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Wed, 23 Jul 2014
When do n and 2n have the same digits?
[This article was published last month on the math.stackexchange blog, which seems to have died young, despite many earnest-sounding promises beforehand from people who claimed they would contribute material. I am repatriating it here.] A recent question on math.stackexchange asks for the smallest positive integer !!A!! for which the number !!2A!! has the same decimal digits in some other order. Math geeks may immediately realize that !!142857!! has this property, because it is the first 6 digits of the decimal expansion of !!\frac 17!!, and the cyclic behavior of the decimal expansion of !!\frac n7!! is well-known. But is this the minimal solution? It is not. Brute-force enumeration of the solutions quickly reveals that there are 12 solutions of 6 digits each, all permutations of !!142857!!, and that larger solutions, such as 1025874 and 1257489 seem to follow a similar pattern. What is happening here? Stuck in Dallas-Fort Worth airport one weekend, I did some work on the problem, and although I wasn't able to solve it completely, I made significant progress. I found a method that allows one to hand-calculate that there is no solution with fewer than six digits, and to enumerate all the solutions with 6 digits, including the minimal one. I found an explanation for the surprising behavior that solutions tend to be permutations of one another. The short form of the explanation is that there are fairly strict conditions on which sets of digits can appear in a solution of the problem. But once the set of digits is chosen, the conditions on that order of the digits in the solution are fairly lax. So one typically sees, not only in base 10 but in other bases, that the solutions to this problem fall into a few classes that are all permutations of one another; this is exactly what happens in base 10 where all the 6-digit solutions are permutations of !!124578!!. As the number of digits is allowed to increase, the strict first set of conditions relaxes a little, and other digit groups appear as solutions. NotationThe property of interest, !!P_R(A)!!, is that the numbers !!A!! and !!B=2A!! have exactly the same base-!!R!! digits. We would like to find numbers !!A!! having property !!P_R!! for various !!R!!, and we are most interested in !!R=10!!. Suppose !!A!! is an !!n!!-digit numeral having property !!P_R!!; let the (base-!!R!!) digits of !!A!! be !!a_{n-1}\ldots a_1a_0!! and similarly the digits of !!B = 2A!! are !!b_{n-1}\ldots b_1b_0!!. The reader is encouraged to keep in mind the simple example of !!R=8, n=4, A=\mathtt{1042}, B=\mathtt{2104}!! which we will bring up from time to time. Since the digits of !!B!! and !!A!! are the same, in a different order, we may say that !!b_i = a_{P(i)}!! for some permutation !!P!!. In general !!P!! might have more than one cycle, but we will suppose that !!P!! is a single cycle. All the following discussion of !!P!! will apply to the individual cycles of !!P!! in the case that !!P!! is a product of two or more cycles. For our example of !!a=\mathtt{1042}, b=\mathtt{2104}!!, we have !!P = (0\,1\,2\,3)!! in cycle notation. We won't need to worry about the details of !!P!!, except to note that !!i, P(i), P(P(i)), \ldots, P^{n-1}(i)!! completely exhaust the indices !!0. \ldots n-1!!, and that !!P^n(i) = i!! because !!P!! is an !!n!!-cycle. Conditions on the set of digits in a solutionFor each !!i!! we have $$a_{P(i)} = b_{i} \equiv 2a_{i} + c_i\pmod R $$ where the ‘carry bit’ !!c_i!! is either 0 or 1 and depends on whether there was a carry when doubling !!a_{i-1}!!. (When !!i=0!! we are in the rightmost position and there is never a carry, so !!c_0= 0!!.) We can then write: $$\begin{align} a_{P(P(i))} &= 2a_{P(i)} + c_{P(i)} \\ &= 2(2a_{i} + c_i) + c_{P(i)} &&= 4a_i + 2c_i + c_{P(i)}\\ a_{P(P(P(i)))} &= 2(4a_i + 2c_i + c_{P(P(i)})) + c_{P(i)} &&= 8a_i + 4c_i + 2c_{P(i)} + c_{P(P(i))}\\ &&&\vdots\\ a_{P^n(i)} &&&= 2^na_i + v \end{align} $$ all equations taken !!\bmod R!!. But since !!P!! is an !!n!!-cycle, !!P^n(i) = i!!, so we have $$a_i \equiv 2^na_i + v\pmod R$$ or equivalently $$\big(2^n-1\big)a_i + v \equiv 0\pmod R\tag{$\star$}$$ where !!v\in{0,\ldots 2^n-1}!! depends only on the values of the carry bits !!c_i!!—the !!c_i!! are precisely the binary digits of !!v!!. Specifying a particular value of !!a_0!! and !!v!! that satisfy this equation completely determines all the !!a_i!!. For example, !!a_0 = 2, v = \color{darkblue}{0010}_2 = 2!! is a solution when !!R=8, n=4!! because !!\bigl(2^4-1\bigr)\cdot2 + 2\equiv 0\pmod 8!!, and this solution allows us to compute $$\def\db#1{\color{darkblue}{#1}}\begin{align} a_0&&&=2\\ a_{P(0)} &= 2a_0 &+ \db0 &= 4\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 0 \\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 1\\ \hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \db0 &= 2\\ \end{align}$$ where the carry bits !!c_i = \langle 0,0,1,0\rangle!! are visible in the third column, and all the sums are taken !!\pmod 8!!. Note that !!a_{P^n(0)} = a_0!! as promised. This derivation of the entire set of !!a_i!! from a single one plus a choice of !!v!! is crucial, so let's see one more example. Let's consider !!R=10, n=3!!. Then we want to choose !!a_0!! and !!v!! so that !!\left(2^3-1\right)a_0 + v \equiv 0\pmod{10}!! where !!v\in{0\ldots 7}!!. One possible solution is !!a_0 = 5, v=\color{darkblue}{101}_2 = 5!!. Then we can derive the other !!a_i!! as follows: $$\begin{align} a_0&&&=5\\ a_{P(0)} &= 2a_0 &+ \db1 &= 1\\ a_{P^2(0)} &= 2a_{P(0)} &+ \db0 &= 2 \\\hline a_{P^3(0)} &= 2a_{P^2(0)} &+ \db1 &= 5\\ \end{align}$$ And again we have !!a_{P^n(0)}= a_0!! as required. Since the bits of !!v!! are used cyclically, not every pair of !!\langle a_0, v\rangle!! will yield a different solution. Rotating the bits of !!v!! and pairing them with different choices of !!a_0!! will yield the same cycle of digits starting from a different place. In the first example above, we had !!a_0 = 2, v = 0010_2 = 2!!. If we were to take !!a_0 = 4, v = 0100_2 = 4!! (which also solves !!