For a fixed N, can we find an optimum function which makes the minimum number of comparisions to sort N elements?

First we should observe that a comparison between R[i] and R[j] will yield the same result as a comparison between A[i] and A[j]. If we can sort R using this optimum function, then we can sort A in the same way.

To find the optimum function, I could go thru all possible functions and test them with all possible values of R, and then measure which one does best. Here, 'best' could be defined in various ways: average run time, worst run time etc. Instead of going thru all functions, maybe I can go about it in a heuristic way.

When we make a comparision between two elements of R, we're classifying S= megaset of R (all possible values of R) into two classes. For example, let's have 4 elements and we decide to compare elements 0 and 1.

S={ 0: abcd 1: abdc 2: acbd 3: acdb 4: adcb 5: adbc 6: bacd 7: badc 8: bcad 9: bcda 10: bdca 11: bdac 12: cbad 13: cbda 14: cabd 15: cadb 16: cdab 17: cdba 18: dbca 19: dbac 20: dcba 21: dcab 22: dacb 23: dabc }The comparison between elements 0 and 1 can result in -1 or 1 (R[0] is less than R[1] or opposite). We can now partition S into two sets: for the first one, the comparison results in -1 and for the second one it's 1.

S_01_L= { 0: abcd 1: abdc 2: acbd 3: acdb 4: adcb 5: adbc 8: bcad 9: bcda 10: bdca 11: bdac 16: cdab 17: cdba } S_01_G={ 6: bacd 7: badc 12: cbad 13: cbda 14: cabd 15: cadb 18: dbca 19: dbac 20: dcba 21: dcab 22: dacb 23: dabc }This would be expressed in our function like this:

if (A[0]<A[1]) { code_01_L } else { code_01_G }Now, if we use element 0 again in code_01_L, we will not divide S_01_L into two equal parts, which we're aiming for. For instance:

S_01_L_02_L= { 0: abcd 1: abdc 2: acbd 3: acdb 4: adcb 5: adbc 9: bcda 10: bdca } S_01_L_02_G= { 8: bcad 11: bdac 16: cdab 17: cdba }Comparing A[0] to A[2] has now splitted the current set 8:4, which is not optimum. We made a comparision, but it didn't give us enough information. Now we have to distinguish between 8 elements within the first set. If we had made an optimum choice, the comparison would give us exactly one bit of information and the split would be 6:6. Let's look at the comparison between element 2 and 3, maybe:

S_01_L_23_L= { 0: abcd 2: acbd 5: adbc 8: bcad 11: bdac 16: cdab } S_01_L_23_G= { 1: abdc 3: acdb 4: adcb 9: bcda 10: bdca 17: cdba }This is a perfect 6:6 split and should require less operations per branch.

In our initial comparison 01, instead of partitioning S we could have done a compare-and-exchange operation and transform S into a smaller set like this:

if (A[0]>=A[1]) A[0] <-> A[1];This would transform S into S_01:

S_01= { 0: abcd 1: abdc 2: acbd 3: acdb 4: adcb 5: adbc 8: bcad 9: bcda 10: bdca 11: bdac 16: cdab 17: cdba }Which is the same thing as S_01_L. After this first exchange, we have forced A to conform to an element of S_01_L. Now we only need to distinguish between the members of S_01_L.

It's not clear that we will always have a perfect split. Let's see if that is the case. Now I'll just use the numbers and see where we can go from our current best position S_01_23, which is the result of two perfect splits. We have 6 possible comparisons:

S_01= { 0 1 2 3 4 5 8 9 10 11 16 17 } S_01_23= { 0 2 5 8 11 16 } 01 already done, no effect 02 S_01_23_02= { 0 2 4 5} 03 S_01_23_03= { 0 2 5 8 11} 12 S_01_23_12= { 0 1 6 7 15} 13 S_01_23_13= { 0 2 8 12} 23 already done, no effectSo our third comp-ex doesn't help as much, it eliminates 1 or 2 possibilities from 6 permutations. Maybe my initial choice of 01_23 wasn't so good after all. Definitely, this requires a search. This was quite obvious when I started (due to the huge literature about the subject) but I didn't understand exactly why. I thought maybe things were getting inefficient because we eventually run into nasty little sets and could do better with different splits at the upper levels. It's not so. Even the upper level sets can't always be partitioned nicely.

I could make a search on this and see which sequence of compare-exchange operations finish the sort fastest. From the Internet, I found this optimal sequence for 4 elements:

S_01= { 0 1 2 3 4 5 8 9 10 11 16 17 } S_01_23= { 0 2 5 8 11 16 } S_01_23_02= { 0 2 4 5 } S_01_23_02_13= { 0 2 } S_01_23_02_13_12= { 0 }Instead of going thru all N! permutations or 2^N bit strings, I thought maybe I can verify a solution symbolicly? Like this:

G[0]= A0 G[1]= A1 G[2]= A2 G[3]= A3 apply (0,1) and (2,3) G[0]= min(A0,A1) G[1]= max(A0,A1) G[2]= min(A2,A3) G[3]= max(A2,A3) apply (0,2) and (1,3) G[0]= min( min(A0,A1), min(A2,A3)) G[1]= min( max(A0,A1), max(A2,A3)) G[2]= max( min(A0,A1), min(A2,A3)) G[3]= max( max(A0,A1), max(A2,A3)) apply (1,2) G[0]= min( min(A0,A1), min(A2,A3)) G[1]= min(min( max(A0,A1), max(A2,A3)), max( min(A0,A1), min(A2,A3))) G[2]= max(min( max(A0,A1), max(A2,A3)), max( min(A0,A1), min(A2,A3))) G[3]= max( max(A0,A1), max(A2,A3))Verifying the 0zix and 3zix elements is straightforward.

G[0]= min( min(A0,A1), min(A2,A3))= min(A0,A1,A2,A3) G[3]= max( max(A0,A1), max(A2,A3))= max(A1,A1,A2,A3)Obviously, G[2] ≥ G[1] since

G[1]= min(U,V) G[2]= max(U,V) so, G[0] ≤ G[1] ≤ G[2] ≤ G[3]I now have the problem of proving that G[i] is unique, resulting in:

G[0] < G[1] < G[2] < G[3]Proving G[0]!=G[3] is trivial, since Ai is unique.

Instead of that, I could name all the intermediate values and establish relations between them.

step 0 1 2 3 relations ----------------------------------- init . . . . - 01 G1 G2 . . G1 < G2 23 G1 G2 G3 G4 G3 < G4 02 G5 G2 G6 G4 G5 < G6, G5= min(G1,G3), G6= max(G1,G3) 13 G5 G7 G6 G8 G7 < G8, G7= min(G2,G4), G8= max(G2,G4) 12 G5 G9 G10 G8 G9 < G10, G9= min(G7,G6), G10= max(G7,G6)Going back to the algebraic approach:

order max(a,b) min(max(a,b),c) min(a,c) min(b,c) max(min(a,c),min(b,c)) ----------------------------------------------------------------------- abc b b a b b acb b c a c c bac a a a b a bca a c c b c cab b c c c c cba a c c c c min( max(a,b), c) == max( min(a,c), min(b,c) ) max( min(a,b), c) == min( max(a,c), max(b,c) ) Here is the code (base64): I2luY2x1ZGUgPHN0ZGlvLmg+CiNpbmNsdWRlIDxzdHJpbmcuaD4KI2luY2x1 ZGUgPHN0ZGxpYi5oPgoKCnN0YXRpYyBpbnQgZmFjdFsyMF07CgpjaGFyIE9P WzIwXTsKCgp2b2lkIGdlbl9tdGhfcGVybShpbnQgTSxpbnQgTixpbnQgKkEp CnsKICBpbnQgaSxqLGVsdCxpZHg7CiAgc3RhdGljIGludCBmYWN0WzIwXTsK CiAgaWYgKCFmYWN0WzBdKQogICB7IGZhY3RbMF09IDE7IGZvcihpPTE7aTwy MDtpKyspIGZhY3RbaV09IGZhY3RbaS0xXSppOyB9CgogIGZvcihpPTA7aTxO O2krKykgQVtpXT0gaTsKCiAgZm9yKGk9MDtpPE47aSsrKQogIHsKICAgIGlk eD0gTSAvIGZhY3RbTi1pLTFdOwogICAgZWx0PSBBW2lkeCtpXTsKICAgIGZv cihqPWlkeCtpO2o+aTtqLS0pIEFbal09IEFbai0xXTsKICAgIEFbaV09IGVs dDsKICAgIE0gJT0gZmFjdFtOLWktMV07CiAgfQoKfQoKdm9pZCBwcmluYXJy KGNoYXIgKm1zZyxpbnQgKkEsaW50IE4pCnsKICBpbnQgaTsKICBwcmludGYo IiVzIixtc2cpOwogIGZvcihpPTA7aTxOO2krKykgcHJpbnRmKCIlYyIsJ2En K0FbaV0pOwogIHByaW50ZigiXG4iKTsKfQoKaW50IHBlcm1fbnVtYmVyKGlu dCBOLGludCAqQSkKewogIGludCBNLGksajsKICBNPSAwOwogIGZvcihpPTA7 aTxOO2krKykKICB7CiAgICBNICs9IEFbaV0qZmFjdFtOLWktMV07CiAgICBm b3Ioaj1pKzE7ajxOO2orKykKICAgICAgaWYgKEFbal0+QVtpXSkgQVtqXS0t OwogIH0KICByZXR1cm4gTTsKfQoKaW50IG1heChpbnQgeCxpbnQgeSkKewog IGlmIChzdHJjaHIoT08seCk+c3RyY2hyKE9PLHkpKSByZXR1cm4geDsKICBy ZXR1cm4geTsKfQoKaW50IG1pbihpbnQgeCxpbnQgeSkKewogIGlmIChzdHJj aHIoT08seCk8c3RyY2hyKE9PLHkpKSByZXR1cm4geDsKICByZXR1cm4geTsK fQoKCmludCBtYWluKGludCBhcmdjLGNoYXIgKiphcmd2KQp7CiAgaW50IEFb MjBdOwogIGludCBOLGksajsKICBpbnQgRj0xOwogIGludCAqUzsKICBpbnQg YSxiLGM7CgogIE49IDM7CgogIGZhY3RbMF09IDE7IAogIGZvcihpPTE7aTwy MDtpKyspIAogICAgIGZhY3RbaV09IGZhY3RbaS0xXSppOyAKCiAgYT0gJ2En OyBiPSAnYic7IGM9ICdjJzsKICBwcmludGYoCiJvcmRlciBtYXgoYSxiKSBt aW4obWF4KGEsYiksYykgbWluKGEsYykgbWluKGIsYykgbWF4KG1pbihhLGMp LG1pbihiLGMpKVxuIik7CnByaW50ZigKIi0tLS0tLS0tLS0tLS0tLS0tLS0t LS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0tLS0t LS0tLS0tXG4iKTsKICBmb3IoaT0wO2k8NjtpKyspCiAgewogICAgZ2VuX210 aF9wZXJtKGksMyxBKTsKICAgIGZvcihqPTA7ajwzO2orKykgT09bal09ICdh JyArIEFbal07CiAgICBPT1szXT0gMDsKcHJpbnRmKAoiJXMgICAgICAlYyAg ICAgICAgICAgJWMgICAgICAgICAgICVjICAgICAgICVjICAgICAgICAgICAg ICVjICAgICAgICAgIFxuIiwKT08sICAgICBtYXgoYSxiKSxtaW4obWF4KGEs YiksYyksbWluKGEsYyksbWluKGIsYyksbWF4KG1pbihhLGMpLG1pbihiLGMp KQopOwogIH0KCiAgcmV0dXJuIDA7Cn0KCgo= Wait, I like this. X=max(a,b) Y=min(X,c) Z=min(a,c) T=min(b,c) U=max(Z,T) order X Y Z T U --------------- abc b b a b b acb b c a c c bac a a a b a bca a c c b c cab b c c c c cba a c c c c I2luY2x1ZGUgPHN0ZGlvLmg+CiNpbmNsdWRlIDxzdHJpbmcuaD4KI2luY2x1ZGUgPH N0ZGxpYi5oPgoKCnN0YXRpYyBpbnQgZmFjdFsyMF07CgpjaGFyIE9PWzIwXTsKCgp2 b2lkIGdlbl9tdGhfcGVybShpbnQgTSxpbnQgTixpbnQgKkEpCnsKICBpbnQgaSxqLG VsdCxpZHg7CiAgc3RhdGljIGludCBmYWN0WzIwXTsKCiAgaWYgKCFmYWN0WzBdKQog ICB7IGZhY3RbMF09IDE7IGZvcihpPTE7aTwyMDtpKyspIGZhY3RbaV09IGZhY3RbaS 0xXSppOyB9CgogIGZvcihpPTA7aTxOO2krKykgQVtpXT0gaTsKCiAgZm9yKGk9MDtp PE47aSsrKQogIHsKICAgIGlkeD0gTSAvIGZhY3RbTi1pLTFdOwogICAgZWx0PSBBW2 lkeCtpXTsKICAgIGZvcihqPWlkeCtpO2o+aTtqLS0pIEFbal09IEFbai0xXTsKICAg IEFbaV09IGVsdDsKICAgIE0gJT0gZmFjdFtOLWktMV07CiAgfQoKfQoKdm9pZCBwcm luYXJyKGNoYXIgKm1zZyxpbnQgKkEsaW50IE4pCnsKICBpbnQgaTsKICBwcmludGYo IiVzIixtc2cpOwogIGZvcihpPTA7aTxOO2krKykgcHJpbnRmKCIlYyIsJ2EnK0FbaV 0pOwogIHByaW50ZigiXG4iKTsKfQoKaW50IHBlcm1fbnVtYmVyKGludCBOLGludCAq QSkKewogIGludCBNLGksajsKICBNPSAwOwogIGZvcihpPTA7aTxOO2krKykKICB7Ci AgICBNICs9IEFbaV0qZmFjdFtOLWktMV07CiAgICBmb3Ioaj1pKzE7ajxOO2orKykK ICAgICAgaWYgKEFbal0+QVtpXSkgQVtqXS0tOwogIH0KICByZXR1cm4gTTsKfQoKaW 50IG1heChpbnQgeCxpbnQgeSkKewogIGlmIChzdHJjaHIoT08seCk+c3RyY2hyKE9P LHkpKSByZXR1cm4geDsKICByZXR1cm4geTsKfQoKaW50IG1pbihpbnQgeCxpbnQgeS kKewogIGlmIChzdHJjaHIoT08seCk8c3RyY2hyKE9PLHkpKSByZXR1cm4geDsKICBy ZXR1cm4geTsKfQoKCiNkZWZpbmUgRVFWQVIoWCxZKSBpbnQgWDsKI2RlZmluZSBFUV NUUihYLFkpICIgIiAjWCAKI2RlZmluZSBFUURFQ0woWCxZKSBwcmludGYoIiAgIiBF UVNUUihYLFkpICI9IiAjWSAiXG4iICk7CiNkZWZpbmUgRVFMSU4oWCxZKSBlcWxpbi hFUVNUUihYLFkpKTsKI2RlZmluZSBFUURPKFgsWSkgIFggPSBZOwojZGVmaW5lIEVR UFJUKFgsWSkgZXFwcnQoWCxFUVNUUihYLFkpKTsKCiNkZWZpbmUgRVFVQVRJT05TIF wKICBFUShYLG1heChhLGIpKSBcCiAgRVEoWSxtaW4oWCxjKSkgXAogIEVRKFosbWlu KGEsYykpIFwKICBFUShULG1pbihiLGMpKSBcCiAgRVEoVSxtYXgoWixUKSkgCgp2b2 lkIGVxbGluKGNoYXIgKnRpdGxlKQp7CiAgaW50IGk7CiAgZm9yKGk9MDt0aXRsZVtp XTtpKyspIHByaW50ZigiLSIpOwp9Cgp2b2lkIGVxcHJ0KGludCB2LCBjaGFyICp0aX RsZSkKewogIGludCBpOwogIGludCBMPSBzdHJsZW4odGl0bGUpOwogIGZvcihpPTA7 aStpPEw7aSsrKSBwcmludGYoIiAiKTsKICBwcmludGYoIiVjIix2KTsgaSsrOwogIG Zvcig7aTxMO2krKykgcHJpbnRmKCIgIik7Cn0KCgppbnQgbWFpbihpbnQgYXJnYyxj aGFyICoqYXJndikKewogIGludCBBWzIwXTsKICBpbnQgTixpLGo7CiAgaW50IGEsYi xjOwoKICBOPSAzOwoKICBmYWN0WzBdPSAxOyAKICBmb3IoaT0xO2k8MjA7aSsrKSAK ICAgICBmYWN0W2ldPSBmYWN0W2ktMV0qaTsgCgojZGVmaW5lIEVRIEVRVkFSCiAgRV FVQVRJT05TCiN1bmRlZiBFUQoKI2RlZmluZSBFUSBFUURFQ0wKICBFUVVBVElPTlMK I3VuZGVmIEVRCgojZGVmaW5lIEVRIEVRU1RSCiAgcHJpbnRmKCJvcmRlciIgRVFVQV RJT05TICJcbiIpOwojdW5kZWYgRVEKCiNkZWZpbmUgRVEgRVFMSU4KICBwcmludGYo Ii0tLS0tIik7IAogIEVRVUFUSU9OUwogIHByaW50ZigiXG4iKTsKI3VuZGVmIEVRCg ogIGE9ICdhJzsgYj0gJ2InOyBjPSAnYyc7CiAgZm9yKGk9MDtpPDY7aSsrKQogIHsK ICAgIGdlbl9tdGhfcGVybShpLDMsQSk7CiAgICBmb3Ioaj0wO2o8MztqKyspIE9PW2 pdPSAnYScgKyBBW2pdOwogICAgT09bM109IDA7CiAgICAgICAgICAgIC8vb3JkZXIK ICAgIHByaW50ZigiJXMgICIsIE9PKTsKI2RlZmluZSBFUSBFUURPCiAgICBFUVVBVE lPTlMKI3VuZGVmIEVRCgojZGVmaW5lIEVRIEVRUFJUCiAgICBFUVVBVElPTlMKI3Vu ZGVmIEVRCiAgICBwcmludGYoIlxuIik7CiAgfQoKICByZXR1cm4gMDsKfQoKCg==So, a new notation is in order to make some manipulations. From now on '.' is minimum and '+' is maximum. '.' may be omitted if unambigous.

