You have been hired for your tech knowledge as a Secret Agent's sidekick to ensure that the good guy can get his job done and the world can be saved.

This is your last mission before retiring with a high paycheck and the gratitude of the whole world. But before you have to disarm the Evil Genius' Big Overly Massive BOMB (seems that the evil genius is a smart ass who likes recursive acronyms). Anyways you and your buddy are in the very core of the Evil Genius secret base, ready to disarm the BOMB that could wipe the continent out. In your previous mission you managed to get the disarm code that to your astonishment is just "PASSWORD_01". You connect your keyboard to the BOMB but when you are ready to go, the Evil Genius' henchmen enter in with a barrage of bullets. Unfortunately one of these bullets impacts on your keyboard. "Finish the job while I distract those d*ckheads!" says your buddy and then begins to fire his gun. But how could you finish the job with only half a keyboard?


  1. Write a program that outputs in any way you want the string PASSWORD_01 (uppercase).
  2. Since your keyboard has been hit by a bullet, you can only use these keys:

    1 2 3 4 5

    Q W E R T

    A S D F G

    < > Z X C

    Ctrl Shift Tab Space

    Using the Shift key, your keyboard allows you to use these characters:

    ! " · $ %

  3. You don't have any other hardware than the keyboard and the screen (a mouse for example), nor a file with the password written in your computer.

  4. You don't have an Internet connection.

  5. You can assume you have the interpreter shell / source editor opened before the bullets came in. Sadly you have not written anything in it before the keyboard was hit.

  6. You don't have a virtual keyboard. In fact the BOMB has a TOO_FUNNY detector that will make it explode if you try to use the Standard loopholes.

  7. Since your buddy is waiting for you to finish soon to escape off the Secret Base, you will have to write the smallest code possible (so it's and !).

Good luck because the count down has begun!

HELP MODE: Also a KEY (only one Key you want from the keyboard, but not the Shift + Key or any other combination) has miraculously survived too. E.g.: You can use =, or 0 or /… This extra key cannot be any of the letters in PASSWORD_01 (so you cannot add P or O). Name that key in your answer. You have a penalization of 10 characters for use of this key regardless of how many times you use the key.

  • 75
    \$\begingroup\$ Can I change the layout of the keyboard to DVORAK? \$\endgroup\$ Commented Jun 11, 2014 at 10:54
  • 45
    \$\begingroup\$ Do we get the enter key at least? \$\endgroup\$
    – Οurous
    Commented Jun 11, 2014 at 11:51
  • 77
    \$\begingroup\$ Whitespace, here is your big chance! \$\endgroup\$
    – german_guy
    Commented Jun 11, 2014 at 13:08
  • 50
    \$\begingroup\$ Why couldn't the password have been stewardesses? \$\endgroup\$ Commented Jun 11, 2014 at 18:07
  • 28
    \$\begingroup\$ Joke answer: I'm a spy, I switch to the backup keyboard hidden in my watch. \$\endgroup\$
    – Nzall
    Commented Jun 11, 2014 at 20:50

30 Answers 30


bash, vim and dc (343)

I'm sitting at a bash prompt which has been configured by the Evil Genius, and of course he has VISUAL=vim in the default environment. Using the default bash binding for edit-and-execute-command (C-x C-e) bash invokes $VISUAL and will execute its buffer contents on exit (ZZ). I type in the following sequence (Note: kbd-mode are commands issued in normal mode and control sequences):

  1. C-xC-e

  2. acat<<<45C-c4C-aaa

    Now the vim buffer contains cat<<<49a. Continuing ...

  3. 45C-c3C-aaa

  4. 54C-c41C-aaa

  5. 55C-c13C-aaa

    Now the vim buffer contains cat<<<49a48a95a68a. Continuing ...

  6. 51C-c31C-aaa

  7. 55C-c24C-aaa

  8. 44C-c43C-aaa

  9. 42C-c41C-aaa

  10. 42C-c41C-aaa

  11. 54C-c11C-aaa

  12. 55C-c25C-aaa>s

    Now the vim buffer contains cat<<<49a48a95a68a82a79a87a83a83a65a80a>s

    Get out of insert mode, save and exit

  13. C-cZZ

    The s file now contains a dc script that generates the desired string on the stack, now we need to add print commands.

  14. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

    Repeat the above command 9 times.

  15. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  16. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  17. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  18. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  19. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  20. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  21. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  22. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  23. C-xC-eadc<<<55C-c25C-aaa>>sC-cZZ

  24. C-xC-eacat<<< f >>s

    Execute the dc script:

  25. C-xC-eadc sC-cZZ



Contents of the s file:

  • 4
    \$\begingroup\$ I was thinking to do a vim script because of the Ctrl character; good thinking to pair it with dc! Also, excellent solution :) \$\endgroup\$
    – nneonneo
    Commented Jun 12, 2014 at 0:28
  • 2
    \$\begingroup\$ How long did it take you to write all those <kbd></kbd> tags? \$\endgroup\$
    – justhalf
    Commented Jun 13, 2014 at 17:13
  • 15
    \$\begingroup\$ @justhalf: I wrote it in vim with <Leader>k bound like this map <Leader>k i<kbd></kbd>^[F>a, so it wasn't as bad as it looks :-). \$\endgroup\$
    – Thor
    Commented Jun 13, 2014 at 23:25
  • 20
    \$\begingroup\$ Now it's apparent that Agent Thor knows his tool thoroughly. You deserve the retirement, sir! \$\endgroup\$
    – justhalf
    Commented Jun 14, 2014 at 1:46

Ruby, 57 + 10 (*) = 67


This answer uses * and % to build the ASCII values of the missing characters (and 0 as Fixnum) and pushes them into $* (ARGV). This array is then used in combination with a format string to generate the correct password, which is printed with $><< ($> is stdout).

