Game theory problem related to poker (academic content)

[Duerig]

Ok I'm so sick of this problem that I'm going back to school work.

[MrWookie47]

Hmm, upon reading your hint, I guess 3 and 4 isn't right, since if A knew of B's ambiguity, a sum of 7 would be clear to him. I guess I'm going to keep thinking after all...

[mmbt0ne]

I'm doing it, answer to follow shortly.

[tijean]

[ QUOTE ]
When B says he doesn't know the numbers, the sum revealed to him has to be a sum that can be attained by two squared numbers in at least two different ways. A knows this, and his sum has to be a sum that can be attained by summing up two numbers in at least two ways (he has just the sum of the numbers so it can be many ways), AND there has to be at least two ways any pair of those numbers can be squared and summed to form the same square sum. When B hears A saying he doesn't know, B knows that... etc

[/ QUOTE ]

This is what I love about math. I'm a writer with an English degree and this paragraph reads like complete gibberish to me. So I'll ignore the 'hint' and just go back to the first post.

Since it stops on the fourth iteration, I'll guess that x = 3 or 4, depending on whether you start at 0 or 1. I'll guess that y = 6, since 4-squared + 6-squared = 52, but I see no way to solve for y with the information given.

If any of that is even remotely correct (and I'm guessing it isn't), then the game plays a lot more like Mastermind than Poker. I'm even worse at Mastermind than I am at Poker.

F*ck math.

[mmbt0ne]

There's soooo much calculating in this problem. I have to be doing it the wrong way, there has to be a simple solution. Ok shortcut, reveal yourself to me!

[mmbt0ne]

Balls. This is pissing me off.

Maybe if I type out a breakdown something will come to me. Let's give A 7 and B 25.

A knows it's either 7,0 ; 6,1 ; 5,2 ; 4,3
B knows it's either 5,0 ; 4,3

B says "I don't know"
A crosses out 7,0 ; 6,1 ; 5,2 because all of those would be obvious answers of 49, 37, and 29 respectively.
A says "I know"

Now, how can I go that many more levels that doesn't involve blind guessing....

[MrWookie47]

Well, if that's level 1, level two is:

Give A 17, and give B 169:
B's set: 5,12 or 0,13
A's set is large

B: I don't know
A crosses off all but 8,15 and 5,12.
A: I don't know
B: I know!

Level 3 is, umm, hard.

[mmbt0ne]

Exactly. We need to find a number with 4(?) possible squares you can add up.

Assuming my geometry hasn't left me, you have the following Pythagorean triplets:

<font class="small">Code:</font><hr /><pre>
1 0 1
3 4 5
5 12 13
7 24 25
9 40 41
11 60 61
13 84 85
15 112 113
17 144 145
19 180 181
21 220 221
23 264 265
25 312 313
27 364 365
29 420 421
31 480 481
</pre><hr />

If you take the first 4, you get B having sqr(225)=50625. Now, I guess you can figure out A from that, but how?

B - I don't know [no [censored]]
A - I don't know [ummmmmmmmmm....]

[Jingleheimer]

B says I don't know. Therefore his square can be expressed in two ways.

If we make a list of all of the squares of all pairs of integers and look at the squares that come up twice, we get
50
65
85
125
130
145
170
185
200
205
221
250
260
265
290
...

[MrWookie47]

Dang. I neglected a lot of my Pythagorean triplets (an infinite number, I suppose, since that list goes on), but you also forgot 8 15 17. Holy schnikes is there a lot of calculating in this problem.