<h3 class="heading">A Lesson in Logic(if: not(playlist: 'background', 'isplaying'))[(playlist: 'background', 'play')]</h3>
It was a rainy morning in 1903 - I cannot precisely remember the date. The landlady had set the table for breakfast. Holmes was absent from his usual armchair next to the fire. With a somewhat uncouth manner no doubt born of my sleepiness, I questioned the landlady as to his whereabouts, who merely muttered something about "crazy".
Having nothing of consequence to do, I picked up the newspaper. I had not read for half a minute when the door opened and Sherlock Holmes walked in - twenty years older, but still very much the same man I had met with Stamford in the chemistry lab.
"I have just employed an energetic morning at the mathematics club, Watson."
"What! I did not know you were a mathematician as well." I said.
"I possess no formal degree, but I am the finest consulting detective London has seen."
"What use is that in mathematics?"
"To speak of logic is to speak of mathematics."
"I have heard you say so on many occasions." I continued, prompting him to prove his point.
"Would you hand me the small black pocketbook from the bookshelf? Thank you.
"You say mathematics and logic are discrete, and I claim they are not. Here is a problem I encountered some years ago - you will need logic, mathematics, and deduction to solve this one, Watson!
"I assume you are familiar with the rules of chess?" he continued.
"Yes, but I am by no means a fantastic player."
"No matter. In chess, white always goes first, yes? And the players take turns to move their pieces - one on each turn."
"Correct."
"Now, consider this modification: Instead of making one move per turn, White and Black both make two moves per turn."
"Interesting."
"Your task is to show that there exists a strategy for White by which he can tie any game or win it."
"Really, Holmes! This is preposterous!" I exclaimed.
"Hardly so, Watson. You demanded a playground - I merely provided it. All I can tell you is that a proof exists."
[["Very well - let me see what I can make of it."|double-chess-problem-2]]
<h3 class="heading">A Lesson in Logic(track: 'paperturn', 'play')</h3>
<p class="question">P1.<br>In a game of double chess, both players make two legal moves per turn. Show that White can always draw the game or better.</p>
<p class="hint">Hint: Prove the //existence// of such a strategy. //Finding// the strategy is inconsequential. <br></p>
[[Solution|double-chess-solution-3]]
(if: $game_completed is 1)[[[Index|index]]]<h3 class="heading">A Lesson in Logic(track: 'paperturn', 'play')</h3>
I struggled with the problem, but I couldn’t think of anything. I began to chew on the end of my pencil instead.
After twenty minutes had passed in this fashion, Holmes looked at me and smirked.
"It is a privilege to see your mind at work, Watson.
"Suppose that Black can win no matter what moves White plays. Then, on his first move, let White pick up any of his knights, and move them to a different square. On the second move of the same turn, let White pick up the same knight, and move him back to the starting square."
"Hah! White has wasted a turn." I said.
"On the contrary. By returning the knight to his original square, all the pieces are now in their home positions - the chessboard is exactly the same as it was when the game began, with one crucial change - it is now Black’s move! Now, according to our assumption, White should be able to win whatever Black plays.
"But both players cannot win a game of chess - there is either one winner or a drawn game. Hence, both White and Black winning is a contradiction, which means our assumption was incorrect. Therefore, Black cannot win no matter whatever moves White plays, which guarantees White a draw or better."
"This did not cross my mind." I said, sheepishly.
"In this problem, we were able to prove a statement with mathematical certainty using only logic - no complex formulae or derivations were necessary. Suffice to say, then, that those things are at least interlinked?" Holmes said mischievously.
"I see your point, Holmes, You are right, as usual."
[[Next|2nx2n-intro-1]]
<h3 class="heading">Mathematics On The Chessboard(track: 'paperturn', 'play')</h3>
"Watson, you have been puzzling over the newspaper for a long time today. I have read through the entire paper, but I have not gleaned anything of interest from it. Is it possible that perhaps you see more than I do?"
I flung the paper away and sighed.
"It’s this infernal riddle, Holmes. I thought it simple enough in the morning, but I am at a loss now, and the problem is as unsolved as ever."
"Toss us the paper, then."
Holmes looked over the riddle.
"Interesting - and I think, solvable!"
He then tore out a long strip from the paper and bent laboriously over it. I resumed my futile task of poring through the remainder of the publication, when five minutes later, Holmes hid the scrap of paper under his book.
"I assume you've got it." I said, somewhat bitterly.
"Yes. I will restate the problem for you:
"Consider a chessboard of dimensions <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mn>2</mn><mi>n</mi></msup><mo>× </mo><msup><mn>2</mn><mi>n</mi></msup></mrow><annotation encoding="application/x-tex">2^{n} \times 2^{n}</annotation></semantics></math>. If one square from this chessboard is removed, is it possible to completely cover the board using these L-shaped dominos?"
"I haven't the faintest clue, but I’ll [[look at the problem again.|2nx2n-problem-2]]" I said.
<h3 class="heading">Mathematics On The Chessboard(track: 'paperturn', 'play')</h3>
<p class="question">P2.<br>Consider a chessboard of dimensions <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msup><mn>2</mn><mi>n</mi></msup><mo>× </mo><msup><mn>2</mn><mi>n</mi></msup></mrow><annotation encoding="application/x-tex">2^{n} \times 2^{n}</annotation></semantics></math>. An arbitrarily chosen square is removed from this board. Is it possible to tile the resulting board with an L-shaped domino?
</p>
<div style="text-align:center;"> <img width=600 height=600 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/08/2nx2n-sol-i1-1.jpg" alt="A larger chessboard with dimensions 2 raised to n by 2 raised to n. The board has one square removed and an L-shaped domino drawn on it."> </div>
<br>
<p class="hint">Hint: Consider using mathematical induction. <br></p>
[[Solution|2nx2n-solution-3]]
(if: $game_completed is 1)[[[Index|index]]]<h3 class="heading">Mathematics On The Chessboard(track: 'paperturn', 'play')</h3>
"I assume you are familiar with mathematical induction?"
"Yes, I am."
"Excellent! Let us begin, then.
"First answer this: If I were given a stepladder, is it possible to climb as high as I want given that I can climb onto the bottom rung, and that I can always climb to the next rung?"
"I don't see what this has to do with the problem, but yes."
"Ah, but this does help us - in fact, not just us - this statement is the basis for some of the most beautiful proofs in number theory!
"To start with, let us demonstrate that it is possible to tile the board for the smallest possible <span class="math inline"><em>n</em></span><span class="math inline">, a 2 × 2</span> board.
"Regardless of what square is removed from the board, an L-shaped domino will cover the remaining three squares independent of their colors. Correct?"