(\star)!!) we would get the same cycle of values of the !!a_i!! but starting from !!4!! instead of from !!2!!, and similarly if we take !!a_0=0, v = 1000_2 = 8!! or !!a_0 = 1, v = 0001_2!!. So we can narrow down the solution set of !!(\star)!! by considering only the so-called bracelets of !!v!! rather than all !!2^n!! possible values. Two values of !!v!! are considered equivalent as bracelets if one is a rotation of the other. When a set of !!v!!-values are equivalent as bracelets, we need only consider one of them; the others will give the same cyclic sequence of digits, but starting in a different place. For !!n=4!!, for example, the bracelets are !!0000, 0001, 0011, 0101, 0111, !! and !!1111!!; the sequences !!0110, 1100,!! and !!1001!! being equivalent to !!0011!!, and so on. ExampleLet us take !!R=9, n=3!!, so we want to find 3-digit numerals with property !!P_9!!. According to !!(\star)!! we need !!7a_i + v \equiv 0\pmod{9}!! where !!v\in{0\ldots 7}!!. There are 9 possible values for !!a_i!!; for each one there is at most one possible value of !!v!! that makes the sum zero: $$\pi \approx 3 $$ $$\begin{array}{rrr} a_i & 7a_i & v \\ \hline 0 & 0 & 0 \\ 1 & 7 & 2 \\ 2 & 14 & 4 \\ 3 & 21 & 6 \\ 4 & 28 & \\ 5 & 35 & 1 \\ 6 & 42 & 3 \\ 7 & 49 & 5 \\ 8 & 56 & 7 \\ \end{array} $$ (For !!a_i=4!! there is no solution.) We may disregard the non-bracelet values of !!v!!, as these will give us solutions that are the same as those given by bracelet values of !!v!!. The bracelets are: $$\begin{array}{rl} 000 & 0 \\ 001 & 1 \\ 011 & 3 \\ 111 & 7 \end{array}$$ so we may disregard the solutions exacpt when !!v=0,1,3,7!!. Calculating the digit sequences from these four values of !!v!! and the corresponding !!a_i!! we find: $$\begin{array}{ccl} a_0 & v & \text{digits} \\ \hline 0 & 0 & 000 \\ 5 & 1 & 512 \\ 6 & 3 & 637 \\ 8 & 7 & 888 \ \end{array} $$ (In the second line, for example, we have !!v=1 = 001_2!!, so !!1 = 2\cdot 5 + 0; 2 = 1\cdot 2 + 0;!! and !!5 = 2\cdot 2 + 1!!.) Any number !!A!! of three digits, for which !!2A!! contains exactly the same three digits, in base 9, must therefore consist of exactly the digits !!000, 125, 367,!! or !!888!!. A warningAll the foregoing assumes that the permutation !!P!! is a single cycle. In general, it may not be. Suppose we did an analysis like that above for !!R=10, n=5!! and found that there was no possible digit set, other than the trivial set Something like this occurs, for example, in the !!n=4, R=8!! case. Solving the governing equation !!(2^5-1)a_0 + v \equiv 0\pmod 8!! yields only four possible digit cycles, namely !!{0,1,2,4}, {1,3,6,4}, {2,5,2,5}!!, and !!{3,7,6,5}!!. But there are several additional solutions: !!2500_8\cdot 2 = 5200_8, 2750_8\cdot 2 = 5720_8, !! and !!2775_8\cdot 2 = 5772_8!!. These correspond to permutations !!P!! with more than one cycle. In the case of !!5720_8!!, for example, !!P!! exchanges the !!5!! and the !!2!!, and leaves the !!0!! and the !!7!! fixed. For this reason we cannot rule out the possibility of an !!n!!-digit solution without first considering all smaller !!n!!. The Large Equals Odd ruleWhen !!R!! is even there is a simple condition we can use to rule out certain sets of digits from being single-cycle solutions. Recall that !!A=a_{n-1}\ldots a_0!! and !!B=b_{n-1}\ldots b_0!!. Let us agree that a digit !!d!! is large if !!d\ge \frac R2!! and small otherwise. That is, !!d!! is large if, upon doubling, it causes a carry into the next column to the left. Since !!b_i =(2a_i + c_i)\bmod R!!, where the !!c_i!! are carry bits, we see that, except for !!b_0!!, the digit !!b_i!! is odd precisely when there is a carry from the next column to the right, which occurs precisely when !!a_{i-1}!! is large. Thus the number of odd digits among !!b_1,\ldots b_{n-1}!! is equal to the number of large digits among !!a_1,\ldots a_{n-2}!!. This leaves the digits !!b_0!! and !!a_{n-1}!! uncounted. But !!b_0!! is never odd, since there is never a carry in the rightmost position, and !!a_{n-1}!! is always small (since otherwise !!B!! would have !!n+1!! digits, which is not allowed). So the number of large digits in !!A!! is exactly equal to the number of odd digits in !!B!!. And since !!A!! and !!B!! have exactly the same digits, the number of large digits in !!A!! is equal to the number of odd digits in !!A!!. Observe that this is the case for our running example !!1042_8!!: there is one odd digit and one large digit (the 4). When !!R!! is odd the analogous condition is somewhat more complicated, but since the main case of interest is !!R=10!!, we have the useful rule that: For !!R!! even, the number of odd digits in any solution !!A!! is equal to the number of large digits in !!A!!. Conditions on the order of digits in a solutionWe have determined, using the above method, that the digits !!{5,1,2}!! might form a base-9 numeral with property !!P_9!!. Now we would like to arrange them into a base-9 numeral that actually does have that property. Again let us write !!A = a_2a_1a_0!! and !!B=b_2b_1b_0!!, with !!B=2A!!. Note that if !!a_i = 1!!, then !!b_i = 3!! (if there was a carry from the next column to the right) or !!2!! (if there was no carry), but since !!b_i=3!! is impossible, we must have !!a_i = 2!! and therefore !!a_{i-1}!! must be small, since there is no carry into position !!i!!. But since !!a_{i-1}!! is also one of !!