Archivename: puzzles/archive/combinatorics
Lastmodified: 17 Aug 1993 Version: 4 See reader questions & answers on this topic!  Help others by sharing your knowledge ==> combinatorics/alphabet.blocks.p <== What is the minimum number of dice painted with one letter on all six sides such that all permutations without repetitions of n letters can be formed by placing n dice together in a line? ==> combinatorics/alphabet.blocks.s <== n= 2 3 4 5 6 (8,4) (9,7) (9,3) (10,7) (11,7) aijklm abcde? acdefg abcde? abkmuz bijklm fghij? bhijkl fghij? bcpwy? cnopqr klmno? cmnopq klmno? cdlnvz dnopqr pqrst? dhnrvy pqrst? deqxu? estuvw uvwxy? eiosvw uvwxy? efmowz fstuvw afkpu? fjptwx afkpu? fgryv? gxyz?? bglqv? gkquxy bglqv? ghnkx? hxyz?? chmrwz lmrstu chmrwz hisuw? dinsxz zab??? dinsxz ijoly? ejotyz jatvx? pqrstz I think I can prove that there is no solution with 11 dice with 9 don't cares or with 10 dice, but I haven't checked all the details, so I might have made a mistake. In any case, that leaves open the case of 11 dice with 8 don't cares; my guess is that it is not possible.  John Rickard (jrickard@eoe.co.uk) ==> combinatorics/coinage/combinations.p <== Assuming you have enough coins of 1, 5, 10, 25 and 50 cents, how many ways are there to make change for a dollar? ==> combinatorics/coinage/combinations.s <== 292. The table is shown below: Amount 00 05 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 Coins .01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 .05 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 .10 1 2 4 6 9 12 16 20 25 30 36 42 49 56 64 72 81 90 100 110 121 .25 1 2 4 6 9 13 18 24 31 39 49 60 73 87 103 121 141 163 187 214 242 .50 1 2 4 6 9 13 18 24 31 39 49 62 77 93 112 134 159 187 218 253 292 The meaning of each entry is as follows: If you wish to make change for 50 cents using only pennies, nickels and dimes, go to the .10 row and the 50 column to obtain 36 ways to do this. To calculate each entry, you start with the pennies. There is exactly one way to make change for every amount. Then calculate the .05 row by adding the number of ways to make change for the amount using pennies plus the number of ways to make change for five cents less using nickels and pennies. This continues on for all denominations of coins. An example, to get change for 75 cents using all coins up to a .50, add the number of ways to make change using only .25 and down (121) and the number of ways to make change for 25 cents using coins up to .50 (13). This yields the answer of 134. ==> combinatorics/coinage/dimes.p <== "Dad wants onecent, twocent, threecent, fivecent, and tencent stamps. He said to get four each of two sorts and three each of the others, but I've forgotten which. He gave me exactly enough to buy them; just these dimes." How many stamps of each type does Dad want? A dime is worth ten cents.  J.A.H. Hunter ==> combinatorics/coinage/dimes.s <== The easy way to solve this is to sell her three each, for 3x(1+2+3+5+10) = 63 cents. Two more stamps must be bought, and they must make seven cents (since 17 is too much), so the fourth stamps are a two and a five. ==> combinatorics/coinage/impossible.p <== What is the smallest number of coins that you can't make a dollar with? I.e., for what N does there not exist a set of N coins adding up to a dollar? It is possible to make a dollar with 1 current U.S. coin (a Susan B. Anthony), 2 coins (2 fifty cent pieces), 3 coins (2 quarters and a fifty cent piece), etc. It is not possible to make exactly a dollar with 101 coins. ==> combinatorics/coinage/impossible.s <== The answer is 77: a) 5c = 1 or 5; b) 10c = 1 or 2 a's (1,2,6,10) c) 25c = 1 or 2 b's + 1 a d) 50c = 1 or 2 c's e) $1 = 1 or 2 d's total penny nickel dime quarter half 5 1 2 1 1 6 3 1 1 1 7 5 1 1 8 4 3 1 9 6 2 1 10 8 1 1 11 10 1 12 7 4 1 13 9 3 1 14 11 2 1 15 13 1 1 16 15 1 17 14 3 18 16 2 19 18 1 20 20 21 5 13 3 22 5 15 2 23 5 17 1 24 5 19 25 10 12 3 26 10 14 2 27 10 16 1 28 10 18 29 15 11 3 30 15 13 2 31 15 15 1 32 15 17 33 20 10 3 34 20 12 2 35 20 14 1 36 20 16 37 25 9 