War Game
by Mark Halpin and Dave Shukan
This is a logic puzzle describing the outcome of the card game Battle Line between North and South. There are 9 columns into which players play up to 3 cards each, and one player (or possibly none) wins the flag associated with each column. One route to solving for the final layout is appears below. For clarity, the solution uses numbers rather than rank names, and poker-hand terminology rather than formation names. Numbers prefaced by G (e.g., G5) refer to the number of the given information in the puzzle.
A | By G4, one side must have all odd cards or all even cards, since otherwise not all sets of 20 cards removed would leave an even sum. (Remove any 20, and if an even sum remains, exchange an odd and even card.) If all-even, the side must have at least 21 cards. If all-odd, the side must have at least 22 cards. (This side will be called the "parity" side.) |
B | The parity side cannot include any straight or straight flush. Thus, by G1, the non-parity side must contain a winning straight flush, a winning straight, and a losing straight. And since a straight flush cannot oppose a straight flush, the four types of losing hands do not include a straight flush. |
C | By B, the non-parity side has at least 3 odd cards and at least 3 even cards (one per straight/straight flush). If the missing rank (G3) is odd, there would not be 22 odd cards left for the parity side (only 30 odd cards in the deck). If the missing rank is even, then because a 6 has been deserted, there would not be 21 even cards left for the parity side. Thus, the missing rank has a different parity than the parity side. |
D | By G8 and G7, North won a 20-or-more point hand against a 1X doubleton. Without Fog, South could have won with a yellow card the next turn. This would not have been a flush, since the X was orange, or a triplet since the doubleton could not be not 1 1. So it was a straight or a sum (Host). If it was a Host, then it would have to be 1 10 10 to beat 20, meaning that the doubleton would have to be 1 10. But if the doubleton were 1 10, then North would be the parity side, and that side could not contain a 10 (because of G2), and could not contain 88 or a 6 (because of G11), thus with even cards the most it could sum to would be 844 = 16, in violation of G8. So, if the South doubleton were 1 10, then North would have to be odd parity, but in that case the sum would be at least 21 (3 odd cards cannot sum to 20). And 1 10 could not beat a 21 (if it tied, it would come later and so would lose). Thus, the hypothetical winning South hand could not be a sum, and must be a straight. And the only straight containing a 1 is 123. But South could not have doubleton 1 3, since the 3 would have to be orange (by G7), and the orange 3 is singleton (also by G7). Thus, South has 1 2O in column 3, and North would have at most a Host without the Fog -- it could not be higher than that, since otherwise South could not have won on the next turn with a yellow card. And because South has 1 2, North must be the parity side. And South must have the Yellow 3 in hand when Fog was played, which must remain as an unplayed card at the end of the game (since South played no more yellow cards after Fog was played). |
E | By G8, South went first and played 1 Tactics card and removed a card from North. North played exactly 2 Tactics cards (could not have played a third until South played a second). Thus, South has 2 more cards than North if North won, and 3 more cards than North if South won. Thus, South has at least 24 cards if he lost or at least 25 cards if he won. |
F | By D, South has a known doubleton 1 2. He cannot have all the doubletons (at least 2) plus the singleton 3, since then he would have a maximum of 23 cards, in violation of E. Thus, North has at least one doubleton (containing a 1) or the singleton 3, which means that North's parity side must be all odds and contain at least 22 cards. South's hand contains at least 25 cards (by E). In fact, North must have exactly 22, since if North had 24 (the next possibility consistent with G4), then South would have at least 26, and the 9 remaining unplayed cards (given that 6O was played and removed) could sum to at most TTTTTT 3 88 = 79 (we know that a yellow 3 is unplayed), which violates G19. (Note that they could not be TTTTTT 3 99 = 81, since having 3 unplayed odd cards would leave fewer than 24 odd cards for the North hand, since South has straight flush/straight/straight plus a doubleton 1.) Thus, North has exactly 22 cards and South has exactly 24 cards if he lost or exactly 25 cards if he won. Since the winning player played a 10 (G14), South won, since North had no even card. Thus, South has 25 cards. |
