## The Next Palindrome

### May 22, 2009

This exercise is much harder than it sounds, due to all the special cases that appear. Here’s one solution; it’s ugly, but it works:

`(define (palin n)`

(let* ((ds (string->list n)) (len (length ds)) (len2 (quotient len 2)))

(cond ((equal? ds (list #\9)) "11")

((= len 1) (list->string (plus1 ds)))

((all-nines? ds)

(let loop ((ds ds) (ps (list #\1)))

(if (null? ds)

(list->string (cons #\1 (cdr ps)))

(loop (cdr ds) (cons #\0 ps)))))

((even? len)

(let* ((left (take len2 ds)) (right (drop len2 ds)))

(if (lt? right (reverse left))

(list->string (append left (reverse left)))

(let ((x (plus1 left)))

(list->string (append x (reverse x)))))))

(else (let* ((left (take len2 ds))

(middle (car (drop len2 ds)))

(right (cdr (drop len2 ds))))

(if (lt? right (reverse left))

(list->string (append left (list middle) (reverse left)))

(if (char=? middle #\9)

(let ((x (plus1 left)))

(list->string (append x (list #\0) (reverse x))))

(list->string (append left (plus1 (list middle)) (reverse left))))))))))

All the one-digit numbers are special cases, so the main body of the code doesn’t have to deal with null left-half and right-half digit lists. Another special case is numbers that consist of only nines, since the result is longer than the input. What’s left are two cases, even-length numbers and odd-length numbers.

Even-length numbers are easier. First, we split them into two halves, left and right. The easy case occurs when the left half is less than the right half when the digits of the left half are reversed; the result is just the left half and its reverse, appended together. Otherwise, the left half must be incremented by one before it is appended to the front of its reverse.

Odd-length numbers are similar, but add the additional wrinkle of a middle digit. In the particular case where the right half is less than the reverse left half and the middle digit is nine, the middle digit becomes zero, and the left half is incremented by one before it is appended to the middle digit and its reverse. The other cases are similar to the even-length numbers.

We treat the numbers as strings because of the requirement to handle very long numbers, as long as a million digits. That means we need some functions from grade-school aritemetic:

`(define (all-nines? ds)`

(cond ((null? ds) #t)

((not (char=? (car ds) #\9)) #f)

(else (all-nines? (cdr ds)))))

`(define (lt? a b)`

(cond ((< (length a) (length b)) #t)

((< (length b) (length a)) #f)

(else (let loop ((a a) (b b))

(cond ((null? a) #f)

((char<? (car a) (car b)) #t)

((char<? (car b) (car a)) #f)

(else (loop (cdr a) (cdr b))))))))

`(define (plus1 ds)`

(let loop ((ds (reverse ds)) (carry 0) (xs '()))

(cond ((null? ds)

(if (zero? carry)

(reverse xs)

(cons #\1 (reverse xs))))

((char=? (car ds) #\9)

(loop (cdr ds) 1 (cons #\0 xs)))

(else (append (reverse (cdr ds))

(list (integer->char

(+ (char->integer (car ds)) 1)))

(reverse xs))))))

Here you can see `palin`

in action:

`> (map palin '("0" "88" "808" "1999" "2133" "9999" "99999"))`

("1" "99" "818" "2002" "2222" "10001" "100001")

Because I had seen so many incorrect solutions, I wanted to be sure my solution was correct. I calculated all the palindromes to one million, by brute force, and compared the result to a loop that calls `palin`

to calculate the next palindrome iteratively:

`(equal?`

(list-of (list->string x)

(n range 1 10000002)

(x is (string->list (number->string n)))

(equal? x (reverse x)))

(let loop ((p "0") (ps '()))

(if (< 7 (string-length p)) (reverse ps)

(let ((x (palin p)))

(loop x (cons x ps))))))

`Palin`

uses `take`

and `drop`

from the Standard Prelude, and the test program uses list comprehensions.

