RFC 1035 Header Structure - dns
I'm studying about dns and would like to understand about this information, because I could not fully understand.
The header contains the following fields:
1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| ID |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
|QR| Opcode |AA|TC|RD|RA| Z | RCODE |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| QDCOUNT |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| ANCOUNT |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| NSCOUNT |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
| ARCOUNT |
+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
I would to know what mean this numbers on top.
The numbers across the top are simply the bit numbers within the 16 bit word, although as is common with the RFC series of documents they're ordered from most significant bit to least, instead of the (more intuitive) other way around.
So, for example, given an array data of octets containing that header, the ID would be:
(data[0] << 8) | data[1]
and the QR bit would be the most significant bit of data[2]
Related
Counting 15's in Cribbage Hand
Background
This is a followup question to my previous finding a straight in a cribbage hand question and Counting Pairs in Cribbage Hand
Objective
Count the number of ways cards can be combined to a total of 15, then score 2 points for each pair. Ace worth 1, and J,Q,K are worth 10.
What I have Tried
So my first poke at a solution required 26 different formulas. Basically I checked each possible way to combine cards to see if the total was 15. 1 way to add 5 cards, 5 ways to add 4 cards, 10 ways to add 3 cards, and 10 ways to add 2 cards. I thought I had this licked until I realized I was only looking at combinations, I had not considered the fact that I had to cap the value of cards 11, 12, and 13 to 10. I initially tried an array formula something along the lines of:
MIN(MOD(B1:F1-1,13)+1,10)
But the problem with this is that MIN takes the minimum value of all results not the individual results compared to 10.
I then tried it with an IF function, which worked, but involved the use of CSE formula even wehen being used with SUMPRODUCT which is something I try to avoid when I can
IF(MOD(B1:F1-1,13)+1<11,MOD(B1:F1-1,13)+1,10)
Then I stumble on an answer to a question in code golf which I modified to lead me to this formula, which I kind of like for some strange reason, but its a bit long in repetitive use:
--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2)
My current working formulas are:
5 card check
=(SUMPRODUCT(--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2))=15)*2
4 card checks
=(SUM(AGGREGATE(15,6,--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2),{1,2,3,4}))=15)*2
=(SUM(AGGREGATE(15,6,--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2),{1,2,3,5}))=15)*2
=(SUM(AGGREGATE(15,6,--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2),{1,2,4,5}))=15)*2
=(SUM(AGGREGATE(15,6,--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2),{1,3,4,5}))=15)*2
=(SUM(AGGREGATE(15,6,--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2),{2,3,4,5}))=15)*2
3 card checks
same as 4 card checks using all combinations for 3 cards in the {1,2,3}.
There are 10 different combinations, so 10 different formulas.
The 2 card check was based on the solution by Tom in Counting Pairs in Cribbage Hand and all two cards are checked with a single formula. (yes it is CSE)
2 card check
{=SUM(--(--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2)+TRANSPOSE(--MID("01020304050607080910101010",1+(MOD(B1:F1-1,13)*2),2))=15))}
Question
Can the 3 and 4 card combination sum check be brought into a single formula similar to the 2 card check?
Is there a better way to convert cards 11,12,13 to a value of 10?