"Evidently."
"Let us //assume// that for some integer <span class="math inline"><em>k</em></span>, the board can be tiled if one square is removed. We do not need to prove this fact yet, however, we will need to prove something soon."
"Fair enough." I said.
"You can now see how our ladder fits in. If I can climb onto the bottom rung and I can climb to one rung from the other–"
"You would be able to climb wherever you wanted - capital!"
Holmes smirked.
"I was sure you would get it in the end. However, we still need to prove that we can climb to the [[next rung of the ladder.|2nx2n-solution-4]]"
<h3 class="heading">Mathematics On The Chessboard(track: 'paperturn', 'play')</h3>
"For the final step of this problem, consider a board with dimensions <span
class="math inline">2<sup><em>k</em> + 1</sup> × 2<sup><em>k</em> + 1</sup></span>. This board is made up of //four// smaller boards of dimensions <span
class="math inline">2<sup><em>k</em></sup> × 2<sup><em>k</em></sup></span> - correct?"
<br>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/08/2nx2n-sol-i4.jpg" alt="an 8x8 chessboard made from four smaller 4x4 boards."> </div>
"Our assumption assists here. It is possible to tile the <span class="math inline">2<sup><em>k</em></sup></span> board with one arbitrarily chosen square removed. So we shall remove one corner square from each of the four smaller boards. We tile the smaller boards and arrange them, by rotating if required, in such a way that the four removed squares become the 2 × 2 center of the bigger board. <br>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/08/2nx2n-sol-i2.jpg" alt="a sample 8x8 board with 4 squares, one for each 4x4 board, removed."> </div>
"What do we do with the 2 × 2 center, then?"
"That is the most beautiful part of the problem. This space, as you recall, can be filled with an additional domino.
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/08/2nx2n-sol-i2-1.jpg" alt="a sample 8x8 board with 4 squares, one for each 4×4 board, removed."> </div>
"Adding this domino leaves us with exactly one empty space, thus solving the problem for a
<span class="math inline">2<sup><em>k</em> + 1</sup> × 2<sup><em>k</em> + 1</sup></span> board!"<br>
"Moreover," Holmes continued, "<span class="math inline"><em>k</em> + 1</span> is not limited to any value - we can take any value of <span class="math inline"><em>k</em> + 1</span>, since we have solved the problem for <span class="math inline"><em>n</em> = 1</span> and <span class="math inline"><em>n</em> = <em>k</em></span>."
"Remarkable indeed, Holmes! A flawless chain of reasoning - and every link rings true! Now the only question to solve is - was this logic, or mathematics?" I joked.
"Ah, Watson, the answer of that question is left as an exercise to you!"
[[Next|retrograde-intro-1]]
[[More Information|2nx2n-alternative-5]]<h3 class="heading">A Checkered Past(track: 'paperturn', 'play')</h3>
A bleak, snowy day saw Holmes and I trapped inside our rooms. With hot cocoa as my companion, I endeavoured to read through the newspaper. A pompous headline caught my attention:
"Derbyshire Murder Solved: Why The Past Is More Important Than The Present"
"Here is a colleague of yours, Holmes! He seems to rely rather heavily on the past to draw conclusions about the present."
"Ah yes. A pretty headline, is it not?"
"I feel inclined to agree."
"In the realm of logic, patterns play a key role, Watson, as you very well know. When you have roughly six hours to spare and another day as dreary as this one, I recommend a nice tour of the Crime section of any library of your choosing. I guarantee that we shall find parallel cases to remind us of Smith’s cheque forgery last year, or the Anderson murder in Camden Town."
"Tell me something, Holmes. Since the past can give you the present in the real world, can we extend this approach to the theoretical world? I should imagine that there exist many problems where "what happened" is of significant consequence, perhaps more so than "what is" - apart, of course, from the world of crime."
Holmes smiled.
"The showman in me says I shall demonstrate. Pray bring me that chessboard, and the pieces it comes with."
"The type of problems we will solve now have nothing to do with the game itself, Watson - that is to say, we are not concerned with the outcome of the game. It is of little consequence if Black wins or White gets stalemated - we are concerned with the past. The players are permitted to play as eccentrically as possible - the only condition being of course, that all of their moves are legal.
"Our job is to merely analyse the moves that occurred in the past, which led to a position in the present."
"I see." I replied.
"These problems are referred to as retrograde analyses. They can tax the mind as much as any real-world problems do. All of them are, however, a fascinating conjunction of inflexible logic and the unchangeable past."
I rubbed my palms together in anticipation.
"I cannot wait to begin!"
"Then I shall not torture you with waiting any longer. Have [[a look at the chessboard|retrograde1-problem-2]], Watson!"
<h3 class="heading">A Colorful Solution(track: 'paperturn', 'play')</h3>
<p class="question">P4.<br><br>In this game, all pieces have stayed true to the color of their home squares. <br>
Pieces that began on white squares have only travelled on white squares, and pieces that began on black squares have only travelled on black squares. Such a game, in which no piece has moved off its original color, is called a monochromatic game.<br>
Given that this game was monochromatic, what square does the bishop belong on?
</p>
<div style="text-align:center;"> <img width=600 height=600 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/08/retrograde1-prob-i1-2.png" alt="chessboard with white pawns on d2 and f2, white king on e1 and black king on g8. a bishop "> </div>
<br><br>
<p class="hint">Hint: Think about square colors and piece colors in general.<br></p>
[[Solution|retrograde2-sol-3]]
(if: $game_completed is 1)[[[Index|index]]]<h3 class="heading">The Tale of the Inferno and the Downpour(track: 'paperturn', 'play')</h3>
Holmes was clearly enjoying himself. My helplessness had culminated in staring at the board with a dumbstruck expression on my face.
"Watson, I once heard a fable when I was a young lad down in Birmingham. There was a raging inferno in the forest, and there was also a torrential downpour at the same time. The people of the town nearby, they did not enter the forest for fear of being burnt alive or washed away - and yet, in the morning, the inferno and the flood had vanished - nothing to be seen of either!"
I looked at Holmes pityingly.
"The poor man has gone mad from solving crime." I muttered.
"I assure you that I retain my grasp over my mental attic, Watson! Suppose the bishop were on the wrong colour, c4.
"Then, given that the game is monochromatic, what piece on the black squares could have captured the last piece which fell on a black square?"
"Perhaps the king-" I began, but Holmes cut me off.
"Certainly not the king, or the pawns, since they never moved! Chess is a game of soldiers, Watson - an army of black soldiers fighting white soldiers. A white soldier cuts a black soldier down, and is in turn defeated by another black soldier. Turn by turn, the two armies will decimate themselves.
