## digitalmars.D - next_permutation and cartesian product for ranges?

- H. S. Teoh (67/67) Oct 09 2012 Recently I've been working on some computational geometry code, and
- Andrei Alexandrescu (11/34) Oct 09 2012 Yes, we need next_permutation. It will be up to you to convince the
- Jonathan M Davis (3/5) Oct 09 2012 Sounds violent. :)
- H. S. Teoh (82/108) Oct 09 2012 Is there much chance of this, if the algorithm only works correctly when
- Timon Gehr (15/52) Oct 09 2012 That's cute. =)
- H. S. Teoh (79/95) Oct 09 2012 Ahahahaha... I knew joiner existed; why didn't I think of it. :P
- Timon Gehr (3/25) Oct 09 2012 I usually use standard function call notation for zip. (but the mail
- Andrei Alexandrescu (7/49) Oct 09 2012 We draw the line at memory safety. If you must assume distinct elements
- H. S. Teoh (22/57) Oct 09 2012 The algo will never transgress into unsafe. It's just that it will
- Andrei Alexandrescu (6/10) Oct 10 2012 I meant the resulting range spans the cross product (all combinations)
- H. S. Teoh (33/43) Oct 10 2012 Hmm. We must be using a different set of terms, because in my
- Andrei Alexandrescu (10/50) Oct 10 2012 Same thing for sets. I agree using Cartesian is more precise.
- Andrei Alexandrescu (3/5) Oct 10 2012 I meant "Multiple save points are NOT necessary."
- bearophile (244/247) Oct 09 2012 I don't like the design of C++ STL here. Instead of a
- Timon Gehr (4/12) Oct 09 2012 I agree that that's better conceptually, but your implementations have
- ixid (1/2) Oct 09 2012 The naming scheme could do with... improvement.
- bearophile (4/5) Oct 09 2012 struct PermutationSwaps then?
- ixid (4/9) Oct 09 2012 I think you're missing what I meant, naming a struct
- Steven Schveighoffer (5/12) Oct 09 2012 Is there any advantage over having a function? I'd think you could easi...
- bearophile (9/11) Oct 09 2012 Generators (that yield lexicographic permutations, permutation
- Don Clugston (4/13) Oct 10 2012 What makes you think that permutations are only ever used in a range
- H. S. Teoh (17/36) Oct 10 2012 [...]

Recently I've been working on some computational geometry code, and noticed that Phobos doesn't have any equivalent of STL's next_permutation. I saw there were some discussions about this back in 2010 -- did anything come of it? If there is any interest in next_permutation, I'd also like to propose a next_even_permutation for ranges without duplicate elements. There is a simple way of extending Narayana Pandita's in-place algorithm for enumerating all permutations such that only even permutations are returned. // Also, I'm interested in Cartesian products of n ranges. Cartesian products tend to be too long to explicitly construct, so it lends itself perfectly to a range interface where the elements of the product are computed on the fly. Now I saw in the dlang.org search results that Cartesian products have been discussed before, but got roadblocked because the naïve implementation of Cartesian product (simply enumerate the first range, then popFront the second range when the first runs out and gets restored to the initial .save point, etc.) doesn't lend itself well to infinite ranges. Here's my proposed solution: we can use a trick that mathematicians use in set theory to prove that the cardinality of the rationals is equal to the cardinality of the natural numbers. Make a table of infinite width and height, with rows representing one copy of the natural numbers and columns another copy, then enumerate all elements by enumerating diagonals. This produces a one-to-one correspondence between the natural numbers and pairs of natural numbers, of which the rationals are a subset (thus one has |N| ≤ |Q| ≤ |N|, proving that |N| = |Q|). For ranges, we don't care about eliminating duplicates like 2/4 = 1/2, since we want all possible pairs. So the infinite table looks like: (0,0) (1,0) (2,0) (3,0) (4,0) ... (0,1) (1,1) (2,1) (3,1) ... (0,2) (1,2) (2,2) ... (0,3) (1,3) ... (0,4) ... ... The diagonals are therefore: (0,0) (0,1) (1,0) (0,2) (1,1) (2,0) (0,3) (1,2) (2,1) (3,0) (0,4) (1,3) (2,2) (3,1) (4,0) ... The characteristic of the n'th diagonal is that the sum of elements in each diagonal is equal to n. For example, the zeroth diagonal has only (0,0) since that's the only pair that sums to 0. The first diagonal is (0,1), (1,0): each pair sums to 1. The second diagonal: each pair sums to 2, and so on. The general form for the i'th element of the n'th diagonal is therefore (i, n-i), for 0 ≤ i ≤ (n-1). So to enumerate the Cartesian product of two infinite ranges, we enumerate its diagonals by interpreting the above pairs as indices into the respective ranges. So, the first range can be a forward range and the second bidirectional (since for each diagonal we need to enumerate the second range from the n'th element down to the first). Of course, if the first range is bidirectional and the second is forward, we can just swap them and reverse the pairs generated. If one or both of the ranges is finite, then we can fall back to the naïve method of enumerating the Cartesian product, in which case we can relax the requirement on the ranges to be just one forward range and one input range (with the infinite range being the input range, if one of them is infinite). To support the Cartesian product of n ranges, we just compose multiple binary Cartesian products. Comments? :) T -- Many open minds should be closed for repairs. -- K5 user