Sample Data
| B | C | D | E | F | POINTS
+----+----+----+----+----+
| 1 | 2 | 3 | 17 | 31 | <= 2 (all 5 add to 15)
| 1 | 2 | 3 | 17 | 32 | <= 2 (Last 4 add to 15)
| 11 | 18 | 31 | 44 | 5 | <= 16 ( 4x(J+5), 4X(5+5+5) )
| 6 | 7 | 8 | 9 | 52 | <= 4 (6+9, 7+8)
| 1 | 3 | 7 | 8 | 52 | <= 2 (7+8)
| 2 | 3 | 7 | 9 | 52 | <= 2 (2+3+K)
| 2 | 4 | 6 | 23 | 52 | <= 0 (nothing add to 15)
Excel Version
Excel 2013
For 5:
=(SUMPRODUCT(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10))=15)*2
For 4:
=SUMPRODUCT(--(MMULT(INDEX(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)*ROW($1:$10)^0,ROW($1:$5),{1,2,3,4;1,2,3,5;1,2,4,5;1,3,4,5;2,3,4,5}),ROW($1:$4)^0)=15))*2
For 3
=SUMPRODUCT(--(MMULT(INDEX(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)*ROW($1:$10)^0,ROW($1:$10),{1,2,3;1,2,4;1,2,5;1,3,4;1,3,5;1,4,5;2,3,4;2,3,5;2,4,5;3,4,5}),ROW($1:$3)^0)=15))*2
For 2:
SUMPRODUCT(--((CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10))+(TRANSPOSE(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)))=15))
All together:
=(SUMPRODUCT(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10))=15)*2+
SUMPRODUCT(--(MMULT(INDEX(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)*ROW($1:$10)^0,ROW($1:$5),{1,2,3,4;1,2,3,5;1,2,4,5;1,3,4,5;2,3,4,5}),ROW($1:$4)^0)=15))*2+
SUMPRODUCT(--(MMULT(INDEX(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)*ROW($1:$10)^0,ROW($1:$10),{1,2,3;1,2,4;1,2,5;1,3,4;1,3,5;1,4,5;2,3,4;2,3,5;2,4,5;3,4,5}),ROW($1:$3)^0)=15))*2+
SUMPRODUCT(--((CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10))+(TRANSPOSE(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)))=15))
For older versions we need to "trick" INDEX into accepting the arrays as Row and Column References:
We do that by using N(IF({1},[thearray]))
=(SUMPRODUCT(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10))=15)*2+
SUMPRODUCT(--(MMULT(INDEX(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)*ROW($1:$10)^0,N(IF({1},ROW($1:$5))),N(IF({1},{1,2,3,4;1,2,3,5;1,2,4,5;1,3,4,5;2,3,4,5}))),ROW($1:$4)^0)=15))*2+
SUMPRODUCT(--(MMULT(INDEX(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)*ROW($1:$10)^0,N(IF({1},ROW($1:$10))),N(IF({1},{1,2,3;1,2,4;1,2,5;1,3,4;1,3,5;1,4,5;2,3,4;2,3,5;2,4,5;3,4,5}))),ROW($1:$3)^0)=15))*2+
SUMPRODUCT(--((CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10))+(TRANSPOSE(CHOOSE(MOD(A1:E1-1,13)+1,1,2,3,4,5,6,7,8,9,10,10,10,10)))=15))
This is a CSE That must be confirmed with Ctrl-Shift-Enter instead of Enter when exiting edit mode.
Adding zeros to a string without generating a new variable
I am trying to add zeros to a string variable in such a way that all levels of the variables have same number of digits (assume 3).
clear
input tina bina str4 pine
1 10 "99"
1 11 "99"
2 11 "99"
2 11 "99"
3 12 "."
4 12 "888"
5 14 "88"
6 15 "777"
7 16 "77"
8 17 "0"
8 18 "7"
end
I managed to do this by generating a new variable which stores the number of digits I need to add to each observation in order to reach 3:
generate pi=3-strlen(pine)
replace pine= ("0"*pi) + pine if strlen(pine)<3
I wonder if there is a way to obtain the same result but without generating the variable?
I tried the following but it does not work :
replace pine= ("0"*(`=3-strlen(pine)')) + pine if strlen(pine)<3
Probably I am not so clear about what happens when I evaluate expressions.
Your approach does not work because it evaluates the expression for the first observation only:
. display `= 3 - strlen(pine)'
1
The single quotes are not required:
replace pine = ("0" * (3-strlen(pine) ) ) + pine if strlen(pine) < 3
+--------------------+
| tina bina pine |
|--------------------|
1. | 1 10 099 |
2. | 1 11 099 |
3. | 2 11 099 |
4. | 2 11 099 |
5. | 3 12 00. |
|--------------------|
6. | 4 12 888 |
7. | 5 14 088 |
8. | 6 15 777 |
9. | 7 16 077 |
10. | 8 17 000 |
|--------------------|
11. | 8 18 007 |
+--------------------+
I know there is already an accepted answer, but I wanted to throw out my suggestion. This is maybe a little bit simpler than the other answer and is straightforward to explain. You just want to replace a string variable of real numbers with leading zeros and keep it as a string. You can easily do this by running: replace pine = string(real(pine),"%03.0f") Depending on your goal this is maybe better than the previous answer, because it maintains your missing value as missing and not add zeros to it. Hopefully this helpful.
Excel Summing Sequences?
