nyuuzyou/wikis huggingface dataset:
Henry Ernest Dudeney/Modern Puzzles/157 - Crossing the Lines

== by : $157$ ==
;Crossing the Lines
:There is a little puzzle about which, for many years, I have perpetually received enquiries as to its possibility of solution.
:You are asked to draw the diagram of Figure $1$ (exclusive of the little crosses) with three continuous strokes of the pencil,
::without removing the pencil from the paper during a stroke, or going over a line twice.
:As generally understood, it is quite impossible.
:Wherever I have placed a cross there is an "odd node", and the law for all such cases is that half as many lines will be necessary as there are odd nodes --
::that is, points from which you can depart in an odd number of ways.
:Here we have, as indicated, $8$ odd nodes, from each of which you can proceed in three directions (an odd number),
::and therefore, four lines will be required.
:But, as I have shown in my book of , it may be solved by a trick, overriding the conditions as understood.
:You first fold the paper, and with a thick lead-pencil draw $CD$ and $EF$, in Figure $2$, with a single stroke.
:Then draw the line from $A$ to $B$ as the second stroke, and $GH$ as the third!
:
:During the last few years this puzzle has taken a new form.
:You are given the same diagram and asked to start where you like and try to pass through every short line comprising the figure,
::once and once only, without crossing your own path.
:Figure $3$ will make quite clear what is meant.
:It is an attempted solution, but it fails because the line from $K$ to $L$ has not been crossed.
:We might have crossed it instead of $KM$, but that would be no better.
:Is it possible?
:Many who write to me about the puzzle say that though they have satisfied themselves as a "pious opinion", that it cannot be done,
::yet they see no way whatever of proving the impossibility, which is quite another matter.
:I will show my way of settling the question.
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $157$. -- Crossing the Lines
* : Combinatorial & Topological Problems: Route & Network Puzzles: $414$. Crossing the Lines
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/157 - Crossing the Lines/Solution

== by : $157$ ==
;Crossing the Lines
== Solution ==
The solution is straightforward by Characteristics of Traversable Graph, but it is interesting to see how wordy could be:
:Let us suppose that we cross the lines by bridges, represented in Figure $1$ by the little parallels.
:
:Now, in Figure $2$, I transform the diagram, reducing the spaces $A$, $B$, $C$, $D$, $E$ to mere points,
::and representing the bridges that connect these spaces by lines or roads.
:This transformation does not affect the conditions,
::for there are $16$ bridges or roads in one case, and $16$ roads or lines in the other,
::and they connect with $A$, $B$, $C$, $D$, $E$ in precisely the same way.
:It will be seen that $9$ bridges or roads connect with the outside.
:Obviously we are free to join these up in pairs in any way we choose, provided the roads do not cross one another.
:The simplest way is shown in Figure $3$, where on coming out from $A$, $B$, $C$, or $E$,
::we immediately return to the same point by the adjacent bridge, leaving one point, $X$, necessarily in the open.
:In Figure $2$ there are $4$ odd nodes, $A$, $B$, $D$, and $X$ (if we decide on the exits and entrances, as in Figure $3$),
::so, as I have already explained, we require $2$ strokes (half of $4$) to go over all the roads,
::proving a perfect solution to be impossible.
:Now, let us cancel the line $AB$.
:Follow the line in Figure $3$ and you will see that this can be done, omitting the line from $A$ to $B$.
:This route the reader will easily transform into Figure $4$ if he says to himself,
::"Go from $X$ to $D$, from $D$ to $E$, from $E$ to the outside and return into $E$," and so on.
:The route can be varied by linking up those outside bridges differently,
::by making $X$ an outside bridge to $A$ or $B$, instead of $D$, and by taking the cancelled line either at $AB$, $AD$, $BD$, $XA$, $XB$, or $XD$.
:In Figure $5$ I make $X$ lead to $B$.
:We still omit $AB$, but we must start and end at $D$ and $X$.
:Transformed in Figure $6$, this will be seen to be the precise example that I gave in the question.
:The reader can now write out as many routes as he likes for himself,
::but he will always find it necessary to omit one line or crossing.
:It is thus seen how easily sometimes a little cunning, like that of the transformation shown,
::will settle a perplexing question of this kind.
== Sources ==
* : Solutions: $157$. -- Crossing the Lines
* : Answers: $414$. Crossing the Lines
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/158 - The Nine Bridges

