Largest palindrome product: Difference between revisions
m (→{{header|Julia}}: timer) |
m (Int128) |
||
| Line 522: | Line 522: | ||
m, sol = 0, (0, 0) |
m, sol = 0, (0, 0) |
||
for i in 1:L |
for i in 1:L |
||
lo = max(i, |
lo = max(Int128(i), Int128(up - (up - L + 1)^2 ÷ (up - i)) + 1) |
||
hi = |
hi = Int128(up - (up - L)^2 ÷ (up - i)) |
||
for j in lo:hi |
for j in lo:hi |
||
I = (up - i) * 10^tail |
I = (up - i) * 10^tail |
||
| Line 568: | Line 568: | ||
15 465 (999999998341069, 999999975838971) 999999974180040040081479999999 |
15 465 (999999998341069, 999999975838971) 999999974180040040081479999999 |
||
16 51 (9999999900000001, 9999999999999999) 99999999000000000000000099999999 |
16 51 (9999999900000001, 9999999999999999) 99999999000000000000000099999999 |
||
62.575515 seconds (241.50 M allocations: 16.491 GiB, 25.20% gc time, 0.07% compilation time) |
|||
</pre> |
</pre> |
||
Revision as of 02:43, 8 November 2021
- Task
Task description is taken from Project Euler (https://projecteuler.net/problem=4)
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.
- Stretch Goal
Find the largest palindrome made from the product of two n-digit numbers, where n ranges from 4 to 7.
- Extended Stretch Goal
Find the largest palindrome made from the product of two n-digit numbers, where n ranges beyond 7,
ALGOL 68
<lang algol68>BEGIN # find the highest palindromic multiple of various sizes of numbers #
# returns n with the digits reversed #
PROC reverse = ( LONG INT v )LONG INT:
BEGIN
LONG INT r := 0 ;
LONG INT n := v;
WHILE n > 0 DO
r *:= 10; r +:= n MOD 10;
n OVERAB 10
OD;
r
END # reverse # ;
LONG INT pow := 10;
FOR n FROM 2 TO 7 DO
LONG INT low := pow * 9;
pow *:= 10;
LONG INT high := pow - 1;
print( ( "Largest palindromic product of two ", whole( n, 0 ), "-digit integers: " ) );
BOOL next n := FALSE;
LONG INT i := high + 1;
WHILE i -:= 1;
i >= low AND NOT next n
DO
LONG INT j = reverse( i );
LONG INT p = ( i * pow ) + j;
# k can't be even nor end in 5 to produce a product ending in 9 #
LONG INT k := high + 2;
WHILE k -:= 2;
IF k < low
THEN FALSE
ELIF k MOD 10 = 5
THEN TRUE
ELIF LONG INT l = p OVER k;
l > high
THEN FALSE
ELIF p MOD k = 0 THEN
print( ( whole( k, 0 ), " x ", whole( l, 0 ), " = ", whole( p, 0 ), newline ) );
next n := TRUE;
FALSE
ELSE TRUE
FI
DO SKIP OD
OD
OD
END</lang>
- Output:
Largest palindromic product of two 2-digit integers: 99 x 91 = 9009 Largest palindromic product of two 3-digit integers: 993 x 913 = 906609 Largest palindromic product of two 4-digit integers: 9999 x 9901 = 99000099 Largest palindromic product of two 5-digit integers: 99979 x 99681 = 9966006699 Largest palindromic product of two 6-digit integers: 999999 x 999001 = 999000000999 Largest palindromic product of two 7-digit integers: 9998017 x 9997647 = 99956644665999
Also showing the maximum for 2 and 4 .. 7 digit numbers. Tests for a better product before testing for palindromicity. <lang algol68>BEGIN # find the highest palindromic multiple of various sizes of numbers #
PROC is pal = ( LONG INT n )BOOL:
BEGIN
STRING x = whole( n, 0 );
INT l := UPB x + 1;
BOOL result := TRUE;
FOR i FROM LWB x WHILE i < l AND result DO
l -:= 1;
result := x[ i ] = x[ l ]
OD;
result
END # is pal # ;
# maximum 2 digit number #
LONG INT max := 99;
# both factors must be >= 10for a 4 digit product #
LONG INT limit start := 10;
FOR w FROM 2 TO 7 DO
LONG INT best prod := 0;
# one factor must be divisible by 11 #
LONG INT limit end = 11 * ( max OVER 11 );
LONG INT second := limit start;
LONG INT first := 1;
# loop from hi to low to find the best result in the fewest steps #
LONG INT n := limit end + 11;
WHILE n -:= 11;
n >= limit start
DO