You can run the code at http://programmingpraxis.codepad.org/oHbzw5WF. For more about the math of palindromic numbers, see sequence A002113 in Neil J. A. Sloane’s Online Encyclopedia of Integer Sequences.

Pages: 1 2

;; It doesn’t need to be that complicated.

;; Below is a simpler solution

(define (next-palindrome n)

(define (next-palindrome* n)

(define digits-of-n (string->list (format “~a” n)))

(cond

[(equal? digits-of-n (reverse digits-of-n)) n]

[else (next-palindrome* (add1 n))]))

(next-palindrome* (add1 n)))

(map next-palindrome ‘(0 88 808 1999 2133 9999 99999)) ==> ‘(1 99 818 2002 2222 10001 100001)

Here’s how I did it:

bisect xs = go xs xs where

go xs [] = ([],[],xs)

go (x:xs) (_:[]) = ([],[x],xs)

go (x:xs) (_:_:ys) = let (fs,m,ss) = go xs ys in (x:fs,m,ss)

increment = foldr go ([],True) where

go ’9′ (xs,True) = (’0′:xs,True)

go x (xs,True) = (succ x:xs,False)

go x (xs,False) = (x:xs,False)

nextPalindrome xs | fhr > secondHalf = firstHalf ++ middle ++ fhr

| otherwise = startInc ++ endInc where

(firstHalf, middle, secondHalf) = bisect xs

fhr = reverse firstHalf

(incremented, carry) = increment (firstHalf ++ middle)

incremented’ | carry = ’1′:incremented

| otherwise = incremented

startInc | carry = init incremented’

| otherwise = incremented’

endInc | null middle = reverse incremented’

| otherwise = tail (reverse incremented’)

It’s a bit fiddly, but there’s only really two cases: first half reversed > second half: take the palindrome starting with the first half. first half reversed <= second half: take the palindrome starting with the first half plus one (being careful to drop a digit if we increase the length in so doing).

;; Hopefully. Better formated

@Jaime Vargas: note the problem requires this to work on inputs 1000000 digits. Try your program on (10^500000 – 1) * (10^500000).

Once more with formatting?

[...] Praxis – The Next Palindrome By Remco Niemeijer Today’s Programming Praxis problem is about palindromic numbers, i.e. numbers that read the same backwards [...]

My Haskell solution (see http://bonsaicode.wordpress.com/2009/05/22/programming-praxis-%E2%80%93-the-next-palindrome/ for a version with comments):

Lilac, my solution still applies eventhough it is inneficient.

Updated version that’s one line shorter and 50% faster:

Remco, please try nextPalindrome 9919.

;; Take 2: Efficient version

New version that works correctly for 9919:

Remco,

I’m curious. In the first version, how did 9919, or any of the smaller numbers where the second half is less than the first half, pass your test? The test looks to me like it should have worked.

They didn’t, because there was no such test. In the 0, 88…99999 list you used the second half is always greater than or equal to the first half.

The other test just recursively calls nextPalindrome, so the input number is always symmetrical, in which case my original algorithm works ok.

Since there was no test for that type of case I never noticed that bug.

@Jaime Vargas: If thermodynamics means it’s impossible for your program to be executed on certain valid inputs (because any computing device would break down due to entropy first), then I would say the solution does not apply. 10^499978 years is plenty of time for that (and that’s assuming an insanely fast computer is used).

Hey I am Happy To see this Post

“Next Palindrome Number”! which was posted on my blog!

And Thanks For Solving This!

Very Nice Collection!

ruby solution (http://codepad.org/cREcLRuP)

This is a solution in c#

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace LeapYear_Trial

{

class Program

{

public bool Palindrome(string number)

{

return number.SequenceEqual(number.Reverse());

}

static void Main(string[] args)

{

Program obj = new Program();

Console.WriteLine(“Input a number “);

int num = Convert.ToInt32(Console.ReadLine());

int nextNum = num + 1;

while(!obj.Palindrome(nextNum.ToString()))

{

nextNum += 1;

}

Console.WriteLine(“The next smallest palindrome larger than {0} is {1}”, num, nextNum);

}

}

}