{5,1,2}!!, and it cannot also be !!1!!, it must be !!2!!. This shows that the 1, unless it appears in the rightmost position, must be to the left of the !!2!!; it cannot be to the left of the !!5!!. Similarly, if !!a_i = 2!! then !!b_i = 5!!, because !!4!! is impossible, so the !!2!! must be to the left of a large digit, which must be the !!5!!. Similar reasoning produces no constraint on the position of the !!5!!; it could be to the left of a small digit (in which case it doubles to !!1!!) or a large digit (in which case it doubles to !!2!!). We can summarize these findings as follows: $$\begin{array}{cl} \text{digit} & \text{to the left of} \\ \hline 1 & 1, 2, \text{end} \\ 2 & 5 \\ 5 & 1,2,5,\text{end} \end{array}$$ Here “end” means that the indicated digit could be the rightmost. Furthermore, the left digit of !!A!! must be small (or else there would be a carry in the leftmost place and !!2A!! would have 4 digits instead of 3) so it must be either 1 or 2. It is not hard to see from this table that the digits must be in the order !!125!! or !!251!!, and indeed, both of those numbers have the required property: !!125_9\cdot 2 = 251_9!!, and !!251_9\cdot 2 = 512_9!!. This was a simple example, but in more complicated cases it is helpful to draw the order constraints as a graph. Suppose we draw a graph with one vertex for each digit, and one additional vertex to represent the end of the numeral. The graph has an edge from vertex !!v!! to !!v'!! whenever !!v!! can appear to the left of !!v'!!. Then the graph drawn for the table above looks like this: A 3-digit numeral with property !!P_9!! corresponds to a path in this graph that starts at one of the nonzero small digits (marked in blue), ends at the red node marked ‘end’, and visits each node exactly once. Such a path is called hamiltonian. Obviously, self-loops never occur in a hamiltonian path, so we will omit them from future diagrams. Now we will consider the digit set !!637!!, again base 9. An analysis similar to the foregoing allows us to construct the following graph: Here it is immediately clear that the only hamiltonian path is !!3-7-6-\text{end}!!, and indeed, !!376_9\cdot 2 = 763_9!!. In general there might be multiple instances of a digit, and so multiple nodes labeled with that digit. Analysis of the !!0,0,0!! case produces a graph with no legal start nodes and so no solutions, unless leading zeroes are allowed, in which case !!000!! is a perfectly valid solution. Analysis of the !!8,8,8!! case produces a graph with no path to the end node and so no solutions. These two trivial patterns appear for all !!R!! and all !!n!!, and we will ignore them from now on. Returning to our ongoing example, !!1042!! in base 8, we see that !!1!! and !!2!! must double to !!2!! and !!4!!, so must be to the left of small digits, but !!4!! and !!0!! can double to either !!0!! or !!1!! and so could be to the left of anything. Here the constraints are so lax that the graph doesn't help us narrow them down much: Observing that the only arrow into the 4 is from 0, so that the 4 must follow the 0, and that the entire number must begin with 1 or 2, we can enumerate the solutions: 1042 1204 2041 2104 If leading zeroes are allowed we have also: 0412 0421 All of these are solutions in base 8. The case of !!R=10!!Now we turn to our main problem, solutions in base 10. To find all the solutions of length 6 requires an enumeration of smaller solutions, which, if they existed, might be concatenated into a solution of length 6. This is because our analysis of the digit sets that can appear in a solution assumes that the digits are permuted cyclically; that is, the permutations !!P!! that we considered had only one cycle each. There are no smaller solutions, but to prove that the length 6 solutions are minimal, we must analyze the cases for smaller !!n!! and rule them out. We now produce a complete analysis of the base 10 case with !!R=10!! and !!n\le 6!!. For !!n=1!! there is only the trivial solution of !!0!!, which we disregard. (The question asked for a positive number anyway.) !!n=2!!For !!n=2!!, we want to find solutions of !!3a_i + v \equiv 0\pmod{10}!! where !!v!! is a two-bit bracelet number, one of !!00_2, 01_2, !! or !!11_2!!. Tabulating the values of !!a_i!! and !!v\in{0,1,3}!! that solve this equation we get: $$\begin{array}{ccc} v& a_i \\ \hline 0 & 0 \\ 1& 3 \\ 3& 9 \\ \end{array}$$ We can disregard the !!v=0!! and !!v=3!! solutions because the former yields the trivial solution !!00!! and the latter yields the nonsolution !!99!!. So the only possibility we need to investigate further is !!a_i = 3, v = 1!!, which corresponds to the digit sequence !!36!!: Doubling !!3!! gives us !!6!! and doubling !!6!!, plus a carry, gives us !!3!! again. But when we tabulate of which digits must be left of which informs us that there is no solution with just !!3!! and !!6!!, because the graph we get, once self-loops are eliminated, looks like this: which obviously has no hamiltonian path. Thus there is no solution for !!R=10, n=2!!. !!n=3!!For !!n=3!! we need to solve the equation !!7a_i + v \equiv 0\pmod{10}!! where !!v!! is a bracelet number in !!{0,\ldots 7}!!, specifically one of !!0,1,3,!! or !!7!!. Since !!7!! and !!10!! are relatively prime, for each !!v!! there is a single !!a_i!! that solves the equation. Tabulating the possible values of !!a_i!! as before, and this time omitting rows with no solution, we have: $$\begin{array}{rrl} v & a_i & \text{digits}\\ \hline 0& 0 & 000\\ 1& 7 & 748 \\ 3& 1 & 125\\ 7&9 & 999\\ \end{array}$$ The digit sequences !!0,0,0!! and !!9,9,9!! yield trivial solutions or nonsolutions as usual, and we will omit them in the future. The other two lines suggest the digit sets !!1,2,5!! and !!4,7,8!!, both of which fails the “odd equals large” rule. This analysis rules out the possibility of a digit set with !!a_0 \to a_1 \to a_2 \to a_1!!, but it does not completely rule out a 3-digit solution, since one could be obtained by concatenating a one-digit and a two-digit solution, or three one-digit solutions. However, we know by now that no one- or two-digit solutions exist. Therefore there are no 3-digit solutions in base 10. !!n=4!!For !!n=4!! the governing equation is !!15a_i + v \equiv 0\pmod{10}!! where !!v!! is a 4-bit bracelet number, one of !!{0,1,3,5,7,15}!!. This is a little more complicated because !!