It's funny because I was class of 2014, but I think I took this class in the fall. I am not even going to address the other point, cause it's that stupid. I think after I've done this, there's now been 4 backtracks (C->D->B->E backtrack order) so it seems like for this simple example with a size 2 domain works (this part I'm a little bit fuzzy on proving how there's 2d solutions, it looks to me like the worst case is when your assignment to E doesn't match your assignment to A, and you have to do 2*d due to having to go through both halves?)Īnyways, I'm fuzzy on CS188, haven't taken that class in years and the only times I've ever had to go back to algorithms was to study for interviews lol. Also, let's see how long it takes before they backtrack, and turn him back into a boy, considering the backlash from Japanese fans, lol. Now, I can continue by assigning Blue to E. How many times do I need to backtrack? well I would have to unassign Blue from D, but now D has no valid domains, so I have to unassign Green from B, however, B has no domains, so now I have to unassign green from E. Therefore, C has no valid elements in its domain so I need to backtrack. Suppose I assigned the value Blue for A, then Green for E, Green for B, Blue for D, and since I assigned Green for B, I have forced C to have blue, but C CANNOT have blue since D now has Blue. However, you might be wondering, why the order A->E->B->D->C will have 2D backtracking. This means that by assigning C, I have actually pretty much solved the problem moving forward with this ordering which means that the worst case scenario is that there will never be any backtracking since by assigning C, I have constrained all the domains on all of my other nodes, and in fact, this will be true for any sized domain (if it works for 2, it will obviously work for 2+ elements in my domain). Then, when I look at D, since C was assigned Blue, I know D must therefore be assigned value Green as well and in this process so on and so forth. ![]() Variations include WPA-2 which is the most secure encryption alternative till date. ![]() In other words you use the old-fashioned method of trial and error to gain access. so I choose Blue, then, I can eliminate from B, Blue from its domain, so B must have the value Green. Efficient cracking of the passphrase of such a network requires the use of a wordlist with the common passwords. That means, I choose an x in (Blue, Green) for C. Then, suppose I begin assigning values for this CSP in this order C->B->D->E->A, (Blue, Green) and my constraint, no node can have the same color as the node to the its right. To redefine, suppose I illustrate this problem as a coloring problem with 5 nodes, A, B, C, D, E.Īnd each node can have this set of domains. This effect only applies to the user who activated the spell. ![]() Looking at the problem, I think I can illustrate how you might arrive at the answers in here. Upon activation, the fog of war is cleared and all units may be seen.
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