Number of palindrome splits

Question

In this task you will take as input a non-negative integer \$n\$, and output the number of pairs of non-negative integers \$a,b\$ such that both are palindromes^*, \$a \leq b\$, and \$a+b = n\$. For example if \$n\$ is \$22\$ then the valid pairs are

\$ \begin{array}{c|c} a & b \\ \hline 0 & 22 \\ 11 & 11 \\ \end{array} \$

So the output is \$2\$.

As another example, if \$n\$ is \$145\$ then the valid pairs are

\$ \begin{array}{c|c} a & b \\ \hline 4 & 141 \\ 44 & 101 \\ \end{array} \$

So the output is 2.

Your submission should be a program or function. Answers will be scored in bytes with fewer bytes being the goal.

Test Cases

\$ \begin{array}{c|c c|c} \mathrm{Input} & \mathrm{Output} & \mathrm{Input} & \mathrm{Output} \\ \hline 0 & 1 & 12 & 5\\ 1 & 1 & 13 & 4\\ 2 & 2 & 14 & 4\\ 3 & 2 & 15 & 3\\ 4 & 3 & 16 & 3\\ 5 & 3 & 17 & 2\\ 6 & 4 & 18 & 2\\ 7 & 4 & 19 & 1\\ 8 & 5 & 20 & 1\\ 9 & 5 & 21 & 0\\ 10 & 5 & 22 & 2\\ 11 & 5 & 23 & 1\\ \end{array} \$

OEIS A260254

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^{* In base 10}

Answer 1

Jelly, 9 bytes

ŻŒḂ€ḋṚ$HĊ

A monadic Link accepting a non-negative integer which yields a non-negative integer.

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How?

Counts all pairs without the \$a\leq b\$ restriction, halves and rounds up. Note that the halved count is only a fraction if \$\frac n 2\$ is a palindrome and in such cases we want to count this \$a=b\$ pair.

ŻŒḂ€ḋṚ$HĊ - Link: integer, n    e.g. 22
Ż         - zero-range               [0,1,2,...,9,10,11,12,...,21,22]
   €      - for each:
 ŒḂ       -   is palindrome (digits) [1,1,1,...,1,0,1,0,...0,1]
      $   - last two links as a monad:
     Ṛ    -   reverse                [1,0,...,0,1,0,1,...,1,1,1]
    ḋ     -   dot-product           3  (=1×1+1×0+...+1×0+1×1+0×1+...+0×1+1×1)
       H  - halve                   1.5
        Ċ - ceil                    2

Answer 2

JavaScript (ES6),  74  73 bytes

n=>(g=a=>a>n-a?0:![a,n-a].some(n=>[...n+''].reverse().join``-n)+g(-~a))``

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Answer 3

Python 2,  73 70  63 bytes

lambda n:sum(`n-v`+`v`==(`v`+`n-v`)[::-1]for v in range(n/2+1))

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Note that:

(string_a == reverse(string_a)) and (string_b == reverse(string_b))

is equivalent to

reverse(string_a + string_b) == (string_b + string_a)

(where + is concatenation)

Answer 4

Prolog (SWI), 99 bytes

N-C:-aggregate_all(count,(between(0,N,A),B is N-A,B=<A,+A,+B),C).
+N:-atom_codes(N,C),reverse(C,C).

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Ungolfed Code

After adding white space, this solution reads very similarly to the challenge specification. It simply asks for the number of pairs palindrome integers within specified bounds that sum to N.

count_splits(N,C) :- 
  aggregate_all(count,(
    between(0,N,A),
    B is N-A, B=<A,
    palindrome(A),
    palindrome(B)
  ),C).

palindrome(N) :-
  atom_codes(N,C),
  reverse(C,C).

Answer 5

05AB1E, 10 bytes

ÝεÂQ}Â*O;î

Port of @JonathanAllan's Jelly answer, so make sure to upvote him as well!

Try it online or verify all test cases.

Explanation:

Ý        # Push a list in the range [0, (implicit) input-integer]
 ε       # Map each value to:
  Â      #  Bifurcate the value; short for Duplicate & Reverse copy
   Q     #  And check if it's equal to the value itself (1 if a palindrome; 0 if not)
 }Â      # After the map: bifurcate the entire list as well
   *     # Multiply the values at the same indices in the lists
    O    # Take the sum of that
     ;î  # And then halve and ceil it
         # (after which the result is output implicitly)

Answer 6

Perl 6, 44 bytes

{+grep {.flip eq[R,] $_},(^$_ Z($_...$_/2))}

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Finds the number of pairs of numbers such that the reverse of the string representation is equal to the string representation of the reversed pair.