# with n falling, the lower limit of m can rise with #
# the best-found-so-far second number. Doing this #
# lowers the iteration count by a lot. #
LONG INT m := max + 2;
WHILE m -:= 2;
IF m < second
THEN FALSE
ELIF LONG INT prod = n * m;
best prod > prod
THEN FALSE
ELIF NOT is pal( prod )
THEN TRUE
ELSE # maintain the best-found-so-far result #
first := n;
second := m;
best prod := prod;
TRUE
FI
DO SKIP OD
OD;
print( ( "Largest palindromic product of two ", whole( w, 0 )
, "-digit numbers: ", whole( first, 0 ), " * ", whole( second, 0 )
, " = ", whole( best prod, 0 )
, newline
)
);
max *:= 10;
max +:= 9;
limit start *:= 10
OD
END</lang>
- Output:
Largest palindromic product of two 2-digit numbers: 99 * 91 = 9009 Largest palindromic product of two 3-digit numbers: 913 * 993 = 906609 Largest palindromic product of two 4-digit numbers: 9999 * 9901 = 99000099 Largest palindromic product of two 5-digit numbers: 99979 * 99681 = 9966006699 Largest palindromic product of two 6-digit numbers: 999999 * 999001 = 999000000999 Largest palindromic product of two 7-digit numbers: 9997647 * 9998017 = 99956644665999
C#
Main Task
<lang csharp>using System; class Program {
static bool isPal(int n) {
int rev = 0, lr = -1, rem;
while (n > rev) {
n = Math.DivRem(n, 10, out rem);
if (lr < 0 && rem == 0) return false;
lr = rev; rev = 10 * rev + rem;
if (n == rev || n == lr) return true;
} return false; }
static void Main(string[] args) {
var sw = System.Diagnostics.Stopwatch.StartNew();
int x = 900009, y = (int)Math.Sqrt(x), y10, max = 999, max9 = max - 9, z, p, bp = x, ld, c;
var a = new int[]{ 0,9,0,3,0,0,0,7,0,1 }; string bs = "";
y /= 11;
if ((y & 1) == 0) y--;
if (y % 5 == 0) y -= 2;
y *= 11;
while (y <= max) {
c = 0;
y10 = y * 10;
z = max9 + a[ld = y % 10];
p = y * z;
while (p >= bp) {
if (isPal(p)) {
if (p > bp) bp = p;
bs = string.Format("{0} x {1} = {2}", y, z - c, bp);
}
p -= y10; c += 10;
}
y += ld == 3 ? 44 : 22;
}
sw.Stop();
Console.Write("{0} {1} μs", bs, sw.Elapsed.TotalMilliseconds * 1000.0);
}
}</lang>
- Output @ Tio.run:
913 x 993 = 906609 245.2 μs
Stretch
<lang csharp>using System;
class Program {
static bool isPal(long n) {
long rev = 0, lr = -1, rem;
while (n > rev) {
n = Math.DivRem(n, 10, out rem);
if (lr < 0 && rem == 0) return false;
lr = rev; rev = 10 * rev + rem;
if (n == rev || n == lr) return true;
} return false; }
static void doOne(int n) {
int ld, c; string bs = "";
string sx = "9" + new string('0', (n - 1) << 1) + "9", sm = new string('9', n);
long x = long.Parse(sx), y = (long)Math.Sqrt(x), oy, max = long.Parse(sm), max9 = max - 9, z, yy, p, bp = x;
var a = new long[] { 0, 9, 0, 3, 0, 0, 0, 7, 0, 1 };
y /= 11;
if ((y & 1) == 0) y--;
if (y % 5 == 0) y -= 2;
y *= 11; oy = y;
while (y <= max) y += 22; y -= 22;
while (y >= oy) {
c = 0;
yy = y * 10;
z = max9 + a[ld = (int)(y % 10)];
//Console.WriteLine("y,z: {0},{1}", y, z);
p = y * z;
while (p >= bp) {
if (isPal(p)) {
if (p > bp) bp = p;
bs = string.Format(" {0,2} {1,10} x {2,-10} = {3}{4}", n, y, z - c, new string(' ', 10 - n), bp); }
p -= yy; c += 10; }
y -= ld == 7 ? 44 : 22; }
Console.WriteLine(bs); }
static void Main(string[] args) {
Console.WriteLine("digs factor factor palindrome");
var sw = System.Diagnostics.Stopwatch.StartNew();
for (int i = 2, h = 1; i <= 10; h = ++i >> 1) {
if ((i & 1) == 0) {
string b = new string('9', i),
a = new string('9', h) + new string('0', (h) - 1) + "1",
c = new string('9', h) + new string('0', i) + new string('9', h);
Console.WriteLine(" {0,2} {1,10} x {2,-10} = {3}{4}", i, a, b, new string(' ', 10 - i), c); }
else doOne(i);
}
sw.Stop();
Console.Write("{0} sec", sw.Elapsed.TotalSeconds);
}
}</lang>
- Output @ Tio.run:
Showing results for 2 through 10 digit factors.
