text:
Fort Worden is humming like a well-oiled Pfaff!
It has been a few years since I've been able to visit this annual gathering of kitemakers and designers. But with the change in dates applied this year, my conflicts were eliminated. Susie and I decided to take the drive up through the jagged Olympic Mountains and along the scenic fjords of Puget Sound. We arrived at the far northwest tip of the United States in time to see the lights of Victoria Canada twinkling across the water.
Back in 1988, the premier issue of American Kite Magazine called Fort Worden a gathering of "artists, engineers, and mystics". The magazine is gone now, but the retreat continues and has improved with age. Now in it's 18th year, it has emerged as the largest of the kitemaking conferences in America.
Movie buffs and romantics will recognize the Fort as the location where Officer and a Gentleman was filmed several years ago. The classic turn of the (last) century wooden buildings resonate with history and are littered with military photos of former residents. It now operates as a Washington State Park and Conference Center.
Classes are run Friday through Sunday. And such classes!!
Robert Trepaniér of Montréal, Quebec teaching his signature class in making life-size "self portrait" kites.
Wolfgang Grimsel of Oldenburg, Germany with plans for a "kite in motion" with rotating parts.
Martin Lester of Bristol, England, famous for his representational soft kites.
Sharon Musto of Winnipeg, Manitoba showing the process to draw and construct a small, elegantly appliquéd della Porta.
José Sainz of San Diego, California teaching his classic techniques for sewing and cutting appliqué.
Kevin Shannon of Carlisle, Pennsylvania making a variation of the ancient Chinese kiteline climber.
Scott Skinner of Monument, Colorado creating exquisitely subtle paper kites by combining paper pieces in patterns or images to form the skin.
Barry Poulter of West Sussex, England discussing concepts and techniques in surface design and shape enhancement.
Bob & Charmayne Umbowers of Gig Harbor, Washington making an efficient, small-but-powerful double parasled.
The Yaripa Group of Medellin, Columbia teaching a class in making Columbia's traditional, hexagonal yaripa and paper or cloth kite.
All together there were more than 40 instructors teaching all aspects of kite design, construction, and decoration.
Most attendees stay in the old dormitories, but Susie and I were invited to share one of the Victorian duplexes built for non-commissioned officers. Susie asked the difference between a commissioned and a non-commissioned officer. Clearly the distinction was the size of the house and possibly the date when plumbing was moved indoors.
Our kitchen featured instruction on how to use the telephone. Unfortunately, I couldn't find a phone or a phone plug in the house or anywhere else. Most people might find that relaxing. Not me! Even my cell phone wouldn't work in these isolated woods. I was finally able to plug in and dial out for email and phone messages in the back of the camp cafeteria...
Funds for Conference expenses are paid by registrants. Costs run from $85 to $200 depending on whether you want optional meals or housing.
Costs for importing instructors are raised by raffling off kites, trinkets, and memorabilia in one of the largest and most intense ticket lotteries in the kite world. Over $7,000 was raised, one ticket at a time.
I only have one complaint about the weekend. Cookies.
There were cookies everywhere, along with candy, brownies, and other assorted treats. Meals were plentiful as well, but it was the snacks that pervaded the classrooms and presented temptation at every turn. I'm not really complaining. The hospitality was overwhelming. But we're planning a trip to Australia in September and I'm worried about fitting into my thong when I reach Bondi Beach.
We pulled out of Fort Worden on Sunday afternoon with a cheesecake raffle prize in the back seat. Our heads were brimming with new ideas and the joy of spending time with good friends. Susie drove, and I wondered how to write an Update that might make a sewing weekend sound as interesting as it really was.
Next week I'm off to the Smithsonian in Washington DC and then on to France a few days later. See you there!!
Back to the Update Index
Back to Gomberg Kite Productions International


Our season came to an end at Walsh Park on Saturday as Kilmacthomas proved the more efficient team in the junior football county final. In truth seven points was a little harsh on us as we had plenty of possession, played with a lot of heart and defended manfully but Kilmac were able to find scores with less effort and that made the difference in the end. It's been a very successful year on the junior football front and credit must go to the management team of Stephen Curley and Tomas Mc Carthy as well as the work done in previous years by Terence Mc Sweeney.
Kilmacthomas were the first to score with a point from play and it was clear that they were going to be a step up in class from our previous opponents. Our captain Kieran Geary opened the scoring for us on nine minutes with a fine point but two minutes later we were dealt a blow when Kilmac were awarded a penalty that most would view as harsh. It was dispatched to the net but Tallow's response was strong with three fine points from play from Kieran Geary, Kevin Curley and Stuart Barry bringing us back level after seventeen minutes. Kilmac responded with a point of their own to retake the lead before we conceded another goal that pushed Kilmacthomas four points up. Michael Curley got a free a minute later that he made no mistake with to reduce the gap but another flowing forward move by Kilmac put them back into a four point advantage. We were not for giving up though and a minute later an accurate Kevin Curley pass found Michael Curley with only the goalkeeper to beat and he made no mistake with his left foot. There was still time for Kilmac to respond with a point before the interval and this left them two points to the good at halftime 2-4 to 1-5.
The second half saw play push up towards the Kilmacthomas half of the field as we gained a lot of possession but this had the effect of leaving our defence stretched and Kilmac were clever in playing the possession they got into the corners and allowing their forwards to run into space. Ten minutes into the second half we had conceded four points without reply as we failed to turn our possession and territory into scores and Kilmac, to their credit, had not given Michael Curley any chances from placed balls. JP and our backs were preforming heroics in dealing with the space in front of them as we pushed forward in search of scores but we couldn't find space up front and were often unlucky as the final ball didn't go to hand. Kilmacthomas stretched their lead seven minutes from time to seven points before Thomas Ryan gave us a glimmer of hope with a mazy run and a fierce shot that just went over the bar. That was to be our final score though and Kilmac added one last point two minutes from time to see out their win.
Best wishes to corner back Eric Hickey who missed the county final through injury.


Karen L Messick Photography: Spring Tulips and the photographers old addage......"Shoot the Vertical right after you shoot the Horizontal."
Spring Tulips and the photographers old addage......"Shoot the Vertical right after you shoot the Horizontal."
This weekend I helped on a volunteer basis, The Baltimore Camera Club with their annual Camera School. I teach composition in the classroom on Saturday and then assist with the field lessons on Sunday. Its a busy weekend but fun! Today we were in Sherwood Gardens and making images of the tulips. I got a chance to do a little personal shooting with the D4, and my 200 mm macro lens fitted with a 52mm extension tube for the first time. The 52 mm extension tube did not seem to fit as well to the body mount but I made the connection anyway and it worked fine. The one really great thing is the ISO capability of the D4. The increased ISO capability for macro is so sweet! At times I was using ISO 1000, that I never before was able to use doing macro work and still get a useable image. I was able to achieve shutter speeds at 1/125 sec with the extension tube connected, a real plus for me, especially since we had a little breeze today.
In my composition class I drive home the idea of making lots of shots when you have great subjects..and always make vertical and horizontal images of the same subject, then decide when you get home which one you might like best! I also cover color theory and this purple pink tulip with a background of out of focus purple azalea shrubs really resonated with the idea of working with analogous colors....but when you pop in the green stem and find the one tulip with the yellow it adds a whole color spectrum including color opposites of green and magenta...and the triad color of yellow!
I always work in horizontal and vertical orientations when making images.
Stunning images! Thanks for the inspiration.
Thank you Sue! Thats really nice of you to say! I appreciate you checking out my blog!
Watermen's Harbor, Kent Island, Md.
New York City IS for Lovers!
New York City Streets~About the Boys!
Old Boats...Kent Island Harbor..Heralding Spring!


CHENNAI: Thiruvottiyur, Manali, Sholinganallur and five other areas added to the city corporation in 2011 can hope for some respite from waterlogging around five years from now, with the civic body preparing to execute a massive Rs 4,034-crore project for a storm water drain network in these neighbourhoods.
The integrated network, covering more than 250 sqkm, will drain rainwater into the city rivers that will carry it to the sea. The corporation, however, has not revealed if the drains will be connected to rainwater harvesting systems to replenish aquifers in the city where the groundwater levels have plummeted.
These areas will also get an underground sewerage network at an estimated Rs 400 crore, the government announced in the assembly on Friday.
"We will start work on the integrated storm water drain system by the end of this calendar year or at the end of this fiscal. It will take five years to complete," a senior corporation official said. Over the period of construction, the cost is likely to escalate to 5,313 crore, according to estimates in a detailed project report, making it one of the corporation's biggest projects.
Residents of the areas expected to benefit from the drain network welcomed the announcement. "This promises great relief to several areas where life comes to a standstill whenever it rains," said Ravindran Narayanan, of Ambattur. "Since this is not a low-lying area, water always flows from here and collects in Mogappair East."
The public works department in 2009 started work on a storm water drain project to prevent flooding with funding of 1,448 crore from the Jawaharlal Nehru National Urban Renewal Mission ( JNNURM). The work is still in progress.
"The project will augment the existing storm water drain network by 1,069.4km," the official said. The project area is divided into four drainage basins, Kosasthalaiyar (originating in Vellore), Cooum, Adyar and Kovalam.


The good news is once you've said "I do" to getting married, you and your spouse-to-be can explore ways to save when it comes to your home insurance and auto insurance, credit cards, and personal finances. There are also new financial products to consider, including getting your first mortgage and obtaining life insurance. Thankfully, there are plenty of ways to save here too.
Related Read: Is Mortgage Life Insurance Or Term Life Insurance The Best Choice For You?
Knowing what to look for can help you save hundreds of dollars a year off your insurance, mortgage and credit card products. The following infographic breaks down areas where you could be saving money and spending smarter.
After all, isn't it time you say, "I do" to savings as well?


text:
using System;
using System.Collections.Generic;

using ZXing.Common;
using ZXing.Core;
using ZXing.PDF417.Internal;
using ZXing.Writer;

namespace ZXing.PDF417
{
   /// <summary>
   /// <author>Jacob Haynes</author>
   /// <author>qwandor@google.com (Andrew Walbran)</author>
   /// </summary>
   public sealed class PDF417Writer : ZXingWriter
   {
      /// <summary>
      /// default white space (margin) around the code
      /// </summary>
      private const int WHITE_SPACE = 30;

      /// <summary>
      /// </summary>
      /// <param name="contents">The contents to encode in the barcode</param>
      /// <param name="format">The barcode format to generate</param>
      /// <param name="width">The preferred width in pixels</param>
      /// <param name="height">The preferred height in pixels</param>
      /// <param name="hints">Additional parameters to supply to the encoder</param>
      /// <returns>
      /// The generated barcode as a Matrix of unsigned bytes (0 == black, 255 == white)
      /// </returns>
      public BitMatrix encode(String contents,
                              BarcodeFormat format,
                              int width,
                              int height,
                              IDictionary<EncodeHintType, object> hints)
      {
         if (format != BarcodeFormat.PDF_417)
         {
            throw new ArgumentException("Can only encode PDF_417, but got " + format);
         }

         var encoder = new Internal.PDF417();
         var margin = WHITE_SPACE;
         var errorCorrectionLevel = 2;

         if (hints != null)
         {
            if (hints.ContainsKey(EncodeHintType.PDF417_COMPACT))
            {
               encoder.setCompact((Boolean) hints[EncodeHintType.PDF417_COMPACT]);
            }
            if (hints.ContainsKey(EncodeHintType.PDF417_COMPACTION))
            {
               encoder.setCompaction((Compaction) hints[EncodeHintType.PDF417_COMPACTION]);
            }
            if (hints.ContainsKey(EncodeHintType.PDF417_DIMENSIONS))
            {
               var dimensions = (Dimensions) hints[EncodeHintType.PDF417_DIMENSIONS];
               encoder.setDimensions(dimensions.MaxCols,
                                     dimensions.MinCols,
                                     dimensions.MaxRows,
                                     dimensions.MinRows);
            }
            if (hints.ContainsKey(EncodeHintType.MARGIN))
            {
               margin = (int)(hints[EncodeHintType.MARGIN]);
            }
            if (hints.ContainsKey(EncodeHintType.ERROR_CORRECTION))
            {
               var value = hints[EncodeHintType.ERROR_CORRECTION];
               if (value is PDF417ErrorCorrectionLevel ||
                   value is int)
               {
                  errorCorrectionLevel = (int)value;
               }
            }
            if (hints.ContainsKey(EncodeHintType.CHARACTER_SET))
            {
#if !SILVERLIGHT || WINDOWS_PHONE
               var encoding = (String)hints[EncodeHintType.CHARACTER_SET];
               if (encoding != null)
               {
                  encoder.setEncoding(encoding);
               }
#else
               // Silverlight supports only UTF-8 and UTF-16 out-of-the-box
               encoder.setEncoding("UTF-8");
#endif
            }
            if (hints.ContainsKey(EncodeHintType.DISABLE_ECI))
            {
               encoder.setDisableEci((bool)hints[EncodeHintType.DISABLE_ECI]);
            }
         }

         return bitMatrixFromEncoder(encoder, contents, width, height, margin, errorCorrectionLevel);
      }

      /// <summary>
      /// Encode a barcode using the default settings.
      /// </summary>
      /// <param name="contents">The contents to encode in the barcode</param>
      /// <param name="format">The barcode format to generate</param>
      /// <param name="width">The preferred width in pixels</param>
      /// <param name="height">The preferred height in pixels</param>
      /// <returns>
      /// The generated barcode as a Matrix of unsigned bytes (0 == black, 255 == white)
      /// </returns>
      public BitMatrix encode(String contents,
                              BarcodeFormat format,
                              int width,
                              int height)
      {
         return encode(contents, format, width, height, null);
      }

