question for the supergeeks! [binary code]
Discussion
anyone got a clue on the following? formatted exactly as it was given to me.
100110111010001
100110111010001
000111001000100
011000110100100
010111001000000
110000101100000
100110111010001
100110111010001
000111001000100
011000110100100
010111001000000
1100001011100000
Hints:
1) There is one extra "bit"
2) The message repeats once
3) The "letters" are interleaved
4) The "word" length is prime
100110111010001
100110111010001
000111001000100
011000110100100
010111001000000
110000101100000
100110111010001
100110111010001
000111001000100
011000110100100
010111001000000
1100001011100000
Hints:
1) There is one extra "bit"
2) The message repeats once
3) The "letters" are interleaved
4) The "word" length is prime
The one extra bit part makes it trickier I think, but does this
help?
http://www1.cs.columbia.edu/~zeph/3261/hw/haystack...
help?
http://www1.cs.columbia.edu/~zeph/3261/hw/haystack...
Silent1 said:
Hi mate! Wot you still doing here, I thought you were on the other side of the planet.
Oops, this is the internet, you could be, I guess...
Back on topic, yeah, work out your own puzzles you tart. It's like coming on here asking technical questions when you're being paid £800 a day as a consultant and you need an answer for your client, which I've seen plenty of times (none from me, may I add...)
Silent1 said:
rumbled!question was asked on a guitar forum i frequent. so far the prize is unclaimed and thought it would be worth a shot here. cheeky i know but i know absolutely nothing about binary!
What does "The "letters" are interleaved" mean exactly? The decoded letters, or the binary? I take it this is 8 bit binary rather than 7 bit (it's a very poorly worded quiz 
).
The word length is eleven, and this is the only bit you need to decode (unrepeated):
100110111010001100110111010001000111001000100011000110100100010111001000000110000101100000
You can visually see where the extra bit is.
).The word length is eleven, and this is the only bit you need to decode (unrepeated):
100110111010001100110111010001000111001000100011000110100100010111001000000110000101100000
You can visually see where the extra bit is.
cyberface said:
Silent1 said:
Hi mate! Wot you still doing here, I thought you were on the other side of the planet.
Oops, this is the internet, you could be, I guess...
Back on topic, yeah, work out your own puzzles you tart. It's like coming on here asking technical questions when you're being paid £800 a day as a consultant and you need an answer for your client, which I've seen plenty of times (none from me, may I add...)
30th of this month young padawan.
Silent1 said:
cyberface said:
Silent1 said:
Hi mate! Wot you still doing here, I thought you were on the other side of the planet.
Oops, this is the internet, you could be, I guess...
Back on topic, yeah, work out your own puzzles you tart. It's like coming on here asking technical questions when you're being paid £800 a day as a consultant and you need an answer for your client, which I've seen plenty of times (none from me, may I add...)
30th of this month young padawan.
I am SO feeling my age...Get a fat pipe or HSPA ASAP, I'll send you stuff packaged over if you need, and looking forward to seeing / hearing the Good Life
Things here are mental, and not in a good way... all kicks off in November so we shall see....Keep in touch mate.
that's a bit annoying, those clues suck.
What do they mean, are the 15 digit nos all one digit short and they are showing parity from the last given 16bit/2byte/example? whats the point of having an "obvious" extra 1 placed in at the 6/7/8 least significant bit on the final word
or are we to assume that the bytes are each 5 digits long so giving a range (0-31) just large enough to cover the alphabet 1=A, 2=B ...... or maybe they are all ascci code nos 65=A 97=a upwards, hmmmmmmmm
also if the word length is prime do they mean the no of bytes in the binary word is prime ie 1 2 3 5 7 11 bytes per word? They dont make it very clear.
What do they mean, are the 15 digit nos all one digit short and they are showing parity from the last given 16bit/2byte/example? whats the point of having an "obvious" extra 1 placed in at the 6/7/8 least significant bit on the final word
or are we to assume that the bytes are each 5 digits long so giving a range (0-31) just large enough to cover the alphabet 1=A, 2=B ...... or maybe they are all ascci code nos 65=A 97=a upwards, hmmmmmmmm
also if the word length is prime do they mean the no of bytes in the binary word is prime ie 1 2 3 5 7 11 bytes per word? They dont make it very clear.
Edited by tpivette on Wednesday 24th September 15:02
100110111010001 line1
100110111010001 line2
000111001000100 line3
011000110100100 line4
010111001000000 line5
11000010 11 00000 < line6 WOW! heres the change
100110111010001 7 same as line 1
100110111010001 8 same as line 2
000111001000100 9 same as line 3
011000110100100 10 same as line 4
010111001000000 11 same as line 5
11000010 111 00000 < line 12, the extra bit is in around the msb of the final byte, either of the three 111's
I think thats what they mean by the message repeats......
and that theres an extra bit.
But the question is, how many binary digits make up the byte and from that a value - dec/hec/oct?.
Im assuming once you get this value you can then x-refer to the ascii code no that each letter/symbol has.
100110111010001 line2
000111001000100 line3
011000110100100 line4
010111001000000 line5
11000010 11 00000 < line6 WOW! heres the change
100110111010001 7 same as line 1
100110111010001 8 same as line 2
000111001000100 9 same as line 3
011000110100100 10 same as line 4
010111001000000 11 same as line 5
11000010 111 00000 < line 12, the extra bit is in around the msb of the final byte, either of the three 111's
I think thats what they mean by the message repeats......
and that theres an extra bit.
But the question is, how many binary digits make up the byte and from that a value - dec/hec/oct?.
Im assuming once you get this value you can then x-refer to the ascii code no that each letter/symbol has.
Edited by tpivette on Wednesday 24th September 15:30
is this what other are thinking. If you decide the byte length =5 digits ie a prime no, then sort the given data minus the duplicate series
10011 01110 10001
10011 01110 10001
00011 10010 00100
01100 01101 00100
01011 10010 00000
11000 01011 100000
then convert those binary nos to decimal nos
19 14 17
19 14 17
3 18 4
12 13 4
11 18 0
24 11 32
then transpose those on a a=0 b=1 c=2 d=4 basis
T O R
T O R
D S E
M N E
L S A
Y L Z
then de interleave
T R O D E N L A L
O T R S M E S Y Z
You get a pile of crap, bugger.
10011 01110 10001
10011 01110 10001
00011 10010 00100
01100 01101 00100
01011 10010 00000
11000 01011 100000
then convert those binary nos to decimal nos
19 14 17
19 14 17
3 18 4
12 13 4
11 18 0
24 11 32
then transpose those on a a=0 b=1 c=2 d=4 basis
T O R
T O R
D S E
M N E
L S A
Y L Z
then de interleave
T R O D E N L A L
O T R S M E S Y Z
You get a pile of crap, bugger.
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