I'll preface this by saying I Don't want the answer, just a nudge in the right direction.
The conversion table for the encryption is as follows:
Input --> Output
0 --> 15
1 --> 14
2 --> 13
3 --> 12
4 --> 0
5 --> 1
6 --> 2
7 --> 3
8 --> 7
9 --> 6
10 --> 5
11 --> 4
12 --> 8
13 --> 9
14 --> 10
15 --> 11
Here's what I've done so far.
Essentially there are 2 patterns and I want to know how to implement them both in a single circuit. I'm completely stumped.
B2 is always negated. If B2 is not set, all other bits are negated.
Seems like an easy enough condition to encode in a circuit.
B2 is acting as a selector, so if you've already developed two circuits, you can select which one to use based on B2: (NOT B2 AND FIRST) OR (B2 AND SECOND).
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I am creating a schedule for a sports league with several dozen teams. I already have all of the games in a set order and now I just need to assign one team to be the "home" team and one to be "away" for each game.
The problem has two constraints:
Each pair of teams must play an equal number of home and away
games against each other. For example, if team A and team B play 4
games, then 2 must be hosted by A and 2 by B. Assume that each pair
of teams plays an even number of games against each other.
No team should have more than three consecutive home games or three
consecutive away games at any point in the schedule.
I have been trying to use brute force in R to solve this problem but I can't get any of my code blocks to solve the issue in a timely fashion. Does anyone have any advice on how to deal with either (or both) of the above constraints algorithmically?
You need to do more research on simple scheduling.
There are a lot of references on line for these things.
Here are the basics for your application. Let's assume a league of 6 teams; the process is the same for any number.
Match 1: Simply write down the team numbers in order, in pairs, in a ring. Flatten he ring into two lines. Matches are upper (home) and lower(away).
1 2 3
6 5 4
Matches 2-5: Team 1 stays in place; the others rotate around the ring.
1 6 2
5 4 3
1 5 6
4 3 2
1 4 5
3 2 6
1 3 4
2 6 5
That's one full cycle. To balance the home-away schedule, simply invert the fixtures every other match:
1 2 3 5 4 3 1 5 6 3 2 6 1 3 4
6 5 4 1 6 2 4 3 2 1 4 5 2 6 5
There's your first full round. Simply replicate this, again switching home-away fixtures in alternate rounds. Thus, the second round would be:
6 5 4 1 6 2 4 3 2 1 4 5 2 6 5
1 2 3 5 4 3 1 5 6 3 2 6 1 3 4
Repeat this pair of rounds as many times as needed to get the length of schedule you need.
If you have an odd quantity of teams, simply declare one of the numbers to be the "bye" in the schedule. I find it easiest to follow if I use the non-rotating team -- team 1 in this example.
Note that this home-switching process guarantees that no team has three consecutive matches either home or away: they get two in a row when rounding the end of the row. However, even the two-in-a-row doesn't suffer at the end of the round: both of those teams break the streak in the first match of the next round.
Unfortunately, for an arbitrary existing schedule, you are stuck with a brute-force search with backtracking. You can employ some limits and heuristics, such as balancing partial home-away fixtures as the first option at each juncture. Still, the better approach is to make your original schedule correct by design.
There's also a slight problem that you cannot guarantee that your existing schedule will fulfill the given requirements. For instance, given the 8-team fixtures in this order:
1 2 3 4
5 6 7 8
1 2 5 6
3 4 7 8
1 3 5 7
2 4 6 8
It is not possible to avoid having at least two teams playing three consecutive home or away matches.
I am trying to insert a textbox into a chartsheet using xlsxwriter in python 3.4 to give more information about the chart. Is there a way to insert a textbox into a chartsheet? I could insert it into a worksheet and not a chartsheet.
Also, is there a way to edit the legend using xlsxwriter? I have a row of 30 values and I want to plot in a batch of 10
batch1 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
I dont want "batch1" to be shown in the legend 3 times. I want to customize it as iteration1,iteration2,iteration3. Is there a way?
Inserting a textbox into a chart, or chartsheet, isn't supported.
In relation to the second question, you can remove items from the legend in XlsxWriter using the delete_series feature of set_legend():