      /// <summary>
      /// Takes encoder, accounts for width/height, and retrieves bit matrix
      /// </summary>
      private static BitMatrix bitMatrixFromEncoder(Internal.PDF417 encoder,
                                                    String contents,
                                                    int width,
                                                    int height,
                                                    int margin,
                                                    int errorCorrectionLevel)
      {
         encoder.generateBarcodeLogic(contents, errorCorrectionLevel);

         const int lineThickness = 2;
         const int aspectRatio = 4;
         sbyte[][] originalScale = encoder.BarcodeMatrix.getScaledMatrix(lineThickness, aspectRatio*lineThickness);
         bool rotated = false;
         if ((height > width) ^ (originalScale[0].Length < originalScale.Length))
         {
            originalScale = rotateArray(originalScale);
            rotated = true;
         }

         int scaleX = width/originalScale[0].Length;
         int scaleY = height/originalScale.Length;

         int scale;
         if (scaleX < scaleY)
         {
            scale = scaleX;
         }
         else
         {
            scale = scaleY;
         }

         if (scale > 1)
         {
            sbyte[][] scaledMatrix =
               encoder.BarcodeMatrix.getScaledMatrix(scale*lineThickness, scale*aspectRatio*lineThickness);
            if (rotated)
            {
               scaledMatrix = rotateArray(scaledMatrix);
            }
            return bitMatrixFrombitArray(scaledMatrix, margin);
         }
         return bitMatrixFrombitArray(originalScale, margin);
      }

      /// <summary>
      /// This takes an array holding the values of the PDF 417
      /// </summary>
      /// <param name="input">a byte array of information with 0 is black, and 1 is white</param>
      /// <param name="margin">border around the barcode</param>
      /// <returns>BitMatrix of the input</returns>
      private static BitMatrix bitMatrixFrombitArray(sbyte[][] input, int margin)
      {
         // Creates the bitmatrix with extra space for whitespace
         var output = new BitMatrix(input[0].Length + 2 * margin, input.Length + 2 * margin);
         var yOutput = output.Height - margin - 1;
         for (int y = 0; y < input.Length; y++)
         {
            var currentInput = input[y];
            var currentInputLength = currentInput.Length;
            for (int x = 0; x < currentInputLength; x++)
            {
               // Zero is white in the bytematrix
               if (currentInput[x] == 1)
               {
                  output[x + margin, yOutput] = true;
               }
            }
            yOutput--;
         }
         return output;
      }

      /// <summary>
      /// Takes and rotates the it 90 degrees
      /// </summary>
      private static sbyte[][] rotateArray(sbyte[][] bitarray)
      {
         sbyte[][] temp = new sbyte[bitarray[0].Length][];
         for (int idx = 0; idx < bitarray[0].Length; idx++)
            temp[idx] = new sbyte[bitarray.Length];
         for (int ii = 0; ii < bitarray.Length; ii++)
         {
            // This makes the direction consistent on screen when rotating the
            // screen;
            int inverseii = bitarray.Length - ii - 1;
            for (int jj = 0; jj < bitarray[0].Length; jj++)
            {
               temp[jj][inverseii] = bitarray[ii][jj];
            }
         }
         return temp;
      }
   }
}

redpajama_set_name: RedPajamaGithub


text:
\section{Introduction}
A \emph{perfect matching} of a graph $G$ is a subset of edges of $G$ such that every vertex of the graph is incident with exactly one edge of the perfect matching.
In this paper, we study perfect matchings of highly connected regular graphs.  It is well-known that if $G$ is a $d$-regular $(d-1)$-edge-connected graph of even order, then $G$ admits a perfect matching~\cite{berge}. Moreover, the following holds according to Plesn\'ik~\cite{plesnik}.

\begin{theorem}\label{plesnik}
Let $G$ be a $d$-regular $(d-1)$-edge-connected graph of even order. If $X$ is a set of edges of $G$ with $|X|\leq d-1$, then $G-X$ admits a perfect matching.
\end{theorem}

Theorem~\ref{plesnik} has been examined in various modifications and generalizations.
For instance, Chartrand and Nebeský considered an existence of a $1$-factor in $(d-2)$-edge-connected graphs \cite{nebesky}, while several authors (see e.g. \cite{liu, katernis, shiu}) inspected an existence of a $k$-factor, for $1 \le k \le d-1$, where a \emph{$k$-factor} is a $k$-regular spanning subgraph. Note that a perfect matching is a $1$-factor, and thus its complement in a $d$-regular graph is $(d-1)$-factor.

The \emph{matching preclusion number} is the minimum number of edges whose deletion
results in a graph without a perfect matching. Using this notion Theorem~\ref{plesnik} reads as ``The matching preclusion number of a $d$-regular $(d-1)$-edge-connected graph is $d$\,''. Cheng et~al.~\cite{cheng1} defined the \emph{conditional matching preclusion number} to be the matching preclusion number where deleted edges cannot be all incident with a single vertex. Cheng et~al. proved in \cite{cheng2} that
if all $d$-edge-cuts in a $d$-regular bipartite graph $G$ separate a single vertex, then the conditional matching preclusion number of $G$ is at least $d+1$. The same statement is not true if we omit the biparticity condition \cite{cheng3}. However if we add an additional condition that $G$ has no independent set of size $|V(G)|/2-1$, then the conditional matching preclusion number of $G$ is at least $d+1$ \cite[Theorem 2.1]{cheng3}.
Besides that, Cheng et al. \cite{cheng3} defined that a graph is \emph{super-$k$-edge-connected of order $l$} if every edge-cut of size at most $k$ separates one large component from the remaining components that together contain at most $l$ edges. In \cite[Theorem~3.1-3.4]{cheng3} Cheng et al. show that super-$(3k-6)$, $(3k-7)$ and $(3k-8)$-connected graphs of order $2$ have the conditional matching preclusion number equal to $2k-3$ under various assumptions (i. e. on the size of the largest independent set).

In this paper, we present strengthening of Theorem~\ref{plesnik} for graphs of higher connectivity. A major obstacle in obtaining such a result is
the fact that maximum edge-connectivity in a $d$-regular graph is $d$. In order to extend the notion of edge-connectivity in a $d$-regular graph beyond $d$, we consider a cyclic edge-connectivity, which is a parameter commonly used in the related area of matching extensions (see e. g. \cite{aldred, holton, lou}). A graph is \emph{cyclically $h$-edge-connected}
if there is no edge-cut containing less than $h$ edges separating
two subgraphs both containing a circuit. If $G$ is $d$-regular
and $h\le d$, then cyclic $h$-edge-connectivity coincides with
$h$-edge-connectivity. Therefore, cyclic edge-connectivity is a natural extension of the notion of
edge-connectivity. Moreover, if a $d$-regular graph is cyclically $(d+1)$-edge-connected, for $d\geq 3$, then every $d$-edge-cut must have a single vertex on at least one side.
As the main result we prove that if we raise the cyclic edge-connectivity of $G$ by $2k$ compared to the requirements of Theorem~\ref{plesnik}, then the set $X$ can contain $k$ extra edges, provided two obvious necessary conditions are satisfied.

Let $H$ be a graph with a vertex $v$ of degree $1$. Let $w$
be the neighbour of $v$ in $H$. Delete the vertices $v$ and $w$ to obtain $H'$. The graph $H'$ has a perfect matching if and only if $H$ has a perfect matching. The operation of deleting $v$ and $w$ from $H$ is called a \emph{leaf matching operation} or an \emph{LM operation}, for short.
The following theorem is the main result of this paper.

\begin{theorem}\label{thm1}
Let $G$ be a $d$-regular cyclically $(d-1+2k)$-edge-connected graph
of even order, for  $k\ge 0$. Let $X$ be a set of $d-1+k$ edges of $G$.
The graph $G-X$ does not admit a perfect matching
if and only if
\begin{enumerate}
\item[\emph{(i)}] an isolated vertex is obtained in $G-X$ by a series of leaf matching operations, or
\item[\emph{(ii)}] $G-X$ has an independent set containing more than half of the vertices of $G$.
\end{enumerate}
Moreover, if (ii) holds, then the vertices that do not belong to the independent set induce at most $k-1$ edges in $G-X$.
\end{theorem}

\noindent Note that the existence of an independent set from (ii) of Theorem~\ref{thm1} would imply that $G$ is close to being a bipartite graph with uneven sizes of partition sets.

For $k=0$, Theorem~\ref{thm1} coincides with Theorem~\ref{plesnik} as neither (i) nor (ii) is possible. Indeed, for (i), it can be proved by induction that the edge-cut whose one side has vertices participating in the LM operations contains at most $2d-2$ edges.
This is not possible as the other part has even number of vertices (for more details see the proof of Theorem~\ref{thm3} which uses similar ideas). For (ii), the moreover part of the theorem would be false if (ii) was true. For $k=1$, we are able to reformulate Theorem~\ref{thm1} as follows.

\begin{theorem}\label{thm3}
Let $G$ be a $d$-regular cyclically $(d+1)$-edge-connected graph of even order, for $d\geq 3$.
Let $X$ be a set of $d$ edges of $G$.
The graph $G-X$ does not admit a perfect matching
if and only if
\begin{enumerate}
\item[\emph{(a)}] \label{c22} all the edges from $X$ are incident with a common vertex, or
\item[\emph{(b)}] \label{c12} $G-X$ is bipartite and the edges from $X$ are incident with vertices from the same partition.
\end{enumerate}
\end{theorem}

\begin{figure}[h]
\begin{center}
\includegraphics[width=8cm]{fig1.pdf}
\caption{Cubic cyclically 4-edge-connected graph without 1-factor avoiding dashed edges}
\label{fig1}
\end{center}
\end{figure}
It is interesting that the two conditions of Theorem~\ref{thm1} and Theorem~\ref{thm3} are not in one-to-one correspondence for every graph.  Figure~\ref{fig1} provides an example of a graph $G$, where the set $X$ is depicted by dashed edges. Note that while the edges of $X$ are not all incident with a common vertex, an isolated vertex can be obtained in $G-X$ by a series of LM operations (the subscript of the vertices indicates the order of LM operations applied in $G-X$).

\bigskip

Let $G$ be a graph and let $X$ be a subset of edges of $G$. If $G$ is sufficiently cyclically edge-connected, then the condition (i) from Theorem~\ref{thm1} is a local condition, which can be often verified easily. Thus, in order to use Theorem~\ref{thm1}, it is crucial to have a method that allows us to verify the condition (ii) efficiently. We provide one such method while we prove several theorems
on \emph{cubic} graphs, which are $3$-regular graphs. Recall that the complement of a perfect matching in a cubic graph is a $2$-factor.
Many important theorems and conjectures can be reformulated in terms of $2$-factors of cubic graphs. For instance, the $4$-colour theorem can be restated as ``Every bridgeless planar cubic graph has a $2$-factor without circuits of odd lengths''. Considering general cubic graphs of high cyclic edge-connectivity, very interesting conjecture of Jaeger and Swart \cite{jaeger} states that every cyclically $7$-edge-connected cubic graph is $3$-edge-colourable, i.~e.~has a $2$-factor without odd circuits. A solution to this conjecture would shed more light on several important conjectures, e.~g.~to prove the $5$-flow conjecture it would be sufficient to consider cubic graphs with cyclic edge-connectivity equal to $6$ \cite{kochol}. However, the progress towards proving the conjecture of Jaeger and Swart is very limited.

Aldreed et al. \cite{aldred2} examined when it is possible to extend an independent set of edges into a $2$-factor. Here, we aim to extend a path $P$ of a given cubic graph into a $2$-factor. This is possible if and only if $G-P$ admits a perfect matching. Note that according to Theorem~\ref{plesnik}, it is always possible to find a perfect matching that avoids a path of length $1$ or $2$ (where the \emph{length of a path} is the number of its edges).  For longer paths, Theorem~\ref{thm1} gives a necessary and sufficient condition when the extension is possible.
Let $G$ be a graph, and let $G_1$ and $G_2$ be two subgraphs of $G$. We define the \emph{distance} of $G_1$ and $G_2$ to be $\min\{ d(v_1,v_2) | v_1\in G_1,v_2\in G_2 \}$, where $d(v_1,v_2)$ is the length of a shortest path between $v_1$ and $v_2$ in $G$. The distance between the prescribed edges is a common prerequisite in theorems on matching extensions (see e.~g.~\cite{aldred}).
We apply Theorem~\ref{thm1} to show, that if we have three paths on $l$ edges that are sufficiently far apart, for a given integer $l$ in a sufficiently cyclically edge-connected graph, one of the paths can be extended into a $2$-factor.

\begin{theorem}\label{thm4}
Let $G$ be a cubic cyclically $\max\{4l-5,2\}$-edge-connected graph and let $P_1$, $P_2$, and $P_3$ be
three paths of $G$ on $l$ edges such that the distance of any two of them
is at least $\max\{8l-17,0\}$. Then there is a $2$-factor of $G$ that contains
$P_1$, $P_2$ or $P_3$.
\end{theorem}

We note that while the cyclic edge-connectivity and the path distance from Theorem~\ref{thm4} may not need to be optimal, the condition on the number of paths cannot be reduced. Let us demonstrate that Theorem~\ref{thm4} does not hold if only two paths are prescribed, even if we assume arbitrary cyclic edge-connectivity and distance between the paths. Let $H$ be a bipartite subcubic graph with partitions $M$ and $N$ such that $M$ contains exactly $18$ vertices of degree $2$ and a couple of vertices of degree $3$, while
$N$ contains only vertices of degree $3$. Partition the set of vertices of degree $2$ into two sets $A=\{a_1, \ldots, a_9\}$ and $B=\{b_1,\ldots, b_9\}$ of size $9$. Add new vertices $x_i$ and $y_i$ to $H$, for $i\in\{ 1,\ldots, 7\}$. Add new edges $a_1x_1$, $a_8x_7$ $a_9x_7$, $b_1y_1$, $b_8y_7$, $b_9y_7$, and for $i\in\{1,\ldots, 6\}$, add edges $a_{i+1}x_i$, $b_{i+1}y_i$, $x_{i+1}x_i$ and $y_{i+1}y_i$. Note that the resulting graph is cubic.
Set $P_1=x_1x_2\cdots x_7$ and $P_2=y_1y_2\cdots y_7$.
The edge-set of $P_1$ cannot be extended into a $2$-factor, because $\{x_1, x_2, \ldots, x_7\} \cup N \cup \{y_1, y_3, y_5, y_7\}$ is an independent set of $G-P_1$ containing more than half of the vertices.
Similarly, the edge-set of $P_2$ cannot be extended into a $2$-factor, because
$\{y_1, y_2, \ldots, y_7\} \cup N \cup \{x_1, x_3, x_5, x_7\}$ is an independent set of $G-P_2$ containing more than half of the vertices. Note that the choice of $H$ can be done so that the resulting graph is arbitrarily cyclically edge-connected, and $P_1$ and $P_2$ are in any arbitrary distance.

