HP - 50g written problem complex number. How to write (3 + i * 3 * sqr (3)) to RPN? (3, i3sqr (3))
Hello everyone, I don't know how to write (i 3, * (3 * sqr (3))) RPN mode.
Can someone help me please? I already searched the gug pdf guide help, without result.
Thank you.
Hello
A few notes in the representation of a complex number as a vector:
(1) the i is omitted, the second number is supposed to be the imaginary part.
(2) the 50 g accepts only decimal numbers for vector representation
Thus, at the entrance there. I guess you have the calculator in complex mode and exact mode (about not checked in "MODE"-> "SAE"), i.e. "C =" displayed at the top of the screen.
RPN mode:
Press 3 then ENTRY (3 is at level 1)
Press ENTER again (3 is now duplicated and on both levels 1 and 2 - one advantage of RPN)
Press ENTER again (3 is now duplicated and on levels 1, 2 and 3)
Press the button of sqrt (x). Level 1 shows a 3 with a sign of the square root.
Press on multiply key, level 1 shows 3· SQRT (3)
Press LeftShift (white), then I (above the TOOL key), level 1 indicates an i
Press on multiply key, level 1 shows 3· SQRT (3)· I have
Press + key, level 1 shows 3 + 3· SQRT (3)· I have (or 3· SQRT (3)· i + 3 according to the parameter Flag - 27)
If you want to keep the symbolic result (with the symbol of the 'SQUARE root'), leave it in this form, if you want to convert to vector format press (red) RightShift, then -> NUM (above the ENTER key).
The result will be a decimal vector shape (3, 5.19615242271).
Kind regards
BartdB
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The first thing he does is take the ascii value of each character in the password and converts them into a binary string. The code it uses is far too complex and can be simplified, but which is not a problem here.
I've recreated the first step of SQL like this...
SQL > ed
A written file afiedt.buf
1 with chr_val like)
2. Select level l
3, dump('password',10,level,1) in the dmp
4, to_number (substrb (dump('password',10,level,1), instr (dump('password',10,level,1),' ', 1) + 1)) as chr_val
5, ascii (substr('password',level,1)) as chr_val - equivalent of extraction of useless dump
6 double
7. connect by level<=>=>
8 )
9, r (l, b, ch, chr_val, result, vtemp) as
10 (select l, 0 b, chr (chr_val), chr_val)
11, cast (null as varchar2 (8)) as a result
12, chr_val as vtemp
13 of chr_val
14 union of all the
15 select l, b + 1, b, ch, chr_val
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17, trunc(vtemp/2) as vtemp
18 r
where the 19 b + 1<=>=>
(20) depth search first by l, defined b seq
21, as)
22 select l, ch, chr_val, str_to_bit result
23 r
where the 24 b = 8
25 arrested by l, seq
26 )
27 select listagg (ch) within the Group (order) as password
28, listagg (chr_val, ',') within the Group (order) byte_vals
29, listagg (str_to_bit) within the Group (order) bit_vals
30 sec.
SQL > /.
PASSWORD BYTE_VALS BIT_VALS
--------------- ---------------------------------------- ----------------------------------------------------------------------------------------------------
password 0111000001100001011100110111001101110111011011110111001001100100 112,97,115,115,119,111,114,100
Then he takes this string binary ("bit_vals" in my example) and does the following:
1. take the first 3 bits of left and to transpose on the right end of the string.
2 chops the resultant bit string upward into sections of 4 bits (which is known as a 'nibble' inside)
3. for each bit in the nibble, he treats the bits in binary reverse to normal and gives them a value of 1,2,4 or 8 from left to right for each bit set to 1
4. for each nibble it adds the value of 1,2,4,8 bits to give a value from 0 to 15
To show that the use SQL...
