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I have some dificult problem i try to solve out.
and it goes like this.
I have 3 numbers:
64 230 4 = 60
129 230 4 = 61
194 230 4 = 62
69 1 108 = 1500
134 1 108 = 1501
199 1 108 = 1502
14 185 34 = 0797
79 185 34 = 0798
128 186 34 = 0799
79 186 98 = 1798
128 187 98 = 1799
Now the 1st number can be from 0 to 16, 64 to 79,128 to 143,192 to 207.
the 2nd number can be from 0 to 255
and the 3rd number can be from 0 to 127.
All this are binary operation inside a cpu. I cannot figure out how it is calculated.
If someone finds a pattern please let me know. Thanks!
I have 3 numbers:
64 230 4 = 60
129 230 4 = 61
194 230 4 = 6269 1 108 = 1500
134 1 108 = 1501
199 1 108 = 150214 185 34 = 0797
79 185 34 = 0798
128 186 34 = 079979 186 98 = 1798
128 187 98 = 1799
That certainly looks like a lot more than 3 numbers to me. What do you mean you have 3 numbers?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Does he perhaps mean that given the 3 numbers in the valid ranges, there are 2 binary operations being performed that gives the result on the right?
The Beginning Of All Things To End.
The End Of All Things To Come.
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Ok, so the spaces indicate an operation. Now is it all one single operation, or can the two be different? If they can be different, can they be used in a different order?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Hallo, thanks for reply.
So i writed more exemples thats why its more than 3 numbers.
64 230 4 = 60 so 1st number is 62, 2nd number is 230 and 3rd number is 4. I know that 1st number can only be from:
0 to 16, 64 to 79,128 to 143,192 to 207, the 2nd number can be from 0 to 255, and the 3rd number can be from 0 to 127. Now tru some operation this 3 numbers must result in 60.
I found that:
129 230 4 = 61 so if 1st munber rise by 65 the result rise by 1
i also finded out that if the first number+65 equals 16, 79,143,207 then the 2nd number rises by 1 and the result rises by 1.
@ricky I dont know how many operations are made, or in what order but i think 1st number is the less semnificative bit, whyle the 3rd number its the most semnificative bit. Also i want to mention that if i write the numbers by chance i get some crc error .
@luca-deltodesco yes there are some binary operations but i dont know how many.
If you need more examples i can provide them.
I tryed everything i know in binary operations but i cant find a patern here, thats why i turned to math forum, maybe someone sees a patern.
thanks!
Now the 1st number can be from 0 to 16, 64 to 79,128 to 143,192 to 207
It may be better to think of the first number as 2 numbers, because each range (except the first??) is 16. And each range starts at a multiple of 64.
Examples:
15 => 0*64 + 15
65 => 1*64 + 1
207 => 3*64 + 15
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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Hi, ok i understand. Now i just found the 00 result.
so
64 192 12 = 0
129 192 12 = 1
194 192 12 = 2
if i follow the pattern here
194 + 55 = 259 but it exceeds the max value it can be(207), what could be the 3 result??
And i am sorry i made a mistake:
Now the 1st number can be from 0 to 15, 64 to 79,128 to 143,192 to 207 (so hex 00 to 0F, 04 to 4F, 80 to 8F, C0 to CF)
Sorry again i made a mistake.
194 + 55 = 259 but it exceeds the max value it can be(207), what could be the 3 result??
that is:
194 + 65 = 259 but it exceeds the max value it can be(207), what could be the 3 result??
64 192 12 =0
129 192 12 =1
194 192 12 =2
3 192 13 =3
68 192 13 =4
133 192 13 =5
198 192 13 =6
7 192 14 =7
72 192 14 =8
137 192 14 =9
64 230 4 =60
129 230 4 =61
194 230 4 =62
3 230 5 =63
68 230 5 =64
133 230 5 =65
198 230 5 =66
7 230 6 =67
72 230 6 =68
137 230 6 =69
i managed to find the 0 to 9 , or from 60 to 69 combination but following this patern i couldnt calculate the 10 or 70, or 59. Can anyone see something i dont see
Thanks.
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