Puzzles and Models - Image intensive!

[Faramond]

Where's my prize?

[truehobbit]

Hehe, Faramond is right, Jny! He did solve the riddle, too, and gave the right answer!
(Or, at least, the same as IS! I haven't figured out the answer myself yet. )

[Jnyusa]

?

Now you have both surpassed my mathematical acumen. I can't translate Faramond's answer.



What's the key?

Jn

[Faramond]

the answer in binary

=

10100101111111

=

8192+2048+256+64+32+16+8+4+2+1

=

10623



I was too obscure.

[Jnyusa]

Aha!

Now I feel stupfid! And how did Hobby figure out it was powers of 2? Why are you teaching English, Hobby? You should be teaching math!

Well, I have to find an extra-special prize for this one. Just as soon as I get back from the supermarket.

<jimmies feet into shoes>

Jn

[truehobbit]

We did a very little messing around with the binary system back in fifth or sixth grade at school - just for logical training, this was way before "binary" meant anything to anyone in computer terms!

I didn't suspect anything before Faramond asked for a prize, but then I thought that his number probably was the same as Idylle's, only in the decimal, rather than binary system.

And - - I didn't even remember how this worked (hey, sixth grade was 25 years ago! ) - but I looked it up - powers of 2, yes (but if you got this from Faramond's post up there, then you should be teaching maths, Jny! Or did you just know that binary meant powers of 2? I had to find me a nice maths site that explained the system - - but I also found this, which did the work more quickly: http://www.mathepower.com/stellenwertsysteme.php
(Dual is German for binary)

Edited to add: must have been in fifth grade - clicking on that link for numeral systems in subjects for fifth grade on that site gives you the page I linked to - *sigh* if you think about all the things you once learned and forgot again...

[Jnyusa]

A prize for Faramond:



And another for Hobby:



It's special pets day!

Hobby, I'm off to look at your links.

Jn

[Jnyusa]

Hobby, your link is German.

I could sort of follow what they were saying, but I wasn't confident that I really understood.

In Faramond's post I just looked at his number series and saw they were powers of two in descending order.

Faramond, could you use that approach to actually solve the puzzle? Or did you solve the puzzle and then convert the answer afterwards?

Also, given the answer you provided ... it would not necessarily have to be a binary answer, would it? It could, for example, be the sum of the powers of three, or five? One would discover that by trial and error? - or is there an easier way to look at that and know the referent number system.

(It's so convenient to have you here to answer math questions. :D
I did find that other equation I was looking for last summer, btw ... did I tell you that the problem had been solved?)

Idylle: For some reason, I spent 90% of my time on the first letter of the first word in the question. It just didn't register.

Hobby: I haven't figured out the answer myself yet.

I didn't start at the beginning, because the permutations are endless and I get headaches easily. I started with the letters p and q - hoping there would be some - because p is the only letter with 4 zeros and q is always followed by u, which makes a rather long series. Lucikly there were both, and I was able to work backwards and forwards from those letters.

Jn

[Alatar]

the answer in binary

=

10100101111111

=

8192+2048+256+64+32+16+8+4+2+1

=

10623



I was too obscure.

Or you could just open Calc, view Scientific, switch to Binary and paste in 10100101111111. Then switch back to Dec and you'll see 10623.

I always go for the easy option.

[truehobbit]

Hobby, your link is German.

I could sort of follow what they were saying, but I wasn't confident that I really understood.

In Faramond's post I just looked at his number series and saw they were powers of two in descending order.

Yes, it's German - I just posted it because of the box where you can have any numbers switched to a different system, not because of any explanation it provided! Sorry about the misunderstanding!

There's a box where you can enter digits, and then you pick a system from which, and a system into which you want to change it - for example an octal (or whatever the word in English might be - a system based on powers of 8 ) "35647" would be "3232213" in a system that uses powers of 4 etc etc. I just posted it to show that even though I looked up how this stuff worked (because I didn't remember from school) I didn't do the actual maths on this one myself!

