On my only two attempts, it never landed on the target card. Weird.
Well you've got about a 4% chance of that happening, or 1 in 25.
The Kruskal Count Card Trick
21–30 of 43 posts
Re: The Kruskal Count Card Trick
#22Earlier quoted context omitted.
> For example, if you have 10 cards on the first row, then there are at most 10 paths to the end, but those can merge In fact they must converge, suppose that the first card on the first row is less than 10, then it will merge with another path on the first move (because it will land on the first row), so it must be 10 to have no merging paths. The next card will merge with another path on the first row if it's less…
> In fact they must converge Are you sure? Is this board not a counter-example if A starts on the right-most column and B starts on the second from the right: x x x x x x x 8 9 x x x x x x 8 x 9 x x x x x 8 x x 9 x x x x 7 x x x 9 x x 7 x x x x x A B x x x x x x
Yes, the claim is 'at least two of the paths taken from the 10 cards on the first row must converge', not 'there are no paths taken from the 10 cards on the first row which do not converge'.
Re: The Kruskal Count Card Trick
#23i couldn’t find how the “trap card” was determined
On the "how it works"[0] page, they explain that they choose a card in the first row and perform the walk. The point of the trick is, it's extremely likely that, regardless of the initial card chosen, the walks will merge (and thus step onto the trap card you chose when setting up).
0: http://faculty.uml.edu/rmontenegro/research/kruskal_count/in...
Re: The Kruskal Count Card Trick
#24On my only two attempts, it never landed on the target card. Weird.
Same here.. tried it twice.. did not end up on the target card.
Just clarifying if how I counted was correct..
5 x x x x 9
Did this, but did not work for me.
Re: The Kruskal Count Card Trick
#25i couldn’t find how the “trap card” was determined
It's explained here albeit vaguely: http://faculty.uml.edu/rmontenegro/research/kruskal_count/in...
You start at the first card and progress until you would fall off the end on the next step. That card you're on is what you pick as the trap.
Re: The Kruskal Count Card Trick
#26On my only two attempts, it never landed on the target card. Weird.
Same here.. tried it twice.. did not end up on the target card. Just clarifying if how I counted was correct.. 5 x x x x 9 Above is a crude illustration.. If I first pick the card with "5", I start counting from 1 from the next card, till 5. And then I would land up on whatever that card is (9 in this example), and then I would repeat. Did this, but did not work for me.
It doesn't really matter which method you use for counting, so long as you're consistent.
Re: The Kruskal Count Card Trick
#27Earlier quoted context omitted.
Same here.. tried it twice.. did not end up on the target card. Just clarifying if how I counted was correct.. 5 x x x x 9 Above is a crude illustration.. If I first pick the card with "5", I start counting from 1 from the next card, till 5. And then I would land up on whatever that card is (9 in this example), and then I would repeat. Did this, but did not work for me.
It doesn't really matter which method you use for counting, so long as you're consistent.
It does if you are using the web version, because the key card there is chosen using a specific method.
Re: The Kruskal Count Card Trick
#28On my only two attempts, it never landed on the target card. Weird.
Same here.. tried it twice.. did not end up on the target card. Just clarifying if how I counted was correct.. 5 x x x x 9 Above is a crude illustration.. If I first pick the card with "5", I start counting from 1 from the next card, till 5. And then I would land up on whatever that card is (9 in this example), and then I would repeat. Did this, but did not work for me.
I also did two passes through the game and did not end up on the trap card. Worried that I was not following the rules correctly I came to the comments to check.
I followed the rules just as you illustrated.
Re: The Kruskal Count Card Trick
#29Earlier quoted context omitted.
If there's a connection I certainly don't see it.
There is a strong connection with rainbow tables. A common misconception is that rainbow tables are a complete set of hash values. What they actually are a method of saving storage space at the cost of making searches slower. The idea is to make a function that maps hash values back to password-like strings. With that function chains like this are built: I=initial password guess P=new password-like string V=hash valu…
Rainbow tables are actually a slight revision on this, because the merging of paths is a bad thing as it leads to false positives. Specifically, I can detect that my hash eventually leads to the end of a chain, but it may very well be that it merged into that chain at some point and so not contained in my table.
Rainbow tables minimize the chance of merges by using a sequence of unhash functions in chains so a merge can only occur if it happens the same point in the sequence of unhash functions.
Re: The Kruskal Count Card Trick
#30Earlier quoted context omitted.
> For example, if you have 10 cards on the first row, then there are at most 10 paths to the end, but those can merge In fact they must converge, suppose that the first card on the first row is less than 10, then it will merge with another path on the first move (because it will land on the first row), so it must be 10 to have no merging paths. The next card will merge with another path on the first row if it's less…
> In fact they must converge Are you sure? Is this board not a counter-example if A starts on the right-most column and B starts on the second from the right: x x x x x x x 8 9 x x x x x x 8 x 9 x x x x x 8 x x 9 x x x x 7 x x x 9 x x 7 x x x x x A B x x x x x x
I may have not made it clear, but like falsedan said, I'm not claiming that all paths must converge, only that at least two paths must converge.
edit: That being said, I made a mistake by following the claim from above that there are 10 possible paths, in fact there are only 9 cards in a row.