Ruby, 62 + 10 (.) = 72, no linebreak

$>.<< "%cASSW%cRD%c%d1".% %w%%<<311%231<<214%135<<321%113<<1%1

Roughly the same principle as the above version, except that here the array is built from an empty array literal (%w%%). Some fiddling with . is needed to get the desired operator precedence.

  • 13
    \$\begingroup\$ "This is your last mission before retiring with a high paycheck and the gratitude of the whole world." Congratulations, we all thank you :) \$\endgroup\$
    – Timtech
    Commented Jun 11, 2014 at 14:01
  • 4
    \$\begingroup\$ The ultimate proof that Perl is one of the biggest inspiration of Ruby. \$\endgroup\$ Commented Jun 11, 2014 at 14:11
  • 11
    \$\begingroup\$ How did you do the linebreak? \$\endgroup\$
    – Cephalopod
    Commented Jun 11, 2014 at 16:21
  • 4
    \$\begingroup\$ @Arian The same way nneonneo runs his Python code after typing it into the REPL ... I'll add a version without linebreak. \$\endgroup\$
    – Ventero
    Commented Jun 11, 2014 at 21:26
  • \$\begingroup\$ Does the evil genius have ruby installed? \$\endgroup\$
    – Mohammad
    Commented Jun 13, 2014 at 10:24

CJam - 24 + 10 (() = 34


No additional key - 39

211 131%c"ASSW"211 132%c"RD"321 113%cT1

Try it at http://cjam.aditsu.net/


Whitespace (148 + 10 = 158)

Enter key needs to be used here.

SS+1010000L // Push code of P
TLSS        // Output character
SS+1000001L // Push code of A
TLSS        // Output character
SS+1010011L // Push code of S
SLS         // Duplicate top stack
SS+1010111L // Push code of W
SS+1001111L // Push code of O
SS+1010010L // Push code of R
SS+1000100L // Push code of D
SS+1011111L // Push code of _
SS+0L       // Push 0
TLST        // Output number
SS+1L       // Push 1
TLST        // Output number
LLL         // End program

My notation's explanation:

  • S, +, 0 are spaces.
  • T, 1 are tabs.
  • L is new line.
  • // starts comment.

Each line is a command in whitespace language.


  • 1
    \$\begingroup\$ Ah man... I was just working on that :( \$\endgroup\$
    – Teun Pronk
    Commented Jun 11, 2014 at 13:09
  • 1
    \$\begingroup\$ Me too, but I gave up on this knowing someone would be faster \$\endgroup\$
    – german_guy
    Commented Jun 11, 2014 at 13:11
  • \$\begingroup\$ I thought 'L' wasn't available? \$\endgroup\$ Commented Jun 12, 2014 at 14:18
  • \$\begingroup\$ @500-InternalServerError: Actually L stands for new line character. Check the Demo for the real code. What I posted is a readable version. \$\endgroup\$ Commented Jun 12, 2014 at 15:07
  • 1
    \$\begingroup\$ Ah, I get it now. \$\endgroup\$ Commented Jun 12, 2014 at 15:17

Python (2200 395 + 10)


I needed the + character (costing +10), which can be obtained from the numpad (or on certain key layouts).

Yeah, the BOMB probably went off while I was typing that.

The basic approach is to construct ever larger character sets by using exec and "%c"%(number). There are four execs nested inside each other. Gradually, my digit set progresses up from

  1. 12345
  2. 1234567 (6 = ASCII 54, 7 = ASCII 55)
  3. 123456789 (8 = ASCII 56, 9 = ASCII 57)
  4. 0x0123456789abcdef

so that in the final iteration it is possible to express any character (so that the innermost version can actually run any program at all).

Around 50% of the program is just quote characters ("%c"%34 is a double quote), because the nested nature of the exec statements demands "escaping" quote characters aggressively.

  • \$\begingroup\$ If you use the "+" from the numeric keypad, you are independent from the keyboard language. \$\endgroup\$
    – celtschk
    Commented Jun 12, 2014 at 20:31
  • \$\begingroup\$ @celtschk: Good point! I will edit that in. \$\endgroup\$
    – nneonneo
    Commented Jun 12, 2014 at 20:32
  • 1
    \$\begingroup\$ Rather than constructing digit sets with nested execs, couldn't you get numbers using arithmetic? Ex. exec '%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c'%(112,114,51+54,55+55,1+115,32,4+35,25+55,11+54,31+52,31+52,32+55,24+55,31+51,13+55,41+54,3+45,4+45,4+35) \$\endgroup\$
    – Kevin
    Commented Jun 17, 2014 at 12:09
  • 2
    \$\begingroup\$ Because you can't get both parens. (I did consider this!) \$\endgroup\$
    – nneonneo
    Commented Jun 17, 2014 at 13:37
  • \$\begingroup\$ Additionally, Python doesn't have a lot of operators that bind more strongly than %. ** is one, but it tends to be fairly useless for constructing most terms. \$\endgroup\$
    – nneonneo
    Commented Jun 17, 2014 at 13:39

Insomnia 39 35 31 29

This language comes up when I was looking around for a language that encodes its instructions with single ASCII character. The language actually operates on the decimal ASCII code of the character, so it is still quite flexible with half of the keyboard destroyed.

Drop it down further to 29 characters, which is possible after reducing memory usage and increases the search space:

FdFzt%dF<rGdt>tt Gtreeet t%4s

I decided to run my tweaked up program on the full set of allowed characters, and cut the solution down to 31 characters:

FdFGt dtreeC>tt FFdx<Fdtr ztq4s

I used a program to search for this solution. It is one among many solutions returned by my program, all with the same size.