\gcd(15,10)\ne 1!!. Tabulating the possible digit sets, we get: $$\begin{array}{crrl} a_i & 15a_i& v & \text{digits}\\ \hline 0 & 0 & 0 & 0000\\ 1 & 5 & 5 & 1250\\ 1 & 5 & 15 & 1375\\ 2 & 0 & 0 & 2486\\ 3 & 5 & 5 & 3749\\ 3 & 5 & 15 & 3751\\ 4 & 0 & 0 & 4862\\ 5 & 5 & 5 & 5012\\ 5 & 5 & 5 & 5137\\ 6 & 0 & 0 & 6248\\ 7 & 5 & 5 & 7493\\ 7 & 5 & 5 & 7513\\ 8 & 0 & 0 & 8624 \\ 9 & 5 & 5 & 9874\\ 9 & 5 & 15 & 9999 \\ \end{array}$$ where the second column has been reduced mod !!10!!. Note that even restricting !!v!! to bracelet numbers the table still contains duplicate digit sequences; the 15 entries on the right contain only the six basic sequences !!0000, 0125, 1375, 2486, 3749, 4987!!, and !!9999!!. Of these, only !!0000, 9999,!! and !!3749!! obey the odd equals large criterion, and we will disregard !!0000!! and !!9999!! as usual, leaving only !!3749!!. We construct the corresponding graph for this digit set as follows: !!3!! must double to !!7!!, not !!6!!, so must be left of a large number !!7!! or !!9!!. Similarly !!4!! must be left of !!7!! or !!9!!. !!9!! must also double to !!9!!, so must be left of !!7!!. Finally, !!7!! must double to !!4!!, so must be left of !!3,4!! or the end of the numeral. The corresponding graph is: which evidently has no hamiltonian path: whichever of 3 or 4 we start at, we cannot visit the other without passing through 7, and then we cannot reach the end node without passing through 7 a second time. So there is no solution with !!R=10!! and !!n=4!!. !!n=5!!We leave this case as an exercise. There are 8 solutions to the governing equation, all of which are ruled out by the odd equals large rule. !!n=6!!For !!n=6!! the possible solutions are given by the governing equation !!63a_i + v \equiv 0\pmod{10}!! where !!v!! is a 6-bit bracelet number, one of !!{0,1,3,5,7,9,11,13,15,21,23,27,31,63}!!. Tabulating the possible digit sets, we get: $$\begin{array}{crrl} v & a_i & \text{digits}\\ \hline 0 & 0 & 000000\\ 1 & 3 & 362486 \\ 3 & 9 & 986249 \\ 5 & 5 & 500012 \\ 7 & 1 & 124875 \\ 9 & 7 & 748748 \\ 11 & 3 & 362501 \\ 13 & 9 & 986374 \\ 15 & 5 & 500137 \\ 21 & 3 & 363636 \\ 23 & 9 & 989899 \\ 27 & 1 & 125125 \\ 31 & 3 & 363751 \\ 63 & 9 & 999999 \\ \end{array}$$ After ignoring !!000000!! and !!999999!! as usual, the large equals odd rule allows us to ignore all the other sequences except !!124875!! and !!363636!!. The latter fails for the same reason that !!36!! did when !!n=2!!. But !!142857!! , the lone survivor, gives us a complicated derived graph containing many hamiltonian paths, every one of which is a solution to the problem: It is not hard to pick out from this graph the minimal solution !!125874!!, for which !!125874\cdot 2 = 251748!!, and also our old friend !!142857!! for which !!142857\cdot 2 = 285714!!. We see here the reason why all the small numbers with property !!P_{10}!! contain the digits !!124578!!. The constraints on which digits can appear in a solution are quite strict, and rule out all other sequences of six digits and all shorter sequences. But once a set of digits passes these stringent conditions, the constraints on it are much looser, because !!B!! is only required to have the digits of !!A!! in some order, and there are many possible orders, many of which will satisfy the rather loose conditions involving the distribution of the carry bits. This graph is typical: it has a set of small nodes and a set of large nodes, and each node is connected to either all the small nodes or all the large nodes, so that the graph has many edges, and, as in this case, a largish clique of small nodes and a largish clique of large nodes, and as a result many hamiltonian paths. OnwardThis analysis is tedious but is simple enough to perform by hand in under an hour. As !!n!! increases further, enumerating the solutions of the governing equation becomes very time-consuming. I wrote a simple computer program to perform the analysis for given !!R!! and !!n!!, and to emit the possible digit sets that satisfied the large equals odd criterion. I had wondered if every base-10 solution contained equal numbers of the digits !!1,2,4,8,5,!! and !!7!!. This is the case for !!n=7!! (where the only admissible digit set is !!{1,2,4,5,7,8}\cup{9}!!), for !!n=8!! (where the only admissible sets are !!{1,2,4,5,7,8}\cup {3,6}!! and !!{1,2,4,5,7,8}\cup{9,9}!!), and for !!n=9!! (where the only admissible sets are !!{1,2,4,5,7,8}\cup{3,6,9}!! and !!{1,2,4,5,7,8}\cup{9,9,9}!!). But when we reach !!n=10!! the increasing number of bracelets has loosened up the requirements a little and there are 5 admissible digit sets. I picked two of the promising-seeming ones and quickly found by hand the solutions !!4225561128!! and !!1577438874!!, both of which wreck any theory that the digits !!1,2,4,5,8,7!! must all appear the same number of times. AcknowledgmentsThanks to Karl Kronenfeld for corrections and many helpful suggestions. [Other articles in category /math] permanent link Sun, 20 Jul 2014
Similarity analysis of quilt blocks
As I've discussed elsewhere, I once wrote a program to enumerate all the possible quilt blocks of a certain type. The quilt blocks in question are, in quilt jargon, sixteen-patch half-square triangles. A half-square triangle, also called a “patch”, is two triangles of fabric sewn together, like this: Then you sew four of these patches into a four-patch, say like this: Then to make a sixteen-patch block of the type I was considering, you take four identical four-patch blocks, and sew them together with rotational symmetry, like this: It turns out that there are exactly 72 different ways to do this. (Blocks equivalent under a reflection are considered the same, as are blocks obtained by exchanging the roles of black and white, which are merely stand-ins for arbitrary colors to be chosen later.) Here is the complete set of 72:
It's immediately clear that some of these resemble one another, sometimes so strongly that it can be hard to tell how they differ, while others are very distinctive and unique-seeming. I wanted to make the computer classify the blocks on the basis of similarity. My idea was to try to find a way to get the computer to notice which blocks have distinctive components of one color. For example, many blocks have a distinctive diamond shape in the center. Some have a pinwheel like this: which also has the diamond in the middle, while others have a different kind of pinwheel with no diamond: I wanted to enumerate such components and ask the computer to list which blocks contained which shapes; then group them by similarity, the idea being that that blocks with the same distinctive components are similar. The program suite uses a compact notation of blocks and of shapes that makes it easy to figure out which blocks contain which distinctive components. Since each block is made of four identical four-patches, it's enough just to examine the four-patches. Each of the half-square triangle patches can be oriented in two ways:
Here are two of the 12 ways to orient the patches in a four-patch:
Each 16-patch is made of four four-patches, and you must imagine that the
four-patches shown above are in the upper-left position in the
16-patch. Then symmetry of the 16-patch block means that triangles with the
same label are in positions that are symmetric with respect to the
entire block. For example, the two triangles labeled Triangles must be colored opposite colors if they are part of the same patch, but other than that there are no constraints on the coloring. A block might, of course, have patches in both orientations: All the blocks with diagonals oriented this way are assigned
descriptors made from the letters Once you have chosen one of the 12 ways to orient the diagonals in the
four-patch, you still have to color the patches. A descriptor like
In each case, all four diagonals run from northwest to southeast. (All other ways of coloring this four-patch are equivalent to one of these under one or more of rotation, reflection, and exchange of black and white.) We can describe a patch by listing the descriptors of the eight triangles, grouped by which triangles form connected regions. For example, the first block above is:
because there's an isolated white The other five
All six have I made up a list of the descriptors for all 72 blocks; I think I did
this by hand. (The work directory contains a and it will only be significant if the in which case you get this distinctive and interesting-looking hook component. There is only one block that includes this distinctive hook component;
it has descriptor and the blocks formed from such patches always have a distinctive half-diamond component on each edge, like this: (The stippled areas vary from block to block, but the blocks with The blocks listed at http://hop.perl.plover.com/quilt/analysis/images/ee.html
all have the Other distinctive components have similar short descriptors. The two pinwheels I
mentioned above are
Descriptor so the full sixteen-patch looks like this: where the stippled parts can vary. A look at the list of blocks with
component
I had made a list of the descriptors for each of the the 72 blocks, and from this I extracted a list of the descriptors for interesting component shapes. Then it was only a matter of finding the component descriptors in the block descriptors to know which blocks contained which components; if the two blocks share two different distinctive components, they probably look somewhat similar. Then I sorted the blocks into groups, where two blocks were in the same group if they shared two distinctive components. The resulting grouping lists, for each block, which other blocks have at least two shapes in common with it. Such blocks do indeed tend to look quite similar. This strategy was actually the second thing I tried; the first thing didn't work out well. (I forget just what it was, but I think it involved finding polygons in each block that had white inside and black outside, or vice versa.) I was satisfied enough with this second attempt that I considered the project a success and stopped work on it. The complete final results were:
And these tabulations of all the blocks with various distinctive components: bd bf bfh bfi cd cdd cdf cf cfi ee eg egh egi fgh fh fi gg ggh ggi gh gi j It may also be interesting to browse the work directory. [Other articles in category /misc] permanent link Fri, 18 Jul 2014
On uninhabited types and inconsistent logics
Earlier this week I gave a talk about the Curry-Howard isomorphism. Talks never go quite the way you expect. The biggest sticking point was my assertion that there is no function with the type a → b. I mentioned this as a throwaway remark on slide 7, assuming that everyone would agree instantly, and then we got totally hung up on it for about twenty minutes. Part of this was my surprise at discovering that most of the audience (members of the Philly Lambda functional programming group) was not familiar with the Haskell type system. I had assumed that most of the members of a functional programming interest group would be familiar with one of Haskell, ML, or Scala, all of which have the same basic type system. But this was not the case. (Many people are primarily interested in Scheme, for example.) I think the main problem was that I did not make clear to the audience what Haskell means when it says that a function has type a → b. At the talk, and then later on Reddit people asked
If you know one of the HM languages, you know that of course it
doesn't; it has type A function which takes an integer and returns a string does not have
the type a → b; it has the type Int → String. You must pass it an
integer, and you may only use its return value in a place that makes
sense for a string. If f has this type, then But if f had
the type a → b, then Say function f had type a → b. Then you would be able to use the
expression
and they would all type check correctly, regardless of the type of x. In the first line, f x would return a number; in the second line f would return a list; in the third line it would return a string, and in the fourth line it would return a boolean. And in each case f could be able to do what was required regardless of the type of x, so without even looking at x. But how could you possibly write such a function f? You can't; it's impossible. Contrast this with the identity function
as long as x has the right type for
as long as Actually I wonder now if part of the problem is that we like to write a → b when what we really mean is the type ∀a.∀b.a → b. Perhaps making the quantifiers explicit would clear things up? I suppose it probably wouldn't have, at least in this case. The issue is a bit complicated by the fact that the function
does have the type a → b, and, in a language with exceptions,
Unfortunately, just as I thought I was getting across the explanation of why there can be no function with type a → b, someone brought up exceptions and I had to mutter and look at my shoes. (You can also take the view that these functions have type a → ⊥, but the logical principle ⊥ → b is unexceptionable.) In fact, experienced practitioners will realize, the instant the type
a → b appears, that they have written a function that never returns.