Oct 09 2012

On 10/9/12 1:52 PM, H. S. Teoh wrote:Recently I've been working on some computational geometry code, and noticed that Phobos doesn't have any equivalent of STL's next_permutation. I saw there were some discussions about this back in 2010 -- did anything come of it? If there is any interest in next_permutation, I'd also like to propose a next_even_permutation for ranges without duplicate elements. There is a simple way of extending Narayana Pandita's in-place algorithm for enumerating all permutations such that only even permutations are returned.Yes, we need next_permutation. It will be up to you to convince the Grand Inquisition Committee of Reviewers that next_even_permutation is necessary and/or sufficient.// Also, I'm interested in Cartesian products of n ranges. Cartesian products tend to be too long to explicitly construct, so it lends itself perfectly to a range interface where the elements of the product are computed on the fly. Now I saw in the dlang.org search results that Cartesian products have been discussed before, but got roadblocked because the naïve implementation of Cartesian product (simply enumerate the first range, then popFront the second range when the first runs out and gets restored to the initial .save point, etc.) doesn't lend itself well to infinite ranges. Here's my proposed solution: we can use a trick that mathematicians use in set theory to prove that the cardinality of the rationals is equal to the cardinality of the natural numbers.[snip] Yah, I call that the Cantor's diagonalization trick - I reckon that's how he proved there are more reals than rationals: http://goo.gl/MTmnQ. We need that badly. Please implement pronto and let us destroy you for having done so. Thanks, Andrei

Oct 09 2012

On Tuesday, October 09, 2012 14:25:13 Andrei Alexandrescu wrote:We need that badly. Please implement pronto and let us destroy you for having done so.Sounds violent. :) - Jonathan M Davis

Oct 09 2012

On Tue, Oct 09, 2012 at 02:25:13PM -0400, Andrei Alexandrescu wrote:On 10/9/12 1:52 PM, H. S. Teoh wrote:[...]Is there much chance of this, if the algorithm only works correctly when the input array has no duplicate elements? AFAIK there's no easy way to ensure unique elements without introducing runtime checks. [...]If there is any interest in next_permutation, I'd also like to propose a next_even_permutation for ranges without duplicate elements. There is a simple way of extending Narayana Pandita's in-place algorithm for enumerating all permutations such that only even permutations are returned.Yes, we need next_permutation. It will be up to you to convince the Grand Inquisition Committee of Reviewers that next_even_permutation is necessary and/or sufficient.[...] Here it is: ------------------------------------------------------------------------------- import std.algorithm; import std.range; // I wrote this 'cos I needed it. Why isn't it in std.range? // A better name for this is probably "denest" or "denestOne", since it doesn't // really flatten arbitrarily-nested ranges. auto flatten(R)(R ror) if (isInputRange!R && isInputRange!(ElementType!R)) { struct Flatten(R) { R ror; ElementType!R r; property bool empty() { if (ror.empty) return true; // This shenanigan is needed because operating on // ror.front directly sometimes doesn't work, e.g. if // ror.front.popFront() works on a copy of ror.front // and so ror is never updated. while (r.empty) { ror.popFront(); if (ror.empty) return true; r = ror.front; } return false; } property auto front() { return r.front; } void popFront() { r.popFront(); while (r.empty) { ror.popFront(); if (ror.empty) return; r = ror.front; } } } auto r = Flatten!R(ror); if (!ror.empty) r.r = ror.front; // get things started return r; } // This is what I really wanted to show. auto cartesianProd(R1,R2)(R1 range1, R2 range2) if (isForwardRange!R1 && isInfinite!R1 && isBidirectionalRange!(typeof(take(range2,1))) && isInfinite!R2) { // Behold the awesomeness of D functional programming! ;-) return flatten( map!((int n) => zip( take(range1.save, n), retro(take(range2.save, n)) ))(sequence!"n"(1)) ); } void main() { import std.conv; import std.stdio; // A simple test case: Cartesian product of the set of natural numbers // with itself. auto N = sequence!"n"(0); auto N2 = cartesianProd(N, N); writefln("N*N = %(%s, %)", map!"[a[0], a[1]]"(N2.take(20))); // Let's prove that it works for more than just the natural numbers auto odds = sequence!"2*n+1"(0); auto evens = sequence!"2*n"(0); writefln("odds * evens = %(%s, %)", map!"[a[0], a[1]]"( cartesianProd(odds, evens).take(20))); } ------------------------------------------------------------------------------- Destroy! T -- Ignorance is bliss... but only until you suffer the consequences!Now I saw in the dlang.org search results that Cartesian products have been discussed before, but got roadblocked because the nave implementation of Cartesian product (simply enumerate the first range, then popFront the second range when the first runs out and gets restored to the initial .save point, etc.) doesn't lend itself well to infinite ranges. Here's my proposed solution: we can use a trick that mathematicians use in set theory to prove that the cardinality of the rationals is equal to the cardinality of the natural numbers.[snip] Yah, I call that the Cantor's diagonalization trick - I reckon that's how he proved there are more reals than rationals: http://goo.gl/MTmnQ. We need that badly. Please implement pronto and let us destroy you for having done so.