digs factor factor palindrome 2 91 x 99 = 9009 3 913 x 993 = 906609 4 9901 x 9999 = 99000099 5 99979 x 99681 = 9966006699 6 999001 x 999999 = 999000000999 7 9997647 x 9998017 = 99956644665999 8 99990001 x 99999999 = 9999000000009999 9 999920317 x 999980347 = 999900665566009999 10 9999900001 x 9999999999 = 99999000000000099999 2.1622142 sec
Wow! how did that go so fast? The results for the even-number-of-digit factors were manufactured by string manipulation instead of calculation (since the pattern was obvious). This algorithm can easily be adapted to BigIntegers for higher n-digit factors, but the execution time is unspectacular.
F#
<lang fsharp> // Largest palindrome product. Nigel Galloway: November 3rd., 2021 let fN g=let rec fN g=[yield g%10; if g>=10 then yield! fN(g/10)] in let n=fN g in n=List.rev n printfn "%d" ([for n in 100..999 do for g in n..999->n*g]|>List.filter fN|>List.max) </lang>
- Output:
906609
FreeBASIC
Version 1
<lang freebasic>function make_pal( n as ulongint ) as ulongint
'turn a number into a palindrom with twice as many digits
dim as string ns, ret
ns = str(n) : ret = ns
for i as uinteger = len(ns) to 1 step -1
ret += mid(ns, i, 1)
next i
return val(ret)
end function
function has_dig( n as ulongint, d as uinteger ) as boolean
'does the number n have d decimal digits? if 10^(d-1)<=n and n<10^d then return true else return false
end function
dim as integer np
for d as uinteger = 2 to 7
for n as ulongint = 10^d - 1 to 10^(d-1) step -1 'count down from 999...
'since the first good number we encounter
'must be the highest
np = make_pal( n ) 'produce a 2d-digit palindrome from it
for f as ulongint = 10^d - 1 to 10^(d-1) step -1 'look for highest d-digit factor
if np mod f = 0 then
if has_dig( np/f, d ) then 'if np/f also has d digits we are done :)
print f;" *";np/f;" =";np
goto nextd
end if
end if
next f
next n
nextd: 'yes, I used a goto. sue me.
next d</lang>
- Output:
99 * 91 = 9009 993 * 913 = 906609 9999 * 9901 = 99000099 99979 * 99681 = 9966006699 999999 * 999001 = 999000000999 9998017 * 9997647 = 99956644665999
Version 2
This version is based on Version 1 with only a few changes and some extra code based on the fact that one divisor can be divided by 11, this speeds it even more up and a option for using goto, exit or continue. highest n is 9, (highest possible for unsigned 64bit integers). <lang freebasic>' version 07-10-2021 ' compile with: fbc -s console
' Now you can choice, no speed changes for all 3 ' 1: use goto ' 2: use exit ' 3: use continue
- Define Option_ 1 ' set option_ to 1, 2 or 3. for all other value's uses 1
Function make_pal( n As UInteger ) As ULongInt
'turn a number into a palindrom with twice as many digits
Dim As String ns = Str(n), ret = ns
For i As UInteger = Len(ns) To 1 Step -1
ret += Mid(ns, i, 1)
Next i
Return ValULng(ret)
End Function
Dim As ULongInt np, tmp Dim As Double t1 =Timer For d As UInteger = 2 To 9
For n As UInteger = 10^d -2 To 10^(d -1) Step -1
np = make_pal( n )
tmp = Sqr(np)
tmp = tmp - (10^d - 1 - tmp)
tmp = tmp - tmp Mod 11
If (tmp And 1) = 0 Then tmp = tmp + 11
For f As UInteger = tmp To 10^d -1 Step 22
If np Mod f = 0 Then
If np \ f > (10^d) Then Continue For
Print f; " * "; np \ f; " = "; np
#If (option_ = 2)
Exit For, For
#ElseIf (option_ = 3)
Continue For, For, For
#Else
GoTo nextd
#EndIf
End If
Next f
Next n
#If (option_ <> 2 Or option_ <> 3)
nextd:
#EndIf
Next d
Print Timer-t1 ' empty keyboard buffer While InKey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
99 * 91 = 9009 993 * 913 = 906609 9999 * 9901 = 99000099 99979 * 99681 = 9966006699 999999 * 999001 = 999000000999 9998017 * 9997647 = 99956644665999 99999999 * 99990001 = 9999000000009999 999980347 * 999920317 = 999900665566009999
Go
18 digit integers are within the range of Go's uint64 type though finding the result for 9-digit number products takes a while - around 15 seconds on my machine. <lang go>package main