min( max(a,b), c) == max( min(a,c), min(b,c) ) (a+b)c = ac+bc // distributive ?? fantastic. (a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd G[1]= min(min( max(A0,A1), max(A2,A3)), max( min(A0,A1), min(A2,A3))) ---------- --------- A0.A1 A2.A3 (A0+A1) (A2+A3) ---------------------------- --------------------------- (A0+A1)(A2+A3) (A0.A1 + A2.A3) (A0A2+A0A3+A1A2+A1A3) ---------------------------------------------------------------- (A0A2+A0A3+A1A2+A1A3) (A0.A1 + A2.A3) = A0.A2.A0.A1 + A0.A2.A2.A3 + A0.A3.A0.A1 + A0.A3.A2.A3 + A1.A2.A0.A1 + A1.A2.A2.A3 + A1.A3.A0.A1 + A1.A3.A2.A3 we have, X.X= X and Y+Y=Y = A0.A1.A2 + A0.A2.A3 + A0.A1.A3 + A0.A2.A3 + A0.A1.A2 + A1.A2.A3 + A0.A1.A3 + A1.A2.A3 G[1]= A0.A1.A2 + A0.A2.A3 + A0.A1.A3 + A1.A2.A3 G[2]= max(min( max(A0,A1), max(A2,A3)), max( min(A0,A1), min(A2,A3))) ---------- ---------- ---------- ---------- (A0+A1) (A2+A3) A0.A1 A2.A3 ---------------------------- ---------------------------- (A0.A2+A0.A3+A1.A2+A1.A3) A0.A1 + A2.A3 = A0.A2 + A0.A3 + A1.A2 + A1.A3 + A0.A1 + A2.A3 = A0.A1 + A0.A2 + A0.A3 + A1.A2 + A1.A3 + A2.A3So, to prove that G[2]≥G[1], we just need to show that max(G[1],G[2]) is G[2]. This is normally obvious from the fact that one is the minimum and the other is the maximum of two elements, but I want to develop the technique. What happens now is that we 'add' (max) the two expressions together, simplify it, and then see that it's equal to G[2].

G[1]= A0.A1.A2 + A0.A2.A3 + A0.A1.A3 + A1.A2.A3 + G[2]= A0.A1 + A0.A2 + A0.A3 + A1.A2 + A1.A3 + A2.A3 ----------------------------------------------------- A0.A1.A2 + A0.A2.A3 + A0.A1.A3 + A1.A2.A3 + A0.A1 + A0.A2 + A0.A3 + A1.A2 + A1.A3 + A2.A3It's time to realize more tools for simplification. Obviously, max(A0.A1+A0.A2, A0.A1.A2) is A0.A1+A0.A2, but I should prove it somehow.

a+a.b = a ... RULE-1 X=min(a,b) Y=max(a,X) order X Y --------- ab a a ba b a A0.A1+ A0.A2 + A0.A1.A2= A0.A1+ (A0.A2) + (A0.A2).A1 -----as above------- = A0.A1 + A0.A2In fact, RULE-1 is obvious if you think about the identity elements for the minimum and maximum operations. Let M be the identity for maximum and W for minimum. Whenever you find the minimum of W and any a, the result is a. This means that W is greater than everything else and similarly M is less than everything else.

a+a.b = a.W + a.b = a.(W+b) = a.W = aI'll go about this in the following fashion. I will keep all expressions in sum-of-simple-products format, i.e:

G[x]= max( min(elements from Ai), min(elements from Ai), ... )For each min() expression, I will use a machine word. Bit i will be set in the word if Ai occurs in the min() expression. Finally, the whole expression will be a sequence of machine words, terminated by a zero word for convenience.

When I want to find the maximum of two expressions, I simply concatenate them together and simplify the resulting word string. Unifying this string will let me eliminate max(X,X) kind of redundancies. Also, using RULE-1 I will further simplify them.

For finding the minimum of two expressions, I will make a cartesian product of the expressions. The product operation between two machine words will be the bitwise OR operation. This will concatenate the argument lists of the two min() functions, eliminating redundant entries at the same time.

Here it is:

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Later on, I will make some sort of search to pass time.

time / level --> | a G[0,0] G[1,0] ... G[F,0] | r G[0,1] G[1,1] ... G[F,1] v r G[0,2] G[F,2] a .. y G[0,N-1] .. G[F,N-1]Anyway, let's say that the final array U[z]= G[F,z] for simplicity. We want to know what U looks like. We already know that the 0zix element should be the minimum and the last element should be the maximum of all objects.

U[0]= min(A0,A1,..A(N-1)) U[N-1]= max(A0,A1,..A(N-1))Also, let's define com(A,z) as the set of z-element combinations of A.

Now, for instance

U[2]= min( x | x= max(a,b,c) | {a,b,c} ∈ com(A,3) )Let's consider two elements k,e of com(A,3). If max(k) is greater than max(e), then k can not be the one which goes to the first three elements. Since max(e) is less than max(k), k contains an element that is larger than all those in e. When we loop through all elements in com(A,3) in this manner, we reach the above formula for U[2].

In other words, U[2] is the maximum of three objects. However, it must be the minimum possible to compute in this way.

When we generalize the above formula, we reach:

U[z]= min( x | x= max(v0, v1, .. vz) | {v0,v1,..vz} ∈ com(A,z+1) )This does generate U[0] properly.

com(A,1)= { {A0}, {A1} , {A2}, .. {A(N-1)} } U[0]= min (max({A0}), max({A1}), max({A2}, ..) U[0]= min (A0, A1, A2, ..)It also generates U[N-1]

has only one element com(A,N)= { {A0,A1, A2, .. A(N-1) } } U[N-1]= min( max({A0,A1,A2,..}) ) U[N-1]= max(A0, A1, A2 .. )Another way to do this is to think about the last p+1 elements of the array instead of the first z+1, where p+z+1= N.

U[z]= max( x | x = min( {v0,v1,..vp} ) | {v0,v1,..vp} ∈ com(A,p+1) )Deriving U[N-1]:

p+z+1= N p+N-1= N p= 0 com(A, 1)= { {A0}, {A1}, {A2}, .. } U[N-1]= max( min({A0}), min({A1}), min({A2}), .. } U[N-1]= max( A0, A1, A2 .. )Deriving U[0]:

p+z+1= N p+0+1= N p= N-1 has one element com(A,N)= { {A0,A1,A2, .. } } U[0]= max( min({A0,A1,A2,..}) ) U[0]= min({A0,A1,..}) U[0]= min(A0,A1,..)This second form, based on max-of-mins, is more compatible with the way I implemented expressions before. So, I'm going to use it.

This formula does generate a sorted array. Since the sorted array is unique, all sorting machines should end up with the same expressions for a given N. So, I could verify my findings by trying out the solutions from the internet for various N. Here is the above program with the extra test added:

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3 0.1/0.2/1.2 4 0.1/2.3/0.2/1.3/1.2 5 0.1/3.4/0.2/1.2/0.3/2.3/1.4/1.2/3.4 6 0.1/1.2/2.3/3.4/4.5/0.1/1.2/2.3/3.4/0.1/1.2/2.3/0.1/1.2/0.1This looks promising, these sorting machines all conform to my formulas.

I initially thought that symbolic validation would be faster than going thru all cases but it's not so. It's even worse. With the try-and-see approach, we only go thru N! or 2^N cases to see whether the machine sorts. In contrast, the symbolic verification requires us to create expressions with combinatorial size. This requires both time and space proportional to 2^N. However, expressions let me reason about how to search around so I'm going to keep them.

Since I know what the final expressions must look like, I could easily expand the network beyond the input size. For instance, I can have 2N lines for a size N input, copying the input values from [0..N-1] to [N..2N-1]. This would allow me to store intermediate values in the bottom part of the network. At the final stage of construction I could search the array G[0..2N-1] to find the expression for each U[i] and that subset would be my output lines.

In any case, I have the initial and the final states for the G[] network.

G[0,z]= Az G[F,z]= max( min(v0,v1,..vp) | {v0,v1,..vp} ∈ com(A,p+1), p+z+1=N)Now I need to go from one to the other.

Let's write out G[F,N-2], the element one before the last.

say that n= N-1 G[F,N-2]= A0.A1 + A0.A2 + A0.A3 ... A0.An + A1.A2 + A1.A3 ... A1.An + A2.A3 ... A2.An + ... A(n-1)An = A0(A1+A2+...An) + A1(A2+A3+...An) + A2(A3+A4+...An) + .. A(n-1)AnI could make a huge circuit for this, and then use the intermediate results to compute other things.

Let's make one for N=6 and try to reason on it. I'm going to skip the A. On the right side, 0 stands for A0 etc.

