are 56,802 and 84,771, where the 84,771 unfortunately exceeds the maximum 65,535.

In addition, since the maximum size of an int is 32²-1 (2,147,483,647) and the result 4,815,162,342 is larger than that, an explicit cast to long, which can hold up to 64²-1 (9,223,372,036,854,775,807), is required.

---

Boring answer would have been 14 bytes without any bonuses:

v->4815162341L

Try it online.

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

Java 8, score: 24 bytes / 2 = 12

v->(long)''*'Ⓥ'*'䧶'

Contains an unprintable character 0x1B.

Try it online.

Explanation:

v-> // Method with empty unused parameter and long return-type
 (long) // Cast the character (and therefore the result) to a long
 '' // 27
 *'Ⓥ' // Multiplied by 9419
 *'䧶' // Multiplied by 18934

In Java, characters can be autoboxed to integers holding their unicode value. Unfortunately, the maximum supported unicode for characters is 65,535, so I can't use just two characters to multiply (since the largest two numbers that divide the expected 4,815,162,342 are 56,802 and 84,771, where the 84,771 unfortunately exceeds the maximum 65,535.

In addition, since the maximum size of an int is 32²-1 (2,147,483,647) and the result 4,815,162,342 is larger than that, an explicit cast to long, which can hold up to 64²-1 (9,223,372,036,854,775,807), is required.

---

Boring answer would have been 14 bytes without any bonuses:

v->4815162341L

Try it online.

share|improve this answer

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

share|improve this answer

share|improve this answer

edited 11 hours ago

edited 11 hours ago

edited 11 hours ago

answered 12 hours ago

Kevin Cruijssen

32.9k554176

answered 12 hours ago

Kevin Cruijssen

32.9k554176

answered 12 hours ago

Kevin Cruijssen

32.9k554176

32.9k554176

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I like this one. Pretty short for being Java :)
  – Emigna
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Thanks. It's too bad it requires the cast to long and doesn't support very large unicode characters. If it weren't for those two restrictions mentioned, just v->'𩦦'*'湡' (15 bytes, score 7.5) would have been enough. But it's still very short indeed. :) Although Java mostly has many, many weaknesses in terms of codegolfing (duhh..), calculating with characters because we aren't allowed to use digits is one of its few strengths. Was also pretty useful in this rather similar answer of mine.
  – Kevin Cruijssen
  11 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I like this one. Pretty short for being Java :)
  – Emigna
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Thanks. It's too bad it requires the cast to long and doesn't support very large unicode characters. If it weren't for those two restrictions mentioned, just v->'𩦦'*'湡' (15 bytes, score 7.5) would have been enough. But it's still very short indeed. :) Although Java mostly has many, many weaknesses in terms of codegolfing (duhh..), calculating with characters because we aren't allowed to use digits is one of its few strengths. Was also pretty useful in this rather similar answer of mine.
  – Kevin Cruijssen
  11 hours ago
  

  
  
  
  

1

1

I like this one. Pretty short for being Java :)
– Emigna
11 hours ago

I like this one. Pretty short for being Java :)
– Emigna
11 hours ago

@Emigna Thanks. It's too bad it requires the cast to long and doesn't support very large unicode characters. If it weren't for those two restrictions mentioned, just v->'𩦦'*'湡' (15 bytes, score 7.5) would have been enough. But it's still very short indeed. :) Although Java mostly has many, many weaknesses in terms of codegolfing (duhh..), calculating with characters because we aren't allowed to use digits is one of its few strengths. Was also pretty useful in this rather similar answer of mine.
– Kevin Cruijssen
11 hours ago

@Emigna Thanks. It's too bad it requires the cast to long and doesn't support very large unicode characters. If it weren't for those two restrictions mentioned, just v->'𩦦'*'湡' (15 bytes, score 7.5) would have been enough. But it's still very short indeed. :) Although Java mostly has many, many weaknesses in terms of codegolfing (duhh..), calculating with characters because we aren't allowed to use digits is one of its few strengths. Was also pretty useful in this rather similar answer of mine.
– Kevin Cruijssen
11 hours ago

add a comment | 

up vote
2
down vote

Neim, 6 5 bytes, 3 2.5 points

Jσς§A

Explanation:

J Push 48
 σ Push 15
 ς Push 16
 § Push 23
 A Push 42
 Implicitly join the contents 
 of the stack together and print

Try it online!

