perm filename LABEL.PUB[DOC,LMM] blob sn#056032 filedate 1973-07-31 generic text, type T, neo UTF8

COMMENT ⊗ VALID 00017 PAGES RECORD PAGE DESCRIPTION 00001 00001 00003 00002 .<< pub macros, declarations >> 00006 00003 . << title page >> 00007 00004 .TITL INTRODUCTION 00011 00005 .TITL PART A: DEFINITIONS 00017 00006 .HD PERMUTATIONS AND PERMUTATION GROUPS 00025 00007 .TITL PART B: SOLUTION TO THE LABELLING PROBLEM 00027 00008 .HD A Better Method 00033 00009 .HD LABEL RECURSION 00036 00010 .HD ORBIT RECURSION 00041 00011 .HD SPECIAL CASES 00045 00012 .HD FINAL TECHNIQUE 00051 00013 .TITL PART C: SUMMARY OF LABELLING STEPS 00054 00014 .TITL PART D: EXTENDED EXAMPLES 00057 00015 .HD |METHOD β(see Figure 11)| 00061 00016 .HD Labellings on the Octahedral skeletal framework. 00068 00017 .TITL ACKNOWLEDGEMENTS 00069 ENDMK ⊗; .<< pub macros, declarations >> .DEVICE XGP . .PAGE FRAME 56 HIGH 75 WIDE .AREA TEXT LINES 4 TO 54 ; .TITLE AREA HEADING LINES 1 TO 3 ; .TITLE AREA FOOTING LINE 56 ; . .FONT 1 "BDR25.FNT[DOC,LMM]"; .FONT 2 "BDI25"; .FONT 3 "NGB25"; .FONT 4 "FIX25"; .FONT 5 "FIX13X"; .AT "β" ⊂ TURN ON "%{" }%1{TURN OFF⊃; .AT "λ" ⊂ TURN ON "%{" }%2{TURN OFF⊃; .AT "α" ⊂ TURN ON "%{" }%3{TURN OFF⊃; .AT "ε" ⊂ TURN ON "%{" }%4{TURN OFF⊃; .AT "~" ⊂ TURN ON "%{" }%5{TURN OFF⊃; . .<< footnote macros >> .FOOTSEP←"------------------"; .COUNT REFERENCE INLINE PRINTING "↑1" .AT "<<" LABEL ":" TEXT ">" ⊂ .LABEL: NEXT REFERENCE!;!; .SEND FOOT ⊂ .PREFACE 1; SPREAD←1; INDENT 0,0; TURNON "{"; {REFERENCE! LABEL}TEXT .TURNOFF .BREAK ⊃ ⊃ . .COUNT FIGGER INLINE PRINTING "Fig. 1" .MACRO FIGURE(NUM,SIZ,FIGHEAD) ⊂ .BEGIN NOFILL;NEXT FIGGER!; .!}α→→→ Figure NUM, FIGHEAD ←←←←←β .END⊃ . .<<label macros >> .MACRO HD (HEADING) ⊂ .BEGIN IF LINES<10 THEN SKIP TO COLUMN 1 ELSE SKIP 2; NOFILL;BREAK; αHEADINGβ .END; ONCE PREFACE 0;⊃ . .MACRO TITL (TITLE) ⊂ .BEGIN IF LINES<10 THEN SKIP TO COLUMN 1 ELSE SKIP 2; BREAK; CENTER; αTITLEβ .END;⊃ . .MACRO GRAF (HDNG) ⊂ αHDNG.β .⊃ . .MACRO I6 ⊂ INDENT 6,6,6; SPREAD←1; PREFACE 1; ⊃ .MACRO SPACE1 ⊂ SPREAD←1; PREFACE 2; BREAK; ⊃ .MACRO SPACE2 ⊂ SPREAD←2; PREFACE 3; BREAK; ⊃ . .TURNON "↑↓[]&→" . . << title page >> .BEGIN CENTER LABELLING OBJECTS HAVING SYMMETRY L. Masinter, N.S. Sridharan, D. H. Smith Computer Science Department Stanford University Stanford, California May, 1973 .END .GROUP SKIP 5 .BEGIN INDENT 6,6,6 ABSTRACT. An algorithm for finding a complete set of non-equivalent labellings of a symmetric object and applications of the algorithm to problems in chemistry are presented. .END . .EVERY FOOTING(,{PAGE},) . .SKIP TO COLUMN 1 .SPACE2; .TITL INTRODUCTION The class of combinatorial problems dealing with finding a complete set of non-isomorphic objects under varying constraints and concepts of isomorphism, has wide applications in a variety of fields. The problem attacked in this paper is one of finding all distinct ways to assign a given number of labels or colors to the parts of a symmetric object when it is also known how many parts get each of the labels or colors. In chemistry, one manifestation of this problem is to make all assignments of ligands (from a fixed set) to a given carbocyclic skeleton<<ML:Masinter, L., Sridharan, N. S., et.al., Applications of Artificial Intelligence for Chemical Inference - XII, submitted, 1973, λJ. Amer. Chem. Soc.β>. Part A of this paper may be read as a brief tutorial on the nature of the problem and an introduction to the terminology found in more technical treatments. Part B is a textual description of a method for the solution of this type of problem. Part C is a summary of the procedure in a more algorithmic form; an even more formal description and a proof of correctness is available elsewhere<<BROWN: Brown, H., Masinter, L., Hjelemeland, L., Constructive Graph Labelling Using Double Cosets, submitted, 1972, λDiscrete Mathematicsβ; also, Stanford Computer Science Memo ??, Stanford University.