I'm trying to do sequential summing on a spreadsheet. The first rows are data by date, and I want do sums by the week. but Excel's autopopulate keeps screwing it up and I don't know how to fix that. Date A ---- 1 | 5 2 | 5 3 | 5 4 | 5 5 | 5 6 | 5 7 | 5 8 | 5 9 | 5 10 | 5 11 | 5 12 | 5 13 | 5 14 | 5 so what I want in another area Week Total 1 | =sum(A1:A7) 2 | =sum(A8:A14) 3 | =sum(A15:A21) 4 | continue like this for 52 weeks but what excel keeps giving me with it's auto populating is Week Total 1 | =sum(A1:A7) #The first iteration 2 | =sum(A2:A8) #auto generated 3 | =sum(A3:A9) #auto generated How can I get excel to give me the results I want here? I've been searching on summing for a while and can't seem to even phrase my question right.
=sum(indirect("A"&(row()*7-6)&":A"&(row()*7)))
pasted in row 1 and below should work
at least in sheets, it does. (and excel docs say indirect works)
Spark: count events based on two columns
I have a table with events which are grouped by a uid. All rows have the columns uid, visit_num and event_num.
visit_num is an arbitrary counter that occasionally increases. event_num is the counter of interactions within the visit.
I want to merge these two counters into a single interaction counter that keeps increasing by 1 for each event and continues to increase when then next visit has started.
As I only look at the relative distance between events, it's fine if I don't start the counter at 1.
|uid |visit_num|event_num|interaction_num|
| 1 | 1 | 1 | 1 |
| 1 | 1 | 2 | 2 |
| 1 | 2 | 1 | 3 |
| 1 | 2 | 2 | 4 |
| 2 | 1 | 1 | 500 |
| 2 | 2 | 1 | 501 |
| 2 | 2 | 2 | 502 |
I can achieve this by repartitioning the data and using the monotonically_increasing_id like this:
df.repartition("uid")\
.sort("visit_num", "event_num")\
.withColumn("iid", fn.monotonically_increasing_id())
However the documentation states:
The generated ID is guaranteed to be monotonically increasing and unique, but not consecutive. The current implementation puts the partition ID in the upper 31 bits, and the record number within each partition in the lower 33 bits. The assumption is that the data frame has less than 1 billion partitions, and each partition has less than 8 billion records.
As the id seems to be monotonically increasing by partition this seems fine. However:
I am close to reaching the 1 billion partition/uid threshold.
I don't want to rely on the current implementation not changing.
Is there a way I can start each uid with 1 as the first interaction num?
Edit
After testing this some more, I notice that some of the users don't seem to have consecutive iid values using the approach described above.
Edit 2: Windowing
Unfortunately there are some (rare) cases where more thanone row has the samevisit_numandevent_num`. I've tried using the windowing function as below, but due to this assigning the same rank to two identical columns, this is not really an option.
iid_window = Window.partitionBy("uid").orderBy("visit_num", "event_num")
df_sample_iid=df_sample.withColumn("iid", fn.rank().over(iid_window))
The best solution is the Windowing function with rank, as suggested by Jacek Laskowski.
iid_window = Window.partitionBy("uid").orderBy("visit_num", "event_num")
df_sample_iid=df_sample.withColumn("iid", fn.rank().over(iid_window))
In my specific case some more data cleaning was required but generally, this should work.
Send DNS data: MSB or LSB first?
I'm implementing a DNS(multicast DNS in fact) in c#. I just want to know if I must encode my uint/int/ushort/... with the LSB on the left or the MSB on the left. And more globally how I could know this? One of this is standard? Because I didn't found anything in the IETF description. I found a lot of things(each header field length, position), but I didn't found this. Thank you!
The answer is in RFC 1035 (2.3.2. Data Transmission Order) Here is the link: http://www.ietf.org/rfc/rfc1035.txt And the interesting part 2.3.2. Data Transmission Order The order of transmission of the header and data described in this document is resolved to the octet level. Whenever a diagram shows a group of octets, the order of transmission of those octets is the normal order in which they are read in English. For example, in the following diagram, the octets are transmitted in the order they are numbered. 0 1 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | 1 | 2 | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | 3 | 4 | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | 5 | 6 | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ Whenever an octet represents a numeric quantity, the left most bit in the diagram is the high order or most significant bit. That is, the bit labeled 0 is the most significant bit. For example, the following diagram represents the value 170 (decimal). 0 1 2 3 4 5 6 7 +-+-+-+-+-+-+-+-+ |1 0 1 0 1 0 1 0| +-+-+-+-+-+-+-+-+ Similarly, whenever a multi-octet field represents a numeric quantity the left most bit of the whole field is the most significant bit. When a multi-octet quantity is transmitted the most significant octet is transmitted first.