3 38 25 11 2 39 25 13 1 40 25 15 41 30 8 3 42 30 10 2 43 30 12 1 44 30 14 45 35 7 3 46 35 9 2 47 35 11 1 48 35 13 49 40 6 3 50 40 8 2 51 40 10 1 52 40 12 53 45 5 3 54 45 7 2 55 45 9 1 56 45 11 57 50 4 3 58 50 6 2 59 50 8 1 60 50 10 61 55 3 3 62 55 5 2 63 55 7 1 64 55 9 65 60 2 3 66 60 4 2 67 60 6 1 68 60 8 69 65 1 3 70 65 3 2 71 65 5 1 72 65 7 73 70 3 74 70 2 2 75 70 4 1 76 70 6 77 can't be done 78 75 1 2 79 75 3 1 80 75 5 81 can't be done 82 80 2 83 80 2 1 84 80 4 85 can't be done 86 can't be done 87 85 1 1 88 85 3 89 can't be done 90 can't be done 91 90 1 92 90 2 9395 can't be done 96 95 1 9799 can't be done 100 100 ==> combinatorics/color.p <== An urn contains n balls of different colors. Randomly select a pair, repaint the first to match the second, and replace the pair in the urn. What is the expected time until the balls are all the same color? ==> combinatorics/color.s <== (n1)^2. If the color classes have sizes k1, k2, ..., km, then the expected number of steps from here is (dropping the subscript on k): 2 k(k1) (j1) (kj) (n1)  SUM (  + SUM  ) classes, 2 1<j<k (nj) class.size=k The verification goes roughly as follows. Defining phi(k) as (k(k1)/2 + sum[j]...), we first show that phi(k+1) + phi(k1)  2*phi(k) = (n1)/(nk) except when k=n; the k(k1)/2 contributes 1, the term j=k contributes (j1)/(nj)=(k1)/(nk), and the other summands j<k contribute nothing. Then we say that the expected change in phi(k) on a given color class is k*(nk)/(n*(n1)) times (phi(k+1) + phi(k1)  2*phi(k)), since with probability k*(nk)/(n*(n1)) the class goes to size k+1 and with the same probability it goes to size k1. This expected change comes out to k/n. Summing over the color classes (and remembering the minus sign), the expected change in the "cost from here" on one step is 1, except when we're already monochromatic, where the handy exception k=n kicks in. One can rewrite the contribution from k as (n1) SUM (kj)/(nj) 0<j<k which incorporates both the k(k1)/2 and the previous sum over j. That makes the proof a little cleaner. ==> combinatorics/full.p <== Consider a string that contains all substrings of length n. For example, for binary strings with n=2, a shortest string is 00110  it contains 00, 01, 10 and 11 as substrings. Find the shortest such strings for all n. ==> combinatorics/full.s <== Knuth, Volume 2 Seminumerical Algorithms, section 3.2.2 discusses this problem. He cites the following results: Shortest length: m^n + n  1, where m = number of symbols in the language. Algorithms: [Exercise 7, W. Mantel, 1897] The binary sequence is the LSB of X computed by the MIX program: LDA X JANZ *+2 LDA A ADD X JNOV *+3 JAZ *+2 XOR A STA X [Exercise 10, M. H. Martin, 1934] Set x[1] = x[2] = ... = x[n] = 0. Set x[i+1] = largest value < n such that substring of n digits ending at x[i+1] does not occur earlier in string. Terminate when this is not possible. If we instead consider the strings as circular, we have a well known problem whose solution is given by any hamiltonian cycle in the de Bruijn (or Good) graph of dimension K. (Or equivalently an eulerian circuit in the de Bruijn graph of dimension K1) As a string of length 2^K is produced, it must be optimal, and any shortest sequence must be an eulerian circuit in a dB graph. The de Bruijn graph Tn has as its vertex set the binary nstrings. Directed edges join nstrings that may be derived by deleting the left most digit and appending a 0 or 1 to the right end. de Bruijn + van ArdenneEhrenfest (in 1951) counted the number of eulerian circuits in Tn. There are 2^(2^(n1)n) of them. Some examples: K=2 1100 K=3 11101000 K=4 1111001011010000 The solution to the above problem (noncircular strings) can be found by duplicating the first K1 digits of the solution string at the end of the string. These are not the only solutions, but they are of minimum length: 2^K + K1. We can obtain a lower bound for the optimal sequence for the general case as follows: Consider first the simpler case of breaking into an answer machine which accepts d+1 digits, values 