"While both sides must sustain casualties, at least one survivor must prevail.
"What could this survivor be? In this case, nothing but the bishop - which must be on b4, a black square!"
"Good heavens!"
"There is a certain elegance in the solutions of the most absurd problems, Watson."
A thought struck me.
"Would this approach work for a Black rook on two different squares - say a1 and b1?"
"That is what makes it so remarkable, Watson! The last piece doesn’t have to be a White bishop - it can just as easily be a Black pawn, or even the Black queen! Moreover, we do not need specifically the mentioned squares, or even two adjacent squares! In fact, we do not need anything but two differently coloured squares and the last piece."
I was awestruck by this answer, employing not complex moves and vigorous analysis but pure mathematical brilliancy.
"It might interest you to know, Watson, that this problem was composed by Moriarty! It carries much of the cunning simplicity his crimes are known for."
"I should have known. This problem is truly diabolical."
Holmes acknowledged my praise with a lazy wave of his hand and reached across the table for his violin.
[[Winter 1914|The Last Problem]]<h3 class="heading">A Checkered Past(track: 'paperturn', 'play')</h3>
<p class="question">P3.<br>No pawn has promoted. Is this position possible?
</p>
<div style="text-align:center;"> <img width=600 height=600 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/08/screenshot-2023-08-15-at-12.27.30-pm.png" alt="chessboard with white pawns on b2 and d2, white bishop on d4, white king on f1, black knight on c5, black pawn on f7 and black king on g8."> </div>
[[Yes | retrograde1-incorrect-3]]
[[No|retrograde1-solution-4]]
(if: $game_completed is 1)[[[Index|index]]]<h3 class="heading">A Checkered Past(track: 'paperturn', 'play')</h3>
"I don't see why not." I said, shrugging.
"Can you not see anything on the board that jumps out at you?" cried Holmes, with unusual intensity.
"Well, all I see are the pawns on b2 and d2, the White king on-"
"The pawns, yes, the pawns! Look at the pawns, Watson!"
[[What about the pawns?|retrograde1-solution-4-incorrect]]<h3 class="heading">A Checkered Past(track: 'paperturn', 'play')</h3>
"Hah! It's not possible." I said triumphantly.
"Pray elaborate."
"The question to answer is; how did White's bishop on d4 get out from its home square, c1? The pawns on b2 and d2 have not moved to let it out. Therefore, the bishop on d4 must be a promoted bishop!"
"Capital, Watson! You show a reasonable measure of genius yourself!"
[[Next|retrograde2-problem-2]]
<h3 class="heading">A Checkered Past(track: 'paperturn', 'play')</h3>
"Enlighten me."
"If a promotion had not occurred, how could the white bishop on d4 have gotten out from its home square? The pawns on b2 and d2 have not moved to let it out. There is no way it can move, and therefore must have been captured on its own square. Hence, the bishop on d4 is clearly a promoted bishop, in which case the statement 'no promotions have taken place' simply holds no water!"
"How foolish of me! It was right there!"
"Right there, and yet hidden to the amateur eye! It is this thread of observation, deduction, and reasoning, my dear doctor, that makes the past so crucial - and permits perfect reconstruction of the past from the present!"
Holmes got up from his seat and began to pace the room in some strong emotion.
"It is crucial that you understand this, Watson! The ability to know what happened in the past given nothing but the present is an invaluable one - so important, in fact, that I depend on it for my bread and butter!
"I think it best if we shift our focus to development of the logical faculties I prize so highly - using the chessboard, of course."
[[Next|retrograde2-problem-2]] <h3 class="heading">Winter 1914(track: 'paperturn', 'play')</h3>
The rest of our days at 221B were spent in other cases. A few ones stand out - notably, the adventure of Josiah Amberley the colourman, and the strange events at Abbas Parva relating to Ms. Ronder.
Soon thereafter, I was called to Russia. Revolution was imminent, and considering I was still army personnel (albeit getting on in years), Col. Pritchard felt that there was an advisory role which I could fulfill. Therefore, I found myself saying goodbye to my close acquaintances, and packing my bags to leave 221B for the foreseeable future.
And so, on a chilly winter morning in 1914, I found myself boarding the Trans-Siberian railway in the capacity of army doctor.
[[The Last Problem|n-queens-intro-1]]
<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbywatson">
1916, Moskva River Slums
My dear Holmes,
I write to you in the fond hope that at least this letter will pass the checks on the border and make it to London. Your absence is painfully noticed for the last two years, and I wait eagerly for the day I step back on English soil.
I have had the pleasure of meeting Pushkin, one of the brigadiers of our little outpost here. When he mentioned chess and logic puzzles, my mind immediately went to you. Pushkin is as round as you are tall, and possesses a measure of intelligence similar to yours!
The other day, he put forth a problem which I was unable to solve. It seems very trivial, but I know that things are not always what they seem to be. I am enclosing the problem with this letter.
"What is the greatest number of independent queens that can be arranged on an 'n' by 'n' board?"
I await any headway you can make. I hope that the War ends soon and I am relieved of my duties.
Yours,
Dr. John Watson,
MD
Army Medical Department
PS: I enclose a photograph of the Kremlin!
<img class="picture" src="https://moneaalhad.files.wordpress.com/2023/08/moskva-1611856310-e1692192167765.png">
</p>
[[Problem 5|n-queens-problem-2]]startingVol: 0.4
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(newplaylist: 'background', "track1", "track2", "track3", "track4", "track5", "track6", "track7")<h5 class="heading">SoME '23</h5>
<h2 class="heading">A Study In Greyscale(newplaylist: 'background', "track1", "track2", "track3", "track4", "track5", "track6", "track7")</h2>
<h4 class="heading">Introduction</h4>[[Hello!|hello-easter]]
Join Sherlock Holmes and Dr. Watson on the chessboard honing the logical faculties Holmes prizes so highly. In this 221B spinoff, we explore the combinatoric aspect of the chessboard - music by Chopin now included! However, Holmes requires that you possess the following knowledge:
* Rules of chess, <a href="https://en.wikipedia.org/wiki/Rules_of_chess#:~:text=from%20the%20board.-,Promotion,-Main%20article%3A" target="_blank">promotions</a> and <a href="https://en.wikipedia.org/wiki/Rules_of_chess#:~:text=that%20is%20revealed.-,Movement,-Basic%20moves" target="_blank">how pieces move</a>.
* That's it.
The project blends Sherlock Holmes and chess to explore strategy and logic. It's a creative nod to classic literature encouraging critical thinking through puzzles.<br>
<h4 class="heading">Instructions</h4>This website is best experienced on a computer with headphones on.