G[F,5]= 0+1+2+3+4+5 G[F,4]= 0(1+2+3+4+5) + 1(2+3+4+5) + 2(3+4+5) + 3(4+5) + 4.5 G[F,3]= 0.1.2 + 0.1.3 + 0.1.4 + 0.1.5 + 0.2.3 + 0.2.4 + 0.2.5 + 0.3.4 + 0.3.5 + 0.4.5 1.2.3 + 1.2.4 + 1.2.5 + 1.3.4 + 1.3.5 + 1.4.5 2.3.4 + 2.3.5 + 2.4.5 + 3.4.5 = 0.1(2+3+4+5) + 0.2(3+4+5) + 0.3(4+5) + 0.4.5 + 1.2(3+4+5) + 1.3(4+5) + 1.4.5 + 2.3(4+5) + 2.4.5 + 3.4.5 G[F,2]= 0.1.2.3+ 0.1.2.4+ 0.1.2.5+ 0.1.3.4+ 0.1.3.5+ 0.1.4.5+ 0.2.3.4+ 0.2.3.5+ 0.2.4.5+ 0.3.4.5+ 1.2.3.4+ 1.2.3.5+ 1.2.4.5+ 1.3.4.5+ 2.3.4.5 = 0.1.2(3+4+5) + 0.1.3(4+5) + 0.1.4.5 + 0.2.3(4+5) + 0.2.4.5 + + 0.3.4.5 + 1.2.3(4+5) + 1.2.4.5 + 1.3.4.5 + 2.3.4.5 + G[F,1]= 0.1.2.3.4+ 0.1.2.3.5+ 0.1.2.4.5+ 0.1.3.4.5+ 0.2.3.4.5+ 1.2.3.4.5 = 0.1.2.3(4+5) + 0.1.2.4.5 + 0.1.3.4.5 + 0.2.3.4.5 + 1.2.3.4.5 G[F,0]= 0.1.2.3.4.5 Here, the sum A4+A5 and its dual A4.A5 occur very frequently, if we set u= A4.A5 and U= A4+A5, maybe we can simplify this a little. G[F,5]= 0+1+2+3+4+5 = 0+1+2+3+U G[F,4]= 0(1+2+3+4+5) + 1(2+3+4+5) + 2(3+4+5) + 3(4+5) + 4.5 = 0(1+2+3+U) + 1(2+3+U) + 2(3+U) + 3.U + u G[F,3]= 0.1(2+3+4+5) + 0.2(3+4+5) + 0.3(4+5) + 0.4.5 + 1.2(3+4+5) + 1.3(4+5) + 1.4.5 + 2.3(4+5) + 2.4.5 + 3.4.5 = 0.1(2+3+U) + 0.2(3+U) + 0.3.U + 0.u + 1.2(3+U) + 1.3.U + 1.u + 2.3.U + 2.u + 3.u G[F,2]= 0.1.2(3+4+5) + 0.1.3(4+5) + 0.1.4.5 + 0.2.3(4+5) + 0.2.4.5 + 0.3.4.5 + 1.2.3(4+5) + 1.2.4.5 + 1.3.4.5 + 2.3.4.5 + = 0.1.2(3+U) + 0.1.3.U + 0.1.u + 0.2.3.U + 0.2.u + 0.3.u + 1.2.3.U + 1.2.u + 1.3.u + 2.3.u G[F,1]= 0.1.2.3(4+5) + 0.1.2.4.5 + 0.1.3.4.5 + 0.2.3.4.5 + 1.2.3.4.5 = 0.1.2.3.U + 0.1.2.u + 0.1.3.u + 0.2.3.u + 1.2.3.u G[F,0]= 0.1.2.3.4.5= 0.1.2.3.u Now, V=3+U and v=3.U seem to be the next best replacement. In order to get rid of the number 3, I will need yet another one x=3.u G[F,5]= 0+1+2+3+U= 0+1+2+V G[F,4]= 0(1+2+3+U) + 1(2+3+U) + 2(3+U) + 3.U + u = 0(1+2+V) + 1(2+V) + 2.V + v + u G[F,3]= 0.1(2+3+U) + 0.2(3+U) + 0.3.U + 0.u + 1.2(3+U) + 1.3.U + 1.u + 2.3.U + 2.u + 3.u = 0.1.(2+V) + 0.2.V + 0.v + 0.u + 1.2.V + 1.v + 1.u + 2.v + 2.u + x = 0.1.(2+V) + 0.2.V + 1.2.V + (u+v)(0+1+2) + x G[F,2] = 0.1.2(3+U) + 0.1.3.U + 0.1.u + 0.2.3.U + 0.2.u + + 0.3.u + 1.2.3.U + 1.2.u + 1.3.u + 2.3.u = 0.1.2.V + 0.1.v + 0.1.u + 0.2.v + 0.2.u + 0.x + 1.2.v + 1.2.u + 1.x + 2.x = 0.1.2.V + (u+v)(0.1+0.2+1.2) + x(0+1+2) G[F,1]= 0.1.2.3.U + 0.1.2.u + 0.1.3.u + 0.2.3.u + 1.2.3.u = 0.1.2.v + 0.1.2.u + 0.1.x + 0.2.x+ 1.2.x = (u+v)0.1.2 + x(0.1+0.2+1.2) G[F,0]= 0.1.2.x OK. Tidy it up: G[F,5]= 0+1+2+V G[F,4]= 0(1+2+V) + 1(2+V) + 2.V + v + u G[F,3]= 0.1.(2+V) + 0.2.V + 1.2.V + (u+v)(0+1+2) + x G[F,2]= 0.1.2.V + (u+v)(0.1+0.2+1.2) + x(0+1+2) G[F,1]= (u+v)0.1.2 + x(0.1+0.2+1.2) G[F,0]= 0.1.2.x Some more simplification is due, there are a lot of common factors here G[F,5]= 0+1+2+V G[F,4]= 0(1+2+V) + 1(2+V) + 2.V + v + u = 0.1+0.2+0.V + 1.2 + 1.V + 2.V + (v+u) = (0.1+0.2+1.2)+ V(0+1+2) + (v+u) G[F,3]= 0.1.(2+V) + 0.2.V + 1.2.V + (u+v)(0+1+2) + x = 0.1.2+ V(0.1+0.2.1.2) + (u+v)(0+1+2) + x G[F,2]= 0.1.2.V + (u+v)(0.1+0.2+1.2) + x(0+1+2) G[F,1]= (u+v)0.1.2 + x(0.1+0.2+1.2) G[F,0]= 0.1.2.x So many things to refactor.. K= 0.1+0.2+1.2 L= 0.1.2 Z= 0+1+2 M= u+v G[F,5]= 0+1+2+V = Z+V G[F,4]= (0.1+0.2+1.2)+ V(0+1+2) + (v+u) = K + V.Z + M G[F,3]= 0.1.2+ V(0.1+0.2.1.2) + (u+v)(0+1+2) + x = L + V.K + M.Z + x G[F,2]= 0.1.2.V + (u+v)(0.1+0.2+1.2) + x(0+1+2) = L.V + M.K + x.Z G[F,1]= (u+v)0.1.2 + x(0.1+0.2+1.2) = M.L + x.K G[F,0]= 0.1.2.x = L.x Pretty simple result, it looks like this: G[F,5]= Z+V G[F,4]= K + V.Z + M G[F,3]= L + V.K + M.Z + x G[F,2]= L.V + M.K + x.Z G[F,1]= M.L + x.K G[F,0]= L.x Now, let's look at what's happening here. K,L and Z are in fact the resulting of sorting A0,A1,A2. Let's look at the definitions: L= 0.1.2 // minimum of 0,1,2 K= 0.1+0.2+1.2 // fits perfectly to our formula for the 1zix // element in a 3-element array Z= 0+1+2 // maximum of 0,1,2 The definitions for x,M and V are similar, but there is a little asymmetry here. u= 4.5 U= 4+5 M= u+v = 4.5 + 3.4 + 3.5 x= 3.u = 3.4.5 V= 3+U = 3+4+5 v= 3.U = 3(4+5)= 3.4+3.5 again, x= 3.4.5 // minimum of 3,4,5 M= 3.4 + 3.5 + 4.5 // mid element of sort(3,4,5) V= 3+4+5 // maximum of 3,4,5If we wanted to make a circuit out of these, it would look like this:

Here the L shaped gates are minmax gates. The upper output is the minimum and the lower output is the maximum for the two inputs on the left.

x1= A1.A2 x2= A1+A2 x3= A0.x1= A0.A1.A2 -> minimum = L x4= A0+x1= A0+A1.A2 x5= x4.x2 = (A0+A1.A2)(A1+A2) = A0.A1 + A0.A2+ A1.A2.A1 + A1.A2.A2 = A0.A1 + A0.A2 + A1.A2 -> mid value= K x6= x4+x2 = A0+A1.A2 + A1+ A2 = A0 + A1 + A2 -> maximum = ZNow, we can replace these gates with compex operations. This is because we no longer need the values A0,A1 and A2 for the rest of the computation. They can simply be replaced by L,K and Z. Also, the size of the circuit is the same as the number of inputs. This makes it possible to replace the circuit with a permuting network.

OK. I'm not really confident with my final equations, they look weird. I'll try to derive them again, but doing the replacements from the very start. Here are the new namings:

a= 0.1.2 d= 3.4.5 b= 0.1+0.2+1.2 e= 3.4+3.5+4.5 c= 0+1+2 f= 3+4+5 G[F,5]= 0+1+2+3+4+5 = c+f [FINAL] G[F,4]=Finally, I got it right this time, I guess:0.1+ 0.2+ 0.3+ 0.4+ 0.5+1.2+ 1.3+ 1.4+ 1.5+ 2.3+ 2.4+ 2.5+ 3.4+ 3.5+ 4.5 = b +0.3 + 0.4 + 0.5 +1.3+ 1.4+ 1.5+2.3+ 2.4+ 2.5+3.4+ 3.5+ 4.5 = b + 0(3+4+5) + 1(3+4+5) + 2(3+4+5) + 3.4+ 3.5+ 4.5 = b +0.f+1.f+2.f+ 3.4 + 3.5 + 4.5 = b + (0+1+2)f + 3.4+3.5+4.5 = b + c.f + e = c.f + b + e [FINAL] G[F,3]=0.1.2+ 0.1.3+ 0.1.4+ 0.1.5+ 0.2.3+ 0.2.4+ 0.2.5+ 0.3.4+ 0.3.5+ 0.4.5+ 1.2.3+ 1.2.4+ 1.2.5+ 1.3.4+ 1.3.5+ 1.4.5+ 2.3.4+ 2.3.5+ 2.4.5+3.4.5= a+d+ 0.1.3+ 0.1.4+ 0.1.5+ 0.2.3+ 0.2.4+ 0.2.5+0.3.4+ 0.3.5+0.4.5+1.2.3+ 1.2.4+ 1.2.5+ 1.3.4+ 1.3.5+ 1.4.5+ 2.3.4+ 2.3.5+ 2.4.5 = a+d+ 0.1.3+ 0.1.4+ 0.1.5+ 0.2.3+ 0.2.4+ 0.2.5+0.(3.4+ 3.5+4.5)1.2.3+ 1.2.4+ 1.2.5+ 1.3.4+ 1.3.5+ 1.4.5+ 2.3.4+ 2.3.5+ 2.4.5 = a+d+0.e+ 0.1.3+ 0.1.4+ 0.1.5+ 0.2.3+ 0.2.4+ 0.2.5+ 1.2.3+ 1.2.4+ 1.2.5+ 1.3.4+ 1.3.5+ 1.4.5+2.3.4+ 2.3.5+2.4.5= a+d+0.e+2.e + 0.1.3+ 0.1.4+ 0.1.5+ 0.2.3+ 0.2.4+ 0.2.5+ 1.2.3+ 1.2.4+ 1.2.5+1.3.4+ 1.3.5+1.4.5= a+d+0.e+2.e +1.e0.1.3+ 0.1.4+ 0.1.5+ 0.2.3+ 0.2.4+ 0.2.5+ 1.2.3+ 1.2.4+ 1.2.5 = a+d+e.(0+1+2)0.1.3+ 0.1.4+ 0.1.5+ 0.2.3+ 0.2.4+ 0.2.5+ 1.2.3+ 1.2.4+ 1.2.5 = a+d+e.c+0.1.3+ 0.1.4+ 0.1.5+0.2.3+ 0.2.4+ 0.2.5+1.2.3+ 1.2.4+ 1.2.5 = a+d+e.c+ 3.(0.1+0.2+1.2) + 0.1.4+ 0.1.5+ 0.2.4+ 0.2.5+ 1.2.4+ 1.2.5 = a+d+e.c + 3.b + 4.b + 5.b = a+d+e.c + (3+4+5).b = a+d+e.c + f.b = a + b.f + c.e + d [FINAL] G[F,2]= 0.1.2.3+ 0.1.2.4+ 0.1.2.5+ 0.1.3.4+ 0.1.3.5+ 0.1.4.5+ 0.2.3.4+ 0.2.3.5+ 0.2.4.5+0.3.4.5+1.2.3.4+ 1.2.3.5+ 1.2.4.5+1.3.4.5+2.3.4.5=(0+1+2).3.4.5 +0.1.2.3+ 0.1.2.4+ 0.1.2.5+ 0.1.3.4+ 0.1.3.5+ 0.1.4.5+ 0.2.3.4+ 0.2.3.5+ 0.2.4.5+ 1.2.3.4+ 1.2.3.5+ 1.2.4.5 = c.d+0.1.2.3+ 0.1.2.4+ 0.1.2.5+ 0.1.3.4+ 0.1.3.5+ 0.1.4.5+ 0.2.3.4+ 0.2.3.5+ 0.2.4.5+ 1.2.3.4+ 1.2.3.5+ 1.2.4.5 = c.d+0.1.2(3+4+5)+0.1.3.4+ 0.1.3.5+ 0.1.4.5+ 0.2.3.4+ 0.2.3.5+ 0.2.4.5+ 1.2.3.4+ 1.2.3.5+ 1.2.4.5 = c.d+ a.f +0.1.3.4+ 0.1.3.5+ 0.1.4.5+0.2.3.4+ 0.2.3.5+ 0.2.4.5+1.2.3.4+ 1.2.3.5+ 1.2.4.5 = c.d+a.f+(0.1+0.2+1.2).3.4 0.1.3.5+ 0.1.4.5+ 0.2.3.5+ 0.2.4.5+ 1.2.3.5+ 1.2.4.5 = c.d+a.f+b.3.40.1.3.5+ 0.1.4.5+0.2.3.5+ 0.2.4.5+1.2.3.5+ 1.2.4.5 = c.d+a.f+b.3.4+b.3.5+ 0.1.4.5+ 0.2.4.5+ = c.d+a.f+b.3.4+b.3.5+b.4.5 = c.d+a.f+b.(3.4+3.5+4.5) = c.d+a.f+b.e G[F,2]= a.f + b.e + cd [FINAL] G[F,1]=0.1.2.3.4+ 0.1.2.3.5+ 0.1.2.4.5+0.1.3.4.5+ 0.2.3.4.5+ 1.2.3.4.5 =a.3.4+ a.3.5+ a.4.5++ 0.1.3.4.5+ 0.2.3.4.5+ 1.2.3.4.5 =a.3.4+ a.3.5+ a.4.5++ 0.1.3.4.5+ 0.2.3.4.5+ 1.2.3.4.5 = a(3.4+3.5+4.5)+ 0.1.3.4.5+ 0.2.3.4.5+ 1.2.3.4.5 = a.e + 0.1.d + 0.2.d + 1.2.d = a.e + d(0.1+0.2+1.2) = a.e + d.b = a.e + b.d [FINAL] G[F,0]= 0.1.2.3.4.5 G[F,0]= a.d [FINAL]

a= 0.1.2 d= 3.4.5 b= 0.1+0.2+1.2 e= 3.4+3.5+4.5 c= 0+1+2 f= 3+4+5 G[F,5]= c+f G[F,4]= b+cf+e G[F,3]= a+bf+ce+d G[F,2]= af+be+cd G[F,1]= ae+bd G[F,0]= adNow we need to find terms x,y such that they appear in both x.y and x+y form in the above equations. When we find them, this means that they are going to be inputs to a minmax gate and their outputs will be used in the above formula.

Here are the pairs:

- c and f
- b and e
- a and d

G[F,5]=As you can see above, not all terms are covered by these gates. For example, bf+ce in G[F,3] doesn't include any of our gate outputs. However, maybe we can manipulate these to get better expressions:c+fG[F,4]=b+cf+eG[F,3]=a+bf+ce+dG[F,2]= af+be+cd G[F,1]= ae+bd G[F,0]=ad

G[F,3]= a+bf+ce+d = a+d + X X= bf+ce since we know that b<c and e<f, b= bc and e= ef X= bc.f+c.ef = cf(b+e) G[F,3]= a+d + cf (b+e)Done! We expressed G[F,3] in terms of the outputs of our three gates as described before. Now let's move on to the others

G[F,2]= af+be+cd = be + af + cd we know that a<c and d<f : a= ac , d= df = be + ac.f + c.df = be + cf(a+d) G[F,1]= ae+bd again, a<b and d<e so, a=ab and d= de =ab.e + b.de = be(a+d)Here are the final results:

G[F,5]= c+f G[F,4]= b+e + cf G[F,3]= a+d + cf(b+e) G[F,2]= be + cf(a+d) G[F,1]= be(a+d) G[F,0]= adNow we observe that, if we put a,d in a gate and obtain the minimum and maximum results, we will no longer need the values a and d anymore. So, this little gate will act like a small sorting network, replacing the input values. Let's say that

ad = T and a+d = U G[F,5]= c+f G[F,4]= b+e + cf G[F,3]= U + cf(b+e) G[F,2]= be + cf.U G[F,1]= be.U G[F,0]= TThe same can be done for other gates as well.

be= K and b+e= L G[F,5]= c+f G[F,4]= L + cf G[F,3]= U + cf.L G[F,2]= K + cf.U G[F,1]= K.U G[F,0]= TThe only remaining pair is cf,

cf= Y and c+f= Z G[F,5]= Z G[F,4]= L + Y G[F,3]= U + Y.L G[F,2]= K + Y.U G[F,1]= K.U G[F,0]= TSo, we can continue constructing our sorting network. We already had two little sorting networks which transformed (A0,A1,A2) to (a,b,c) and (A3,A4,A5) to (d,e,f). I'll start from the output of those:

These three operations don't use each other's values so they can be done in one step.

Here note that T and Z are used in only one place in the equations as the values of G[F,0] and G[F,5]. These are no longer necessary since we have already calculated G[F,0] and G[F,5]. Those lines will continue to the end of the machine, untouched.

We're left with four equations now. Unfortunately, there is no relation between Y and U. Their counterparts have gone outside our scope:

K<L G[F,4]= L + Y G[F,3]= U + Y.L G[F,2]= K + Y.U G[F,1]= K.U

This is 6 levels with 12 comparators. I'll now try this with my validator:

6 1.2/4.5/0.1/3.4/1.2/4.5/0.3/1.4/2.5/2.4/1.3/2.3It does work. How about I try it without the last comparison:

WARNING: invalid network 6 1.2/4.5/0.1/3.4/1.2/4.5/0.3/1.4/2.5/2.4/1.3 WARNING: invalid networkThis one didn't work, so I did have to do the last one. Maybe later I'll try to figure out what exactly got stored in G[F-1,2] and G[F-1,3].

When I got stuck, there wasn't much I could do anyway, I couldn't resolve the situtation with just one level of comparators. So this is indeed the best that could be done in this situation.

My initial choice of factoring by 3 elements turned out to be poor. According to Wikipedia, my network has one too many layers. I got the number of comparators right. 12 is listed as optimum for 6 inputs.

First, I wrote down the equations for the final stage of the network. i.e. what do the final values of G[] look like. From there, I tried to find variables x and y such that both x.y and x+y occur in the expressions. These were to be our very first gate inputs.

After that, I used the input variables Ax,Ay to find max(Ax,Ay) and min(Ax,Ay) and then expressed the final values in terms of these values.

By repeating the above process, I reached a point where I couldn't do it any more. At this point, I simply calculated what was easy to calculate and moved on.

The last stages of my construction could be thought of something like a search. However, it's not really a blind search, we have a lot of information about the values in G[x] at that point.

