# Am I a Rude Number?

For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one person, with two hands of five fingers each), assume we continue with a third hand, with additional hands added as required.

Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.

## Input:

An integer between 0 and 109 (inclusive).

## Output:

A truthy/falsy value that indicates whether the input is a rude number.

## Test Cases:

Input:    Output:
0   --->  falsy
3   --->  falsy
4   --->  truthy
25  --->  falsy
36  --->  truthy
127 --->  falsy
131 --->  truthy


## Scoring:

This is , so the lowest score in bytes wins.

• assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work. – Veskah Feb 27 at 3:24
• @Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers. – Gryphon Feb 27 at 3:26
• It's worse if you're British - 6 is rude too then! – Matthew Feb 27 at 7:07
• Is it OK to take input in a different base than 10? – wastl Feb 27 at 20:08
• 5 seems fairly rude too. Not sure anyone would say "Oh she had her thumb out, that's perfectly polite" – ale10ander Mar 1 at 1:43

# APL (dzaima/APL), 5 bytes

4∊32⊤


Try it online!

4∊ is 4 a member of

32⊤ to-base-32?

• s/bytes/characters though? – tripleee Feb 27 at 10:34

# Regex (ECMAScript), 37 bytes

Input is in unary, as the length of a string of xs.

^((?=(x+)(\2{31}x*))\3)*(x{32})*x{4}$ Try it online! ^ ( (?=(x+)(\2{31}x*)) # \2 = floor(tail / 32); \3 = tool to make tail = \2 \3 # tail = \2 )* # Loop the above as many times as necessary to make # the below match (x{32})*x{4}$             # Assert that tail % 32 == 4

• I thought I knew regex, but apparently not. – CT Hall Feb 28 at 4:16

# JavaScript (SpiderMonkey), 23 bytes

f=x=>x&&x%32==4|f(x>>5)


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This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.

# JavaScript (SpiderMonkey), 26 bytes

x=>x.toString(32).match(4)


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It's interesting that /4/.test(...) cost one more byte than ....match(4).

# Japt, 5 bytes

sH ø4


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### Explanation

      // Implicit input
sH    // To a base-H (=32) string
ø  // Contains
4 // 4 (JavaScript interprets this as a string)


## Ruby, 36 19 bytes

->n{n.to_s(32)[?4]}


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Saved 17 bytes with @tsh's method.

• This returns true for 2207, which has a binary representation of 100010011111 – Embodiment of Ignorance Feb 27 at 4:10
• @EmbodimentofIgnorance That is the correct result, is it not? The second hand is 00100. – Doorknob Feb 27 at 4:16
• I don't speak Ruby. But why not ->n{n.to_s(32)=~/4/}? – tsh Feb 27 at 4:19
• @tsh because I'm not as clever as you :) – Doorknob Feb 27 at 4:23
• Forgive me if I'm not understanding the question, but isn't the first hand of 2207 10001, the second 00111, and the third 11? None of them have their middle finger only up – Embodiment of Ignorance Feb 27 at 4:28

# APL+WIN, 10 bytes

Prompts for input of integer

4∊(6⍴32)⊤⎕


Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.

# Perl 6, 16 bytes

{.base(32)~~/4/}


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Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.

You can prove this by the fact that $$\2^5 = 32\$$ so each digit is the state of each hand.

# Python 2, 34 32 bytes

f=lambda a:a%32==4or a>0<f(a/32)


Try it online!

2 bytes thanks to tsh

• :| you can edit posts, you know that right – ASCII-only Feb 27 at 7:37
• Yes; I know. It was an accident! An accident, I tell ya! – Chas Brown Feb 27 at 8:52
• @tsh Oh nice, forgot that shortcircuits – ASCII-only Feb 27 at 10:24

# x86 Machine Code, 17 bytes

6A 20 59 85 C0 74 09 99 F7 F9 83 FA 04 75 F4 91 C3


The above bytes define a function that takes the number as input in the EAX register, and returns the result as a Boolean value in the EAX register (EAX == 0 if the input is not a rude number; EAX != 0 if the input is a rude number).

