Print all lexicographically increasing numbers under 10000

Question

A lexicographically increasing number is an integer whose digits are in strictly increasing order. Print all lexicographically increasing numbers under 10000.

Here are lines of the expected output:

0 1 2 3 4 5 6 7 8 9 12 13 14 15 16 17 18 19 23 24 25 26 27 28 29 34 35 36 37 38 39 45 46 47 48 49 56 57 58 59 67 68 69 78 79 89 123 124 125 126 127 128 129 134 135 136 137 138 139 145 146 147 148 149 156 157 158 159 167 168 169 178 179 189 234 235 236 237 238 239 245 246 247 248 249 256 257 258 259 267 268 269 278 279 289 345 346 347 348 349 356 357 358 359 367 368 369 378 379 389 456 457 458 459 467 468 469 478 479 489 567 568 569 578 579 589 678 679 689 789 1234 1235 1236 1237 1238 1239 1245 1246 1247 1248 1249 1256 1257 1258 1259 1267 1268 1269 1278 1279 1289 1345 1346 1347 1348 1349 1356 1357 1358 1359 1367 1368 1369 1378 1379 1389 1456 1457 1458 1459 1467 1468 1469 1478 1479 1489 1567 1568 1569 1578 1579 1589 1678 1679 1689 1789 2345 2346 2347 2348 2349 2356 2357 2358 2359 2367 2368 2369 2378 2379 2389 2456 2457 2458 2459 2467 2468 2469 2478 2479 2489 2567 2568 2569 2578 2579 2589 2678 2679 2689 2789 3456 3457 3458 3459 3467 3468 3469 3478 3479 3489 3567 3568 3569 3578 3579 3589 3678 3679 3689 3789 4567 4568 4569 4578 4579 4589 4678 4679 4689 4789 5678 5679 5689 5789 6789

This is a code golf challenge! Shortest answer wins!

(P.S. looking for a python solution)

do we need to print them on separate lines or is space-separated OK? — Giuseppe
– Giuseppe, Commented Nov 16, 2018 at 19:29
Welcome to PPCG! Nice first challenge. For future challenges, I can recommend using the Sandbox to refine a challenge and get meaningful feedback before posting it to main. — AdmBorkBork
– AdmBorkBork, Commented Nov 16, 2018 at 19:38
To expand on @Giuseppe's question, can we output separated by commas, spaces, in array format [0,1,...], etc. or must we output each number on a separate line? — ETHproductions
– ETHproductions, Commented Nov 16, 2018 at 19:42
Do the numbers need to be in a specific order, or do they just need to all exist? — Kamil Drakari
– Kamil Drakari, Commented Nov 16, 2018 at 19:56
@VarunPatro, please update the challenge to explicitly state that each number by on a separate line (although I'd recommend against that requirement) and make sure to inform any existing solutions that don't do so. — Shaggy
– Shaggy, Commented Nov 16, 2018 at 20:32

xnor · Accepted Answer · 2018-11-16 20:42:18Z

32

Python 2, 56 bytes

for n in range(9999): if eval('<'.join(`n`))**n:print n

Try it online!

Converts each number like 124 to an expression 1<2<4 and evaluates it to check if the digits are sorted,

A hiccup happens for one-digit numbers giving an expression that just is the number itself. This causes 0 to evaluate to a Falsey value even though it should be printed. This is fixed by a trick suggested by Erik the Outgolfer of doing **n, which gives truthy value 0**0 for n=0 and doesn't affect the truth value otherwise.

edited Nov 16, 2018 at 20:42

answered Nov 16, 2018 at 20:01

xnor

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\$\begingroup\$ What does the ` do in `n`? \$\endgroup\$