But, yes, like you said, it could be based on any system.

So - not sure if anyone still needs an explanation, but here goes (and my apology to the resident mathematician for the probable crudeness of this ):

We happen to use a system based on 10 different digits, so that we only need to add a new digit for each power of ten:
1
10
100
1000

When you have the decimal number 523 that just means you have 5 "hundreds", 2 "tenners" and 3 "ones".

If you had only four digits to use, you'd have to figure out how many "fours", "sixteenths" etc you have in the number instead.

If you have two digits to use, you need to add a digit for each next power of two - I've always found it easiest to imagine that the one is "yes" and the 0 is "no" on a string of numbers.
1 = 1
10 = 2
11 = 2 + 1 = 3
100 = 4
101 = 5
etc.

At school we had a little piece of cardboard on which we'd written:

1024 512 256 128 64 32 16 8 4 2 1

So if the task was to write down the number 523 we could just look at it and write:
1000001011

To transfer any number back into decimal, you have to start multiplying and adding from right to left:
The digit on the right is the rest, so it's just that digit.
The second one is whatever system you are using multiplied by whatever number it is: n*x
The third is n*n*x
etc.
For example, in a system of eight:
35647
is
(8*8*8*8*3)+(8*8*8*5)+(8*8*6)+(8*4)+7 = 15271 in decimal.

Also, given the answer you provided ... it would not necessarily have to be a binary answer, would it? It could, for example, be the sum of the powers of three, or five? One would discover that by trial and error? - or is there an easier way to look at that and know the referent number system.

That's a good question! I guess if you see no digit higher than 3 you could guess that it's a system of 4 that's being used - but of course it might just as well be a coincidence that there was no higher digit in a decimal number.
However, other systems than binary and decimal aren't really in use (apart from hexadecimal, of course).

I didn't start at the beginning, because the permutations are endless and I get headaches easily. I started with the letters p and q - hoping there would be some - because p is the only letter with 4 zeros and q is always followed by u, which makes a rather long series. Lucikly there were both, and I was able to work backwards and forwards from those letters.

Wow, thanks - yes, that should be helpful! After 15 minutes I had enough variables to wear out my patience! Very clever way of approach!

[Faramond]

Oooh! My prize looks dangerous!



Faramond, could you use that approach to actually solve the puzzle? Or did you solve the puzzle and then convert the answer afterwards?

You have to leave it in binary if you want to solve the puzzle. If you converted that giant number from the puzzle into decimal you would lose the proper separation, in a sense. You couldn't work with it. I don't know how to explain it. Well, take a simple combo, ABA. That's 1101, which comes out to 13 in decimal. You can't break apart 13 to come up with ABA, in fact it would look more like AC, if A=1 and C=3.


Also, given the answer you provided ... it would not necessarily have to be a binary answer, would it? It could, for example, be the sum of the powers of three, or five? One would discover that by trial and error? - or is there an easier way to look at that and know the referent number system.

I'm not sure I understand ... a series of ones and zeros does not have to be a binary number, of course. Without being told the number 10011100110 could be any number system, of course. But I felt it was implied here. I think I'm not understanding your question!

(It's so convenient to have you here to answer math questions. :D
I did find that other equation I was looking for last summer, btw ... did I tell you that the problem had been solved?)

Oh cool! Sorry I couldn't be more help with that!


I didn't start at the beginning, because the permutations are endless and I get headaches easily. I started with the letters p and q - hoping there would be some - because p is the only letter with 4 zeros and q is always followed by u, which makes a rather long series. Lucikly there were both, and I was able to work backwards and forwards from those letters.

Yes, exactly what I did. I also found some 'h' for some reason, I think because the other thing they could be was unlikely, and I just had to keep in mind that where I saw an 'h' wasn't necessarily an 'h'. I pasted the series into a word processor and used find-replace to hunt down some combos, as well. Is that considered cheating? I had to tell it what to hunt down!