On the other hand, if the paths have fewer than $6$ edges, then it is possible to prove analogous result to Theorem~\ref{thm4} even for two paths. We analyse the cases when $l=3$ and $l=4$ and provide the complete characterisation when  none of these two paths can be extended into a $2$-factor. The following theorem is a corollary of the case when $l=4$.

\begin{corollary} \label{cor7}
Let $G$ be a cubic cyclically $7$-edge-connected graph and let $v\in V(G)$. Then there exists a $2$-factor of $G$ such that $v$ does not belong to its $7$-circuit.
\end{corollary}
Corollary \ref{cor7} can be viewed as a (very) weakened version of the conjecture of Jaeger and Swart: we do not forbid circuits of all odd lengths, only of the length $7$ and we forbid it only for the circuit passing through a given vertex $v$. Despite the apparent simplicity of
Corollary \ref{cor7}, we are not aware of any significantly simpler approach on how to prove it than the one used in this paper.
Moreover, we are not aware of any result that would imply a version of
Corollary~\ref{cor7} where all occurrences of the number $7$ is replaced by a larger integer.

The organisation of the paper is as follows. In Section~\ref{s2} we prove Theorem~\ref{thm1} and Theorem~\ref{thm3}. In Section~\ref{s3} we prove Theorem~\ref{thm4}. In Section~\ref{s4} we deal with extensions of two paths of the length 3 and 4 into a $2$-factor, and we prove Corollary~\ref{cor7}.

\section{Perfect matchings without preselected edges}\label{s2}

For a graph $G$ and a vertex $v\in V(G)$, let $\deg_G(v)$ denote the degree
of $v$ in $G$. For a set $W\subseteq V(G)$ let $\partial_G(W)$ denote the set of edges of $G$ with exactly one end in~$W$. For a subgraph $H$ of a graph $G$, we abbreviate $\partial_G(V(H))$ to $\partial_G(H)$.
Let ${\text{oc}}(G)$ denote the number of odd components of $G$.
For two sets $A$ and $B$, $A\Delta B$ denotes the \emph{symmetric difference} of $A$ and $B$, i. e. $A\Delta B = (A-B) \cup (B-A)$.
Now we proceed with a proof of Theorem~\ref{thm1}.

\begin{proof}[Proof of Theorem~\ref{thm1}]
If either of the conditions (i) and (ii) is satisfied, then $G-X$ admits no perfect matching.
Therefore, we need to prove only the converse statement.

Let $G'=G-X$ and suppose that $G'$ has no perfect matching.
Due to Tutte's perfect matching theorem \cite[Theorem 2.2.1]{diestel} there is a vertex-set $S$, such that
${\text{oc}}(G'-S)>|S|$. Since $G'$ has even number of vertices, we conclude that ${\text{oc}}(G'-S)-|S|\ge 2$.
We choose $S$ so that ${\text{oc}}(G'-S)-|S|$ is maximal and subject to this condition
$|S|$ is maximal. We distinguish three cases.
\medskip

\noindent
\underline{Case 1:} There exists a component $H$ of $G'-S$ such that $G'-H$ is acyclic.
\smallskip

\noindent
We prove that an isolated vertex can be produced  by a series of LM operations.
We prove the statement by induction.
As the induction hypothesis is different from the statement of Theorem~\ref{thm1}
we formulate the statement in a separate lemma.

\begin{lemma}\label{lemaD}
Let $G'$ be a graph, let $S$ be a subset of $V(G')$, and let $H$ be a component of
$G'-S$. Suppose that
\begin{itemize}
\item $G'-H$ is acyclic, and
\item ${\text{oc}}(G'-S)-|S| \ge 2$.
\end{itemize}
Then an isolated vertex can be created in $G'$ by a series of LM operations.
Moreover, all LM operations remove only vertices from $G'-H$
and the  created isolated vertex is also from $G'-H$.
\end{lemma}

\begin{proof}
We proceed by induction on the number of vertices in $G'$.

\noindent
If $|S|=0$, then $G'$ has a component other than $H$ that is a tree containing odd number
of vertices. In this tree we can create an isolated vertex by a series of LM operations.
Therefore, the lemma holds in this case.

Suppose $|S|>0$. We have ${\text{oc}}(G'-H-S)-|S| \ge 1$.
Let $D_1$, $D_2$, \dots, $D_c$ be the components of $G'-H$.
Then
$$
\sum_{i=1}^c {\text{oc}}(D_i-S)-|D_i \cap S| ={\text{oc}}(G'-H-S)-|S| \ge 1.
$$
Therefore, there is a component $D^*$ of $G'-H$ such that
\begin{eqnarray}
{\text{oc}}(D^*-S)-|D^* \cap S| \ge 1. \label{dstar}
\end{eqnarray}
If $D^*$ is a single vertex, then by (\ref{dstar}) this vertex is not in $S$.
Then it is an isolated vertex in $G'$ from $G'-H$, proving the lemma.
Therefore, $D^*$ has at least two vertices.

Let $d_i$ be the number of vertices from $D^* \cap S$
of degree $i$ in $D^*$.
As $D^*$ is a tree, the number of components of $D^*-S$
is at most $1+\sum_{i=2}^d (i-1)\cdot d_i$. On the other hand
$D^*$ has at least $2+\sum_{i=3}^d (i-2)\cdot d_i$ leafs.
Suppose, for the contradiction, that all leafs of $D^*$ are in $S$.
Then $D^*\cap S$ has at least
$2+\sum_{i=2}^d (i-2)\cdot d_i+ \sum_{i=2}^d d_i$ vertices. It follows that
$$
{\text{oc}}(D^*-S)-|D^* \cap S| \le \left(1+\sum_{i=2}^d (i-1)\cdot d_i\right)
 - \left(2+\sum_{i=2}^d (i-1)\cdot d_i\right) = -1,
$$
which contradicts (\ref{dstar}). Therefore $D^*-S$ contains a leaf $v$ of
$D^*$.

Let $w$ be the neighbour of $v$.
We set  $G''=G'-\{v, w\}$, $S'=S-\{w\}$ (note that $S$ may not contain $w$)
and $H'=H$. The prerequisites of the lemma are satisfied for $G''$, $S'$ and $H'$ taken as $G'$, $S$ and $H$, respectively. By the induction hypothesis
we can create an isolated vertex by LM operations in $G''$.
On the other hand, we can create $G''$ from $G'$ by an LM operation because
$1=\deg_{D^*}(v)=\deg_{G'}(v)$.
\end{proof}
\smallskip

\noindent
\underline{Case 2:} The graph $G'-H'$ contains a circuit for every component $H'$ of $G'-S$ and there exists at least one component $H$ of $G'-S$ that contains a circuit.
\smallskip

\noindent
Each of the graphs $H$ and $G'-H$ contains a circuit, hence by the cyclic edge-connectivity assumption $|\partial_G(H)|\ge d-1+2k$.
For every other odd component $H'$ of $G'-S$ we have $|\partial_G(H')|\ge d$ since $G$ is $d$-regular (this holds even when $k=0$, since if $d$ is even, all edge-cuts are even and if $d$ is odd, then an edge-cut separating subgraphs with odd number of vertices must be odd).
Let ${\cal H}$ be the set consisting of odd components of $G'-S$ and $H$. It follows that
$$
\sum_{H' \in {\cal H}}  |\partial_G(H')| \ge d \cdot ({\text{oc}}(G'-S)-1)+d-1+2k.
$$
This translates to
$$
\sum_{H' \in {\cal H}}  |\partial_{G'}(H')| \ge d \cdot ({\text{oc}}(G'-S)-1)+d-1+2k-2(d-1+k).
$$
Since all these edges have the second end-vertex in $S$ and $|\partial_{G'}(S)| \le d |S|$, it follows that
$$
d \cdot {\text{oc}}(G'-S) - 2d+1 \le d|S|.
$$
Thus ${\text{oc}}(G'-S) < |S|+2$, which contradicts ${\text{oc}}(G'-S)-|S|\ge 2$.
\medskip

\noindent
\underline{Case 3:} The graph $G'-H'$ contains a circuit for every component $H'$ of $G'-S$ and each component of $G'-S$ is a tree.
\smallskip

\noindent
Suppose first that one of the components of $G'-S$, say $H$, contains more than one vertex, and let $w$ be a neighbour of a leaf $u$ of $H$. Let $S'=S\cup\{w\}$. If $H$ is even, then ${\text{oc}}(G'-S')\geq {\text{oc}}(G'-S)+1$, since at least one new odd component is created (the one that contains $u$). If $H$ is odd, then again ${\text{oc}}(G'-S')\geq {\text{oc}}(G'-S)+1$, because at least two new odd components are created (one of them contains $u$, the other one exists since $|H-\{u,w\}|$ is odd). If ${\text{oc}}(G'-S')\geq {\text{oc}}(G'-S)+2$, then we obtain a contradiction with the maximality of ${\text{oc}}(G'-S)-|S|$. Otherwise, we obtain a contradiction with the maximality of $|S|$.

Suppose now that each component of $G'-S$ is an isolated vertex. Since $|S|<{\text{oc}}(G'-S)$, the vertices
from $G'-S$ form an independent set containing more than half of the vertices of $G'$, satisfying the condition (ii) of the theorem.

We are left to prove that the set $S$ induces at most $k-1$ edges.
Since ${\text{oc}}(G'-S)-|S|\ge 2$, it follows that
$$
\sum_{v\in G'-S} \deg_G(v)=d\cdot{\text{oc}}(G'-S)\geq 2d +d|S|.
$$
In $G'$, this is
$$
\sum_{v\in G'-S} \deg_{G'}(v) \ge 2d +d|S| - 2(d-1+k),
$$
and all these edges have the second end-vertex in $S$.
Suppose for the contrary, that the vertices of $S$ induce at least $k$ edges.
Then $|\partial_{G'}(S)|\le d|S|-2k$. Combining the last two inequalities we conclude
$$
2d +d|S| - 2(d-1+k) \le d|S|-2k,
$$
a contradiction.
\end{proof}

We finish the section by proving Theorem~\ref{thm3}.
\begin{proof}[Proof of Theorem~\ref{thm3}]
If either of the conditions (a) and (b) is satisfied, then $G-X$ admits no perfect matching. Therefore, we need to prove only the converse statement.

Suppose that $G-X$ has no perfect matching. According to Theorem~\ref{thm1}, either we create an isolated vertex in $G-X$ by a series of LM operations or $G-X$ is bipartite and edges from $X$ are incident with vertices from the same partition. If the latter one occurs, we are done. Therefore, we may assume that an isolated vertex is created by a series of LM operations in $G-X$. Suppose that we have obtained an isolated vertex after $s$ LM operations. Let $v_i$ be a vertex of degree 1 that we remove during an $i$-th LM operation. Let $w_i$ be a neighbour of $v_i$ that is deleted together with $v_i$ during the $i$-th LM operation. Let $V_i=\bigcup_{j=1}^i\{v_j\}$, $W_i=\bigcup_{j=1}^i\{w_j\}$ and $U_i=V_i\cup W_i$.

If $s=0$, then $G-X$ contains an isolated vertex, and the condition (a) holds.
Assume that $s=1$. Note that in order to create a vertex of degree 1 in $G-X$, exactly $d-1$ edges of $X$ must be incident with the vertex $v_1$. Since $s=1$, deleting $U_1$ from $G-X$ results in a graph that contains an isolated vertex $u$. Thus either
every edge incident with $u$ has the second end-vertex in $U_1$,
or  there is exactly one edge $e=uu'$ incident with $u$ that does not have the second end-vertex in $U_1$ and $e\in X$.
In the former case, since $|\partial_G(U_1)| \le 2d-2$, $|\partial_G(U_1\cup\{u\})| \le d-2$, we obtain a contradiction with the cyclic edge-connectivity of $G$. In the latter case, $|\partial_G(U_1\cup\{u\})| \le (2d-2)-(d-1)+1=d$. If $|\partial_G(U_1\cup\{u\})| \le d-1$ or if $|V(G)-U_1-\{u\}|\geq 2$, then we again obtain a contradiction with the cyclic edge-connectivity of $G$. Thus $|\partial_G(U_1\cup\{u\})|=d$ and $|V(G)-U_1-\{u\}|=1$. It follows that $V(G)=\{v_1,w_1,u,u'\}$. Note that $e$ must be the only edge between $u$ and $u'$, since $u$ is an isolated vertex in $(G-X)-U_1$. Thus, $|\partial_G(\{u,u'\})| =2d-2$, and since $G$ is $d$-regular there exists exactly one edge between $v_1$ and $w_1$, and it does not belong to $X$. Since $v_1$ is a vertex of degree $1$ in $G-X$, it follows that every other edge incident with $v_1$ is from $X$ and has an end-vertex in $\{u,u'\}$. Therefore $G-X$ is a bipartite graph with partitions $\{w_1\}$ and $\{v_1,u,u'\}$. Moreover, no edge from $X$ is incident with $w_1$, implying the condition (b) of the theorem. Thus we may assume that $s\geq 2$.

