//-------------------------------------------------------------------
// enigma1345.c (c) 2005 by Charles Petzold (www.charlespetzold.com)
//
// "Score? Date?" from New Scientist, 18 June 2005, page 54.
//
// The premise of this puzzle is too complex to be easily summarized. 
// It involves a grid showing the results of four "football" teams
// playing each other. But the scores and dates of upcoming matches
// are coded similarly. Determine which cells are scores.
//------------------------------------------------------------------- 

#include <stdio.h>

int main(void)
{
	// These variables represent the results of each match. 
	// For example, AB is the result of team A playing B.
	// If AB is 0, the match was a draw. If AB is 1, then
	// A won, and if AB is 2, then B won.

	int AB, AC, AD, BC, BD, CD;

	// These variables represent end-of-season points for 
	// each team. Each team gets 1 point for each draw and 
	// 3 points for each win.

	int A, B, C, D;

	// Points can be easily calculated by indexing these
	// two arrays with the variables, AB, AC, etc.
	// Use Points1 for the first team in each pair and
	// Points2 for the second team. For example, for AB,
	// Points1[AB] is A's points for the match, and
	// Points2[AB] is B's points.

	int Points1[] = { 1, 3, 0 };
	int Points2[] = { 1, 0, 3 };

	// A total of 6 matches are played. Each team plays every
	// other team just once. These loops go through all the
	// possibilities.

	for (AB = 0; AB < 3; AB++)
	for (AC = 0; AC < 3; AC++)
	for (AD = 0; AD < 3; AD++)
	for (BC = 0; BC < 3; BC++)
	for (BD = 0; BD < 3; BD++)
	for (CD = 0; CD < 3; CD++)
	{
		// In the grid, C-D is the pair 3-4. That must
		// be a score because the equivalent date would
		// be 3 April, and we know the matches took 
		// place in January to March. So, we only want 
		// possibilities where CD equals 2 (ie, D won).

		if (CD != 2)
			continue;

		// For A-C, the grid shows 2-3, and for A-D the
		// grid shows 1-3. If those are both dates, the
		// dates are 1 March and 2 March, which would 
		// means A would have to play two matches in a 
		// row, which is disallowed. At least one of these
		// must be a score. Only consider cases where 
		// AC or AD equals 2 (that is, A lost).

		if (AC != 2 && AD != 2)
			continue;

		// For both A-B and B-C, the grid shows 4-1.
		// They can't both be dates because that would
		// require B to play two matches on the same day.
		// Only consider cases when A beats B or B beats C.

		if (AB != 1 && BC != 1)
			continue;

		// Calculate the season-ending points.

		A = Points1[AB] + Points1[AC] + Points1[AD];
		B = Points2[AB] + Points1[BC] + Points1[BD];
		C = Points2[AC] + Points2[BC] + Points1[CD];
		D = Points2[AD] + Points2[BD] + Points2[CD];

		// We are told the total points of all teams is
		// odd. Only consider cases where the total is odd.

		if (((A + B + C + D) & 1) == 0)
			continue;

		// We are also told that one team accumulated
		// 5 points. Reject cases where noone has 5 points.

		if ((A != 5) && (B != 5) && (C != 5) && (D != 5))
			continue;

		// Print out the results.

		printf ("AB=%i AC=%i AD=%i BC=%i BD=%i CD=%i\n", 
			AB, AC,AD, BC, BD, CD);
	}
}

// The program does *not* take account of other criteria mentioned in the
// problem: Because the teams take place at a specific time each day,
// no team can play two matches the same day. The problem also states 
// that no team plays on two consecutive days.
//
// Without these criteria, the program prints the following three possibilities:
//
//	AB=0 AC=0 AD=2 BC=1 BD=0 CD=2
//	AB=0 AC=2 AD=0 BC=1 BD=0 CD=2
//	AB=1 AC=2 AD-0 BC=0 BD=0 CD=2
//
// There are no draws in the grid, so a value of zero in these results
// indicates that the cell in the grid is a date.
//
// The third line implies that B-C and B-D are both dates, but if so,
// they indicate consecutive days, which is not allowed.
//
// Therefore B-D is a date and A-B is a date. We already know that
// C-D is a score. If A-B is a date then B-C must also be a score.
//
// Either A-C or A-D is a score, and the other is a date. Let's 
// reason it out. 
//
// A-B is a date, and that date is 4 January, which means that this 
// particular grid was published on 3 January or earlier. Matches 
// B-C and C-D must have occurred on January 1, 2, or 3. Both 
// involve team C, so they can't be played on consecutive days. One
// must have been played on the 1st and the other on the 3rd. But
// B-C couldn't be played on the 3rd because B is playing A on the 4th.
// Therefore, B-C occurred on 1 January and C-D on 3 January, and
// A-B will occur on 4 January.
// 
// If A-C is a score, the match must have been played during this period.
// But it can't because then C would be playing twice the same day
// or on consecutive days.
//
// Therefore A-D must be the score, and the match occurred on 1 January.
//
// The scores are A-D, B-C, C-D.
//----------------------------------------------------------------------