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# quick way to determine the array is irdered?

 P: n/a Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) Thanks! With best regards, Roman Mashak. E-mail: mr*@tusur.ru Nov 14 '05 #1
14 Replies

 P: n/a On 2005-06-17 21:53:36 -0400, "Roman Mashak" said: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? No. -- Clark S. Cox, III cl*******@gmail.com Nov 14 '05 #2

 P: n/a On Sat, 18 Jun 2005 10:53:36 +0900, "Roman Mashak" wrote in comp.lang.c: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) Thanks! With best regards, Roman Mashak. E-mail: mr*@tusur.ru There are other ways, but not likely to be quicker. You could use a second array of the same type and size (defined or allocated) and copy the first array into the second. Then you could sort the second with qsort() or a sort function you write yourself. Then you could compare the two arrays element by element and if you find a difference, the first array was not ordered. But as I said, not likely to be quicker. -- Jack Klein Home: http://JK-Technology.Com FAQs for comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html comp.lang.c++ http://www.parashift.com/c++-faq-lite/ alt.comp.lang.learn.c-c++ http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html Nov 14 '05 #3

 P: n/a i guess, one simple way ( rather than going for quick sort ) of doing it at o(n) is,(if ur goal is just to check ascending or descending property) Check the difference between the consecutive elements, for the whole array. The array is ascending or descending, depending on the difference.. (more blah-blah: if an array is in descending order, an element will be always greater than its successor... just check this for whole array.. and the other way for asceding order..) Regards, Devaraj Rangasamy Nov 14 '05 #4

 P: n/a but do remember that you should scan the whole array, any how.. ;> eager to know possible further optimizations.. Nov 14 '05 #5

 P: n/a Roman Mashak wrote on 18/06/05 : Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? if (X[0] == searched) else if X[1] == searched <...> else if X[2] == searched <...> else if X[3] == searched <...> <...> else if x[N-1] == searched <...> If you don't like this code, use ... a loop ... ! In real world, the loop is more or less hidden by some lookup functions such as the standard qsort()/bsearch() couple or some handmade functions... -- Emmanuel The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html The C-library: http://www.dinkumware.com/refxc.html I once asked an expert COBOL programmer, how to declare local variables in COBOL, the reply was: "what is a local variable?" Nov 14 '05 #6

 P: n/a "Roman Mashak" wrote Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) No way of doing what you wnat in less than O(N) time. However if you know the propeties of your array you can do a "good enough" test by taking the start, the end, the middle, and the second and third quartiles. The chance of these being in order by chance is relatively low. (5!, or 1 in 120) In addition if the middle is very approximately the mean of the middle three, and you know the distribution is either unform or with a symmetrical central peak, then it is pretty certain that the array is ordered. What the test won't detect is slight deviations from orderedness, for instnace by swapping one pair of elements. These could be malicious or they could be because ordering is not random. However the chance of them arising from a random distribution is vanishingly small. Nov 14 '05 #7

 P: n/a On Sat, 18 Jun 2005 10:53:36 +0900, Roman Mashak wrote: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? Note that comp.lang.c is for discussing the C programming language itself, a good place to discuss algorithms is comp.programming. If you know nothing about the array then you probably won't do better than this, you clearly need to access evenry element of the array to determine this, any element you don't access might be out of order and your algorithm couldn't detect it. OTOH it might be better to deal with this when you build the array, e.g. build it ordered or test ordering while you build it. In that case the determining step becomes trivial. Lawrence Nov 14 '05 #8

 P: n/a Roman Mashak wrote: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) No. You have to check. What will you do if, after checking, X is not ordered? Will you then order it? If so, don't check at all, simply order the array with a simple insertion sort. If X were already ordered only checking takes place. -- Joe Wright "Everything should be made as simple as possible, but not simpler." --- Albert Einstein --- Nov 14 '05 #9

 P: n/a Le 18/06/2005 17:01, dans x4********************@comcast.com, «*Joe Wright*» a écrit*: Roman Mashak wrote: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) No. You have to check. What will you do if, after checking, X is not ordered? Will you then order it? If so, don't check at all, simply order the array with a simple insertion sort. If X were already ordered only checking takes place. And if it was not, insertion sort is a snail, bad advice I think. Nov 14 '05 #10