; Determines whether the specified number is a "rude" number.
; Input:    The number to check, in EAX
; Output:   The Boolean result, in EAX (non-zero if rude; zero otherwise)
; Clobbers: ECX, EDX
IsRudeNumber:
push    32           ; \ standard golfing way to enregister a constant value
pop     ecx          ; /  (in this case: ECX <= 32)
CheckNext:
test    eax, eax     ; \ if EAX == 0, jump to the end and return EAX (== 0)
jz      TheEnd       ; /  otherwise, fall through and keep executing
cdq                  ; zero-out EDX because EAX is unsigned (shorter than XOR)
idiv    ecx          ; EAX <= (EAX / 32)
; EDX <= (EAX % 32)
cmp     edx, 4       ; \ if EDX != 4, jump back to the start of the loop
jne     CheckNext    ; /  otherwise, fall through and keep executing
xchg    eax, ecx     ; store ECX (== 32, a non-zero value) in EAX
TheEnd:
ret                  ; return, with result in EAX


Try it online!

• Interesting idea to use idiv, though. I don't see any incremental improvements to this. But see my answer : 14 bytes for a shift loop that uses MOV/AND/SUB/JZ to check the low 5 bits for rudeness. – Peter Cordes Mar 4 at 0:25

# Julia 1.0, 25 bytes

f(n)=n%32==4||n>0<f(n>>5)


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### Julia 1.0, 26 bytes

Alternative that is 1 character shorter, but 1 byte longer, too bad that ∈ takes 3 bytes in unicode.

n->'4'∈string(n,base=32)


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• could you use n->n%32... for your first answer for 2 bytes shorter? – Giuseppe Feb 28 at 22:48
• @Giuseppe, unfortunately no, that function is recursive. – Kirill L. Mar 1 at 7:34

# 05AB1E, 5 bytes

32B4å


Explanation:

32B    # Convert the (implicit) input to Base-32
4å  # And check if it contains a 4
# (output the result implicitly)

• Too bad ₆ is 36, not 32. – Magic Octopus Urn Feb 28 at 19:51

# Catholicon, 4 bytes

ǔ?QǑ


Takes a number as a base-256 string.

Try it online!

Test suite

• Hm, if this is allowed, then is it allowed to accept numbers in base 32 instead? – recursive Feb 27 at 19:47
• @recursive You can surround the numbers in << and >> and it allows for numbers greater than 255 in those, as shown in the test suite. – Okx Feb 28 at 13:11
• It was intended as a question about the challenge but it wasn't very clear. – recursive Feb 28 at 15:44

# C# (Visual C# Interactive Compiler), 31 bytes

n=>{for(;n>0;n/=n%32==4?0:32);}


Outputs by throwing an exception. The way you convert one number from decimal to another base is to divide the decimal number by that base repeatedly and take the remainder as a digit. That is what we do, and we check if any of the digits have a value of 4 in base-32;

Try it online!

• 27? as a bonus it doesn't output in a weird way – ASCII-only Feb 27 at 6:13
• also what is tplig – ASCII-only Feb 27 at 6:15
• @ASCII-only n>31 -> n>0 – tsh Feb 27 at 6:17
• 25 bytes – Kevin Cruijssen Feb 27 at 12:57
• Whether or not a program halts is not an allowed output method. Output via exception is allowed. – Deadcode Feb 28 at 22:58

# J, 12 bytes

4 e.32#.inv]


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# R, 50 48 bytes

any(2^(0:4)%*%matrix(scan()%/%2^(0:34)%%2,5)==4)


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Uses a neat matrix-based approach now (courtesy of @Giueseppe). It generates a 5x7 matrix of bits, converts this to a series of base 32 integers, and checks for any 4s.