>. Part D gives examples and applications of this algorithm in both organic and inorganic chemistry. This problem is encountered in a wide range of applications beyond chemistry-- within many areas of graph theory and combinatorics, for example. It has been known how to compute the number of solutions<<DEBRUIJN: De Bruijn, N. G., Generalization of Polya's Fundamental Theorem in Enumerative Combinatorial Analysis, λNedarl. Akad. Wentesh. Proc. Ser. A, α62β, (1959), 59-69>↑, <<POLYA:De Bruijn, "Polya's Theory of Counting," λApplied Combinatorial Mathematicsβ, E. Beckenbach (ed.), Wiley, New York, 1964, pp 144-184.>, but an efficent method of actually constructing the solutions has not previously been published<<PERLMAN:see, however, Perlman, D. M., Isomorph Rejection on Power Sets, (unpublished), University of California San Diego.>. .TITL PART A: DEFINITIONS .HD SYMMETRY AND ITS RELATIONSHIP TO LABELLING Consider the special case of the general problem: suppose all of the labels are distinct. This means that, for example, one wishes to number the faces of a cube with the numbers {1, 2, 3, 4, 5, 6}, or the "nodes" (atom positions in a graph) of the decalin skeleton (Fig. 1) with numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. There are n! (n factorial) ways of labelling where n is the number of parts. If the object has no symmetry then each of these n! ways are distinct from the rest. However for the decalin skeleton (where n! = 10! = 3,628,800 ways) there is some symmetry. First one picks, arbitrarily, one of the numberings as a point of reference (Fig 2a). Some of the 10! ways are different (Fig 2b); some of them are essentially the same (Fig 2c). .Figure 1,9,The Decalin Skeleton .Figure 2,9,Three of the 10! numberings of the Decalin Skeleton. Intuitively, Figs 2a and 2c are equivalent because one could take 2a, flip it over the 3-8 axis, and get 2c. There is another way of determining the "sameness" of such numberings which is easier in the more complicated cases and is in general more suited to computer application: .BEGIN I6 αDEFINITION:β two numberings of the nodes of a graph are λequivalentβ if the connection tables with the respective numberings are identical when the node numbers are written in ascending order and each "connected to" list is in ascending order. .END Table I contains the respective "connection tables" of the structures in Fig. 2. Note that the connection table for Fig 2c is identical to that of Fig 2a; that of Fig 2b is different. .BEGIN VERBATIM GROUP SELECT 4 __________________________________________________________________ Table I. Structure (2a) Structure (2b) Structure (2c) node|connected | node|connected | node|connected | to | | to | | to | | 1 2,1O | 1 8,9 | 1 2,1O 2 1,3 | 2 7,3 | 2 1,3 3 2,4,8 | 3 2,6 | 3 2,4,8 4 3,5 | 4 6,8,1O | 4 3,5 5 4,6 | 5 9,1O | 5 4,6 6 5,7 | 6 3,4 | 6 5,7 7 6,8 | 7 2,8 | 7 6,8 8 3,7,9 | 8 1,4,7 | 8 3,7,9 9 8,1O | 9 1,5 | 9 8,1O 1O 1,9 | 1O 4,5 | 1O 1,9 ____________________________________________________________________ .END This definition means, among other things, that properties such as valency are preserved: If two numberings are equivalent and in the first, node 1 is trivalent, then in the second, node 1 is trivalent. If there are other properties of the nodes (they are already colored or labelled, for example), then this definition can be extened to include the preserving of those properties. For example, suppose there are atom names associated with (some of) the nodes of the graph. Then one can define equivalent numberings to be those which not only have identical connection tables, but also the same atom names for the corresponding nodes. Then Fig 3a would still be equivalent to Fig 3c but no longer to Fig 3b since, although the connection tables of 3a and 3b are identical, node 1 in Fig 3a is labelled with an "N" while in 3b it unlabelled. .Figure 3,9,Partially labelled graphs reduce the equivalencies. .HD PERMUTATIONS AND PERMUTATION GROUPS Given a numbering of a structure, one can use a condensed notation to write down other numberings in terms of the first (Table II). In the 2b case, the row of numbers means that, in sequence, the node numbered 1 in the reference numbering 2a is now numbered 2, the node originally numbered 2 is now numbered 10, and so on. All of these are written down with respect to the original "reference" numbering of figure 2a. .BEGIN NOFILL GROUP SELECT 4 _________________________________________________________________ Table II Condensed Notation