Answer 7

Charcoal, 18 bytes

ＩＬΦ⊕⊘θ⬤Ｉ⟦ι⁻θι⟧⁼λ⮌λ

Try it online! Link is to verbose version of code. The halved input has to be incremented because a needs to vary over the inclusive range from 0 to n/2. Explanation:

     θ              Input `n`
    ⊘               Halved
   ⊕                Incremented
  Φ                 Filter over implicit range
        ⟦           Begin list
         ι          Current index `a`
           θ        Input `n`
            ι       Current index `a`
          ⁻         Subtracted (i.e. `b`)
             ⟧      End list
       Ｉ            Vectorised cast to string
      ⬤             Both strings satisfy
                 λ  Current string
                ⮌   Reversed
               λ    Current string
              ⁼     Are equal
 Ｌ                  Length
Ｉ                   Cast to string for implicit print

Answer 8

Python 3.8, 89 87 68 66 bytes

Adapting Jonathan's clever answer to Python 3 (with f format strings) and removing [] for a generator expression instead of list comprehension, we shave 21 bytes with

lambda n:sum(f"{i}{n-i}"==f"{n-i}{i}"[::-1]for i in range(n//2+1))

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My old answer:

lambda n:len([i for i in range(n//2+1)if(s:=str(i))==s[::-1]and(t:=str(n-i))==t[::-1]])

You can try it online

Answer 9

JavaScript (Node.js),  70 69  67 bytes

-2 thanks to Arnauld (change each string cast from x+''+y to x+[y])

I don't really know much JavaScript, I based this around Arnauld's answer, any advice is very welcome!

n=>(g=a=>(v=n-a)<a?0:(v+[a]==[...a+[v]].reverse().join``)+g(-~a))``

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Note that:

(string_a == reverse(string_a)) and (string_b == reverse(string_b))

is equivalent to

reverse(string_a + string_b) == (string_b + string_a)

(where + is concatenation)

Answer 10

PHP, 83 86 85 73 bytes

for(;$i<=$argn/2;$i++)$k+=strrev($j=$argn-$i)==$j&&strrev($i)==$i;echo$k;

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-13 bytes and bug fix thanks to @640KB

Answer 11

C (gcc), ^{114 \$\cdots\$ 105} 98 bytes

Saved 7 bytes thanks to rtpax!!!

i;m;p(n){for(i=0,m=n;i=i*10+n%10,n/=10;);n=i==m;}r;a;f(n){for(a=r=0;a<=n/2;)r+=p(a)*p(n-a++);n=r;}

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Answer 12

Vyxal l, 63 bits^v2, 7.875 bytes

½ʀ2('Ḃ⁼;-

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Bitstring:

000100100000010110000001100101111001000100100001100010110111001

Explained

½ʀ2('Ḃ⁼;-­⁡​‎‎⁡⁠⁡‏⁠‎⁡⁠⁢‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁣‏⁠‎⁡⁠⁤‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁢⁡‏⁠‎⁡⁠⁢⁢‏⁠‎⁡⁠⁢⁣‏⁠‎⁡⁠⁢⁤‏‏​⁡⁠⁡‌⁤​‎‎⁡⁠⁣⁡‏‏​⁡⁠⁡‌⁢⁡​‎‎⁡⁠⁣⁡‏‏​⁡⁠⁡‌⁢⁢​‎‏​⁢⁠⁡‌­
½ʀ         # ‎⁡Push the range [0, input / 2] for later
  2(       # ‎⁢Twice:
    'Ḃ⁼;   # ‎⁣  Filter the top of the stack by is palindromic
        -  # ‎⁤  And subtract that from the input.
        -  # ‎⁢⁡  This acts as a check for pairs, by keeping numbers that are palindromes before and after subtraction
# ‎⁢⁢The l flag outputs the length of the top of the stack
💎

Created with the help of Luminespire.

Only took me around 3 and a half years to finally write this answer. Things sure have changed since then.

Answer 13

MathGolf, 14 bytes

)rmÑ_x^mÅε*Σ)½

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Explanation:

)               # Increase the (implicit) input-integer by 1
 r              # Pop and push a list in the range [0, input+1)
  mÑ            # Check for each value whether it's a palindrome (1 if truthy; 0 if falsey)
    _           # Duplicate this list
     x          # Reverse the copy
      ^         # Zip the two together to create pairs
       m        # Map over each pair,
        Å       # using the following two commands:
         ε      #  Reduce by:
          *     #   Multiplying
           Σ    # Then take the sum of this list
            )   # Increase this sum by 1
             ½  # And integer-divide it by 2
                # (after which the entire stack joined together is output implicitly)

Answer 14

Perl 5 -p, 51 bytes

$\+=($,==reverse$,)&&$_==reverse;++$,<=--$_&&redo}{

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