• @Giuseppe Oops, completely missed that. Should work now, though disappointingly 19 bytes longer. I don't think there's an inverse function to strtoi other than for hexadecimal and octal in base R – Nick Kennedy Feb 28 at 17:45
• 48 bytes with some matrix magic. I believe the bit conversion is longer than intToBits but then we can work with ints instead of raw which ends up saving a byte -- see for instance this with intToBits – Giuseppe Feb 28 at 18:04
• @Giuseppe it's a completely different (and neat) solution to mine - do you want me to update mine or you post your own? – Nick Kennedy Feb 28 at 18:18
• you're free to take it. :-) – Giuseppe Feb 28 at 18:24
• of course, porting one of the many answers that tests for the presence of a digit 4 in a base-32 number is, oh, 29 bytes. – Giuseppe Feb 28 at 22:46

# Python 3, 43 bytes

Checks every 5-bit chunk to see if it is rude (equal to 4).

lambda n:any(n>>5*i&31==4for i in range(n))


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• *5-bit chunk... – ASCII-only Feb 27 at 7:38

# C (gcc), 34 bytes

f(i){return i?i&31^4?f(i/32):1:0;}


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• 32 bytes by rearranging the expression. – Arnauld Feb 27 at 10:11
• or 25 bytes by abusing the way gcc is compiling this. – Arnauld Feb 27 at 10:11

# Charcoal, 6 bytes

№⍘Ｎ³²4


Try it online! Link is to verbose version of code. Outputs -s according to how rude the number is. Explanation:

  Ｎ     Input as a number
⍘      Convert to base as a string
³²   Literal 32
№       Count occurrences of
4  Literal string 4


I use string base conversion to avoid having to separate the numeric literals for 32 and 4.

# Tidy, 18 bytes

{x:4∈base(32,x)}


Try it online! Checks if 4 is an element of base(32,x) (base conversion).

elem 9.(mapM(:[6..36])[0..5]!!)


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# Cubix, 26 bytes

u!@-W14;OIS%/\;;,p;?wO@u/s


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Wraps onto a cube with edge length 3 as follows

      u ! @
- W 1
4 ; O
I S % / \ ; ; , p ; ? w
O @ u / s . . . . . . .
. . . . . . . . . . . .
. . .
. . .
. . .


Watch it run

A fairly basic implementation, without all the redirects it does :

• IS initiates the program by pushing the input and 32 to the stack
• %4-! gets the remainder and checks if it is 4 by subtraction
• 1O@ output 1 if it was 4 and halt
• ;;, clean up the stack and do integer divide
• p;? clean up bottom of the stack and check div result for 0
• O@ if div result zero output and halt
• s swap the top of stack and start back at step 2 above

# MATL, 8 bytes

32YA52=a


Try it online!

• @Luis I can definitely drop the G (not sure why I included it in the first place) but that's only one byte (thanks for spotting that!). Changing 32YA52 to 32_YA4 is the same number of bytes right? – Sanchises Mar 1 at 7:06
• Ah, yes, I can't count – Luis Mendo Mar 1 at 10:26
• @Luis Count? Who needs to count when you can '32_YA4'n'32YA52'n- – Sanchises Mar 1 at 11:56

## Batch, 77 45 bytes

@cmd/cset/a"m=34636833,n=%1^m*4,(n-m)&~n&m*16


Based on these bit twiddling hacks. Explanation: Only 6 hands need to be checked due to the limited range (30 bits) of the input that's required to be supported. The magic number m is equivalent to 111111 in base 32, so that the first operation toggles the rude bits in the input number. It then remains to find which of the 6 hands is now zero.

# x86 machine code, 14 bytes

(same machine code works in 16-bit, 32-bit, and 64-bit. In 16-bit mode, it uses AX and DI instead of EAX and EDI in 32 and 64-bit mode.)

Algorithm: check low 5 bits with x & 31 == 4, then right-shift by 5 bits, and repeat if the shift result is non-zero.

