Find the path has specified length on the 2d matrix - graph

I have a problem about algorithm.
Given 2d matrix such:
2, 1, 2, 5, 5, 0
1, 4, 0, 1, 0, 8
2, 8, 4, 1, 7, 1
5, 6, 4, 9, 7, 9
8, 7, 9, 6, 2, 5
6, 6, 7, 4, 8, 3
Question: use "up", "left", "right", "down" moving to find path has length is 10 (can't revisit a node).
Example:
2, 1, 2, 5, 5, 0
1, 4, 0, [1], 0, 8
2, 8, 4, [1], [7], [1]
5, 6, 4, 9, 7, 9
8, 7, 9, 6, 2, 5
6, 6, 7, 4, 8, 3
More specifically, the algorithm needs to answer the question: exist or not exist such a way

I've solved the problem by dividing into 2 small problems:
subset sum problem
check path exists through a set of nodes

Related

How to properly set rendering Rmarkdown to pdf?

I use Rmarkdown to generate reports and if my line is too long it is usually cut after rendering.
Is there a way to fix it?
I attach a screenshot in order better explain my issue.
You can use the chunk option tidy=TRUE to automatically insert line breaks in the code.
---
output: pdf_document
---
```{r, tidy = TRUE}
c(1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0)
```
The linebreaks are inserted by formatR::tidy_source(). See https://yihui.org/knitr/options/#code-decoration for more details.
chunk_content <- "c(1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0)"
formatR::tidy_source(text = chunk_content, width.cutoff = 30)
#> c(1, 2, 3, 4, 5, 6, 7, 8, 9, 0,
#> 1, 2, 3, 4, 5, 6, 7, 8, 9,
#> 0, 1, 2, 3, 4, 5, 6, 7, 8,
#> 9, 0, 1, 2, 3, 4, 5, 6, 7,
#> 8, 9, 0)

How to convert a stem and leaf plot into a data set in R?

The stem and leaf plot that I need to convert is given below-
24|9
23|
22|1
21|7
20|2, 2, 5, 5, 6, 9, 9, 9
19|0, 0, 0, 0, 0, 1, 1, 2, 4, 4, 5, 8
18|0, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 9, 9, 9
17|1, 1, 1, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 9
16|0, 0, 1, 1, 1, 1, 2, 4, 5, 5, 6, 6, 8, 8, 8, 8
15|0, 1, 1, 1, 1, 1, 1, 5, 5, 5, 5, 6, 6, 6, 7, 7, 8, 9
14|0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 9, 9
13|0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 9
12|1, 1, 1, 2, 2, 2, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9
11|0, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 9, 9
10|0, 2, 3, 3, 3, 4, 4, 5, 7, 7, 8
9|0, 0, 9
8|6
Here's maybe one way. If your data looks like this
stem <- "24|9
23|
22|1
21|7
20|2, 2, 5, 5, 6, 9, 9, 9
19|0, 0, 0, 0, 0, 1, 1, 2, 4, 4, 5, 8
18|0, 1, 1, 2, 2, 2, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 9, 9, 9
17|1, 1, 1, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 9
16|0, 0, 1, 1, 1, 1, 2, 4, 5, 5, 6, 6, 8, 8, 8, 8
15|0, 1, 1, 1, 1, 1, 1, 5, 5, 5, 5, 6, 6, 6, 7, 7, 8, 9
14|0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 7, 7, 8, 9, 9
13|0, 0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 9
12|1, 1, 1, 2, 2, 2, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9
11|0, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 9, 9
10|0, 2, 3, 3, 3, 4, 4, 5, 7, 7, 8
9|0, 0, 9
8|6"
Then we can split up the rows and for each row we split by the pipe. Then we split the right side by commas and join each of those values to the value to the left of the pipe.
rows <- strsplit(stem,"\n")[[1]]
values <- unlist(lapply(strsplit(rows,"\\|"), function(x) {
end_digits <- strsplit(x[2], ", ")[[1]]
if (!all(is.na(end_digits))) {
paste0(x[1], end_digits)
} else {
NULL
}
}
))
This will return character values, but you could convert to numeric with
as.numeric(values)
Here is a different approach. Using #MrFlick's stem and rows objects:
rows <- strsplit(stem,"\n")[[1]]
rows.lst <- strsplit(rows,"\\|")
tens <- as.numeric(sapply(rows.lst, "[", 1)) * 10
ones <- sapply(strsplit(sapply(rows.lst, "[", 2), ","), as.numeric)
vals <- unlist(mapply("+", tens, ones))
vals <- vals[!is.na(vals)]

Is it possible to limit forecasts made by bsts to positive values only?

