Tackle This One

[Hamish McBagpipe]

This is from the test for exceptional intelligence here: Link (http://www.highiqsociety.org/noflash/nonmembers/iqtests.htm)

I've done a few of the other questions on this difficult test and am pretty confident of my answers. They take quite a bit of time, for me anyway, and most seem to be at least within the realm of possibility of being solved. This one has been driving me nuts. I like this kind of puzzle, I've been looking at it at work a bit, but I see no way to tackle this except through some kind of mass time-destroying elimination/substitution strategy. I'm not looking for an answer, rather I'm hoping someone has an idea of how to tackle this one that I haven't seen.

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Each letter has an associated numerical value attached to it, and the total of all the letters equals the physicist's total value. For example, if the letters G, L, A, S, E, and R had the values 12, 7, 9, 14, 21, and 5, respectively, American physicist Glaser would have a numerical value of 68.

Your objective is to figure out what the last physicist--Feynman--should be valued.

http://www.highiqsociety.org/img/gif/iqtests/uh1q.gif


That's the whole thing. It doesn't even say if the same number can be assigned more than once but I've been going on the assumption that each one would be different. There are a couple of things that will pop out at you if you look at it for a while but it gets mind-crushing after that.

They do ask that in the interests of maintaining the integrity of the test, no solutions or partial solutions are posted anywhere on the web.

[Hamish McBagpipe]

I haven't even tried to cheat by googling it. I'd be surprised to find anything anyway. I think you know what I'm asking.

The assumption that each number is different is indeed correct. You can go under the assumption that letters are numbered 1-26. Do you see y?

When it comes to physicists, Curie was odd, and even Stern should tell you something. One does not follow the other, however you look at it.

[SheetWise]

If CURIE = 69 and FERMI =57
then
CU - FM = 12

[ QUOTE ]
They do ask that in the interests of maintaining the integrity of the test, no solutions or partial solutions are posted anywhere on the web.


[/ QUOTE ] Well, [censored] em. Mensa wannabes.

This actually can't be solved, given the information available. Do you see why? I'll elaborate.

This should be able to be systematically answered by setting up a series of 25 equations.
r + o + e + n + t + g + e + n = 104
l + o + r + e + n + t + z = 102
etc.

You have 25 equations and 24 variables (J and Q are not used), so there will always be a solution for each letter. You can do a lot of simple algebra, or there are probably some linear algebra tricks to solve this more quickly, but it's just a question of pushing around letters.

Okay, so we can solve for each of the letters listed in the names in blue. We can just plug them in to FEYNMAN to get the answer, right? No. If you look closely, you'll see there's no Y in any of the other names, leaving that value unspecified.

Can we just assume that the numbers 1 through 26 are assigned to all the letters, and then solve for everything other than Y and assign the remaining value to Y? No. Q and J are also unspecified. The best we could do is narrow it down to 3 possibilities.

Feel smart now?

[Dr. StrangeloveX]

There may be an obvious way of assigning values though, in which case the Y value could be found.

Like if you solved for the 23 used letters and got

a=1
b=2
c=3
d=4
etc.

Then they probably mean for you to make an educated guess as to Y's value. The actual pattern will probably be more complicated though.

I think these tests are not particularly interesting. Consider this question:

[ QUOTE ]
Consider two breeding strategies of the fictional Furble. Dominator Furbles can fight for a breeding territory, and if they win, will be able to rear 10 offspring. An alternative is to share territory with another Furble which will allow each to rear 5 offspring. Sharers who attempt to share with dominators will be forced out of the territory, although they will be able to find a new territory. Assume sharers become extra cautious after encountering a dominator and so will always find another territory to share the next time around, but due to lost time will only be able to produce 3 offspring. Dominators are always able to force sharers out of the territory and rear 10 young. Dominators who meet dominators will win 50% of the time. When they lose, they are not able to reproduce that season due to sustained injuries. Individual Furbles cannot switch strategies.

With a total population of 2000 dominator and sharer Furbles, how many would you expect to be dominators?

[/ QUOTE ]

On the face of it, it seems like there should be some kind of equilibrium, where the more dominators there are, the more they interfere with each others' breeding. There isn't. For each individual, it is always advantageous to be a dominator. If you meet another dominator, if you're a dominator, you'll have 10 offspring 50% of the time, for EV = 5. If you're a sharer, you'll have 3 offspring. If you meet a sharer, a dominator will have 10 offspring and a sharer will have 5. Therefore, regardless of the distribution of the population, dominators always have an advantage, and the equilibrium value is 100% dominators.

I guess it's an okay question, but they could have made it a lot more interesting just by stipulating that when dominators meet sharers, the sharer has 6 offspring instead of 3. This creates a situation where domination is less viable than sharing in a population with 99% dominators, and actually makes you figure out a real equilibrium value.

[durron597]

I eventually did get this one. I can assure you that the solution is the only one that makes sense.

No, I will not tell you what it is, as they have asked for the solutions not to be posted online.

[fnord_too]

Isn't this just a simple set of linear equations? Why can't you just set up the matrix and solve it? (No, I haven't tried, I don't have the time to trudge through it, but it seems like just a tedious exercize in linear algebra.)

[purnell]

I remember this problem. I looked at it for a while and decided that either there is no one solution, or that I'm not smart enough to solve it (that is, I can't find a way to approach the problem that will allow me to crack it). Oh, well, I never thought I was the smartest guy in the world, or even close.

[Hamish McBagpipe]

[ QUOTE ]
This actually can't be solved, given the information available. Do you see why?

[/ QUOTE ]

I've gotten further than this. But you are right, of course, in your explanation. However, I'd think there is a solution.

[Hamish McBagpipe]

[ QUOTE ]
I can assure you that the solution is the only one that makes sense.


[/ QUOTE ]

Yes, I'll keep pushing it around until I get it.

[Georgia Avenue]

23, of course.

Yeah, I'd guess you solve for the numbers that you can solve for, and then there's some pattern in the distribution of numbers to letters that you can figure out to find y.