G | By F, there are 12 unplayed Troop cards at the end (60 - 25 - 22 - 1), one of which is the 3Y. We know the missing rank is even (by C, since parity side is odd), and we have seen a 2, 6, and 10, so the missing rank must be 4 or 8. If 4 was the unplayed rank, the remaining sum is at most 444444 3 TTTTT < 81. Thus, the missing rank is 8. |
H | Since South has 25 cards and at least one doubleton, he must have exactly 7 full hands and 2 doubletons (= 25 cards). Thus, we know at least 6 odd cards that are not in North's hand -- 3 in the straights/straight flushes, the unplayed yellow 3, and the two 1's in the doubletons. Since North needs 22 odd cards, South cannot have an odd triplet. Thus, the 777GBV of G6 is North's. |
I | South has a nonwinning: doubleton, doubleton, straight, and Host (the last by G16). But South won by claiming 5 flags, and thus did not have a losing flush or triplet. Thus, South's column 1 hand, which is higher than a straight (G6), must beat North's 777, and thus must be 10/10/10ROY or a straight flush. |
J | Column 2 has the highest hand, and there is a winning SF, so South Column 2 is a SF. The highest it could be is 567, since there is no 8. If 567, cannot be GVB (the 7's of those colors are in Column 1), or O (by G8), or Y (by G15), so would have to be R. But in that case there would be no available SF for Column 1 -- it would have to be one of ROY to be different from the opposing 7's, but the ROY 3's appear elsewhere, and cannot be 456 because the 6O and 5Y and (under our hypothesis) R 567 are gone. So, if Column 2 is a 567 straight flush, then Column 1 is a triplet. But if Column 2 is a 456 straight flush or lower, South Column 1 still could not be a lower straight flush, since the needed ROY 3's are gone. Thus, in any case, South Column 1 is a triplet, which (by I) must be 10/10/10ROY.
***Column 1 (North, South) = 777GBV, 10 10 10 ROY |
K | South's Column 2 must be a GBorV straight flush of at most 456 (since GBV 7's are used in Column 1, and cannot match the ROY in South's Column 1). |
L | By G17, South has a flush that has exactly one odd card (sum must be odd, and there is not enough room for a flush to have 3 odd cards, see H). This odd card, plus the 6 others from H, leaves room for at most one more odd card in South's hand (to leave 22 for North). |
M | By G13, North's Column 4 hand cannot be a flush (and by parity cannot be a straight). And by G16, it cannot be a losing Host. Thus, it is a winning Host or a triplet. It cannot be a losing triplet, since there is only one triplet outside of Column 1, and a North triplet could only be beaten by a South triplet. South must have at least 2 different values, since otherwise it would be triplet vs triplet or North would lose a Host, which cannot be the case. South also must have an even number of odd cards (for the total sum of the column to be odd), and this number cannot be 2 (too many odd cards for South), so South has zero odd cards. North, therefore, has a triplet, one of which is orange, since South Column 3 has an orange 2 and thus South Column 4 cannot have an orange. And since the 3O is singleton, the triplet must be 1, 5, 7, or 9. But 7 or 9 would render the column unable to sum to 25 (with an all-even South), so must be 1 or 5. If a 1, then by G10 North's Column 3 would have a 1, but then could not sum to at least 20, so cannot be a triplet 1, so must be a triplet 5.
Thus, North Column 4 is 555, one O, and South Column 4 is 442 or 226, with one Y (since 5Y is in another column and thus cannot be in North Column 4). By G5, since there is a strictly lowest card, this must be 442, and by G5 the 2 is green.
Thus, Column 4 is: 5O, 5, 5 4Y, 4, 2G |
N | By G14, North's hand lengths (to get to exactly 22) are either 333331222 or 333333112. It cannot be the first one, since that would be 4 (incomplete) hands each having an orange, so they could not be next to each other. But we know that North has won Columns 3 and 4 (by D and M), and North's Column 9 has 3 cards (by G12), and there is no way to divide four non-touching incomplete hands in rows 1, 2, 5, 6, 7, 8. Thus, North's counts have to be the second set (in some order).