import "fmt"
func reverse(n uint64) uint64 {
r := uint64(0)
for n > 0 {
r = n%10 + r*10
n /= 10
}
return r
}
func main() {
pow := uint64(10)
nextN:
for n := 2; n < 10; n++ {
low := pow * 9
pow *= 10
high := pow - 1
fmt.Printf("Largest palindromic product of two %d-digit integers: ", n)
for i := high; i >= low; i-- {
j := reverse(i)
p := i*pow + j
// k can't be even nor end in 5 to produce a product ending in 9
for k := high; k > low; k -= 2 {
if k % 10 == 5 {
continue
}
l := p / k
if l > high {
break
}
if p%k == 0 {
fmt.Printf("%d x %d = %d\n", k, l, p)
continue nextN
}
}
}
}
}</lang>
- Output:
Largest palindromic product of two 2-digit integers: 99 x 91 = 9009 Largest palindromic product of two 3-digit integers: 993 x 913 = 906609 Largest palindromic product of two 4-digit integers: 9999 x 9901 = 99000099 Largest palindromic product of two 5-digit integers: 99979 x 99681 = 9966006699 Largest palindromic product of two 6-digit integers: 999999 x 999001 = 999000000999 Largest palindromic product of two 7-digit integers: 9998017 x 9997647 = 99956644665999 Largest palindromic product of two 8-digit integers: 99999999 x 99990001 = 9999000000009999 Largest palindromic product of two 9-digit integers: 999980347 x 999920317 = 999900665566009999
Julia
<lang julia>using Primes
function twoprodpal(factorlength)
maxpal = Int128(10)^(2 * factorlength) - 1
dig = digits(maxpal)
halfnum = dig[1:length(dig)÷2]
while any(halfnum .!= 0)
prodnum = evalpoly(Int128(10), [reverse(halfnum); halfnum])
facs = twofac(factorlength, prodnum)
if !isempty(facs)
println("For factor length $factorlength, $(facs[1]) * $(facs[2]) = $prodnum")
break
end
halfnum = digits(evalpoly(Int128(10), halfnum) - 1)
end
end
function twofac(faclength, prodnum)
f = [one(prodnum)]
for (p, e) in factor(prodnum)
f = reduce(vcat, [f * p^j for j in 1:e], init=f)
end
possiblefacs = filter(x -> length(string(x)) == faclength, f)
for i in possiblefacs
j = prodnum ÷ i
j ∈ possiblefacs && return sort([i, j])
end
return typeof(prodnum)[]
end
@Threads.threads for i in 2:12
twoprodpal(i)
end
</lang>
- Output:
For factor length 2, 91 * 99 = 9009 For factor length 3, 913 * 993 = 906609 For factor length 4, 9901 * 9999 = 99000099 For factor length 5, 99681 * 99979 = 9966006699 For factor length 6, 999001 * 999999 = 999000000999 For factor length 7, 9997647 * 9998017 = 99956644665999 For factor length 8, 99990001 * 99999999 = 9999000000009999 For factor length 9, 999920317 * 999980347 = 999900665566009999 For factor length 10, 9999986701 * 9999996699 = 99999834000043899999 For factor length 11, 99999943851 * 99999996349 = 9999994020000204999999 For factor length 12, 999999000001 * 999999999999 = 999999000000000000999999
Faster version
<lang julia>""" taken from https://leetcode.com/problems/largest-palindrome-product/discuss/150954/Fast-algorithm-by-constrains-on-tail-digits """
const T = [Set([(0, 0)])]
function double(it)
arr = empty(it)
for p in it
push!(arr, p, reverse(p))
end
return arr
end
""" Construct a pair of n-digit numbers such that their product ends with 99...9 pattern """ function tails(n)
if length(T) <= n
l = Set()
for i in 0:9, j in i:9
I = i * 10^(n-1)
J = j * 10^(n-1)
it = collect(tails(n - 1))
I != J && (it = double(it))
for (t1, t2) in it
if ((I + t1) * (J + t2) + 1) % 10^n == 0
push!(l, (I + t1, J + t2))
end
end
end
push!(T, l)
end
return T[n + 1]
end
""" find the largest palindrome that is a product of n-digit numbers """ function largestpalindrome(n)
m, tail = 0, n ÷ 2
head = n - tail
up = 10^head
for L in 1 : 9 * 10^(head-1)
# Consider small shell (up-L)^2 < (up-i)*(up-j) <= (up-L)^2, 1<=i<=L<=j
m, sol = 0, (0, 0)
for i in 1:L
lo = max(Int128(i), Int128(up - (up - L + 1)^2 ÷ (up - i)) + 1)
hi = Int128(up - (up - L)^2 ÷ (up - i))
for j in lo:hi
I = (up - i) * 10^tail
J = (up - j) * 10^tail