Under these assumptions we prove the following lemma by induction.

\begin{lemma}\label{LemCor2}
Let $G$ be a $d$-regular cyclically $(d+1)$-edge-connected graph of even order. Let $X$ be a set of $d$ edges of $G$. Assume that a first isolated vertex in $G-X$ is created after $s$ LM operations, where $s\geq 2$. Then
\begin{itemize}
\item[(i)] every edge of $X$ is incident with a vertex of $V_2$,
\item[(ii)] $|\partial_G(U_t)| \le 2d$ for each $2\leq t\leq s$, and $|\partial_G(U_1)| \le 2d-2$, and
\item[(iii)] the edges incident with the vertices of $V_t$ in $G$ either belong to $X$ or they connect a vertex of $V_t$ with a vertex of $W_t$, for each $1\leq t\leq s$.
\end{itemize}
\end{lemma}

\begin{proof}[Proof of Lemma \ref{LemCor2}.]
First we prove (i), the second part of (ii) and (iii) for $t=1$. Note that the latter one is obvious, because $w_1$ is the only neighbour of $v_1$ in $G-X$. Since $s\geq 2$, in order to obtain an isolated vertex after two LM operations, there must be exactly $d-1$ edges of $X$ incident with $v_1$. As there is an edge connecting $v_1$ and $w_1$ in $G$, $|\partial_G(U_1)|\leq 2d-2$, implying the second part of (ii). Let $e$ be the edge of $X$ that is not incident with $v_1$. Assume to the contrary with (i) that $e$ is not incident with $v_2$. Since $v_2$ is a vertex of degree $1$ after the first LM operation in $G-X$, exactly $d-1$ edges incident with $v_2$ have the other end-vertex in $U_1$. Since the remaining edge incident with $v_2$ has $w_2$ as the other end-vertex, $|\partial_G(U_1\cup v_2)| \le (2d-2)-(d-1)+1=d$. This is a contradiction with the cyclic edge-connectivity of $G$, because each side of the edge-cut has at least two vertices - one side being $U_1\cup\{v_2\}$ and the other side containing $w_2$ and a resulting isolated vertex.

We prove the first part of (ii) and (iii) by induction on $t$. For the basis of induction, let $t=2$. Recall that $v_2$ is a vertex of degree 1 in $(G-X)-U_1$ and $e$ is incident with $v_2$. Let us count $|\partial_G(U_2)|$.
There are at most $2d-2$ edges in $\partial_G(U_1)$, at least $d-2$ out of these are incident with $v_2$.
As $G$ has an edge between $v_2$ and $w_2$,
there is at most one edge in $\partial_G(U_2)$ incident with $v_2$ and at most $d-1$ edges in $\partial_G(U_2)$
incident with $w_2$. We conclude that $|\partial_G(U_2)|\leq (2d-2)-(d-2)+1+(d-1)=2d$, and (ii) holds. Moreover, every edge incident with a vertex of $V_2$ either belongs to $X$ or is incident with a vertex of $W_2$ as requested by (iii).

For the induction step, let us assume that the lemma holds for every $j<t$, we are going to prove it for $t$.
Let us count $|\partial_G(U_t)|$. There are at most $2d$ edges in $\partial_G(U_{t-1})$, out of these exactly $d-1$ are incident with $v_{t}$. There are at most $d-1$ edges incident with $w_{t}$ that are not incident with $v_{t}$. Therefore $|\partial_G(U_{t})|\leq 2d-(d-1)+(d-1)$, which proves (ii).

To prove (iii), we only need to prove the statement for the edges incident with $v_{t}$. Since $v_{t}$ is a vertex of degree 1 in $(G-X)- U_{t-1}$, and $v_{t}w_{t}\in E(G)$, every edge incident with $v_t$ has the other end-vertex in $V_{t-1}\cup W_t$. Suppose that there is an edge $v_{t}v_j$ for some $1\leq j\leq t-1$. But the vertex $v_j$ is adjacent only with the vertices of $W_j$ or incident with edges from $X$ by the induction hypothesis. This is a contradiction, and hence every edge incident with  $v_t$ has the other end-vertex in $W_t$ and (iii) holds as well.
\end{proof}

We are ready to finish the proof of Theorem~\ref{thm3}. Let $u$ be a first isolated vertex, which is created after $s$ steps of LM operations, where $s\geq 2$. By Lemma~\ref{LemCor2} (ii), $|\partial_G(U_{s})|\leq 2d$. Thus $d$ out of these $2d$ edges are incident with $u$. Since $G$ is cyclically $(d+1)$-edge-connected, the other $d$ edges must be adjacent to a single vertex, say $u_2$, which is another isolated vertex of $(G-X)- U_s$. In this case, $G-X$ has an independent set $A$ containing more than half of the vertices of $G$, namely $V_s\cup\{u,u_2\}$. Theorem~\ref{thm1} with $k=1$ implies that $G-X$ is bipartite with partitions $A$, and $B=G-A$. Note that $|A|\geq |B|+2$, since $|V(G)|$ is even.

It remains to prove that the edges from $X$ have both ends in $A$. Suppose to the contrary that at most $d-1$ edges of $X$ have both ends in $A$. Then $|\partial_G(A)|\geq d(|B|+2)-2(d-1)$. On the other hand, $|\partial_G(A)|\leq d|B|$. The two inequalities imply that $0\geq 2$, which is a contradiction. Thus (b) holds for $s\geq 2$ as well. This concludes the proof of the theorem.
\end{proof}

\section{Paths contained in $2$-factors}\label{s3}

Let $G$ be a cubic graph.
Let $P=p_0p_1\cdots p_m$ be a path of $G$ of length $m$, $m\ge 3$.
Assume that $G$ is cyclically $(2m-1)$-edge-connected. This, up to several small graphs that do not have cycle-separating edge-cut, implies that $G$ has no circuit of length less than $m$.
Here, $K_2^3$ is a cubic graph on two vertices and three parallel edges.

\begin{lemma}\label{lemaCyklicka}\cite[Proposition 1]{nedela}
A cubic cyclically $c$-edge connected graph $G$ contains no circuit of length less than $c$ unless $G$ is isomorphic to $K_2^3$, $K_4$ or $K_{3,3}$.
\end{lemma}

Our goal is to extend $P$ to a $2$-factor. In $K_2^3$, $K_4$ or $K_{3,3}$ it is possible to extend every path into a $2$-factor. Thus assume that $G$ is not isomorphic to any of these graphs.
Further, assume that $P$ cannot be extended to a $2$-factor of $G$, hence $G-E(P)$ does not admit a perfect matching. According to Theorem~\ref{thm1} applied with $k=m-2$
a series of LM operations in $G-E(P)$ produces an isolated vertex or $G-E(P)$ has an independent set containing more than half of the vertices.

The only vertices of degree $1$ in $G-E(P)$ are $p_1, \ldots, p_{m-1}$.
Let $p'_1, \ldots, p'_{m-1}$ be the neighbours of $p_1, \ldots, p_{m-1}$ outside $P$, respectively.
These vertices are all distinct, otherwise $G$ would have a circuit of length at most $m$,
contradicting Lemma~\ref{lemaCyklicka}. We perform LM operations removing vertices $p_i$ and $p'_i$,
for $i\in\{1,\dots, m-1\}$. If a vertex $p''$ of degree at most $1$ is created, this would imply that
$G$ has a circuit of length at most $m+2$, which is less than $2m-1$ when $m>3$, contradicting Lemma~\ref{lemaCyklicka}.
Thus no further LM operation can be performed since there are no further vertices of degree $1$. Therefore the condition (i) of Theorem~\ref{thm1} does not occur in this case. Theorem~\ref{thm1} (ii) implies that there exists a large independent set $I(P)$ in $G-E(P)$.
If $m=3$, then instead of using Theorem~\ref{thm1}, we use Theorem~\ref{thm3}. In both cases the following holds.

\begin{lemma}\label{lemaLM}
Let $m\ge 3$, and let $G$ be a cyclically $(2m-1)$-edge-connected cubic graph.
Let $P$ be a path of length $m$ in $G$. Assume that $G-E(P)$ has no perfect matching.
Then $G-E(P)$ has an independent set containing more than half of the vertices.
\end{lemma}
\noindent In what follows we describe how to recognise this situation using signed graphs.
\smallskip

A \emph{signed graph} $(G,\Sigma)$ is a graph $G$ whose edges are equipped with a positive or a negative sign. The set $\Sigma$ is the set of \emph{negative edges}, while $E(G)-\Sigma$ is the set of \emph{positive} ones. The \emph{switching} at a vertex $v$ of $G$ is inverting the sign of every edge incident with $v$. Two signed graphs $(G,\Sigma_1)$ and $(G,\Sigma_2)$ are \emph{equivalent} if $(G,\Sigma_1)$ can be obtained from $(G,\Sigma_2)$ by a series of switchings. According to Zaslavsky~\cite{zaslavsky}, two signed graphs $(G,\Sigma_1)$ and $(G,\Sigma_2)$ are equivalent if and only if the symmetric difference $\Sigma_1\Delta \Sigma_2$ is an edge-cut of $G$.
\smallskip

Let $I(P)$ be an independent set of vertices in $G-E(P)$ of size greater than $|V(G)|/2$ and among all such sets choose one that has biggest intersection with $P$. Assume that there is $i$, $0<i<m$, such that $p_i$ is not in $I(P)$.
Let $q_i$ be the neighbour of $p_i$ that is not in $\{p_{i-1},p_{i+1}\}$. If $q_i \not \in I(P)$, then we can add $p_i$ to $I(P)$, contradicting the choice of $I(P)$. If $q_i \in I(P)$, then we add $p_i$ to $I(P)$ and remove $q_i$ from $I(P)$. Note that due to cyclic edge-connectivity,
$q_i\neq p_j$ for any $j\in \{0,\ldots,m\}$, which contradicts the choice of $I(P)$. Thus, $p_i$ is in $I(P)$, for each $i\in\{1,\ldots,m-1\}$.

The set of edges $Y(P)$ is defined as a union of all edges induced by $V(G)-I(P)$ and all edges of $P$ between two vertices of $I(P)$ (thus all inner edges of $P$ are in $Y(P)$). Note that $G-Y(P)$ is bipartite with partitions $I(P)$ and $V(G)-I(P)$. The edges of $Y(P) \cap E(P)$ have both end-vertices in $I(P)$, while the edges of $Y(P)-E(P)$ have both end-vertices in $V(G)-I(P)$. Let $(G, Y(P))$ be a signed graph. Switch at the vertices of $I(P)$ and note that the resulting graph has only negative edges. Thus $(G, Y(P))$ is equivalent with $(G, E(G))$. In summary, if the path $P$ cannot be extended to a 2-factor of $G$, then $(G, Y(P))$ is equivalent to an all-negative signed graph.

Let $P$ and $Q$ be two paths of $G$ such that none of them can be extended into a $2$-factor of $G$.
Then $(G, Y(P))$ and $(G, Y(Q))$ are both equivalent to $(G, E(G))$ and thus they are equivalent.
Therefore, according to Zaslavsky~\cite{zaslavsky}, there is a partition of $V(G)$ into $V_1$ and $V_2$ such that
$E(V_1, V_2)=Y(P) \Delta Y(Q)$, where $E(V_1,V_2)=\{v_1v_2\in E(G)|v_1\in V_1, v_2\in V_2\}$.

The following lemma thus holds not only if $m \ge 4$ but even for $m=3$ as we can use Theorem~\ref{thm3} instead of Theorem~\ref{thm1}.
\begin{lemma}\label{lemaZas}
Let $m \ge 3$ and let $G$ be a cyclically $(2m-1)$-edge-connected cubic graph.
Let $P$ and $Q$ be two paths of length $m$ in $G$ such that none of them can be extended into a $2$-factor of $G$. Then there is a partition of $V(G)$ into $V_1$ and $V_2$ such that $E(V_1, V_2)=Y(P) \Delta Y(Q)$.
\end{lemma}

Assuming even higher cyclic edge-connectivity of $G$, only one of the components of $G-E(V_1, V_2)$ contains a cycle. This observation is used in proofs of the remaining theorems.
We proceed by proving Theorem~\ref{thm4}.

\begin{proof}[Proof of Theorem~\ref{thm4}]
By Theorem~\ref{plesnik}, the result follows for $l\leq 2$. Moreover, if $G\in\{K_2^3,K_4,K_{3,3}\}$, then the theorem holds.
Thus for the rest of the proof, we assume that $l\geq 3$ and that $G\notin\{K_2^3,K_4,K_{3,3}\}$. This allows us to use Lemma~\ref{lemaCyklicka} freely, and apply Lemmas~\ref{lemaLM} and \ref{lemaZas}, because $4l-5 \geq 2l-1$.

Assume for the contrary that none of $P_1$, $P_2$, and $P_3$ can be extended into a $2$-factor of $G$.
Then by Lemma~\ref{lemaLM}, $G-E(P_i)$ has an independent set containing more than half of the vertices, for $i\in\{1, 2, 3\}$.

Let $e\in E(G)$ and let $P$ be a path of $G$. We say that $e$ is \emph{close to $P$} (\emph{distant from $P$}, respectively) if the distance of $e$ from $P$ is at most $4l-9$ (at least $4l-8$, respectively).
Since the distance between any two paths from the theorem statement is at least $8l-17$, we have the following claim.