== by : $158$ ==
;The Nine Bridges
:The diagram represents the map of a district with a peculiar system of irrigation.
:The lines are waterways enclosing the four islands $A$, $B$, $C$, and $D$, each with its house,
::and it will be seen that there are nine bridges available.
:
:Whenever Tompkins leaves his house to visit his friend Johnson, who lives in one of the others,
::he always carries out the eccentric rule of crossing every one of the bridges once, and once only,
::before arriving at his destination.
:How many different routes has he to select from?
:You may choose any house you like as the residence of Tompkins.
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $158$. -- The Nine Bridges
* : Combinatorial & Topological Problems: Route & Network Puzzles: $415$. The Nine Bridges
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/158 - The Nine Bridges/Solution

== by : $158$ ==
;The Nine Bridges
== Solution ==
Tompkins, and likewise Johnson, lives at either $B$ or $D$.
He has $132$ routes to choose from.
== Proof ==
First we transform the map into a graph, by representing each island by a vertex and each bridge by an edge.
The outer mainland is represented by $E$.
The resulting graph is shown below.
:
It remains to count the number of different Eulerian trails.
== Sources ==
* : Solutions: $158$. -- The Nine Bridges
* : Answers: $415$. The Nine Bridges
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/159 - The Five Regiments

== by : $159$ ==
;The Five Regiments
:The diagram represents a map of a certain district.
:The dots and circles are towns and the lines are roads.
:
:During a war five regiments marched to new positions in the night.
:The body stationed at the upper $A$ marched to the lower $A$,
::that at the upper $B$ to the lower $B$,
::that at the upper $C$ to the lower $C$,
::that at the upper $D$ to the lower $D$,
::and the regiment at the left-hand $E$ marched to the right-hand $E$.
:Yet no regiment saw anything of any other regiment.
:Can you mark out the route taken by each so that no two regiments ever go along the same road anywhere?
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $159$. -- The Five Regiments
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/159 - The Five Regiments/Solution

== by : $159$ ==
;The Five Regiments
== Solution ==
:
== Also see ==
* Henry Ernest Dudeney: Puzzles and Curious Problems $270$ - Planning Tours: exactly the same puzzle as this, couched in a different scenario.
== Sources ==
* : Solutions: $159$. -- The Five Regiments
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/15 - Sawing Logs

== by : $15$ ==
;Sawing Logs
:"Your charge," said Mr. Grigsby, "was $30$ shillings for sawing up $3$ cords of wood made up of logs $3$ feet long,
::each log to be cut into pieces $1$ foot in length."
:"That is so," the man replied.
:"Well, here are $4$ cords of logs, all of the same thickness as before,
::only they are in $6$-feet lengths, instead of $3$ feet.
:What will your charge be for cutting them all up into similar $1$-foot lengths?"
:It is curious that they could not at once agree as to the fair price for the job.
:What does the reader think the charge ought to be?
=== Click here for solution ===
== Sources ==
* : Arithmetical and Algebraical Problems: Money Puzzles: $15$. -- Sawing Logs
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/15 - Sawing Logs/Solution

== by : $15$ ==
;Sawing Logs
== Solution ==
$50$ shillings.
== Proof ==
The workman charges for the number of cuts per log.
A cord is a unit of volume.
$4$ cords of $6$-foot logs is the same number of logs as $2$ cords of $3$-foot logs.
The key point is that:
:it takes $2$ cuts to cut a $3$-foot log into $1$-foot pieces
but:
:it takes $5$ cuts to cut a $6$-foot log into $1$-foot pieces.
So the workman should charge:
:$30 \times \dfrac 2 3 \times \dfrac 5 2 = 50 \shillings$
== Sources ==
* : Solutions: $15$. -- Sawing Logs
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/160 - Going to Church

== by : $160$ ==
;Going to Church
:A man living in the house shown as $H$ in the diagram wants to know what is the greatest number of different routes by which he can go to the church at $C$.
:
:The possible roads are indicated by the lines, and he always walks either due $N$, due $E$, or $N.E.$;
::that is, he goes so that every step brings him nearer to the church.
:Can you count the total number of different routes from which he may select?
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $160$. -- Going to Church
* : Combinatorial & Topological Problems: Route & Network Puzzles: $417$. Going to Church
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/160 - Going to Church/Solution