it = collect(tails(tail))
I != J && (it = double(it))
for (t1, t2) in it
val = (I + t1) * (J + t2)
s = string(val)
if s == reverse(s) && val > m
sol = (I + t1, J + t2)
m = val
end
end
end
end
if m > 0
println(lpad(n, 2), " ", lpad(m % 1337, 4), " $sol $(sol[1] * sol[2])")
return m % 1337
end
end
return 0
end
@time for k in 1:16
largestpalindrome(k)
end
</lang>
- Output:
1 9 (9, 1) 9 2 987 (91, 99) 9009 3 123 (993, 913) 906609 4 597 (9901, 9999) 99000099 5 677 (99979, 99681) 9966006699 6 1218 (999001, 999999) 999000000999 7 877 (9998017, 9997647) 99956644665999 8 475 (99990001, 99999999) 9999000000009999 9 1226 (999980347, 999920317) 999900665566009999 10 875 (9999986701, 9999996699) 99999834000043899999 11 108 (99999943851, 99999996349) 9999994020000204999999 12 378 (999999000001, 999999999999) 999999000000000000999999 13 1097 (9999999993349, 9999996340851) 99999963342000024336999999 14 959 (99999990000001, 99999999999999) 9999999000000000000009999999 15 465 (999999998341069, 999999975838971) 999999974180040040081479999999 16 51 (9999999900000001, 9999999999999999) 99999999000000000000000099999999 62.575515 seconds (241.50 M allocations: 16.491 GiB, 25.20% gc time, 0.07% compilation time)
Paper & Pencil
<lang>find two 3-digit factors, that when multiplied together, yield the highest 6-digit palindrome.
lowest possible 6 digit palindrome starting with 9 is 900009 floor(sqrt(900009)) = 948 one factor must be an odd multiple of 11 floor(948 / 11) = 86 it must not be even or a multiple of 5, so use 83 83 * 11 = 913 <- this is the lowest possible first factor the last digit of the second factor must multiply with the last digit of the first factor to get 9 the highest suitable second factor (for 913) is 993 913 x 993 = 906609, a palindrome, now check suitable higher first factors 913 + 22 = 935, an unsuitable multiple of 5, so skip it and use 913 + 44 = 957 957 x 997 = 954129, not a palindrome, so continue (just subtract 9570) 957 x 987 = 944559, not a palindrome, so continue 957 x 977 = 934989, not a palindrome, so continue 957 x 967 = 925429, not a palindrome, so continue 957 x 957 = 915849, not a palindrome, so continue 957 x 947 = 906279, not a palindrome, and less than the best found so far, so stop and continue to the next suitible first number, 957 + 22 = 979 979 x 991 = 970189, not a palindrome, so continue (just subtract 9790) 979 x 981 = 960399, not a palindrome, so continue 979 x 971 = 950609, not a palindrome, so continue 979 x 961 = 940819, not a palindrome, so continue 979 x 951 = 931029, not a palindrome, so continue 979 x 941 = 921239, not a palindrome, so continue 979 x 931 = 911449, not a palindrome, so continue 979 x 921 = 901659, not a palindrome, and less than the best found so far, so stop done because 979 + 22 = 1001</lang>
Perl
<lang perl>use strict; use warnings; use feature 'say'; use ntheory 'divisors';
for my $l (2..7) {
LOOP:
for my $p (reverse map { $_ . reverse $_ } 10**($l-1) .. 10**$l - 1) {
my @f = reverse grep { length == $l } divisors $p;
next unless @f >= 2 and $p == $f[0] * $f[1];
say "Largest palindromic product of two @{[$l]}-digit integers: $f[1] × $f[0] = $p" and last LOOP;
}
}</lang>
- Output:
Largest palindromic product of two 2-digit integers: 91 × 99 = 9009 Largest palindromic product of two 3-digit integers: 913 × 993 = 906609 Largest palindromic product of two 4-digit integers: 9901 × 9999 = 99000099 Largest palindromic product of two 5-digit integers: 99681 × 99979 = 9966006699 Largest palindromic product of two 6-digit integers: 999001 × 999999 = 999000000999 Largest palindromic product of two 7-digit integers: 9997647 × 9998017 = 99956644665999
Phix
Translated from python by Lucy_Hedgehog as found on page 5 of the project euler discussion page (dated 25 Sep 2011), and further optimised as per the C# comments (on this very rosettacode page). You can run this online here.