Old version constructed by hand. I remember staying up till morning to do this.


Here is the Insomnia interpreter for testing.

  • 5
    \$\begingroup\$ +1 for the "staying up till morning" match with your programming language choice. \$\endgroup\$
    – justhalf
    Commented Dec 4, 2014 at 2:36

Vim + PHP on Some Window Managers

65 keys in 44 strokes, with a 10-point penalty for using =.


A breakdown:

  • Alt+Tab appears to work like Esc with Vim, or perhaps with this terminal emulator. Thanks for teaching me something new!
  • aEACE<Alt+Tab>: Enter append mode and insert “EACE”. Exit.
  • z=5<Alt+Tab>: Perform spelling correction. Select 5, “PEACE”. Exit. (Esc seems to work as Enter here!)
  • $xxxxaASSWRD<Alt+Tab>: Go to the end of the line. Delete 4 characters and add ASSWRD, resulting in PASSWRD. Back to normal mode. (No, 4x won’t work.)
  • z=1<Alt+Tab>: PASSWORD is probably going to be the first correction for this. Select it.
  • $a$aA<Alt+Tab>: Go to the end of the line, and add a $aA. Back to normal mode.
  • ==: Format this line of PHP nicely, changing the camelCase $aA to $a_a. This also moves the cursor back to the start of the line.
  • wxx: Go forward a word; the cursor is now before $. Delete two characters – the $ and the a – to make PASSWORD_a.
  • $r1: Go to the end of the line and replace the current character (i.e. a) with 1, resulting in PASSWORD_1.
  • ^X: Decrement the integer under the cursor, resulting in PASSWORD_0.
  • a1: Finally, append 1, for PASSWORD_01!
  • \$\begingroup\$ I can't replicate == command, it just toggles large indentation mode in my vi setup (debian 7, v7.3.547) \$\endgroup\$
    – Cengiz Can
    Commented Jun 19, 2014 at 14:34

Python 2, 75889 bytes

No extra character!

Unfortunately, the code is super long. So instead, here's the code to generate the program:

x=[2**n for n in xrange(16)]
print('exec"""exec'+x[7]*'%'+'c""'+x[6]*'%'+'cr'+x[-2]*'%'+'c'+x[-1]*'%'+'ct'+x[8]*'%'+'c'+x[-4]*'%'+'cASSW'+x[-5]*'%'+'cRD'+x[-3]*'%'+'c'+x[0]*'%'+'s1'+x[10]*'%'+'c '+x[9]*'%'+'c""'+x[y]*'%'+x[1]*'%'+'s'+x[y]*'%'+x[2]*'%'+'s'+x[y]*'%'+x[3]*'%'+'s'+x[y]*'%'+x[4]*'%'+'s'+x[y]*'%'+x[5]*'%'+'s"""'+'%`1>>1`%`2531>>5`%`321>>2`%`1521>>4`%`211>>1`%`221>>1`%112%34%34%34%34')

Try it online

(to run the code, change the outer print into an exec. Or, generate the program, then paste and run.)


The goal is to execute:


To avoid using ()+ chars, we have to use multiple string format operations chained together. This means that each step added effectively doubles the number of % characters needed for each previous step:


This is not enough though, because some of the numbers required cannot be created, and neither can we create print. So we add a level of abstraction with exec for printing, and a level for the ASCII code points we cannot create to format with. The following is essentially my answer, but with each string of % reduced to a single one (note that the %s%s%s%s%s is not an accurate representation, failing to account for the literal %'s required before each):

exec"""exec%c""%cr%c%ct%c%cASSW%cRD%c%s1%c %c""%s%s%s%s%s""" % `1>>1`%`2531>>5`%`321>>2`%`1521>>4`%`211>>1`%`221>>1`%112%34%34%34%34

The First Step: Outermost exec:

exec""" ... ASSW%cRD ... %s%s%s%s%s""" % `1>>1` % `2531>>5` % `321>>2` % `1521>>4` % `211>>1` % `221>>1`
  1. 1>>1: 0, used for the 0 in PASSWORD_01
  2. 2531>>5: 79, used for inner exec to create O
  3. 321>>2: 80, used for inner exec to create P
  4. 1521>>4: 95, used for inner exec to create _
  5. 211>>1: 105, used for inner exec to create i
  6. 221>>1: 110, used for inner exec to create n

Second Step: Prepare Inner exec

After the above, the inner exec is something like this:

exec%c""%cr%c%ct%c%cASSWORD%c01%c %c""%79%80%95%105%110

Except that you wouldn't quite get that. Still wtih the simplifications, you'd get this:

exec%c""%cr%c%ct%c%cASSWORD%c01%c %c""798095105110

The literal %'s need to be included. So basically I had to add a bunch of them before each %s to end up with literal %'s left after all the formatting. 2**11 of them. Times five.

The rest of the outer exec string format operations are %112%34%34%34%34. The 112 is for p, and the others are quotation marks. After applying those, the result is something like this:

exec"""pr%c%ct"%cASSW%cRD%c01" """%79%80%95%105%110

But in reality it has a bunch more %. It's this:

exec"""pr%%%%%%%%c%%%%%%%%%%%%%%%%ct"%%cASSW%cRD%%%%c01" """%79%80%95%105%110

The final step is to simply run that, after which you get the necessary output.