Such an example was directly responsible for my own initial interest
in functional programming and type systems; I read a 1992 paper (“An
anecdote about ML type
inference”)
by Andrew R. Koenig in which he described writing a merge sort
function, whose type was reported (by the SML type inferencer) as Any discussion of the Curry-Howard isomorphism, using Haskell as an example, is somewhat fraught with trouble, because Haskell's type logic is utterly inconsistent. In addition to the examples above, in Haskell one can write
and as a statement of logic, !!(a\to a)\to a!! is patently false. This might be an argument in favor of the Total Functional Programming suggested by D.A. Turner and others. [Other articles in category /CS] permanent link Thu, 17 Jul 2014
Guess what this does (solution)
A few weeks ago I asked people to predict, without trying it first, what this would print:
(If you haven't seen this yet, I recommend that you guess, and then test your guess, before reading the rest of this article.) People familiar with Perl guess that it will print So far this is a little bit odd, but not excessively odd; it's the
sort of thing you expect from programming languages, all of which more
or less suck. For example, Python's behavior, although different, is
about equally peculiar. Although Python does require that the strings
However, this is not what the program does do. The explanation of two paragraphs earlier is totally wrong. Instead, the program prints nothing, and the reason is incredibly convoluted and bizarre. First, you must know that
The former prints out the list The answer to this question is further complicated by the fact that
the
Perl's method to decide whether a particular
Here the intention is that the So for example, in
the
the In
Perl has a puzzle: Now we return the the original example.
Here Perl sees the unquoted string
and obtains the value This value, the string
then the output would have been sent to the filehandle as usual.
Printing to a non-open filehandle elicits an optional warning from
Perl, but as I mentioned, I have not enabled warnings, so the Had I enabled those optional warnings, we would have seen a plethora of them:
(The first four are compile-time warnings; the last three are issued
at execution time.) The crucial warning is the one at the end,
advising us that the output of [ Addendum 20140718: I keep thinking of the following remark of Edsger W. Dijkstra:
But my intent is different than what Dijkstra describes. His programmer is proud, but I am discgusted. Incidentally, I believe that Dijkstra was discussing APL here. ] [Other articles in category /prog/perl] permanent link Wed, 16 Jul 2014Here's a Perl quiz that I confidently predict nobody will get right. Without trying it first, what does the following program print?
(I will discuss the surprising answer tomorrow.) [Other articles in category /prog/perl] permanent link Tue, 15 Jul 2014
Types are theorems; programs are proofs
[ Summary: I gave a talk Monday night on the Curry-Howard isomorphism; my talk slides are online. ] I sent several proposals to !!con, a conference of ten-minute talks. One of my proposals was to explain the Curry-Howard isomorphism in ten minutes, but the conference people didn't accept it. They said that they had had four or five proposals for talks about the Curry-Howard isomorphism, and didn't want to accept more than one of them. The CHI talk they did accept turned out to be very different from the one I had wanted to give; it discussed the Coq theorem-proving system. I had wanted to talk about the basic correspondence between pair types, union types, and function types on the one hand, and reasoning about logical conjunctions, disjunctions, and implications on the other hand, and the !!con speaker didn't touch on this at all. But mathematical logic and programming language types turn out to be the same! A type in a language like Haskell can be understood as a statement of logic, and the statement will be true if and only if there is actually a value with the corresponding type. Moreover, if you have a proof of the statement, you can convert the proof into a value of the corresponding type, or conversely if you have a value with a certain type you can convert it into a proof of the corresponding statement. The programming language features for constructing or using values with function types correspond exactly to the logical methods for proving or using statements with implications; similarly pair types correspond to logical conjunction, and union tpyes to logical disjunction, and exceptions to logical negation. I think this is incredible. I was amazed the first time I heard of it (Chuck Liang told me sometime around 1993) and I'm still amazed. Happily Philly Lambda, a Philadelphia-area functional programming group, had recently come back to life, so I suggested that I give them a longer talk about about the Curry-Howard isomorphism, and they agreed. I gave the talk yesterday, and the materials are online. I'm not sure how easy they will be to understand without my commentary, but it might be worth a try. If you're interested and want to look into it in more detail, I suggest you check out Sørensen and Urzyczyn's Lectures on the Curry-Howard Isomorphism. It was published as an expensive yellow-cover book by Springer, but free copies of the draft are still available. [Other articles in category /CS] permanent link Sat, 26 Apr 2014Last night I gave a talk for the New York Perl Mongers, and got to see a number of people that I like but don't often see. Among these was Michael Fischer, who told me of a story about myself that I had completely forgotten, but I think will be of general interest.
