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Using the XOR swap algorithm to exchange nibbles between variables without the use of temporary storage. In computer programming, the exclusive or swap (sometimes shortened to XOR swap) is an algorithm that uses the exclusive or bitwise operation to swap the values of two variables without using the temporary variable which is normally required.
However, using two temporary registers, two processors executing in parallel can swap two variables in two clock cycles: Step 1 Processor 1: temp_1 := X Processor 2: temp_2 := Y Step 2 Processor 1: X := temp_2 Processor 2: Y := temp_1 More temporary registers are used, and four instructions are needed instead of three.
Take an array of numbers "5 1 4 2 8", and sort the array from lowest number to greatest number using bubble sort. In each step, elements written in bold are being compared. Three passes will be required; First Pass ( 5 1 4 2 8 ) → ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps since 5 > 1.
Used in Python 2.3 and up, and Java SE 7. ... fast multiplication of two numbers ... Xor swap algorithm: swaps the values of two variables without using a buffer;
A map of the 24 permutations and the 23 swaps used in Heap's algorithm permuting the four letters A (amber), B (blue), C (cyan) and D (dark red) Wheel diagram of all permutations of length = generated by Heap's algorithm, where each permutation is color-coded (1=blue, 2=green, 3=yellow, 4=red).
The algorithm described by Durstenfeld is more efficient than that given by Fisher and Yates: whereas a naïve computer implementation of Fisher and Yates' method would spend needless time counting the remaining numbers in step 3 above, Durstenfeld's solution is to move the "struck" numbers to the end of the list by swapping them with the last ...
The simplest form goes through the whole list each time: procedure cocktailShakerSort(A : list of sortable items) is do swapped := false for each i in 0 to length(A) − 1 do: if A[i] > A[i + 1] then // test whether the two elements are in the wrong order swap(A[i], A[i + 1]) // let the two elements change places swapped := true end if end for if not swapped then // we can exit the outer loop ...
push 1L (the number one with type long) onto the stack ldc 12 0001 0010 1: index → value push a constant #index from a constant pool (String, int, float, Class, java.lang.invoke.MethodType, java.lang.invoke.MethodHandle, or a dynamically-computed constant) onto the stack ldc_w 13 0001 0011 2: indexbyte1, indexbyte2 → value