Someone Just Moved! Who Was It?

or, Apter's Puzzle

Eemcd@aol.com

**The Puzzle**

Stevan Apter recently proposed this puzzle to the online K group: Given a list of distinct items, and a second list changed by moving only one item of the original, find which item has been moved. There was a fair amount of discussion about this, and a number of proposed solutions, a surprising number of which were erroneous. This paper gives a solution that I believe works properly in all cases. The greater part of the paper, however, discusses the reverse problem: Given a solution, find the list that gave rise to it

**Rotated Infix Permutations**

First, to solve Apter's problem: Given the two lists A and B:

A 3 1tell which item in A was moved in order to produce B. It couldn't have been the 5 or 9 or 2 because they form an infix in the order of the original. The 4, which had preceded 5 and 9 and 2 is the one that is out of order: it is now at their right: it has been moved from index 2 to index 5.45 9 2 6 B 3 1 5 9 246

The problem is simplified if the items are replaced by their indices. Since the items are distinct, A can be replaced by the identity permutation, and B by the indices of its items in the origin. For example:

C =: i. # A D =: A i. B C ,: D 0 1 2 3 4 5 6 0 1 3 4 5 2 6C and D contain all the information needed to solve the original A and B problem. In fact, D is all that is needed: the identity permutation C can be understood. In what follows, I'll assume that a problem list is given in this D form. Thus an argument to the solution program might be:

D 0 1 3 4 5 2 6Comparing D to C shows that some of the items remain in their original positions, but others have been moved. The following shows the items that have been moved in bold. These form a rotated infix; because of this I'll call D a rotated infix permutation, which for convenience will be abbreviated to rip.

0 1The nonzero items of D-C show which have been moved:3 4 5 26

D-C 0 0 1 1 1 _3 0Those which have been displaced by the move produce a 1 in the difference; the one moved produces _3. But an item can be moved to a lower position as well as to a higher. Suppose we are given to solve:

E 0 1The items in bold again represent a 1-rotation, but this time it is to the right. If we subtract C from this we get:5 2 3 46

E-C 0 0 3 _1 _1 _1 0Here the value corresponding to the moved item is positive value, and the displaced items produce _1s. Considering both cases, it is evident that the moved item is determined by finding the maximum magnitude in the difference between the list to solve and the identity permutation. There are two difficulties to discuss. The first difficulty: suppose we move an item just one position to the right or left: what results?

F=.0 1The magnitudes of F-C show two possible results. It's unclear whether the 3 has been moved to position 2, or the 2 has been moved to position 3. Either one can be chosen. The rule used here is: Among equal maxima, choose the one occurring first. This is easiest because of the way J's3 24 5 6 F-C 0 0 1 _1 0 0 0

Therankorindexof an element c`b is called the bindex ofc and is defined as the smallest value of i such that c = b_{i}.

Thus, for the rip F above, 3 will be identified as the item that has been moved -- even though it might in fact have been produced by moving 2 to position 3.

The second difficulty arises from the possibility, not excluded in Apter's statement of the problem, of moving an item from its original position to its original position. For example, the list:

G 0 1 2 3 4 5 6if proposed as a problem to solve, might be the result of moving any of the seven items back to its original position. What we get if we subtract C from G is:

G - C 0 0 0 0 0 0 0Again following the rule

**Canonical Specifications**

A rip such as D, E, F, or G can be represented by a list of three integers: its length L, the initial position I of the moved item, and its final position F. For example, the rip D is specified by 7 2 5; E by 7 5 2; F by 7 3 3; and G by 7 0 0. I'll call such a list the rips's canonical representation, or casp.