\medskip

\noindent
\textbf{Claim $1$.} No edge of $G$ is close to both $P_i$ and $P_j$ where $i,j \in \{1,2,3\}$ and $i \neq j$.

\medskip

For each path $P_i$, $i \in \{1, 2, 3\}$, we partition the edges of $Y(P_i)$ into three subsets:
\begin{itemize}
\item $B_i$ are edges of $P_i$,
\item $C_i$ are edges close to $P_i$ that do not belong to $B_i$, and
\item $D_i$ are edges distant from $P_i$.
\end{itemize}

\medskip

Due to the definition of $Y(P_i)$, $B_i$, $C_i$ and $D_i$, the following holds.

\medskip

\noindent
\textbf{Claim 2.} No vertex incident with an edge from $B_i$ is incident with an edge from $C_i\cup D_i$,  for each $i\in\{1,2,3\}$.
\medskip

The definition of $I(P_i)$ implies the following.
\medskip

\noindent
\textbf{Claim 3.} $|B_i|\in\{l-2,l-1,l\}$, for $i\in\{1,2,3\}$ .
\medskip

According to the last part of Theorem~\ref{thm1} where $d=3$ and $k=l-2$
(or according to Theorem~\ref{thm3}
for $l=3$), there are at most $k-1=l-3$ edges of $Y(P_2)$ that are not path edges.
Therefore the following holds.

\medskip

\noindent
\textbf{Claim 4.} $|C_i\cup D_i|\leq l-3$.
\medskip

Without loss of generality, suppose that $P_1$ is a path out of $\{P_1,P_2,P_3\}$ with the minimum number of distant edges. By Claim 1, no edge is close to both $P_2$ and $P_3$. Thus, without loss of generality, we may assume that at most $|D_1|/2$ edges of $D_1$ are close to $P_2$, and denote these edges by $D_1'$. According to Claim~4 and the choice of $P_1$, $|C_2|=|C_2\cup D_2|-|D_2|\leq (l-3)-|D_1|$. Since $2|D'_1|\leq |D_1|$ by the choice of $P_2$, we obtain

\begin{eqnarray}
2\cdot |D'_1| + |C_2| \le l-3. \label{eqd1c2}
\end{eqnarray}

\noindent We are going to show that $|B_2| \leq 2\cdot |D'_1| + |C_2|$, which will be a contradiction with~(\ref{eqd1c2}) by Claim~3.
\medskip

Since none of $P_1$ and $P_2$ can be extended into a $2$-factor of $G$, signed graphs $(G, Y(P_1))$ and $(G, Y(P_2))$ are both equivalent to the signed graph $(G, E(G))$, and thus they are equivalent to each other. Therefore, according to Lemma~\ref{lemaZas} there is a partition of $V(G)$ into $(V_1, V_2)$ such that $E(V_1, V_2)=Y(P_1) \Delta Y(P_2)$. By Claim 3 and Claim 4, $|E(V_1, V_2)|=|Y(P_1) \Delta Y(P_2)|\leq |Y(P_1)|+|Y(P_2)|\leq |B_1|+|C_1\cup D_1|+|B_2|+|C_2\cup D_2|\leq l+(l-3)+l+(l-3)=4l-6$.
\medskip

\noindent
\textbf{Claim 5.} $|E(V_1, V_2)|\leq 4l-6$.
\medskip

Due to the to small size of $E(V_1, V_2)$ and the condition on the cyclic edge-connectivity of $G$, the components of $G-E(V_1, V_2)$ are mostly acyclic.
\medskip

\noindent
\textbf{Claim 6.} There exists exactly one component of $G-E(V_1, V_2)$ that is cyclic.

\begin{proof}[Proof of Claim 6.]
By Claim 5, $|E(V_1, V_2)|\leq 4l-6$. Since $G$ is cyclically $(4l-5)$-edge-connected, at most one of the components of $G-E(V_1, V_2)$, may contain a cycle. Let us show that $G-E(V_1, V_2)$ contains a cyclic component. There are at most $8l-16$ vertices in the acyclic components of $G-E(V_1, V_2)$, $4l-8$ of them on each side of the edge-cut. However, the graph $G$ has girth at least $4l-5$
due to Lemma~\ref{lemaCyklicka}, and thus it has at least $3\cdot 2^{2l-3}-3$ vertices due to the Moore bound~\cite{moore}. Since $3\cdot 2^{2l-2}-2\geq 8l-16$ for any $l\geq 3$, at least one of the components must be large and cyclic.
\end{proof}

We denote the cyclic component of $G-E(V_1, V_2)$ by $H$. Let $W\subseteq V(G)-V(H)$. Let $E^+(W)$ denote the union of $\partial_G(W)$ with the set of edges induced by $W$ in $G$. If $S$ is an induced subgraph of $G$, then we abbreviate $E^+(V(S))$ to $E^+(S)$.
\medskip

\noindent
\textbf{Claim 7.} Let $W\subseteq V(G)-V(H)$. If $ \partial_G(W)\subseteq E(V_1,V_2)$, then $E^+(W)$ induces an acyclic graph.

\begin{proof}[Proof of Claim 7.]
By Claim 5, $|\partial_G(H)| \leq 4l-6$. Since $G$ is cyclically $(4l-5)$-edge-connected, any subgraph of $G-V(H)$ is acyclic. Thus the graph induced by $W$ is also acyclic. Suppose that $E^+(W)$ induces a cycle $C$. Then for each edge $e\in\partial_G(W) \cap C$ there is another edge $f\in\partial_G(W) \cap C$ such that $e$ and $f$ are incident with a common vertex $w$ from $V(G)-W$. Note that no cycle of $H$ contains $w$, since either $w\notin H$ or $w$ is a vertex of degree $1$ in $H$. Thus $|\partial_G(W\cup V(C))|\leq |\partial_G(W)|\leq|E(V_1,V_2)|\leq 4l-6$, and $\partial_G(W\cup V(C))$ separates $C$ from the cycle of $H$, which is a contradiction with the cyclic edge-connectivity of~$G$.
\end{proof}

As $H$ is a component of  $G-E(V_1,V_2)$, $H$ is an induced subgraph, which together with Claim 7 implies the following.
\medskip

\noindent
\textbf{Claim 8.} Each edge from $E(G)$ is either in $E(H)$ or in exactly one set $E^+(S)$, where $S$ is an acyclic component of $G-V(H)$.
\medskip

To bound $|B_2|$ from above, we are going to distinguish whether an edge of $B_2$ belongs to $E(H)$ or to $E^+(S)$ for some acyclic component $S$ of $G-V(H)$. First consider the edges of $H$. For each $e\in B_2\cap E(H)$, $e\notin E(V_1,V_2)$, and thus $e\notin Y(P_1) \Delta Y(P_2)$. Therefore, as $e\in Y(P_2)$,
we have $e\in Y(P_1)$. According to Claim 1 and the definition of $D'_1$, $e\in D'_1$, and
\begin{eqnarray}
|B_2 \cap E(H)| \leq |D'_1 \cap E(H)|. \label{eqinH}
\end{eqnarray}

Assume now that $S$ is an (acyclic) component of $G-V(H)$. Due to the definition of $H$, all vertices of $H$ belong either to $V_1$ or to $V_2$.
Note that the edges of $\partial_G(S)$ have one end in $S$ and one end in $H$.  Hence $\partial_G(S)\subseteq E(V_1,V_2)=Y(P_1) \Delta Y(P_2)$
and all vertices of $S$ incident with an edge from $\partial_G(S)$ belong either to $V_1$ or to $V_2$.
The following lemma bounds the number of edges in $|B_2 \cap E^+(S)|$.

\begin{lemma}\label{lemmathm3}
Let $T$ be an induced subtree of $G-V(H)$ such that $ \partial_G(T)\subseteq Y(P_1) \Delta Y(P_2)$. Assume further that all vertices of $T$ incident with edges from $\partial_G(T)$ belong either to $V_1$ or to $V_2$.
Then $|B_2 \cap E^+(T)| \leq 2 |D'_1 \cap E^+(T)| + |C_2 \cap E^+(T)|$.
\end{lemma}

\begin{proof}[Proof of Lemma~\ref{lemmathm3}.] The proof is by complete induction on $|E(T) \cap (Y(P_1) \Delta Y(P_2))|$.

If $|B_2 \cap E^+(T)| = 0$, then the lemma holds. Assume for the rest of the proof that $|B_2 \cap E^+(T)| \geq 1$.

By the assumption of the lemma and by Claim 5,
$|\partial_G(T)| \le |Y(P_1) \Delta Y(P_2)|=|E(V_1,V_2)| \le 4l-6$.
Thus $T$ has at most $4l-8$ vertices and the edges of $E^+(T)$ are close to $P_2$. By Claim 1, they are distant from $P_1$.
Thus if an edge from  $E^+(T)$ belongs to $Y(P_1)$, then it must be from $D_1'$. On the other hand if an edge from  $E^+(T)$ belongs to $Y(P_2)$, then it is either from $B_2$ or from $C_2$.
\smallskip

If $T$ is an isolated vertex, then either $|\partial_G(T)\cap B_2|=1$ or $|\partial_G(T)\cap B_2|=2$. Let $e\in \partial_G(T)- B_2$.  By Claim 2, $e\notin C_2\cup D_2$. Thus, $e\in Y(P_1)$, and hence $e\in D'_1$. Therefore,

\begin{eqnarray}
|B_2 \cap E^+(T)| \leq 2|D'_1 \cap E^+(T)| \nonumber
\end{eqnarray}

\noindent
and the lemma holds.

Assume for the rest of the proof that $T$ is not an isolated vertex. If $|\partial_G(T)|\geq 2l$, then since $|B_2|\leq l$, the following holds.

\begin{eqnarray}
|B_2 \cap \partial_G(T)| \leq |D'_1 \cap \partial_G(T)| + |C_2 \cap \partial_G(T)|. \label{eqinHpart}
\end{eqnarray}
If, on the other hand, $|\partial_G(T)|< 2l$, then $P_2 \cap E^+(T)$ induces a connected graph in $G$.
Indeed, if $P_2 \cap E^+(T)$ induces a graph with two components there exists a subpath $Q$ of $P_2$ of length at most $l$
that connects two vertices of $T$ and for which $E(Q)\cap E(T)=\emptyset$.
As $T$ is connected, endvertices of $Q$ can be connected by a path $Q'$ in $T$. Since $|\partial_G(T)|< 2l$ and $T$ is a tree, $|V(T)|<2l-2$ and $Q'$ has has at most $2l-4$ edges.
The paths $Q$ and $Q'$ form a cycle of total length at most $3l-4$.
As $G$ is cyclically $(4l-5)$-edge-connected, this contradicts Lemma~\ref{lemaCyklicka}.
Thus $E(P_2) \cap E^+(T)$ induces a connected graph in $G$.
By Claim 7, $E^+(T)$ is acyclic, hence $|B_2 \cap \partial_G(T)| \le 2$. But $|\partial_G(T)|-|B_2| \ge 2$ as $T$ is not an isolated vertex. Thus the inequality~(\ref{eqinHpart}) holds.
\medskip

Let $T_0$ be a union of such components of $T-(Y(P_1) \Delta Y(P_2))$ that contain a vertex of $T$ that is incident with an edge from $\partial_G(T)$. Recall that the vertices of $T$ incident with an edge from $\partial_G(T)$ all belong either to $V_1$ or to $V_2$, say $V_1$. Thus the same holds for the vertices of $T_0$ that are incident with $\partial_G(T)$. Therefore all vertices of $T_0$ must also belong to $V_1$, and so $T_0$ is an induced subgraph of $G$. Hence if $e\in E(T_0)\cap B_2$, then $e\in Y(P_1)$ and so $e\in D_1'$.  Thus
\begin{eqnarray}
|B_2 \cap E(T_0)| \leq |D'_1 \cap E(T_0)|. \label{eqinHpart2}
\end{eqnarray}

Consider a component $T'$ of $T-V(T_0)$. Let us show that $T'$ satisfies the conditions of this lemma. It is easy to see that $T'$ is an induced subtree of $G-V(H)$. By the definition of $T_0$, $\partial_G(T')\subseteq Y(P_1) \Delta Y(P_2)$ and all edges of $\partial_G(T')$ must have one end-vertex in $T_0$ and the other end-vertex in $T'$. Since the first one belongs to $V_1$, the second one must be in $V_2$. Since $T'$ does not contain $\partial_G(T')$, $|E(T') \cap (Y(P_1) \Delta Y(P_2))|< |E(T) \cap (Y(P_1) \Delta Y(P_2))|$. Applying induction to $T'$,
\begin{eqnarray}
|B_2 \cap E^+(T')| \leq 2 |D'_1 \cap E^+(T')| + |C_2 \cap E^+(T')| \label{eqinHpart3}.
\end{eqnarray}
The inequalities (\ref{eqinHpart}), (\ref{eqinHpart2}), together with
the inequality  (\ref{eqinHpart3}) applied to each component $T'$ of $T-V(T_0)$ imply the statement of the lemma.
\end{proof}

By Claim~8, it follows from (\ref{eqinH}) and from Lemma~\ref{lemmathm3} that $|B_2|\le 2|D'_1|+|C_2|$. This is a contradiction with (\ref{eqd1c2}) by Claim 3, which concludes the proof of the theorem.
\end{proof}

\section{On $3$-paths and $4$-paths}\label{s4}
\begin{lemma}\label{lemaFarb}
Let $G$ be a graph, let $A, B\subseteq E(G)$ such that $G-A$ and $G-B$ are bipartite with partitions $V^A_1, V^A_2$ and $V^B_1,V^B_2$, respectively.
Then there exists a partition of $V(G)$ into $V_1$ and $V_2$ such that
\begin{enumerate}
\item $E(V_1,V_2)=A\Delta B$, and
\item $V^B_1=V^A_1\Delta V_1 = V^A_2\Delta V_2$ and $V^B_2=V^A_2\Delta V_1 = V^A_1\Delta V_2$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let us consider the signed graphs $(G,A)$ and $(G,B)$. Note that switching at vertices of $V^A_1$ in $(G,A)$ results in an all-negative signed graph $(G,E(G))$. Similarly, $(G,B)$ can be switched to $(G,E(G))$, and hence $(G,A)$ and $(G,B)$ are equivalent.
Thus we can obtain $(G,B)$ from $(G,A)$ by switching at the vertices
$V^A_1\Delta V^B_1$. Let $V_1=V^A_1\Delta V^B_1$ and $V_2=V(G)-V_1$ ($=V^A_2\Delta V^B_1$).
As $V^A_1\Delta V^B_1=V^A_2\Delta V^B_2$ and $V^A_2\Delta V^B_1=V^A_1\Delta V^B_2$ the statement 2 of this lemma holds.