Oct 09 2012

On 10/09/2012 10:46 PM, H. S. Teoh wrote:On Tue, Oct 09, 2012 at 02:25:13PM -0400, Andrei Alexandrescu wrote:That's cute. =) flatten is in std.algorithm and is called joiner. Also, you need to be careful with index types. I'd suggest: import std.algorithm, std.range; auto cartesianProd(R1,R2)(R1 range1, R2 range2) if (isForwardRange!R1 && isInfinite!R1 && isBidirectionalRange!(typeof(take(range2,1))) && isInfinite!R2) { return sequence!"n"(cast(size_t)1).map!( (size_t n)=>range1.save.take(n) // need to specify param type. DMD bug. .zip(range2.save.take(n).retro), ).joiner; }On 10/9/12 1:52 PM, H. S. Teoh wrote:[...]Is there much chance of this, if the algorithm only works correctly when the input array has no duplicate elements? AFAIK there's no easy way to ensure unique elements without introducing runtime checks. [...]If there is any interest in next_permutation, I'd also like to propose a next_even_permutation for ranges without duplicate elements. There is a simple way of extending Narayana Pandita's in-place algorithm for enumerating all permutations such that only even permutations are returned.Yes, we need next_permutation. It will be up to you to convince the Grand Inquisition Committee of Reviewers that next_even_permutation is necessary and/or sufficient.[...] Here it is: [snip code snippet] Destroy! TNow I saw in the dlang.org search results that Cartesian products have been discussed before, but got roadblocked because the nave implementation of Cartesian product (simply enumerate the first range, then popFront the second range when the first runs out and gets restored to the initial .save point, etc.) doesn't lend itself well to infinite ranges. Here's my proposed solution: we can use a trick that mathematicians use in set theory to prove that the cardinality of the rationals is equal to the cardinality of the natural numbers.[snip] Yah, I call that the Cantor's diagonalization trick - I reckon that's how he proved there are more reals than rationals: http://goo.gl/MTmnQ. We need that badly. Please implement pronto and let us destroy you for having done so.