 P: n/a Le 18/06/2005 17:04, dans BE******************************@laposte.net, «*Jean-Claude Arbaut*» a écrit*: Le 18/06/2005 17:01, dans x4********************@comcast.com, «*Joe Wright*» a écrit*: Roman Mashak wrote: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) No. You have to check. What will you do if, after checking, X is not ordered? Will you then order it? If so, don't check at all, simply order the array with a simple insertion sort. If X were already ordered only checking takes place. And if it was not, insertion sort is a snail, bad advice I think. BTW, doing a check and sorting with heapsort is still better asymptotically :-) Nov 14 '05 #11

 P: n/a On Sat, 18 Jun 2005 10:53:36 +0900, Roman Mashak wrote: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) Thanks! With best regards, Roman Mashak. E-mail: mr*@tusur.ru Maybe you can throw the array into a struct along with a flag indicating whether the array is sorted. Of course it will then be necessary to keep the flag up to date in all cases. struct data { int *X; int is_sorted; }; --Mac Nov 14 '05 #12

 P: n/a In article , Jean-Claude Arbaut wrote: Le 18/06/2005 17:01, dans x4********************@comcast.com, «*Joe Wright*» a écrit*: Roman Mashak wrote: Hello, All! Is there an easy way to determine that array e.g. int X[N] contains ordered items (for example, ascending), except running loop with comparison of items? It would be good to provide me with some useful link :) No. You have to check. What will you do if, after checking, X is not ordered? Will you then order it? If so, don't check at all, simply order the array with a simple insertion sort. If X were already ordered only checking takes place. And if it was not, insertion sort is a snail, bad advice I think. Should run in O (k*N) if k elements in an initially sorted array of N elements have been changed. Kind of hard to beat for small k. Nov 14 '05 #13

 P: n/a Jean-Claude Arbaut wrote: Le 18/06/2005 17:04, dans BE******************************@laposte.net, « Jean-Claude Arbaut » a écrit :Le 18/06/2005 17:01, dans x4********************@comcast.com, « Joe Wright » a écrit :Roman Mashak wrote:Hello, All!Is there an easy way to determine that array e.g. int X[N] contains ordereditems (for example, ascending), except running loop with comparison ofitems?It would be good to provide me with some useful link :) No. You have to check. What will you do if, after checking, X is notordered? Will you then order it? If so, don't check at all, simply orderthe array with a simple insertion sort. If X were already ordered onlychecking takes place.And if it was not, insertion sort is a snail, bad advice I think. You think? I expect that 'int X[N]' is maintained in order. Calling an insertion sort on X when X is already ordered is fast as lightning. BTW, doing a check and sorting with heapsort is still better asymptotically :-) My point was to sort without checking, knowing the result is an ordered array. The choice of which sort algorithm is up to the user. -- Joe Wright "Everything should be made as simple as possible, but not simpler." --- Albert Einstein --- Nov 14 '05 #14

 P: n/a # >>> Is there an easy way to determine that array e.g. int X[N] contains ordered # >>> items (for example, ascending), except running loop with comparison of # >>> items? Sheesh, the answer is there is no library function, but it's easy to code with one loop. int increasing = 1,i; for (i=1; i and >= for strictly decreasing and decreasing. To check monotonictity int monotonic = 1,cc = 0,i; for (i=1; i0) { for (; i /*>=*/ X[i]; } For N=0 and N=1, X is reported as (strictly) monotonic, increasing and decreasing. If there are no other constraints on X besides being integers, you need N-1 binary comparison. Picture it as a binary tree with N leaves: no matter how you arrange the tree, you will always have N-1 internal nodes. On a parallel machines you can divide the tree in halves and halves again recursively and thus complete it in lg N steps, but you will need N/2 processors in parallel. -- SM Ryan http://www.rawbw.com/~wyrmwif/ The whole world's against us. Nov 14 '05 #15

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