The front end of the story is this: Michael first met me at some conference, shortly after the publication of Higher-Order Perl, and people were coming up to me and presenting me with copies of the book to sign. In many cases these were people who had helped me edit the book, or who had reported printing errors; for some of those people I would find the error in the text that they had reported, circle it, and write a thank-you note on the same page. Michael did not have a copy of my book, but for some reason he had with him a copy of Oulipo Compendium, and he presented this to me to sign instead. Oulipo is a society of writers, founded in 1960, who pursue “constrained writing”. Perhaps the best-known example is the lipogrammatic novel La Disparition, written in 1969 by Oulipo member Georges Perec, entirely without the use of the letter e. Another possibly well-known example is the Exercises in Style of Raymond Queneau, which retells the same vapid anecdote in 99 different styles. The book that Michael put in front of me to sign is a compendium of anecdotes, examples of Oulipan work, and other Oulipalia. What Michael did not realize, however, was that the gods of fate were handing me an opportunity. He says that I glared at him for a moment, then flipped through the pages, found the place in the book where I was mentioned, circled it, and signed that. The other half of that story is how I happened to be mentioned in Oulipo Compendium. Back in the early 1990s I did a few text processing projects which would be trivial now, but which were unusual at the time, in a small way. For example, I constructed a concordance of the King James Bible, listing, for each word, the number of every verse in which it appeared. This was a significant effort at the time; the Bible was sufficiently large (around five megabytes) that I normally kept the files compressed to save space. This project was surprisingly popular, and I received frequent email from strangers asking for copies of the concordance. Another project, less popular but still interesting, was an anagram
dictionary. The word list from Webster's Second International
dictionary was available, and it was an easy matter to locate all the
anagrams in it, and compile a file. Unlike the Bible concordance,
which I considered inferior to simply running
And ends:
The cross-references are to save space. When two words are anagrams of one another, both are listed in both places. But when three or more words are anagrams, the words are listed in one place, with cross-references in the other places, so for example:
saves 52 characters over the unabbreviated version. Even with this optimization, the complete anagram dictionary was around 750 kilobytes, a significant amount of space in 1991. A few years later I generated an improved version, which dispensed with the abbreviation, by that time unnecessary, and which attempted, sucessfully I thought, to score the anagrams according to interestingness. But I digress. One day in August of 1994, I received a query about the anagram dictionary, including a question about whether it could be used in a certain way. I replied in detail, explaining what I had done, how it could be used, and what could be done instead, and the result was a reply from Harry Mathews, another well-known member of the Oulipo, of which I had not heard before. Mr. Mathews, correctly recognizing that I would be interested, explained what he was really after:
Mathews wanted, not exactly an anagram dictionary, but a list of words acceptable for the English version of "ulcerations". They should contain only the letters a d e h i l n o r s t, at most once each. In particular, he wanted a word containing precisely these eleven letters, to use as the translation of "ulcerations". Producing the requisite list was much easier then producing the anagram dictionary iself, so I quickly did it and sent it back; it looked like this:
The leftmost column is the alphabetical list of letters. This is so that if you find yourself needing to use the letters 'a d e h s' at some point in your poem, you can jump to that part of the list and immediately locate the words containing exactly those letters. (It provides somewhat less help for discovering the shorter words that contain only some of those letters, but there is a limit to how much can be done with static files.) As can be seen at the end of the list, there were three words that each used ten of the eleven required letters: “hortensial”, “threnodial”, “disenthral”, but none with all eleven. However, Mathews replied:
(A threnody is a hymn of mourning.) A few years later, the Oulipo Compendium appeared, edited by Mathews, and the article on Threnodials mentions my assistance. And so it was that when Michael Fischer handed me a copy, I was able to open it up to the place where I was mentioned. [ Addendum 20140428: Thanks to Philippe Bruhat for some corrections: neither Perec nor Mathews was a founding member of Oulipo. ] [Other articles in category /lang] permanent link Sat, 01 Mar 2014This week there has been an article floating around about “What happens when placeholder text doesn't get replaced. This reminds me of the time I made this mistake myself. In 1996 I was programming a web site for a large company which sold cosmetics and skin care products in hundreds of department stores and malls around the country. The technology to actually buy the stuff online wasn't really mature yet, and the web was new enough that the company was worried that selling online would anger the retail channels. They wanted an a web page where you would put in your location and it would tell you where the nearby stores were. The application was simple; it accepted a city and state, looked them
up in an on-disk hash table, and then returned a status code to the
page generator. The status code was for internal use only. For
example, if you didn't fill in the form completely, the program would
return the status code If the form was filled out correctly, but there was no match in the
database, the program would return a status code that the front end
translated to a suitably apologetic message. The status code I
selected for this was Which was all very jolly, until one day there was a program bug and
some user in Podunk, Iowa submitted the form and got back a page with
Anyone could have seen that coming; I have no excuse. [Other articles in category /oops] permanent link Intuitionistic logic is deeply misunderstood by people who have not studied it closely; such people often seem to think that the intuitionists were just a bunch of lunatics who rejected the law of the excluded middle for no reason. One often hears that intuitionistic logic rejects proof by contradiction. This is only half true. It arises from a typically classical misunderstanding of intuitionistic logic. Intuitionists are perfectly happy to accept a reductio ad absurdum proof of the following form: $$(P\to \bot)\to \lnot P$$ Here !!\bot!! means an absurdity or a contradiction; !!P\to \bot!! means that assuming !!P!! leads to absurdity, and !!(P\to \bot)\to \lnot P!! means that if assuming !!P!! leads to absurdity, then you can conclude that !!P!! is false. This is a classic proof by contradiction, and it is intuitionistically valid. In fact, in many formulations of intuitionistic logic, !!\lnot P!! is defined to mean !!P\to \bot!!. What is rejected by intuitionistic logic is the similar-seeming claim that: $$(\lnot P\to \bot)\to P$$ This says that if assuming !!\lnot P!! leads to absurdity, you can conclude that !!P!! is true. This is not intuitionistically valid. This is where people become puzzled if they only know classical logic. “But those are the same thing!” they cry. “You just have to replace !!P!! with !!\lnot P!! in the first one, and you get the second.” Not quite. If you replace !!P!! with !!\lnot P!! in the first one, you do not get the second one; you get: $$(\lnot P\to \bot)\to \lnot \lnot P$$ People familiar with classical logic are so used to shuffling the !!\lnot !! signs around and treating !!\lnot \lnot P!! the same as !!P!! that they often don't notice when they are doing it. But in intuitionistic logic, !!P!! and !!\lnot \lnot P!! are not the same. !!\lnot \lnot P!! is weaker than !!P!!, in the sense that from !!P!! one can always conclude !!\lnot \lnot P!!, but not always vice versa. Intuitionistic logic is happy to agree that if !!\lnot P!! leads to absurdity, then !!\lnot \lnot P!!. But it does not agree that this is sufficient to conclude !!P!!. As is often the case, it may be helpful to try to understand intuitionistic logic as talking about provability instead of truth. In classical logic, !!P!! means that !!P!! is true and !!\lnot P!! means that !!P!! is false. If !!P!! is not false it is true, so !!\lnot \lnot P!! and !!P!! mean the same thing. But in intuitionistic logic !!P!! means that !!P!! is provable, and !!\lnot P!! means that !!P!! is not provable. !!\lnot \lnot P!! means that it is impossible to prove that !!P!! is not provable. If !!P!! is provable, it is certainly impossible to prove that !!P!! is not provable. So !!P!! implies !!\lnot \lnot P!!. But just because it is impossible to prove that there is no proof of !!P!! does not mean that !!P!! itself is provable, so !!\lnot \lnot P!! does not imply !!P!!. Similarly, $$(P\to \bot)\to \lnot P $$ means that if a proof of !!P!! would lead to absurdity, then we may conclude that there cannot be a proof of !!P!!. This is quite valid. But $$(\lnot P\to \bot)\to P$$ means that if assuming that a proof of !!P!! is impossible leads to absurdity, there must be a proof of !!P!!. But this itself isn't a proof of !!P!!, nor is it enough to prove !!P!!; it only shows that there is no proof that proofs of !!P!! are impossible. [Other articles in category /math] permanent link Thu, 27 Feb 2014