BruceWayne
– BruceWayne

2018-11-16 21:58:24 +00:00
Commented Nov 16, 2018 at 21:58
1

\$\begingroup\$ @BruceWayne It takes the string representation. This was removed in Python 3. \$\endgroup\$

xnor
– xnor

2018-11-16 22:00:49 +00:00
Commented Nov 16, 2018 at 22:00
5

\$\begingroup\$ @BruceWayne Note that it's the same as the repr() function, not the str() function. They aren't always the same. Here's an example. \$\endgroup\$

mbomb007
– mbomb007

2018-11-16 22:39:23 +00:00
Commented Nov 16, 2018 at 22:39
1

\$\begingroup\$ @mbomb007 thanks for that comment! I would have thought it was str() equivalent. \$\endgroup\$

BruceWayne
– BruceWayne

2018-11-16 22:48:12 +00:00
Commented Nov 16, 2018 at 22:48
2

\$\begingroup\$ We can handle the 0 case with a bit of trickery. \$\endgroup\$

xsot
– xsot

2018-11-17 16:04:25 +00:00
Commented Nov 17, 2018 at 16:04

| Show 2 more comments

xsot · Accepted Answer · 2018-11-17 15:59:59Z

14

Python 2, 55 bytes

i=0 exec"print i\ni+=1\nif eval('<'.join(`i`)):1;"*7000

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answered Nov 17, 2018 at 15:59

xsot

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xnor · Accepted Answer · 2018-11-17 01:57:13Z

Haskell, 50 bytes

unlines[s|s<-show<$>[0..6^5],s==scanl1(max.succ)s]

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Outputs a multiline string. We check that the number s increasing using s==scanl1(max.succ)s, a variant of the usual sortedness check s==scanl1 max s that ensures strict sortedness by incrementing each digit character before taking the maximum of it and the next digit.

Ourous saved a byte by using 6^5 as the upper bound in place of a 4-digit number.

Dennis · Accepted Answer · 2018-11-16 22:22:38Z

Jelly, 7 bytes

9ŒPḌḣ⁹Y

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How it works

9ŒPḌḣ⁹Y Main link. No arguments. 9 Set the return value to 9. ŒP Powerset; promote 9 to [1, ..., 9] and generate all subsets. Ḍ Undecimal; map the subsets of digits to the integers they represent. ⁹ Yield 256. ḣ Dyadic head; take the first 256 elements of the integer list. Y Separate the result by linefeeds.

I'm trying to figure out how 0 gets included here but I don't know Jelly. Am I correct that Jelly's powerset includes the empty array which then gets converted to 0 when "undecimaled"? — Shaggy
– Shaggy, Commented Nov 17, 2018 at 0:21

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

Japt `-R`, 12 11 8 bytes

L²Ç¶ìüÃð

Test it

L :100 ² :Squared Ç :Map the range [0,L²) ì : Split to a digit array ü : For the sake of simplicity*, let's say: Sort & deduplicate : Implicitly rejoin to an integer ¶ : Test for equality with original number Ã :End map ð :Get 0-based indices of truthy elements :Implicitly join with newlines and output

*Or, to offer a better explanation: the ü method sorts an array and splits it into equal elements (e.g., [8,4,8,4].ü() -> [[4,4],[8,8]]) and then, in what seems to be a strange quirk and hopefully not a bug, the ì method, when converting the array back to a number, takes the first element of each nested array, rather than first flattening the array, which is what I expected when I tried this trick (e.g., [[4,4],[8,8]].ì() -> 48).

I have to say, that ü trick you guys used is genius :-) @Oliver — ETHproductions
– ETHproductions, Commented Nov 16, 2018 at 19:45
@Oliver, you must have posted that as I was updating; great minds ... :) — Shaggy
– Shaggy, Commented Nov 16, 2018 at 19:48
@ETHproductions, like most things, I tried it on a whim - amazed it works. — Shaggy
– Shaggy, Commented Nov 16, 2018 at 19:48

Giuseppe · Accepted Answer · 2018-11-16 21:20:11Z

R, 62 49 bytes

`[`=write;0[1];for(i in 1:4)combn(1:9,i)[1,i,,""]

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Because combn iterates through its input in the order given, it's easy to create all the lexicographically increasing integers, printing them out in order. write prints them each i-digit number in lines of width i, neatly fulfilling the newline requirement as well.

