S2T2

Jim Randell 8:42 am on 2 December 2021 Permalink Reply
Tags: by: Stephen Hogg ( 62 )   

• 

  Jim Randell 8:43 am on 2 December 2021 Permalink | Reply

  We can express the the total number of crenels in three different ways (assuming no mistakes were made in the counting):

  t = 5n + 10 (where 10 ≤ n ≤ 95, so 60 ≤ t ≤ 485)
  t = p1 + p2 (where p1 and p2 are consecutive primes)
  t = q1 × q2 × q3 (where q1, q2, q3 are primes)

  This Python program looks at consecutive primes, until it finds a total that satisfies the remaining two expressions.

  It runs in 47ms.

  Run: [ @replit ]

  from enigma import (primes, tuples, div, printf)
  
  # consider consecutive pairs of primes (primes.between(29, 242) is sufficient)
  for (p1, p2) in tuples(primes.between(2, 485), 2):
    # form the total
    t = p1 + p2
    if t < 60: continue
    if t > 485: break
  
    # check it can be expressed as 5n + 10
    n = div(t - 10, 5)
    if n is None: continue
  
    # check it is the product of three prime numbers
    qs = primes.factor(t)
    if len(qs) != 3: continue
  
    # output solution
    printf("{t} = +({n}..{n4}) = {p1} + {p2} = *{qs}", n4=n + 4)
  

  Solution: There were 410 crenels in total.

  So we have:

  410 = 80 + 81 + 82 + 83 + 84
  410 = 199 + 211
  410 = 2 × 5 × 41

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• 

  Frits 10:59 pm on 2 December 2021 Permalink | Reply

  More analysis and using the 6n + 1 or 6n – 1 rule (just for fun).

  Other than 2 and 3, prime numbers must be of the form 6n + 1 or 6n – 1.

  Enigma function is_prime() is called only once.
  Function islice() is borrowed from enigma function first().

    
  from itertools import islice
  from enigma import is_prime
  
  # t = 5n + 10 (where 10 <= n <= 95, so 60 <= t <= 485)
  # t = p1 + p2 (where p1 and p2 are consecutive primes) ==> t is even
  # t = q1 × q2 × q3 (where q1, q2, q3 are primes and q1 <= q2 <= q3) 
  # t is even ==> q1 = 2, n is even and 30 <= q2 x q3 <= 242
  
  # n is even ==> n = 2 * m and 5 <= m <= 47
  # t = p1 + p2 = 2 * (q2 * q3) = 10m + 10 = (m + 1) * 10
  
  # t must end on a zero ==> q2 must be 5 and 6 <= q3 <= 48
  # t = 10 * q3 ==> q3 = m + 1
  
  # list of prime numbers from 6 up to 48 (used for q3)
  P = {3, 5, 7}
  P |= {x for x in range(11, 49, 2) if all(x % p for p in P)}
  P = {x for x in P if x > 5}
  
  # try to express next prime as 6k - 1 or 6k + 1
  def nextprime(n):
    # in this puzzle n always ends on a 5
    return list(islice([p for x in range(n // 6 + 1, 81) for i in [-1, 1] 
                if is_prime(p := 6 * x + i) and p > n], 0, 1))[0]
  
  # list of prime numbers from 31 (prime before 5 x 7) up to 235 (5 x 47)
  P2 = {3, 5, 7, 11, 13, 17}
  P2 |= {x for x in range(19, 236, 2) if all(x % p for p in P2)}
  
  # try to express previous prime as 6k - 1 or 6k + 1
  def prevprime(n):
    # in this puzzle n always ends on a 5
    return list(islice([p for x in range(n // 6, 0, -1) for i in [1, -1] 
                if (p := 6 * x + i) in P2 and p < n], 0, 1))[0]
  
  bef35 = prevprime(35) # prime number before 5 x lowest q3 prime
  P2 = {x for x in P2 if x >= bef35}
  
  # add one more prime (to be found by nextprime_P2() )
  np = nextprime(max(P2))
  P2.add(np)
  P2limit = np // 6 + 1  # variable to be used in nextprime_P2()
  
  # try to express next prime as 6k - 1 or 6k + 1
  def nextprime_P2(n):
    # in this puzzle n always ends on a 5
    return list(islice([p for x in range(n // 6 + 1, P2limit) for i in [-1, 1]
                if (p := 6 * x + i) in P2 and p > n], 0, 1))[0]   
  
  # t = q1 x q2 x q3 = 2 x 5 x q3
  for q3 in P:
    t = 10 * q3  
    # find prime before half of t
    p1 = prevprime(5 * q3) 
    
    # t = p1 + p2 (where p1 and p2 are consecutive primes)
    if t - p1 not in P2: continue
    
    if nextprime_P2(5 * q3) != t - p1: continue
    
    print(t, "crenels in total")
  

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  • 

    Jim Randell 4:30 pm on 3 December 2021 Permalink | Reply

    I’m considering adding [[ Primes.before(n) ]] and [[ Primes.after() ]] to the prime sieve class, which would give you the largest prime less than n and the smallest prime greater than n.

    [Note: These routines are now available in enigma.py]

    The puzzle could then be solved with:

    Run: [ @replit ]

    from enigma import (primes, printf)
    
    primes.expand(252)
    
    for q in primes.between(6, 48):
      m = 5 * q
      p1 = primes.before(m)
      p2 = primes.after(m)
      t = 2 * m
      if p1 + p2 == t:
        n = 2 * q - 2
        # output solution
        printf("t={t} = +({p1}, {p2}) = *(2, 5, {q}) = +({n}..{n4})", n4=n + 4)
    

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