The function CfR below solves Apter's problem, yielding the casp from the rip:

CfR =: monad define"1 NB. casp from rip R=.y. F=.(i.>./)|(-i.@#)R L=.#R I=.F{R L,I,F )Applying it to the sample rips:

CfR D 7 2 5 CfR E 7 5 2 CfR F 7 3 2 CfR G 7 0 0For aficionados of the one liner:

OLCfR =. # , (([ , ~ {) ~ (i. >. /) @ ([: | ] - [: i. #)) ~A function inverse to CfR would take a casp L,I,F and yield the rip which gave rise to it. It can be described informally like this:

lay out a row of L numbered blocks

+-+-+-+-+-+-+-+ |0|1|2|3|4|5|6| +-+-+-+-+-+-+-+remove the Ith block (which has the number I on it) and set it aside

+-+-+ +-+-+-+-+ |0|1| |3|4|5|6| +-+-+ +-+-+-+-+ +-+ |2| +-+collapse the remaining blocks over the empty space

+-+-+-+-+-+-+ |0|1|3|4|5|6| +-+-+-+-+-+-+separate these into two parts, the first F and the rest

+-+-+-+-+-+ +-+ |0|1|3|4|5| |6| +-+-+-+-+-+ +-+and insert the removed block in the space made (which is F)

+-+-+-+-+-+-+-+ |0|1|3|4|5|2|6| +-+-+-+-+-+-+-+This function RfC which, given a casp, yields a rip, is:

RfC =: monad define"1 'L I F'=.y. M=.I-.~i.L (F{.M),I,(F}.M) )Here are some uses of RfC:

RfC 7 2 5 0 1 3 4 5 2 6 RfC 7 5 2 0 1 5 2 3 4 6 RfC 7 3 2 0 1 3 2 4 5 6 RfC 7 0 0 0 1 2 3 4 5 6Roughly speaking, CfR and RfC are inverses, and thus we'd like to be able to say that:

P -: RfC CfR Pand

C -: CfR RfC CThe function MC below, given a positive integer argument, yields a 3-column table of all the casps for rips of that length. For a list of length n, there are n2 rips, one for each initial position versus each final position. A variation of the odometer function is required. Given a length L, form

MC=:13 : 'y.,.(2#y.)#:i.*:y.' ]C3=.MC 3 3 0 0 3 0 1 3 0 2 3 1 0 3 1 1 3 1 2 3 2 0 3 2 1 3 2 2Next, I'll get the corresponding rips:

R3 =. RfC C3And finally, put C3 next to R3 so we can see the correspondences:

2 2 2 6 2 2": C3,.R3 3 0 0 0 1 2 3 0 1 1 0 2 3 0 2 1 2 0 3 1 0 1 0 2 3 1 1 0 1 2 3 1 2 0 2 1 3 2 0 2 0 1 3 2 1 0 2 1 3 2 2 0 1 2The number of distinct results is not L2, but less than that. When I and F are the same the results are the same: each of 3 0 0, 3 1 1, and 3 2 2 yields 0 1 2. From these three solutions we get only one casp, so we can reduce the initial 9 by 2. In general, the diminishment coming from this case is L-1. Next, the results are the same when I and F are adjacent numbers, as in 3 0 1 and 3 1 0, and 3 1 2 and 3 2 1. From these four casps we get only two rips, so we can reduce the total by another 2. In general, the diminishment coming from this case is also L-1, one for each adjacent pair. The 9 results are thus reduced by another 2, leaving just 5. The general formula for the number of distinct rips is

The function NR below, given an integer argument, yields the number of distinct rips of that length:

NR=.2 _2 1&p. (],.NR)>:i.10 1 1 2 2 3 5 4 10 5 17 6 26 7 37 8 50 9 65 10 82

**The Anatomy of Rips**

I've been in the habit of checking sequences such as the second column above against the entries in Neil Sloane's invaluable collection, which appeared in print first in his *A Handbook of Integer Sequences* (Academic Press, 1973). Recently the book has been supplemented by an ever growing collection on his web site, *On-Line Encyclopedia of Integer Sequences*, at:

http://www.research.att.com/~njas/sequences/

The series 1 2 5 10 is ID Number A002522, which describes it in offset 0 as n2+1. My series is offset 1, which implies the formula previously given, namely n2-2n+2. It also notes that this sequence is the "Left edge of A055096" and if you refer to this sequence you will find that it is a triangle like Pascal's, where the entries are sums of distinct squares:

5 10 13 17 20 25 26 29 34 41 37 40 45 52 61The left edge is a truncated version of our series, lacking its first two items. Pursuing this any further will take me on too wide a detour from my main goal: the analysis of rips; but I shall be referring often to Sloane's Encyclopedia in what follows.

