{"id":20247,"date":"2022-11-09T04:02:23","date_gmt":"2022-11-08T22:32:23","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-printing-n-pairs-of-prime-numbers-c\/"},"modified":"2022-11-09T04:02:23","modified_gmt":"2022-11-08T22:32:23","slug":"solved-printing-n-pairs-of-prime-numbers-c","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-printing-n-pairs-of-prime-numbers-c\/","title":{"rendered":"[Solved] Printing n pairs of prime numbers, C++"},"content":{"rendered":"

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Here is the top-level simple pseudo-code for such a beast:<\/p>\n

def printTwinPrimes(count):\n    currNum = 3\n    while count > 0:\n        if isPrime(currNum) and isPrime(currNum + 2):\n            print currnum, currnum + 2\n            count = count - 1\n        currNum = currNum + 2\n<\/code><\/pre>\n
It simply starts at 3<\/code> (since we know 2,4<\/code> is impossible as a twin-prime pair because 4<\/code> is composite). For each possibility, it checks whether it constitutes a twin-prime pair and prints it if so.<\/p>\n
So all you need to do (other than translating that into real<\/em> code) is to create isPrime()<\/code>, for which there are countless examples on the net.<\/p>\n
For completeness, here’s a simple one, by no means the most efficient but adequate for beginners:<\/p>\n
def isPrime(num):\n    if num < 2:\n        return false\n    root = 2\n    while root * root <= num:\n        if num % root == 0:\n            return false\n        root = root + 1\n    return true\n<\/code><\/pre>\n
Though you could<\/em> make that more efficient by using the fact that all primes other than two or three are of the form 6n\u00b11, n >= 1<\/code>(a)<\/sup>:<\/p>\n
def isPrime(num):\n    if num < 2: return false\n    if num == 2 or num == 3: return true\n    if num % 2 == 0 or num % 3 == 0: return false\n    if num % 6 is neither 1 nor 5: return false\n\n    root = 5\n    adder = 2                   # initial adder 2, 5 -> 7\n    while root * root <= num:\n        if num % root == 0:\n            return false\n        root = root + adder     # checks 5, 7, 11, 13, 17, 19, ...\n        adder = 6 - adder       # because alternate 2, 4 to give 6n\u00b11\n    return true\n<\/code><\/pre>\n
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In fact, you can use this divisibility trick to see if an arbitraily large number stored as a string is likely to be a prime. You just have to check if the number below it or above it is divisible by six. If not, the number is definitely not a prime. If so, more (slower) checks will be needed to fully ascertain primality.<\/p>\n
A number is divisible by six only if it is divisible by both two and three. It’s easy to tell the former, even numbers end with an even digit.<\/p>\n
But it’s also reasonably easy to tell if it’s divisible by three<\/em> since, in that case, the sum of the individual digits will also<\/em> be divisible by three. For example, lets’ use 31415926535902718281828459<\/code>.<\/p>\n
The sum of all those digits is 118<\/code>. How do we tell if that’s<\/em> a multiple of three? Why, using exactly the same trick recursively:<\/p>\n
118: 1 + 1 + 8 = 10\n 10: 1 + 0     = 1\n<\/code><\/pre>\n
Once you’re down to a single digit, it’ll be 0<\/code>, 3<\/code>, 6<\/code>, or 9<\/code> if the original number was a multiple of three. Any other digit means it wasn’t (such as in this case).<\/p>\n
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(a)<\/sup> If you divide any non-negative number by six and the remainder is 0, 2 or 4, then it’s even and therefore non-prime (2 is the exception case here):<\/p>\n
6n + 0 = 2(3n + 0), an even number.\n6n + 2 = 2(3n + 1), an even number.\n6n + 4 = 2(3n + 2), an even number.\n<\/code><\/pre>\n
If the remainder is 3, then it is divisible by 3 and therefore non-prime (3 is the exception case here):<\/p>\n
6n + 3 = 3(2n + 1), a multiple of three.\n<\/code><\/pre>\n
That leaves just the remainders 1 and 5, and those numbers are all of that form 6n\u00b11<\/code>.<\/p>\n<\/p><\/div>\n
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solved Printing n pairs of prime numbers, C++ <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Here is the top-level simple pseudo-code for such a beast: def printTwinPrimes(count): currNum = 3 while count > 0: if isPrime(currNum) and isPrime(currNum + 2): print currnum, currnum + 2 count = count – 1 currNum = currNum + 2 It simply starts at 3 (since we know 2,4 is impossible as a twin-prime … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[324],"class_list":["post-20247","post","type-post","status-publish","format-standard","hentry","category-solved","tag-c"],"yoast_head":"\n[Solved] Printing n pairs of prime numbers, C++ - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-printing-n-pairs-of-prime-numbers-c\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Printing n pairs of prime numbers, C++ - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Here is the top-level simple pseudo-code for such a beast: def printTwinPrimes(count): currNum = 3 while count > 0: if isPrime(currNum) and isPrime(currNum + 2): print currnum, currnum + 2 count = count - 1 currNum = currNum + 2 It simply starts at 3 (since we know 2,4 is impossible as a twin-prime ... 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