# Be the First 1 (leave only the first Truthy)

### Intro

Every year, Dyalog Ltd. holds a student competition. The challenge there is to write good APL code. This is a language agnostic edition of this year's eighth problem.

I have explicit permission to post this challenge here from the original author of the competition. Feel free to verify by following the provided link and contacting the author.

### Problem

Given a Boolean* list, "turn off" all the Truthies after the first Truthy.

No Truthies? No problem! Just return the list unmodified.

### Examples

[falsy,truthy,falsy,truthy,falsy,falsy,truthy][falsy,truthy,falsy,falsy,falsy,falsy,falsy]

[][]

[falsy,falsy,falsy,falsy][falsy,falsy,falsy,falsy]

* All your truthies must be identical, and all your falsies must be identical. This includes output.

• Can we use bit lists or other truthy/falsy list representations that are more natural in our language of choice? – Martin Ender May 7 '17 at 20:32
• Well yeah, if you talk about "truthy" and "falsy" in the challenge instead of "booleans", "true" and "false". ;) – Martin Ender May 7 '17 at 20:36
• I'm not clear on the booleans. Can we use 0/1 even if our language has True/False? – xnor May 7 '17 at 20:59
• @xnor Ah, good point. I think it would be fair to allow choosing input, but output must match, don't you think so? – Adám May 7 '17 at 21:12
• @xnor I hear you, but if Haskell cannot treat numbers as Booleans, or cannot do arithmetic on Booleans, then that is a real limitation in the golfing power of Haskell, and ought to be reflected in the byte count by necessitating conversions or other work-arounds. What do you think of the footnote formulation? – Adám May 7 '17 at 21:19

# APL, 2 bytes

<\


Evaluates to the function "scan using less-than". Try it online!

## Explanation

In APL, the operator \ (scan) reduces each nonempty prefix of an array from the right using the provided function. For example, given the array 0 1 0, it computes 0 (prefix of length 1), 0<1 (prefix of length 2) and 0<(1<0) (prefix of length 2) and puts the results into a new array; the parentheses associate to the right. Reducing by < from the right results in 1 exactly when the last element of the array is 1 and the rest are 0, so the prefix corresponding to the leftmost 1 is reduced to 1 and the others to 0.

• Finally! I have been wondering. – Adám May 8 '17 at 10:25
• Now I suppose you can answer in J too, no? – Adám May 8 '17 at 10:31
• @Adám Yes, in J it's 3 bytes: </\ Jelly probably has an analogous 2-byte solution too. – Zgarb May 8 '17 at 10:33
• No, I don't think so, because Jelly is left-to-right. – Adám May 8 '17 at 10:38
• You should post separate language answers as separate posts. – Adám May 8 '17 at 10:43

# Python 2, 35 bytes

while 1:b=input();print b;True&=b<1


Try it online! Input and output are lines of True/False.

Based on Dennis's solution. Redefines the variable True to be False after a True input is encountered. That way, any further inputs of True will evaluate to False and be printed as such. Python 3 no longer allows True and False to be redefined, so this answer uses Python 2.

The redefinition is True&=b<1, i.e. True = True & (b<1). When the input b is True, then (b<1) is False (since True==1), so True becomes False. See Kevin Cruijssen's answer for a shorter way to handle this redefinition.

• You can redefine True??? This deserves a +1 just because hax >_> – hyper-neutrino May 8 '17 at 1:33
• @HyperNeutrino Yes, but not in Python 3. (Which is fine because the language here is Python 2.) – Brian McCutchon May 8 '17 at 2:40
• @BrianMcCutchon Okay thanks. That is just weird though... – hyper-neutrino May 8 '17 at 2:57
• @HyperNeutrino It's probably worth mentioning that you can do True, False = False, True. – Brian McCutchon May 8 '17 at 2:59
• @HyperNeutrino - nope. Builtins still return the 'real' value, it's just 'True' that you type that changes. (Or modules, in some cases...). So bool(1) return True, but bool(1) == True returns False. – TLW May 8 '17 at 3:12