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Okx

12.2k27100

add a comment | 

up vote
2
down vote

Neim, 6 5 bytes, 3 2.5 points

Jσς§A

Explanation:

J Push 48
 σ Push 15
 ς Push 16
 § Push 23
 A Push 42
 Implicitly join the contents 
 of the stack together and print

Try it online!

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Okx

12.2k27100

add a comment | 

up vote
2
down vote

up vote
2
down vote

Neim, 6 5 bytes, 3 2.5 points

Jσς§A

Explanation:

J Push 48
 σ Push 15
 ς Push 16
 § Push 23
 A Push 42
 Implicitly join the contents 
 of the stack together and print

Try it online!

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Okx

12.2k27100

Neim, 6 5 bytes, 3 2.5 points

Jσς§A

Explanation:

J Push 48
 σ Push 15
 ς Push 16
 § Push 23
 A Push 42
 Implicitly join the contents 
 of the stack together and print

Try it online!

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Okx

12.2k27100

share|improve this answer

share|improve this answer

edited 6 hours ago

edited 6 hours ago

edited 6 hours ago

answered 6 hours ago

Okx

12.2k27100

answered 6 hours ago

Okx

12.2k27100

answered 6 hours ago

Okx

12.2k27100

12.2k27100

add a comment | 

add a comment | 

up vote
1
down vote

JavaScript, 143 bytes (not sure how to score)

(g=`$2*2`)=>g.repeat(6).replace(/(.)/g,(m,p,i,k='')=>
 (k=m*[g-3,g-2,g,g,+g+2,g*3-1][i]
 ,RegExp(`$g-2|$g`).test(i)?k-1:i==+g+1?k-(g/2):k))

Try it online!

Start with six 4's, multiply, add, subtract by, to, from 4 to derive output.

share|improve this answer

answered 19 hours ago

guest271314

264211

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Ok, you were trying to get the bonus, but with a code this size, it didn't compensate. Why not simply '4815162342'?
  – Eduardo Hoefel
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoHoefel Do not gather the "score" or "bonus" system, did not try to get the bonus, just tried to not use any of the numbers required at output. The code outputs the numbers without hardcoding any of the numbers. The number 4, with addition, subtraction, multiplication and the index of the number 4 within a string (or array) can be used to derive the required numbers.
  – guest271314
  19 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your score is 143*0.7=100.1
  – Jo King
  19 hours ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

JavaScript, 143 bytes (not sure how to score)

(g=`$2*2`)=>g.repeat(6).replace(/(.)/g,(m,p,i,k='')=>
 (k=m*[g-3,g-2,g,g,+g+2,g*3-1][i]
 ,RegExp(`$g-2|$g`).test(i)?k-1:i==+g+1?k-(g/2):k))

Try it online!

Start with six 4's, multiply, add, subtract by, to, from 4 to derive output.

share|improve this answer

answered 19 hours ago

guest271314

264211

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Ok, you were trying to get the bonus, but with a code this size, it didn't compensate. Why not simply '4815162342'?
  – Eduardo Hoefel
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoHoefel Do not gather the "score" or "bonus" system, did not try to get the bonus, just tried to not use any of the numbers required at output. The code outputs the numbers without hardcoding any of the numbers. The number 4, with addition, subtraction, multiplication and the index of the number 4 within a string (or array) can be used to derive the required numbers.
  – guest271314
  19 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your score is 143*0.7=100.1
  – Jo King
  19 hours ago
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript, 143 bytes (not sure how to score)

(g=`$2*2`)=>g.repeat(6).replace(/(.)/g,(m,p,i,k='')=>
 (k=m*[g-3,g-2,g,g,+g+2,g*3-1][i]
 ,RegExp(`$g-2|$g`).test(i)?k-1:i==+g+1?k-(g/2):k))

Try it online!

Start with six 4's, multiply, add, subtract by, to, from 4 to derive output.

share|improve this answer

answered 19 hours ago

guest271314

264211

JavaScript, 143 bytes (not sure how to score)

(g=`$2*2`)=>g.repeat(6).replace(/(.)/g,(m,p,i,k='')=>
 (k=m*[g-3,g-2,g,g,+g+2,g*3-1][i]
 ,RegExp(`$g-2|$g`).test(i)?k-1:i==+g+1?k-(g/2):k))

Try it online!