== by : $160$ ==
;Going to Church
== Solution ==
$321$ different routes.
== Proof ==
Starting from $H$, there is one way to get to each of the points in the northerly direction, and one way to those due east.
Taking the second column, there are as many ways to get to each point as the sum of those to each of the points from which you can reach that point directly.
That is, let $R_{m, n}$ be the number of routes to the point in row $m$ and column $n$.
Then:
:$R_{m, n} = R_{m - 1, n} + R_{m, n - 1} + R_{m - 1, n - 1}$
on the understanding that if $n - 1$ or $m - 1$ is outside the grid, its contribution is $0$.
Hence we can build an array of the number of routes, where each number is $R_{m, n}$ as defined:
:$\begin{array}{ccccc}
1 & 9 & 41 & 129 & 321 \\
1 & 7 & 25 & 63 & 129 \\
1 & 5 & 13 & 25 & 41 \\
1 & 3 & 5 & 7 & 9 \\
1 & 1 & 1 & 1 & 1
\end{array}$
and the result follows.
== Also see ==
* Definition:Pascal's Triangle, which is what you get if you can't go diagonally.
== Sources ==
* : Solutions: $160$. -- Going to Church
* : Answers: $417$. Going to Church
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/161 - A Motor-Car Puzzle

== by : $161$ ==
;A Motor-Car Puzzle
:A traveller starts in his car from the point $A$ and wishes to go as far as possible while making only $15$ turnings, and never going along the same road twice.
:The dots represent towns and are one mile apart.
:
:Supposing, for example, that he went straight to $B$, then straight to $C$, then to $D$, $E$, $F$, and $G$,
::then you will find that he has gone $37$ miles in five turnings.
:How far can he go in $15$ turnings?
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $161$. -- A Motor-Car Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/161 - A Motor-Car Puzzle/Solution

== by : $161$ ==
;A Motor-Car Puzzle
:A traveller starts in his car from the point $A$ and wishes to go as far as possible while making only $15$ turnings, and never going along the same road twice.
:The dots represent towns and are one mile apart.
:
:Supposing, for example, that he went straight to $B$, then straight to $C$, then to $D$, $E$, $F$, and $G$,
::then you will find that he has gone $37$ miles in five turnings.
:How far can he go in $15$ turnings?
== Solution ==
The roads unused have been rendered as dotted lines:
:
== Also see ==
* $269$ - Sixteen Straight Runs from his collection, which is a similar problem.
== Sources ==
* : Solutions: $161$. -- A Motor-Car Puzzle
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/162 - The Fly and the Honey

== by : $162$ ==
;The Fly and the Honey
:I have a cylindrical cup four inches high and six inches in circumference.
:On the inside of the vessel, one inch from the top, is a drop of honey,
::and on the opposite side of the vessel, one inch from the bottom on the outside, is a fly.
:Can you tell exactly how far the fly must walk to reach the honey?
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $162$. -- The Fly and the Honey
* : Geometrical Problems: Solid Geometry Puzzles: $325$. The Fly and the Honey
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/162 - The Fly and the Honey/Solution

== by : $162$ ==
;The Fly and the Honey
== Solution ==
:$5$ inches.
== Proof ==
Let $F$ be the starting position of the fly and $H$ be the position of the honey.
:
We unroll the mug and lay it flat so as to see the geometry easily.
The shortest distance between $F$ and $H$ is found by Heron's Principle of Reflection to be via the point $A$.
We extend $FA$ to $B$ where $B$ is the reflection of $H$ in the rim of the cup.
We see that the fly needs to move:
:$3$ inches horizontally.
:$3$ inches vertically and another inch vertically for a total of $4$ inches vertically.
The shortest distance from $F$ to $H$ is then found by Pythagoras's Theorem, where $FB$ is seen to be the hypotenuse of the $3$-$4$-$5$ triangle.
== Sources ==
* : Solutions: $162$. -- The Fly and the Honey
* : Answers: $325$. The Fly and the Honey
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/163 - The Russian Motor-Cyclists

== by : $163$ ==
;The Russian Motor-Cyclists
:Two Russian Army motor-cyclists, on the road $A$, wish to go to $B$.
:Now Pyotr said: "I shall go to $D$, which is $6$ miles, and then take the straight road to $B$, another $15$ miles.
:But Sergei thought he would try the upper road by way of $C$.
:Curiously enough, they found on reference to their odometers that the distance either way was exactly the same.
:
:This being so, they ought to have been able easily to answer the General's simple question,
::"How far is it from $A$ to $C$?"
:it can be done in the head in a few moments, if you only know how.
:Can the reader state correctly the distance?
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $163$. -- The Russian Motor-Cyclists
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $281$. The Russian Motorcyclists
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/163 - The Russian Motor-Cyclists/Solution