To navigate, click on the blue links. Once you visit a few passages, arrows will show up next to the title. You can [[click|easter-egg-2]] those to go back and forth.
Links are highlighted in blue. All links lead to [[new passages.|easter-egg]]
To turn off the background music, hit the arrow on the left and select 'Mute'. There's also a volume slider.
If you'd like to read the entire thing at your own leisure and not in a browser, you can <a href="https://github.com/some3-encroissant/some3-encroissant.github.io/blob/main/En-Croissant_SoME3.pdf" target="_blank">download a PDF</a> from our GitHub.<br>
(align:"=><=")+(box:"X")[<h4 class="heading">''[[Start|double-chess-intro-1]]''</h4>]<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="question">P5.<br>What is the maximum number of independent queens that can be placed on an <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>×</mo><mi>n</mi></mrow><annotation encoding="application/x-tex">n \times n</annotation></semantics></math> board?
</p>
Independent arrangement of queens refers to a position in which no queen attacks any other.
<p class="hint">Hint: Is an arrangement of <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>n</mi><mo>+</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">n+1</annotation></semantics></math> queens possible? What about <math display="inline" xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mi>n</mi><annotation encoding="application/x-tex">n</annotation></semantics></math>?</p>
[[Solution|n-queens-sol-3]]
(if: $game_completed is 1)[[[Index|index]]]<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbyholmes">
1916, 221B Baker St.
Dear Watson,
It is a relief to hear from you. Things are not as well as they can be - yet I am sure I shall find a way out. I am but a vessel for my intellect, Watson. I think you will agree with me when I say both vessel and intellect are very formidable.
I apologise for boring you with my problems and my lack of seemly modesty, but I shall endeavour to solve the problem. My hand is laid up at the moment so I have taken the liberty of having it typed.
[[Next|n-queens-solution-3]]
Yes, like this one.
Now go read the [[rest of the instructions.|Instructions]]It's funny when something just "clicks", huh?
I thought it was hilarious. Never mind.
[[Instructions]] <h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbyholmes">There can’t be more than one queen in any column or row, giving us the upper bound as <span class="math inline"><em>n</em></span>. In adjacent columns, the next queen cannot be placed either in the same row, or the adjacent rows.
Therefore, the general strategy is going to be to put two queens a knight’s distance away. Note that only one queen can be arranged on a <span class="math inline">2 × 2</span> board that is independent. Similarly, at most two queens can be placed on a <span class="math inline">3 × 3</span> board.
Here, we shall define the concept of a positive diagonal and a negative diagonal. Pictures are enclosed for your convenience.
A positive diagonal is any diagonal with a positive slope. We call the longest such diagonal as the great positive diagonal. Similarly, a negative diagonal is any diagonal with a negative slope.
<div style="text-align:center"> <img src="https://moneaalhad.files.wordpress.com/2023/08/untitled-design-1-844005314-e1692213675265.png"></div>
<p class="handwrittenbyholmes"><strong>Observation: </strong>For squares lying on the same negative diagonals, the sum of the coordinates is the same; for those lying on the same positive diagonals, the difference is the same.
[[Next|n-queens-solution-4]]
<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbyholmes">
<strong>Claim</strong>: for <span class="math inline"><em>n</em> ≥ 4</span>, it is possible to arrange <span class="math inline"><em>n</em></span> independent queens on an <span class="math inline"><em>n</em> × <em>n</em></span> board. <br>
<strong>Proof</strong>:<br>
<strong>Case 1</strong>: <span class="math inline"><em>n</em></span> is an even integer, <span class="math inline"><em>n</em> = 2<em>k</em>.</span>
Start by putting a queen on the square <span class="math inline">(1, 2)</span>, and put the next queen a knight’s distance away to the upper right side of the previous one, until you reach the <span class="math inline"><em>k</em><sup><em>t</em><em>h</em></sup></span> column.
Then start again from the first row of <span class="math inline">(<em>k</em>+1)<sup><em>t</em><em>h</em></sup></span> column and repeat the process. This is to say that put a queen on <span class="math inline">(<em>i</em>, 2<em>i</em>), 1 ≤ <em>i</em> ≤ <em>k</em></span> and on <span class="math inline">(<em>k</em>+<em>j</em>, 2<em>j</em>−1), 1 ≤ <em>j</em> ≤ <em>k</em></span>.
No queen lies on the same row or the column by construction. Now it remains to show that no two queens lie on the same diagonal. This holds trivially for the positive diagonals as the queens are placed such that the next one lies immediately above the diagonal of the previous one. Additionally, the first <span class="math inline"><em>k</em></span> queens lie above the great positive diagonal and the rest <span
class="math inline"><em>k</em></span> queens lie below it. Therefore, there is no common positive diagonal between any of the queens, concluding the argument for the positive diagonals.
Note again that two squares lie on the same negative diagonal <span class="math inline"><em>i</em><em>f</em><em>f</em></span> the sum of their coordinates is the same.
[[Next|n-queens-solution-5]]<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbyholmes">
The first <span class="math inline"><em>k</em></span> queens are placed on <span class="math inline">(<em>i</em>, 2<em>i</em>), 1 ≤ <em>i</em> ≤ <em>k</em></span> and the next <span class="math inline"><em>k</em></span> are placed on <span class="math inline">(<em>k</em>+<em>j</em>, 2<em>j</em>−1), 1 ≤ <em>j</em> ≤ <em>k</em></span>. Thus we seek the solutions to the equation <span class="math inline"><em>i</em> + 2<em>i</em> = <em>k</em> + <em>j</em> + 2<em>j</em> − 1</span>, or <span class="math inline">6(<em>i</em>−<em>j</em>) + 2 = <em>n</em></span>, as <span class="math inline"><em>n</em> = 2<em>k</em></span>. This is possible only when <span class="math inline"><em>n</em></span> leaves a remainder of 2 when divided by 6. Therefore, the arrangement described above won’t work for those <span class="math inline"><em>n</em></span> which leave remainder 2 when divided by 6.
Other cases in which n is even and leaves remainder 0 or 4 when divided by 6 are not problematic. Since <span class="math inline"><em>n</em></span> is even, it can only leave remainder 0, 2, or 4 when divided by 6.
To conclude, an arrangement described as above for remainders 0 and 4 is independent.
