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Integer overflow can be demonstrated through an odometer overflowing, a mechanical version of the phenomenon. All digits are set to the maximum 9 and the next increment of the white digit causes a cascade of carry-over additions setting all digits to 0, but there is no higher digit (1,000,000s digit) to change to a 1, so the counter resets to zero.
Although it is possible to get around this problem using conversion code and larger data types, it makes using Java cumbersome for handling unsigned data. While a 32-bit signed integer may be used to hold a 16-bit unsigned value losslessly, and a 64-bit signed integer a 32-bit unsigned integer, there is no larger type to hold a 64-bit unsigned ...
The minimum size for char is 8 bits, the minimum size for short and int is 16 bits, for long it is 32 bits and long long must contain at least 64 bits. The type int should be the integer type that the target processor is most efficiently working with.
Java 5 Update 5 (1.5.0_05) is the last release of Java to work on Windows 95 (with Internet Explorer 5.5 installed) and Windows NT 4.0. [36] Java 5 was first available on Apple Mac OS X 10.4 (Tiger) [37] and was the default version of Java installed on Apple Mac OS X 10.5 (Leopard). Public support and security updates for Java 1.5 ended in ...
Can all-pairs shortest paths be computed in strongly sub-cubic time, that is, in time O(V 3−ϵ) for some ϵ>0? Can the Schwartz–Zippel lemma for polynomial identity testing be derandomized? Does linear programming admit a strongly polynomial-time algorithm? (This is problem #9 in Smale's list of problems.)
This limitation is overcome in modern algorithms, which can solve to optimality (in the sense of finding solutions with minimum waste) very large instances of the problem (generally larger than encountered in practice [8] [9]). The cutting-stock problem is often highly degenerate, in that multiple solutions with the same amount of waste are ...
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LCS(R 1, C 1) is determined by comparing the first elements in each sequence. G and A are not the same, so this LCS gets (using the "second property") the longest of the two sequences, LCS(R 1, C 0) and LCS(R 0, C 1). According to the table, both of these are empty, so LCS(R 1, C 1) is also empty, as shown in the