== by : $163$ ==
;The Russian Motor-Cyclists
== Solution ==
:$3 \tfrac 1 3$ miles.
The "pretty little rule" as so patronisingly puts it is applied thus:
:... divide $15$ by $6$ and add $2$, which gives us $4 \tfrac 1 2$.
:Now divide $15$ by $4 \tfrac 1 2$, and the result $3 \tfrac 1 3$ miles) is the required distance between the two points.
Thus if $a + x$ and $b$ are the legs of the right triangle:
:$x = \dfrac b {\dfrac b a + 2}$
which is actually a pretty unwieldy little rule considering how rarely one would end up using it.
== Proof ==
Let $x$ miles be the distance from $A$ to $C$.
Let $y$ miles be the distance from $C$ to $B$.
We have:
The general case can be explored as:
as we were to show.
== Sources ==
* : Solutions: $163$. -- The Russian Motor-Cyclists
* : Answers: $281$. The Russian Motorcyclists
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/164 - Those Russian Cyclists Again

== by : $164$ ==
;Those Russian Cyclists Again
:Here is another little experience of the two Russian Army motor-cyclists from the last puzzle.
:In the section from a map given in the diagram we are shown three long straight roads, forming a right-angled triangle.
:
:The General asked the two men how far it was from $A$ to $B$.
:Pyotr replied that all he knew was that riding round the triangle, from $A$ to $B$,
::from there to $C$ and home to $A$, his odometer showed exactly $60$ miles,
::while Sergei could only say that he happened to know that $C$ was exactly $12$ miles from the road $A$ to $B$ --
::that is, to the point $D$, as shown by the dotted line.
:Whereupon the General made a very simple calculation in his head and declared that the distance from $A$ to $B$ must be ...
:Can the reader discover so easily how far it was?
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $164$. -- Those Russian Cyclists Again
* : Geometrical Problems: Triangle, Square & Other Polygon Puzzles: $282$. Those Russian Cyclists Again
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/164 - Those Russian Cyclists Again/Solution

== by : $164$ ==
;Those Russian Cyclists Again
== Solution ==
:$25$ miles.
== Proof ==
Let $BC = a$, $AC = b$ and $AB = c$.
Let $CD = d$.
Let $AD = c_1$ and $BD = c_2$.
We have:
Then we have:
Setting $a + b + c = 60$ and $d = 12$ gives the result.
Hence 's General simply squared $60$ and divided by $2 \times \paren {60 + 12}$:
:$c = \dfrac {60^2} {2 \times \paren {12 + 60} } = \dfrac {3600} {144} = 25$
== Sources ==
* : Solutions: $164$. -- Those Russian Cyclists Again
* : Answers: $282$. Those Russian Cyclists Again
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/165 - The Despatch-Rider in Flanders

== by : $165$ ==
;The Despatch-Rider in Flanders
:A despatch-rider on horseback, somewhere in Flanders, had to ride with all possible speed from $A$ to $B$.
:The distances are marked on the map.
:
:Now, he can ride just twice as as fast over the soft turf (the shaded bit) as he can ride over the loose sand.
:Can you show what is the quickest possible route for him to take?
:This is just one of those practical problems with which the soldier is faced from day to day when on active service.
:Important results may hang on the rider taking the right or the wrong route.
:Which way would you have gone?
:Of course, the turf and the sand extend for miles to the right and the left with the same respective depths of three miles and two miles,
::so there is no trick in the puzzle.
=== Click here for solution ===
== Sources ==
* : Unicursal and Route Problems: $165$. -- The Despatch-Rider in Flanders
* : Geometrical Problems: Solid Geometry Puzzles: $311$. The Dispatch Rider in Flanders
Category:Henry Ernest Dudeney/Modern Puzzles
Henry Ernest Dudeney/Modern Puzzles/165 - The Despatch-Rider in Flanders/Solution