I began my study of rips by finding the *anagram index* of a fair number of them, using J's A. primitive, and listing on a sheet of paper those of length seven in order by length of rotated infix, and within that by highest maximum I and F. Here are those with rotated infix of length 3, and the associated anagram index (I've put the disordered infix items in **bold** type).

I<F L I F rip A. 7 4 6 0 1 2 35 6 43 7 3 5 0 1 24 5 36 8 7 2 4 0 13 4 25 6 30 7 1 3 02 3 14 5 6 144 7 0 21 2 03 4 5 6 840 I>F L I F rip A. 7 6 4 0 1 2 36 4 54 7 5 3 0 1 25 3 46 12 7 4 2 0 14 2 35 6 48 7 3 1 03 1 24 5 6 240 7 2 02 0 13 4 5 6 1440

The obvious way of entering the data thus gathered was to identify the rows with I and the columns with F, and to put the anagram index in item for I,F in row I, column F. This was unsatisfactory, since the numbers were going in what seemed the wrong direction, so instead I made tables where the rows were numbered by how far the right edge of the infix was from the right end of the table, and the columns were numbered by the length of the infix. Thus the anagram index of 30 where the L I F is 7 2 4, which has the infix 2 spaces from the right edge and has length 3 would be entered at row 2 column 3 of the I<F table.

I<FStudying these tables convinced me that I could see the rule determining the value of an entry. For the I<F table, column 1 would always be zero; an infix of length one meant that I and F were the same, so the result would be the identity permutation, which is permutation 0 for permutations of all lengths. Column 2 would always be !(1+row number). Column j for j>2 would be sums of (j-1) successive items of column 2. For example, the entries in column 3 would be 1+2, 2+6, 6+24; in column 4 would be 1+2+6, 2+6+24, 6+24+120, and so forth. This can be shown using J's Infix adverb, x u\y, where the function u (in our case +/) is applied over successive length-x infixes of y :123456700 1 3 9 33 153 87310 2 8 32 152 87220 6 30 150 87030 24 144 86440 120 84050 72060 I>F123456700 1 4 18 96 600 432010 2 12 72 480 360020 6 48 360 288030 24 240 216040 120 144050 72060

]fs=.!i.10 1 1 2 6 24 120 720 5040 40320 362880 2+/\fs 2 3 8 30 144 840 5760 45360 403200 3+/\fs 4 9 32 150 864 5880 46080 408240 4+/\fs 10 33 152 870 5904 46200 408960For the

I was able to verify my conjectures for both tables after a few experiments, and wrote two functions that would give me the value for any entry in either table:

ILF =: 13 : '+/!(x.+1)+i.y.-1'"0 IGF =: 13 : '(y.-1)*!(y.-1)+x.'"0 3 ILF 4 864 3 IGF 4 2160The

d16=: by=: ' '&;@,.@[,.] d17=: over=: ({.;}.)@":@, a18=: tab=: 1 :'[ by ]over x./' (i.7) ILF tab 1+i.7x +-+--------------------------------------------------+ | |1 2 3 4 5 6 7| +-+--------------------------------------------------+ |0|0 1 4 18 96 600 4320| |1|0 2 12 72 480 3600 30240| |2|0 6 48 360 2880 25200 241920| |3|0 24 240 2160 20160 201600 2177280| |4|0 120 1440 15120 161280 1814400 21772800| |5|0 720 10080 120960 1451520 18144000 239500800| |6|0 5040 80640 1088640 14515200 199584000 2874009600| +-+--------------------------------------------------+ (i.7) IGF tab 1+i.7x +-+----------------------------------------------+ | |1 2 3 4 5 6 7| +-+----------------------------------------------+ |0|0 1 3 9 33 153 873| |1|0 2 8 32 152 872 5912| |2|0 6 30 150 870 5910 46230| |3|0 24 144 864 5904 46224 409104| |4|0 120 840 5880 46200 409080 4037880| |5|0 720 5760 46080 408960 4037760 43954560| |6|0 5040 45360 408240 4037040 43953840 522955440| +-+----------------------------------------------+In both ILF and IGF the arguments are modified by adding or subtracting 1. I wondered whether I could get any further insights by writing versions of ILF and IGF in which the arguments were not offset.

