Pandigital prime: Difference between revisions

===Using digits 1 to n===

This makes use of the optimization in the Factor entry.

var t = a[i]

a[i] = a[j]

a[j] = t

t = a[i]

a[i] = a[j]

a[j] = t

System.print("The largest pandigital decimal prime which uses all the digits 1..n once is:")

for (n in [7, 4]) {

var perms = permutations.call((1..n).toList)

for (i in perms.count - 1..0) {

if (perms[i][-1] % 2 == 0 || perms[i][-1] == 5) continue

var p = Num.fromString(perms[i].join())

if (Int.isPrime(p)) {

Fmt.print("$,d", p)

return

</pre>

===Using digits 0 to n===

Applying the Factor logic for the digits 0..n, we only need to check 8 and 5 digit cases as it can't be any of the others.

<lang ecmascript>import "/math" for Int

import "/fmt" for Fmt

// generates all permutations in lexicographical order

var permutations = Fn.new { |input|

var perms = [input]

var a = input.toList

var n = a.count - 1

for (c in 1...Int.factorial(n+1)) {

var i = n - 1

var j = n

while (a[i] > a[i+1]) i = i - 1

while (a[j] < a[i]) j = j - 1

var t = a[i]

a[i] = a[j]

a[j] = t

j = n

i = i + 1

while (i < j) {

t = a[i]

a[i] = a[j]

a[j] = t

i = i + 1

j = j - 1

}

perms.add(a.toList)

}

return perms

}

System.print("The largest pandigital decimal prime which uses all the digits 0..n once is:")

for (n in [7, 4]) {

var perms = permutations.call((0..n).toList)

for (i in perms.count - 1..0) {

if (perms[i][0] == "0" || perms[i][-1] % 2 == 0 || perms[i][-1] == 5) continue

var p = Num.fromString(perms[i].join())

if (Int.isPrime(p)) {

Fmt.print("$,d", p)

return

}

}

}</lang>

{{out}}

<pre>