Find relevant digit groupings

Question

Recently, my reputation was 25,121. I noticed that each digit grouping (i.e. the numbers separated by commas) was a perfect square.

Your challenge is, given a non-negative integer N and a unary boolean Black Box Function f : Z^* → B , yield a truthy value if each value of f applied to the digit groupings of N is truthy, and falsey otherwise.

One can find the digit groupings by splitting the number into groups of 3, starting from the right side. The leftmost group may have 1, 2, or 3 digits. Some examples:

12398123  -> 12,398,123    (3 digit groupings)
10        -> 10            (1 digit grouping)
23045     -> 23,045        (2 digit groupings)
100000001 -> 100,000,001   (3 digit groupings)
1337      -> 1,337         (2 digit groupings)
0         -> 0             (1 digit grouping)

Additional rules

• This function can map to either booleans (e.g. true and false), 1s and 0s, or any truthy/falsey value. Please specify which format(s) are supported by your answer.
• You may take an integer as input, or an integer string (i.e. a string composed of digits).
• You may write a program or a function.
• When passing the digital groups to the function f, you should trim all unnecessary leading zeroes. E.g., f, when applied to N = 123,000 should be executed as f(123) and f(0).

Test cases

Function notation is n -> f(n), e.g., n -> n == 0. All operators assume integer arithmetic. (E.g., sqrt(3) == 1)

function f
integer N
boolean result

n -> n == n
1230192
true

n -> n != n
42
false

n -> n > 400
420000
false

n -> n > 0
0
false

n -> n -> 0
1
true

n -> sqrt(n) ** 2 == n
25121
true

n -> sqrt(n) ** 2 == n 
4101
false

n -> mod(n, 2) == 0
2902414
true

n -> n % 10 > max(digits(n / 10))
10239120
false

n -> n % 10 > max(digits(n / 10))
123456789
true

Answer 1

Brachylog, 8 bytes

ḃ₁₀₀₀↰₁ᵐ

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The blackbox function goes on the second line (or the "Footer" on TIO) and the integer is read from STDIN. Prints true. or false. accordingly.

ḃ₁₀₀₀      Compute the base-1000 digits of the input.
     ↰₁ᵐ   Map the blackbox predicate over each digit. We don't care about the
           result of the map, but the predicate must succeed for each digit,
           otherwise the entire map fails.

Answer 2

APL (Dyalog), 16 13 bytes

3 bytes saved thanks to @Adám

∧/⎕¨1e3⊥⍣¯1⊢⎕

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How?

1e3⊥⍣¯1⊢⎕ - input the number and encode in base 1000

⎕¨ - input the function and apply on each

∧/ - reduce with logical and

Answer 3

Jelly, 5 bytes

bȷÇ€Ạ

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The command-line argument is the number. The line above the line this function resides in is the main line of the rest of the program, that is, the code that gets called for each of the groups. Be careful not to refer to the line bȷÇ€Ạ is in! The example used here is the 5th test case.

Answer 4

Python 2, 46 bytes

g=lambda f,n,k=1000:f(n%k)and(n<k or g(f,n/k))

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Answer 5

Haskell, 42 40 38 bytes

f#n=f(mod n 1000)&&(n<1||f#div n 1000)

The blackbox function must return True or False.

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Edit: -2 -4 bytes thanks to @ovs.

Answer 6

C (gcc), 66 58 48 bytes

-10 bytes thanks to @Neil!

t=1000;g(f,i)int f();{i=f(i%t)&(i<t||g(f,i/t));}

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Answer 7

Wolfram Language (Mathematica), 30 bytes

And@@#/@#2~IntegerDigits~1000&

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Answer 8

JavaScript (ES6), 40 36 bytes

f=>g=i=>f(i%1e3)&(i<1e3||g(i/1e3|0))

Takes the function and value by currying and returns 0 or 1. Edit: Saved 4 bytes thanks to @Shaggy.

Answer 9

Pyth, 9 bytes

.AyMjQ^T3

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Assumes the black-box function is named y. You can declare such a function using L (argument: b), as shown on TIO. I will implement all the test cases later, if I have time.

Answer 10

Stax, 8 bytes

Vk|Eym|A

Stax programs do not have function calls or arguments, so we store a block in the Y register that consumes and produces a single value. This can be done before the program code.