  • 8
    \$\begingroup\$ Impressive! You typed all that by hand before the BOMB exploded? Without making any mistakes (no backspace key)? You, sir or madam, deserve your pension. ;D \$\endgroup\$
    – DLosc
    Commented Feb 16, 2017 at 2:17

bash + vim (56)

Borrowing the Ctrl-XCtrl-E bash trick from Thor's solution, here is how I would do it in bash+vim:

C-XC-E starts default editor (usually vim)

a starts insert mode

.space ASSW

C-Vx4f inserts O


C-Vx5f inserts _


C-3 is equivalent to escape (not just in vim, but anywhere in a terminal)

C-X subtracts 1 from the 1 i just typed

a insert mode again

1 C-3

buffer content is now . ASSWORD_01

<< unindent line (a no-op, since line is not indented) and move cursor to 1st column


C-X start word completion

C-V complete with ex command

C-V 9 more times selects the entry Print

C-3 back to normal mode

XXXxx deletes rint 

< < back to column 1

s delete ., start insert mode

e c C-X C-V ex command completion once again, entry echo already selected because of the ec i just typed

space C-3 buffer content now echo PASSWORD_01

Z Z save buffer, close vim, bash executes file content, i.e. echo PASSWORD_01

By the way: C-3 has many useful brethren: C-J is Enter, C-I is Tab, C-H is Backspace, C-2 is C-@ (i.e. a null-byte). And for emacs users it is nice to know that Escape followed by another key is equivalent to Alt + that key. So, even without Escape and Alt you can still type Meta-x like this: C-3x


Perl, (31+10 / 41+10 / 64+10)

(^ key used, 3 ways)

exec"easswsrdcqw"^"5    <  <AF"                   # exec "PASSWORD_01"

  • Valid if running as a system command qualifies as "output."

exec"QQZSqeasswsrdcqw"^"422<Q5    <  <AF"         # exec "echo PASSWORD_01"
  • Valid if I'm allowed to use external/OS commands. (Run here)

exec"EQAXqwQqQZQxqeasswsrdcqwxqq"^q!5434QZ4S534$S5    <  <AF$SS! 
                  # exec "perl -e\"die\\\"PASSWORD_01\\\"\""
  • Valid if I can execute Perl via the system. (Run here) <-- this one uses print instead of die

The scripts use Perl's bitwise string XOR operator ^, which does an XOR on the character bits of the two strings. This lets me recreate the missing characters for PASSWORD_01, and create the characters for the system command.

I made the three variants depending on how lenient the rules are. I assume that since variants 2 and 3 actually output the words on Ideone, the solutions are valid. I removed the little story I had here since I figured no one would read it, but you can read it in the edits if you're curious!


MS-DOS .COM 8086 machine code (72 bytes)

Time is precious, so no time to dilly-dally with compilers or interpreters! Simply open your text editor of choice and type in the following (replace \t with TAB):


As a hexdump:

00000000 : 34 35 31 44 3C 31 44 41 51 58 31 44 31 31 44 33 : 451D<1DAQX1D11D3
00000010 : 41 51 58 34 63 31 44 44 51 58 34 3C 53 5A 21 47 : AQX4c1DDQX4<SZ!G
00000020 : 71 09 47 71 33 57 71 45 45 45 45 45 53 58 35 09 : q.Gq3WqEEEEESX5.
00000030 : 09 32 21 3C 3C 3C 3C 3C 3C 3C 3C 3C 65 41 53 53 : .2!<<<<<<<<<eASS
00000040 : 57 7A 52 44 3C 31 31 24                         : WzRD<11$

Save as a .COM file and run it to save the day.

The code assumes certain start values for registers, so might not work for all DOS flavours. Just hope that someone risked being fired by not buying IBM.

Slightly more understandable representation:

org 0x100
bits 16
cpu 8086

; Assumes the following start values for registers (as per http://www.fysnet.net/yourhelp.htm):
;   AX = 0
;   BX = 0
;   CX = 0x00ff
;   SI = 0x100

; Change e => P and z => O
    xor al, 0x35
    xor [si+0x3c], ax
    xor [si+0x41], ax

; Fix int 0x21 and ret
    push cx
    pop ax              ; AX = 0xFF
    xor [si+0x31], ax
    xor [si+0x33], ax

; Change <1 => _0
    inc cx              ; CX = 0x100
    push cx
    pop ax              ; AX = 0x100
    xor al, 0x63        ; AX = 0x163
    xor [si+0x44], ax

; We're using DOS interrupt 0x21, function 0x09 to output the final string.

; AH must equal 0x09 to select the function.
; DS:DX must point to the $-terminated string we want to output.

; Get DX to the correct value (0x13c)
    push cx
    pop ax              ; AX = 0x100
    xor al, 0x3c        ; AX = 0x13c
    push bx
    pop dx              ; DX = 0

; We use part of the Program Segment Prefix to temporarily store the value,
; since MOVing or PUSHing AX is impossible.
; The part in question is the second File Control Block, which we will not use anyway.
    and [bx+0x71], ax
    or [bx+0x71], ax
    xor dx, [bx+0x71]   ; DX = 0x13c

; NOPs to get int 0x21 and ret on high enough addresses
    inc bp
    inc bp
    inc bp
    inc bp
    inc bp

; Set AH to the correct value. AL is set too, but is ignored by the DOS interrupt.
    push bx
    pop ax              ; AX = 0
    xor ax, 0x0909      ; AX = 0x0909

; What will become instructions int 0x21 and ret
    db 0x32, 0x21
    db 0x3c

; Padding to have the password at an address we can refer to.
    times 60-($-$$) db '<'

; Password
    pw  db  "eASSWzRD<11$"

oOo code (300)


Easy. (Linefeeds are optional, and just here to make the code "more readable") Code generator used:

o=lambda b,c:"".join(i.upper()if j else i for i,j in zip(__import__('itertools').cycle(c),map(int,"".join((3-len(i))*'0'+i for i in(bin('><[]-+.,'.index(i))[2:]for i in b)))))

The same code in a less silly encoding:


brainfuck, 400 + 10 (+) = 410 bytes


Try it online!

