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) and then the nine and four of diamonds. 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, one of 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.
This article is copyright ©2008 by Michael Keller. All rights reserved.