I am learning to use various forecasting packages available in R, and came across bsts(). The data I deal with is a time series of demands.
data=c(27, 2, 7, 7, 9, 4, 3, 3, 3, 9, 6, 2, 6, 2, 3, 8, 6, 1, 3, 8, 4, 5, 8, 5, 4, 4, 6, 1, 6, 5, 1, 3, 0, 2, 6, 7, 1, 2, 6, 2, 8, 6, 1, 1, 3, 2, 1, 3, 1, 6, 3, 4, 3, 7, 3, 4, 1, 7, 5, 6, 3, 4, 3, 9, 2, 1, 7, 2, 2, 9, 4, 5, 3, 4, 2, 4, 4, 8, 6, 3, 9, 2, 9, 4, 1, 3, 8, 1, 7, 7, 6, 0, 1, 4, 8, 9, 2, 5)
ts.main=ts(data, start=c(1910,1), frequency=12)
ss <- AddLocalLinearTrend(list(), y=ts.main)
ss <- AddSeasonal(ss, y=as.numeric(ts.temp), nseasons=12)
model <- bsts(as.numeric(ts.temp),
state.specification = ss,
niter = 1000)
pred <- predict(model, horizon = 12)
Is there way I can restrict pred$mean from becoming negative?
Since your data are a time series of counts, you need to take that into account rather than assume Gaussian errors; for some discussion on this and elaboration of some approaches, see for example Brandt et al 2000 and Brandt and Williams 2001. Luckily, the bsts package has a built-in functionality for this, the family option (see pages 24 to 26 of the documentation).
So, you can just do this
model <- bsts(as.numeric(ts.main),
state.specification = ss,
family = 'poisson',
niter = 1000)
so that the bsts() function correctly considers the data as counts, which will solve your issue, since the draws from the posterior predictive distribution will then be non-negative by definition.

Creating a histogram with appropriate counts and labels in R

I have a dataset (dat), which I am hard-coding in here:
dat = c(5, 9, 5, 6, 5, 6, 8, 4, 6, 4, 6, 6, 4, 6, 4, 6, 5, 5, 6, 5, 6, 7, 4, 5, 4, 4, 6, 4, 4, 5, 7, 6, 3, 5, 5, 5, 5, 4, 6, 3, 6, 5, 4, 6, 5, 8, 4, 8, 5, 5, 4, 4, 6, 6, 4, 6, 4, 7, 4, 1, 4, 6, 3, 6, 3, 4, 6, 6, 3, 6, 6, 2, 5, 5, 4, 7, 6)
table(dat)
By doing the table function above on the data, I see that there should be a count of 1 for values of 1, and count of 1 for values of 2. However, when I plot the data using hist, I get a count of 2.
hist(dat, col="lightgreen", labels = TRUE, xlim=c(0,10), ylim=c(0,27))
This is the first problem. The other problem is that I am trying to plot the x label value for the corresponding bin (where there should be 11 bins, labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10). Even though I have no 0 values or 10 values, I would like to illustrate that they had a count of 0, and have their bins - like the rest- labeled. How can I accomplish that?
Thanks.
am = hist(dat, col="lightgreen", labels = TRUE,
breaks=seq(min(dat)-2,max(dat)),
axes=F)
axis(2)
axis(1,at=am$mids,seq(min(dat)-1,max(dat)))
Did you mean like this:
hist(dat, col="lightgreen", labels = TRUE,
xlim=c(0,10), ylim=c(0,27), breaks = 0:10, at=0:10)

Error reading dataset in R

I have problem in reading a dataset
My code :
require(igraph)
g <- graph(c(0, 1, 1, 2, 2, 0, 1, 3, 3, 4,
4, 5, 5, 3, 4, 6, 6, 7, 7, 8,
8, 6, 9, 10, 10, 11, 11, 9))
Error :
Error in graph(c(0, 1, 1, 2, 2, 0, 1, 3, 3, 4, 4, 5, 5, 3, 4, 6, 6, 7, :
At structure_generators.c:84 : Invalid (negative) vertex id, Invalid vertex id
The problem seems to be vertex of name 0
yourgraph <- c(0, 1, 1, 2, 2, 0, 1, 3, 3, 4,
4, 5, 5, 3, 4, 6, 6, 7, 7, 8,
8, 6, 9, 10, 10, 11, 11, 9)
g <- graph(yourgraph + 1)

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