  • 2
    \$\begingroup\$ You can remove the < at the end. \$\endgroup\$ Commented Jun 18, 2020 at 18:35

Poetic, 319 bytes

war restarts.a test.a craze.a few desecrated areas.a few deceased staff.a sad defeat.waters red.we start a retreat after a war.a carcass wasted.a war act started warfare.war gets screwed.crew defeated.targets set.war affects qatar.a red war creates regret.a war deterred far extra starred staff.we scatter.gee!a regress

Try it online!

Poetic is an esolang I made in 2018 for a class project. It's basically brainfuck with word-lengths instead of symbols.

It was an interesting challenge to write a piece of text with certain word lengths without the right side of the keyboard (or many spaces or punctuation marks). I can only hope that it was worth it.


Emacs (running in a terminal), 26 bytes (possibly 24 bytes)

The program itself (the commas delimit the boundaries between bytes in the on-the-wire format that Emacs uses to accept its input):

W, S, Ctrl-3, $, D, A, Ctrl-T, S, R, W, D, Ctrl-3, $, 5, Ctrl-Q, 5, f, Ctrl-X, r, Ctrl-Space, 1, Ctrl-X, r, g, 1, Ctrl-T

Encoding of this as raw bytes, to prove byte count:

00000000: 5753 1b24 4441 1453 5257 441b 2435 1135  WS.$DA.SRWD.$5.5
00000010: 6618 7200 3118 7267 3114                 f.r.1.rg1.

(Note: some of the details may vary based on how Emacs is configured; this answer requires quoted-char-radix to be set to 16, and for Emacs to use the spellcheck dictionary that's default on my British English system. Both of these seem like reasonable configuration settings, but it's possible that your copy of Emacs may be configured differently. A different dictionary would likely still give a 26-byte program, but slightly different misspellings might need to be used so that the corrections we wanted can be accepted by non-bullet-ridden keys.)

Note about scoring

In addition to the keys specified in the question, I originally thought this answer required either Esc or Alt to work. (Esc is on the half of the keyboard that's still intact, so it morally feels like it should be allowed, but it isn't explicitly permitted.) However, user2845840's answer points out that (with the usual encoding of terminal control codes) Esc can alternatively be typed as Ctrl-3, and that is explicitly permitted by the question. (An aside: while testing this, I learned some other interesting ways to type control codes, with Ctrl-2 typing NUL, and Ctrl-4…7 typing character codes 28…31 respectively. Note that the default configuration of many terminals is to interpret a character code 28 coming from user input as a request to crash the currently running program, so be careful testing this.)

Alt is also on the intact half of the keyboard but isn't mentioned in the question. Although Alt and Esc have different encodings in some terminals, they are encoded the same way in other terminals, and (presumably to work around these compatibility issues) Emacs interprets the two keys (and their encodings) as entirely equivalent to each other. If we could use Alt, though, it would save two bytes (because it has a shorter encoding than Esc does).


I'm not sure whether it should have any influence on the editor wars, but Emacs seems to beat vim at least in the case of this question. Emacs is pretty suited to editor golf measured in bytes, because it relies heavily on chords which take up multiple keypresses but only a single byte (thus an Emacs program is often slower to type than the equivalent Vim program, but shorter on disk). Additionally, most of the most important Emacs commands are in the bottom-left corner of the keyboard, to be close to Ctrl, very helpful with a question like this one.

"You can assume you have the interpreter shell / source editor opened before the bullets came in. Sadly you have not written anything in it before the keyboard was hit.", so I'm assuming that we have an empty file open in Emacs and need to type the password into it. (We'd need to save the file afterwards, and probably exit Emacs, but the bytes for that aren't being counted in other people's answers so I'm not counting them here either. It's totally doable using the left hand side of the keyboard, though, Ctrl-X, Ctrl-S, Ctrl-X, Ctrl-C.)

Taking it a command (or block of similar commands) at a time:

  • W, S: Enter WS into the document.
  • Ctrl-3, $: Invoke the spellchecker. WS isn't a real word, but it finds lots of similar two-letter words. (The usual way to invoke the spellchecker is Alt-$; here, we're using Ctrl-3 as a stand-in for Esc, which in turn is a stand-in for Alt.)
  • D: Using the spellchecker, correct WS to PS. (When the spellchecker is invoked using Alt-$, as happened here, it only checks one word, so it deactivates after doing this.)
  • A: insert A, giving PSA.
  • Ctrl-T: Swap the previous two characters, giving PAS.
  • S, R, W, D: Type SRWD, giving PASSRWD.
  • Ctrl-3, $, 5: We invoke the spellchecker again, because we want to correct our misspelled PASSRWD into PASSWORD. Note that we can't have it guess the word we want on our first try, like it would with PASSWRD, because the key to accept the closest real word is 0 which we can't press. As a result, the slightly more extreme misspelling PASSRWD is used to push the word we want into position 5, where we can accept it.
  • Ctrl-Q, 5, f: Insert the character with character code U+5f, i.e. _. The document now reads PASSWORD_ (or will when we start typing the next command; before then, the underscore doesn't appear in case we type another hex digit).
  • Ctrl-X, r, Ctrl-Space, 1: Store the current cursor position (relative to the start of the file) in register 1. For some bizarre reason, this is 1-indexed, so (having written 9 characters so far) the cursor is at position 10.
  • Ctrl-X, r, g, 1: Copy the content of register 1 into the document. It now reads PASSWORD_10.
  • Ctrl-T: Swap the two characters before the cursor. We now have PASSWORD_01, like the question asks for.