0 to n1. We wish to find the minimal universal code that will allow us access to any such answering machine. Let us construct a digraph G = (V,E), where the n^d vertices are labelled with a d sequence of digits. Notation: let [v_{i,1},v_{i,2},...,v_{i,d}] denote the labelling on node v_i. An edge e = (v_i, v_j) is in E iff for k in 1, ..., d1: v_{i,k+1} = v_{j,k}, i.e., the last d1 digits in the labelling of the initial vertex of e is identical with the first d1 digits in the labelling of the terminal vertex of e. We associate with each edge a value, t(e) = v_{j,d}, the last digit in the labelling of the terminal vertex. The intuition goes as follows: we are going to perform a Euler circuit of the digraph, where the label on the current vertex gives the last d digits in the output sequence so far. If we make a transition on edge e, we output the tone/digit t(e) as the next output value, thus preserving the invariant on the labelling. How do we know that a Euler circuit exists? Simple: a connected digraph has an Euler circuit iff for all vertices v: indegree(v) = outdegree(v). This property is trivially true for this digraph. So, in order to generate a universal code for the AM, we simply output 0^d (to satisfy the precondition for being in vertex [0,...,0]), and perform an Euler circuit starting at node [0,...,0]. Now, the total length of the universal sequence is just the number of edges traversed in the Euler circuit plus the initial precondition sequence, or n^d * n + d (number of vertices times the outdegree) or n^{d+1} + d. That this is a minimal sequence is obvious. Next, let us consider the machine AM' where the security code is of the form [0...n1]^d [0...m1], i.e., d digits ranging from 0 to n1, followed by a terminal digit ranging from 0 to m1, m < n. We build a digraph G = (V, E) similar to the construction above, except for the following: an edge e is in E iff t(e) in 0 to m1. This digraph is clearly nonEulerian. In particular, there are two classes of vertices: (1) v is of the form [0...n1]^{d1} [0...m1] (``fat'' vertices) and (2) v is of the form [0...n1]^{d1} [m...n1] (``thin'' vertices) Observations: there are (n^{d1} * m) fat vertices, and (n^{d1} * (nm)) thin vertices. All vertices have outdegree of m. Fat vertices have indegrees of n, and thin vertices have indegrees of 0. Color all the edges blue. The question now becomes: can we put a bound on how many new (red) edges must we add to G in order to make a blue edge covering path possible? (Instead of thinking of edges being traversed multiple times in the blue edge covering path, we allow multiple edges between vertices and allow each edge to be traversed once.) Note that, in this procedure, we add edges only if it is allowed (the vertex labelling constraint). We will first obtain a lower bound on the length of a blue covering circuit, and then transform it into a bound for arbitrary blue covering paths. Clearly, we must add at least (nm)*(n^{d1}*m) edges incident from the fat vertices. [ We need (nm) new outgoing edges for each of (n^{d1}*m) vertices to bring the outdegree up to the indegree. ] Let us partition our vertices into sets. Denote the range [0..m1] by S, the range [m..n1] by L, and the range [0..n1] by X. Let S_0 = { v: v = [X^{d1}S] }. S_0 is just the set of fat vertices. Define in(S_0) = number of edges from vertices not in S to vertices in S. Define out(S_0) in the corresponding fashion, and let excess(S_0) = in(S_0)out(S_0). Clearly, excess(S_0) = n^{d1}m(nm) from the argument above. Generalizing the requirement for Eulerian digraphs, we see that we must add excess(S_0) edges from S_0 if the blue edges connected to/within S_0 are to be covered by some circuit (edges may not be travered multiple times  we add parallel edges to handle that case). In particular, edges from S_0 will be incident on vertices of the form [X^{d2}SX]. Furthermore, they can not be [X^{d2}SS] since that is a subset of S_0 and adding those edges will not help excess(S_0). [Now, these edges may be needed if we are to have