Oct 09 2012

On Tue, Oct 09, 2012 at 11:24:27PM +0200, Timon Gehr wrote: [...]That's cute. =) flatten is in std.algorithm and is called joiner.Ahahahaha... I knew joiner existed; why didn't I think of it. :PAlso, you need to be careful with index types. I'd suggest: import std.algorithm, std.range; auto cartesianProd(R1,R2)(R1 range1, R2 range2) if (isForwardRange!R1 && isInfinite!R1 && isBidirectionalRange!(typeof(take(range2,1))) && isInfinite!R2) { return sequence!"n"(cast(size_t)1).map!( (size_t n)=>range1.save.take(n) // need to specify param type. DMD bug. .zip(range2.save.take(n).retro), ).joiner; }[...] Ack! UFCS overload! ;-) (just kidding... it's actually a lot easier to read with UFCS.) And I think that trailing comma should be a ')'. Alright, now to take care of the case where one of the ranges is not infinite: ------------------------------------------------------------------------------- import std.algorithm; import std.range; auto cartesianProd(R1,R2)(R1 range1, R2 range2) { // I decided to hoist the conditions in here 'cos the cases overlap. // Probably still need to factor them and put them in the sig // constraint. But hey, this works for now. static if (isForwardRange!R1 && isInfinite!R1 && isBidirectionalRange!(typeof(take(range2,1))) && isInfinite!R2) { // Courtesy of Timon Gehr return sequence!"n"(cast(size_t)1) .map!((size_t n) => range1.save.take(n) .zip(range2.save.take(n).retro)) .joiner; } else static if (isInputRange!R1 && !isInfinite!R1 && isForwardRange!R2) { // A one-liner! Gotta love D's functional programming features! return range2.map!((ElementType!R2 a) => zip(range1.save, repeat(a))) .joiner; } else static if (isInputRange!R2 && !isInfinite!R2 && isForwardRange!R1) { // I could've aliased this to the previous case with tuples // swapped, but why bother when you can do it the direct way? return range1.map!((ElementType!R1 a) => zip(repeat(a), range2.save)) .joiner; } else static assert(0); } unittest { // A simple test case: Cartesian product of the set of natural numbers with // itself. auto N = sequence!"n"(0); auto N2 = cartesianProd(N, N); assert(map!"[a[0],a[1]]"(N2.take(10)).equal([[0, 0], [0, 1], [1, 0], [0, 2], [1, 1], [2, 0], [0, 3], [1, 2], [2, 1], [3, 0]])); // Let's prove that it works for more than just the natural numbers auto odds = sequence!"2*n+1"(0); auto evens = sequence!"2*n"(0); assert(map!"[a[0],a[1]]"(cartesianProd(odds, evens).take(10)) .equal([[1, 0], [1, 2], [3, 0], [1, 4], [3, 2], [5, 0], [1, 6], [3, 4], [5, 2], [7, 0]])); // Test finite cases auto arr = [1,2,3]; assert(cartesianProd(arr,N) .take(10) .map!"[a[0],a[1]]" .equal([[1, 0], [2, 0], [3, 0], [1, 1], [2, 1], [3, 1], [1, 2], [2, 2], [3, 2], [1, 3]])); assert(cartesianProd(N,arr) .take(10) .map!"[a[0],a[1]]" .equal([[0, 1], [0, 2], [0, 3], [1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3], [3, 1]])); auto barr = [4,5,6]; assert(cartesianProd(arr,barr) .map!"[a[0],a[1]]" .equal([[1, 4], [2, 4], [3, 4], [1, 5], [2, 5], [3, 5], [1, 6], [2, 6], [3, 6]])); } // To keep Andrei happy: // vim:set ts=4 sw=4 expandtab: ------------------------------------------------------------------------------- Destroy!! T -- Nothing in the world is more distasteful to a man than to take the path that leads to himself. -- Herman Hesse

Oct 09 2012

On 10/10/2012 01:13 AM, H. S. Teoh wrote:On Tue, Oct 09, 2012 at 11:24:27PM +0200, Timon Gehr wrote: [...]I usually use standard function call notation for zip. (but the mail client wraps around the code, so I inserted a manual line break.)That's cute. =) flatten is in std.algorithm and is called joiner.Ahahahaha... I knew joiner existed; why didn't I think of it. :PAlso, you need to be careful with index types. I'd suggest: import std.algorithm, std.range; auto cartesianProd(R1,R2)(R1 range1, R2 range2) if (isForwardRange!R1 && isInfinite!R1 && isBidirectionalRange!(typeof(take(range2,1))) && isInfinite!R2) { return sequence!"n"(cast(size_t)1).map!( (size_t n)=>range1.save.take(n) // need to specify param type. DMD bug. .zip(range2.save.take(n).retro), ).joiner; }[...] Ack! UFCS overload! ;-) (just kidding... it's actually a lot easier to read with UFCS.)

Oct 09 2012

On 10/9/12 4:46 PM, H. S. Teoh wrote:On Tue, Oct 09, 2012 at 02:25:13PM -0400, Andrei Alexandrescu wrote:We draw the line at memory safety. If you must assume distinct elements and state that assumption, you're fine as long as you don't transgress into unsafe. One alternative would be to run a O(N) check with probability 1/N, trick that assumeSorted() does.On 10/9/12 1:52 PM, H. S. Teoh wrote:[...]Is there much chance of this, if the algorithm only works correctly when the input array has no duplicate elements? AFAIK there's no easy way to ensure unique elements without introducing runtime checks.If there is any interest in next_permutation, I'd also like to propose a next_even_permutation for ranges without duplicate elements. There is a simple way of extending Narayana Pandita's in-place algorithm for enumerating all permutations such that only even permutations are returned.Yes, we need next_permutation. It will be up to you to convince the Grand Inquisition Committee of Reviewers that next_even_permutation is necessary and/or sufficient.[...]crossProduct.[...] Here it is: ------------------------------------------------------------------------------- import std.algorithm; import std.range; // I wrote this 'cos I needed it. Why isn't it in std.range? // A better name for this is probably "denest" or "denestOne", since it doesn't // really flatten arbitrarily-nested ranges. auto flatten(R)(R ror)Now I saw in the dlang.org search results that Cartesian products have been discussed before, but got roadblocked because the nave implementation of Cartesian product (simply enumerate the first range, then popFront the second range when the first runs out and gets restored to the initial .save point, etc.) doesn't lend itself well to infinite ranges. Here's my proposed solution: we can use a trick that mathematicians use in set theory to prove that the cardinality of the rationals is equal to the cardinality of the natural numbers.[snip] Yah, I call that the Cantor's diagonalization trick - I reckon that's how he proved there are more reals than rationals: http://goo.gl/MTmnQ. We need that badly. Please implement pronto and let us destroy you for having done so.if (isInputRange!R&& isInputRange!(ElementType!R))Tab detected. System overload