# Delete/hide series index 0 and 2 from the legend.
chart.set_legend({'delete_series': [0, 2]})
See the docs on set_legend()
Say I have data that look like this:
level start end
1 1 133.631 825.141
2 2 133.631 155.953
3 3 146.844 155.953
4 2 293.754 302.196
5 3 293.754 302.196
6 4 293.754 301.428
7 2 326.253 343.436
8 3 326.253 343.436
9 4 333.827 343.436
10 2 578.066 611.766
11 3 578.066 611.766
12 4 578.066 587.876
13 4 598.052 611.766
14 2 811.228 825.141
15 3 811.228 825.141
or this:
level start end
1 1 3.60353 1112.62000
2 2 3.60353 20.35330
3 3 3.60353 8.77526
4 2 72.03720 143.60700
5 3 73.50530 101.13200
6 4 73.50530 81.64660
7 4 92.19030 101.13200
8 3 121.28500 143.60700
9 4 121.28500 128.25900
10 2 167.19700 185.04800
11 3 167.19700 183.44600
12 4 167.19700 182.84600
13 2 398.12300 418.64300
14 3 398.12300 418.64300
15 2 445.83600 454.54500
16 2 776.59400 798.34800
17 3 776.59400 796.64700
18 4 776.59400 795.91300
19 2 906.68800 915.89700
20 3 906.68800 915.89700
21 2 1099.44000 1112.62000
22 3 1099.44000 1112.62000
23 4 1100.14000 1112.62000
They produce the following graphs:
As you can see there are several time intervals at different levels. The level-1 interval always spans the entire duration of the time of interest. Levels 2+ have time intervals that are shorter.
What I would like to do is select the maximum number of non-overlapping time intervals covering each period that contain the maximum number of total time within them. I have marked in pink which ones those would be.
For small dataframes it is possible to brute force this, but obviously there should be some more logical way of doing this. I'm interested in hearing some ideas about what I should try.
EDIT:
I think one thing that could help here is the column 'level'. The results come from Kleinberg's burst detection algorithm (package 'bursts'). You will note that the levels are hierarchically organized. Levels of the same number cannot overlap. However levels successively increasing e.g. 2,3,4 in successive rows can overlap.
In essence, I think the problem could be shortened to this. Take the levels produced, but remove level 1. This would be the vector for the 2nd example:
2 3 2 3 4 4 3 4 2 3 4 2 3 2 2 3 4 2 3 2 3 4
Then, look at the 2s... if there are fewer than or only one '3' then that 2 is the longest interval. But if there are two or more 3's between successive 2's, then those 3s should be counted. Do this iteratively for each level. I think that should work...?
e.g.
vec<-df$level %>% as.vector() %>% .[-1]
vec
#[1] 2 3 2 3 4 4 3 4 2 3 4 2 3 2 2 3 4 2 3 2 3 4
max(vec) #4
vec3<-vec #need to find two or more 4's between 3s
vec3[vec3==3]<-NA
names(vec3)<-cumsum(is.na(vec3))
0 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 8 8
2 NA 2 NA 4 4 NA 4 2 NA 4 2 NA 2 2 NA 4 2 NA 2 NA 4
vec3.res<-which(table(vec3,names(vec3))["4",]>1)
which(names(vec3)==names(vec3.res) & vec3==4) #5 6
The above identifies rows 5 and 6 (which equate to rows 6 and 7 in original df) as having two fours that lie between 3's. Perhaps something using this sort of approach might work?
OK here is a stab using your second data set to test. This might not be correct in all cases!!
library(data.table)
dat <- fread("data.csv")
dat[,use:="maybe"]
make.pass <- function(dat,low,high,the.level,use) {
check <- dat[(use!="no" & level > the.level)]
check[,contained.by.above:=(low<=start & end<=high)]
check[,consecutive.contained.by.above:=
(contained.by.above &
!is.na(shift(contained.by.above,1)) &
shift(contained.by.above,1)),by=level]
if(!any(check[,consecutive.contained.by.above])) {
#Cause a side effect where we've learned we don't care:
dat[check[(contained.by.above),rownum],use:="no"]
print(check)
return("yes")
} else {
return("no")
}
}
dat[,rownum:=.I]
dat[level==1,use:=make.pass(dat,start,end,level,use),by=rownum]
dat
dat[use=="maybe" & level==2,use:=make.pass(dat,start,end,level,use),by=rownum]
dat
dat[use=="maybe" & level==3,use:=make.pass(dat,start,end,level,use),by=rownum]
dat
#Finally correct for last level
dat[use=="maybe" & level==4,use:="yes"]
I wrote these last steps out so you can trace in your own interactive session to see what's happening (see the print to get an idea) but you can remove the print and also condense the last steps into something like lapply(1:dat[,max(level)-1], function(the.level) dat[use=="maybe" & level==the.level,use:=make.pass......]) In response to your comment if there are an arbitrary number of levels you will definitely want to use this formalism, and follow it with a final call to dat[use=="maybe" & level==max(level),use:="yes"].
Output:
> dat
level start end use rownum
1: 1 3.60353 1112.62000 no 1
2: 2 3.60353 20.35330 yes 2
3: 3 3.60353 8.77526 no 3
4: 2 72.03720 143.60700 no 4
5: 3 73.50530 101.13200 no 5
6: 4 73.50530 81.64660 yes 6
7: 4 92.19030 101.13200 yes 7
8: 3 121.28500 143.60700 yes 8
9: 4 121.28500 128.25900 no 9
10: 2 167.19700 185.04800 yes 10
11: 3 167.19700 183.44600 no 11
12: 4 167.19700 182.84600 no 12
13: 2 398.12300 418.64300 yes 13
14: 3 398.12300 418.64300 no 14
15: 2 445.83600 454.54500 yes 15
16: 2 776.59400 798.34800 yes 16
17: 3 776.59400 796.64700 no 17
18: 4 776.59400 795.91300 no 18
19: 2 906.68800 915.89700 yes 19
20: 3 906.68800 915.89700 no 20
21: 2 1099.44000 1112.62000 yes 21
22: 3 1099.44000 1112.62000 no 22
23: 4 1100.14000 1112.62000 no 23
level start end use rownum
On the off chance this is correct, the algorithm can roughly be described as follows:
Mark all the intervals as possible.