-- demo\rosetta\Largest_palindrome_product.exw with javascript_semantics requires("1.0.1") -- (mpz_fdiv_qr(), mpz_si_sub() added to mpfr.js, mpz_mod_ui(), mpz_fdiv_q_ui(), mpz_fdiv_r(), mpz_fdiv_ui() fixed) include mpfr.e function ispalindrome(mpz x) string s = mpz_get_str(x) return s == reverse(s) end function function inverse(mpz x, integer m) -- Compute the modular inverse of x modulo power(10,m). -- Return -1 if the inverse does not exist. -- This function uses Hensel lifting. integer a = {-1, 1, -1, 7, -1, -1, -1, 3, -1, 9}[mpz_fdiv_ui(x,10)+1] if a!=-1 then mpz ax = mpz_init() while true do mpz_mul_si(ax,x,a) {} = mpz_mod_ui(ax,ax,m) if mpz_cmp_si(ax,1)==0 then exit end if mpz_si_sub(ax,2,ax) mpz_mul_si(ax,ax,a) a = mpz_fdiv_q_ui(ax,ax,m) end while end if return a end function function pal2(integer n) assert(n>1) mpz {best,factor,y,r} = mpz_inits(4) if even(n) then // (as per the C# comments) mpz_ui_pow_ui(factor,10,n/2) mpz_sub_si(factor,factor,1) mpz_ui_pow_ui(best,10,n/2*3) mpz_mul(best,best,factor) mpz_add(best,best,factor) assert(ispalindrome(best)) mpz_ui_pow_ui(factor,10,n) mpz_sub_si(factor,factor,1) assert(ispalindrome(factor)) else // Get a lower bound: integer k = floor(n/2) mpz {maxf,maxf11,minf,x,t,maxy,p} = mpz_inits(7) while true do mpz_ui_pow_ui(maxf,10,n) mpz_sub_si(maxf,maxf,1) mpz_sub_si(maxf11,maxf,11) {} = mpz_fdiv_q_ui(maxf11,maxf11,22) mpz_mul_si(maxf11,maxf11,22) mpz_add_si(maxf11,maxf11,11) mpz_ui_pow_ui(minf,10,n-k) mpz_sub(minf,maxf,minf) mpz_add_si(minf,minf,2) mpz_mul(best,minf,minf) mpz_set_si(factor,0) // This palindrome starts with k 9's. // Hence the largest palindrom must also start with k 9's and // therefore end with k 9's. // Thus, if p = x * y is the solution then // x * y + 1 is divisible by m. integer m = power(10,k) -- (should not exceed 1e8) mpz_set(x,maxf11) while mpz_cmp_si(x,1)>=0 do mpz_mul(t,x,maxf) if mpz_cmp(t,best)=-1 then exit end if integer ry = inverse(x, m) if ry!=-1 then mpz_add_si(maxy,maxf,1-ry) mpz_mul(p,maxy,x) while mpz_cmp(p,best)>0 do if ispalindrome(p) then mpz_set(best,p) mpz_set(factor,x) end if mpz_mul_si(t,x,m) mpz_sub(p,p,t) end while end if mpz_sub_si(x,x,22) end while if mpz_cmp_si(factor,0)!=0 then exit end if k -= 1 end while end if mpz_fdiv_qr(y,r,best,factor) assert(mpz_cmp_si(r,0)=0) return {best, factor, y} end function constant fmt = "Largest palindromic product of two %d-digit integers: %s = %s x %s (%s)\n" for n=2 to 12 do atom t1 = time() mpz {p,x,y} = pal2(n) string sp = mpz_get_str(p), sx = mpz_get_str(x), sy = mpz_get_str(y), e = elapsed(time()-t1) printf(1,fmt,{n,sp,sx,sy,e}) end for
- Output:
Largest palindromic product of two 2-digit integers: 9009 = 99 x 91 (0s) Largest palindromic product of two 3-digit integers: 906609 = 913 x 993 (0s) Largest palindromic product of two 4-digit integers: 99000099 = 9999 x 9901 (0s) Largest palindromic product of two 5-digit integers: 9966006699 = 99979 x 99681 (0s) Largest palindromic product of two 6-digit integers: 999000000999 = 999999 x 999001 (0s) Largest palindromic product of two 7-digit integers: 99956644665999 = 9997647 x 9998017 (0.0s) Largest palindromic product of two 8-digit integers: 9999000000009999 = 99999999 x 99990001 (0s) Largest palindromic product of two 9-digit integers: 999900665566009999 = 999920317 x 999980347 (0.8s) Largest palindromic product of two 10-digit integers: 99999000000000099999 = 9999999999 x 9999900001 (0s) Largest palindromic product of two 11-digit integers: 9999994020000204999999 = 99999996349 x 99999943851 (0.1s) Largest palindromic product of two 12-digit integers: 999999000000000000999999 = 999999999999 x 999999000001 (0s)