Maybe if I do this right from the beginning, I can end up with slightly better results. Let's do that. Try this with 2 variables at the beginning instead of 3. I did it, and you get stuck much earlier than the 3-variable version.

The problem is that, as you make more and more gates, you lose the symmetry in the equations. This causes some variables to occur many more times than the others and it becomes impossible to derive a clean gate expression from the equations.

- Is the reverse of a sorting network also a sorting network?
- Is it possible to move one compex gate to another place and still end up with a sorting network (possibly with less levels)? Naturally, we don't consider moves within the same level; they don't really have an ordering there.
- Some networks are equivalent to each other thru an input rename operation. i.e. if your not-so-sorting network sorts everything else correctly, but the 3rd and the last output are exchanged. In this case, you simply exchange the names of the 3rd and the last lines (you replace their names in the compex gates) and you get a sorting network. Therefore it's sufficient to show that G[F] contains all the necessary final expressions; they can easily be rearranged into the correct order.

If that is the case, then we don't really need to go thru the gate. The gate can be a NOP or just a shuffle operation. Like this:

Let U: (C,D) = COMPEX(A,B) be our gate, with index(A)=index(C)<index(B)=index(D) where index(V) is the name of the line. i.e. 0zix line, 1zix line etc. if A= MAX(A,B) U is a shuffle operation. A and B were already sorted, but in reverse order. If we rewrite all compex operations, up to this point, to replace A with B and B with A simultaneously, then A and B will occur in the correct order at this point of the network. Problem solved, we don't need the gate. if B= MAX(A,B) U is a NOP. We can remove it without any other operation.Useless gates may or may not occur in algorithmically generated sorting networks, it's a point to be investigated. However, there is another application for this.

Let's have a sorting network W. If we prefix this sorting network with a new compex operation U, then the resulting network K=U|W will still be a sorting network. W already sorts its input regardless of how they occur in the input. Adding U doesn't change that fact. Now, we may end up with some useless gates in K-U.

If that is the case, then we can remove the useless gates and potentially may end up with a sorting network with less levels. If there are multiple useless gates, then we have reduced the number of gates which is also good.

It turns out that the reverse of a sorting network is **not** always
a sorting network. I think I can still create a reproduction operator:

reproduce(A,B): C= A + reverse(B) loop C -= useless(C) R = reverse(C) if R is a sorting network C = R until no useless gates are found return CThis operator seems to be asymmetric since the child would probably look a lot like A rather than B. It's worth a try, but I should first make a proper network implementation before trying all these new ideas.

The useless operator simply looks at the inputs and outputs of each gate and determines whether it's really a NOP. The uselessness of a gate depends on the structure preceding it. So, modifying the beginning of a network will eventually cause some gates in the middle or end of the network to become useless. This doesn't happen so often at the beginning of the network since there isn't enough structure there to make gates redundant.

I have a variation on the useless operator, which also takes into account
the structure **after** the gate in question. It doesn't involve anything
fancy, it simply removes a gate from a sorting network and checks whether
it still sorts. It's somewhat expensive but it can do things
which are hard to prove correct by symbolic manipulation.

There is a variant to the prefixing operation too. If the last level contains a single gate, this is very inefficient. The single gate occupies a whole level by itself and you can't make it more efficient by putting more gates in that level because they will be useless. My rmlast operator tries every possible prefix gate and determines whether adding any of these gates to the start of the network makes that last gate useless. Failing that, it finds the prefix which makes the most number of gates useless. Even if the new prefix gate creates a new level for itself, it still makes sense to do it. The last level is going to be deleted when you find out that the last gate in it is useless. So, the number of levels will still be the same, but we will have a single gate in a level by itself at the beginning of the network rather than the end. The beginning of the network is much more open to manipulation because there is no structure there yet. Almost nothing you add there will be useless.

Another fun one is the 'densify' operator. This one inserts random gates troughout the sorting network. It does this according to these rules:

- The resulting network should still be a sorting network.
- The added gates can not create new levels. They must fit into the levels they were created into.

For densify to work, there should be empty slots in the levels. There we counter the problem of partitioning the network into levels. You can divide a network into levels in many ways, you can agressively pull everything towards one side, as far as they can go, or you can try to balance the levels by minimizing the difference in size between neighbouring levels.

In my implementation I did only two, aggressive pull left and aggressive pull right. These are called normalize and rightnormalize. When you 'normalize' a network, some empty slots will be created at the end of the network. densify then can fill in these slots randomly and make some inefficient gates into useless gates. If you do a rightnormalize, space will open up in the beginning of the network. If you then apply densify in this area, the new gates can affect the rest of the network succeeding them. It's quite hit and miss I can't give a good guideline for this.

I also tried out the reproduction stuff. It does work as I expected but the result is usually worse than what you start with. It has the effect of thinning out the last levels of a network. If it's densely populated by bubble-sort style gates at the end, applying this will remove quite a bit of them at the expense of adding more levels. When you stop the reproduction and cut the network from where it has already sorted the input, the resulting network has more levels at the end than the original one. Needless to say, this operator has very little effect on the beginning of the network.

In my current implementation this operator has some bug which stops me from investigating it further. If I can fix that, maybe it can help thin out the number of the gates at the end, which is a quite difficult task for other operators.

In any case, if you start searching from an already well made network like a mergesort network, it takes a lot of time to reach the optimal solution because the network resists change so much. If you start with a completely random solution or something like bubble sort, then it converges faster and has more chance of getting to the optimum. Here is my current implementation as of 20160725.

In addition to this, I fixed the bug in the reproduction operator. repromax does the reproduction as explained previously. It does the flip-remove-useless step as many times as possible. When I was observing it by doing one step at a time, I was under the impression that the operator would thin out the later stages. It's not so. Sometimes it results in very dense stages with almost all slots filled in. It can quickly go from a single gate at the end to a vertical wall.

I also tried to find shuffle gates with no avail. Shuffle gates are similar to useless gates. In useless gates, there is no difference between the input and the output, the inputs arrive at the gate already sorted. In shuffle gates, the inputs arrive sorted in reverse order and this can be shown algebraically. I haven't seen any shuffle gates while I was playing around.

Anyway, I have some operators to work on now, at least. They converge very fast towards the optimum (and also away) but it takes a long time to reach the actual optimum solution. Here is the latest implementation.

Instead of this, now I append a whole vertical wall to the end
of the network. A wall is a level which contains compex(2i,2i+1)
for all i&leN/2-1. This can remove a lot of gates and potentially
merge some levels to recover the level we just added. For this
purpose, the operator **ScanFwd** can be used. This is just
like TryDel but scans from the front of the network to the end.
Currently, ScanFwd skips the first level and can't remove anything
from the front but I will fix that later.

Similarly you can add a wall to the front of the network. This will also create some deletable nodes in the same manner. ScanFwd was in fact created for this operation; try to remove gates except for the ones at the front (we just added them).

In some cases, it's possible to swap two last gates and still end up with a sorting network. Like this for example:

----o------- --------o---- | | ----o----o-- to ---o----o---- | | ---------o-- ---o---------In such a case, I suspect that the middle line is already the mid value of the three and these two gates are there just to compare the first and the last lines. If that's the case, then we can replace the whole thing by one gate. This needs to be investigated.

I did this and found that I was wrong. Even if the swapped network sorts,
both gates still have to be present. Since this configuration doesn't
resulting in anything being removed, I generalized the swap operation to
all gates. The **SwapSimple** operator swaps two gates at random (which
are not in the same level) and checks whether the resulting network sorts.
Surprisingly, this operator finds a swappable pair most of the time. Sometimes
it has to go thru several iterations but it succeeds still. The effect of
this operator is to change the number of levels for a given number of gates.
Occasionally it can expose some useless gates if you have too many to start
with.

This operator can be used in a search as follows:

try_swap(N): P= swap(N) if P.levels < N.levels return P return remove_useless(densify(P))When the search sets a record for number of nodes, I could try doing all the swaps to see whether there's a network with the same number of nodes and less number of levels.

Halving a sorting network results in a sorting network for less number
of inputs. Sometimes you can even reach the optimum this way. For the
converse, I implemented a **Double** operation. If the number of inputs
was N, I first create a network of size 2N. Then I sort two halves using
the current network. After that, I sort the middle N values using the
current network. Finally, I sort the two N-sized halves using the current
network again. The final network sorts if N was even but it doesn't work
for odd N. This is understandable. For odd N, the 'middle N' can be determined
in two ways and you end up with some unprocessed line whichever you choose.

The third installment of the sorting network tool is here.

This is the plan: I will maintain the expressions for the outputs in my algebraic format. Together with it, I will maintain a relation between input lines. This relation will tell me which line is known to be lesser than which. When I add a gate to the network I'll do two things: rewrite the expressions in terms of the new values and update the relation.

Let's assume that I insert a gate (a,b)=>(a',b') into the network where a'=min(a,b) and b'=max(a,b). I will rewrite each expression in U (the final stage of G[] above). Let's consider the expression U[i]. I'll do the following:

- If a term in U[i] contains a.b, then I'll replace that with a'. We already know that a'=a.b. We're done with this term now.
- If a term doesn't contain a or b, I'll leave it alone.
- I'll list terms which contain a only (from {a,b}) and those with b only. If the intersection of these sets is K, then the sum of the terms in two lists will be K.(a+b)= K.b'. I'll put this in the new expression.
- I'll subtract K from the two lists. If there are remaining elements which contain only a or only b, I don't know how I will resolve it. Maybe I can use some tricks I used above. One option is to abort this gate.

a'= a.b b'= a+b 1: a' < b' 2: k<a ∧ k<b ⇒ k<a' ∧ k<b' 3: k>a ∧ k>b ⇒ k>a' ∧ k>b' 4: a<k ⇒ a'<k 5: k<b ⇒ k <b' 6: k<a ⇒ k ?a' but k <b' 7: b<k ⇒ b'?k but a'<kFrom the above equations, we can see that the number of entries in the relation will increase by one each time we add a gate.