Start with a given level. Pick a particular interval (by=rownum) say called X. With X in mind, subset a copy of the data to all higher-level intervals.
Mark any of these that are contained in X as "contained in X".
If consecutive intervals at the same level are contained in X, X is no good b/c it wastes intervals. In this case label X's "use" variable as "no" so we'll never think about X again. [Note: if it's possible that non-consecutive intervals are contained in X, or that containing multiple intervals across levels could ruin X's viability, then this logic might need to be changed to count contained intervals instead of finding consecutive ones. I didn't think about this at all, but it's just occurring to me now, so use at your own risk.]
On the other hand, if X passed the test, then we've already established it's good. Mark it as a "yes." But importantly, we also have to mark any single interval contained in X as "no," or else when we iterate the step it will forget that it was contained inside a good interval and mark itself as "yes" as well. This is the side effect step.
Now, iterate, ignoring any results that we've already determined.
Finally any "maybe"s leftover at the highest level are automatically in.
Let me know what you think of this--this is a rough draft and some aspects might not be correct.
This question already has answers here:
How to sum a variable by group
(18 answers)
Closed 7 years ago.
I don't know how to word the title exactly, so I will just do my best to explain below... Sorry in advance for the .csv format.
I have the following example dataset:
print(data)
ID Tag Flowers
1 1 6871 1
2 2 6750 1
3 3 6859 1
4 4 6767 1
5 5 6747 1
6 6 6261 1
7 7 6750 1
8 8 6767 1
9 9 6812 1
10 10 6746 1
11 11 6496 4
12 12 6497 1
13 13 6495 4
14 14 6481 1
15 15 6485 1
Notice that in Lines 2 and 7, the tag 6750 appears twice. I observed one flower on plant number 6750 on two separate days, equaling two flowers in its lifetime. Basically, I want to add every flower that occurs for tag 6750, tag 6767, etc throughout ~100 rows. Each tag appears more than once, usually around 4 or 5 times.
I feel like I need to apply the unlist function here, but I'm a little bit lost as to how I should do so.
Without any extra packages, you can use function aggregate():
res<-aggregate(data$Flowers, list(data$Tag), sum)
This calculates a sum of the values in Flowers column for every value in the Tag column.
I have been searching for long but unable to find a solution for this.
My question is "Suppose you have n street lights(cannot be moved) and if you get any m from them then it should have atleast k working.Now in how many ways can this be done"
This seems to be a combination problem, but the problem here is "m" must be sequential.
Eg:
1 2 3 4 5 6 7 (Street lamps)
Let m=3
Then the valid sets are,
1 2 32 3 43 4 54 5 65 6 7Whereas,1 2 4 and so are invalid selections.
So every set must have atleast 2 working lights. I have figured how to find the minimum lamps required to satisfy the condition but how can I find the number of ways in it can be done ?
There should certainly some formula to do this but I am unable to find it.. :(
Should always be (n-m)+1.
E.g., 10 lights (n = 10), 5 in set (m = 5):
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
6 7 8 9 10
Gives (10-5)+1 = 6 sets.
The answer should always be m choose k for all values of n where n > m > k. I'll try to explain why;
Given, for example, the values m = 10, n = 4, k = 2, you can start by generating all possible permutations of 1s and 0s for sets of 4 lights, with exactly 2 lights on;
1100
0110
0011
1001
0101
1010
As you can see, there are 6 permutations, because 4 choose 2 = 6. You can choose any of these 6 permutations to be the first 4 lights. You then continue the sequence until you get n (in this case 10) lights, ensuring that you only ever add a zero if you must in order to keep the condition true of having 2 lights on for every 4. What you will find is that the sequence simply repeats; for example:
1100 -> next can be 1, so 11001
Next can still be 1 and meet the condition, so 110011.
The next must now be a zero, giving 1100110, and then again -> 11001100. This simply continues until the length is n : 1100110011. Given that the starting four can only be one of the above set, you will only get 6 different permutations.
Now, since the sequence will repeat exactly the same for any value of n, it means that the answer will always be m choose k.
For your example in your comment of 6,3,2, I can only find the following permutations:
011011
110110
101101
Which works, because 3 choose 2 = 3. If you can find more, then I guess I'm wrong and I've probably misunderstood again :D but from my understanding of this problem, I'm certain that the answer will always be m choose k.