Start with six 4's, multiply, add, subtract by, to, from 4 to derive output.

share|improve this answer

answered 19 hours ago

guest271314

264211

share|improve this answer

share|improve this answer

answered 19 hours ago

guest271314

264211

answered 19 hours ago

guest271314

264211

answered 19 hours ago

guest271314

264211

264211

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Ok, you were trying to get the bonus, but with a code this size, it didn't compensate. Why not simply '4815162342'?
  – Eduardo Hoefel
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoHoefel Do not gather the "score" or "bonus" system, did not try to get the bonus, just tried to not use any of the numbers required at output. The code outputs the numbers without hardcoding any of the numbers. The number 4, with addition, subtraction, multiplication and the index of the number 4 within a string (or array) can be used to derive the required numbers.
  – guest271314
  19 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your score is 143*0.7=100.1
  – Jo King
  19 hours ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Ok, you were trying to get the bonus, but with a code this size, it didn't compensate. Why not simply '4815162342'?
  – Eduardo Hoefel
  19 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EduardoHoefel Do not gather the "score" or "bonus" system, did not try to get the bonus, just tried to not use any of the numbers required at output. The code outputs the numbers without hardcoding any of the numbers. The number 4, with addition, subtraction, multiplication and the index of the number 4 within a string (or array) can be used to derive the required numbers.
  – guest271314
  19 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your score is 143*0.7=100.1
  – Jo King
  19 hours ago
  
  

  
  
  
  

2

2

Ok, you were trying to get the bonus, but with a code this size, it didn't compensate. Why not simply '4815162342'?
– Eduardo Hoefel
19 hours ago

Ok, you were trying to get the bonus, but with a code this size, it didn't compensate. Why not simply '4815162342'?
– Eduardo Hoefel
19 hours ago

@EduardoHoefel Do not gather the "score" or "bonus" system, did not try to get the bonus, just tried to not use any of the numbers required at output. The code outputs the numbers without hardcoding any of the numbers. The number 4, with addition, subtraction, multiplication and the index of the number 4 within a string (or array) can be used to derive the required numbers.
– guest271314
19 hours ago

@EduardoHoefel Do not gather the "score" or "bonus" system, did not try to get the bonus, just tried to not use any of the numbers required at output. The code outputs the numbers without hardcoding any of the numbers. The number 4, with addition, subtraction, multiplication and the index of the number 4 within a string (or array) can be used to derive the required numbers.
– guest271314
19 hours ago

Your score is 143*0.7=100.1
– Jo King
19 hours ago

Your score is 143*0.7=100.1
– Jo King
19 hours ago

add a comment | 

up vote
1
down vote

PHP, 35/2=17.5

<?=zzzzzzzzzzzzzzz^NVBVKOVKLVHIVNH;

a digital approach: 40*.7=28

<?=2+2,_,5+3,_,17-2,_,17-1,_,17+6,_,7*6;

no digits, no strings: 68/2 = 34

<?=$p++,!$p++,$p+=$p,_,$p+=$p,_,~-$q=$p+$p,_,$q,_,--$p+$q,_,$p*~-$p;

---

Try them online.

share|improve this answer

answered 15 hours ago

Titus

12.7k11237

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Or just 14 bytes for <?=4815162342;
  – Jo King
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  OP hasn´t replied to wether we can omit the delimiters or not; but yeah. Why not just 10 bytes: 4815162342. Or <?=~ + 10 unprintables -> 15/2=7.5
  – Titus
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

PHP, 35/2=17.5

<?=zzzzzzzzzzzzzzz^NVBVKOVKLVHIVNH;

a digital approach: 40*.7=28

<?=2+2,_,5+3,_,17-2,_,17-1,_,17+6,_,7*6;

no digits, no strings: 68/2 = 34

<?=$p++,!$p++,$p+=$p,_,$p+=$p,_,~-$q=$p+$p,_,$q,_,--$p+$q,_,$p*~-$p;

---

Try them online.