a circuit, but we do not consider them since they do not help excess(S_0).] So, we are forced to add excess(S_0) edges from S_0 to S_1 = { v: v = [X^{d2}SL] }. Color these newly added edges red. Let us define in(S_1), out(S_1) and excess(S_1) as above for the modified digraph, i.e., including the red excess(S_0) edges that we just added. Clearly, in(S_1) = out(S_0) = n^{d1}m(nm), and out(S_1) = m*S_1 = m*n^{d2}m(nm), so excess(S_1) = n^{d2}m(nm)^2. Consider S_0 union S_1. We must add excess(S_1) edges to S_0 union S_1 to make it possible for the digraph to be covered by a circuit, and these edges must go from {S_0 union S_1} to S_2 = { v: v = [X^{d3}SL^2] } by a similar argument as before. Repeating this partitioning process, eventually we get to S_{d1} = { v: v = [SL^{d1}] }, where union of S_0 to S_{d1} will need edges to S_d = { v: v = [L^d] }, where this process terminates. Note that at this time, excess(union of S_0 to S_{d1}) = m(nm)^d, but in(S_d) = 0 and out(S_d) = m(nm)^d, and the process terminates. What have we shown? Adding up blue edges and the red edges gives us a lower bound on the total number of edges in a blueedges covering circuit (not necessarily Eulerian) in the complete digraph. This comes out to be n^{d+1}(nm)^{d+1} edges. Next, we note that if we had an optimal path covering all the blue edges, we can transform it into a circuit by adding d edges. So, a minimal path can be no more than d edges shorter than the minimal circuit covering all blue edges. [Otherwise, we add d extra edges to make it into a shorter circuit.] So the shortest blue covering path through the digraph is at least n^{d+1}{nm}^{d+1}d. With an initial precondition sequence of length d (to establish the transition invariant), the shortest universal answering machine sequence is of length at least n^{d+1}(nm)^{d+1}. While this has not been that constructive, it is easy to see that we can achieve this bound. If we looked at the vertices in each of the S_i's, we just add exactly the edges to S_{i+1} and no more. The resultant digraph would be Eulerian, and to find the minimal path we need only start at the vertex labelled [{n1}^d], find the Euler circuit, and omit the last d edges from the tour. ==> combinatorics/gossip.p <== n people each know a different piece of gossip. They can telephone each other and exchange all the information they know (so that after the call they both know anything that either of them knew before the call). What is the smallest number of calls needed so that everyone knows everything? ==> combinatorics/gossip.s <== 1 for n=2 3 for n=3 2n4 for n>=4 This can be achieved as follows: choose four people (A, B, C, and D) as the "core group". Each person outside the core group phones a member of the core group (it doesn't matter which); this takes n4 calls. Now the core group makes 4 calls: AB, CD, AC, and BD. At this point, each member of the core group knows everything. Now, each person outside the core group calls anybody who knows everything; this again requires n4 calls, for a total of 2n4. The solution to the "gossip problem" has been published several times: 1. R. Tidjeman, "On a telephone problem", Nieuw Arch. Wisk. 3 (1971), 188192. 2. B. Baker and R. Shostak, "Gossips and telephones", Discrete Math. 2 (1972), 191193. 3. A. Hajnal, E. C. Milner, and E. Szemeredi, "A cure for the telephone disease", Canad Math. Bull 15 (1976), 447450. 4. Kleitman and Shearer, Disc. Math. 30 (1980), 151156. 