# Aceto, 19 17 bytes

### New version (17 bytes):

This new version takes the characters one at a time and is best executed with the -F option. It works similar, but not identical to the previous solution:

 >,
Op0
p|1u
,ip^


|p1u
iOp<
|!X
rd!r


This is the first Aceto answer that highlights what it can do relatively well, I would say. The "lists" are input streams, with one input per line, "1" for true, and "0" for false, with an empty string signifying the end of the list. Aceto programs run on a Hilbert curve, starting on the bottom left, and ending on the bottom right. First, we read a string, duplicate, and negate (!) it, turning empty strings into True, everything else into False. Then there's a conditional horizontal mirror (|): If the top element on the stack is truthy, mirror horizontally. This happens when the string was empty. If we do the mirroring, we land on the X, which kills the interpreter.

Otherwise, we convert the remaining copy on the stack to an integer and do another conditional horizontal mirror: This time, because 1 is truthy and 0 is falsy, we mirror if we see the (first) true value. If we don't mirror (so we saw a 0) we print what's on the stack (since the stack is empty, a zero) and jump to the Origin of the curve, where we started, starting the whole process again.

Otherwise, when we saw a 1, we mirror and land on the u, which reverses the direction we move on the Hilbert curve. 1p prints a 1, and now we go on the same O we would have gone if we had seen a 0, but since we're in "reversed mode", our origin is at the bottom right, so we jump there.

Now we read another string, and negate it. If the string was empty, and therefore the top stack element is truthy,  will not escape the next command (X), making us quit.

Otherwise (if the string wasn't empty), we do escape the X and ignore it. In that case, we go to the left (<), print 0 (because the stack is empty), and jump back to the Origin.

• Congratulations on your first proper challenge solved in Aceto. – Adám May 7 '17 at 21:26
• Looks at diagram. Right… – Adám May 7 '17 at 21:58
• @Adám It probably won't help (if you don't know Aceto) on its own, but I thought it might be good to see alongside the text to be able to follow it better. – L3viathan May 7 '17 at 22:03

# Java8, 24 19 Bytes

Long::highestOneBit


Hope this is legal; I got the impression the input / output doesn't have to evaluate as true/false in the language. Takes a long as input and gives one as output, with ones being true and zeroes being false in the binary representation. For example, binary 00101 is 5 and would return binary 00100 which is 4.

Five bytes thanks to @puhlen

• Nice approach. Java being competitive‽ – Adám May 8 '17 at 11:04
• Wow, JAVA as a competitive answer‽ – Adalynn May 8 '17 at 11:06
• Not entirely sure if this is valid for codegolf rules, but this could be improved to 19 chars by using a method reference: Long::highestOneBit which produces the identical result with a shorter syntax – puhlen May 8 '17 at 15:58
• @puhlen expressions evaluating to anonymous functions are allowed. – Cyoce May 8 '17 at 17:27
• @NathanMerrill The java.lang package is imported by default. From the language spec "A compilation unit automatically has access to all types declared in its package and also automatically imports all of the public types declared in the predefined package java.lang." – JollyJoker May 9 '17 at 7:21

## Retina, 6 bytes

1>1
0


Try it online!

Input is a list of 0s (for False) and 1s (for True).

Matches all 1 and replaces each except the first one (1>) with a 0.

• I can see it now. You working in an office on some OS. A manager comes over and yells at you for writing an entire OS with regex. – Christopher May 15 '17 at 21:04
• I was wondering who left that ^ dumb comment then realized it was me... – Christopher Apr 12 '20 at 16:35

# V, 7 bytes

f1òf1r0


Try it online!

My first V submission! \o/

## How it works

f1òf1r0
f1       "go to the next occurence of 1
ò      "repeat the following until end:
f1    "    go to the next occurence of 1
r0  "    replace with 0

• How does this work? – Brian McCutchon May 8 '17 at 2:43
• @BrianMcCutchon Explanation added. – Leaky Nun May 8 '17 at 2:45
• This fails for a 1 in the first position :( – nmjcman101 May 9 '17 at 2:17
• @nmjcman101 fixed. – Leaky Nun May 9 '17 at 3:08
• Since you changed the input format, you can swap r0 with <C-x> to decrement the ones and save a byte. – nmjcman101 May 9 '17 at 13:54

# Python 2, 33 bytes

while 1:b=input();print b;True^=b


Port of @xnor's Python answer, minus 2 bytes. Normally I would just suggest it as a golf in the comments, but since @xnor has specifically stated we can outgolf any of his answers for a bounty, here it is as a separated answer.