If we're allowed to use Alt, we can probably encode the "invoke spellchecker" command as the single byte a4 rather than spelling it out as 1b 24; it appears twice, so that leads to two bytes of savings. (Most modern terminals use 1b 24 as the encoding for Alt-$ to avoid clashes with UTF-8, but the a4 encoding is also encountered from time to time, sometimes available as a configuration option.)

Possible byte savings probably involve golfier misspellings to correct. PSASWRD would be a byte shorter to type, but unfortunately, the spellchecker doesn't seem capable of gleaning PASSWORD out of that, so PASSRWD is the best approach I've found so far. The register-based approach to gaining 10 is also ridiculously unwieldy, but there aren't many ways of creating numbers from nowhere in Emacs, and 0 is a painful character to get hold of otherwise. (At least there were a couple of amazingly useful coincidences: the cursor just happening to end up in position 10, which contains a 0, right when needed; and the fact that Emacs accepts the redundant g register operation to insert the contents of a register into the document, in addition to the more intuitive i.)

  • \$\begingroup\$ I can use QASS??5 than WS??DA?S to input PASS and PASSRD??3 for PASSWORD, does it work on other emacs? \$\endgroup\$
    – l4m2
    Commented Apr 9 at 8:21
  • \$\begingroup\$ @l4m2: On my dictionary, PASSRD does indeed resolve to "PASSWORD" at position 3, which would save a byte. I won't add that to my answer because I didn't think of it myself, but it's a good improvement and worth preserving in the comments. \$\endgroup\$
    – ais523
    Commented Apr 9 at 22:46

ferNANDo, 179 + 10 = 189 bytes

2 1
1 2 1 2 1 1 1 1
1 2 1 1 1 1 1 2
1 2 1 2 1 1 2 2
1 2 1 2 1 1 2 2
1 2 1 2 1 2 2 2
1 2 1 1 2 2 2 2
1 2 1 2 1 1 2 1
1 2 1 1 1 2 1 1
1 2 1 2 2 2 2 2
1 1 2 2 1 1 1 1
1 1 2 2 1 1 1 2

Try it online!

Enter used. I'd use 0 and 1 to make the code more readable but I can't use 0.

  • \$\begingroup\$ I think enter is allowed. \$\endgroup\$ Commented Sep 28, 2016 at 22:19
  • 3
    \$\begingroup\$ OP didn't say Enter was allowed. \$\endgroup\$
    – acrolith
    Commented Sep 28, 2016 at 23:35

JS-Forth, 103 bytes

The string will be returned on the stack as an array of characters.

12544 1 3 << >> 24 1 << 1521 4 >> 34 1 << 41 1 << 2531 5 >> 351 2 >> 332 2 >> 32 2 >> 131 1 >> 321 2 >>

Try it online - Commented version


I first found the list of allowed words. Essentially, the only things I could use that would be useful are:

  • 12345 Numeric constants
  • << Shift left
  • >> Shift right
  • s>f Push single to float stack
  • f>d Pop double from float stack
  • fsqrt Square root on float stack

So I could use numeric constants, bit-shift, and compute a square root using the following (>> replaces drop, shifting by 0 to remove the 0):

s>f fsqrt f>d >>

Fortunately, I found possible bit-shifts for every constant I needed to create that were shorter than using square roots. For the most part, I searched by simply printing each number squared, or a larger power of two if any of them contained a digit 6-0. Then, I realized I could use loss of precision from f>d to find more possibilities. (I could add to the number, but still get the same integer square root.) A bit later, I started using bit-shifting to find some, and then all, of the constants. The greater the >> bit-shift, the more I could add to the "magic number" and still get the same result. So I found the smallest bit-shifts I could use to get the necessary results. Even numbers could sometimes use <<, odds had to use >>.

Commented code (\ starts a comment):

\ 49
12544 1 3 << >>

\ 48
24 1 <<

\ 95
1521 4 >>

\ 68
34 1 <<

\ 82
41 1 <<

\ 79
2531 5 >>

\ 87
351 2 >>

\ 83 (x2)
332 2 >> 332 2 >>

\ 65
131 1 >>

\ 80
321 2 >>

gForth does not have the words << or >>. Instead it has lshift and rshift, which I could not use.


Decimal, 190 180 + 10 = 190 bytes


Try it online! Decimal is an esoteric language that uses nothing but decimals and the letter D. However, I needed the +10 penalty because almost every command uses 0.

Luckily, all of the commands needed to print PASSWORD_01 don't require any numbers over 6:

  • 12...D - push a character
  • 2 - pop a value
  • 41D - pop top two stack values, multiply and push result
  • 301 - print DSI (default stack index)

By repeatedly pushing character values that only use the digits 1 through 5, then adding them, I can make the character values for each letter in PASSWORD_01.

Ungolfed and commented:

12055D     ; push char 55 to stack
12025D     ; push char 25 to stack
41D        ; pop [DSI] and [DSI-1], add and push result
3012       ; print from stack to output, pop
12055D     ; push char 55 to stack
12010D     ; push char 10 to stack
41D        ; pop top two stack values, add and push result
3012       ; print from stack to output, pop
12055D     ; push char 55 to stack
12025D     ; push char 25 to stack
41D        ; pop x2, add, push
12003D     ; push char 3 to stack
41D        ; pop x2, add, push
301301     ; print, print
12004D     ; push char 4 to stack
41D        ; pop x2, add, push
3012       ; print, pop
12055D     ; push char 55 to stack
12024D     ; push char 24 to stack
41D        ; pop x2, add, push
301        ; print
12003D     ; push char 4 to stack
41D        ; pop x2, add, push
3012       ; print, pop
12055D     ; push char 55 to stack
12013D     ; push char 13 to stack
41D        ; pop x2, add, push
3012       ; print, pop
12055D     ; push char 55 to stack
12040D     ; push char 40 to stack
41D        ; pop x2, add, push
3012       ; print, pop
12045D     ; push char 45 to stack
12003D     ; push char 3 to stack
41D        ; pop x2, add, push
301        ; print
12001D     ; push char 1 to stack
41D        ; pop x2, add, push
301        ; print

Stax, 91+10 bytes

"W"EEEd$"$"1 !$Re]"$""ASSW"R"$""a"EEE$"$"a$Re]R"$""RD"R"$""a"EEEd$"$""5"Re]R"$"1 !$R"$""1"R

Try it online!