[Faramond]

Or you could just open Calc, view Scientific, switch to Binary and paste in 10100101111111. Then switch back to Dec and you'll see 10623.

I always go for the easy option.

Indeed, that's what I actually did when I wrote down my answer in decimal!

Laziness is me.

[truehobbit]

I haven't yet said thanks for my prize! They are adorable, Jny!
I wish I really had them and could cuddle them now!

[Jnyusa]

Hobby, I know! All those prizes at the Lutefisk festival ... it was so unfair that they were only virtual.

Faramond: I'm not sure I understand ... a series of ones and zeros does not have to be a binary number

I meant the number 10623. That would not have to be the decimal equivalent of a binary number. But I was foolish not to assume so because it was a binary puzzle. When I looked at it I thought it might be base 6, and that didn't work out to anything logical so I assumed instead that you had posted a random number as a joke.

I pasted the series into a word processor and used find-replace to hunt down some combos, as well. Is that considered cheating? I had to tell it what to hunt down!

I did that too. For the letter 'e'. But there was no rule you had to solve it in the most difficult way possible.

Jn

[Faramond]

oh, I get it now.

Well, you're right, I didn't specify.

Though 10623 couldn't be base 6 because the digit 6 is in it. The lowest base it could be is base 7!

The way to really make the puzzle evil would be to randomly assign letters to the first 26 binary numbers and then write the message that way.

[Jnyusa]

The lowest base it could be is base 7!

Yes, of course. See? I told you I was stupfid.

I had a kind of weird math education ... Nothing at all in numbers theory ... no trig, no physics. When I moved into economics I got flooded with calculus and statistics and more statistics and very advanced statistics. Then my field ... rather, my interest .. took a sharp right turn and everything I'm reading now is from physics. So I sat down last summer and started to teach myself trig and physics. Still working at it. Things like number systems ... I don't intuit anything, and have to think about every explanation for a long time before I understand it.

The way to really make the puzzle evil would be to randomly assign letters to the first 26 binary numbers and then write the message that way.

I'm going to do it with a short quote from Tolkien and see who can solve it. Combination cryptogram and binary.

Jn

[Jnyusa]

Notoriously easy to create! ... and I added a 'clue' to this one which will make it notoriously easy to solve, I'm afraid. But ... we'll have some quick fun.

OK. The following is a very short quote from Tolkien, taken from RotK. The quote has 18 letters in it, and uses exactly half the letters in the alphabet.

Each letter of the alphabet, with the exception of three, has been assigned a number at random, and that number has been translated into binary. Three letters are not assigned randomly; they are the three letters that appear more than once have been assigned to a binary code that is particularly distinctive. (That's the clue.)

Here's the catch. The numbers to which letters are assigned are not 1-26. They are 26 numbers that form a familiar series, and the series has been given a little twist.

(I do think the clue will allow Faramond and Idylle to solve this in about 10 seconds. But let's see how others fare!)

1101100000100000100001000101110000010011100001100110000010010110101111010111111011000


Jn

(Hope I haven't made any translation mistakes!!! When you get it, post that you've got it and send me the answer via PM. More prizes are promised.)

[truehobbit]

I don't really get that.

You mean that instead of a = 1, b = 10 etc, the letters are mixed up, so that it might be m = 1, f = 10 etc?

[Jnyusa]

Yes, that's right.

But the three letters that appear more than once have a distinctive signature, that is, the binary code attached to them is readily identified. If you translate those three sequences to their corresponding decimal equivalent, it should give you a clue to the number series used, and that will allow you to figure out the binary code for the other numbers in the series. Once you've got all the numbers, it becomes an ordinary cryptogram, and a very short one from Tolkien that you should be able to solve quickly. (Assuming I didn't make any coding mistakes.)

Jn

[Faramond]

Oh no, I can't get it in 10 seconds!

Without the extra clues such a short passage would be impossible with the randomized binary code, I think. You really need lots and lots of letters to have any hope of beginning to identify which letters go to what combinations.

With all the clues, I think this one must be possible, though.