# Project Euler #4: Largest Palindrome Product

Posted on 2017-03-08 by nbloomf

Spoiler alert! This page is part of a series on solutions to Project Euler problems. If you prefer to solve problems yourself, do not read on!

This post is literate Haskell; you can load the source into GHCi and play along.

First some boilerplate.

module ProjectEuler004 where

import Data.List
import Data.Maybe
import System.Exit

Problem 4 from Project Euler:

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

“Palindromeness” is very different from most interesting properties of numbers; it is an artifact of one of many possible representations of the number. In this case, the representation in base 10.

Let’s start by nailing down a definition for “palindrome”.

Suppose $$A = \sum_{k=0}^{t-1} a_k 10^k$$ is a $$t$$-digit number base 10 (that is, the $$a_k$$ are integers between 0 and 9 (inclusive) and $$a_{t-1}$$ is not zero). We say $$A$$ is a palindrome if $$a_{t-1-k} = a_k$$ for all $$0 \leq k < t$$.

The first thing I’ll do is write helper functions to convert a number to its base 10 digits and back.

-- base 10 digits of n in 'little endian' order
-- (least significant digits first)
toDigits :: Integer -> [Integer]
toDigits n = if n <= 0
then []
else (nrem10) : toDigits (nquot10)

fromDigits :: [Integer] -> Integer
fromDigits = sum . zipWith (*) (map (10^) [0..])

numDigits :: Integer -> Integer
numDigits n = sum $map (const 1)$ toDigits n

Sanity check:

test_digits :: Integer -> Bool
test_digits n = n == (fromDigits $toDigits n) And a test: $> all test_digits [1..1000000]
True

Note that our lists of digits come out “backward”; that is, least significant first.