Uses ] to make singletons.


Let's see how "P" is generated and how the "ASSW" is appended without using the usual +.

"W"                             Push "W", or [87]
   E                            Convert to a scalar, 87
    E                           Convert to an array of digits, [8,7]
     E                          Push 8 on the stack, then push 7.
      d                         Pop the 7 and discard it.

                                Start making components of regex substitution
       $                        Pop 8, push "8"
        "$"                     Push "$"
           1 !                  Push 0. The space is necessary because normally one doesn't do 1! to obtain 0
              $                 Pop 0, push "0"
               R                Do regex substitution 's/$/0/g' on the string "8"

                                Basically doing "8"+"0" without using +
                e               Pop "80", push 80
                 ]              Singleton. Pop 80, push [80], or "P"
                  "$""ASSW"R    Use the same trick again to append "ASSW" to "P"

The rest are just repeating the same trick.

"a"EEE$"$"a$Re]R swaps the digits of "a" or [97] to make [79] or "O" and appends it to the string.

"$""RD"R appends "RD".

"$""a"EEEd$"$""5"Re]R uses "a" again to obtain "9" and discards the "7", then combines the "9" with a literal "5" to form [95] or "_", and appends it to the string.

"$"1 !$R obtains a zero by logical not of 1 and appends it to the string.

"$""1"R appends the final "1" and we are done.

Implicit output.

  • 1
    \$\begingroup\$ O_O ... this is clever. \$\endgroup\$
    – recursive
    Commented Feb 25, 2018 at 7:20
  • \$\begingroup\$ Thank you for your appreciation. I believe it can be made shorter (perhaps by choosing a different key?) though. \$\endgroup\$ Commented Feb 25, 2018 at 7:37

Whitespace, score: 111 (101 bytes + 10 for Enter)

[S S T  T   T   T   T   T   N
_Push_-31_1][S S T  T   S S S S S N
_Push_-32_0][S S S T    T   T   T   N
_Push_15__][S S T   T   T   S S N
_Push_-12_D][S S S T    S N
_Push_2_R][S S T    T   N
_Push_-1_O][S S S T T   T   N
_Push_7_W][S S S T  T   N
_Push_3_S][S N
S _Duplicate_3_S][S S T T   T   T   T   N
_Push_-15_A][S S S N
_Create_Label_LOOP][S S S T S T S S S S N
_Push_80][T S S S _Add][T   N
S S _Output_as_character][N

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).

Explanation in pseudo-code:

This solution is shorter than the existing Whitespace answer (@n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ asked me to post it as a separated answer when I suggested it as golf) by utilizing this Whitespace tip of mine.

Push the code-points for the characters of "10_DROWSSAP" minus constant 80 to the stack
Start LOOP:
  Add 80 to the current code-point
  Pop and output it as character
  Go to the next iteration of LOOP

The constant 80 has been generated with this Java program, which I've created and used for some previous Whitespace answers of mine.


Easyfuck, 145+10(.)


This was really challenging as there were no ways of manipulating cells other than the modulo operation and setting them to some multiples of 16. Fortunately the letters W and Z were present letting me wait for a certain duration and set the current cell to seconds elapsed. I had to also include . as it's the only way of printing characters in easyfuck. All in all, one of the most fun challenges I partook in. I sure hope my buddy can distract them for ~96 more seconds...


Befunge-98, 43 +10 = 53 bytes


Try it online!

The 10 byte penalty is for the use of the , key (not sure why this couldn't be accomplished with Shift+<, but that seems to be the rule). And although the source is technically 40 characters, the three · characters contribute an additional three bytes because of their UTF-8 encoding.


05AB1E, 87 86 bytes


Try it online!

Maybe the longest 05AB1E code ever?

Unfortunately the language doesn't support things like "Q"< for "decrement Q to get P"

My extra key is the V.

Thanks to @Kevin Cruijssen for -1 byte.


Build the string ".b". This is the code to turn integers to upper case letters:
"."             push "."
A               push the lower case alphabet
1.$             drop the first letter of that
R               reverse the remaining string 
24.$            drop the first 24 letters of that, so only "b" remains
AT<.$R15>.$     do the same trick for "j"
A21<.$R5.$      and for "u" 
.V              run "u" as code to turn "j" to "J"
.V              run "J" as code, to join ".", "b" to ".b"

Save that string for later use:
D               duplicate
21              push 21
s               switch the top two elements [".b", 21, ".b"]
.V              run ".b" as code, to get "U"
.V              run "U" as code, to save ".b" in X

Build password string:
15>X.V          push the 16th letter of the alphabet ("P")
"ASSW"          push "ASSW"
15X.V           push "O"
"RD"            push "RD"
122D<s2X.V      push 121, 122, "B"
.V              run that as code to push "_"
TR              push 01
TX.V            push "J"
.V              run that as code, to join everything to the desired string 

For comparison: The shortest way to print "PASSWORD_01" is 6 bytes long

Try it online!