\$\begingroup\$ great idea exploiting combn ! \$\endgroup\$

digEmAll
– digEmAll

2018-11-17 14:36:13 +00:00
Commented Nov 17, 2018 at 14:36 — digEmAll
– digEmAll, Commented Nov 17, 2018 at 14:36
\$\begingroup\$ Extremely clever aliasing! \$\endgroup\$

J.Doe
– J.Doe

2018-11-18 12:15:41 +00:00
Commented Nov 18, 2018 at 12:15 — J.Doe
– J.Doe, Commented Nov 18, 2018 at 12:15

AdmBorkBork · Accepted Answer · 2018-11-19 13:24:02Z

PowerShell, 42 40 bytes

0..1e4|?{-join("$_"|% t*y|sort -u)-eq$_}

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Loop from 0 to 1e4 (i.e., 10000). Pull out those objects where |?{...} the number as a string $_ is -equal to the number cast toCharArray and then sorted with the -unique flag. In other words, only numbers that are the same as their sorted and deduplicated strings. Each of those are left on the pipeline and output is implicit.

Sean · Accepted Answer · 2018-11-26 17:43:33Z

6

Perl 6, 25 bytes

[<](.comb)&&.say for ^1e4

-1 byte thanks to nwellnhof

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.comb produces a list of the digits of each number, and [<] does a less-than reduction on that list, equivalent to: digit1 < digit2 < ... < digitN.

edited Nov 26, 2018 at 17:43

answered Nov 16, 2018 at 19:54

Sean

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2

\$\begingroup\$ [<](.comb)&&.say saves a byte. \$\endgroup\$

nwellnhof
– nwellnhof

2018-11-17 09:56:13 +00:00
Commented Nov 17, 2018 at 9:56
\$\begingroup\$ This is surprisingly readable. (I already know a little Perl 6, but still...) \$\endgroup\$

J-L
– J-L

2018-11-19 20:38:44 +00:00
Commented Nov 19, 2018 at 20:38

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nimi · Accepted Answer · 2018-11-17 02:31:28Z

5

Haskell, 56 55 bytes

Edit: -1 byte thanks to @Ourous

mapM print$filter(and.(zipWith(<)<*>tail).show)[0..6^5]

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edited Nov 17, 2018 at 2:31

answered Nov 16, 2018 at 19:38

nimi

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Dennis · Accepted Answer · 2018-11-17 14:13:50Z

Pyth, 10 bytes

jiRThc2yS9

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How it works

jiRThc2yS9 S9 Yield [1, 2, 3, 4, 5, 6, 7, 8, 9]. y Take all 512 subsets. c2 Split the array of subsets into 2 pieces (256 subsets each). h Head; take the first piece. iRT Convert each subset from base 10 to integer. j Separate by newlines.

DLosc · Accepted Answer · 2022-07-02 05:13:46Z

Regenerate `-a`, 72 bytes

0|([1-9])(([2-9])(){$3-$1-1}(([3-9])(){$6-$3-1}(([4-9])(){$9-$6-1})?)?)?

Appropriately, outputs the numbers in lexicographic order. Attempt This Online!

Explanation

0|

Match 0, or:

([1-9])

Match a digit 1 through 9 (capture group 1).

(...)?

Either stop there, or continue:

([2-9])

Match a digit 2 through 9 (capture group 3), and:

(){$3-$1-1}

Match an empty group X times, where X equals the second digit minus the first digit minus 1. If the second digit is not larger than the first digit, this quantity is negative, and using a negative number as a repetition count causes the match to fail. Thus, only matches where the second digit is larger than the first digit are included.

(...)?

Either stop there, or continue:

([3-9])

Match a digit 3 through 9 (capture group 6), and:

(){$6-$3-1}

Match an empty group Y times, where Y equals the third digit minus the second digit minus 1 (ensuring that the third digit is larger than the second digit).

(...)?

Either stop there, or continue:

([4-9])

Match a digit 4 through 9 (capture group 9), and:

(){$9-$6-1}

Match an empty group Z times, where Z equals the fourth digit minus the third digit minus 1 (ensuring that the fourth digit is larger than the third digit).