2banner, which tells you when someone else is looking at the same web page
I was able to release a pretty nice piece of software today, courtesy
of my employer, ZipRecruiter. If you have a family of web pages, and
whenever you are looking at one you want to know when someone else is
looking at the same page, you can use my package. The package is
called A typical use case would be a customer service organization. Say your
users create requests for help, and the the customer service reps have
to answer the requests. There is a web page with a list of all the
unserviced requests, and each one has a link to a page with details
about what is requested and how to contact the person who made the
request. But it might sometimes happes that Fred and Mary
independently decide to service the same request, which is at best a
waste of effort, and at worst confusing for the customer who gets
email from both Fred and Mary and doesn't know how to respond. With
You can similarly trick out the menu page itself, to hide the menu items that someone is already looking out. I wanted to use someone else's package for this, but I was not able to find one, so I wrote one myself. It was not as hard as I had expected. The system comprises three components:
Often a project seems easy but the more I think about it the harder it seems. This project was the opposite. I thought it sounded hard, and I didn't want to do it. It had been an outstanding request of the CS people for some time, but I guess everyone else thought it sounded hard too, because nobody did anything about it. But the longer I let it percolate around in my head, the simpler it seemed. As I considered it, one difficulty after another evaporated. Other than the jQuery stuff, which I had never touched before, the hardest part was deciding what to do about the API server. I could easily have written a standalone, special-purpose API server, but I felt like it was the wrong thing, and anyway, I didn't want to administer one. But eventually I remembered that our main web site is driven by Catalyst, which is itself a system for replying to HTTP requests, which already has access to our database, and which already knows which user is making each request. So it was natural to say that the API was to send HTTP requests to certain URLs on our web site, and easy-peasy to add a couple of handlers to the existing Catalyst application to handle the API requests, query the database, and send the right replies. I don't know why it took me so long to think of doing the API server with Catalyst. If it had been someone else's suggestion I would probably feel dumb fror not having thought of it myself, because in hindsight it is so obvious. Happily, I did think of it, because it is clearly the right solution for us. It was not too hard to debug. The three components are largely independent of one another. The draft version of the API server responded to GET requests, which are easier to generate from the browser than the POST requests that it should be getting. Since the responses are in JSON, it was easy to see if the API server was replying correctly. I had to invent techniques for debugging the jQuery stuff. I didn't know the right way to get diagnostic messages out of jQuery, so I put a big text area on the web page, and had the jQuery routines write diagnostic messages into it. I don't know if that's what other people do, but I thought it worked pretty well. JavaScript is not my ideal language, but I program in Perl all day, so I am not about to complain. Programming the front end in JavaScript and watching stuff happen on the page is fun! I like writing programs that make things happen. The package is in ZipRecruiter's GitHub repository, and is available under a very lenient free license. Check it out. (By the way, I really like working for ZipRecruiter, and we are hiring Perl and Python developers. Please send me email if you want to ask what it is like to work for them.) [Other articles in category /tech] permanent link Mon, 10 Feb 2014You sometimes hear people claim that there is no perfectly efficient machine, that every machine wastes some of its input energy in noise or friction. However, there is a counterexample. An electric space heater is perfectly efficient. Its purpose is to heat the space around it, and 100% of the input energy is applied to this purpose. Even the electrical energy lost to resistance in the cord you use to plug it into the wall is converted to heat. Wait, you say, the space heater does waste some of its energy. The coils heat up, and they emit not only heat, but also light, which is useless, being a dull orange color. Ah! But what happens when that light hits the wall? Most of it is absorbed, and heats up the wall. Some is reflected, and heats up a different wall instead. Similarly, a small fraction of the energy is wasted in making a quiet humming noise—until the sound waves are absorbed by the objects in the room, heating them slightly. Now it's true that some heat is lost when it's radiated from the outside of the walls and ceiling. But some is also lost whenever you open a window or a door, and you can't blame the space heater for your lousy insulation. It heated the room as much as possible under the circumstances. So remember this when you hear someone complain that incandescent light bulbs are wasteful of energy. They're only wasteful in warm weather. In cold weather, they're free. [Other articles in category /physics] permanent link Sat, 08 Feb 2014I wrote some time ago about Moonpig's use of GUIDs: every significant object was given a unique ID. I said that this was a useful strategy I had only learned from Rik, and I was surprised to see how many previously tricky programming problems became simpler once the GUIDs were available. Some of these tricky problems are artifacts of Perl's somewhat limited implementation of hashes; hash keys must be strings, and the GUID gives you an instantaneous answer to any question about what the keys should be. But it reminds me of a similar maxim which I was thinking about just yesterday: Every table in a relational database should have a record ID field. It often happens that I am designing some table and there is no obvious need for such a field. I now always put one in anyway, having long ago learned that I will inevitably want it for something. Most recently I was building a table to record which web pages were being currently visited by which users. A record in the table is naturally identified by the pair of user ID and page URL; it is not clear that it needs any further keys. But I put in a record ID anyway, because my practice is to always put in a record ID, and sure enough, within a few hours I was glad it was there. The program I was writing has not yet needed to use the record IDs. But to test the program I needed to insert and manipulate some test records, and it was much easier to write this:
than this:
If you ever end up with two objects in the program that represesent record sets and you need to merge or intersect them synthetically, having the record ID numbers automatically attached to the records makes this quite trivial, whereas if you don't have them it is a pain in the butt. You should never be in such a situation, perhaps, but stranger things have happened. Just yesterday I found myself writing
which nobody should have to do either, and yet there I was. Sometimes programming can be a dirty business. During the bootstrapping of the user-url table project some records
with bad URLs were inserted by buggy code, and I needed to remove
them. The URLs all ended in
So the rule is: giving things ID numbers should be the default, because they are generally useful, like handles you can use to pick things up with. You need a good reason to omit them. [Other articles in category /IT] permanent link |
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