for Numberings Figure 2a numbers: 1 2 3 4 5 6 7 8 9 10 Figure 2b numbers: 2 7 8 1 9 5 10 4 6 3 Figure 2c numbers: 5 4 3 2 1 10 9 8 7 6 _________________________________________________________________ .END One can conceptualize a numbering as a transformation or as a function: The transformation π for 2c is π↓2↓c(1)=5, π↓2↓c(2)=4, π↓2↓c(3)=3, ... π↓2↓c(10)=6. These transformations are λpermutationsβ: one to one mappings from the integers {1,2,...,n} to themselves. The transformation for the "reference" numbering is the identity i(x)=x. It can be shown that whatever the graph, the set of transformations satisfying the "equivalency" requirement above satisfies the property of a group. One may then say: .BEGIN I6 The λsymmetry groupβ of a graph is the set of all transformations which yield identical connection tables. (If there are other properties to be considered, one may include them as part of the connection table). For the decalin skeleton there are 4 permutations in the symmetry group. .END .BEGIN NOFILL GROUP SELECT 4 _____________________________________________________________________ Table III The Symmetry Group of the Decalin Skeleton π↓i 1 2 3 4 5 6 7 8 9 10 π↓v 5 4 3 2 1 10 9 8 7 6 π↓h 10 9 8 7 6 5 4 3 2 1 π↓[180] 6 7 8 9 10 1 2 3 4 5 ____________________________________________________________________ .END These correspond directly to the intuitive geometric symmetries π↓i=identity, π↓h=reflection about horizontal axis, π↓v=reflection about vertical axis, π↓[180] = rotation 180 degrees. It is not too difficult for a computer program to figure out the symmetry group of a graph given its connection table<<GROUPCALC: this needs to be described>. This definition deals with the symmetry of the λnodesβ of a graph. In many cases, one might wish to label the λedgesβ (interconnecting lines) of a graph. In this case, the symmetry group on the edges rather than on the nodes is required. It is very easy to calculate this group from the group on the nodes. Let the numbering for each edge in the graph be the unordered pair of numbers of the nodes that form the end-points. Then the corresponding permutations can be obtained as follows: .BEGIN NOFILL π↓i{n↓1,n↓2}={π↓i(n↓1),π↓i(n↓2)} .END This is most easily shown by way of an example. (Table IV). .BEGIN NOFILL GROUP SELECT 4 __________________________________________________________________ Table IV Permutation Group of Decalin Skeleton acting on the edges π↓i 1-2 2-3 3-8 3-4 4-5 5-6 6-7 7-8 8-9 9-10 1-10 π↓v 4-5 3-4 3-8 2-3 1-2 1-10 9-10 8-9 7-8 6-7 5-6 π↓h 9-10 8-9 3-8 7-8 6-7 5-6 4-5 3-4 2-3 1-2 1-10 π↓[180] 6-7 7-8 3-8 8-9 9-10 1-10 1-2 2-3 3-4 4-5 5-6 ____________________________________________________________________ .END Finding the group of an object is a special kind of labelling problem. Given one way of numbering (labelling with distinct labels) the parts of the object, one finds all other ways which are equivalent. The labelling problem attacked in this paper is the converse: to find a maximal list of labellings, none of which are equivalent to each other. In general the set of all posssible labellings can be split into subsets, such that: .BEGIN I6 1) If two labellings are in different subsets, they are not equivalent. 2) If two labellings are in the same subset, they are equivalent. .END These subsets are called λequivalence classesβ of labellings. The relationship of finding the group, and of finding labellings where all the labels are distinct, can be viewed as follows: Take the n! possible labellings and lay them out in a matrix: each row will contain one equivalence class. (It can be shown that in this special cases where the labels are distinct, all of the equivalence classes are of the same size). To find the group, one needs to find a row. To find the labellings, one needs to pick one element from each row. .Figure 4,10,"Equivalence classes, Groups, and Labellings" .TITL PART B: SOLUTION TO THE LABELLING PROBLEM .HD A Naive Method An obvious method to find the distinct labellings would be to generate all n! of the possible ones, and for each one to check if an equivalent one was previously generated. Unfortunately, this method takes an exhorbitant amount of computation (proportional