Note that $E(V_1,V_2)=\partial_G(V_1)=\partial_G(V^A_1\Delta V^B_1)=
\partial_G(V^A_1) \Delta \partial(V^B_1)$.
Since switching at all vertices of a set $W$ changes signs exactly of the edges in $\partial(W)$, from the fact that switching at $V^A_1$ in $(G, A)$ produces $(G,E(G))$
and switching at $V^B_1$ in $(G, B)$ produces $(G,E(G))$
we derive
$A \Delta \partial_G(V^A_1)=E(G)=B \Delta \partial_G(V^B_1)$
and thus
$\partial_G(V^A_1) \Delta \partial_G(V^B_1)=A \Delta B$
which concludes the proof of the statement 1 of this lemma.
\end{proof}

\begin{figure}[p]
\begin{center} \includegraphics[width=14cm]{fig34.pdf}
\caption{Positions $\mathcal{P}_1$, $\mathcal{P}_2$, \ldots, $\mathcal{P}_8$ of paths related to Theorem~\ref{thm5} and Theorem~\ref{thm6}. The respective paths are depicted by dashed and dotted lines.}
\label{fig34}
\end{center}
\end{figure}

Let $G$ be a graph and let $P_1=v_1v_2v_3v_4$ and $P_2=w_1w_2w_3w_4$ be two paths of $G$ of length $3$. We say that $P_1$ and $P_2$ are in the \textit{position $\mathcal{P}_1$} if $v_2v_3=w_2w_3$, $|\{v_1,v_4,w_1, w_4\}|=4$, and $G-\{v_2,v_3\}$ admits a proper (vertex) $2$-colouring $c$ such that $c(v_1)=c(v_4)\neq c(w_1)=c(w_4)$. This position is depicted on the top of Figure~\ref{fig34}.

\begin{theorem} \label{thm5}
Let $G$ be a cubic cyclically $5$-edge-connected graph and let $P_1$ and $P_2$ be
two distinct paths of $G$ of length $3$. Then there is a $2$-factor of $G$ that contains $P_1$ or $P_2$ if and only if $P_1$ and $P_2$ are not in the position $\mathcal{P}_1$.
\end{theorem}

\begin{proof}
If $G\in\{K_2^3,K_4,K_{3,3}\}$, then the theorem holds. Thus we assume that $G\notin\{K_2^3,K_4,K_{3,3}\}$, and we can use Lemma~\ref{lemaCyklicka} without considering these exceptions.
Let $i\in\{1,2\}$. Suppose that $P_i$ cannot be extended into a $2$-factor of $G$. Equivalently, $G-E(P_i)$ has no perfect matching. By Theorem~\ref{thm3}, $G-E(P_i)$ is bipartite and edges from $E(P_i)$ are incident with vertices from the same partition.
Let $p_i$ be an edge of $P_i$ incident with an end-vertex of $P_i$, and let $q_i$ be an edge of $G-E(P_i)$ that is incident with the central vertex of $P_i-p_i$. Note that $G-\{p_i,q_i\}$ is bipartite.

By the first statement of Lemma~\ref{lemaFarb}, $\{p_1,q_1\}\Delta \{p_2,q_2\}$ is an edge-cut of $G$. Since $G$ is cyclically $5$-edge-connected and cubic, $|\{p_1,q_1\}\Delta \{p_2,q_2\}|\neq 2$. Thus, $\{p_1,q_1\}\Delta \{p_2,q_2\}$ is a $4$-edge-cut, and, again by cyclic edge-connectivity, one side of the edge-cut $\{p_1,q_1\}\Delta \{p_2,q_2\}$ induces an edge, which we denote by $e$.

The edges $p_i$ and $q_i$ are not adjacent, because $G$ does not contain any triangle according to Lemma~\ref{lemaCyklicka}. Again by Lemma~\ref{lemaCyklicka}, $G$ has no $4$-circuit and therefore, $e$ is the only edge that is neighbour of $p_i$ and $q_i$ both. Thus $e\in P_i$.

If $p_1$ is adjacent to $q_2$, then $p_2$ is adjacent to $q_1$. It follows that $P_1=P_2$, which contradicts that $P_1$ and $P_2$ are distinct.
Thus $p_1$ is adjacent to $p_2$, $q_1$ is adjacent to $q_2$,
 and $E(P_i)=\{p_i,e,q_{3-i}\}$, for $i\in\{1,2\}$.  Let $P_1=v_1v_2v_3v_4$, $P_2=w_1w_2w_3w_3$ with the notation chosen in such a way that $p_1=v_1v_2$ and $p_2=w_1w_2$ (and thus $q_1=w_3w_4$ and $q_2=v_3v_4$).

Denote the partitions of $G-\{p_1,q_1\}$ by $V_1^{P_1}$ and $V_2^{P_1}$ in such a way that $v_1 \in V_1^{P_1}$.
Denote the partitions of $G-\{p_2,q_2\}$ by $V_1^{P_2}$ and $V_2^{P_2}$ in such a way that $v_1 \in V_1^{P_2}$. Let $V_1$ and $V_2$ be a partition of $V(G)$ guaranteed by Lemma~\ref{lemaFarb}. Thus $E(V_1,V_2)=\{p_1,q_1,p_2,q_2\}$ is an edge-cut of $G$.
Since $v_1 \in V_1^{P_1} \cap V_1^{P_2}$, the second statement of Lemma~\ref{lemaFarb}
implies that $v_1 \in V_2$. Therefore $V_1=\{v_2,v_3\}$,
$V^{P_2}_1=V^{P_1}_1\Delta \{v_2,v_3\}$ and
$V^{P_2}_2=V^{P_1}_2\Delta \{v_2,v_3\}$.

Thus
$v_1 \in V_1^{P_1} \cap V_1^{P_2}$ implies that
$v_2 \in V_1^{P_1} \cap V_2^{P_2}$,
$w_1 \in V_2^{P_1} \cap V_2^{P_2}$,
$v_3 \in V_2^{P_1} \cap V_1^{P_2}$,
$v_4 \in V_1^{P_1} \cap V_1^{P_2}$, and
$w_4 \in V_2^{P_1} \cap V_2^{P_2}$.
We define a proper vertex $2$-colouring of $G-\{v_2,v_3\}$.
Colour the vertices of $V_1^{P_1}-\{v_2\}$ with white and vertices of $V_2^{P_1}-\{v_3\}$ with black.  For this colouring $c$ we have $c(v_1)=c(v_4)\neq c(w_1)=c(w_4)$.
Thus $P_1$ and $P_2$ are in the position $\mathcal{P}_1$.

For the converse, suppose that $P_1$ and $P_2$ are in the described position, and let $i\in\{1,2\}$. Then the vertices of the colour used for end-vertices of $P_i$ together with the end-vertices of $e$ form a large independent set satisfying Theorem~\ref{thm3} (b) for $P_i$. Thus, there is no perfect matching avoiding $P_i$, and consequently, $P_i$ does not belong to any $2$-factor.
\end{proof}

Let $G$ be a graph and let $P_1$ and $P_2$ be two paths of $G$ of length $4$. We say that $P_1$ and $P_2$ are in the \textit{position $\mathcal{P}_j$}, for $j\in\{2,3,4,5,6,7,8\}$, if their positions correspond to the respective part of Figure~\ref{fig34} and $G$ admits a (vertex) $2$-colouring that is proper except of three monochromatic edges depicted on the figure. The paths are depicted by dashed and dotted lines and all the displayed vertices are distinct.

\begin{theorem} \label{thm6}
Let $G$ be a cubic cyclically $7$-edge-connected graph and let $P_1$ and $P_2$ be two paths of $G$ of length $4$. Then there is a $2$-factor of $G$ that contains $P_1$ or $P_2$ if and only if any subpaths of $P_1$ and $P_2$ on $3$ edges are not in the position $\mathcal{P}_1$ and the paths $P_1$ and $P_2$ are not in any of the positions in $\{\mathcal{P}_2,\mathcal{P}_3,\mathcal{P}_4,\mathcal{P}_5, \mathcal{P}_6, \mathcal{P}_7, \mathcal{P}_8\}$.
\end{theorem}

\begin{proof}
If $G\in\{K_2^3,K_4,K_{3,3}\}$, then the theorem holds. Thus we assume that $G\notin\{K_2^3,K_4,K_{3,3}\}$, and we can use Lemma~\ref{lemaCyklicka} without considering these exceptions.
Let $i\in\{1,2\}$. Suppose first that $P_i$ cannot be extended into a $2$-factor of $G$. Equivalently, $G-E(P_i)$ has no perfect matching. By Lemma~\ref{lemaLM}, $G-E(P_i)$ has an independent set containing more than half of the vertices. Let $I(P_i)$ be defined in the same way as in Section~3; it is an independent set in $G-E(P_i)$ that contains more than half of the vertices and that has maximum intersection with $V(P_i)$ among all such sets. Due to this definition all inner vertices of $P_i$ belong to $I(P_i)$. Let us show that at least one end-vertex of $P_i$ also belongs to $I(P_i)$.

Let $J(P_i):=V(G)-I(P_i)$, and let $s$ ($t$, respectively) be the number of edges in $G$ induced by $I(P_i)$ ($J(P_i)$, respectively). Note that $s\leq 4$, because $I(P_i)$ is an independent set in $G-E(P_i)$ and $|E(P_i)|=4$. Then $\partial_G(I(P_i))=3|I(P_i)|-2s=3|J(P_i)|-2t=\partial_G(J(P_i))$. If $|I(P_i)|\geq |J(P_i)|+4$, then $s\geq 6+t$, which is impossible since $s\leq 4$. Since $|I(P_i)|>|J(P_i)|$, $|I(P_i)|= |J(P_i)|+2$ and $s=t+3$. Hence either $s=3$ or $s=4$, and thus either $|V(P_i)\cap I(P_i)|=4$ or $|V(P_i)\cap I(P_i)|=5$, respectively. In the former case, $J(P_i)$ is an independent set, and $G-E(P_i)$ is bipartite. In the latter case, $J(P_i)$ induces a graph with one edge, denoted by $r_i$.

Consider now the case when $|V(P_i)\cap I(P_i)|=5$.
Note that the end-vertices of $r_i$ belong to a different partition of $G-(E(P_i)\cup\{r_i\})$ than the vertices of $P_i$.

\smallskip
\noindent
\textbf{Claim 1.} If $|V(P_i)\cap I(P_i)|=5$, then $r_i$ is not adjacent to any edge of $E(P_i)$, for $i\in\{1,2\}$.
\medskip

Let $P_1=v_1v_2v_3v_4v_5$, and let $v_2v'_2$ and $v_4v'_4$ be edges of $G-E(P_1)$. If $|V(P_1)\cap I(P_1)|=4$, then assume, without loss of generality, that $V(P_1)\cap I(P_1)=\{v_1,v_2,v_3,v_4\}$, and let $X_1=\{v_2v'_2,v_3v_4\}$.
Note that $G-X_1$ is bipartite with one of the partition sets being $J(P_1)\cup\{v_2\}$. If $|V(P_1)\cap I(P_1)|=5$, then
let $X_1=\{v_2v'_2,v_4v'_4,r_1\}$.
The graph $G-X_1$ is bipartite with one of the partition sets being $J(P_1)\cup\{v_2,v_4\}$. Moreover, all end-vertices of the edges from $X_1$ belong to the same partition of $G-X_1$.

The same argument holds for $P_2$.
Let $P_2=w_1w_2w_3w_4w_5$, and let $w_2w'_2$ and $w_4w'_4$ be edges of $G-E(P_2)$. Moreover, let $X_2=\{w_2w'_2,w_4w'_4\}$ if $|V(P_2)\cap I(P_2)|=4$ and let $X_2=\{w_2w'_2,w_4w'_4,r_2\}$ if $|V(P_2)\cap I(P_2)|=5$. We obtain the following claims.

\smallskip
\noindent
\textbf{Claim 2.}
If $|V(P_i)\cap I(P_i)|=4$, then the graph $G-X_i$ is bipartite, for $i\in\{1,2\}$.
\smallskip

\smallskip
\noindent
\textbf{Claim 3.}
If $|V(P_i)\cap I(P_i)|=5$, then the graph $G-X_i$ is bipartite. All end-vertices of the edges from $X_i$ connect vertices from the same partition of $G-X_i$, for $i\in\{1,2\}$.
\smallskip

We consider four cases.
\smallskip

\textbf{Case 1.} $|V(P_1)\cap I(P_1)|=4$ and $|V(P_2)\cap I(P_2)|=4$.
\\ The paths $v_1v_2v_3v_4$ and $w_1w_2w_3w_4$ fulfil assumptions of Theorem~\ref{thm5}. It follows that the paths $v_1v_2v_3v_4$ and $w_1w_2w_3w_4$ are in the position $\mathcal{P}_1$.
\smallskip

\textbf{Case 2.} $|V(P_1)\cap I(P_1)|=4$ and $|V(P_2)\cap I(P_2)|=5$.
\\By Claims 2 and 3 and by Lemma~\ref{lemaFarb}, $X_1\Delta X_2$ is an edge-cut of $G$. Since $G$ is cyclically $7$-edge-connected and cubic, $|X_1\Delta X_2|\neq 1$. Thus, either $|X_1 \Delta X_2|=3$ or $|X_1\Delta X_2|=5$.