share|improve this answer

answered 15 hours ago

Titus

12.7k11237

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Or just 14 bytes for <?=4815162342;
  – Jo King
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  OP hasn´t replied to wether we can omit the delimiters or not; but yeah. Why not just 10 bytes: 4815162342. Or <?=~ + 10 unprintables -> 15/2=7.5
  – Titus
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

PHP, 35/2=17.5

<?=zzzzzzzzzzzzzzz^NVBVKOVKLVHIVNH;

a digital approach: 40*.7=28

<?=2+2,_,5+3,_,17-2,_,17-1,_,17+6,_,7*6;

no digits, no strings: 68/2 = 34

<?=$p++,!$p++,$p+=$p,_,$p+=$p,_,~-$q=$p+$p,_,$q,_,--$p+$q,_,$p*~-$p;

---

Try them online.

share|improve this answer

answered 15 hours ago

Titus

12.7k11237

PHP, 35/2=17.5

<?=zzzzzzzzzzzzzzz^NVBVKOVKLVHIVNH;

a digital approach: 40*.7=28

<?=2+2,_,5+3,_,17-2,_,17-1,_,17+6,_,7*6;

no digits, no strings: 68/2 = 34

<?=$p++,!$p++,$p+=$p,_,$p+=$p,_,~-$q=$p+$p,_,$q,_,--$p+$q,_,$p*~-$p;

---

Try them online.

share|improve this answer

answered 15 hours ago

Titus

12.7k11237

share|improve this answer

share|improve this answer

answered 15 hours ago

Titus

12.7k11237

answered 15 hours ago

Titus

12.7k11237

answered 15 hours ago

Titus

12.7k11237

12.7k11237

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Or just 14 bytes for <?=4815162342;
  – Jo King
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  OP hasn´t replied to wether we can omit the delimiters or not; but yeah. Why not just 10 bytes: 4815162342. Or <?=~ + 10 unprintables -> 15/2=7.5
  – Titus
  7 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Or just 14 bytes for <?=4815162342;
  – Jo King
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  OP hasn´t replied to wether we can omit the delimiters or not; but yeah. Why not just 10 bytes: 4815162342. Or <?=~ + 10 unprintables -> 15/2=7.5
  – Titus
  7 hours ago
  

  
  
  
  

1

1

Or just 14 bytes for <?=4815162342;
– Jo King
13 hours ago

Or just 14 bytes for <?=4815162342;
– Jo King
13 hours ago

1

1

OP hasn´t replied to wether we can omit the delimiters or not; but yeah. Why not just 10 bytes: 4815162342. Or <?=~ + 10 unprintables -> 15/2=7.5
– Titus
7 hours ago

OP hasn´t replied to wether we can omit the delimiters or not; but yeah. Why not just 10 bytes: 4815162342. Or <?=~ + 10 unprintables -> 15/2=7.5
– Titus
7 hours ago

add a comment | 

up vote
1
down vote

Japt, 10 bytes / 2 = 5

"óT"mc w

Test it

share|improve this answer

answered 11 hours ago

Shaggy

17.6k21663

add a comment | 

up vote
1
down vote

Japt, 10 bytes / 2 = 5

"óT"mc w

Test it

share|improve this answer

answered 11 hours ago

Shaggy

17.6k21663

add a comment | 

up vote
1
down vote

up vote
1
down vote

Japt, 10 bytes / 2 = 5

"óT"mc w

Test it

share|improve this answer

answered 11 hours ago

Shaggy

17.6k21663

Japt, 10 bytes / 2 = 5

"óT"mc w

Test it

share|improve this answer

answered 11 hours ago

Shaggy

17.6k21663

share|improve this answer

share|improve this answer

answered 11 hours ago

Shaggy

17.6k21663

answered 11 hours ago

Shaggy

17.6k21663

answered 11 hours ago

Shaggy

17.6k21663

17.6k21663

add a comment | 

add a comment | 

up vote
1
down vote

JavaScript (SpiderMonkey), 67 bytes / 2 = 33.5 60 bytes / 2 = 30 58 bytes / 2 = 29 48 bytes / 2 = 24

-7 bytes/3.5, -2 bytes/1 courtesy of @JoKing, -10 bytes/5 courtesy of @tsh

print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)