Callable from C with char isrude(unsigned n); according to the x86-64 System V calling convention. 0 is truthy, non-0 is falsy (this is asm, not C1).

 line   addr    code bytes
num
1                             ; input:  number in EDI
2                             ; output: integer result in AL: 0 -> rude, non-zero non-rude
3                             ; clobbers: RDI
4                         isrude:
5                         .check_low_bitgroup:
6 00000000 89F8               mov    eax, edi
7 00000002 241F               and    al, 31          ; isolate low 5 bits
8 00000004 2C04               sub    al, 4           ; like cmp but leaves AL 0 or non-zero
9 00000006 7405               jz    .rude            ; if (al & 31 == 4) return 0;
10
11 00000008 C1EF05             shr    edi, 5
12 0000000B 75F3               jnz   .check_low_bitgroup
13                             ;; fall through to here is only possible if AL is non-zero
14                         .rude:
15 0000000D C3                 ret

16          0E             size:  db $- isrude  This takes advantage of the short-form op al, imm8 encoding for AND and SUB. I could have used XOR al,4 to produce 0 on equality, but SUB is faster because it can macro-fuse with JZ into a single sub-and-branch uop on Sandybridge-family. Fun fact: using the flag-result of a shift by more than 1 will be slow on P6-family (front-end stalls until the shift retires), but that's fine. Footnote 1: This is an assembly language function, and x86 asm has both jz and jnz, so as per meta I can choose either way. I'm not intending this to match C truthy/falsy. It happened to be convenient to return in AL instead of EFLAGS, so we can describe the function to a C compiler without a wrapper, but my choice of truthy/falsy isn't constrained by using a C caller to test it. # ES6, 3130 26 bytes b=>b.toString(32).match4  Feel free to say ideas on how to reduce this further, if any. • Welcome to PPCG! – Laikoni Feb 28 at 19:23 • You don't need to count the name of your function, and although I think you can save a byte by using test, you can actually save two bytes by matching against 4 as a number and letting match convert that into a string and then a RegExp for you. – Neil Mar 1 at 0:53 # Retina 0.8.2, 31 bytes .+$*
+(1+)\1{31}
\$1;
\b1111\b


Try it online! Link includes test cases. Outputs zero unless the number is rude. Works by converting the input to unary and then to unary-encoded base 32 and counting the number of 4s in the result.

# Java 8, 40 33 bytes

n->n.toString(n,32).contains("4")


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Explanation:

n->                 // Method with Integer parameter and boolean return-type
n.toString(n,32)  //  Convert the input to a base-32 String
.contains("4")   //  And check if it contains a "4"


# ><>, 28 bytes

Outputs 4 for rude numbers throws an exception for non-rude numbers.

:1(?^:" ":\
,&-v?=4:%&/
;n<


Try it online!

• An exception is acceptable, the C# answer does it – ASCII-only Mar 1 at 11:22

# Wolfram Language (Mathematica), 37 bytes36 bytes 29 bytes

-2 bytes by Jonathan Frech

#~IntegerDigits~32~MemberQ~4&


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31-byte solution:

MemberQ[IntegerDigits[#,32],4]&


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• Hello and welcome to PPCG. As it stands, your expression is a single Boolean value. Please fix your answer to either be a full program or a function (...#...& is often used in Mathematica). – Jonathan Frech Feb 27 at 9:22
• Hello. Is this what you mean? – Rainer Glüge Feb 27 at 9:32
• pls use tio.run/#mathematica instead of W|A to make sure it's valid mathematica code :P and you don't need the [n] at the end, just the &. Also since posts have edit history, it's fine to leave out previous entries, and the convention for old scores is <s>40</s> <s>36</s> – ASCII-only Feb 27 at 10:22
• Yes. This is what I meant. 29 bytes. – Jonathan Frech Feb 27 at 13:17
• I guess I have to get used to the functional programming style. – Rainer Glüge Feb 27 at 13:34

# Java 8, 2822 21 bytes

n->n%32==4|n>>5%32==4


Inspired by @kevin-cruijssen's answer. Only works for 2 hands.

Try it online!

Explanation:

n->                 // Method with int parameter and boolean return-type
n%32              // Only consider right 5 bytes (fingers)
==4               // Middle finger
| ... n>>5       // Repeat with shifted bits for other hand
`
• I'm pretty sure answers have to work with both hands – Embodiment of Ignorance May 9 at 3:41