Solution to Giza deal number 2112
There are a number of difficulties in solving this deal. All of the eights in the pyramid must be removed using fives from the columns. The locations of the nines and fours are badly intertwined with the sixes and sevens, and the two jacks and two in the pyramid are overlapping, so all three have to be removed using column cards.
Start by discarding the ten of hearts from the pyramid with the three of spades in column 1, the eight of clubs with the five of spades from column 5, and the nine and four of diamonds from the pyramid. The two exposed queens in the pyramid can be discarded with the two exposed aces in the columns, followed by the king of spades:
Although it appears dangerous to discard the jack of diamonds in column 4 with the two of spades in column 2, we need to do that to free up more fours so that we can discard the nines in the lower left part of the pyramid. The nine of hearts can now be removed with the four of spades in column 2, allowing two six-seven pairs to be removed from the pyramid (the king of diamonds in column 2 can also be discarded). We can see that the nine of clubs must be removed along with the four of clubs (discarding nine and four of hearts leaves a blocked position with the last four covering the last nine), so we discard the eight of spades with the five of clubs in column 1, followed by the nine and four of clubs (and king of hearts):
Now only one five remains available to remove the eight of hearts, so we must discard the three of diamonds in column 4 and the ten of spades in column 5, freeing the five of diamonds in column 5 to discard the eight of hearts (the king of clubs goes too). The rest is not too hard: ten of diamonds and three of hearts, jack of hearts and two of diamonds, both queen-ace pairs, six of spades and seven of diamonds, three and ten of clubs, two of hearts and jack of clubs, nine of spades and four of hearts, seven of clubs and six of hearts, jack of spades and two of clubs, and finally five of hearts and eight of diamonds.
Solution to Giza deal number 15849
At first this deal doesn't seem too difficult; you can discard thirty cards or so pretty easily, but eventually subtle traps appear. Unless the sixes in the pyramid are handled carefully, the six of clubs will be blocked (the ten of spades may also prove difficult to remove). The nines and fours may also crossblock with the jacks and twos if you don't pay attention to the order of play.
Let's look at the sixes first. There are three in the pyramid stacked on top of each other. The last one is in column two covering one of its sevens. In order to remove all of the sixes, all of the sevens must be uncovered; in particular the ten of diamonds in the middle of column four must be removed before the six of clubs. But this means we cannot save the ten of diamonds to discard the three of diamonds deep in the pyramid, and two other tens cover the three. So we are going to have to get well into the center of the pyramid while leaving the right side (the ten of clubs and the cards it covers) intact until we can reach the three of diamonds.
At some point we know we want to remove the six of spades with one of the sevens, so we might as well start there, and uncover a five as well so we can start removing the 8-5-8 stack in the pyramid. Our first play is to discard the seven of clubs and six of spades, followed by the five of diamonds and eight of spades, jack of clubs and two of diamonds, ten and three of hearts, seven of hearts and six of diamonds, king of clubs, eight of diamonds and five of hearts, reaching the position below:
If we carelessly discard the four of diamonds with the nine of spades, we will find that all of the remaining fours are covered by nines. We must get to the nine of diamonds to remove the four, so we discard the three of spades and ten of diamonds (since we are saving the ten of clubs). We then discard the seven of diamonds and six of hearts (always prefer a card which is covering something else to the last card in a column), and the three of clubs and ten of spades. Since the queen of spades is covering an ace, we remove it along with the ace of diamonds, followed by the seven of spades and six of clubs (the last of each, so we know it is a safe match), and the king of diamonds. We know we want to get to a five to discard the eight of hearts, but which one? Since the ten of clubs is eventually going to take off the three of diamonds, we need to save the two of clubs to remove the jack of diamonds (since the two of spades cannot do so). So we discard the queen of diamonds with the ace of spades (we have two queens left which can eventually take off the ace of clubs), followed by eight of hearts and five of clubs, jack of diamonds and two of clubs, and finally ten of clubs and three of diamonds. The king of spades comes off, and the rest of the pyramid is essentially symmetric (ignoring the king of hearts), so it does not matter which nine we remove with the four of clubs. One way to finish is: nine and four of clubs, jack of spades and two of hearts, four of spades and nine of hearts, king of hearts, queen and ace of clubs, jack of hearts and two of spades, four of hearts and nine of spades, eight of clubs and five of spades, and queen and ace of hearts.
Solution to Giza deal number 50832
There are lots of traps in this deal. The 4-9-9-4 stack in the pyramid has to be taken off in sequence by the other two nines and fours. The four of diamonds has to be discarded with one of the column nines, then the fours in the pyramid have to be discarded with the nines in the pyramid. The ace of hearts in the pyramid covers three of the queens, so it must be discarded with the queen of diamonds. We won't be able to reach either ten in the pyramid until after we discard the three of spades (which allows the nines and fours to be removed), so we need both of the other tens to remove the threes of hearts and spades.
So we start by discarding the jack of diamonds and two of hearts, then the jack of spades and two of diamonds (in order to remove a pair of cards where one covers the other, we use this technique of combining them with another discard of the same combination of ranks). Now the queen of spades and ace of hearts are discarded, then the ten of diamonds and three of hearts, and king of diamonds. We cannot use the five of clubs to remove the eight of hearts, because that would create a crossblock: the four of clubs would block both eights which could be used to remove the five of hearts, which in turn covers the nine of hearts which we will eventually discard with the four of clubs. So it is necessary to discard the eight and five of hearts together, and we must reach the nine of clubs in column four in order to discard the four of diamonds. We discard the king of spades, seven of clubs and six of spades, queen of spades and ace of clubs, seven and six of diamonds, and finally the nine of clubs and four of diamonds. Now we discard the queen of hearts and ace of diamonds, ten of hearts and three of spades, and the rest is straightforward: nine of spades and four of hearts, nine of hearts and four of clubs, eight of clubs and five of diamonds, jack and two of clubs, king of hearts, ten and three of clubs, seven and six of hearts, eight and five of spades, eight of diamonds and five of clubs, queen of clubs and ace of spades, jack of hearts and two of spades, nine of diamonds and four of spades, ten of spades and three of diamonds, seven of spades and six of clubs, and finally the king of clubs.Back to Solitaire Laboratory main page
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