Assume first that $|X_1\Delta X_2|=3$. Due to cyclic edge-connectivity of $G$, one side of $G-(X_1\Delta X_2)$ must be a vertex $a$. Up to symmetry of $w_2w'_2$ and $w_4w'_4$, there are four options: $v_3v_4=w_2w'_2$, $v_2v'_2=w_2w'_2$, $v_3v_4=r_2$, $v_2v'_2=r_2$. In the first two cases, by Claim~1, $a=w'_4$. By Lemma~\ref{lemaCyklicka}, none of $\{w_1,w_2,w'_2,w_3,w_5\}$ is incident with an edge from $X_1\Delta X_2$. Since one of the edges from $\{v_3v_4,v_2v'_2\}$ is an edge from $X_1\Delta X_2$ while the other one equals $w_2w'_2$, the set of edges $\{v_2v'_2,v_2v_3,v_3v_4,w_2w_3,w_3w_4,w_4w'_4\}$ induces a short cycle, contradicting Lemma~\ref{lemaCyklicka}.
In the latter two cases, since $w_2w'_2$ are $w_4w'_4$ are both incident with $a$, there is a short cycle as well, a contradiction with Lemma~\ref{lemaCyklicka}.

Assume now that $|X_1\Delta X_2|=5$. Due to cyclic edge-connectivity of $G$, one side of $G-(X_1\Delta X_2)$ must induce a path $abc$ on three vertices $a,b,c$. Note that $w_2w'_2$ and $w_4w'_4$ cannot be incident either with the same vertex or with two adjacent vertices, as otherwise $G$ would contain a $4$-circuit or a $5$-circuit, respectively, a contradiction with Lemma~\ref{lemaCyklicka}. Hence, without loss of generality, we may assume that $w_2w'_2$ is incident with $a$ and $w_4w'_4$ is incident with $c$. By Claim 1, $w_2\neq a$ and $w_4\neq c$. We conclude that the path $P_2$ together with edges $ab$ and $bc$ forms either a $4$-circuit or a $6$-circuit, a contradiction.
\smallskip

\textbf{Case 3.} $|V(P_1)\cap I(P_1)|=5$ and $|V(P_2)\cap I(P_2)|=4$.
\\This case is equivalent to Case~2.
\smallskip

\textbf{Case 4.} $|V(P_1)\cap I(P_1)|=5$ and $|V(P_2)\cap I(P_2)|=5$.
\\By Claim~3 and by Lemma~\ref{lemaFarb}, $X_1\Delta X_2$ is an edge-cut. Since $G$ is cyclically $7$-edge-connected and cubic, $|X
_1\Delta X_2|\neq 2$. Thus, either $|X
_1\Delta X_2|=4$ or $|X
_1\Delta X_2|=6$.

Assume first that $|X_1\Delta X_2|=4$. Due to cyclic edge-connectivity of $G$, one side of $G-(X_1\Delta X_2)$ contains a single edge $ab$. If $v_2v'_2$ ($w_2w'_2$, respectively) is incident with $a$ or $b$, then $v_4v'_4$ ($w_4w'_4$, respectively) is incident neither with $a$ nor with $b$ and vice versa, since otherwise $G$ contains a cycle of length at most 5, contradicting Lemma~\ref{lemaCyklicka}. Hence, we may assume, without loss of generality, that $X_1\Delta X_2=\{v_2v'_2, w_2w'_2,r_1,r_2\}$. Thus $v_4v'_4=w_4w'_4$.
Let $V_1^{X_1}$ and $V_2^{X_1}$ ($V_1^{X_2}$ and $V_2^{X_2}$, respectively) be partitions of $G-X_1$ ($G-X_2$, respectively) such that $a\in V_1^{X_1}$ ($a\in V_2^{X_2}$, respectively). Let $V_1$ and $V_2$ be a partition of $V(G)$ guaranteed by Lemma~\ref{lemaFarb}. Note that $E(V_1,V_2)=X_1\Delta X_2$ and $V_1=\{a,b\}$. Thus $b\in V_2^{X_1} \cup V_1^{X_2}$. Since $a$ and $b$ belong to different partitions of $G-X_1$, by Claim~3, it follows that $v_2v'_2$ and $r_1$ are adjacent to the same vertex, say $a$. Then $w_2w'_2$ and $r_2$ are adjacent to $b$. By Claim~1, $a=v'_2$ and $b=w'_2$. Let $r_1:=aa'$ and $r_2=bb'$.
Let $U:=\{v_1,v_2,v_3,w_1,w_2,w_3, a,b,a',b'\}$.
Note that the vertices of $U$ must be all distinct by Lemma~\ref{lemaCyklicka}. Moreover, again by Lemma~\ref{lemaCyklicka}, $v_4\notin U$ and $w_4\notin U$. However, it is possible that $v_4=w_4$.

Assume first that $v_4\neq w_4$. Since $v_4v'_4=w_4w'_4$, $v_4=w'_4$ and $w_4=v'_4$. By definition of $v'_4$ and $w'_4$ and by Lemma~\ref{lemaCyklicka}, $v_5, w_5\notin U\cup\{v_4,w_4\}$ and $v_5\neq w_5$. Colour the vertices according to the partitions $V_1^{X_1}$ and $V_2^{X_1}$. Lemma~\ref{lemaFarb} allows us to determine the colouring of the vertices from $U\cup \{v_4,v'_4,v_5,v'_5\}$. This shows that $P_1$ and $P_2$ are in the position $\mathcal{P}_2$.

Assume now that $v_4=w_4$. Thus $v_4$ is incident with $v_3$ and $w_3$. Since $v_4v'_4=w_4w'_4$ and by definition of $v'_4$ and $w'_4$, $v_4v'_4\notin P_1\cup P_2$ and hence $v'_4=w'_4$ is the third vertex incident with $v_4$. Therefore, $v_5=w_3$ and $w_5=v_3$. Colour the vertices according to the partitions $V_1^{X_1}$ and $V_2^{X_1}$. Lemma~\ref{lemaFarb} allows us to determine the colouring of the vertices from $U\cup \{v_4,v'_4\}$. This shows that $P_1$ and $P_2$ are in the position $\mathcal{P}_3$.

It remains to consider that $|X_1\Delta X_2|=6$.  Due to cyclic edge-connectivity of $G$, one side of $G-(X_1\Delta X_2)$ is either a path on $4$ vertices or a star with three pendant vertices or two isolated vertices.

Let one side of $G-(X_1\Delta X_2)$ be a path on $4$ vertices $a$, $b$, $c$ and $d$. By Lemma~\ref{lemaCyklicka}, $v_2v'_2$ and $v_4v'_4$ are not adjacent. By Claim~3, $v_2v'_2$, $v_4v'_4$ and $r_1$ must be incident with $a$ and $c$, or with $b$ and $d$, say, without loss of generality, $v_2v'_2$ and $r_1$ are incident with $a$ and $v_4v'_4$ is incident with $c$. By Claim~1, $a\neq v_2$. This is a contradiction with Lemma~\ref{lemaCyklicka}, since $\{v_2,v_3,v_4,a,b,c\}$ induces a short cycle.

Let one side of $G-(X_1\Delta X_2)$ be a star with a central vertex $a$ and three pendant vertices $b,c$ and $d$. Note that by Lemma~\ref{lemaCyklicka}, $v_2v'_2$ and $v_4v'_4$ as well as $w_2w'_2$ and $w_4w'_4$ cannot be adjacent. Let, without loss of generality, $v_2v'_2$ and $w_2w'_2$ be incident with $b$, let $v_4v'_4$ be incident with $c$. Note that $a=v_3=w_3$ by Lemma~\ref{lemaCyklicka}, because $v_2v'_2$ and $v_4v'_4$ ($w_2w'_2$ and $w_4w'_4$, respectively) belong to a path of length 4. Thus $b=v_2=w_2$, $c=v_4$, $v_1=w'_2$ and $w_1=v'_2$.

Let us first assume that $w_4w'_4$ is incident with $c$. Thus $c=w_4$,  $v_5=w'_4$ and $w_5=v'_4$. Moreover, two neighbours of $d$ different from $a$ are distinct from vertices $v_1,v_2,v_3,v_4,v_5,w_1,w_5$. Since $G-X_1$ is bipartite, it follows that $P_1$ and $P_2$ are in the position~$\mathcal{P}_4$.

Let us now assume that $w_4w'_4$ is not incident with $c$, then $d=w_4$, $r_2=v_4v_5$ and $r_1=w_4w_5$. Since $G-X_1$ is bipartite, it follows that $P_1$ and $P_2$ are in the position~$\mathcal{P}_5$.

Finally, let one side of $G-(X_1\Delta X_2)$ be two isolated vertices $a$ and $b$. By Lemma~\ref{lemaCyklicka}, $v_2v'_2$ and $v_4v'_4$ as well as $w_2w'_2$ and $w_4w'_4$ cannot be adjacent. Thus, without loss of generality, we may assume that $v_2v'_2$, $w_2w'_2$ and $r_1$ are incident with $a$ while $v_4v'_4$, $w_4w'_4$ and $r_2$ are incident with $b$. By Claim~1, $a\neq v_2$ and $b\neq w_4$. Let $r_1=aa'$ and $r_2=bb'$. We distinguish cases according to positions $v_4$ and $w_2$.

First, assume that $w_2\neq a$ and $v_4\neq b$. Then $v_3$, which is a neighbour of $v_2$ and $v_4$, is different from $a=v'_2$ and $b=v'_4$. Similarly, $w_3$, which is a neighbour of $w_2$ and $w_4$, is different from $a=w'_2$ and $b=w'_4$. Moreover, $v_1$, $w_1$, $v_5$ and $w_5$ is the third vertex adjacent to $v_2$, $w_2$, $v_4$ and $w_4$, respectively. Due to Lemma~\ref{lemaCyklicka}, the vertices $v_1,v_2,v_3,v_4,v_5,w_1,w_2,w_3,w_4,w_5, a,a',b,b'$ are all distinct. Let $V_1^{X_1}$ and $V_2^{X_1}$ be the partitions of $G-X_1$ such that $a\in V_1^{X_1}$. By Claim~3, this is the position $\mathcal{P}_6$.

Second, assume that $w_2\neq a$ and $v_4=b$. Then $w_3$, which is a neighbour of $w_2$ and $w_4$, is different from $a=w'_2$ and $b=w'_4$. Note that $b=v_4=w'_4$ is adjacent to $v'_4,w_4$ and $b'$. The vertex, $v_3$, which is a neighbour of $v_2$ and $v_4$, is different from $v'_4$ by the definition of $v'_4$. Moreover, $v_3$ is different from $w_4$, as otherwise $G$ contains a $5$-cycle $v_2v'_2w_2w_3w_4v_2$, contradicting Lemma~\ref{lemaCyklicka}. Thus, $v_3=b'$ and therefore $v_5=w_4$. Note that by Lemma~\ref{lemaCyklicka}, $v_1,v_2,v'_4,w_1,w_2,w_3,w_4,w_5,a,a',b,b'$ are distinct vertices. Let $V_1^{X_1}$ and $V_2^{X_1}$ be the partitions of $G-X_1$ such that $a\in V_1^{X_1}$. By Claim~3, this is the position $\mathcal{P}_7$.

The case when $w_2=a$ and $v_4\neq b$ is similar to the previous one. This yields the position $\mathcal{P}_7$ of the paths.