South has lengths = 333333322, with a nonwinning doubleton, doubleton, straight, Host North has lengths = 333333112, with a nonwinning singleton, singleton, doubleton, flush, triplet 777. |
O | By M and G10, North Column 3 has a 5. Thus, the other 2 cards must total at least 16 to get to 21 (G8 plus fact that North has all odds). But by G11, it cannot have 2 cards in this column totaling 16, so in fact the 2 other cards have to total at least 18, making them 9, 9. |
P | By G12, North Column 9 is 993G (neither 9 could become 7G, nor could the 5, since 7G is taken by Column 1). Thus, the ROYG 3's are accounted for and not available for straights or straight flushes. |
Q | Yellow 5 is in a flush, and North Columns 3 and 4 have four more 5's, leaving at most one 5 for the straight flush/straight/straight in South. And by P there are only two 3's available for straights. Thus, these 3 cards -- 5, 3, 3, -- must be spread among the South straight flush/straight/straight. Thus, the only 3 that can be in any flush is 3R and the only 5 that can be in any flush is 5Y.
Since there is only room for one more unaccounted odd card in the South hand (beyond L), at least two of the straights must start with even numbers. And at most 1 can start with a 4 or higher, due to only one 5 remaining. Thus, the straights are: 567, 234, 234 -- or -- 456, 234, 123 -- or -- 456, 234, 234 |
R | (a) One of R flush and Y flush is a losing flush (G15), and must belong to North because South does not have a losing flush. North must also have a winning flush (of some color), since South has a losing straight that does not lose to trips (all three trips have been accounted for) or to a straight or straight flush, which by parity North cannot have.
(b) What color can North's winning flush be? Leaving out the 3R and 5Y (both of which are in flushes), there are only two 3's and one 5 unaccounted for -- and these must be distributed among each of South's straights/straight flush, each of which must have at least one 3 or 5. Thus, no flush has a 3 or 5 other than 3R or 5Y, and thus any non-red or non-yellow flush in North must be exactly 971. But BGV 7's are in column 1, and North has three incomplete hands which each take an odd orange, leaving only 2 odd orange numbers left. Thus, the only available North flushes are R and Y, and North must have one of each. Note that a flush must be in Column 5, or else the 5O in Column 4 would be next to an orange (incomplete hand) in Column 5; and the Column 5 flush must be the losing flush, or else North would have a 3-in-a-row win in Columns 345, which is impossible since North lost. Thus, North's winning flush must be in Column 7, or else again 2 incomplete (and orange) hands would be adjacent. And South's losing straight must therefore be in Column 7 as well, since North's winning flush is the only undetermined North hand it can lose to. (c) North's column 7 flush must have a 9 (in order for 2 cards in that column to add to 16), and thus North's Column 5 flush cannot have a 9 (since four non-orange 9's are needed elsewhere -- 2 in column 3, 2 in column 9 -- and an orange 9 does no good for either of North's flushes). (d) Is it possible that one or both of those R and Y flushes are not the ones in G15, but instead that South has the 3R flush or the 5Y flush? No. Suppose South had the 3R flush, which must be a win, since South does not have a losing flush. Then North would have to have a losing red flush in column 5. But it could not contain a 5 (all 5's are needed elsewhere), or a 3 (3R is in South), or a 9 (by 3, above) -- and 1,7 is not sufficient. Similarly, If South had a (winning) 5Y flush, North's losing Yellow flush could not contain a 3Y (unplayed), 5Y (in South), or 9Y (by 3, above). Thus, South cannot have an 3R or 5Y flush, and both of those flushes must be in North. From what has come before, North's incomplete hands are in Columns 2, 6, 8, and South's are in 3, 9. |
S | North's flush in Column 7 must include a 9, since 2 cards in that column must sum to 16, and nothing else could be higher than a 7 (since facing a straight lower than 8). Thus, no 9 is available for the flush in North's column 5 (the four 9's in Columns 3 and 9 are not orange, since they are opposite incomplete hands, and the orange 9 cannot go in the red or yellow flush). Thus, the North Column 5 flush does not have a 9. But 2 cards in that Column need to sum to 16. Thus, South Column 5 cannot be a triplet (since no 8's), so must be a higher flush, and that flush must be 10 6 X. And it must be GB or V, since the ROY 10's are gone. |
T | If North's Column 5 flush is R, then it must be 731 (it cannot contain a 9, and 5R is not in a flush). Then we know the yellow flush contains 95, and 3Y is unplayed, and cannot be 1Y, since then the straight that it beats would have to be 567 (in order to have 2 cards in the column sum to 16), but then 5 opposes 5. So, the yellow flush would be 957.