to n! squared). Below a method is discussed which takes an amount of time roughly proportional to the number of solutions (i.e. the number of equivalence classes of labellings) and requires only knowledge of the symmetry group. Thus it is useful for labelling objects using their geometric symmetry .reference! ml} as well as the topological symmetry defined above. .HD A Better Method .GRAF Special case: distinguish 1 node First consider the special case where there are only two types of labels such that there is one label of the first type and all the rest of the second. (E.g., color one red, and the rest green, or label one N and the rest C.) Intuitively, for the decalin skeleton, one can see that there are three classes of symmetric nodes, or λorbitsβ, marked with "⊗", "+" and "&" in Fig. 5, and that each distinct labelling corresponds to selecting one node from each type (see Fig 6.) ~(CHECK TO MAKE SURE THAT SYMBOLS ARE APPROPRIATE TO FIGURE)β .Figure 5,9,Orbits in the Decalin Skeleton .Figure 6,9,Three Labellings of Decalin with 1 N and 9 C's. Thus there are three distinct labellings (the ten possible labellings split into three equivalalence classes). .GRAF Computing orbits In general, the parts of a symmetry object split into different orbits (sometimes there is only one type, as in the faces of a cube, or the nodes of the cyclohexane skeleton). To label the parts of an object such that one is distinguished, it is necessary only to find the orbits and then, for each type, pick one of the parts in that type arbitrarily. Note that if the object has no symmetry each type has exactly one part in it. It is very simple to find the different types from the table of the symmetry group: if one writes out the group, as in Table III, with each permutation as a row, then the numbers in each column, taken as a set, form an orbit. The orbits of the group in Table III are: {1,5,6,10}, {2,4,7,9}, {3,8}. After finding the set of orbits, one then can do the special labelling described above (distinguishing only one node): the number of distinct labellings is the number of orbits. Each corresponds to selecting an element from a corresponding orbit and labelling it. For a convention, the first element of each orbit should be chosen (i.e. the one with the smallest number in the reference numbering). .GRAF The reduced symmetry group Once a group has been calculated for a structure, many times one wants to know what the group is after some labels have been attached. The group of a labelled structure is always a λsubsetβ of the group of the unlabelled structure. Thus one needs to know which permutations in the group must be thrown out. To do this, write the "labels" associated with each node next to the node number in the permutation table as in Table II. If in any column, there is an element which has a different label than the label in the "reference" numbering (identity permutation), then that row can be discarded. That is, a permutation is valid if and only if the structure "looks the same" after its node numbers are permuted. Only permutations in the original group can possibly yield structures which do look the same; however, because of the labelling, some of these permutations may yield disimilar structures. These permutations are the ones that must be discarded. .GRAF Reduction techniques In the general labelling problem, there are two important techniques used to reduce the problem. The first is called λlabel recursionβ <<RECUR:A λrecursiveβ technique is one which is defined in terms of itself. It is only necessary that at each step the problem is reduced, and that a terminating condition is eventually met. Here the general solution of the labelling problem is described in terms of less complicated labellings. Several special cases (terminating conditions) are also defined.> and the second λorbit recursionβ. The idea behind label recursion is that one can do one label at a time. The idea behind orbit recursion is that one can label one orbit at a time. These reductions are discussed in detail below. .HD LABEL RECURSION If one is given many (more than 2) kinds of labels, say n↓1 of type 1, n↓2 of type 2, ... n↓k of type k, one may proceed as follows: Solve the labelling problem for n↓1 of one type of label and n↓2+n↓3+...