Try it online!

share|improve this answer

edited 7 hours ago

answered 15 hours ago

guest271314

264211

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)
  – tsh
  14 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Or just print(4815162342) for 17 bytes
  – Jo King
  13 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

JavaScript (SpiderMonkey), 67 bytes / 2 = 33.5 60 bytes / 2 = 30 58 bytes / 2 = 29 48 bytes / 2 = 24

-7 bytes/3.5, -2 bytes/1 courtesy of @JoKing, -10 bytes/5 courtesy of @tsh

print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)

Try it online!

share|improve this answer

edited 7 hours ago

answered 15 hours ago

guest271314

264211

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)
  – tsh
  14 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Or just print(4815162342) for 17 bytes
  – Jo King
  13 hours ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript (SpiderMonkey), 67 bytes / 2 = 33.5 60 bytes / 2 = 30 58 bytes / 2 = 29 48 bytes / 2 = 24

-7 bytes/3.5, -2 bytes/1 courtesy of @JoKing, -10 bytes/5 courtesy of @tsh

print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)

Try it online!

share|improve this answer

edited 7 hours ago

answered 15 hours ago

guest271314

264211

JavaScript (SpiderMonkey), 67 bytes / 2 = 33.5 60 bytes / 2 = 30 58 bytes / 2 = 29 48 bytes / 2 = 24

-7 bytes/3.5, -2 bytes/1 courtesy of @JoKing, -10 bytes/5 courtesy of @tsh

print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)

Try it online!

share|improve this answer

edited 7 hours ago

answered 15 hours ago

guest271314

264211

share|improve this answer

share|improve this answer

edited 7 hours ago

edited 7 hours ago

edited 7 hours ago

answered 15 hours ago

guest271314

264211

answered 15 hours ago

guest271314

264211

answered 15 hours ago

guest271314

264211

264211

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)
  – tsh
  14 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Or just print(4815162342) for 17 bytes
  – Jo King
  13 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)
  – tsh
  14 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Or just print(4815162342) for 17 bytes
  – Jo King
  13 hours ago
  

  
  
  
  

1

1

print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)
– tsh
14 hours ago

print(a=-~-~-~-~,a+=a,b=a+~-a,a+a,a+b,--b+b+b)
– tsh
14 hours ago

Or just print(4815162342) for 17 bytes
– Jo King
13 hours ago

Or just print(4815162342) for 17 bytes
– Jo King
13 hours ago

add a comment | 

up vote
0
down vote

Python 2, 16 bytes

print 4815162342

Try it online!

share|improve this answer

answered 16 hours ago

Vedant Kandoi

1166

add a comment | 

up vote
0
down vote

Python 2, 16 bytes

print 4815162342

Try it online!

share|improve this answer

answered 16 hours ago

Vedant Kandoi

1166

add a comment | 

up vote
0
down vote

up vote
0
down vote

Python 2, 16 bytes

print 4815162342

Try it online!

share|improve this answer

answered 16 hours ago

Vedant Kandoi

1166

Python 2, 16 bytes

print 4815162342

Try it online!

share|improve this answer

answered 16 hours ago

Vedant Kandoi

1166

share|improve this answer

share|improve this answer

answered 16 hours ago

Vedant Kandoi

1166

answered 16 hours ago

Vedant Kandoi

1166

answered 16 hours ago

Vedant Kandoi

1166

1166

add a comment | 

add a comment | 

up vote
0
down vote

perl -M5.010 -Mre=eval, 32/2 == 16 bytes

This program is mostly unprintable characters -- characters which aren't even Unicode characters. Here is a hexdump of the program:

$ od -x solution
0000000 2727 7e3d 277e c0d7 8c84 869e cbd8 c7d3
0000020 ced3 d3ca c9ce cdd3 d3cc cdcb 82d8 27d6
0000040 
$

And here's the program to create the solution:

#!/opt/perl/bin/perl

use 5.026;

use strict;
use warnings;
no warnings 'syntax';

use experimental 'signatures';

my $q = ~"(?say'4,8,15,16,23,42')";
print "''=~~'$q'";