<div style="text-align:center"> <img width=600 src="https://moneaalhad.files.wordpress.com/2023/08/12queens.png" alt="A 12x12 board with the positions of the queens marked."><p class="handwrittenbyholmes">A 12x12 board with the positions of the queens marked.</div>
[[Next|n-queens-solution-6]]<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbyholmes">For <span class="math inline"><em>n</em> = 6<em>m</em> + 2</span>, consider the following arrangement: In the <span class="math inline"><em>k</em> − 3</span> columns starting from the <span class="math inline">2<sup><em>n</em><em>d</em></sup></span> column and ending with <span class="math inline">(<em>k</em>−2)<sup><em>n</em><em>d</em></sup></span>, put a queen at a knight’s distance away upwards in every other row starting from row 3 and moving right, viz. put a queen on <span class="math inline">(<em>r</em>, 2<em>r</em>−1), 2 ≤ <em>r</em> ≤ <em>k</em> − 2</span>.
Now, in the <span class="math inline"><em>k</em> − 3</span> columns starting from the <span class="math inline">(<em>k</em>+3)<sup><em>r</em><em>d</em></sup></span> till the <span class="math inline">(<em>n</em>−1)<sup><em>s</em><em>t</em></sup></span>, put a queen at a knight’s distance away moving up and to the right in every other row starting from row 6 and ending at row <span class="math inline"><em>n</em> − 2</span>, viz. put a queen on <span class="math inline">(<em>k</em>+3+<em>r</em>, 6+2<em>r</em>), 0 ≤ <em>r</em> ≤ <em>k</em> − 4</span>.
This leaves columns <span class="math inline">1, <em>k</em> − 1, <em>k</em>, <em>k</em> + 1, <em>k</em> + 2, </span> & <span class="math inline"> <em>n</em></span> and rows <span
class="math inline">1, 2, 4, <em>n</em> − 3, <em>n</em> − 1,</span><span class="math inline"></span> & <span class="math inline"><em>n</em></span> where there is no queen. Out of those, place the queens on the following squares:
<span class="math inline">(1, <em>n</em>−3), (<em>k</em>−1, 1), (<em>k</em>, <em>n</em>−1), (<em>k</em>+1, 2), (<em>k</em>+2, <em>n</em>), (<em>n</em>, 4)</span>.
<div style="text-align:center"> <img width=600 src="https://moneaalhad.files.wordpress.com/2023/08/chess14x14-2833243024-e1692276794295.png"><p class="handwrittenbyholmes">Arrangement of queens on a 14 × 14 board using <span class="math inline"><em>6m + 2</em></span> method.</div>
By construction, it’s evident that no two queens lie on the same row or column.
Now to check for the diagonals, assign the numbers 1 to <span class="math inline"><em>n</em></span> to squares<span class="math inline"> (<em>n</em>, 1), (<em>n</em>−1, 1), ⋯, (1, 1)</span> respectively, and the numbers <span class="math inline"><em>n</em> + 1</span> to <span class="math inline">2<em>n</em> − 1</span> to squares <span class="math inline">(1, 2), (1, 3), ⋯, (1, <em>n</em>)</span> respectively. Then, label each positive diagonal according to the numbered square that belongs to it.
<p class="handwrittenbyholmes">Similarly, assign numbers 1 to <span
class="math inline"><em>n</em></span> to the squares <span
class="math inline">(1, 1), (2, 1), ⋯, (<em>n</em>, 1)</span> respectively and the numbers <span
class="math inline"><em>n</em> + 1</span> to <span
class="math inline">2<em>n</em> − 1</span> to the squares <span
class="math inline">(<em>n</em>, 2), (<em>n</em>, 3), ⋯, (<em>n</em>, <em>n</em>)</span> and label the negative diagonals.
<div style="text-align:center"> <img src="https://moneaalhad.files.wordpress.com/2023/08/fef.png" width=800 alt="n × n boards labelled with positive and negative diagonals."><p class="handwrittenbyholmes">n × n boards labelled with positive and negative diagonals.</div>
[[Next|n-queens-solution-7]]<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbyholmes">Also label the queens by the number of the columns on which they are placed. <span class="math inline"><em>r</em><sup><em>t</em><em>h</em></sup></span> queen is placed on the <span class="math inline"><em>r</em><sup><em>t</em><em>h</em></sup></span> column.
Then one can observe from the queen arrangement described above that the ordered pairs describing queen’s label and the positive diagonal’s label on which the queen is placed, <span class="math inline">(<em>q</em><em>u</em><em>e</em><em>e</em><em>n</em>−<em>l</em><em>a</em><em>b</em><em>e</em><em>l</em>, <em>p</em><em>o</em><em>s</em><em>i</em><em>t</em><em>i</em><em>v</em><em>e</em>−<em>d</em><em>i</em><em>a</em><em>g</em><em>o</em><em>n</em><em>a</em><em>l</em>−<em>l</em><em>a</em><em>b</em><em>e</em><em>l</em>)</span>, will be as follows:
<span
class="math inline">(1, 2<em>n</em>−4), (2, <em>n</em>+1), (3, <em>n</em>+2), (4, <em>n</em>+3), ⋯, (<em>k</em>−3, 3<em>k</em>−4), (<em>k</em>−2, 3<em>k</em>−3), (<em>k</em>−1, <em>k</em>+2),</span>
<span
class="math inline">(<em>k</em>, 3<em>k</em>−1), (<em>k</em>+1, <em>k</em>+1), (<em>k</em>+2, 3<em>k</em>−2), (<em>k</em>+3, <em>k</em>+3), (<em>k</em>+4, <em>k</em>+4), (<em>k</em>+5 <em>k</em>+5), ⋯, (<em>n</em>−1, <em>n</em>−1), (<em>n</em>, 4).</span>
No two queens will have the same positive diagonal provided that <span class="math inline">4 < <em>k</em> + 1</span> and <span class="math inline">4<em>k</em> − 4 > 3<em>k</em> − 1</span>, or equivalently, <span class="math inline"><em>k</em> > 3</span>, or equivalently, <span class="math inline"><em>n</em> > 6</span>.