+n↓k of another type. (Call the second type of label "blank"). The result will be a list of partially labelled objects (along with their reduced symmetry groups). Take each of the results and label the "blank" parts with n↓2 labels of one kind, n↓3 of another, ... ,n↓k of another. It has been proved .REFERENCE! BROWN} that the results will be a list of all distinct solutions to the original problem. For example, to label the decalin skeleton with 1 label "N", 1 label "S" and 8 labels "C", one first labels with 1 "N" and 9 "blanks" obtaining the 3 results in figure 7a. (Fig 7a1, 7a2, 7a3). There are now 3 new problems: to label the "blanks" of Figs. 7a1-3 under their respective reduced symmetry, with 1 "S" and 8 "C"'s. In Figs 7a1 and 7a2, there is no symmetry left and so each of the "blanks" has its own orbit; thus there are 9 distinct labellings apiece. In Fig. 7a3 there are 5 orbits under the reduced symmetry, and thus there are 5 additional possiblities (Fig 7b). .Figure 7,15,|Labellings with 1 N, 1 S, and 8 C's.| .HD ORBIT RECURSION There are 3 cases in the problem of 1 N, 9 C on the decalin skeleton. .BEGIN NOFILL (case 1) 1 N goes to orbit {1,5,6,10}. (case 2) 1 N goes to orbit {2,4,7,9}, (case 3) 1 N goes to orbit {3,8}. .END There is only one possibility in each of these cases. Suppose there were 3 N's. Then there would be 9 cases. (Table IV). .BEGIN VERBATIM GROUP SELECT 4 _________________________________________________________________ Table IV. (3 N's on a Decalin) # N's going to case orbit orbit orbit number {1,5,6,10} {2,4,7,9} {3,8} 1 3 0 0 2 2 1 0 3 2 0 1 4 1 2 0 5 1 1 1 6 1 0 2 7 0 3 0 8 0 2 1 9 0 1 2 _________________________________________________________________ .END In some of these cases there are more than one possibility (cases 2, 3, 4, 5 and 8). However, every labelling fits into one of these cases, and labellings from different cases cannot be equivalent. Thus, each of these cases can be done independently, and the results collected together. To do any one of the cases, the labellings of the orbits can be done sequentially. That is, the rows of Table IV can be done independently, and for each row the columns can be done from left to right. Considering case 5, one can first label one of {1,5,6,10} with one N, (and the rest "blanks"). Since {1,5,6,10} is an orbit, one can chose node 1; the result is Fig 8. .Figure 8,9,|1 N to orbit {1,5,6,10}| Second, one labels {2,4,7,9} with 1 N (and 3 blanks). Note that {2,4,7,9} is no longer an orbit under the reduced group. Stick to the original plan-- it is necessary to find λnewβ orbits under the reduced group to label {2,4,7,9}. Since there is no symmetry left, each of {2,4,7,9} falls into its own orbit. The "special case" described below under "No Group" can be used directly to find the 4 solutions (Fig 9). .Figure 9,12,Second step of case 5 Then, for each of these solutions, {3,8} must be labelled with 1 N (and 1 blank). The final result is 8 possibilities for case 5, none of which has any remaining symmetry. .HD SPECIAL CASES There are several special cases of labellings in which the problem can be reduced or solved immediately. Although these special cases may be amenable to the more general algorithm, their solution is computationally simpler. .GRAF Only one type of label If the number of labels (of a given type) is the same as the number of objects, then there is exactly one way of doing the labelling. This check for this special case is necessary, since in orbit recursion, often the sub-problem is of this form; n↓1=n, and n↓2=0 or vice versa. .graf Only one of a given label Already mentioned was the case where there was one label of one kind and n-1 of the other; it was only necessary to find the orbits, and label one element arbitrarily from each of the orbits. One might note here that the order in which one applies the labels in label recursion is arbitrary and thus if there is only one of any of the labels, then this special case can be applied immediately. .graf No group When there is no symmetry left and there are two label types (n↓1 of the first type, n↓2 of the second, n↓1+n↓2=n, the number of objects to be labelled), the labelling reduces to a simple combinatorial problem: given n distinct objects, find all ways of selecting n↓1 of them. This can be done by the following recursive algorithm: .BEGIN NOFILL To find all selections of k objects out of set S whose size is n: 1) if k = 0 then there is only one selection, the empty set. 2) if k > n then there are no possible selections. 