![]() ![]() They also don't have good trackpads, and their keyboard + trackpad alignment is wonky (it's off-center in a lot of cases! How weird is that?) Most non-Apple laptops don't have very good color accuracy. It's painful to do so, given how invested I am in the ecosystem, but we're already beyond the threshold that many of us would have left earlier in the decade.Įdit - It's also really hard as a designer + developer + would-be researcher in the making to find a good computer. As they've chosen to.Īpple can now make their machine reject a new, third-party repair part like a bad transplant. Should they choose to.Īpple is now restricting what other OSes you can boot into. ![]() Of course Apple's Craig Federighi assures us that the people making such assertions are "tools" (, timestamp 53:33) and they have no intention whatsoever of taking away our ability to do general compute on the machines we buy and own.Īpple can already decide what binaries you can execute. Apple already has the ability to restrict whats apps I can run (they can simply toggle a switch for all users to "no unsigned binaries"), and congrats! Apple is the sole decider of what we get to use on our computers. Īnd this is why I won't be moving to Apple silicon. It's fair to say that I was mostly dismissed. Set j = last-2 and find first j such that a =.So yesterday I wrote about the blurring lines of ownership, and people came back with some fairly disparate responses. So 6 is next larger and 2345(least using numbers other than 6) Find lexicogrpahically least way to extend the new aĬonsider example array state of for sorted Īfter 56432(treat as number) ->nothing larger than 6432(using 6,4,3,2) beginning with 5.Increase a by smallest feasible amount.Find largest j such that a can be increased.There are n! permutations at most and hasNextPermutation(.) runs in O(n) time complexity This is the asymptotically optimal way O(n*n!) of generating permutations after initial sorting. It was enough when I needed it, but it's no itertools.permutations by a long shot. This method is non-recursive, but it is slightly slower on my computer and xrange raises an error when n! is too large to be converted to a C long integer (n=13 for me). NewPermutation.append(available.pop(index)) This way the numbers 0 through n!-1 correspond to all possible permutations in lexicographic order. You have n choices for the first item, n-1 for the second, and only one for the last, so you can use the digits of a number in the factorial number system as the indices. I used an algorithm based on the factorial number system- For a list of length n, you can assemble each permutation item by item, selecting from the items left at each stage. Indices, indices = indices, indicesĪnd another, based on itertools.product: def permutations(iterable, r=None):įor indices in product(range(n), repeat=r): If len(elements) AB AC AD BA BC BD CA CB CD DA DB DC Use itertools.permutations from the standard library: import itertoolsĪdapted from here is a demonstration of how itertools.permutations might be implemented: def permutations(elements):
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