SQL > byte (select ' 0111000001100001011100110111001101110111011011110111001001100100' as pieces of double)
2, swap3bit as (-take the binary string and put the first bits (high) 3 as a (low) bit of the right hand side)
3. Select bytes.bits
4, substr (bit 4) | substr (bits, 1, 3) as init_substr
5 bytes
6 )
7, split4 as (-chop the string of bits nibbles (half bytes - 4 bits))
8. Select level l
9, substr (init_substr, ((level-1) * 4) + 1, 4) as a nibble
swap3bit 10
11. connect by level<=>=>
12 )
13, bitpowers (select l
14, snack
15, to_number (substr(nibble,1,1)) * power (2, 1-1) as bitval1
16, to_number (substr(nibble,2,1)) * power (2, 2-1) as bitval2
17, to_number (substr(nibble,3,1)) * power (2, 3-1) as bitval3
18, to_number (substr(nibble,4,1)) * power (2, 4-1) as bitval4
19, to_number (substr(nibble,1,1)) * power (2, 1-1) +.
20 to_number (substr (nibble, 2, 1)) * power (2, 2-1) +.
21 to_number (substr (nibble, 3, 1)) * power (2, 3-1) +.
22 to_number (substr (nibble, 4, 1)) * power (2, 4-1) as total_val
23 of split4
24 )
25 select * from bitpowers
26.
L NIBB BITVAL1 BITVAL2 BITVAL3 BITVAL4 TOTAL_VAL
---------- ---- ---------- ---------- ---------- ---------- ----------
1 1000 1 0 0 0 1
2 0011 0 0 4 8 12
3 0000 0 0 0 0 0
4 1011 1 0 4 8 13
5 1001 1 0 0 8 9
6 1011 1 0 4 8 13
7 1001 1 0 0 8 9
8 1011 1 0 4 8 13
9 1011 1 0 4 8 13
10 1011 1 0 4 8 13
11 0111 0 2 4 8 14
12 1011 1 0 4 8 13
13 1001 1 0 0 8 9
14 0011 0 0 4 8 12
15 0010 0 0 4 0 4
16 0011 0 0 4 8 12
16 selected lines.
These final values are then re-combination as strings in reverse order so that you get then:
'12' |' 4'||' 12' |' 9'||' 13'... and so on.
In SQL...
SQL > byte (select ' 0111000001100001011100110111001101110111011011110111001001100100' as pieces of double)
2, swap3bit as (-take the binary string and put the first bits (high) 3 as a (low) bit of the right hand side)
3. Select bytes.bits
4, substr (bit 4) | substr (bits, 1, 3) as init_substr
5 bytes
6 )
7, split4 as (-chop the string of bits nibbles (half bytes - 4 bits))
8. Select level l
9, substr (init_substr, ((level-1) * 4) + 1, 4) as a nibble
swap3bit 10
11. connect by level<=>=>
12 )
13, bitpowers (select l
14, snack
15, to_number (substr(nibble,1,1)) * power (2, 1-1) as bitval1
16, to_number (substr(nibble,2,1)) * power (2, 2-1) as bitval2
17, to_number (substr(nibble,3,1)) * power (2, 3-1) as bitval3
18, to_number (substr(nibble,4,1)) * power (2, 4-1) as bitval4
19, to_number (substr(nibble,1,1)) * power (2, 1-1) +.
20 to_number (substr (nibble, 2, 1)) * power (2, 2-1) +.
21 to_number (substr (nibble, 3, 1)) * power (2, 3-1) +.
22 to_number (substr (nibble, 4, 1)) * power (2, 4-1) as total_val
23 of split4
24 )
25 select listagg (to_char (total_val)) the Group (order of the desc) as pwd
26 of bitpowers
27.
PWD
-------------------------------------------------------------------------------------------------------------------
12412913141313139139130121
Now, the problem of decryption is that these numbers are concatenated without padding for a fixed number of digits by value, so you don't know if it was
'12' |' 4'||' 12' |' 9'||' 13'... and so forth as we did it, or whether he was
'1'||' 2'||' 4'||' 12' |' 9'||' 1'||' 3'... and so on, or any other combination of values from 0 to 15
There is essentially no information to allow you to divide the string upwards in the correct components to allow the whole process be reversed.
So, you are out of luck... no chance of decrypting it.
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