Try it online.

The difference? True&=b<1 has been changed to True^=b.

Why does this work?

First we get some potential False inputs (b=input();). Each of those will be printed as is (print b;). And the True^False will remain True, so 'variable' True remains truthy.

Then we encounter the first True input, which will also be printed. After that, the True^True will evaluate to False, so the 'variable' True is from now on falsey.

After that it doesn't matter what our input is anymore, since True is redefined as falsey. So the print b; will always print False at this point. And in addition, the True^=b can now only be False^False, so 'variable' True will remain falsey.

• Wow, TIL True can be reassigned. Nice golf! – Surculose Sputum Apr 6 '20 at 13:25
• @SurculoseSputum Yeah, I also didn't knew before seeing xnor's answer. Then again, I think you can reassign most constants in Python. I think I've also seen someone user and overwrite the id constant in another challenge for example. – Kevin Cruijssen Apr 6 '20 at 13:28
• @KevinCruijssen Nice find, bounty posted! I like how elegant the ^= update is with it becoming harmless after the first True. What makes your find more impressive is that a lot of voters on my answer apparently didn't think of this, in addition to me. – xnor Apr 6 '20 at 19:15
• Funny, because I was thinking about this @xnor answer :p turns out this must be a recurrent trick, as id is the shortest predefined and reassignable name in Python – RGS Apr 10 '20 at 10:19
• @jaaq That's evil! XD – Kevin Cruijssen Apr 13 '20 at 8:40

# 05AB1E, 6 bytes

Code:

ā<s1kQ


Explanation:

ā         # External enumeration, get a and push [1 .. len(a)]
<        # Decrement each
s       # Swap to get the input
1k     # Get the first index of 1
Q    # Check for equality with the enumeration array


Uses the 05AB1E encoding. Try it online!

# Jelly, 4 bytes

+\=a


Try it online!

Here's a rather different algorithm to most of the other golfing language solutions (although after I posted it, I noticed that the R solution also uses this algorithm), and tying with the current Jelly record holder.

## Explanation

+\=a
+\    Cumulative sum of the input list
=   Compare corresponding elements with the input
a  Logical AND corresponding elements with the input


As long as all elements to the left of an element are 0, the cumulative sum up to an element will equal the element itself. To the right of the first 1, the two are different (because we're now adding the nonzero total of the elements to the left). Thus, +\= gives us a list containing 1 (i.e. true) up to and including the first truthy element. Finally, logical AND with the original list will give us a 1 for only the first truthy element.

Anonymous function taking and returning a list of Bools.

Use as (foldr(\x l->x:map(x<)l)[])[False,True,False,False].

foldr(\x l->x:map(x<)l)[]


Try it online!

# How it works

• Folds over a list from the right, prepending new elements and possibly modifying those following.
• x is the element to be prepended to the sublist l.
• Uses that False compares less than True, so map(x<)l will turn any Trues in l into False if x is True.

## JavaScript (ES6), 33 26 bytes

a=>a.map(e=>e&!(i-=e),i=1)


I/O is in arrays of 0s and 1s.

# Turing machine simulator, 39 bytes

0 0 0 r 0
0 1 1 r 1
1 0 0 r 1
1 1 0 r 1


Try it online!

• Essentially the same thing I'm doing in my answer. – L3viathan May 8 '17 at 7:36

# brainfuck, 55 bytes

+>,[[->+>[->-<]>+[-<+>]<<<]>>[-<-<<[->>+<<]>>>]<.[-]<,]


Try it online!

## J, 3 bytes

</\


Defines a monadic verb. This is a trivial port of my APL answer. Try it online!