3) Otherwise, pick an element x from S. a) Find all selections of k objects out of the set S-{x}. b) Find all selections of k-1 objects out of the set S-{x}; to each of these, add the element x. .END A k-subset of a set S (a k-subset is a subset with k elements) either contains an element x or not. In case 3a, one finds those selections which do not contain x. In case 3b, one finds those selections which do. However, each of these cases are simpler than the original selection problem; the size of the set S reduces, as well as the value of k. The terminating conditions, k=0, or k greater than the size of the set S, insure that the process will halt. .HD FINAL TECHNIQUE It has been now shown that every problem can be reduced to cases where there are only two types of labels, n↓1 of the first, and n↓2 of the second, (n↓1+n↓2=n, the number of nodes to be labelled), both n↓1 and n↓2 > 1, and there is only one orbit. No more simple reductions are left. One solution, however, is another trick. Pick the first node and label it, calculating the reduced symmetry group and new orbits. Label the rest of the nodes (under the reduced group) with n↓1-1 labels of type 1 and n↓2 labels of type 2. The result will contain a representative of each equivalence class of labellings; however, if n↓1>2 then there may be some duplicates (i.e., two or more of the results may actually be equivalent). .hd special kludges for fast programs by Larry Masinter .begin nofill (this section to be written?) (this section to be written?) (this section to be written?) (this section to be written?) (this section to be written?) (this section to be written?) (this section to be written?) .end For example, the cyclohexane skeleton has a group of order twelve (has twelve permutations), and there is only one orbit. To label it with three N's, one labels node 1 with a N, calculates the reduced group and new orbits; then finds the various cases for distributing the remaining two N's among those orbits. (Table V.) Then for each case, do each orbit sequentially. Cases 1, 3, 4 and 5 are fairly straightforward; in case 2, first label {1,2} with 1 N. (Figure 9). Now the group reduces even further, and one gets the two results depicted in Figure 9b. Note that cases 1 and 2a are equivalent as well as 2b, 3, and 5. What to do? Fortunately, there is a good way of throwing out the impostors without having to check each of the results against each of the others for equivalency. If there is a permutation π in the group, such that .BEGIN NOFILL π (labelled set) > labelled set .END; CONTINUE then the labelled set is an impostor-- throw the bum out. Furthermore, every impostor is detected this way. All that is necessary is that when doing these "lower level" labellings, that one is careful to pick the "first" element of each orbit to label and the "first orbit" when there are a choice of orbits. .BEGIN GROUP NOFILL SELECT 4 ____________________________________________________________ Table V cases for 2 N's on {2,3,4,5,6} of cyclohexane case {2,6} {3,5} {4} 1 2 - - 2 1 1 - 3 1 - 1 4 - 2 - 5 - 1 1 ____________________________________________________________ .END Fortunately, this technique is rarely necessary -- usually in the course of a computation, the "special cases" catch almost everything. For example, to label the decalin skeleton, it is never needed since even when one is labelling say orbit {2,4,7,9}, there is either 1, 2, n-1 or n-2 labels to be attached. For the cyclohexane skeleton, it is only needed if there are 3 of one label and 3 of another; if there were 3,2 and 1 of three respective label types for example, just do the single label first -- the group will then reduce and again this "final technique" will not be necessary. Only in cases where there is an orbit with at least 6 elements and there are at least 3 of each of the label types is this technique required. .TITL PART C: SUMMARY OF LABELLING STEPS .BEGIN SPREAD←1; PREFACE 1; INDENT 0,3,3 1) Calculate the group, if not previously calculated. 