Finally assume that $w_2=a$ and $v_4=b$. Since $v_3$ is a neighbour of $v_2$ and $v_4$, and by Lemma~\ref{lemaCyklicka}, either $v_2w_4\in E(G)$ or $v_2b'\in E(G)$. Similarly, since $w_3$ is a neighbour of $w_2$ and $w_4$, and by Lemma~\ref{lemaCyklicka}, either $v_2w_4\in E(G)$ or $w_4a'\in E(G)$. Therefore, either $v_2w_4\in E(G)$ or $\{v_2b',w_4a'\}\subseteq E(G)$. Since $v_2aa'w_4bb'v_2$ would be a $6$-cycle, contradicting Lemma~\ref{lemaCyklicka}, $v_2w_4$ must be an edge of $G$. Thus $w_3=v_2$ and $v_3=w_4$. Consequently $a'=w_1$ and $b'=v_5$. By Lemma~\ref{lemaCyklicka}, $v_1,v_2,v_3,v_4,v_5,v'_4,w_1,w_2,w_5,w'_2$ are all distinct. Let $V_1^{X_1}$ and $V_2^{X_1}$ be the partitions of $G-X_1$ such that $a\in V_1^{X_1}$. By Claim~3, this is the position $\mathcal{P}_8$.
\smallskip

For the converse, if any subpaths of length 3 of $P_1$ and $P_2$ are in the position $\mathcal{P}_1$, by Theorem~\ref{thm5} $G$ does not have a $2$-factor containing either of $P_1$ and $P_2$. If the paths $P_1$ and $P_2$ are in the position $\mathcal{P}_j$, for $j\in\{2,3,4,5,6,7,8\}$, then we show that $G-E(P_1)$ and $G-E(P_2)$ contain large independent sets, contradicting Theorem~\ref{thm1}. Consider the vertex colouring depicted in Figure~\ref{fig34}. Note that $G-E(P_1)$ contains a large independent set formed by white vertices if we recolour the vertices with label $1$ to the opposite colour. Similarly, if we recolour the vertices with label $2$ to the opposite colour, we obtain a large independent set in $G-E(P_2)$ formed by white vertices in the positions   $\mathcal{P}_2$ and $\mathcal{P}_3$, and by black vertices in the remaining positions.
`\end{proof}

Let $v$ be a vertex of a cubic cyclically $7$-edge-connected graph. Then there exist two different paths on $4$ edges with one end-vertex at $v$ and with the other end-vertex at a vertex of distance $4$ from $v$. Hence Theorem~\ref{thm6} has the following corollary.

\begin{corollary*}\textbf{\ref{cor7}.}
Let $G$ be a cubic cyclically $7$-edge-connected graph and let $v\in V(G)$. Then there exists a $2$-factor of $G$ such that $v$ does not belong to a $7$-circuit.
\end{corollary*}

\section*{Acknowledgements}
\noindent The first author acknowledges support from the
VEGA grant No. 1/0876/16, and from the project APVV-15-0220. The work of the second author was partially supported by the
project GA17-04611S of the Czech Science Foundation and by the project
LO1506 of the Czech Ministry of Education, Youth and Sports.

redpajama_set_name: RedPajamaArXiv


Real time news for people who live, work or play in the 'celebrity playground' of WeHo & nearby areas of Beverly Hills & Hollywood. These tweets are not funny.
The web's leading Gay Adult entertainment company. Official Next Door Studios Twitter account.
Currently fighting for the right to party! I scream, sing and twerk in @BUTCHERBABIES. Y'all need Lilith! ??? Click that Link Bruh!


text:
Hawaii Football
Hawaii football: An early look at Wisconsin
Does Hawaii have a chance against Wisconsin? Probably not.
By DontWarriorBoutIt@warriorboutit Aug 20, 2015, 1:29pm PDT
Share All sharing options for: Hawaii football: An early look at Wisconsin
Jeff Hanisch-USA TODAY Sports
In week 4 Hawaii will fly on over to Madison, Wisconsin, to take on the Wisconsin Badgers. The 'Bows and the Badgers have met six times before with Wisconsin holding a 5-1 series edge with an average score of 38-14. The last meeting, a 51-10 ass-whoopin' at Aloha Stadium, ended the Warriors' bid to go bowling in '09. Just as before, this one looks to be a blowout by the Badgers.
On offense Wisconsin returns senior quarterback Joel Stave who passed for 2,494 yards with 22 touchdowns as a sophomore in 2013. Stave would go on to lose the starting job early on in 2014 to Tanner McEvoy when he showed up to camp and inexplicably lost the ability to throw. Stave would serve as the backup for the first five games of the season before being handed the reins again in October and would go on to finish with 1,350 yards, nine touchdowns and 10 interceptions.
Despite his mediocre final line he was solid over his seven regular season starts, throwing for 928 yards with 7 touchdowns and just one interception before getting absolutely stifled by Ohio State in the Big Ten championship game. With Tanner McEvoy being converted to safety, Stave seems to be the unquestioned starter heading into 2015.
As for the rest of the offense:
At running back the Badgers return junior Corey Clement. Clement rushed for 949 yards on 147 attempts and 9 touchdowns while backing up Heisman finalist Melvin Gordon. The Badgers also feature a trio of highly touted freshmen in Taiwan Deal, Bradrick Shaw, and Jordan Stevenson.
At receiver Alex Erickson returns for his senior season after catching 55 passes for 772 yards and 3 touchdowns. He is joined by fellow senior Jordan Fredrick who put up 126 yards on 13 catches in 2014.
At tight end sophomore Troy Fumagalli should take on the starting role after catching 14 passes for 187 yards in his freshman campaign.
On the offensive line Wisconsin returns only two starters from 2014; center Dan Voltz and left tackle Tyler Marz.
On defense the Badgers return seven of their starters from a unit that ranked fourth in total defense a year ago. Last year the Badgers also ranked 4th in passing defense, 23rd in rushing defense, 17th scoring defense, and 3rd in third down percentage.
The rest of the defense:
At defensive end Wisconsin returns Chikwe Obasih and former starting noseguard, now defensive end, Arthur Goldberg.
Both starting outside linebackers return in Joe Schobert (69 tackles 13.5 for loss, 3 sacks, 2 forcedfumbles, one fumble recovery, 7 pass breakups) and Vince Biegel (56 tackles, 16.5 for loss, 7.5 sacks)
The secondary returns cornerbacks Sojourn Shelton (33 tackles, 1 for loss, 6 pass breakups) and Second team All-Big Ten selection Darius Hillary (41 tackles, 3 for loss, 5 pass breakups) as well as Second team All-Big Ten safety Michael Caputo (106 tackles, 6 for loss, 1 sack, 6 pass breakups, 1 interception)
Wisconsin, again, looks like a team to be reckoned with in the Big Ten. In their preseason power rankings for the Big Ten ESPN placed Wisconsin third in the conference and hailed them as the "West Division favorite." So does Hawaii really stand a chance in this contest? Maybe if Joel Stave suddenly forgets how to throw a football again. But even then probably not.
Wisconsin: A lot
Hawaii: A Little

redpajama_set_name: RedPajamaCommonCrawl


Like I said in an earlier post, Kathleen has announced that she'll release a new polish every year for her birthday, and she kept to her word this year by releasing Das Esspensive XXVI for her 26th birthday. So let's take a look at this new iteration on a birthday theme.
As you can see from the above photo, it's a pretty light color. I'd consider it more of a nail polish topper than a polish that could be worn as a polish on its own. Below, I used it as a nail topper over Miss Honey.
So this polish came off way easier than the original Das Esspensive. It peeled off almost like a gel. And it chipped almost instantly. Ugh. As a topper is stayed on much longer, so I'll probably just use it for that going forward.
Let me know what you think of this shade if you've gotten a chance to pick it up!


Novenas: December 16 to 24, we get together with family and friends every night to pray and sing Christmas songs to baby Jesus. And, of course, we eat our Christmas treats (Natilla, Buñuelos y Hojuelas).
El niño Jesus: Baby Jesus is the one who brings presents to Colombian kids, not Santa Claus, maybe it's too far for him 🙂 . On December 24 baby Jesus comes at 12:00 am and leaves presents around the kid's beds and the next day the kids find them.
Noche Buena: On December 24th, families get together for Noche Buena, our Christmas Eve gathering. There is a lot of food, dancing and presents.
Fiesta de Fin de Año: On December 31st families get together to celebrate the end of the year, with food, dancing and fireworks. In my family it was a special day because it was also my grandmother's birthday.
Place the flour and sugar in a medium bowl and using an electric hand mixer, mix well. Add the butter and continue mixing at low speed for about 20 seconds.
Add the orange zest and start adding the orange juice. Keep mixing at low speed while you are adding the juice.
Once the mixture has pulled together, mix at a slightly higher speed to fully incorporate the ingredients.
Roll the dough into ball, wrap in plastic wrap, and chill for about 20 minutes.
Sprinkle some flour onto your work surface and roll the dough to 1/4-inch thick. Cut into 1 x 3 ½ inches strips. Add more flour to the work surface if necessary and again roll each strip to about 1/8 inch thick.
Place a heavy pot over medium heat and fill it about one-third of the way with vegetable oil. Heat until a deep-fry thermometer registers 340°F.
Add the pieces and fry until golden brown. Using a slotted spoon transfer to a plate lined with paper towel. Sprinkle with sugar and serve warm.
These look great! One of my great aunts used to make something similar during holiday time. I know in Italian they are called zeppole! Whatever they are called in any language, they are GOOD!
It's great learning some of your traditions. Sometimes these simple desserts are the best.
In Colombia we start on December 7th with "El Dia de las Velitas" prior to Dec. 8th The Inmaculate Conception catholic holiday.
Lovely post..great to know about christams time in Colombia :)..
I'd be popular if I made this on Christmas morning…. I should save the recipe!
I'm from Tunja, Colombia in the Boyaca department. I still remember that the one that was bringing presents to us was Papá Noel, not baby Jesus. Maybe that is another part of Colombia. After "El Dia de las Velitas" family and neighbors get together for the Novena that Erica already explained up above.
there is an italian christmas treat that is exactly the same and the name escapes me right now but I grew up eating them…since then I found out there are varieties of the same in other european countries as well…shoot…it bums me out the name isn't coming to me…I'm gonna wait to send this for a minute.
oh wow i love that Jesus brings the gifts thats spot on, send me a few of these babies!!
There would be no way that I could eat just a few of these-I would eat them all. Fried pastry is just so darn good. Love it.
Such fun to hear about all the Christmas traditions! These cookies look delicious!
I was also born in Colombia. However, I came to the U.S at the age of 7. I do remember all these wonderful traditions and I'm very fortunate to have my entire family here in the States. Actually, I was remembering that they were my favorite and yes, my Mamita made them for me as well. WONDERFUL BLOG!
I enjoyed not only the recipe but the back ground on your cultural and family traditions as well.
Thanks for sharing your wonderful traditions with us! And the hojuelas look scrumptious, I would love to try those!
I loved reading about your traditions! These look like a delicious treat!
What beautiful family traditions Erica! I especially love the novenas, what a beautiful time that must be 🙂 Thanks for sharing!!
It's great to know your family traditions in Columbia. I agree with Baby Jesus more than Santa Claus. Thanks Erica.
My apartment is freezing. Please pass me a plate of hojuelas and hot chocolate.
It is great to find these recipes with background of traditions; novenas were a beautiful time, and yes! These are great with "cafecito". All this is just about baby Jesus, Let's rejoice and shine the love of Jesus! Thank you, Erica.
Thank you guys for all the wonderful comments! Happy holidays!
How many hojuelas does this recipe make?
Hi Erica! These look so great!!! Thanks for posting this, I will give them a try and let you know how they turned out. I also just started my own blog and would love it if you could visit it worldof-jackie.blogspot. Thanks!
I made this 4 times last year for Christmas! Everyone I served it to adored it…but none more that my family. Easy to make, pretty to serve and delicious!
My aunt used to make something called Hojaldras. They seem similar, but I feel like she put cheese in them and did not top with powdered sugar. In fact I used to put butter on them! Know anything about this?? She was from Cali.
I was born in Colombia and grew up in Bogota, we also had our presents delivered by Baby Jesus… but we also did the novenas in front of the manger until January 6th or 7th, when the three kings reached the baby. Some wonderful days. I am very happy to see this recipe, at home we called them arepuelas.
While made from scratch are ALWAYS the best, my mom would use flour tortillas and fry them as a shortcut.
Nostalgia. Siento mucha nostalgia en este tiempo, más de 24 años que no pasó fiestas de fin de año en mi tierra. Gracias por tus deliciosas recetas, muy fáciles de seguir!
Hi I'm making this recipe for my Spanish project and I wanted to know how many it made.
Me morí! Por fin una receta sin huevo! Mil gracias! Soy de Medellín y vivo en Alemania hace unos eternos 13 años, y cada diciembre me carcome la nostalgia y me pongo a hacer natilla, bandeja paisa, y ahora POR FIN HOJUELAS! Soy vegana, y por eso es difícil encontrar recetas veganas de comida tradicional colombiana… Aún no he intentado hacer buñuelos con queso falso, pero lo intentaré este año! Mil gracias por compartir la receta!


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We can also assist Tenants in ensuring Landlord consent is obtained and also Contract Administrate the works on site.


The European Union (EU) Trade Commissioner, Karel De Gucht, and Singapore's Minister for Trade and Industry, Lim Hng Kiang, have officially launched the first round of negotiations for a free trade agreement (FTA) between the EU and Singapore.
It was said that the EU-Singapore FTA (EUSFTA) marks a milestone in the EU's longstanding relations with Singapore. De Gucht was of the opinion that "the proposed free-trade agreement will strengthen economic ties between Singapore and the EU, provide new opportunities for traders and consumers alike, and contribute to generating growth in our economies."
Lim Hng Kiang added: "The EU is an important economic partner of Singapore. This FTA is a major undertaking and when concluded, will further strengthen the excellent economic relations between Singapore and the EU. Singapore looks forward to working closely with the EU to develop an ambitious agreement that will benefit businesses from both sides."
The first round of negotiations is scheduled to take place from March 8-12, 2010, in Singapore. At the end of their meeting, Lim Hng Kiang and Karel De Gucht expressed the commitment of both Singapore and the EU to work towards an expeditious conclusion of a high-quality FTA.
Singapore and the EU enjoy close economic ties that, it is hoped, will be boosted further by the EUSFTA. The EU is Singapore's largest trading partner and foreign investor.
Bilateral trade exceeded EUR55bn (USD39bn) in 2008. Conversely, Singapore is the EU's most important trading partner in the Asian region and, particularly, its largest trading partner among the ten countries of the Association of Southeast Asian Nations (ASEAN) that, overall, have an annual bilateral trade in goods and services of some EUR175bn with the EU.
Both sides also enjoy robust bilateral investment ties. Over the years, Singapore and the EU have invested over EUR100bn in each other's economies. Already, it was added, EU companies in many sectors have chosen Singapore as a hub from which to serve the Pacific Rim region.
EU had pursued talks on a region-to-region FTA with ASEAN from 2007 until their pause in March 2009. Later last year, the EU member states gave the green light to pursue negotiations towards FTAs with the individual countries within ASEAN.
While opening FTA negotiations with Singapore, De Gucht disclosed that the EU's ultimate goal is still to achieve an agreement within a regional framework; the bilateral agreements with single ASEAN countries becoming building blocks towards that objective.
"Although Singapore is the 'first one in,' our door remains open for other ASEAN countries interested in negotiating a comprehensive free trade agreement with us," he stated. While previously in Hanoi, he had been able to announce that Vietnam and the EU would also start negotiations on a bilateral agreement.