# Jelly, 4 bytes

A port of my 05AB1E answer.

i1=J


Explanation (argument α):

i1        # Index of 1 (1-indexed) in α
=       # Check for equality with the array:
J      # [1 .. len(α)]


Try it online!

# R, 24 bytes

cumsum(T<-scan(,F))==T&T


Try it online!

Example:

For input FALSE TRUE TRUE FALSE
cumsum(T<-scan(,F))==T returns TRUE TRUE FALSE FALSE. The F in the scan ensures logical input.
FALSE TRUE TRUE FALSE and TRUE TRUE FALSE FALSE is FALSE TRUE FALSE FALSE. A single & does an elementwise comparison.

• @rturnbull unfortunately the input format has to be the same as the output. – MickyT May 10 '17 at 17:17
• cumsum(scan(,F))==1 works for 19 bytes. – Robin Ryder Apr 6 '20 at 13:44

# Octave, 23 bytes

@(a)diff([0 cummax(a)])


Try it online!

First difference of cumulative max of the list.

# Perl 5, 20 bytes

sub{map$_&&!$x++,@_}


Truthy is 1 and falsey is '' (an empty string).

Explanation:

map loops over elements of the list it @_, the arguments passed to the subroutine, setting each element to $_ locally and returning an array of the return values it computes from each element. $_&&!$x++ outputs $_ if $_ is falsey and !$x++ if it is truthy. (Note that && is short-circuiting, so !$x++ is not executed until the first truthy value is reached). $x++ returns 0 (which is falsey) the first time it is run and then increments every time (and so remains truthy). The ! negates $x++, and so it returns truthy the first time it is encountered and falsey thereafter. # Python, 58 bytes lambda x:[x[i]and x.index(x[i])==i for i in range(len(x))]  If x[i] is false, the output is false; otherwise, it gives whether or not the element is the first occurence in the array of itself. # PHP, 37 bytes foreach($_GET as$v)echo$v*!v++,' ';


# Pyth - 9 bytes

.e&b!s<Qk


Try it here

.e&b!s<Qk
.e          # Python's 'enumerate' (i.e., for each index k and each element b at that index)
<Qk   # The first k elements of the input
s      # 'Sum' these first k elements (with booleans, this is a logical 'or')
&b!       # The value of the output at index k is [value of input @ index k]&&[the negation of the 'sum']

• It seems to be more efficient to use a variable and just map over it normally: m&!~|Z. – FryAmTheEggman May 8 '17 at 0:28

# Python 2, 45 36 bytes

r=0
while 1:n=input();print n>r;r+=n


Input and output are one Boolean (True or False) per line.

Try it online!

# Idris, 98 bytes

The code is a bit longer, but on the plus side you get the compile time guarantee that the output has the same size as the input!

import Data.Vect
f:Vect n Bool->Vect n Bool
f[]=[]
f(x::y)=x::if x then replicate _ False else f y


I'm going to give an extensive explanation to have you understand the basic workings of Idris.

### Explanation

import Data.Vect


The type Vect : Nat -> Type -> Type is not imported by default

f : Vect n Bool -> Vect n Bool


Contrary to Haskell, Idris uses a single colon : to specify types and also requires you to specify the type, as the dependent type checker can't possibly infer it in all cases. Since Vect takes a Nat (natural number) and a Type, we provide just that, n being a natural number and Bool being the type. Even though we didn't specify the type of n explicitly, Idris can infer it to be Nat. Since n occurs both in the argument and the resulting type, they need to be the same.

f [] = []


As the base case, the empty list returns the empty list. This is the only possible implementation, anything else such as f[]=[True] would give a compiler error, since the type of [] is Vect 0 Bool but the type of [True] is Vect 1 Bool.

f (x :: y) = x :: if x then


:: is used as the cons operator in Idris, similar to : in Haskell. We put the first element in the resulting list, unchanged.

  replicate _ False else


If the first element is true, we want the rest of the list to be false in any case, so we just replicate the value False. Since Idris knows that the rest of the list has to have the same size as y, it can infer the value of _ (namely length y).

  f y


If the first element is false, we just move on by recursively calling the function on the rest.