If North's Column 5 flush is Y, then it must be 751 (no 9Y or 3Y available). Then the red flush would be 93, and 5R is unavailable, and 931 would sum to same as 751 (violating G15), so red flush would be 973. Thus, North's flushes are either: (a-group) 731R and 957Y -- or -- (b-group) 751Y and 973R. Suppose the b-group. Then the straight opposite North's winning 973 flush must be 456 (since no 3 or 7). That means the South hands opposite North's 3 incomplete hands are: straight-flush containing-a-3, straight-containing-a-3, and flush. But North's singleton 3 could not appear opposite any of those (because it cannot face a 3 or lose to a flush). Thus, North's flushes must be the a-group. |
U | We know the values of all of North's complete hands, and 5O is gone, and 3O is singleton. Can the 1O be the other singleton? If so, then we need to find a space for either the 7O or 9O elsewhere in North, or somewhere in South, or among the unplayed cards.
But 7O cannot take the place of any other 7, since all those colors are determined to be not orange. And 7O cannot be in any South straight (5O is gone) or the South flush with the odd card, since there are no available orange cards to accompany it. And it cannot remain among the unplayed cards, since then there would be exactly two unplayed odd cards (7O and 3Y), and that would be an even sum, not 81. Similarly, 9O cannot take the place of any other 9, since 4 of them appear opposite incomplete hands (that contain an orange card) and the fifth is Y. And it cannot go in the South hand for the same reasons or be among the unplayed cards for the same reasons as 7O. Thus, 1O cannot be singleton (and by G6 cannot be in a doubleton), and 5O is already accounted for, so the orange cards in the North incomplete hands are 3O, 7O, and 9O. |
V | The South Column 5 flush 10 6 X can only differ from one of North's (a-group) flushes by 2 (G17) if the X is a 3 or 7, none of which are available, which would require South to have a Column 6 or Column 8 flush with sum 9, 13, 19, or 23.
What odd card can go in the South Column 6 or Column 8 flush? There are no 3's or 5's available for flushes other than the two North flushes. And the 7's and 9's are all accounted for (two 7's in flushes and 777 in Column 1 and a 7 in an incomplete hand; four 9s in Columns 3 and 9, 9Y in flush, and 9 in incomplete hand). Thus, the odd card in the South Column 6 or Column 8 flush is a 1. |
W | Since North Column 5 is red (by T) and South Column 4 has GY (by M), North Column 4 must be 555BVO, and South Column 4 must be 442RYG.
***Column 4 (North, South) = 555BVO, 444RYG |
X | What color is the 1 accompanying the North doubleton? R1 and Y1 are accounted for elsewhere, and O1 cannot be in a doubleton, so must be B, G, or V. So, it cannot go in Column 2 (next to the 777BGV). So, the North doubleton is in Column 6 or 8. Since the South Column 6 or 8 flush contains a 1 (by V), it cannot be across from the doubleton. But the singleton 3 did not lose to the flush, so the singleton 3 must be in Column 2 (it is not the doubleton or across from the South flush, and that covers both Columns 6 and 8).