After that it starts to struggle a bit:
Largest palindromic product of two 13-digit integers: 99999963342000024336999999 = 9999996340851 x 9999999993349 (40.4s) Largest palindromic product of two 14-digit integers: 9999999000000000000009999999 = 99999999999999 x 99999990000001 (0s) Largest palindromic product of two 15-digit integers: 999999974180040040081479999999 = 999999998341069 x 999999975838971 (1 minute and 12s) Largest palindromic product of two 16-digit integers: 99999999000000000000000099999999 = 9999999999999999 x 9999999900000001 (0s)
Python
Original author credit to user peijunz at Leetcode. <lang python>""" taken from https://leetcode.com/problems/largest-palindrome-product/discuss/150954/Fast-algorithm-by-constrains-on-tail-digits """
T=[set([(0, 0)])]
def double(it):
for a, b in it:
yield a, b
yield b, a
def tails(n):
Construct pair of n-digit numbers that their product ends with 99...9 pattern
if len(T)<=n:
l = set()
for i in range(10):
for j in range(i, 10):
I = i*10**(n-1)
J = j*10**(n-1)
it = tails(n-1)
if I!=J: it = double(it)
for t1, t2 in it:
if ((I+t1)*(J+t2)+1)%10**n == 0:
l.add((I+t1, J+t2))
T.append(l)
return T[n]
def largestPalindrome(n):
""" find largest palindrome that is a product of two n-digit numbers """
m, tail = 0, n // 2
head = n - tail
up = 10**head
for L in range(1, 9*10**(head-1)+1):
# Consider small shell (up-L)^2 < (up-i)*(up-j) <= (up-L)^2, 1<=i<=L<=j
m = 0
sol = None
for i in range(1, L + 1):
lo = max(i, int(up - (up - L + 1)**2 / (up - i)) + 1)
hi = int(up - (up - L)**2 / (up - i))
for j in range(lo, hi + 1):
I = (up-i) * 10**tail
J = (up-j) * 10**tail
it = tails(tail)
if I!=J: it = double(it)
for t1, t2 in it:
val = (I + t1)*(J + t2)
s = str(val)
if s == s[::-1] and val>m:
sol = (I + t1, J + t2)
m = val
if m:
print("{:2d}\t{:4d}".format(n, m % 1337), sol, sol[0] * sol[1])
return m % 1337
return 0
if __name__ == "__main__":
for k in range(1, 14):
largestPalindrome(k)
</lang>
- Output:
1 9 (9, 1) 9 2 987 (91, 99) 9009 3 123 (993, 913) 906609 4 597 (9901, 9999) 99000099 5 677 (99979, 99681) 9966006699 6 1218 (999001, 999999) 999000000999 7 877 (9998017, 9997647) 99956644665999 8 475 (99990001, 99999999) 9999000000009999 9 1226 (999980347, 999920317) 999900665566009999 10 875 (9999986701, 9999996699) 99999834000043899999 11 108 (99999943851, 99999996349) 9999994020000204999999 12 378 (999999000001, 999999999999) 999999000000000000999999 13 1097 (9999999993349, 9999996340851) 99999963342000024336999999
Raku
<lang perl6>use Inline::Perl5; my $p5 = Inline::Perl5.new(); $p5.use: 'ntheory'; my &divisors = $p5.run('sub { ntheory::divisors $_[0] }');
.say for (2..12).map: {.&lpp};
multi lpp ($oom where {!($_ +& 1)}) { # even number of multiplicand digits
my $f = +(9 x $oom); my $o = $oom / 2; my $pal = +(9 x $o ~ 0 x $oom ~ 9 x $o); sprintf "Largest palindromic product of two %2d-digit integers: %d × %d = %d", $oom, $pal div $f, $f, $pal
}
multi lpp ($oom where {$_ +& 1}) { # odd number of multiplicand digits
my $p;
(+(1 ~ (0 x ($oom - 1))) .. +(9 ~ (9 x ($oom - 1)))).reverse.map({ +($_ ~ .flip) }).map: -> $pal {
for my @factors = divisors("$pal")».Int.grep({ .chars == $oom }).sort( -* ) {
next unless $pal div $_ ∈ @factors;
$p = sprintf("Largest palindromic product of two %2d-digit integers: %d × %d = %d", $oom, $pal div $_, $_, $pal);
last;
}
last if $p;
}
$p
}</lang>
