Problem:
A prime gap of length n is a run of n-1 consecutive composite numbers between two successive primes
We will write a function gap with parameters:
- g (integer >= 2) which indicates the gap we are looking for
- m (integer > 2) which gives the start of the search (m inclusive)
- n (integer >= m) which gives the end of the search (n inclusive)
- n won't go beyond 1100000.
In the example above gap(2, 3, 50) will return [3, 5] or (3, 5) or {3, 5} which is the first pair between 3 and 50 with a 2-gap.
So this function should return the first pair of two prime numbers spaced with a gap of g between the limits m, n if these numbers exist otherwise `nil or null or None or Nothing (or ... depending on the language).
Examples:
- gap(2, 5, 7) --> [5, 7] or (5, 7) or {5, 7}
- gap(2, 5, 5) --> nil. In C++ {0, 0}. In F# [||]. In Kotlin, Dart and Prolog return []`
- gap(4, 130, 200) --> [163, 167] or (163, 167) or {163, 167}
([193, 197] is also such a 4-gap primes between 130 and 200 but it's not the first pair)
- gap(6,100,110) --> nil or {0, 0} or ... : between 100 and 110 we have 101, 103, 107, 109 but 101-107is not a 6-gap because there is 103in between and 103-109is not a 6-gap because there is 107in between.
- You can see more examples of return in Sample Tests.
public class GapInPrimes {
public static void main(String[] args) {
List<Long> result = LongStream
.of(gap(4, 130, 200))
.boxed()
.collect(Collectors.toList());
System.out.println(result);
}
public static long[] gap(long g, long m, long n) {
return LongStream
.iterate(m % 2 == 0 ? m + 1 : m, l -> l + 2)
.limit((n - m) / 2)
.filter(l -> BigInteger
.valueOf(l)
.isProbablePrime(5) && BigInteger
.valueOf(l + g)
.isProbablePrime(5))
.filter(l -> {
return LongStream
.iterate(l + 2, c -> c + 2)
.limit((g - 2) / 2)
.parallel()
.filter(c -> BigInteger
.valueOf(c)
.isProbablePrime(5))
.mapToObj(c -> false)
.findAny()
.orElse(true);
})
.mapToObj(l -> new long[]{l, l + g})
.findFirst()
.orElse(null);
}
}
Output:
[163, 167]
Step by step:
for gap(6.100.110)
LongStream
.iterate(m % 2 == 0 ? m + 1 : m, l -> l + 2)
.limit((n - m) / 2)
[101, 103, 105, 107, 109]
.filter(l -> BigInteger
.valueOf(l)
.isProbablePrime(5)
[101, 103, 107, 109]
&& BigInteger
.valueOf(l + g)
.isProbablePrime(5)
[101, 103, 107]
.filter(l -> {
return LongStream
.iterate(l + 2, c -> c + 2)
.limit((g - 2) / 2)
.parallel()
.filter(c -> BigInteger
.valueOf(c)
.isProbablePrime(5))
.mapToObj(c -> false)
.findAny()
.orElse(true);
})
[]
for gap(4.130.200)
LongStream
.iterate(m % 2 == 0 ? m + 1 : m, l -> l + 2)
.limit((n - m) / 2)
[131, 133, 135, 137, 139, 141, 143, 145, 147, 149, 151, 153, 155, 157, 159, 161, 163, 165, 167, 169, 171, 173, 175, 177, 179, 181, 183, 185, 187, 189, 191, 193, 195, 197, 199]
.filter(l -> BigInteger
.valueOf(l)
.isProbablePrime(5)
[131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199]
&& BigInteger
.valueOf(l + g)
.isProbablePrime(5)
[163, 193]
.filter(l -> {
return LongStream
.iterate(l + 2, c -> c + 2)
.limit((g - 2) / 2)
.parallel()
.filter(c -> BigInteger
.valueOf(c)
.isProbablePrime(5))
.mapToObj(c -> false)
.findAny()
.orElse(true);
})
[163, 193]
.mapToObj(l -> new long[]{l, l + g})
.findFirst()
.orElse(null);
[163, 167]
Maybe need time to fully understand, but it'll definitely worth for streams.
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