$> toDigits 12345 [5,4,3,2,1] Then we can detect whether a given number is a palindrome in base 10. is_palindrome_10 :: Integer -> Bool is_palindrome_10 n = let ds = toDigits n in ds == (reverse ds) Now we’re not just looking for the largest palindrome of a given length; that would be easy – the string of all 9s is the largest palindrome with a given number of digits. Instead, we want the largest palindrome that is the product of two 3-digit numbers. The most obvious solution is to list all the products of two 3-digit numbers, filter for the palindromes, and find the max. -- the triple (a,b,a*b) which yields the largest -- palindrome product a*b among the pairs of -- t-digit numbers a and b pe4' :: Integer -> (Integer, Integer, Integer) pe4' t = let thd (_,_,x) = x min = 10^(t-1) max = 10^t - 1 in maximumBy (\x y -> compare (thd x) (thd y))$
filter (is_palindrome_10 . thd) $[(a,b,a*b) | a <- [min..max], b <- [a..max]] This pe4' works well enough: $> pe4' 1
(3,3,9)
$> pe4' 2 (99,91,9009)$> pe4' 3
(993,913,906609)
$> pe4' 4 (9901,9999,99000099) Two observations. First, there appears to be a pattern emerging in these results. Second, unfortunately I can’t feasibly explore that pattern further (yet) because as t gets larger, pe4' t gets way slow. It’s not hard to see why; pe4' t constructs a list of $\binom{9 \cdot 10^{t-1}}{2} = \frac{(9 \cdot 10^{t-1})(9 \cdot 10^{t-1} - 1)}{2}$ candidates to check by brute force. To see if we can find a better way, let’s look at the simplest version of this problem: finding the largest palindrome product among products of 1-digit numbers. To this end, consider the multiplication table for 1-digit numbers below. $\begin{array}{c|ccccccccc} & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline 9 & 81 & 72 & 63 & 54 & 45 & 36 & 27 & 18 & 9 \\ 8 & & 64 & 56 & 48 & 40 & 32 & 24 & 16 & 8 \\ 7 & & & 49 & 42 & 35 & 28 & 21 & 14 & 7 \\ 6 & & & & 36 & 30 & 24 & 18 & 12 & 6 \\ 5 & & & & & 25 & 20 & 15 & 10 & 5 \\ 4 & & & & & & 16 & 12 & 8 & 4 \\ 3 & & & & & & & 9 & 6 & 3 \\ 2 & & & & & & & & 4 & 2 \\ 1 & & & & & & & & & 1 \\ \end{array}$ The first thing to jump out at me is that the vast majority of these products are not palindromes. That’s not really surprising I guess – there are $$9 \cdot 10^{2t-1}$$ different $$2t$$-digit numbers, but only $$9 \cdot 10^{t-1}$$ of them are palindromes, so to a first approximation the probability that an arbitrary $$2t$$-digit number is a palindrome is about $$10^{-t}$$. So maybe instead of searching among the products of $$t$$-digit numbers for palindromes, it would be faster to search among the palindromes for products of $$t$$-digit numbers. ## A Digression on Palindromes The rarity of palindromes highlights a disturbing possibility. The problem of finding the largest palindrome product implicitly assumes that at least one palindrome product must exist, but this is not obvious to me. After some experimenting, I am convinced that palindrome products exist among pairs of $$t$$-digit numbers for any $$t$$. I got a little distracted thinking about this probem, so I’ll just dump the results here. The first result allows us to recognize a family of palindromes constructed from other palindromes. Let $$t,u,v \geq 1$$ be natural numbers, and let $$A = \sum_{k=0}^{t-1} a_k 10^k$$ and $$B = \sum_{k=0}^{u-1} b_k 10^k$$ be $$t$$- and $$u$$-digit palindromes, respectively. That is, $$a_k = a_{t-1-k}$$ when $$0 \leq k < t$$ and $$b_k = b_{u-1-k}$$ when $$0 \leq k < u$$. Then $M = A + 10^{t+v} B + 10^{t+u+2v} A$ is a $$(2t+u+2v)$$-digit palindrome. Note that $\begin{eqnarray*} M & = & A + 10^{t+v} B + 10^{2t+u+2v} A \\ & = & \sum_{k=0}^{t-1} a_k 