Largest palindromic product of two 2-digit integers: 91 × 99 = 9009 Largest palindromic product of two 3-digit integers: 913 × 993 = 906609 Largest palindromic product of two 4-digit integers: 9901 × 9999 = 99000099 Largest palindromic product of two 5-digit integers: 99681 × 99979 = 9966006699 Largest palindromic product of two 6-digit integers: 999001 × 999999 = 999000000999 Largest palindromic product of two 7-digit integers: 9997647 × 9998017 = 99956644665999 Largest palindromic product of two 8-digit integers: 99990001 × 99999999 = 9999000000009999 Largest palindromic product of two 9-digit integers: 999920317 × 999980347 = 999900665566009999 Largest palindromic product of two 10-digit integers: 9999900001 × 9999999999 = 99999000000000099999 Largest palindromic product of two 11-digit integers: 99999943851 × 99999996349 = 9999994020000204999999 Largest palindromic product of two 12-digit integers: 999999000001 × 999999999999 = 999999000000000000999999
Ring
<lang ring>? "working..."
prod = 1 bestProd = 0 // maximum 3 digit number max = 999 // both factors must be >100 for a 6 digit product limitStart = 101 // one factor must be divisible by 11 limitEnd = 11 * floor(max / 11) second = limitStart iters = 0
// loop from hi to low to find the best result in the fewest steps for n = limitEnd to limitStart step -11
// with n falling, the lower limit of m can rise with
// the best-found-so-far second number. Doing this
// lowers the iteration count by a lot.
for m = max to second step -2
prod = n * m
if isPal(prod)
iters++
// exit when the product stops increasing
if bestProd > prod
exit
ok
// maintain the best-found-so-far result
first = n
second = m
bestProd = prod
ok
next
next
put "The largest palindrome is: " ? "" + bestProd + " = " + first + " * " + second ? "Found in " + iters + " iterations" put "done..."
func isPal n
x = string(n)
l = len(x) + 1
i = 0
while i < l
if x[i++] != x[l--]
return false
ok
end
return true</lang>
- Output:
working... The largest palindrome is: 906609 = 913 * 993 Found in 6 iterations done...
Wren
The approach here is to manufacture palindromic numbers of length 2n in decreasing order and then see if they're products of two n-digit numbers. <lang ecmascript>var reverse = Fn.new { |n|
var r = 0
while (n > 0) {
r = n%10 + r*10
n = (n/10).floor
}
return r
}
var pow = 10 for (n in 2..7) {
var low = pow * 9
pow = pow * 10
var high = pow - 1
System.write("Largest palindromic product of two %(n)-digit integers: ")
var nextN = false
for (i in high..low) {
var j = reverse.call(i)
var p = i * pow + j
// k can't be even nor end in 5 to produce a product ending in 9
var k = high
while (k > low) {
if (k % 10 != 5) {
var l = p / k
if (l > high) break
if (p % k == 0) {
System.print("%(k) x %(l) = %(p)")
nextN = true
break
}
}
k = k - 2
}
if (nextN) break
}
}</lang>
- Output:
Largest palindromic product of two 2-digit integers: 99 x 91 = 9009 Largest palindromic product of two 3-digit integers: 993 x 913 = 906609 Largest palindromic product of two 4-digit integers: 9999 x 9901 = 99000099 Largest palindromic product of two 5-digit integers: 99979 x 99681 = 9966006699 Largest palindromic product of two 6-digit integers: 999999 x 999001 = 999000000999 Largest palindromic product of two 7-digit integers: 9998017 x 9997647 = 99956644665999
XPL0
<lang XPL0>func Rev(A); \Reverse digits int A, B; [B:= 0; repeat A:= A/10;
B:= B*10 + rem(0);
until A = 0; return B; ];
int Max, M, N, Prod; [Max:= 0; for M:= 100 to 999 do
for N:= M to 999 do
[Prod:= M*N;
if Prod/1000 = Rev(rem(0)) then
if Prod > Max then Max:= Prod;
];
IntOut(0, Max); ]</lang>
- Output:
906609