__END__

share|improve this answer

answered 8 hours ago

Abigail

39716

add a comment | 

up vote
0
down vote

perl -M5.010 -Mre=eval, 32/2 == 16 bytes

This program is mostly unprintable characters -- characters which aren't even Unicode characters. Here is a hexdump of the program:

$ od -x solution
0000000 2727 7e3d 277e c0d7 8c84 869e cbd8 c7d3
0000020 ced3 d3ca c9ce cdd3 d3cc cdcb 82d8 27d6
0000040 
$

And here's the program to create the solution:

#!/opt/perl/bin/perl

use 5.026;

use strict;
use warnings;
no warnings 'syntax';

use experimental 'signatures';

my $q = ~"(?say'4,8,15,16,23,42')";
print "''=~~'$q'";

__END__

share|improve this answer

answered 8 hours ago

Abigail

39716

add a comment | 

up vote
0
down vote

up vote
0
down vote

perl -M5.010 -Mre=eval, 32/2 == 16 bytes

This program is mostly unprintable characters -- characters which aren't even Unicode characters. Here is a hexdump of the program:

$ od -x solution
0000000 2727 7e3d 277e c0d7 8c84 869e cbd8 c7d3
0000020 ced3 d3ca c9ce cdd3 d3cc cdcb 82d8 27d6
0000040 
$

And here's the program to create the solution:

#!/opt/perl/bin/perl

use 5.026;

use strict;
use warnings;
no warnings 'syntax';

use experimental 'signatures';

my $q = ~"(?say'4,8,15,16,23,42')";
print "''=~~'$q'";

__END__

share|improve this answer

answered 8 hours ago

Abigail

39716

perl -M5.010 -Mre=eval, 32/2 == 16 bytes

This program is mostly unprintable characters -- characters which aren't even Unicode characters. Here is a hexdump of the program:

$ od -x solution
0000000 2727 7e3d 277e c0d7 8c84 869e cbd8 c7d3
0000020 ced3 d3ca c9ce cdd3 d3cc cdcb 82d8 27d6
0000040 
$

And here's the program to create the solution:

#!/opt/perl/bin/perl

use 5.026;

use strict;
use warnings;
no warnings 'syntax';

use experimental 'signatures';

my $q = ~"(?say'4,8,15,16,23,42')";
print "''=~~'$q'";

__END__

share|improve this answer

answered 8 hours ago

Abigail

39716

share|improve this answer

share|improve this answer

answered 8 hours ago

Abigail

39716

answered 8 hours ago

Abigail

39716

answered 8 hours ago

Abigail

39716

39716

add a comment | 

add a comment | 

up vote
0
down vote

Red, 50 bytes / 2 = 25

foreach c"abcdcefgaf"[prin index? find"cfgade.b"c]

Try it online!

Prints the numbers without separator

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Galen Ivanov

5,46211031

add a comment | 

up vote
0
down vote

Red, 50 bytes / 2 = 25

foreach c"abcdcefgaf"[prin index? find"cfgade.b"c]

Try it online!

Prints the numbers without separator

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Galen Ivanov

5,46211031

add a comment | 

up vote
0
down vote

up vote
0
down vote

Red, 50 bytes / 2 = 25

foreach c"abcdcefgaf"[prin index? find"cfgade.b"c]

Try it online!

Prints the numbers without separator

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Galen Ivanov

5,46211031

Red, 50 bytes / 2 = 25

foreach c"abcdcefgaf"[prin index? find"cfgade.b"c]

Try it online!

Prints the numbers without separator

share|improve this answer

edited 8 hours ago

answered 8 hours ago

Galen Ivanov

5,46211031

share|improve this answer

share|improve this answer

edited 8 hours ago

edited 8 hours ago

edited 8 hours ago

answered 8 hours ago

Galen Ivanov

5,46211031

answered 8 hours ago

Galen Ivanov

5,46211031

answered 8 hours ago

Galen Ivanov

5,46211031

5,46211031

add a comment | 

add a comment | 

up vote
0
down vote

APL(Dyalog Unicode), 18/2 = 9 bytes

×/⎕UCS'𩦦湡'

Just boring old character multiplication.

Try it online!

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Quintec

1,095517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The byte count is 18 in UTF-8. Neither 𩦦 nor 湡 exist in any APL code page, as far as I know.
  – Dennis♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis Thanks.
  – Quintec
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

APL(Dyalog Unicode), 18/2 = 9 bytes

×/⎕UCS'𩦦湡'

Just boring old character multiplication.