Similarly, the ordered pairs for the queen number and its respective negative diagonal number, <span class="math inline">(<em>q</em><em>u</em><em>e</em><em>e</em><em>n</em>−<em>l</em><em>a</em><em>b</em><em>e</em><em>l</em>, <em>n</em><em>e</em><em>g</em><em>a</em><em>t</em><em>i</em><em>v</em><em>e</em>−<em>d</em><em>i</em><em>a</em><em>g</em><em>o</em><em>n</em><em>a</em><em>l</em>−<em>l</em><em>a</em><em>b</em><em>e</em><em>l</em>)</span>, are as follows:
<span
class="math inline">(1,<em> n</em>−3), (2, 4), (3, 7), (4, 10), ⋯, (<em>k</em>−3, 3<em>k</em>−11), (<em>k</em>−2, 3<em>k</em>−8), (<em>k</em>−1, <em>k</em>−1), </span>
<span
class="math inline">(<em>k</em>, 3<em>k</em>−2), (<em>k</em>+1, <em>k</em>+2), (<em>k</em>+2, 3<em>k</em>+1), (<em>k</em>+3, <em>k</em>+8), (<em>k</em>+4, <em>k</em>+11), (<em>k</em>+5, <em>k</em>+14), ⋯,
(<em>n</em>−1, 2<em>n</em>−4), (<em>n</em>, <em>n</em>+3).</span>
Here the dots denote terms of an arithmetic progression of the second coordinate with difference 3. The numbers 4, 7, 10, … , <span class="math inline">3<em>k</em> − 8, 3<em>k</em> − 2, 3<em>k</em> + 1</span> all leave remainder 1 on division by 3 //(Why?// <span class="math inline"><em>n</em> = 6<em>m</em> + 2.</span>) while the numbers <span class="math inline"><em>k</em> − 1, <em>k</em> + 2, <em>k</em> + 8</span>, … , <span class="math inline">2<em>n</em> − 4</span> are all divisible by 3.
The numbers <span class="math inline"><em>n</em> − 3</span> and <span class="math inline"><em>n</em> + 3</span> leave remainder 2 on division by 3.
Thus, no numbers occur more than once in the above two lists, proving that no two queens lie on the same diagonals. <br>
<strong>Case 2</strong>: <span class="math inline"><em>n</em></span> is an odd integer, <span class="math inline"><em>n</em> ≥ 5</span>.
<strong>Observation: </strong> In all the arrangements constructed for even <span class="math inline"><em>n</em></span> as above, there are no queens on the great positive diagonal.
Note that <span class="math inline"><em>n</em> − 1</span> is even for odd <span class="math inline"><em>n</em></span>. Also, for <span class="math inline"><em>n</em> ≥ 5</span>, we have <span class="math inline"><em>n</em> − 1 ≥ 4</span>.
The bigger <span class="math inline"><em>n</em> × <em>n</em></span> board contains a smaller <span class="math inline">(<em>n</em>−1) × (<em>n</em>−1)</span> subsquare. In that subsquare starting from the bottom left square, we place <span class="math inline"><em>n</em> − 1</span> queens following the above scheme. The very last queen, <span class="math inline"><em>n</em><sup><em>t</em><em>h</em></sup></span> queen, is placed on the top right corner.
Independence is guaranteed since this <span
class="math inline"><em>n</em><sup><em>t</em><em>h</em></sup></span> queen is independent of the previous <span class="math inline"><em>n</em> − 1</span> queens.
[[At last!|n-queens-closing-8]]<h3 class="heading">The Last Problem(track: 'paperturn', 'play')</h3>
<p class="handwrittenbyholmes">And now, Watson, after this exhaustive proof (in a literal sense), I really must take your leave. There is a rascal in the army who has been threatening Lord Davis for quite some time who needs my attention.
I await the day you step on English soil, Watson.
Best wishes,
Sherlock Holmes
Consulting Detective
221B, Baker Street
PS: I have had the honour of visiting Col. Pritchard, who gave me a copy of this photograph. I attach it here as per his instructions - it is the Berkshire regiment!
<div style="text-align:center"><img class="holmes" width=400 src="https://moneaalhad.files.wordpress.com/2023/08/untitled-design-4-1109691213-e1692295019731.png"></div>
</p>
[[Closing]]
<h3 class="heading">Closing</h3>
That was the last letter I ever received from Holmes. The years of the War passed by in a hazy blur, and so did the revolution. My garrison was slowly either killed, sent back to England or shot by insurgents - and now, fifteen years later, I sit here by myself in a lodge provided by the Embassy.
My hand is unsteady, and my memory fails me sometimes. The Jezail bullet, fired so long ago in Afghanistan, still plagues me. The revolution is over long since, and there is no place for me in Russia or England.
I cannot help but reminisce about the past. I would pay to see a chowkidar from India or one of the lads from Berkshire who served alongside me - even Moriarty would be a welcome face.
My heart yearns for the warm English sun, the fields full of daffodils, the hey-ho of London, even with all of the rascals and pickpockets that roam the streets, the trains and most of all, the occupant of my lodgings at 221B Baker Street - Sherlock Holmes.
[[What now?|end]]
<h5 class="heading">SoME '23(set: $game_completed to 1)(playlist: 'background', 'pause')</h5>
<h2 class="heading">Fin.</h2>
Our submission for SoME 2023 ends here. During the process of making this explainer, we've learnt a lot of cool stuff - namely, programming, discrete maths, graph theory and how to center content inside a div.
We've proofread this thing a couple of times. However, if you find bugs, errors, inaccuracies, anachronisms or canonical errors, please email us at <strong>somecroissant@gmail.com</strong>.<br><br>
<h4 class="heading">Extra Problems</h4>
Given the time constraint, we decided to skip over [[this problem|the-hardest-checkmate]]. However, don't let that deter you from checking it out – it's a great exercise in logical thinking. <br><br>
<h4 class="heading">References</h4>
Any progress in the field of science is a team effort. We're standing on the shoulders of these giants...
* Engel A, Problem-Solving Strategies; Springer: Problem Books in Mathematics, p. 364.
* Matousek J; Nesetril, J, Invitation to Discrete Mathematics, Oxford: Second Edition, p. 23.
* Smullyan R, The Chess Mysteries of Sherlock Holmes, p. XI, 26.
* Yaglom A M; Yaglom I M, Challenging Mathematical Problems with Elementary Solutions (Volume 1), p. 12.<br><br>
<h4 class="heading">Resources Used</h4>
And these giants as well!
* <a href="https://www.twinery.org" target="_blank" rel="noopener noreferrer">Twine</a> to generate interactive stories:
* <a href="https://www.canva.com" target="_blank" rel="noopener noreferrer">Canva</a> to edit images
* <a href="https://www.figma.com" target="_blank" rel="noopener noreferrer">Figma</a> to create frames, mockups and prototypes
* <a href="https://www.calligraphr.com/" target="_blank" rel="noopener noreferrer">Calligraphr</a> to generate our own font, 'Bymoriarty'
* <a href="https://pandoc.org" target="_blank" rel="noopener noreferrer">Pandoc</a> to alleviate headaches caused by LaTeX and HTML mismatches
* The <a href="https://github.com/ChapelR/harlowe-audio" target="_blank" rel="noopener noreferrer">Harlowe Audio Library</a> by ChapelR for background music<br><br>
Besides these, there's a whole lot of online forums, documentation, helpdesks, tutorials and obscure Reddit posts that we'd like to thank.
We're grateful to Grant Sanderson and 3Blue1Brown for making knowledge free, accessible, and easy to understand.