10^k + 10^{t+v} \sum_{k=0}^{u-1} b_k 10^k + 10^{2t+u+2v} \sum_{k=0}^{t-1} a_k 10^k \\ & = & \sum_{k=0}^{t-1} a_k 10^k + \sum_{k=0}^{u-1} b_k 10^{t+v+k} + \sum_{k=0}^{t-1} a_k 10^{t+u+2v+k} \\ & = & \sum_{k=0}^{t-1} a_k 10^k + \sum_{k=t+v}^{t+u+v-1} b_{k-t-v} 10^k + \sum_{k=t+u+2v}^{2t+u+2v-1} a_{k-t-u-2v} 10^k \\ & = & \sum_{k=0}^{2t+u+2v-1} e_k 10^k \end{eqnarray*}$ where $e_k = \left\{ \begin{array}{ll} a_k & \mathrm{if}\ 0 \leq k < t \\ 0 & \mathrm{if}\ t \leq k < t+v \\ b_{k-t-v} & \mathrm{if}\ t+v \leq k < t+u+v \\ 0 & \mathrm{if}\ t+u+v \leq k < t+u+2v \\ a_{k-t-u-2v} & \mathrm{if}\ t+u+2v \leq k < 2t+u+2v. \end{array} \right.$ Certainly $$M$$ has $$2t+u+2v$$ digits. To see that $$M$$ is a palindrome, we need to check that $e_{2t+u+2v-1-k} = e_k$ for $$0 \leq k < 2t+u+2v$$. We will break this interval into 5 subintervals. 1. Suppose $$0 \leq k < t$$. Then $$0 \leq t-1-k < t$$, and so $$t+u+2v \leq 2t+u+2v-1-k < 2t+u+2v$$. So $e_{2t+u+2v-1-k} = a_{2t+u+2v-1-k-t-u-2v} = a_{t-1-k} = a_k = e_k.$ 2. Suppose $$t \leq k < t+v$$. Then $$0 \leq t+v-1-k < v$$, and so $$t+u+v \leq 2t+u+2v-1-k < t+u+2v$$. So $e_{2t+u+2v-1-k} = 0 = e_k.$ 3. Suppose $$t+v \leq k < t+u+v$$. Then $$0 \leq t+u+v-1-k < u$$, and so $$t+v \leq 2t+u+2v-1-k < t+u+v$$. So $\begin{eqnarray*} e_{2t+u+2v-1-k} & = & b_{2t+u+2v-1-k-t-v} \\ & = & b_{t+u+v-1-k} \\ & = & b_{u-1-t-u-v+1-k} \\ & = & b_{k-t-v} \\ & = & e_{k}. \end{eqnarray*}$ 4. Suppose $$t+u+v \leq k < t+u+2v$$. Then $$0 \leq t+u+2v-1-k < v$$, and so $$t \leq 2t+u+2v-1-k < t+v$$. So $e_{2t+u+2v-1-k} = 0 = e_k.$ 5. Suppose $$t+u+2v \leq k < 2t+u+2v$$. Then $$0 \leq 2t+u+2v-1-k < t$$. So $\begin{eqnarray*} e_{2t+u+2v-1-k} & = & a_{2t+u+2v-1-k} \\ & = & a_{t-1-2t-u-2v+i+k} \\ & = & a_{k+t-u-2v} \\ & = & e_k. \end{eqnarray*}$ Thus $$M$$ is a $$(2t+u+2v)$$-digit palindrome. This result lets us recognize some palindromes. The next result lets us construct new palindrome products from old ones. Let $$A$$ and $$B$$ be $$t$$-digit numbers such that both $$AB$$ and $$2AB$$ are $$2t$$-digit palindromes. Let $$v$$ be a positive integer, and define $H_v(A) = A(1 + 10^{2t+v})\ \mathrm{and}\ K_v(B) = B(1 + 10^{2t+v}).$ Then $$H_v$$ and $$K_v$$ are $$(3t+v)$$-digit numbers and $$H_vK_v$$ is a $$(6t+2v)$$-digit palindrome. Expanding $$H_vK_v$$, we have $H_vK_v = AB + 10^{2t+v}(2AB) + 10^{2t+2t+2v}AB.$ The previous theorem applies, and $$H_vK_v$$ is a $$(6t+2v)$$-digit palindrome. Let’s make $$H_v$$ and $$K_v$$ executable. h_ :: Integer -> Integer -> Integer h_ v a = a * (1 + 10^(2*t+v)) where t = numDigits a k_ :: Integer -> Integer -> Integer k_ v b = b * (1 + 10^(2*t+v)) where t = numDigits b And the next result gives us a concrete family of palindrome products with factors of even digit length. Let $$t$$ and $$m$$ be positive natural numbers, and define $$Q_{t,m} = \sum_{k=0}^{t-1} 10^{mk}$$. Now define $A_{t,m} = Q_{2t,m},$ $B_{t,m} = 1 + 10^{tm}(10^m-1)Q_{t,m},$ and $C_t = Q_{t,m}(1 + 10^{3tm}).$ Then $$A_t$$ is a $$(2tm - m + 1)$$-digit number, and $$B_t$$ is a $$2tm$$-digit number, and both $$A_tB_t$$ and $$2A_tB_t$$ are $$(4tm - m + 1)$$-digit palindromes. Note that $$A_{t,m} = Q_{t,m}(1 + 10^{tm})$$ and $Q_{t,m} = \frac{10^{tm} - 1}{10^m-1}.$ Expanding $$A_{t,m}B_{t,m}$$, then, we have $A_{t,m}B_{t,m} = Q_{t,m}(1 + 10^{tm})(1 - 10^{tm} + 10^{2tm}) = C_{t,m}$ as needed. To see the digit counts, note that $$Q_{t,m}$$ has $$(tm-m+1)$$ digits. Note also that all the digits of $$C_{t,m}$$ are 1, so that $$2C_{t,m}$$ is also a palindrome. Let’s make $$Q_{t,m}$$, $$A_{t,m}$$, and $$B_{t,m}$$ executable. q_ :: Integer -> Integer -> Integer q_ t m = sum$ map (\k -> 10^(m*k)) [0..(t-1)]