Try it online!

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Quintec

1,095517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The byte count is 18 in UTF-8. Neither 𩦦 nor 湡 exist in any APL code page, as far as I know.
  – Dennis♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis Thanks.
  – Quintec
  7 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

APL(Dyalog Unicode), 18/2 = 9 bytes

×/⎕UCS'𩦦湡'

Just boring old character multiplication.

Try it online!

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Quintec

1,095517

APL(Dyalog Unicode), 18/2 = 9 bytes

×/⎕UCS'𩦦湡'

Just boring old character multiplication.

Try it online!

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Quintec

1,095517

share|improve this answer

share|improve this answer

edited 7 hours ago

edited 7 hours ago

edited 7 hours ago

answered 8 hours ago

Quintec

1,095517

answered 8 hours ago

Quintec

1,095517

answered 8 hours ago

Quintec

1,095517

1,095517

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The byte count is 18 in UTF-8. Neither 𩦦 nor 湡 exist in any APL code page, as far as I know.
  – Dennis♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis Thanks.
  – Quintec
  7 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The byte count is 18 in UTF-8. Neither 𩦦 nor 湡 exist in any APL code page, as far as I know.
  – Dennis♦
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis Thanks.
  – Quintec
  7 hours ago
  

  
  
  
  

The byte count is 18 in UTF-8. Neither 𩦦 nor 湡 exist in any APL code page, as far as I know.
– Dennis♦
7 hours ago

The byte count is 18 in UTF-8. Neither 𩦦 nor 湡 exist in any APL code page, as far as I know.
– Dennis♦
7 hours ago

@Dennis Thanks.
– Quintec
7 hours ago

@Dennis Thanks.
– Quintec
7 hours ago

add a comment | 

up vote
0
down vote

Whitespace, score: 49 bytes / 2 = 24.5

[S S S T T S T T T S T T T T S S S T S N
_Push_56802][S S S T S T S S T S T T S S T S S S T T N
_Push_84771][T S S N
_Multiply][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Pseudo-code:

Integer i = 56802
Integer j = 84771
Integer k = i * j
Print k as number to STDOUT

share|improve this answer

answered 6 hours ago

Kevin Cruijssen

32.9k554176

add a comment | 

up vote
0
down vote

Whitespace, score: 49 bytes / 2 = 24.5

[S S S T T S T T T S T T T T S S S T S N
_Push_56802][S S S T S T S S T S T T S S T S S S T T N
_Push_84771][T S S N
_Multiply][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Pseudo-code:

Integer i = 56802
Integer j = 84771
Integer k = i * j
Print k as number to STDOUT

share|improve this answer

answered 6 hours ago

Kevin Cruijssen

32.9k554176

add a comment | 

up vote
0
down vote

up vote
0
down vote

Whitespace, score: 49 bytes / 2 = 24.5

[S S S T T S T T T S T T T T S S S T S N
_Push_56802][S S S T S T S S T S T T S S T S S S T T N
_Push_84771][T S S N
_Multiply][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Pseudo-code:

Integer i = 56802
Integer j = 84771
Integer k = i * j
Print k as number to STDOUT

share|improve this answer

answered 6 hours ago

Kevin Cruijssen

32.9k554176

Whitespace, score: 49 bytes / 2 = 24.5

[S S S T T S T T T S T T T T S S S T S N
_Push_56802][S S S T S T S S T S T T S S T S S S T T N
_Push_84771][T S S N
_Multiply][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Pseudo-code:

Integer i = 56802
Integer j = 84771
Integer k = i * j
Print k as number to STDOUT

share|improve this answer

answered 6 hours ago

Kevin Cruijssen

32.9k554176

share|improve this answer

share|improve this answer

answered 6 hours ago

Kevin Cruijssen

32.9k554176

answered 6 hours ago

Kevin Cruijssen

32.9k554176

answered 6 hours ago

Kevin Cruijssen

32.9k554176

32.9k554176

add a comment | 

add a comment | 

up vote
0
down vote

Z80Golf, 17 bytes * 0.5 = 8.5

00000000: 1b17 1e1a 1e19 1d1c 1b1d 2676 0a03 c5ee ..........&v....
00000010: 2f /

Try it online!