Thus, South's straight flush in Column 2 must not contain a 3 and thus must contain a 5. The only 5's available are G and R, but cannot be R since south Column 1 is ROY, so must be G. Since G3 and G7 are spoken for, South Column 2 must be 456GGG. ***Column 2 (North, South) = 3O / 456GGG Because North Column 4 is BVO, and the 5G is now gone, North Column 3 must be 995GRR. Column 3 (North, South) = 995GRR / 21OV or OB -- since the 1 cannot be ROYG The only non-1, non-8 orange remaining available to pair with the 1 in South Column 9's doubleton is 4O. ***Column 9 (North, South) = 993VBG / 41OY (9O must be single or double, only VB are left; 1 cannot be VBGO, and R is taken elsewhere) Since the North doubleton 1 is B, G, or V, it cannot go in Column 8 (next to BGV Column 9), so must be Column 6. Then, the South flush with the odd card cannot be Column 6, since it contains a 1, so must be Column 8. This leaves Columns 6 and 7 for the South straights, which both contain a 3. Column 5 (North, South) = 731RRR / T6(2, 4) (VorB flush -- since 6G is gone) Column 7 (North, South) = 975YYY / (123, 234)noY Column 6 (North, South) = (9,7)1O(GVorB) / 234 All 1's except the 1O are accounted for (3 doubletons, 2 flushes). All North cards are accounted for, and the 1O cannot be unplayed, since that (along with the unplayed 3Y) would render the sum of the unplayed cards even. (There are no more unaccounted odd cards.) The only place the 1O could go in the South hand is in the Column 7 straight as a 123. Thus, Column 7 (North, South) = 975YYY / 123Oxx |
Y | The only 3's and 4's unaccounted for are B and V. The 3's must go in South's Column 6 and 7 straights; one 4 goes in South's Column 6 straight and one goes in South's Colum 5 or Column 8 flush.
What color is the 3 in the 123 straight in Column 7? Suppose it is B. Then the 4B cannot go in Column 8 and must go in Column 5. That means that the 4V goes in the 234 straight in Column 6 along with the 3V. Similarly, the 2B cannot now go in Column 6 or 8 (both next to a B card), so must go in Column 7. Since South Column 6 must be the winning straight and already has a 3V and 4V, it cannot have an 2V (which would make it a straight flush), thus must have the 2 R or Y. Column 8 has a 1, which cannot be a 1B (which cannot go next to the 3B in Column 7), and cannot be a 1G, since there are not 2 available cards to accompany it in a flush. Thus, Column 8 is the V flush, with a 1 and 2 other cards. Since all cards in the layout are known except those 2 other cards, it is known that those 2 other cards must sum to 12 in order for the unplayed cards to sum to 81. These can only be 10V and 2V. The 1 in South Column 3 is then B, and the 1 in the North doubleton in Column 6 is G, which is the only remaining 1. Now, instead suppose that the 3 in the 123 flush in Column 7 is V. The analysis in the previous paragraph can be repeated, switching V and B. One of these must be the case. And the layout where the 123 straight has a B3 has more blues than the layout where the 123 straight has a 3V, so by G5 the more-blue layout must be the case. ***Column 7 (North, South) = 975YYY / 123OBB ***Column 5 (North, South) = 731RRR / T64BBB ***Column 3 (North, South) = 995GRR / 21OB |
Z | The only numbers not yet placed are 9O and 7O. The 9O must go in Column 8, since if the 7O went there the sum of all single-color hands would be 90, in violation of G18.
***Column 8 (North, South) = 9O / T21VVV The only color not yet placed is the 2 in South Column 6, which can be R or Y (by conclusion Y). If Y, then R and Y would be tied for fewest number of cards played (7 each), in violation of G3. Thus, the 2 in South Column 6 must be R. ***Column 6 (North, South) = 71OG / 234RVV |
The final layout is: Taking the differences between the sums of the winning and losing hands (see flavor text), this reads ILIAD from the winner's left to right using the winner's flags, and POET from the loser's left to right using the loser's flags. ILIAD POET is HOMER, which is the solution. |
The final answer is: HOMER.