{...}Yd     store a block in the Y register that executes ...
Vk|E        get "digits" of input using base 1000
    ym      map "digits" to array using y as mapping function
      |A    all elements are truthy?

Here's an example using the perfect square function.

Answer 11

Clean, 54 bytes

import StdEnv
$n=if(n<1)True($(n/1000))&&f(n rem 1000)

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Defines the function $ :: Int -> Bool, expecting a function f :: Int -> Bool to be defined elsewhere.

Answer 12

Java (OpenJDK 9), 94 bytes

f->n->{int r=0,l=n.length();for(;l>0;l-=3)r+=f.test(n.substring(l<4?0:l-3,l))?0:1;return r<1;}

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Answer 13

Common Lisp, 73 72 bytes

(defun g(x f)(and(funcall f(mod x 1000))(or(< x 1e3)(g(floor x 1e3)f))))

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Answer 14

05AB1E, 8 bytes

₄вεI.V}P

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Explanation

₄в         # convert first input to base-1000
  ε   }    # apply to each element
   I.V     # execute second input as code
       P   # product of the resulting list

Takes the number as the first line of input and the function as the second.
Outputs 1 for truthy and 0 for falsy.

Answer 15

Excel VBA, 79 bytes

An anonymous VBE immediate window function that takes input, n as type integer from range [A1], and the name of a publically defined VBA function from range [B1].

t=1:n=[A1]:While n:t=t*-Application.Run(""&[B1],n Mod 1E3):n=Int(n/1E3):Wend:?t

Usage example

In a public module, the input function, in this case f() is defined.

Public Function f(ByVal n As Integer) As Boolean
    Let f = (n Mod 2 = 0)
End Function

The input variables are set.

[A1]=2902414    ''  Input Integer
[B1]="f"        ''  input function

The immediate window function is then called.

t=1:n=[A1]:While n:t=t*-Application.Run(""&[B1],n Mod 1E3):n=Int(n/1E3):Wend:?t
 1              ''  Function output (truthy)

Answer 16

Ruby, 37 bytes

g=->f,n{f[n%x=1000]&&(n<x||g[f,n/x])}

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A recursive lambda, taking function and integer and returning boolean.

36 bytes (only positive n)

g=->f,n{n>0?f[n%k=1000]&&g[f,n/k]:1}

This version returns 1 for truthy, false for falsey. Unfortunately it can fail when n = 0

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Answer 17

Appleseed, 51 bytes

(lambda(n f)(all(map f(or(to-base 1000 n)(q(0))))))

Anonymous lambda function that takes a number and a function and returns a boolean value.

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(lambda (n f)         ; Function with parameters n and f
 (all                 ; Return true if all elements of this list are truthy:
  (map f              ; Map the function f to each element of
   (or                ; This list if it is nonempty:
    (to-base 1000 n)  ; Convert n to a list of "digits" in base 1000
    (q (0))           ; Or if that list is empty (when n=0), then use the list (0) instead
   ))))

Answer 18

Add++, 15 bytes

L,1000$bbbUª{f}

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Requires a function f to be declared in the TIO header.

How it works

D,f,@,0.5^i2^A=	; Declares a function 'f' to check if a perfect square
		; E.g. 25 -> 1; 26 -> 0

L,		; Declare the main lambda function
		; Example argument: 		[25121]
	1000$bb	; Convert to base 1000	STACK = [[25 121]]
	bUª{f}	; Is 'f' true for all?	STACK = [1]

Answer 19

Husk, 7 bytes

`ΛmrC_3

Try it online! Takes N as a string and f as a function object. Given a number, f can return any truthy or falsy value.

Answer 20

Japt, 9 bytes

Had to sacrifice 3 bytes and take input as a string in order to handle the n=>n>0 test case when n=0. I could get them back if it was permitted to pass the groupings to the function as strings and convert them to integers within the function, when necessary

ò3n)mn eV

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The function, as allowed by consensus, is preassigned to variable V on the second line of the header - the empty first line is essential.

Here are all the test case functions rewritten for Japt.

n -> n == n                           X{X==X}
n -> n != n                           X{X!=X}
n -> n > 400                          X{X>400}
n -> n > 0                            X{X>0}
n -> n -> 0                           X{X{0}}
n -> sqrt(n) ** 2 == n                X{Xq v1}
n -> mod(n, 2) == 0                   X{X%2==0}
n -> n % 10 > max(digits(n / 10))     X{X%10>Xz10 ì rw}