== by : $165$ ==
;The Despatch-Rider in Flanders
== Solution ==
The rider needs to ride across the sand to a point on the turf / sand boundary exactly $1$ mile from the western boundary of the map, then straight across the turf to $B$.
== Proof ==
Let $G$ be the point on the turf / sand boundary to which he must aim.
Let $x$ be the distance from that point to the western boundary of the map $E$.
:
There appears to be no quick and easy way to solve this quartic, unless one were to plough through Ferrari's Method.
However, if we are sharp we note that:
:$3 - 42 + 179 + 56 - 196 = 0$
and so from Coefficients of Polynomial add to 0 iff 1 is a Root we have that $x = 1$ is a root of that quartic.
We can factor that quartic if we like:
:$\paren {x - 1} \paren {3 x^3 - 39 x^2 + 140 x + 196} = 0$
but we already have the solution $x = 1$ which adequately fulfils the conditions of the problem.
We check the geometry:
and all is well.
== Sources ==
* : Solutions: $165$. -- The Despatch-Rider in Flanders
* : Answers: $311$. The Dispatch Rider in Flanders
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation

== by : $166$ ==
;Picture Presentation
:A wealthy collector had ten valuable pictures.
:He proposed to make a presentation to a public gallery, but could not make up his mind as to how many he would give.
:So it amused him to work out the exact number of different ways.
:You see, he could give any one picture, any two, any three, and so on, or give the whole ten.
:The reader may think it a long and troublesome calculation,
::but I will give him a little rule that will enable him to get the answer in all such cases without any difficulty and only trivial labour.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $166$. -- Picture Presentation
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $445$. Picture Presentation
Category:Henry Ernest Dudeney/Modern Puzzles
Category:Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation
Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation/Solution

== by : $166$ ==
;Picture Presentation
== Solution ==
:$1023$, or $1024$ if you include the possibility of not giving any of his pictures at all.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Solutions: $166$. -- Picture Presentation
* : Answers: $445$. Picture Presentation
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation
Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation/Solution/Proof 1

== by : $166$ ==
;Picture Presentation
== Solution ==
== Proof ==
For each picture, the collector can choose to give it away or to keep it.
He has to make this choice $10$ times in a row.
Hence he would have $2^{10} = 1024$ options, including not giving away any.
== Sources ==
* : Solutions: $166$. -- Picture Presentation
* : Answers: $445$. Picture Presentation
Category:Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation
Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation/Solution/Proof 2

== by : $166$ ==
;Picture Presentation
== Solution ==
== Proof ==
The number of ways you can pick $r$ pictures from a collection of $10$ is equal to the binomial coefficient $\dbinom n r$, which is given by:
:$\dbinom n r = \dfrac {n!} {r! \paren {n - r}!}$
Hence the total number of ways the collector can make part or all of his collection available to the public is:
:$\ds \sum_{k \mathop = 1}^n \dbinom {10} k$
which is equal to:
:$\ds \sum_{k \mathop = 0}^n \dbinom {10} k - \dbinom {10} 0$
From Sum of Binomial Coefficients over Lower Index:
:$\ds \sum_{k \mathop = 0}^n \dbinom {10} k = 2^{10} = 1024$
while from Binomial Coefficient with Zero:
:$\dbinom {10} 0$
Hence if the collector wants to consider the option of not giving any of his paintings after all, the number of options he has is $2^{10} = 1024$.
If he is committed to giving at least $1$, then the option of giving none is off the table, so the answer is $1023$.
Category:Henry Ernest Dudeney/Modern Puzzles/166 - Picture Presentation
Henry Ernest Dudeney/Modern Puzzles/167 - A General Election

== by : $167$ ==
;A General Election
:In how many different ways may a Parliament of $615$ members be elected if there are only $4$ parties:
::Conservatives, Liberals, Socialists, and Independents?
:You see you might have $\text C. 310$, $\text L. 152$, $\text S. 150$, $\text I. 3$;
::or $\text C. 0$, $\text L. 0$, $\text S. 0$, $\text I. 615$;
::or $\text C. 205$, $\text L. 205$, $\text S. 205$, $\text I. 0$; and so on.
:The candidates are indistinguishable, as we are only concerned with the party numbers.
=== Click here for solution ===
== Sources ==
* : Combination and Group Problems: $167$. -- A General Election
* : Combinatorial & Topological Problems: Miscellaneous Combinatorial Puzzles: $446$. A General Election
Category:Henry Ernest Dudeney/Modern Puzzles