\$\begingroup\$ Very clever. But -36 bytes! \$\endgroup\$

Deadcode
– Deadcode

2022-07-02 23:10:10 +00:00
Commented Jul 2, 2022 at 23:10 — Deadcode
– Deadcode, Commented Jul 2, 2022 at 23:10

Deadcode · Accepted Answer · 2022-07-03 00:26:24Z

Regenerate `-a`, 36 34 bytes

0|(([2-9])$4{$1/$2}|$3![1-9]()){4}

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This was directly inspired by the language's creator's answer, but I figured this big optimization building on that idea was worth its own post.

 0 # Zero must be matched as a special case, because no other # numbers have a leading '0' digit. | ( ([2-9]) # \2 = a digit intended to be greater than the previous # one ($1), but until being validated below, it can # be any digit in [2-9]. # We could actually do ([0-9]) here and it'd generate the # same output, but would run slightly less efficiently. $4{$1/$2} # Repeat a nonexistent group $1/$2 number of times. $1 and $2 # are the literal contents of those captures interpreted as # numeric values upon which arithmetic can be done in a # quantifier, which we do here. $1/$2 will evaluate to # zero iff $2 is greater than $1, which is the only thing # that will allow it to match, because if it repeats more # than zero times, it tries to match the nonexistent group # $4 which can't match. | $3 # If the above fails to match, due to $1 containing the # digit '9', the regex engine will try other matches, even # if they use short-circuiting alternation. So we put a # dummy match here, which can only match on iterations # other than the first one. Once this matches, it will be # the only thing that can match in all subsequent # iterations, because the attempted math "$2-$1-1" will # fail still - now due to $1 no longer being a number (it # will be blank). # Short-circuiting alternation - never try the following unless the all of # the above failed to match. ! [1-9] # Match the first digit. This can only happen on the first # iteration, thanks to the short-circuiting alternation. () # $3 = empty capture - lets subsequent iterations know that # they are no longer the first iteration ) # $1 = whatever digit was matched above, for use by the next # iteration {4} # Repeat the above loop for exactly 4 iterations. This # allows a variable number of digits, due to the "$3" dummy # alternative.

Alternative 34 bytes:

0|(([2-9])$3{$1/$2}|{#1}![1-9]){4}

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It might not be intended that this is possible, but Regenerate allows a quantifier to be preceded by nothing. This is what's happening with {#1} - it's impossible for it to match on the first iteration because $1 hasn't been captured yet and doesn't have a length. On subsequent iterations, a match of nothing gets repeated one time (the length of $1).

Oh wow, nice work. The "repeat nothing" idiom was originally unintentional, I think, but once I discovered it was useful, I decided it was a feature rather than a bug. :) — DLosc
– DLosc, Commented Jul 5, 2022 at 18:14

DJMcMayhem · Accepted Answer · 2018-11-16 20:21:49Z

V, 41 bytes

7000ïÎaÛ Îy$úúP Ç^¨ä*©±$/d ÎãlD çˆ±/d HO0

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Hexdump:

00000000: 3730 3030 efce 61db 0ace 7924 fafa 500a 7000..a...y$..P. 00000010: c75e a8e4 2aa9 b124 2f64 0ace e36c 440a .^..*..$/d...lD. 00000020: e788 b12f 640a 484f 30 .../d.HO0

Neil · Accepted Answer · 2018-11-16 20:59:43Z

Charcoal, 19 bytes

ΦＥＸχ⁴Ｉι¬Φι∧μ¬‹§ι⊖μλ

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

 χ Predefined variable 10 Ｘ To the power ⁴ Literal 4 Ｅ Map over implicit range ι Current value Ｉ Cast to string Φ Filter over strings where ι Current string Φ Filtered over characters μ Character index (is nonzero) ∧ And μ Character index ⊖ Decremented § Indexed into ι Current string ¬ Is not ‹ Less than λ Current character ¬ Results in an empty string Implicitly print matches on separate lines