ILFx=:13 : '+/!x.+i.y.'"0 ILFx tab~i.7x +-+----------------------------------------+ | |0 1 2 3 4 5 6| +-+----------------------------------------+ |0|0 1 2 4 10 34 154|3422 |1|0 1 3 9 33 153 873|7489 |2|0 2 8 32 152 872 5912|54116 |3|0 6 30 150 870 5910 46230|54117 |4|0 24 144 864 5904 46224 409104|54118 |5|0 120 840 5880 46200 409080 4037880| |6|0 720 5760 46080 408960 4037760 43954560| +-+----------------------------------------+ 4 142 1048 IGFx=:13 : 'y.*!y.+x.'"0 IGFx tab~i.7x +-+--------------------------------------------------+ | |0 1 2 3 4 5 6| +-+--------------------------------------------------+ |0|0 1 4 18 96 600 4320|1563 |1|0 2 12 72 480 3600 30240|18931 |2|0 6 48 360 2880 25200 241920|52571 |3|0 24 240 2160 20160 201600 2177280|52520 |4|0 120 1440 15120 161280 1814400 21772800|52557 |5|0 720 10080 120960 1451520 18144000 239500800|52521 |6|0 5040 80640 1088640 14515200 199584000 2874009600| +-+--------------------------------------------------+ 4 142 52849 52560 52578 52648I've put numbers to the right of those rows and at the bottom of those columns which correspond to entries in Sloane's Encyclopedia. The encyclopedia doesn't contain entries for all the rows and columns of ILFx and IGFx, but they are easily obtained. The functions below give the first ten items of the indicated row or column:

NB. x items of row y of ILFx ILFI=.13 : '+/\!y.+i.x:x.' NB. x items of column y of ILFx ILFJ=:13 : 'y.+/\!i.<:y.+x:x.' NB. x items of row y of IGFx IGFI=:13 : '(!y.+i.10x)*i.x:x.' NB. x items of column y of IGFx IGFJ=.13 : 'y.*!y.+i.x:x.'And here they are, applied to 10 items of row 3 and column 3 for each table:

10 ILFI 3 6 30 150 870 5910 46230 409110 4037910 43954710 522956310 10 ILFJ 3 4 9 32 150 864 5880 46080 408240 4032000 43908480 10 IGFI 3 0 24 240 2160 20160 201600 2177280 25401600 319334400 4311014400 10 IGFJ 3 18 72 360 2160 15120 120960 1088640 10886400 119750400 1437004800Sloane contains tables and triangles as well as linear sequences. Table ILFx is closely related to Sloane's sequence 54115:

1 1 1 1 2 3 1 6 8 9 1 24 30 32 33 1 120 144 150 152 151The formula for this triangular array T is given as:

Triangular array generated by its row sums:

T(n,0)=1 for n>=1 T(n,1)=r(n-1) T(n,k)=T(n,k-1)+r(n-k) for k=2,3,...,n, n>=2 r(h)=sum of the numbers in row h of T.

The rows of this triangle are derived from ILFx by reading the counterdiagonals from left to right. We can get the counterdiagonals in J by using J's Box and Oblique:

,.7{.</.|:ILFx/~i.7 +-------------------------+ |0 | +-------------------------+ |0 1 | +-------------------------+ |0 1 2 | +-------------------------+ |0 2 3 4 | +-------------------------+ |0 6 8 9 10 | +-------------------------+ |0 24 30 32 33 34 | +-------------------------+ |0 120 144 150 152 153 154| +-------------------------+Here it is in triangular form:

0 0 1 0 1 2 0 2 3 4 0 6 8 9 10 0 24 30 32 33 34 0 120 144 150 152 153 154Table IGFx is closely related to Sloane's 51683:

1 2 4 6 12 18 24 48 72 96 120 240 360 480 600 720 1440 2160 2880 3600 4320The rule for this triangle is given in 1-offset as Table:

The J *Phrases* book gives only a limited number of functions concerning rips.