We're strong proponents of free, open-source, and easy to use software. We encourage you to use our submission for education wherever you wish to with proper attribution.
Here's some more information [[about us|about-us]].
[[This link will take you back to the title page.|Instructions]]
Want to jump to a problem? Here's the [[index.|index]]<h3 class="heading">Extra Problems</h3>
These will be uploaded soon. Keep checking back!
[[This link will take you back to the title page.|Instructions]]
[[Back to Fin.|end]]<h3 class="heading">About Us</h3>
<div style="text-align:center"><img class="picture" src="https://moneaalhad.files.wordpress.com/2023/08/img_189636286759-1.jpeg"></div>
[[Fin.|end]] <h3 class="heading">Index Page</h3>
This page allows you to jump to any problem provided you've visited it atleast once.
* [[Double Chess|double-chess-problem-2]] <br>
* [[2^^n^^ by 2^^n^^ Board|2nx2n-problem-2]]<br>
* [[Promoted Bishop|retrograde1-problem-2]]<br>
* [[Bishop Square Color|retrograde2-problem-2]]<br>
* [[//n// Independent Queens|n-queens-problem-2]]<br>
* [[White To Play, Mate in Two|the-hardest-checkmate]]<br>
[[Return to Conclusion|end]]
<div style="text-align:center"><img width=500 src = "https://moneaalhad.files.wordpress.com/2023/08/helo-fish-.jpeg"></div>
[[Instructions]] <h3 class="heading">Blackfriars Chess Club(if: not(playlist: 'background', 'isplaying'))[(playlist: 'background', 'play')](track: 'paperturn', 'play')</h3>
It was a fine evening for a stroll around town. Winter had lifted her cold embrace from the land, and the smoke from the chimneys had long since drifted away on the strong spring breeze. Holmes had just returned from the boxing ring. He tossed a letter onto the table.
"It’s an invitation from Alfred Newman at the Blackfriars Chess Club - prizes offered to anyone who can solve a series of chess puzzles-"
"Say no more. I shall get my coat, and my hat."
"Your enthusiasm is commendable - off we go, then."
Five minutes later, we were sitting in a brougham speeding away from Baker Street.
"What sort of a person is Alfred Newman? The name eludes me."
"Oh, he is a showman. By trade, he edits the Times. He has made quite a few friends in high places, and is well off - the reason, you see, for him organising this particular event. Newman is a huge chess enthusiast - indeed, I may go so far as to call him a fanatic. He is obsessed with "uncovering the next hidden talent". I have no doubt that at this event, we shall find numerous agents ready to shower potential candidates with contracts."
"Have you met him?"
"Once, long back in connection with one of my cases. I had found a misprint in the Times."
I raised an eyebrow.
"Ah, Watson, you must not blame me for enjoying what little praise I get as a consulting detective. In the beginning of my career, I felt strongly that I should be credited with my work, and Newman conveniently forgot to print my name next to Lestrade. It was this misprint I was going to correct.
"Be prepared, Watson, for devilish problems. The man is shrewd as a fox, and the problems are likely to be confusing."
There was a considerable crowd at the chess club. The line of cabs waiting outside was disheartening, but Holmes slipped a shilling to the man at the door, who let us in without much fuss.
The arrangement inside was spectacular. There were twelve tables, each suffocated by a large assortment of people - save for one, where only two people were engaged in heated discussion. One of them, a portly man with a moustache, was evidently very invested in the debate.
"I tell you, it’s not possible," he said.
Holmes and I walked to Table 12, where [[this position|hardest-checkmate-problem-3]] was laid out.
<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
<p class="question"><br>It is White's move. Can he mate in two?</p>
<div style="text-align:center;"> <img width=600 height=600 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/fc-2-castled-1.png" alt="chessboard for problem"> </div>
<p class="hint">Hint: Account for pieces on the board as well as off the board. It is strongly recommended to follow this solution with a chessboard.
[[Next|hardest-checkmate-2]]
(if: $game_completed is 1)[[[Index|index]]]<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
"Good evening. Would you care to try Mr. Newman’s hardest puzzle yet?"
Holmes looked over the board again.
"Fiendish - and familiar! But first - who was that gentleman?"
"That is Lord Sabitzer, from the Black Forest. He claimed that this position was impossible, and there is no mate in 2."
"Interesting, very interesting!" Holmes remarked, rubbing his palms together.
"What is the prize for solving this particular puzzle?" I enquired.
The supervisor smiled.
"As this is Mr. Newman’s first event in Blackfriars, a bottle of champagne and thirty pounds are offered to whoever can find a solution. Moreover, whoever manages to also solve the puzzle on Table 6 alongside this one will get an additional ten pounds."
"Excellent! I would like to try." I said.
The table supervisor left me with a notepad. I found two ways to mate in four moves almost instantly. But try as I might, I simply could not find a mate-in-two. The Black king always found ways to escape. For instance, <span class="chessmove">Qd6!</span> is a great move - but countered by castling queenside.
I also calculated some trapping positions to no avail.
Finally, after almost forty minutes, I wiped my brow and left Table 12. Holmes was discussing something animatedly with Lord Sabitzer, and I motioned him over.
"Well, Watson? How goes the fight for thirty pounds?"
"Dismal. I was able to find a mate in four moves, but that does not help me."
"I see. It seems [[I shall have to try."|hardest-checkmate-solution-4]]<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
Leaving Holmes to his own devices, I chatted with various acquaintances for a while. Almost half an hour later, there was a smattering of applause from Table 12.
"Why, you’ve done it, Holmes!" I said.
"Always the tone of surprise, my dear Watson - but it’s quite all right! I have solved the position, and am twenty-eight pounds richer if you account for our cab fare back home."
"Pray tell me how."
Around this time, the crowd from some of the other tables was beginning to shift in our direction, and some of the supervisors themselves were walking over.
"Like all of you here, I immediately found mate in four moves - and also in five, and seven moves. But that is of no consequence - let us look at the moves I played out.
"There is no sequence where we can check and mate on the next move. Therefore, the first move must be a trapping move - that is, it must threaten checkmate on the next move.
"What pieces can deliver mate? The strongest piece is clearly the queen. Therefore, <span class="chessmove"> Qd6!</span> seems like a good idea, threatening <span class="chessmove">Qxe7#</span>. The Black queen is pinned by the rook, and capturing with the pawn gives <span class="chessmove">Ng7#</span>.
However, Black has a sneaky trick up his sleeve - castling counters //Qd6// and it is no longer a checkmate in two moves.
