WordWalk Puzzles

WordWalk Puzzles for 2023-01-28 at https://wordwalk.net Yesterday's puzzles: English rootword was "GRADUALLY". German rootword was "ZWEIFLER". Spanish rootword was "DESCUIDA". Kids rootword was "HUMAN". #wordwalkpuzzles   Here's the solution for yesterday's English puzzle: First subword,  "DRUG":  Second and Third subwords, "DAY" and "LAY": After final subword, "GUY":   Just before placing the final letter of the rootword, "GRADUALLY":       Upon completion:   wordwalk.net  

Today's German puzzle is tough, but you can logically eliminate some stuff by just looking at the arrows and the subwords, before even trying to place a subword. Doing that led me to try the subword 'HIER' first. Das heutige deutsche Puzzle ist schwierig, aber Sie können einige Dinge logischerweise eliminieren, indem Sie sich nur die Pfeile und die Teilwörter ansehen, bevor Sie überhaupt versuchen, ein Teilwort zu platzieren. Das führte dazu, dass ich zuerst das Teilwort „HIER“ versuchte.  

Currently, WordWalk starts you off with 0 points and then you get 10 points for each correctly placed subword and an extra 100 for the rootword (which also counts a subword so, in fact, you 110 points for the rootword).  So, for a five-subword puzzle you can get a maximum of 160 points. Moving forward (possibly starting tomorrow) the points system will be different.  You will start out with a certain number of points (say, 100, for discussion purposes).  Then you will get 10 points added to your points stash for each new correctly guessed letter and also 10 points for each new correctly guessed arrow.  These points will not be added, however, until after you've completely and successfully placed the subword in the puzzle graph.  Once that is done you will get some extra points (say, 25) for the subword itself and the new points for new letters and arrows that were obtained while placing the subword. If you fail while trying to place a subword, you won't lose rem...

In mathematical graph theory, a directed graph is a structure consisting of two sets, a set of vertices and a set of edges.  (In WordWalk we sometimes call these nodes and arrows.)   For any given vertex, we can ask how many edges lead into the vertex and how many lead away from the vertex.  These two numbers have special names in graph theory: in-degree of a vertex = number of edges (arrows) pointing toward the vertex out-degree of a vertex = number of edges pointing away from the the vertex.  In WordWalk, knowing these two numbers can help you make intelligent guesses regarding the likelihood of a given unknown node being one letter vs another. For instance, the letter E, is very common, and if you look at all the subwords that contain E followed by another letter that should be approximately one less than the out-degree of that node (one less because it may also happen in the unknown rootword).  Likewise, the number of times E is preceded by some other letter...

Sometimes none of the subwords begin with the same letter as the rootword. Consider this puzzle:     Here the rootword begins with P, but none of the subwords do.  What to do? You'll have to use intelligent guessing that eliminates possibilities.  Suppose we try for the first subword "ITS".  Since I is a vowel, it will have to go into one of the two diamond shapes.   But which one? Sometimes it is useful to count the number of arrows going into and out of each shape.  The top left diamond has seven such arrows while the right diamond has 6.  Since I is more common than the other vowel, it is more likely that the left one takes the I and the right one the U.  That's the correct guess, and so we now get: Now, where does the T go?  T follows I in at least two of the subwords and I follows T in at least of the subwords.  So, we can expect an arrow coming from the I to the T and vice versa, which mean we can certainly rule out...

 Here's a snippet from today's English word.  I'm trying to find the word "COIN" as the first word. I attempted this word first, because I was given the "C" as the first letter of the rootword, but then the "O" was the only diamond shape (i.e., a vowel) that "C" pointed to.   So that had to be second.  By a process of elimination, however, the only remaining diamond shape had to be the only remaining vowel, "I".   I've just entered the "I" but now need to decide which of the blank ellipses should get the "N".  It can't be the lower left one, because there isn't an arrow to it from "I".  Same is true for the rightmost ellipse.  Hence it must be the lower right one. Indeed, this is so: Then I'd guess the rightmost blank bubble must be "K", since it points to "N" and "KNOCK" is a word.  So that should be easy to get.  Again by a process of eliminati...

It is important for two reasons (but perhaps a little tedious) to find all of the subwords before guessing the rootword.   The first reason is that (as of the current implementation) you get an extra 10 points for each subword correctly guessed.   The other reason is that you can now take advantage of the arrows that have not yet been used by any of the subwords thus far and hence must be arrows belonging to the rootword.  Two nodes connected by an arrow indicates that the digram (two consecutive letters in a word) must be present somewhere in the rootword.  This can give you a major clue as to what the rootword is. Consider the following German WordWalk Puzzle in which all of the subwords have been found but the rootword still needs to be guessed:   You'll notice that there is only one thin black arrow that has not yet turned green. Otherwise, all the nodes and edges (aka arrows) have been traversed at least once.  This tells us that the digram "EN" m...