In case the gate is aborted, we can try another. This will constitute a search, which is nice. The nice thing is that, as I add more gates, the final expressions will become simpler, taking up less space. If I can somehow find a way to compress the initial expressions, maybe I can reduce the whole memory requirement by a magnitude. I still don't see how I could make a 64-input network using this but maybe the relation by itself will provide enough information to get by.

Each time I successfully make a gate, I will replace a' by a and b' by b, keeping the number of lines constant. When I do that, I will increase a counter for each a and b, which represents the complexity of the value. When looking for new gates to make, I can combine the least complex values so that the network proceeds in a balanced fashion.

When I get completely stuck, I can fall back to my original trick: just compute whatever is easy to compute and try to recover from that. I could even stop the generation process there and continue with a purely search based approach.

In order to speed up the process, I could even start out with a fixed layer such as a wall. Then, I could generate the final expressions based on the outputs of the wall in the first place, reducing the expression size considerably. It is even possible to do things in parallel. The expressions around the middle of the network are the biggest and they could be split up into chunks, which would be processed in isolation at first.

I had designed the ordering relation as a secondary tool to keep track of which lines are less then which lines. However, as I experiment with the program, I realize that the relation is much more useful than I thought especially for rewriting the final expressions.

In my manual generation effort, I had used the following simple rules for rewriting the final expressions. I implemented these in the program, and was able to deduce one further rule. Here are the simple ones:

Let us make a gate (a,b) => (a',b') and C be a term (minimum of some elements) which doesn't contain a or b. We then can rewrite:

After renaming, these two expressions become C.a and C.b respectively.Now I have thought of a new rule, let's look at it:

We have C.x.b where x < a according to R we can rewrite this as: C.x.a.b since x = x.a (because x<a) this is equal to: C.x.a' and after renaming: C.x.a ( this a is not the same thing as the previous a. this one is the value of a after the gate has finished running). Let's think about this: x.b a' x.a' -------------- x<a<b x a x [[ since x<a and a<b => x<b ]] x<b<a x b x b<x<a b b bI will call this RULE-4. The same thing applies to C.x.a:

We have C.x.a x < b C.x.a = C.x.b.a = C.x.a' after renaming, it's C.x.a again.Let's call this RULE-5 in the code.

OK one more rule (RULE-6). This was actually something I had used when I was manually building a sorting network. The idea is to find a common factor between two terms and combine them, just like I did for the simple RULE-3. In this case, the common factor doesn't occur naturally inside the two terms. It has to be derived from the terms by adding redundant elements.

rule6(C.a, D.b, a, b) F= maximize(C.a) G= maximize(D.b) R= F & G if R contains C and D, then rewrite C.a+D.b as R.b' This is called R.b after renaming.Consider the following case:

Ordering Relation 012345 0.<<... 1>..... 2>..... 3....<< 4...>.. 5...>.. rewriting for gate(0,3): 04+13 Here, neither 04 nor 13 can be rewritten using previous techniques. However, we can do the following according to RULE-6: 04 = 014 13= 134 04+13= 014 + 134 = 14(0+3) = 14.3'It's worth noting that R should be reduced to its minimum, in order to avoid making expressions bigger than they need to be.

When applied to a partial expression, I will try all pairs of terms in order. When two terms get combined like this, I will record the result and mark both terms as removable. However, I won't actually remove them from the expression since one term may combine with two or more terms and might be needed later.

After all terms are processed in this way I remove the ones marked as removable.

Another transformation I apply to expresssions is as follows:

if E = A + B + C and B > A, then E= B + CThis is called RULE-8. Finding out whether a complex term B is greater than A is not trivial. I do it in the following way:

- Let A = a
_{0}.a_{1}...a_{n} - Let B = b
_{0}.b_{1}...b_{m} - For b
_{j}, if there is some a_{i}such that b_{j}>a_{i}, then we can append b_{j}to A without changing its meaning, i.e. A' = A.b_{j}= A - If b
_{j}is included in A, then we don't need to find some such a_{i}, since we can still append b_{j}to A without affecting its meaning. - When you do the above for all b
_{j}, then we will have A' = A.B = A. - By RULE-1, B + A.B = B.

Right now the program is in a state similar to a manual-search engine. There is no backtracking but all candidate gates are considered until I find a successful one.

If I make a search based on this algebraic approach, two factors will affect the space searched and the order it's searched. First, the set of simplification rules will determine which gates can be created from analysing the final expressions. As you add more rules, the final expressions become simpler. These simpler expressions lend themselves more easily to creation of gates when compared to more complicated expressions.

Second, the order in which the candidate gates are tried affects the order the space is searched. I've tried a couple of criteria for this. If I prioritize gates for which the distance between the inputs is smallest, the program ends up with bubble sort. If I do the reverse, but also insist that the inputs be of the lowest complexity (has gone thru minimal number of gates so far), then the program creates the beginnings of merge sort, but then deviates.

Even the algorithm used for sorting the candidate gates affects the results. An unstable sort seems to do better because a stable one will leave the candidate gates in the order it finds them and that tends to concentrate the effort on the top side of the network, leaving the rest mostly unmodified. This quickly results in irreducible final expressions.

The sorting criteria for candidate gates could be varied as the network progresses. I could first favor gates with distant input lines and then switch to favoring gates near the edges etc.

Doing a search using the expressions as a guide seems pretty reasonable, but it would be quite slow since each gate entails a lot of rewriting work. Even if I make a much larger set of simplification rules, a search is mandatory because often the networks created have too many levels. Up to 8 inputs, the output has optimal size, but the program overshoots by one gate for 9 inputs. I haven't solved the 10 input case with it, this requires backtracking.

The relation also contains a lot of information. Most of this information isn't available from the expressions themselves because the expressions in the network or in the outputs get transformed radically with each gate. Processing them algebraically without the aid of the relation becomes more and more fruitless as they get simpler. Maybe it could be a good idea to perform a search using the relation as the only guide. I could forgo the whole algebraic system and proceed with the entries in the relation only.

Here is an open question, QUESTION-1. Starting from an empty network and an empty relation, if we apply a sequence of gates to the relation according to RELATRAN and end up with a full relation, does the sequence of gates necessarily form a sorting network? I have suspicion that not-sorting networks would also create full relations occasionally. If that wasn't the case then verifying that a network sorts would become a polynomial problem rather than an exponential one and somebody would have found it by now. This is still worth consideration, maybe I could enumerate all networks which end up with a full relation and see if any of them doesn't sort, for small N of course, like 5 or 6.

So, first thing is to make a backtracking search program based on the relation only. I will use the expression based verifier. Initially, I will use the search program to find ALL networks with full relations, just to progress the above idea. Later on, I'll use pruning to skip over things when they have too many gates/layers. In any case, here is the latest version with all the algebraic tricks.

Let's look at an example: For 5 inputs, the program creates a network with 10 gates. After that, it can't find anything shorter and exhausts the search space. Quite unusual. When I later apply the TryDel operator I discussed above, I find a deletable gate. Depending on the way I sort gate candidates, I end up with different sorting networks with the same number of gates. Some of these networks can be processed by TryDel to achieve an optimal sized network. Some don't. When I find an M-sized network, I skip all other networks that are longer than or equal to M-gates. This is the reason why I can't find such an optimizable network every time I change the ordering of the candidate gates.

In any case, search using only the relation isn't very productive, but there is a positive outcome. It seems that the answer to QUESTION-1 is "Yes?". Up to 11 inputs I never found any network which provides a full relation but doesn't sort. At 16 inputs I got one but I have doubts about the verifier because the expressions get quite large and I might have some memory issues there. It just seems obvious that a full relation should imply a sorting network but I need to prove it before I can use it as a verifier. I still have doubts, if it was this simple somebody would have thought of it by now. There must be something I'm missing. Maybe the 16-input case was correctly identified? I don't know I should look into this.

**Update:** I found a bug in the verifier. I wasn't computing number of
combinations correctly for N≥13 because the factorials overflowed. Now I
do it with proper factorization etc. and it seems that the answer to QUEST-1
is more likely to be YES. I have run the program with different number of
inputs for quite some time and haven't seen any network where the relation
is full but the network doesn't sort. This update isn't included in sor5 below.

Since searching by relation only isn't very productive, I have to go back to
my algebraic approach. In the manual search using the algebraic approach,
I had one more algorithm which I didn't mention. After forming a gate, I
look at the expressions in the **network** (not the final expressions)
to find out which line is less than which line. By doing these manipulations,
I was able to find out more entries for the relation. I just made a test with
6 inputs and found that 12 entries in the relation were recovered in this
way. Keeping a network is somewhat expensive because the expressions in it
become more and more complicated as the network progresses. However, this
might be the only way to move forward.

The relation-only search program is here. I have another idea for this avenue. It's highly unlikely to yield anything but I could give it a try if run out of options. Instead of overwriting elements a and b, I could append new elements a' and b' into the relation and keep a and b as is. Maybe this could eliminate some of the information loss.

