% From John Kalman % % (3) The input file %i3 below, due to Mark Waddell and me, for % solving the cryptarithmetic puzzle ab x cd = efgh, where a, b, c, d, % e, f, g, and h are distinct digits. % We wish to find all 8-tuples (a,b,c,d,e,f,g,h) of distinct digits % such that the product % % a b % % c d % _______ % % e f g h % _______ % % is valid. For example, since 72 x 95 = 6840, (7,2,9,5,6,8,4,0) % is one such 8-tuple. % It is understood that a, c, and e are non-zero. % Since multiplication is commutative, we may as well assume that a < c; % for example, we do not need to find the 8-tuple (9,5,7,2,6,8,4,0) % as well as the 8-tuple (7,2,9,5,6,8,4,0). % % The computation of the above product may be represented as follows: % % a b % % c d % ___________ % % y3 y2 y1 % % z3 z2 z1 % ___________ % % e f g h % ___________ % % We use the predicate CS(u,x,y,z,v) to express that % % x % % y % _ % % z % _ % % is a possible column in the addition of an integer in which x % occurs as a digit to an integer in which y occurs as a digit; % u is the carryover from the previous column and v is the carryover % to the next column. % For example, CS(0,6,8,4,1) and CS(1,3,4,8,0) are true, and occur % in the computation of the product 72 x 95. % Note that, in such an addition, u is 0 or 1, and v is 0 or 1. % % We use the predicate CP(u,x,y,z,v) to express that % % ... y ... % % x % ___________ % % ... z ... % ___________ % % is a possible step in the multiplication of an integer in which % y occurs as a digit by the one-digit integer x; u is the carryover % from the previous step and v is the carryover to the next step. % For example, CP(1,5,7,6,3) is true, and represents a step in % the multiplication 72 x 95. % Note that, in such a multiplication, u and v are at most 8 (since % 9 x 9 = 81), or, if x and y are required to be distinct, at most 7 % (since 9 x 8 = 72). [In fact, there is not much gain to be made from % using the fact that carryovers of 9 and 8 are impossible. Experimental % runs took 39.34 secs with L(0) to L(7), and 42.48 secs using arbitrary % digits.] set(hyper_res). set(print_lists_at_end). assign(stats_level,1). % Output a few important statistics list(usable). K(0). % Sum carryover (at most 1) K(1). L(0). % Product carryover (at most 7 for distinct digits) L(1). L(2). L(3). L(4). L(5). L(6). L(7). D(0). % Digits D(1). D(2). D(3). D(4). D(5). D(6). D(7). D(8). D(9). end_of_list. list(sos). -K(u) | -D(x) | -D(y) | CS(u,x,y,$MOD($SUM($SUM(u,x),y),10),$DIV($SUM($SUM(u,x),y),10)). -L(u) | -D(x) | -D(y) | CP(u,x,y,$MOD($SUM(u,$PROD(x,y)),10),$DIV($SUM(u,$PROD(x,y)),10)). -CP(0,xd,xb,y1,x1) | -CP(x1,xd,xa,y2,y3) | -CP(0,xc,xb,z1,x2) | -CP(x2,xc,xa,z2,z3) | -$LT(xa,xc) | -CS(0,y1,0,xh,x3) | -CS(x3,y2,z1,xg,x4) | -CS(x4,y3,z2,xf,x5) | -CS(x5,0,z3,xe,0) | S(xa,xb,xc,xd,xe,xf,xg,xh). end_of_list. list(passive). % The following three clauses subsume "solutions" with a = 0, c = 0, or e = 0 S(0,xb,xc,xd,xe,xf,xg,xh). S(xa,xb,0,xd,xe,xf,xg,xh). S(xa,xb,xc,xd,0,xf,xg,xh). % The following 28 clauses subsume "solutions" in which distinct % letters are represented by the same digit S(x1,x1,x3,x4,x5,x6,x7,x8). S(x1,x2,x1,x4,x5,x6,x7,x8). S(x1,x2,x3,x1,x5,x6,x7,x8). S(x1,x2,x3,x4,x1,x6,x7,x8). S(x1,x2,x3,x4,x5,x1,x7,x8). S(x1,x2,x3,x4,x5,x6,x1,x8). S(x1,x2,x3,x4,x5,x6,x7,x1). S(x1,x2,x2,x4,x5,x6,x7,x8). S(x1,x2,x3,x2,x5,x6,x7,x8). S(x1,x2,x3,x4,x2,x6,x7,x8). S(x1,x2,x3,x4,x5,x2,x7,x8). S(x1,x2,x3,x4,x5,x6,x2,x8). S(x1,x2,x3,x4,x5,x6,x7,x2). S(x1,x2,x3,x3,x5,x6,x7,x8). S(x1,x2,x3,x4,x3,x6,x7,x8). S(x1,x2,x3,x4,x5,x3,x7,x8). S(x1,x2,x3,x4,x5,x6,x3,x8). S(x1,x2,x3,x4,x5,x6,x7,x3). S(x1,x2,x3,x4,x4,x6,x7,x8). S(x1,x2,x3,x4,x5,x4,x7,x8). S(x1,x2,x3,x4,x5,x6,x4,x8). S(x1,x2,x3,x4,x5,x6,x7,x4). S(x1,x2,x3,x4,x5,x5,x7,x8). S(x1,x2,x3,x4,x5,x6,x5,x8). S(x1,x2,x3,x4,x5,x6,x7,x5). S(x1,x2,x3,x4,x5,x6,x6,x8). S(x1,x2,x3,x4,x5,x6,x7,x6). S(x1,x2,x3,x4,x5,x6,x7,x7). end_of_list.