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Java Forum / General / August 2006generating a positive rand number in range (0, 2^63]
Please correct me if I am wrong but as per my understanding java.util.Random.nextLong() can return a negative number as well. I need to get a positive long only. What would be the best way? (Math.abs wont work since it cannot convert LONG.MIN_VALUE to a positive long). Suggestions? Thanks! Varun Kacholia wrote: > Please correct me if I am wrong but as per my understanding > java.util.Random.nextLong() can > return a negative number as well. I need to get a positive long only. > What would be the best way? > (Math.abs wont work since it cannot convert LONG.MIN_VALUE to a > positive long). r.nextLong() >>> 1 comes to mind. Or you could use ((long)r.next(31) << 32) + r.next(32).  SignatureEric Sosman esosman@acm-dot-org.invalid > r.nextLong() >>> 1 comes to mind. Or you could use that could still result in 0. (r.nextLong() >>> 1 + 1 wont help either) > ((long)r.next(31) << 32) + r.next(32). 0? Varun Kacholia wrote: >> r.nextLong() >>> 1 comes to mind. Or you could use > [quoted text clipped - 3 lines] > > 0? I apologize; I didn't notice your subject line. Note that a `long' cannot cover the range you've asked for; 2^63 is larger than Long.MAX_VALUE. A `double' can cover the range, but not with 63-bit accuracy. If you really need (0,2^63] I think you'll need to use BigInteger values. The easiest way is probably to use r.next() twice to generate a 63-bit value in the range [0,2^63), generate a BigInteger from that (use the BigInteger(byte[]) constructor), and add BigInteger.ONE to the result. SignatureEric Sosman esosman@acm-dot-org.invalid > I apologize; I didn't notice your subject line. my bad as well. i intend it to be between (0, 2^63) (excluding 0 and 2^63). suggestions? > Note that a `long' cannot cover the range you've asked for; > 2^63 is larger than Long.MAX_VALUE. A `double' can cover the [quoted text clipped - 4 lines] > BigInteger(byte[]) constructor), and add BigInteger.ONE to the > result. Varun Kacholia wrote: >> I apologize; I didn't notice your subject line. > [quoted text clipped - 10 lines] >> BigInteger(byte[]) constructor), and add BigInteger.ONE to the >> result. How about: long result; do{ result = r.nextLong() >>> 1; }while(result == 0) [Based on one of Eric's earlier suggestions] Patricia > Varun Kacholia wrote: >>> I apologize; I didn't notice your subject line. my >> bad as well. i intend it to be between (0, 2^63) [quoted text clipped - 18 lines] > > [Based on one of Eric's earlier suggestions] How about just: long result; do { result = r.nextLong(); } while (result <= 0); >>Varun Kacholia wrote: > [quoted text clipped - 29 lines] > } > while (result <= 0); It's a "micro-pessimization." Roughly speaking, Patricia's code will average one nextLong() call per delivered result, while Jussi's will average two. (Patricia accepts the nextLong() value with probability (2^63-1)/2^63 = 1-2^(-63), while Jussi accepts each value with probability (2^63-1)/2^64 = 0.5-2^(-64).) SignatureEric Sosman esosman@acm-dot-org.invalid ... >>> How about: >>> [quoted text clipped - 17 lines] > with probability (2^63-1)/2^63 = 1-2^(-63), while Jussi accepts > each value with probability (2^63-1)/2^64 = 0.5-2^(-64).) Thanks. I didn't realize her code was _that_ efficient. I should explain why I asked about the alternative at all. The reason is that my version is rather obviously still uniform (all the accepted numbers are equally probable in the underlying generator) while with your/Patricia's version something more subtle is going on, and I get nervous when there is something subtle going on. Let me see. Assume we have a uniform distribution for the 2^64 bit patterns before the shift. When we lop off the rightmost bit, we are left with 2^63 bit patterns, each of which occurred once with 0 and once with 1 to its right, so they are all equally probable. And we reject both occurrences of 63 zeroes. I think I see it now. Thanks again. The range you want could be generated with this code below: long x = 0; long y = Math.round(Math.pow(2, 63)); long randomNumber = (long) (Math.random() * (y - x - 1)) + 1; System.out.println(x + " " + y); System.out.println(randomNumber); The range is (x, y] ;-) flávio > > I apologize; I didn't notice your subject line. > [quoted text clipped - 10 lines] > > BigInteger(byte[]) constructor), and add BigInteger.ONE to the > > result. > The range you want could be generated with this code below: > > long x = 0; > long y = Math.round(Math.pow(2, 63)); This is a bit indirect, but equivalent to: long y = Long.MAX_VALUE; "If the argument is positive infinity or any value greater than or equal to the value of Long.MAX_VALUE, the result is equal to the value of Long.MAX_VALUE" (Math.round API documentation). > long randomNumber = (long) (Math.random() * (y - x - > 1)) + 1; There are 2^53 possible results of Math.random(), the same as the number of possible different results from Random's nextDouble() method. The (0,2^63) range has (2^63)-1 possible values. About one in every 1024 positive longs has probability 2^-53. The others all have probability zero. I believe the OP wanted a uniform distribution over all the longs in the range. Patricia Varun Kacholia wrote: > Please correct me if I am wrong but as per my understanding > java.util.Random.nextLong() can [quoted text clipped - 5 lines] > > Thanks! If you intend to exclude zero, and include 2^63, as indicated by the range in the subject, it cannot be done because the largest long, Long.MAX_VALUE, is 2^63-1. Patricia Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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