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

J, 26 bytes

,.(#~(-:/:~@~.)@":"0)i.1e4

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explanation

,. (#~ (-: /:~@~.)@":"0) i.1e4 i.1e4 NB. list 0 to 9999 "0 NB. for each number in the input list @":"0 NB. convert it to a string and (#~ ( ) NB. remove any not passing this test: -: NB. the string of digits matches @~. NB. the nub of the digits (dups removed) /:~ NB. sorted ,. NB. ravel items: take the result of all that NB. and turn it into a big column

Renzo · Accepted Answer · 2018-11-17 10:08:13Z

Common Lisp, 74 72 bytes

(dotimes(i 7e3)(format(apply'char<(coerce(format()"~d"i)'list))"~d~%"i))

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-2 bytes thank to @Shaggy!

Mr. Xcoder · Accepted Answer · 2018-11-17 14:41:38Z

05AB1E (legacy), 8 bytes

4°ÝD€êÃ»

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Works in the new version of 05AB1E as well but is painfully slow for some reason.

How?

 4°ÝD€êÃ» – Full program. 4°Ý – Push [0 ... 10000]. D€ê – Push each integer in [0 ... 10000] sorted and deduplicated at the same time. Ã» – And join the interection of the two lists by newlines.

Nice answer. Better than the 9 byter I had (which only works in the legacy). — Kevin Cruijssen
– Kevin Cruijssen, Commented Nov 19, 2018 at 13:54

Xcali · Accepted Answer · 2019-04-17 19:14:37Z

3

Perl 5, 47 bytes

map{$_&&s/.(?=(.))/$1-$&/gre=~/0|-/||say}0..1E4

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Older:

52 bytes

edited Apr 17, 2019 at 19:14

answered Nov 16, 2018 at 20:24

Xcali

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Chas Brown · Accepted Answer · 2018-11-16 20:04:53Z

2

Python 2, 61 bytes

for i in range(9999): if list(`i`)==sorted(set(`i`)):print i

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answered Nov 16, 2018 at 20:04

Chas Brown

9,7771 gold badge14 silver badges39 bronze badges

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Triggernometry · Accepted Answer · 2018-11-16 20:14:07Z

2

Python 2, 64 61 bytes

lambda:[x for x in range(9999)if sorted(set(`x`))==list(`x`)]

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Gets the unique characters of the integer's string representation, sorts them, and compares the result to the original number.

edited Nov 16, 2018 at 20:14

answered Nov 16, 2018 at 19:44

Triggernometry

8314 silver badges11 bronze badges

\$\begingroup\$ You can save a byte by using range(9999) or any other number between 6790 and 9999. Our solutions are almost identical BTW :) \$\endgroup\$

DJMcMayhem
– DJMcMayhem

2018-11-16 19:47:24 +00:00
Commented Nov 16, 2018 at 19:47
\$\begingroup\$ @DJMcMayhem But then it wouldn't check ALL numbers under 10,000....:P Thanks! Sometimes I get too literal with these challenges. \$\endgroup\$

Triggernometry
– Triggernometry

2018-11-16 19:50:44 +00:00
Commented Nov 16, 2018 at 19:50

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ETHproductions · Accepted Answer · 2018-11-17 17:24:48Z

Jelly, 13 9 8 bytes

Saved 5 bytes thanks to @Dennis

9œcⱮ4ẎŻY

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Explanation

Generates all lexicographically increasing numbers below 10000 by taking the digits [1...9] and finding all combinations of length ≤ 4.

9œcⱮ4ẎŻY Main link. Arguments: none 9 Yield 9. Ɱ4 For each n in [1...4]: œc Yield the combinations of the range [1...9] of length n. Ẏ Tighten; dump each of the 4 lists generated into the main list. Ż Prepend a 0 to the list. Y Join on newlines.

Jelly, 11 10 9 bytes

Saved a byte thanks to @EriktheOutgolfer

ȷ4Ḷ<ƝẠ$ƇY

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Explanation

Filters through the range, keeping the numbers that are lexicographically increasing.