Oct 09 2012

On Tue, Oct 09, 2012 at 05:46:41PM -0400, Andrei Alexandrescu wrote:On 10/9/12 4:46 PM, H. S. Teoh wrote:The algo will never transgress into unsafe. It's just that it will produce incomplete results if the array has duplicate elements (it will fail to produce some permutations). [...]On Tue, Oct 09, 2012 at 02:25:13PM -0400, Andrei Alexandrescu wrote:We draw the line at memory safety. If you must assume distinct elements and state that assumption, you're fine as long as you don't transgress into unsafe. One alternative would be to run a O(N) check with probability 1/N, trick that assumeSorted() does.On 10/9/12 1:52 PM, H. S. Teoh wrote:[...]Is there much chance of this, if the algorithm only works correctly when the input array has no duplicate elements? AFAIK there's no easy way to ensure unique elements without introducing runtime checks.Actually, this one is to prove that |Q|=|N|. The more famous Cantor diagonalization is the one where he proves that the reals are uncountable. But actually, that proof wasn't the original proof; it was something that came later. The original proof used a different method. In any case, I think we don't have to worry about Cartesian products of uncountable ranges. :-PYah, I call that the Cantor's diagonalization trick - I reckon that's how he proved there are more reals than rationals:Please see other post. [...]http://goo.gl/MTmnQ. We need that badly. Please implement pronto and let us destroy you for having done so.I was an idiot. I knew about std.algorithm.joiner but for some reason didn't think of it at the time. In any case, crossProduct doesn't make any sense to me... I don't see what it's got to do with flattening nested ranges.// I wrote this 'cos I needed it. Why isn't it in std.range? // A better name for this is probably "denest" or "denestOne", since it doesn't // really flatten arbitrarily-nested ranges. auto flatten(R)(R ror)crossProduct.// vim:set ts=4 sw=4 expandtab: :-) T -- Always remember that you are unique. Just like everybody else. -- despair.comif (isInputRange!R&& isInputRange!(ElementType!R))Tab detected. System overload

Oct 09 2012

On 10/9/12 7:25 PM, H. S. Teoh wrote:I was an idiot. I knew about std.algorithm.joiner but for some reason didn't think of it at the time. In any case, crossProduct doesn't make any sense to me... I don't see what it's got to do with flattening nested ranges.I meant the resulting range spans the cross product (all combinations) of the passed-in ranges. On another subject, I think this can be done with only input ranges - no need for bidirectional. Andrei

Oct 10 2012

On Wed, Oct 10, 2012 at 09:41:34AM -0400, Andrei Alexandrescu wrote:On 10/9/12 7:25 PM, H. S. Teoh wrote:Hmm. We must be using a different set of terms, because in my understanding, a cross product is a binary operation on 3D vectors that produces a perpendicular vector. The product that produces all combinations of a set is known as a Cartesian product to me.I was an idiot. I knew about std.algorithm.joiner but for some reason didn't think of it at the time. In any case, crossProduct doesn't make any sense to me... I don't see what it's got to do with flattening nested ranges.I meant the resulting range spans the cross product (all combinations) of the passed-in ranges.On another subject, I think this can be done with only input ranges - no need for bidirectional.[...] In the case of finite ranges, one of the ranges can be an input range, but the other must at least be a forward range, since otherwise you could only traverse it once, so it would be impossible to combine each element with every element from the other range. When one of the ranges is infinite, it's obviously not enough for the other range to be an input range, since you'd have to produce an infinite number of combinations per element (of the second range) before moving on to the next element, so you'd never get to it. It should be possible to do it with both ranges being forward ranges, but it may not be very efficient -- think about it, for every element in one range, you need to combine it with every element in the other range, so when you're at, say, the 100th element of the first range, you still need to go back to the 1st, 2nd, 3rd, etc., elements in the second range. But since you can only combine it with a finite number of elements from the second range (otherwise you'll never get to the 101th element), when you get to the 101th element, you still haven't produced some of the combinations of the 100th element yet. So you'll end up needing an increasing number of save points in order to get to every possible combination -- possible but not efficient. Using the Cantor method we can allow one range to be a forward range, and the second a range whose take(n) is a reversible range (I was wrong before: the second range doesn't need to be bidirectional, it works as long as take(n) returns a reversible range). This gives us constant space complexity. I'd say this is a pretty good solution. T -- Век живи - век учись. А дураком помрёшь.