a_ :: Integer -> Integer -> Integer
a_ t m = q_ (2*t) m

b_ :: Integer -> Integer -> Integer
b_ t m = 1 + (10^(t*m))*(10^m - 1)*(q_ t m)

c_ :: Integer -> Integer -> Integer
c_ t m = (q_ t m)*(1 + 10^(3*t*m))

Sanity check:

test_abc :: Integer -> Integer -> Bool
test_abc t m =
let
a = a_ t m; b = b_ t m; c = c_ t m
in and
[ a*b == c
, (numDigits a) == 2*t*m - m + 1
, (numDigits b) == 2*t*m
, (numDigits c) == 4*t*m - m + 1
, is_palindrome_10 c
, is_palindrome_10 (2*c)
]

And a test:

$> and [test_abc t m | t <- [1..10], m <- [1..10]] True This gives an infinite family of palindrome products. Note that if $$m = 1$$, then both factors have the same number of digits – $$2t$$ – and the product has $$4t$$ digits. Here are the first 5 examples. $$t$$ $$A_{t,m}$$ $$B_{t,m}$$ $$A_{t,m}B_{t,m}$$ 1 11 91 1001 2 1111 9901 11000011 3 111111 999001 111000000111 4 11111111 99990001 1111000000001111 5 1111111111 9999900001 11111000000000011111 Note that row $$k$$ in this table has factors $$A$$ and $$B$$ with $$2k$$ digits, and the (palindrome) product has $$4k$$ digits. Taking the first row, $$A = 11$$ and $$B = 91$$, we can use these in our $$H_v$$ and $$K_v$$, with $$v = 1$$, we get another family of palindrome products. $$v$$ $$H_v(A_{1,1})$$ $$K_v(B_{1,1})$$ $$H_v(A_{1,1})K_v(B_{1,1})$$ 1 1100011 9100091 10010200201001 3 110000011 910000091 100100020020001001 5 11000000011 91000000091 1001000002002000001001 7 1100000000011 9100000000091 10010000000200200000001001 Note that if $$v = 2k+1$$ then this table gives two $$(2k+7)$$-digit numbers whose (palindrome) product has $$4k+14$$ digits. Along with the observation that $$993 \cdot 913 = 906609$$ and $$11011 \cdot 91091 = 1003003001$$, we can say the following. Let $$k \geq 2$$. Then among the $$k$$-digit numbers, there exists a pair $$A$$ and $$B$$ such that $$AB$$ is a $$2k$$-digit palindrome. In particular, the largest such palindrome has $$2k$$ digits. Right, because the whole point of this digression was to establish this fact about largest palindrome products. ## Meanwhile Our alternate strategy for this problem was to search among the palindromes for $$k$$-digit factorizations; and now we know that it’s enough to look among the $$2k$$-digit palindromes – we are guaranteed to find a factorization there. This simplifies the search space a bit. So the new strategy is to generate the $$2k$$-digit palindromes in reverse order and look for the first one that factors are a product of two $$k$$-digit numbers. -- the 2k-digit palindromes palindromes :: Integer -> [Integer] palindromes k = map (fromDigits . make) (digits k) where -- turn a list into a palindrome make ds = ds ++ reverse ds digits :: Integer -> [[Integer]] digits k = foo k [] where foo 0 _ = [[]] foo k z = do d <- [9,8,7,6,5,4,3,2,1] ++ z ds <- foo (k-1)  return (d:ds) Now given a $$2k$$-digit palindrome, we want to know whether it factors as a product of two $$k$$-digit numbers. Note that if $$N = AB$$, then without loss of generality $$A \leq \sqrt{N}$$ and $$\sqrt{N} \leq B$$. So it is enough to search for a factorization of $$N$$ where $$10^{t-1} \leq A \leq \lfloor \sqrt{N} \rfloor$$. Here’s a utility function to find $$\lfloor \sqrt{N} \rfloor$$ by bisection. -- find t such that t^2 <= n < (t+1)^2 floor_sqrt :: Integer -> Integer floor_sqrt n = let bisect (a,b) = if b-a <= 1 then a else let m = (b+a)quot2 in if m^2 <= n then bisect (m,b) else bisect (a,m) in if n < 0 then error "floor_sqrt: negative argument" else bisect (1,n) Sanity check: test_floor_sqrt :: Integer -> Bool test_floor_sqrt n = let q = floor_sqrt n in (q^2 <= n) && (n < (q+1)^2) And a test: %> all test_floor_sqrt [1..10000] True Also, if we wish to factor a $$2k$$-digit number as a product of $$k$$ digit numbers, we should search “down” from $$\lfloor \sqrt{N} \rfloor$$ to $$10^{t-1}$$ rather than the reverse. This is because smaller factors are likely to give quotients with too many digits. -- search for a factorization of n into t-digit factors does_factor :: Integer -> Integer -> Maybe (Integer, Integer) does_factor t n = let check m | m < 10^(t-1) = Nothing | nremm == 0 = if numDigits (nquotm) == t then Just m else check (m-1) | otherwise = check (m-1) in case check (floor_sqrt n) of Nothing -> Nothing Just m -> Just (m, nquotm) pe4'' :: Integer -> (Integer, Integer, Integer) pe4'' k = let (a,b) = head$ catMaybes $map (does_factor k) (palindromes k) in (a,b,a*b) This pe4'' is still slow, as we still check an exponential number of cases. But we can squeeze out a couple more values. $> pe4'' 2
(91,99,9009)
$> pe4'' 3 (913,993,906609)$> pe4'' 4
(9901,9999,99000099)
$> pe4'' 5 (99681,99979,9966006699) Interesting! It looks like the form of the largest palindrome product depends on whether the factors have an even or odd number of digits. Anyway, the final answer is: pe4 :: Integer pe4 = let (_,_,x) = pe4'' 3 in x main :: IO () main = do let success = and [ all test_digits [1..1000000] , and [ test_abc t m | t <- [1..10], m <- [1..10] ] , all test_floor_sqrt [1..10000] ] if success then putStrLn$ show pe4
else exitFailure