ȷ4Ḷ<ƝẠ$ƇY Main link. Arguments: none ȷ4 Yield 10^4 (10000). Ḷ Generate the range [0...10000). Ƈ Filter; yield only the numbers where this link return a truthy value. $ Run these two links and yield the result. Ɲ For each pair of items (digits) in the number: < Check whether the left digit is less than the right digit. Ạ All; check that every comparison yielded true. This yields whether the digits are strictly increasing. Y Join the filtered list on newlines.

Kelly Lowder · Accepted Answer · 2018-11-19 02:58:35Z

Wolfram Language (Mathematica), 36 bytes

After I wrote this, it was clarified that each number must be on a new line, so +7 bytes for the Print/@.

This method takes advantage of the fact that the Subsets function 1) doesn't replicate any digits and 2) sorts the output by set size and set contents. FromDigits assembles each list of digits.

-1 byte thanks to @Mr.Xcoder

Print/@FromDigits/@Range@9~Subsets~4

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Funny, I thought of that and just didn't do it because I thought ~ had higher precedence than @ — Kelly Lowder
– Kelly Lowder, Commented Nov 19, 2018 at 2:55

mkst · Accepted Answer · 2018-11-20 08:23:52Z

K (ngn/k) / K (oK), 32 30 26 bytes

Solution:

`0:$&&/'1_'>':'" ",'$!9999

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Explanation:

`0:$&&/'1_'>':'" ",'$!9999 / the solution !9999 / range 0..9998 (could use !6890) $ / string " ",' / prepend " " to each (lower than "0" in ascii) >:' / greater-than each-previous? 1_' / drop first result from each &/' / max (&) over (/) & / indices where true $ / convert to string `0: / print to stdout

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

JavaScript REPL, 64 bytes

A bit of pub golf so probably far from optimal.

(f=n=>n&&f(n-1)+([...n+``].every(x=>y<(y=x),y=0)?` `+n:``))(7e3)

Try it online

Yes, doing it without an IIFE would be a few bytes shorter but that throws an overflow error when called, which would normally be fine as we can assume infinite memory for the purposes of code golf but, to me, doesn't seem to be in the spirit of KC challenges.

Is this a function submission or a full program? If not, you should either count the console.log or relabel your submission as JavaScript REPL. — Dennis
– Dennis, Commented Nov 21, 2018 at 2:58

gastropner · Accepted Answer · 2019-03-11 02:50:56Z

2

C (gcc), 97 89 81 bytes

Thanks to ceilingcat for -8 bytes.

Another -8 thanks to Dennis

g(n){n=!n||n/10%10<n%10&&g(n/10);}f(i){for(i=-1;++i<7e3;g(i)&&printf("%u\n",i));}

Try it online!

edited Mar 11, 2019 at 2:50

answered Nov 17, 2018 at 23:38

gastropner

4,4831 gold badge17 silver badges20 bronze badges

\$\begingroup\$ 81 bytes \$\endgroup\$

Dennis
– Dennis

2018-11-27 13:22:38 +00:00
Commented Nov 27, 2018 at 13:22
\$\begingroup\$ not gonna update this? \$\endgroup\$

ASCII-only
– ASCII-only

2019-03-11 01:53:48 +00:00
Commented Mar 11, 2019 at 1:53
\$\begingroup\$ @ASCII-only Done. Sorry if this oversight upset you. \$\endgroup\$

gastropner
– gastropner

2019-03-11 02:51:23 +00:00
Commented Mar 11, 2019 at 2:51
\$\begingroup\$ 80 bytes \$\endgroup\$

c--
– c--

2022-07-03 02:24:44 +00:00
Commented Jul 3, 2022 at 2:24

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Unrelated String · Accepted Answer · 2019-03-11 05:34:20Z

Brachylog, 12 bytes

7jj⟦{ẹ<₁cẉ}ˢ

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 ẉ Write on its own line { }ˢ every c concatenated <₁ strictly increasing ẹ list of digits of ⟦ numbers from the range from 0 to 7jj 7777.