Oct 10 2012

On 10/10/12 10:27 AM, H. S. Teoh wrote:On Wed, Oct 10, 2012 at 09:41:34AM -0400, Andrei Alexandrescu wrote:Same thing for sets. I agree using Cartesian is more precise. http://www.transtutors.com/math-homework-help/set-theory/cartesian-product.aspxOn 10/9/12 7:25 PM, H. S. Teoh wrote:Hmm. We must be using a different set of terms, because in my understanding, a cross product is a binary operation on 3D vectors that produces a perpendicular vector. The product that produces all combinations of a set is known as a Cartesian product to me.I was an idiot. I knew about std.algorithm.joiner but for some reason didn't think of it at the time. In any case, crossProduct doesn't make any sense to me... I don't see what it's got to do with flattening nested ranges.I meant the resulting range spans the cross product (all combinations) of the passed-in ranges.Yah.On another subject, I think this can be done with only input ranges - no need for bidirectional.[...] In the case of finite ranges, one of the ranges can be an input range, but the other must at least be a forward range, since otherwise you could only traverse it once, so it would be impossible to combine each element with every element from the other range.When one of the ranges is infinite, it's obviously not enough for the other range to be an input range, since you'd have to produce an infinite number of combinations per element (of the second range) before moving on to the next element, so you'd never get to it.Something like that.It should be possible to do it with both ranges being forward ranges, but it may not be very efficient -- think about it, for every element in one range, you need to combine it with every element in the other range, so when you're at, say, the 100th element of the first range, you still need to go back to the 1st, 2nd, 3rd, etc., elements in the second range. But since you can only combine it with a finite number of elements from the second range (otherwise you'll never get to the 101th element), when you get to the 101th element, you still haven't produced some of the combinations of the 100th element yet. So you'll end up needing an increasing number of save points in order to get to every possible combination -- possible but not efficient. Using the Cantor method we can allow one range to be a forward range, and the second a range whose take(n) is a reversible range (I was wrong before: the second range doesn't need to be bidirectional, it works as long as take(n) returns a reversible range). This gives us constant space complexity. I'd say this is a pretty good solution.I'm still thinking of Cantor's method, just a different schedule of spanning the triangle. Multiple save points are necessary. Consider you span the triangle in squares of increasing side. For each side size you first walk along one dimension, then walk along the other. Andrei

Oct 10 2012

On 10/10/12 10:59 AM, Andrei Alexandrescu wrote:I'm still thinking of Cantor's method, just a different schedule of spanning the triangle. Multiple save points are necessary.I meant "Multiple save points are NOT necessary." Andrei

