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1,333 comments, last by RedZep 18 years, 6 months ago
I was looking around at the google results for "Nine Bows" -- almost all of which had to do with Egyptian mythology. Egypts enemies where called the "Nine Bows," because the number nine (The number of base terms at /two) represented the "Pluralty of Pluralties" -- three times three (The number of numbers after + signs, and the number of numbers after - signs at /two). I'm really not sure if we are supposed to find this out, or if this has ANYTHING at all to do with the puzzle, though...
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On the theme of them not being mathematical operators I was investigating the idea that the + and - signs might actually push and pop a stack. That would give something with answers on 4 different levels within the stack, perhaps an IP or short word. Once I've got the numbers into this stack idea though I can't yet find any way of combining them that produces a useful result, so this might be a complete red herring.
could the digits after +- be right/left shifts on the first digit: 27 26 30. I came up with this when I noticed the first numbers don't have sign.

also note that there are 15 digits in total, maybe it's a clue telling that the numbers are actually base 16. If they are bit shifts then +1 23 wont get cancelled by -1 23.
[size="2"]I like the Walrus best.
Well, these are the nine bows:

"The Sed-Festival pavilion of Amenhotep III portrayed each of the Bows as a captive figure with his arms tied behind his back. Next to each is an oval medallion upon which the name of that people which each represents is given. The faces vary in their portrayal according to their locality. The list is considerably different from previous examples. There is no mention of either half of Egypt. Each is represented as a nation that the pharaoh has had to subdue in battle. They are listed as (1) the Hau-Nebut, (2) the Shat, (3) the Ta Shema, (4) the Sekhet-lam, (5) the Ta Mehu, (6) the Pedtiu-Shu, (7) the Tehennu, (8) the Iuntiu-Seti, and (9) the Mentiu-nu-Setet."

That source also quotes a lot of bible verses... could we be looking for a divine answer?
does this means something to someone?:

positivedecimal to binary27 26 30	     11011 11010 11110   +3 11	               11 1011+1 15 22	     	  1 1111 10110   +1 23	               1 10111re-grouped110111 101011 110111 0111111 110110 110111to ASCII7 + 7 ? 6 7negativedecimal to binary-1 23		    1 10111-1 22		    1 10110   -2		         10re-grouped110111 110110 10to ASCII7 6 
[size="2"]I like the Walrus best.
But does it lead somewhere if you translate each character in the sequnze to it´s decimal number and then do something with them? This would give us for the first characters 50|55|32(space)|50|54|32(space)|51|48|43(the '+')|51|32(space)|....
Bessi
this is what i came up with:
 00101011        +  00101101        - 27 26 30	     11011 11010 11110   +3 11	          101011 11 1011+1 15 22	     101011 1 1111 10110   +1 23	          101011 1 10111-1 23		  101101  1 10111-1 22		 101101   1 10110   -2		  101101       1001101111  01011110  10101111   10111010   11111111   01101010   11110111   10110111   01111011   01110110  10110110o         n                                          j                                (          f    



maybe we are supposed to convert the whole string into binary then parse it from there? and that would include converting the whitespaces into binary? it's a thought....

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Nobody's figured this puzzle out yet? Wtf, where are all of the people with time to waste? geez...
"I study differential and integral calculus in my spare time." -- Karl Marx
Alright, so I tried solving the level two problem, and thought that the "27 26 30+3 11+1 15 22+1 23-1 23-1 22-2" might be reverse Polish notation.

template <class type>class stack_t{private:    type *s;    int n;public:    stack_t(int max)    {        s = new type[max];        n = 0;    }        int empty() const    {        return n == 0;    }        void push(type t)    {        s[n++] = t;    }        type pop()    {        return s[--n];    }};void PostfixEvaluation(char *a){    int n = strlen(a);    stack_t <int> st(n);    for(int i = 0; i < n; i++)    {        if(a == '+')            st.push(st.pop() + st.pop());        if(a == '-')            st.push(st.pop() - st.pop());        if(a == '*')            st.push(st.pop() * st.pop());        if(a == '/')            st.push(st.pop() / st.pop());        if((a >= '0') && (a <= '9'))            st.push(0);        while((a >= '0') && (a <= '9'))            st.push(10*st.pop() + (a[i++] - '0'));    }    cout << st.pop() << endl;}int main(){    PostfixEvaluation("27 26 30+3 11+1 15 22+1 23-1 23-1 22-2");    return 0;}


But the final answer is 2. So I dunno if this is a step in the right direction or what, heh.

-david-
In this world gone mad, we won't spank the monkey; the monkey will spank us.
Doesn't Unicode use + and - notation to change what byte in the multibyte character to use?

Just a shot in the dark, I don't really know for sure.

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