7jj⟦{ṫ⊆Ị&ẉ}ˢ also works.

dana · Accepted Answer · 2020-02-08 10:15:32Z

C# (Visual C# Interactive Compiler), 102 101 73 ... 72 bytes

-12 and -4 thanks @Dennis!

for(var i=0;i<7e3;i++)if((i+"").Aggregate((a,b)=>a<b?b:':')<58)Print(i);

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Each integer from 0 to 7k tested by first converting it into a string. Leveraging the fact that C# treats strings as character enumerables and LINQ, an aggregate is calculated for each character enumerable as follows:

compare the accumulated value with the current character
if the current character is greater than the accumulation, return the current character
otherwise return : which is greater than 9

If the result of this is less than : (ASCII 58), then the number has lexicographically increasing digits.

Doesn't the challenge state that all numbers from 0-10000 must be printed? I'm pretty sure this prints numbers 0-7000 — Gymhgy
– Gymhgy, Commented Feb 4, 2019 at 4:34
I believe the largest valid number is 6789? This is less than 7000, so you don't have to go any higher. — dana
– dana, Commented Feb 4, 2019 at 4:47
Not foolish at all :) I'm pretty sure I borrowed it from someone else's answer and I was scratching my head as to why they did it. — dana
– dana, Commented Feb 4, 2019 at 5:02

MarcMush · Accepted Answer · 2022-07-02 15:42:58Z

2

Julia, 44 bytes

0:6^5 .|>x->issorted("$x",lt=<=)&&println(x)

\$(6^5 = 7776)\$

Attempt This Online!

answered Jul 2, 2022 at 15:42

MarcMush

6,88515 silver badges18 bronze badges

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naffetS · Accepted Answer · 2022-07-02 17:04:41Z

2

Vyxal `jM`, 5 bytes

k2'Þ⇧

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Flagless:

Vyxal, 7 bytes

k2ʀ~Þ⇧⁋

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edited Jul 2, 2022 at 17:04

answered Jul 2, 2022 at 15:12

naffetS

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Glory2Ukraine · Accepted Answer · 2025-02-23 21:39:26Z

Husk, 10 bytes

¶mdΣMṖŀ5ḣ9

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Also: 10 bytes ¶mdṁ`Ṗḣ9ŀ5 and 11 bytes ¶fo<5LmdṖḣ9

 MṖ -- make all subsequences, ŀ5 -- having 0 to 4 elements ḣ9 -- of all digits 1 to 9 -- (the result is grouped into sublists) Σ -- unlist md -- join digits together ¶ -- join by newlines

Print all lexicographically increasing numbers under 10000

53 Answers 53

Python 2, 56 bytes

Python 2, 55 bytes

Haskell, 50 bytes

Jelly, 7 bytes

How it works

Japt -R, 12 11 8 bytes

R, 62 49 bytes

PowerShell, 42 40 bytes

Perl 6, 25 bytes

Haskell, 56 55 bytes

Pyth, 10 bytes

How it works

Regenerate -a, 72 bytes

Explanation

Regenerate -a, 36 34 bytes

V, 41 bytes

Charcoal, 19 bytes

J, 26 bytes

explanation

Common Lisp, 74 72 bytes

05AB1E (legacy), 8 bytes

How?

Perl 5, 47 bytes

Python 2, 61 bytes

Python 2, 64 61 bytes

Jelly, 13 9 8 bytes

Explanation

Jelly, 11 10 9 bytes

Explanation

Wolfram Language (Mathematica), 36 bytes

K (ngn/k) / K (oK), 32 30 26 bytes

JavaScript REPL, 64 bytes

C (gcc), 97 89 81 bytes

Brachylog, 12 bytes

C# (Visual C# Interactive Compiler), 102 101 73 ... 72 bytes

Julia, 44 bytes

Vyxal jM, 5 bytes

Vyxal, 7 bytes

Husk, 10 bytes

Linked

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Regenerate `-a`, 72 bytes

Regenerate `-a`, 36 34 bytes

Vyxal `jM`, 5 bytes