Oct 10 2012

Andrei Alexandrescu:Yes, we need next_permutation. It will be up to you to convince the Grand Inquisition Committee of Reviewers that next_even_permutation is necessary and/or sufficient.I don't like the design of C++ STL here. Instead of a next_permutation(), what about ranges that yield all permutations, all adjacent permutations, all combinations? Permutations: import std.algorithm, std.conv, std.traits; struct Permutations(bool doCopy=true, T) if (isMutable!T) { private immutable size_t num; private T[] items; private uint[31] indexes; private ulong tot; this (/*in*/ T[] items) /*pure*/ nothrow in { static enum string L = text(indexes.length); // impure assert(items.length >= 0 && items.length <= indexes.length, "Permutations: items.length must be >= 0 && < " ~ L); } body { static ulong factorial(in uint n) pure nothrow { ulong result = 1; foreach (i; 2 .. n + 1) result *= i; return result; } this.num = items.length; this.items = items.dup; foreach (i; 0 .. cast(typeof(indexes[0]))this.num) this.indexes[i] = i; this.tot = factorial(this.num); } property T[] front() pure nothrow { static if (doCopy) //return items.dup; // not nothrow return items ~ []; // slower else return items; } property bool empty() const pure nothrow { return tot == 0; } void popFront() pure nothrow { tot--; if (tot > 0) { size_t j = num - 2; while (indexes[j] > indexes[j + 1]) j--; size_t k = num - 1; while (indexes[j] > indexes[k]) k--; swap(indexes[k], indexes[j]); swap(items[k], items[j]); size_t r = num - 1; size_t s = j + 1; while (r > s) { swap(indexes[s], indexes[r]); swap(items[s], items[r]); r--; s++; } } } } Permutations!(doCopy,T) permutations(bool doCopy=true, T) (/*in*/ T[] items) /*pure*/ nothrow if (isMutable!T) { return Permutations!(doCopy, T)(items); } unittest { import std.bigint; foreach (p; permutations([BigInt(1), BigInt(2), BigInt(3)])) assert((p[0] + 1) > 0); } version (permutations2_main) { void main() { import std.stdio, std.bigint; foreach (p; permutations!false([1, 2, 3])) writeln(p); alias BigInt B; foreach (p; permutations!false([B(1), B(2), B(3)])) {} } } ---------------------------------- Adjacent permutations by swapping (to be converted in a range): import std.algorithm, std.array, std.typecons, std.range; struct Spermutations { const int n; alias Tuple!(int[], int) TResult; int opApply(int delegate(ref TResult) dg) { int result; TResult aux; int sign = 1; alias Tuple!(int, int) Pair; auto p = iota(n).map!(i => Pair(i, i ? -1 : 0))().array(); aux = tuple(p.map!(pp => pp[0])().array(), sign); result = dg(aux); if (result) goto END; while (p.canFind!(pp => pp[1])()) { // Failed using std.algorithm here, too much complex auto largest = Pair(-100, -100); int i1 = -1; foreach (i, pp; p) if (pp[1]) { if (pp[0] > largest[0]) { i1 = i; largest = pp; } } int n1 = largest[0], d1 = largest[1]; sign *= -1; int i2; if (d1 == -1) { i2 = i1 - 1; swap(p[i1], p[i2]); if (i2 == 0 || p[i2 - 1][0] > n1) p[i2][1] = 0; } else if (d1 == 1) { i2 = i1 + 1; swap(p[i1], p[i2]); if (i2 == n - 1 || p[i2 + 1][0] > n1) p[i2][1] = 0; } aux = tuple(p.map!(pp => pp[0])().array(), sign); result = dg(aux); if (result) goto END; foreach (i3, ref pp; p) { auto n3 = pp[0], d3 = pp[1]; if (n3 > n1) pp[1] = (i3 < i2) ? 1 : -1; } } END: return result; } } void main() { import std.stdio; foreach (n; [3, 4]) { writefln("\nPermutations and sign of %d items", n); foreach (tp; Spermutations(n)) writefln("Perm: %s Sign: %2d", tp.tupleof); } } ---------------------------------- Combinations (to be converted in a range): ulong binomial(long n, long k) pure nothrow in { assert(n > 0, "binomial: n must be > 0."); } body { if (k < 0 || k > n) return 0; if (k > (n / 2)) k = n - k; ulong result = 1; foreach (ulong d; 1 .. k + 1) { result *= n; n--; result /= d; } return result; } struct Combinations(T, bool copy=true) { // Algorithm by Knuth, Pre-fascicle 3A, draft of // section 7.2.1.3: "Generating all partitions". T[] items; int k; size_t len = -1; // computed lazily this(in T[] items, in int k) in { assert(items.length, "combinations: items can't be empty."); } body { this.items = items.dup; this.k = k; } property size_t length() /*logic_const*/ { if (len == -1) // set cache len = cast(size_t)binomial(items.length, k); return len; } int opApply(int delegate(ref T[]) dg) { if (k == items.length) return dg(items); // yield items auto outarr = new T[k]; if (k == 0) return dg(outarr); // yield [] if (k < 0 || k > items.length) return 0; // yield nothing int result, x; immutable n = items.length; auto c = new uint[k + 3]; // c[0] isn'k used foreach (j; 1 .. k + 1) c[j] = j - 1; c[k + 1] = n; c[k + 2] = 0; int j = k; while (true) { // The following lines equal to: //int pos; //foreach (i; 1 .. k +1) // outarr[pos++] = items[c[i]]; auto outarr_ptr = outarr.ptr; auto c_ptr = &(c[1]); auto c_ptrkp1 = &(c[k + 1]); while (c_ptr != c_ptrkp1) *outarr_ptr++ = items[*c_ptr++]; static if (copy) { auto outarr2 = outarr.dup; result = dg(outarr2); // yield outarr2 } else { result = dg(outarr); // yield outarr } if (j > 0) { x = j; c[j] = x; j--; continue; } if ((c[1] + 1) < c[2]) { c[1]++; continue; } else j = 2; while (true) { c[j - 1] = j - 2; x = c[j] + 1; if (x == c[j + 1]) j++; else break; } if (j > k) return result; // End c[j] = x; j--; } } } Combinations!(T,copy) combinations(bool copy=true, T) (in T[] items, in int k) in { assert(items.length, "combinations: items can't be empty."); } body { return Combinations!(T, copy)(items, k); } ---------------------------------- Bye, bearophile