5. R. T. Bumby, "A problem with telephones", Siam J. Disc. Meth. 2 (1981), 1318. ==> combinatorics/grid.dissection.p <== How many (possibly overlapping) squares are in an mxn grid? Assume that all the squares have their edges parallel to the edges of the grid. ==> combinatorics/grid.dissection.s <== Given an n*m grid with n > m. Orient the grid so n is its width. Divide the grid into two portions, an m*m square on the left and an (nm)*m rectangle on the right. Count the squares that have their upper righthand corners in the m*m square. There are m^2 of size 1*1, (m1)^2 of size 2*2, ... up to 1^2 of size m*m. Now look at the nm columns of lattice points in the rectangle on the right, in which we find upper righthand corners of squares not yet counted. For each column we count m new 1*1 squares, m1 new 2*2 squares, ... up to 1 new m*m square. Combining all these counts in summations: m m total = sum i^2 + (n  m) sum i i=1 i=1 (2m + 1)(m + 1)m (n  m)(m + 1)m =  +  6 2 = (3n  m + 1)(m + 1)m/6  David Karr ==> combinatorics/permutation.p <== Compute the nth permutation of k numbers (or objects). ==> combinatorics/permutation.s <== #include <stdio.h> /* adapted from 'Notation as a Tool of Thought', by K.E.Iverson Radix Representation of Permutations */ /* direct from radix; of given order */ dfr(short direct[],short radix[],long order) { if (order) { direct[0] = radix[0]; dfr (direct+1, radix+1, order1); while (order) direct[order] += direct[order] >= direct[0]; } } /* radix representation; of given order and given index */ rr(short radix[], long order, long index) { int i; for (i=1; i<=order; i++) { radix[orderi] = index % i; index = index/i; } } show(short perm[],long order) { while(order) printf("%hd ",*perm++); printf("\n"); } short parity(short radix[],long order) { long p=0; while(order) p+=*radix++; return p%2; } void usage(char *name) { fprintf(stderr,"usage: %s order number_of_permutation\n",name); exit(1); } main(int argc, char *argv[]) { #define MAX_ORDER 512 short radix[MAX_ORDER], direct[MAX_ORDER]; long order, nth; if (argc!=3) usage(argv[0]); order = atol(argv[1]); nth = atol(argv[2]); rr(radix, order, nth1); /* where 0 is the first permuatation */ dfr(direct, radix, order); printf("radix "); show(radix,order); printf("direct "); show(direct,order); printf("parity %d\n",parity(radix,order)); }  J. Henri Schueler, H&h Software, Toronto +1 416 698 9075 jhs@ipsa.reuter.com ==> combinatorics/subsets.p <== Out of the set of integers 1,...,100 you are given ten different integers. From this set, A, of ten integers you can always find two disjoint nonempty subsets, S & T, such that the sum of elements in S equals the sum of elements in T. Note: S union T need not be all ten elements of A. Prove this. ==> combinatorics/subsets.s <== There are 2^10 = 1,024 subsets of the 10 integers, but there can be only 901 possible sums, the number of integers between the minimum and maximum sums. With more subsets than possible sums, there must exist at least one sum that corresponds to at least two subsets. Call two subsets with equal sums A and B. Let C = A intersect B; define S = A  C, T = B  C. Then S is disjoint from T, and sum(S) = sum(AC) = sum(A)  sum(C) = sum(B)  sum(C) = sum(BC) = sum(T). QED Addendum: 9 integers suffice. This was part of my Westinghouse project in 1981 (the above problem was in Martin Gardner's Scientific American column not long before). The argument is along the same lines, but a bit more complicated; for starters you only work with the subsets consisting of 3, 4, 5, or 6 of the 9 elements. Let M(n) be the smallest integer such that there exists an nelement set {a1,a2,a3,...,an=M(n)} of positive integers all 2^n of whose subsums are distinct. The pigeonhole argument of subsets.s shows that M(n)>2^n/n, and it is known that M(n)>c*2^n/sqrt(n) for some c>0. It is still an unsolved problem (with an Erdos bounty) whether there is a positive constant c such that M(n)>c*2^n for all n. Noam D. Elkies (elkies@zariski.harvard.edu) Dept. of Mathematics, Harvard University ==> combinatorics/transitions.p <== How many nbit binary strings (0/1) have exactly k transitions (an adjacent pair of dissimilar bits, i.e., a 01 or a 10)? ==> combinatorics/transitions.s <== A transition can occur at an adjacent pair (i,i+1) where 1<=i<i+1<=n. Since there are k transitions, there are C(n1,k) total number of ways that transitions can occur. But the string may start with a 1 or a 0 (after which its transitions uniquely determine the string). So there are a total of 2C(n1,k) such strings. User Contributions:
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