NB. Phrases from 7.A NB. Rotate last three to the left m7 =: 3&A. NB. Rotate last three right m8 =: 4&A. NB. Rotate last x to the left d9 =: ([: +/[:![:}.[:i.[) A. ] NB. Rotate last x to the right d10=: (!@[ - !@<:@[) A. ]The functions given here expand the possibilities significantly. They are of dubious practical value, because the factorials grow so large so quickly that they really aren't practical to generate rips of any great length. For that, the function RfC is far more practical.

**Contracurrency**

All the while I was working on this material it had been bothering me that the Anagram index grew with respect to the end of the rip, not the beginning. For example,

A.The Anagram index is the same regardless of the length of the rip, when the infix is the same distance from the right end. When the infix is the same distance from the front end, it varies:1 2 03 A. 02 3 13 A. 0 13 4 23 A. 0 1 24 5 33

A.J supports contracurrent indexing, which is right-end oriented: the last item has contracurrent index _1, the penultimate has _2, and the antepenultimate has _3, regardless of the number of items. The previous tables were labeled by length of infix and distance from right edge, with the direction of 1-rotation used to distinguish two separate tables, as produced by the functions ILF and IGF. The function below produces a table labeled by contracurrent indices:1 2 03 A.1 2 03 8 A.1 2 03 4 30 A.1 2 03 4 5 144

RfMC=: 13 : '(-y.)]\A.RfC MC y.'This forms the table of all casps for rips of length y., then obtains the rips, next obtains the Anagram indices, and last, reshapes this into a square table.

] q =. |.->:i.8 _8 _7 _6 _5 _4 _3 _2 _1 q by q over RfMC 8 +--+----------------------------------------+ | | _8 _7 _6 _5 _4 _3 _2 _1| +--+----------------------------------------+ |_8| 0 5040 5760 5880 5904 5910 5912 5913| |_7| 5040 0 720 840 864 870 872 873| |_6|10080 720 0 120 144 150 152 153| |_5|15120 1440 120 0 24 30 32 33| |_4|20160 2160 240 24 0 6 8 9| |_3|25200 2880 360 48 6 0 2 3| |_2|30240 3600 480 72 12 2 0 1| |_1|35280 4320 600 96 18 4 1 0| +--+----------------------------------------+Entry (i,j) in this table is the Anagram index of a rip of any length in which item i has been moved to index j, where i and j are contracurrent indices.

Here are functions to convert between direct casps and contracurrent casps (ccasps):

NB. contracurrent casp from direct CCfD =: -~`,`:3"1 NB. direct casp from contracurrent DfCC =: (|@<. + 0: , ,)/"1Here is a little experiment with the above two functions which shows the limitations of the ccasp form:

MC3 =. MC 3 MCC3 =. CCfD MC3 MC3x =. DfCC MCC3 MCC3x =. CCfD MC3x format =. 2 2 2 5 3 5 2 2 5 3 format ": MC3,.MCC3,.MC3x,.MCC3x 3 0 0 _3 _3 3 0 0 _3 _3 3 0 1 _3 _2 3 0 1 _3 _2 3 0 2 _3 _1 3 0 2 _3 _1 3 1 0 _2 _3 3 1 0 _2 _3 3 1 1 _2 _2 2 0 0 _2 _2 3 1 2 _2 _1 2 0 1 _2 _1 3 2 0 _1 _3 3 2 0 _1 _3 3 2 1 _1 _2 2 1 0 _1 _2 3 2 2 _1 _1 1 0 0 _1 _1A ccasp needs only two items: the contracurrent indices of the item moved and where it is moved to. This loses the information of the length of the rip involved, and so the conversion from ccasp back to casp is not exact: the length imputed is the magnitude of the minimum item. For example, 3 2 1 is converted to _1 _2, but the back conversion is 2 1 0, since the magnitude of the minimum item _2 is 2. However, 2 1 0 is converted exactly back to _1 _2.