2) If there are more than 2 different node types, do the entire labelling process with 1 of the label types, and the rest "blanks"; then for each of the solutions obtained, label the "blanks" with the remaining label types using the reduced symmetry group and collect the results. 3) Otherwise, .BEGIN INDENT 3,6,6 a) Find the orbits. b) If more than one orbit, then .BEGIN INDENT 6,9,9 1) Find the different "cases" -- ways of distributing the labels among the orbits. 2) For each case, .BEGIN INDENT 9,12,12 a) Label the first orbit. b) Label the rest of the orbits using the reduced symmetry group obtained from a), and collect the results. .END END c) Otherwise (only 1 orbit and 2 label types): .BEGIN INDENT 6,9,9 1) If there is only one of one of the label types, pick the "first" node and label it with that label type. This is the only distinct possibility. 2) Otherwise if there are two of one of the label types, label the first node with that label type, calculate the reduced symmetry group and new orbits, and from each of the new orbits, pick the "first" node. The solutions consist of those possibilities (one for each new orbit). 3) Otherwise, (3 or more of each label type, and one orbit) label the "first" node, calculate the reduced symmetry group, label the rest of the nodes under the reduced group, and for each of the results, check if for every permutation π in the original group that .BEGIN NOFILL π(labelled set)≥labelled set .END .CONTINUE If this is not true of a labelled set, discard it as a solution. The result is every such labelling that satisfies this property. .END END END .TITL PART D: EXTENDED EXAMPLES .HD |Study of Diels-Alder Ringsβ<<SIMEK:proposed by Jan Simek, Chemistry .Department, Stanford University. .Research Proposal (unpublished), February, 1973.>| The labelling algorithm has been used to define the scope and boundaries of the Diels-Alder reaction, a well-known and commonly used synthetic reaction. The reaction is shown in Figure 10, and is defined as the 4+2 cycloaddition of an olefin, termed the dienophile, with a conjugated diene, leading to the formation of a cyclohexene-type of ring system (Diels-Alder Ring). .FIGURE 10,10,Diels-Alder reaction The method used by the program to generate Diels Alder Rings is described below, followed by an example of the labelling procedure. The program generated 1176 Diels-Alder Rings using any combination of C,N,O and S. Other atoms such as P could have been included but were deemed not interesting to us. A comparison of the computer print-out with the Ring Index (which covers the literature through 1963) revealed that only 224 (about 19% of 1176) are "known" systems. A ring system was said to be known if the same sequence of atoms had been reported regardless of number or position of unsaturations. The complete list of Diels-Alder Rings is richly suggestive to the synthetic chemists and may serve to increase the information on the scope of the Diels-Alder reaction. .Figure 11,20,Diagram of Method of Generation .HD |METHOD β(see Figure 11)| .graf Step 1 The initial pot of atoms consisted of C↓6N↓6O↓4S↓4. The number of oxygens and sulfurs was limited to four, because no Diels-Alder ring can be made with five or six bivalent atoms, owing to valence restrictions. A list of all possible 76 compositions of 6 atoms was produced using a purely combinatorial procedure. .graf Step 2 Eleven of the 76 compositions were eliminated, again owing to valence limitations. An example of the eleven compositions eliminated is O↓3S↓3. .graf Step 3 For each of the 65 remaining compositions, the Diels-Alder ring skeleton was labelled with the atoms, while ensuring valence constraints were satisfied. .graf Step 4 The results from all labelling steps were collected, without needing to check for duplicates. The labelling algorithm guarantees that the lists were irredundant. .graf Example of labelling An example of a valid composition is C↓4O↓2. Diels-Alder rings formed with this composition can only have carbons double bonded to