## C#, 77 bytes

a=>{var b=1<0;for(int i=0;i<a.Length;){a[i]=b?1<0:a[i];b|=a[i++];}return a;};


Compiles to a Func<bool[], bool[]>. Nothing clever really, just a straight forward solution.

# sed, 16 19 bytes

15 18 bytes sourcecode + 1 byte for -r flag (or -E flag for BSD sed).

:
s/1(0*)1/1\10/
t


Try it online!

Edit: Thanks Riley for pointing out a mistake.

• @Riley Thanks for pointing that out! It looks like TIO has a version of sed that is different from mine (BSD). I can't leave the labels empty. Good to know this. – Maxim Mikhaylov May 8 '17 at 13:19
• Yeah, sorry. TIO uses GNU sed. It's a bug turned feature. – Riley May 8 '17 at 13:20

# Jelly, 4 bytes

TḊṬ^


Try it online!

### How?

This does what was asked in a pretty literal sense:

TḊṬ^ - Main link: list a   e.g. [0,1,0,1,0,0,1]  or  [0,1,0,1,0,1,0]
T    - get the truthy indexes   [  2,  4,    7]      [  2,  4,  6  ]
Ḋ   - dequeue                  [      4,    7]      [      4,  6  ]
T  - make a boolean array     [0,0,0,1,0,0,1]      [0,0,0,1,0,1  ]
^ - XOR that with a          [0,1,0,0,0,0,0]      [0,1,0,0,0,0,0]


# c (with gcc builtins), 40

A slightly different approach:

f(n){return!n?0:1<<31-__builtin_clz(n);}


This may be ruled invalid - in which case I will happily mark this as non-competing.

Input and output "arrays" are 32-bit unsigned integers - this limits the input list size to be exactly 32 - this may be a disqualifier. If the input is less than 32 bits long, then it may be padded with zero bits at the end.

# x86 assembly instructions, 12 bytes

31 c0 0f bd cf 74 04 ff c0 d3 e0 c3


Or in gcc assembly:

    .globl  f
f:
xor     %eax, %eax
bsrl    %edi, %ecx
je  .L2
inc     %eax
sall    %cl, %eax
.L2:
ret


This is a translation of my c answer and has the same I/O specs.

## Batch, 85 73 bytes

:a
@(if %1.==. exit)&set/ar=(1-f)*%1
@echo %r%&set/af^|=%1&shift&goto a


Takes input as command line arguments. For eample: 1.bat 0 1 0 1 0 0 1

Previous version

@set f=1
:a
@(if %1.==. exit)&set/ar=f*%1
@echo %r%&(if %1==1 set f=)&shift&goto a


# Brain-Flak, 230 bytes

([]){{}({}[()]<>)<>([])}{}<>([]){{}({}<>)<>([])}{}<>({<({}<>)<>>()}<(())>){({}[()]<<>({}<>)>)}{}(([])<{{}(({})())({<{}>{}((<()>))}<{}{}>)({}<>)<>([])}<>>){({}[()]<({}<>)<>>)}{}<>([]){{}({}<>)<>([])}{}<>{}{}([]){{}({}<>)<>([])}{}<>


I will explain soon but my mom cooked me some fried potatoes

([]){{}({}[()]<>)<>([])}{}<>([]){{}({}<>)<>([])}{}<> Subtracts one from every item

({<({}<>)<>>()}<(())>){({}[()]<<>({}<>)>)}{} Loops down stack until current item is zero and adds one

(([])<{{} (({})())({<{}>{}((<()>))}<{}{}>) ({}<>)<>([])}<>>){({}[()]<({}<>)<>>)}{}<> On every item of stack if it is 0 do nothing and if it is -1 add one

([]){{}({}<>)<>([])}{}<> Flip stack

{}{} Remove the two zeros at top of stack

([]){{}({}<>)<>([])}{}<> Flip stack back
`

Try it online!

## Special thanks

Special thanks to Wheat Wizard and Riley for helping me a ton with code!