Assembly:

db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '8'	;17 rla
db 0x2F ^ '1'	;1e ld e,
db 0x2F ^ '5'	;1a 0x1a
db 0x2F ^ '1'	;1e ld e, 
db 0x2F ^ '6'	;19 0x19
db 0x2F ^ '2'	;1d dec e
db 0x2F ^ '3'	;1c inc e
db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '2'	;1d dec e
ld h, 0x76	;halt

ld a, (bc)	;load the next digit. The first char in addr in 0x0
inc bc		;get next digit
push bc		;return to the next digit which is basically a nop
xor 0x2F	;decode the digit
		;fall through into putchar. Putchar (0x8000), prints the char in register a

Assembly

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edited 4 hours ago

answered 6 hours ago

Logern

59535

add a comment | 

up vote
0
down vote

Z80Golf, 17 bytes * 0.5 = 8.5

00000000: 1b17 1e1a 1e19 1d1c 1b1d 2676 0a03 c5ee ..........&v....
00000010: 2f /

Try it online!

Assembly:

db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '8'	;17 rla
db 0x2F ^ '1'	;1e ld e,
db 0x2F ^ '5'	;1a 0x1a
db 0x2F ^ '1'	;1e ld e, 
db 0x2F ^ '6'	;19 0x19
db 0x2F ^ '2'	;1d dec e
db 0x2F ^ '3'	;1c inc e
db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '2'	;1d dec e
ld h, 0x76	;halt

ld a, (bc)	;load the next digit. The first char in addr in 0x0
inc bc		;get next digit
push bc		;return to the next digit which is basically a nop
xor 0x2F	;decode the digit
		;fall through into putchar. Putchar (0x8000), prints the char in register a

Assembly

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Logern

59535

add a comment | 

up vote
0
down vote

up vote
0
down vote

Z80Golf, 17 bytes * 0.5 = 8.5

00000000: 1b17 1e1a 1e19 1d1c 1b1d 2676 0a03 c5ee ..........&v....
00000010: 2f /

Try it online!

Assembly:

db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '8'	;17 rla
db 0x2F ^ '1'	;1e ld e,
db 0x2F ^ '5'	;1a 0x1a
db 0x2F ^ '1'	;1e ld e, 
db 0x2F ^ '6'	;19 0x19
db 0x2F ^ '2'	;1d dec e
db 0x2F ^ '3'	;1c inc e
db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '2'	;1d dec e
ld h, 0x76	;halt

ld a, (bc)	;load the next digit. The first char in addr in 0x0
inc bc		;get next digit
push bc		;return to the next digit which is basically a nop
xor 0x2F	;decode the digit
		;fall through into putchar. Putchar (0x8000), prints the char in register a

Assembly

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Logern

59535

Z80Golf, 17 bytes * 0.5 = 8.5

00000000: 1b17 1e1a 1e19 1d1c 1b1d 2676 0a03 c5ee ..........&v....
00000010: 2f /

Try it online!

Assembly:

db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '8'	;17 rla
db 0x2F ^ '1'	;1e ld e,
db 0x2F ^ '5'	;1a 0x1a
db 0x2F ^ '1'	;1e ld e, 
db 0x2F ^ '6'	;19 0x19
db 0x2F ^ '2'	;1d dec e
db 0x2F ^ '3'	;1c inc e
db 0x2F ^ '4'	;1b dec de
db 0x2F ^ '2'	;1d dec e
ld h, 0x76	;halt

ld a, (bc)	;load the next digit. The first char in addr in 0x0
inc bc		;get next digit
push bc		;return to the next digit which is basically a nop
xor 0x2F	;decode the digit
		;fall through into putchar. Putchar (0x8000), prints the char in register a

Assembly

share|improve this answer

edited 4 hours ago

answered 6 hours ago

Logern

59535

share|improve this answer

share|improve this answer

edited 4 hours ago

edited 4 hours ago

edited 4 hours ago

answered 6 hours ago

Logern

59535

answered 6 hours ago

Logern

59535

answered 6 hours ago

Logern

59535

59535

add a comment | 

add a comment | 

 

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