each other (Figure 12). The atoms remaining to be assigned are C↓2O↓2 and the ring positions are numbered 1,2,3,4. .FIGURE 12,12,Diels-Alder Example A .GRAF Verification by Enumeration The results of of the labelling procedure can be verified by combinatorial counting techniques .REFERENCE! POLYA}↑, .REFERENCE! DEBRUIJN}. The following derivation follows closely from that in Liu<<LIU:ββLiu, Introduction to Comb. Math, McGraw-Hill, 1968>. .begin nofill SELECT 4 Cycle Index of Group=PG = 1/2(y↑4&↓1+ y↑2&↓1) Pattern Inventory is 1/2 (x↑1&↓1+ x↑1&↓2)↑4+ (x↑2&↓1+ x↑2&↓2)↑2 = 1/2 (x↑4&↓1+ x↑4&↓2+ 4x↑3&↓1x↓2+ 6x↑2&↓1x↑2&↓1+4x↓1x↑3&↓2) + (x↑4&↓1+ x↑4&↓2+ 2x↑2&↓1x↑2&↓2) = 1/2 2x↑4&↓1+ 2x↑4&↓2+ 8x↑2&↓1x↑2&↓2+ 4x↑3&↓1x↓2+ 4x↑3&↓2x↓1 = x↓1&↑4+ x↓2&↑4 + 4 x↓1&↑2x↓2&↑2 + 2x↓1&↑3x↓2+ 2x↓2&↑3x↓1 showing that there are precisely these number of labellings: labels #labellings C C C C 1 S S S S 1 C C S S 4 C S S S 2 C C C S 2 .end The third entry in the table verifies our labelling with C↓2S↓2 in four ways. .HD Labellings on the Octahedral skeletal framework. This example is concerned with the geometric isomers of structures with octahedral symmetry. (Figure 13). .FIGURE 13,8,Octahedral Skeletal Framework .begin nofill _________________________________________________________________ Labellings on Octahedral Skeletal Framework Group is: i (1 2 3 4 5 6) r↓1 (1 3 5 4 6 2) r↓2 (1 5 6 4 2 3) r↓3 (1 6 2 4 3 5) v↓1 (4 5 3 1 2 6) v↓2 (4 2 6 1 5 3) v↓3 (4 3 2 1 6 5) v↓4 (4 6 5 1 3 2) Orbits: {1 4}, {2 3 5 6}. Example with labels (A A A A B B) Number of labels A = 4 Number of labels B = 2 Partitioning Labels into Orbits: Case number of A's assigned to orbit: # {1 4} {2 3 5 6} 1 2 0 2 1 1 3 0 2 Case 1. To map A A --> {1 4} and A A B B --> {2 3 5 6} Map A A --> {1 4} (trivial case) New group i, r↓1, r↓2, r↓3, v↓1, v↓2, v↓3, v↓4 New orbits {2 3 5 6} To map A A B B --> {2 3 5 6} Case 1.1. Assign A --> 2 New group i,?? ???????? fill in New orbits {5},{3 6} Case 1.1.1. Assign A --> 5 Remaining B B --> {3 6} 1 labelling (A A B A A B) Case 1.1.2. Assign A --> 3 Remaining B B --> {5 6} 1 labelling (A A A A B B) Case 2. To map A B --> {1 4} and A A A B --> {2 3 5 6} Assign A --> 1 and B --> 4 New group i, r↓1, r↓2, r↓3 New orbits {2 3 5 6} To map A A A B --> {2 3 5 6} Case 2.1. Assign B --> 2 To map A A A --> {3 5 6} 1 labelling (A B A B A A) Case 3. To map B B --> {1 4} and A A A A --> {2 3 5 6} Assign B --> and B --> 4 New group i, r↓1, r↓2, r↓3, v↓1, v↓2, v↓3, v↓4 New orbit {2 3 5 6} To map A A A A --> {2 3 5 6} 1 labelling (B A A B A A) .FIGURE X,9,|Four labellings of Octahedral Skeleton with Labels (A A A A B B)| Example with Labels A B C D E F Case 1. A --> orbit {1 4} A --> 1 new group i, r↓1, r↓2, r↓3 new orbits A1 = {2 3 5 6} A2 = {4} Assign label B Case 1.1. B --> orbit A1 B --> 2 new group i new orbits {3} {4} {5} {6} CDEF --> 3,4,5,6 4! = 24 labelling generable with no further involvement of permutation group. Case 1↓2. B --> orbit A2 B --> 4 new group i, r1, r2, r3 new orbits AA1 = {2 3 5 6} Case 1.2.1 C --> orbit AB1 C --> 2 new group i new orbits {3} {5} {6} DEF --> 3 5 6 6 labellings Case 2. A --> orbit {2 3 5 6} A --> 2 new group i, new orbits B1 = {1 4} B2 = {5} B3 = {3 6} Case 2.1. B --> orbit B1 B --> 1 new group i new orbits {3} {4} {5} {6} CDEF --> 3 4 5 6 24 labellings Case 2.2. B --> orbit B2 B --> 5 new group i,? ???? fill in new orbits BB1 = {1 4} BB2 = {3 6} Case 2.2.1. C --> orbit BB1 C --> 1 new group i new orbits {3} {4} {6} DEF --> 3 4 6 6 labellings Case 2.2.2. C --> orbit BB2 C --> 3 new group i new orbit {1} {4} {6} DEF --> 1 4 6 6 labellings Case 2.3. B --> B3 B --> 3 new group i new orbit {1} {4} {5} {6} CDEF 1 4 5 6 24 labellings 90 unique labelings all together. Verification: When all labels are different the total number of labellings (# labels)! 6! = -------------- = ---- = 90 labellings (size of group) 8 .end .TITL ACKNOWLEDGEMENTS We gratefully acknowledge the contributions of Jan Simek, who proposed the application in Part D, Larry Hjelmeland who formulized the initial problem, Professor Harold Brown, who proved the correctness of the original algorithms, and Professor Joshua Lederburg, whose advice and guidance has always been an inspiration.