  • \$\begingroup\$ "Maybe the longest 05AB1E code ever?" Nah, this one of mine is almost 6 times as large. ;p Nice answer regardless. It's hard to only use a certain amount of allowed ASCII character. So many convenient builtins aren't allowed for this challenge. :( \$\endgroup\$ Commented Sep 30, 2019 at 12:57
  • 1
    \$\begingroup\$ -1 byte by using TR instead of 1<1 for the 01 part of the output. \$\endgroup\$ Commented Sep 30, 2019 at 13:03

Lenguage, 814814390776628060509605577373181765864505940171473720140916436885644483746988990322785194756 bytes

Converted from this program:


Try BF code online!


Width, 23 + 10 (m) = 33 bytes


Try it online!

  • \$\begingroup\$ This language reminds me of the CUNEIFORM NUMERIC SIGN NINE SHAR2: 𒐫, an awesome character that is awesome because it's a single character \$\endgroup\$ Commented Apr 6, 2020 at 14:52

Knight, 63 62 + 10 (+) bytes

E++A+55 24+A34+A+55 25+"ASSW"+A+55 24+"RD"+A+54 41+A+45 3 1A34

Try it online!

As noted on Encode your program with the fewest distinct characters possible, Knight needs a minimum of AE +1 to be fully usable (even if it is more torturous than Brainfuck). The + is required to do ASCII arithmetic.

This program generates the string O"PASSWORD_01" by assembling codepoints, and EVALs it with E.

XXX: check if SUBSTITUTE is beneficial here

Knight, 113 110 bytes

E S S S S S S S"ASSWRD1"%21 14F A34%21 15F A%152 52%21 15F A%321A"q"4F A%211 132F!1A%211 131F!1A34F!1A%211 132

Try it online!

Even without +, we can still do this. SUBSTITUTE can allow us to insert strings into other strings, and we can still generate single chars with A%x y. But I think + is worth it, no?

Generates the same string.

For the reference, the magic numbers are:

%21 14   -> 7
%21 15   -> 6
%152 52  -> 48 ('0')
%321A"q" -> %321 113 -> 95 ('_') # used to prevent token collision
%211 132 -> 79 ('O')
%211 131 -> 80 ('P')
  • 3 bytes: use !1 instead of F to avoid token collisions

Julia 1.5, 38 +10 (ENTER) = 48 "bytes"

The keystrokes not available are, of course, PO_0. For Julia, the others are all simple, and simply outputting the string "PASSWORD_01" should count by the rules. Also required: ENTER, as the alternate way to process that in Julia's REPL is Ctrl-J, and J isn't available.

Fortunately, one of the inbuilt commands that you can use TAB to autocomplete is DEPOT_PATH, which provides access to three of the characters. And shortcuts can handle deleting the undesired letters and positioning the cursor to add the missing ones. To get the 0, the easiest method I've found is to use autocomplete to get frexp, and then exp10.


Type: frTABCtrl-ACtrl-DCtrl-DCtrl-E1TAB

Result: exp10

Type: Ctrl-ADETAB

Result: DEPOT_PATHexp10

Type: Ctrl-A"Ctrl-DCtrl-DCtrl-FASSWCtrl-FCtrl-DRDCtrl-FCtrl-DCtrl-DCtrl-DCtrl-DCtrl-DCtrl-DCtrl-DCtrl-FCtrl-T"ENTER

Result: "PASSWORD_01" as output (rather than command) on hitting enter.


Autocomplete with TAB will work as long as there is only one known autocomplete option (otherwise, double-tab will bring up a list of possible completions).

Ctrl-A moves the cursor to the start of the line, Ctrl-D deletes the next character, Ctrl-F advances the cursor forward one space, and Ctrl-T switches the positions of the characters before and after the cursor (and moves the cursor forward one place).


Pyth, 39 + 10 (+) = 49 bytes

r1++++++C+55 25"ASSW"e<G15"RD"C+54 41Z1

Try it online!

This could probably be golfed a bit more but I give up. :) The code adds a series of strings to the stack, appends them all, and then translates the whole thing to uppercase. That leaves one string on the stack which is printed automatically when the code exits.

The Pyth specific tricks used here are related to getting the characters not available on the limited keyboard. The P and _ characters are generated by computing the codepoint of the character and using the Pyth C function. And the o character is extracted from the G variable, which is pre-defined to hold the alphabet. The r1 translation is needed just for that, since the alphabet constant is lowercase. That turns out to be the same number of characters as the codepoint method, so left both in the code.

I had to add the + character so the code gets a +10 penalty.

         +55 25                         - push 75 on the stack
        C                               - convert codepoint to "P"
               "ASSW"                   - push "ASSW" to the stack
       +                                - append top two stack entries
                       G                - push "abc...xyz" to the stack
                      < 15              - take 15 char substring
                     e                  - take last char on string "o"
      +                                 - append top two stack entries
                          "RD"          - push "RD" into the stack
     +                                  - append top two stack entries
                               +54 45   - push 95 on the stack
                              C         - convert codepoint to "_"
    +                                   - append top two stack entries
                                     Z  - push 0 onto the stack
   +                                    - append top two stack entries
                                      1 - push 1 onto the stack
  +                                     - append top two stack entries
r1                                      - convert to uppercase

Deadfish~, 154 + 10 bytes


Try it online!

No {} :(

Needs i, you can't do anything without it...

  • \$\begingroup\$ @Wasif See Help Mode. \$\endgroup\$
    – emanresu A
    Commented Apr 9, 2021 at 10:01

HexEdit, 20


* mean Tab, which switch between Hex mode and Direct input mode

> goes to end of file, aka. after the auto padded 0


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