Am I a Rude Number?

Question

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.

Your Task:

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 code-golf, so the lowest score in bytes wins.

Answer 1

APL (dzaima/APL), 5 bytes

4∊32⊤

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4∊ is 4 a member of

32⊤ to-base-32?

Answer 2

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}$

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^
(
    (?=(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

Answer 3

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).

Answer 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)

Answer 5

Ruby, 36 19 bytes

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

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Saved 17 bytes with @tsh's method.

Answer 6

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.

Answer 7

Python 2, 34 32 bytes

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

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2 bytes thanks to tsh

Answer 8

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.

Answer 9

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).

In human-readable assembly mnemonics:

; 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

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Answer 10

J, 12 bytes

4 e.32#.inv]

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Answer 11

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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Answer 12

05AB1E, 5 bytes

32B4å

Port of @Adám's APL (dzaima/APL) answer.

Try it online or verify all test cases.

Explanation:

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

Answer 13

Catholicon, 4 bytes

ǔ?QǑ

Takes a number as a base-256 string.

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Test suite

Answer 14

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;

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Answer 15

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.

Answer 16

ES6, 31 30 26 bytes

b=>b.toString(32).match`4`

Feel free to say ideas on how to reduce this further, if any.

Answer 17

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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Answer 18

C (gcc), 34 bytes

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

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Answer 19

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.

Answer 20

Tidy, 18 bytes

{x:4∈base(32,x)}

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

Answer 21

Haskell, 31 bytes

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

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Answer 22

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

Answer 23

MATL, 8 bytes

32YA52=a

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Answer 24

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.

Answer 25

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.

Answer 26

Nibbles, 4.5 bytes

?`@32$4

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?`@32$4
?        Find the first index of
      4  4
 `@32    in the base 32 representation of
     $   input

Answer 27

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.

Answer 28

Java 8, 40 33 bytes

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

Port of @Adám's APL (dzaima/APL) answer.

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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"

Answer 29

><>, 28 bytes

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

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

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Answer 30

Wolfram Language (Mathematica), 37 bytes 36 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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