S= { 0: 0000 1: 0001 2: 0010 3: 0011 4: 0100 5: 0101 6: 0110 7: 0111 8: 1000 9: 1001 10: 1010 11: 1011 12: 1100 13: 1101 14: 1110 15: 1111 } S_01= { 0, 1, 3, 4, 5, 7, 8, 9, 11, 12, 13, 15 } S_01_23= { 0, 1, 3, 4, 5, 7, 12, 13, 15 } S_01_23_02= { 0, 1, 3, 5, 7, 9, 13, 15 } S_01_23_02_13= { 0, 1, 3, 5, 7, 15 } S_01_23_02_13_12= { 0, 1, 3, 7, 15 }The result is interesting. It consists of words of type 0

S_01_23_02_03= { 0, 1, 3, 5, 7, 9, 13, 15 }OK. It looks like I messed up doing it manually. Here is the real run for the optimal network:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 1 0 2 3 4 6 7 8 10 11 12 14 15 2 3 0 2 3 8 10 11 12 14 15 0 2 0 2 6 8 10 12 14 15 1 3 0 8 10 12 14 15 1 2 0 8 12 14 15 -1 -1 0000 1000 1100 1110 1111After we got S_01_23_02, we could do the following moves: 01, 02, 03, 12, 13, 23.Let's look at all these:

S_01_23_02 = 0 2 6 8 10 12 14 15 S_01_23_02_01= 0 2 6 8 10 12 14 15 S_01_23_02_02= 0 2 6 8 10 12 14 15 S_01_23_02_03= 0 2 6 8 10 12 14 15 S_01_23_02_12= 0 4 6 8 12 14 15 S_01_23_02_13= 0 8 10 12 14 15 S_01_23_02_23= 0 2 8 10 12 14 15Here, the optimal gate stands out very nicely. Some gates are seemingly useless and the other sub-optimal ones reduce the set less. I wonder if this could be carried on to higher number of inputs.

The ordering relation looks like the following after we reach S_01_23_02:

0123 0.<<< 1>... 2>... 3>...The uselessness of the gates involving input 0 is obvious from the relation.

Now I'm thinking of making a search based on this. I will use the relation as the guide again but this time, I'm hoping that useless gates will be found immediately and their subtrees will not be searched.

Now I have a new open question: QUESTION-2.

If a candidate gate (a,b) doesn't modify any of the input values, can we insert a<b into R?Anyway, back to the augmented search. The search will be guided by the partial ordering relation but as I add gates, I will also compute the set of input integers after the gate. For instance, after the gate 01, no input values will have 1 for the 0zix bit and 0 for the 1zix bit. So, the possible values for this pair will be 00, 10 and 11.

If I start the network with a vertical wall, these values will apply to all pairs within the wall. Therefore, I will have only 3^8=6561 different input values past the wall for 16 input lines. I can even use this for verification, which will be much faster.

It turns out that my pruning implementation for level permutations was incorrect. When I turned that off, the search finds optimum networks for 5,6 and 8 inputs almost instantly. For 7 inputs, it gets stuck at 17 gates 7 levels for a long time. I didn't continue searching further.

The networks found for 6 and 8 inputs are interesting. They are composed of optimum sub-networks, followed by some merging network. In fact, the 8 input network is completely symmetric with regards to the middle line of the network, drawn horizontally. The only starting gates were that of a wall. Now I wonder if I could start with two optimal sub-networks and then search for the merging part only. This would reduce the number of valid inputs past the starting condition even further.

If we had 16 inputs, we would divide it into two halves of 8 inputs each. These have 9 distinct outputs. So, the start of the merging network would have only 9*9=81 input values. This would make for a super fast verifier.

Also, the answer to QUESTION-2 seems to be YES. I did insert the mentioned ordering into the relation when I saw the condition and it didn't break anything. Typically, when I do something unproven, it results in a lot of networks which should sort but don't. In this case, this never happened in the area I covered. Of course this needs further investigation but I can use it for now.

Here is the 6th installment, 15 September 2016.

Given some fixed gates at the beginning, we can reduce the number of necessary input sequences to test. This results in a very fast verifier. I have implemented one case of fixed gates: a vertical wall. Another case is the combination of two optimal sub-networks.

So, maybe it's time I give up completely on algebra based networks. The input
set based
verifier can be used just as effectively, provided that I prefix the network
with some gates. Even if the prefix is empty, the worst case is the same as
the algebra based verifier: 2^{N} space and time.

I implemented a new prefix for search: subnet. The input lines are divided into half and each half is sorted by a known optimal network. After that, I search for the rest of the network. This has yielded a very nice result for 7 inputs. For larger input values, this typically approaches the optimum faster, but still requires a lot of search to actually get there. For instance, the 8-input problem finds the optimum solution almost immediately when starting from a wall. However, it struggles to get there with this new prefix. This is interesting because the wall-based solution already contains two 4-sorters at the very beginning.

My previous algorithm for pruning permutations within each level was incorrect and skipped over many legitimate areas to search. Therefore, it failed to find the optimum solutions. I have a new one. When I add a gate to the network, I check whether this one forms a new level. If so, I continue with the search. Otherwise, I check whether the gates in the level are sorted according to their upper input line. If not, then I abort this gate because the current level is just another permutation of something we already saw or will see. In this check, I don't include the first gate in a level because if we allow the first gate to change, then the level allocation for the whole network could change, making it impossible to reach some configurations.

In any case, currently I'm searching K different orderings for each level. Without this pruning I was searching K! orderings. This is quite a speed up if not perfect. The program exhausts the 7-input search space, prefixed with subnets, in a very short time. For others the effect isn't so visible.

The latest program is here, dated 19 September 2016.

If you remove the top or bottom line (or both) from a sorting network, the result is also a sorting network.This is quite obvious: just assume that you feed the original network with an element less than every other value possible at position 0. The 0zix element would never move and the rest would become sorted since the network sorts EVERYTHING. If you now simply remove the first line, then the rest of the network is still a sorting network.

When you apply this to an **optimal** sorting network, a strange thing
happens: the result is also optimal, after removing useless gates. I verified
this from 8 down. The thing is, although an optimal network reduces to another
optimal network, the reverse isn't necessarily true. If I start with a random
optimal network for N inputs, I can't be sure that I'll end up with an optimal
network for N+1 inputs if I simply add gates to it. Maybe the ones which
give this result are in a seperate class of optimal networks.

So, here is the new deal: I will try to find an optimal network by adding a new line to a lesser network. This will require me to do a search on all levels. i.e. the network no longer changes only at the end, but everywhere. In order to do this efficiently, I should implement my inputset based verifier.

I was too quick to generalize my observation above. If start with a random optimal network, removing the top or bottom of the network doesn't necessarily form another optimal network. The reverse isn't true either. If you start with a random optimal network, add an input line but limit the depth of the network, you aren't even guaranteed to find a sorting network, let alone an optimal one. Here is the 8th version implementing this idea.

I've found that even though adding one line to an optimal network
may not yield useful results, adding **two** lines sometimes does.
However, this results in a much larger search space, especially if I
also add a level in the process. This search takes too long, going thru
all networks is just not feasible. So I have to prune by gate numbers here also.

The current method of generating levels is fine, but adding pruning into there might be problematic because it's already quite complicated. So, I need a new way of generating the subsets. I think I should also prune useless gates, it will cut down a lot of search time. I could use my relation technique for this. Although incomplete, it provides quite a bit of information. When you find out that a gate is useless (thru processing the input for it) and then add that information to the relation, then the relation also has full information (enough to be a verifier).

So here is my dilemma, if I make a level-based search I can skip all the permutations of the level. This is a must and I can't go back. However, the relation works on a gate-by-gate basis. I have to somehow generate the levels consistent with both models. Instead of simply storing numbers for gates, I could make more complicated data structures for them which could hold the relation stuff also. When I add the relations, the levels become dependent on each other since now they affect each others valid search space. This is good but complicates the search. I think the best way to go about this would be to incrementally generate all combinations for a level when a previous level changes. So, here is the new idea:

If we have a network of levels L0, L1, .. Lm, then we proceed as follows.

- Step Lm and check for sorting network.
- If Lm can't be stepped any more, step L(m-1) and reset Lm accordingly.
- If L(m-1) can't be stepped either, go back and step L(m-2).
- If you go back as far as L0 and can't step any more, search has ended.
- If you are able to step at any Lk, then init L(k+1) .. L(m) accordingly.

- Each level will be an array of gate objects.
- Each gate object will contain the relation resulting from the execution of the gate.
- If we have gates A,B,C,D,E available they will be used in the
following order, assuming a maximum count of 3:
- A, AB, ABC, ABD, ABE, AC, ACD, ACE, AD, ADE, AE,
- B, BC, BCD, BCE, BD, BDE, BE,
- C, CD, CDE, CE,
- D, DE

- When I decide to go forward with a gate list for the level, I will compute the linelist resulting from it. If I find that the last gate is useless, then I will abandon this gate list as well as all super-sequences. For instance, if I find out that 'B' is useless when trying to proceed with 'AB', then all the subtree starting with AB (ABC, ABD, ABE etc) will be cancelled.
- The maximum count for a level will be determined from the
limit for the search.
ngates(L)= number of gates for level L nfixed(L)= number of fixed gates for level L maxcount(Li)= limit - ngates(0) .. - ngates(L(i-1)) - nfixed(Li) .. - nfixed(Lm)

- I don't think I will ever see this, but if a fixed gate turns out to be useless, then I abandon this level completely and go back one level. I'm interested only in networks which contain the initial state. If I started modifying fixed gates, then this would deteriorate to a full-space search, which I'm trying to avoid here.

So, here is my new level generator algorithm:

- The main operation is the addition of a gate to the level.
- This will fail occasionally, for two reasons:
- I might have run out of gates to choose from.
- I might reach the maximum count for the level.

- When such a failure occurs, I move the last element forward.
- If this fails also, I will remove the last element and try to move the previous element forward.