Oct 09 2012

On 10/09/2012 11:23 PM, bearophile wrote:Andrei Alexandrescu:I agree that that's better conceptually, but your implementations have overflow issues. (And it is not too obvious how to fix that, given the existing range design.)Yes, we need next_permutation. It will be up to you to convince the Grand Inquisition Committee of Reviewers that next_even_permutation is necessary and/or sufficient.I don't like the design of C++ STL here. Instead of a next_permutation(), what about ranges that yield all permutations, all adjacent permutations, all combinations? [snip code]

Oct 09 2012

struct Spermutations {The naming scheme could do with... improvement.

Oct 09 2012

ixid:The naming scheme could do with... improvement.struct PermutationSwaps then? Bye, bearophile

Oct 09 2012

On Tuesday, 9 October 2012 at 22:06:42 UTC, bearophile wrote:ixid:I think you're missing what I meant, naming a struct "Sperm"-utations seemed a bit unpleasant, I'm not actually bothered about your naming scheme. =)The naming scheme could do with... improvement.struct PermutationSwaps then? Bye, bearophile

Oct 09 2012

On Tue, 09 Oct 2012 17:23:07 -0400, bearophile <bearophileHUGS lycos.com> wrote:Andrei Alexandrescu:Is there any advantage over having a function? I'd think you could easily build a range based on the function, no? -SteveYes, we need next_permutation. It will be up to you to convince the Grand Inquisition Committee of Reviewers that next_even_permutation is necessary and/or sufficient.I don't like the design of C++ STL here. Instead of a next_permutation(), what about ranges that yield all permutations, all adjacent permutations, all combinations?

Oct 09 2012

Steven Schveighoffer:Is there any advantage over having a function? I'd think you could easily build a range based on the function, no?Generators (that yield lexicographic permutations, permutation swaps, combinations, etc) are quite more handy, you can compose them with the other ranges and higher order functions. Adding to Phobos few ranges that build on a next_permutation(), next_combination(), etc, is possible, but I don't see a big need for such C++-style functions, the ranges are enough. Bye, bearophile

Oct 09 2012

On 10/10/12 00:22, bearophile wrote:Steven Schveighoffer:Is there any advantage over having a function? I'd think you could easily build a range based on the function, no?Generators (that yield lexicographic permutations, permutation swaps, combinations, etc) are quite more handy, you can compose them with the other ranges and higher order functions.Adding to Phobos few ranges that build on a next_permutation(), next_combination(), etc, is possible, but I don't see a big need for such C++-style functions, the ranges are enough.What makes you think that permutations are only ever used in a range context? Do you have any evidence for that? Sounds like an abstraction inversion to me.

Oct 10 2012

On Wed, Oct 10, 2012 at 10:59:32AM -0400, Andrei Alexandrescu wrote:On 10/10/12 10:27 AM, H. S. Teoh wrote:[...]On Wed, Oct 10, 2012 at 09:41:34AM -0400, Andrei Alexandrescu wrote:[...]On another subject, I think this can be done with only input ranges - no need for bidirectional.[...] On Wed, Oct 10, 2012 at 11:00:43AM -0400, Andrei Alexandrescu wrote:Using the Cantor method we can allow one range to be a forward range, and the second a range whose take(n) is a reversible range (I was wrong before: the second range doesn't need to be bidirectional, it works as long as take(n) returns a reversible range). This gives us constant space complexity. I'd say this is a pretty good solution.I'm still thinking of Cantor's method, just a different schedule of spanning the triangle. Multiple save points are necessary. Consider you span the triangle in squares of increasing side. For each side size you first walk along one dimension, then walk along the other.On 10/10/12 10:59 AM, Andrei Alexandrescu wrote:[...] In the interest of not stalling the inclusion of a Cartesian product in Phobos any longer (since the original discussion was apparently some *years* ago), I propose that we check in the current implementation of the Cantor method first, and then think about how we can improve it. Since the improved version will *relax* the requirement on the argument ranges, any code based on the current method shouldn't break with the improved version once we figure out how to do it. T -- He who sacrifices functionality for ease of use, loses both and deserves neither. -- SlashdotterI'm still thinking of Cantor's method, just a different schedule of spanning the triangle. Multiple save points are necessary.I meant "Multiple save points are NOT necessary."

Oct 10 2012