<div style="text-align:center;"> <img width=600 height=600 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/fc-1-problem-1.png" alt="chessboard for problem"> </div>
[[What now?|hardest-checkmate-solution-5]]
<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
"I was stumped. There seemed to be no solution indeed." Holmes said.
"However, Watson will tell you that I have a method: if you have eliminated all possible outcomes, then whatever remains, however absurd, must be the truth. I had considered all moves where mating was possible. But mate problems only look at legal moves, not illegal ones!
"If I could prove that either the King or the Black rook had moved in the match sometime prior, I could prove that castling is illegal in this situation - and therefore there is no escape from <span class="chessmove">Qd6!</span>
"Now, castling is legal provided three conditions are met: the rook should not have moved, the king should not have moved and the king cannot come into check while castling.
"And so I set out to find out if Black can castle."
[[Can Black castle?|hardest-checkmate-solution-6]]
<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
"For the rest of this solution, we will refer to the rooks, bishops, and knights as pieces, while the royalty will retain their titles.
"White is missing two pawns, and three pieces - a rook, the light square bishop and a knight. Black is missing a pawn, a rook, both bishops and both knights."
"Let us look first at the pawn on //g2//. It has made four captures coming from <span class="chessmove">c7</span>. Moreover, the captures have been on squares <span class="chessmove">d6</span>, <span class="chessmove">e5</span>, <span class="chessmove">f4</span>, and <span class="chessmove">g3</span> respectively - which are all dark squares. Therefore, the pawns on <span class="chessmove">h4</span> and <span class="chessmove">g6</span> must be from <span class="chessmove">h7</span> and <span class="chessmove">g7</span>.
"What could have been Black’s last move? If it was with the rook or the king, we can play <span class="chessmove">Qd6</span>, since castling is now illegal, and we threaten mate in two. So, we come up with four possible moves:
0. It could be <span class="chessmove">g6</span> from <span class="chessmove">g7</span>.
0. It could be <span class="chessmove">g2</span>, coming from <span class="chessmove">g3</span>.
0. The queen could have moved to <span class="chessmove">f8</span> without a capture. This means that White’s rook captured a piece on <span class="chessmove">g8</span> and delivered check, which was blocked by the queen.
0. The queen could have come from //g7// and captured a white piece on <span class="chessmove">f8</span> thus getting pinned by the rook on <span class="chessmove">g8</span>.
"Let us consider all of these moves in turn - but before, here are some observations I made quickly:
* One of the four White pawns from <span class="chessmove">e4, f3, g5</span> and <span class="chessmove">h6</span> is from <span class="chessmove">d2</span>, as there are two pawns on the //h// file.
* Additionally, all the White pawns have collectively captured four Black pieces.
* Black is missing six pieces. One of these - the light squared bishop - fell on its home square. This proves that neither of White’s pawns from <span class="chessmove">a2</span> and <span class="chessmove">b2</span> have made more than one capture (to tally with 4 captures made by the pawns and the fallen bishop).
* White is missing two pawns and three pieces. Four of these five pieces were captured by the pawn marching from c7 all the way to <span class="chessmove">g3</span> via <span class="chessmove">g6</span> and <span class="chessmove">e5</span>. Of these captures, none can be the pawns from <span class="chessmove">a2</span> and <span class="chessmove">b2</span> (since the //b// pawn will need two captures to get to the //d// column, and we proved that both pawns made collectively one capture).
Therefore, at least one of the pawns from <span class="chessmove">a2</span> and <span class="chessmove">b2</span> has promoted and both pawns made only one capture between them.
[[Next|hardest-checkmate-solution-7]]
<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
"First, let us look at case 3 - the queen on //g7// moving to //f8// to block check. The White rook captured a piece on //g8// to deliver check. This means another Black piece was captured - bringing the total up to five. Now, if this is true, the pawns from //a2// and //b2// could not have made any captures at all.
"Just a moment. How can you be sure?" I interrupted.
"Because Black is missing exactly five pieces. If the rook captured a piece on //g8//, and the black pawn on //g7// made four captures, there is no piece left for the pawns from //a2// and// b2 //to capture." Holmes said patiently.
"Next, consider case 1 - //g6//, coming from //g7//. This would mean that Black’s dark square bishop was captured on its home square, namely //f8//, because neither of the pawns moved to let it out.
"In either of these cases, the pawns from //a2 //and //b2// have made no captures.
"If the pawns did not make captures, are not on the board and one of them still promoted, it must have been the pawn from //a2//-"
"Wait," I said, interrupting thoughtlessly again, "Why could the pawn from //b2// not promote?"
"My dear Watson, the pawn on //b7// has not moved and blocks the// b2// pawn from promoting." Holmes replied with the hint of a smile.
"Pray continue." I said, feeling very flushed.
"So the// a2// pawn has promoted. It cannot make a capture, and therefore can only promote on //a8//. However, if the rook was still on //a8//, the pawn could never have promoted - and therefore, when the pawn promoted, the rook was not on //a8// - hence, the rook has moved at least once in the game.
[[Next Cases|hardest-checkmate-solution-8]]
<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
"Now for cases 2 and 4; //g2// coming from //g3//, and the queen coming from //g7// capturing a White piece on //f8//, getting pinned by the rook in the process.
"Suppose the queen has captured a White piece on //f8//. Then combined with the four White pieces fallen on dark squares, both the pawns need to promote - else, there is one piece less left to capture, and White has the logical impossibility of losing less pieces than the ones which were captured.
"Similarly, if the pawn on //g2// has just moved from //g3//, all of its captures were made on dark squares. Now, at first sight, this might appear to be true, and therefore stonewall us again. But ha! Look at this!" said Holmes energetically, moving towards the board.
"The one piece which can never be captured on a dark square is White’s light square bishop - which is not on the board, therefore forcing both //a2// and //b2// to promote. Both the pawns have made exactly one capture between the two of them - and this capture must have been made by the //b2// pawn, since the pawn on //b7// blocks promotion on //b8//.
"We ask ourselves: what square did this pawn promote on? The answer is, of course, a8. But that must mean that for both promotions, the rook on //a8// was absent - and therefore, has moved at least once in the game."
"The rest of the puzzle is now prosaic, my good friends!" he continued, striding towards the other side of the board. He picked the queen up and put her on //d6//.
"I play queen to //d6//. Having proved that you cannot castle, any move you play now ends in checkmate! If the queen is captured on //d6// by the pawn, //Ng7// delivers checkmate thunderously! The king’s only escape square is now guarded by the dark square bishop on //f6//.
"If the queen is not captured, it is of no consequence what you play. Black can play //g8=Q, Ra2, Rd8// - all of them would not make a pennyworth of difference to the unstoppable //Qxe7#// - the Black queen, the sole defender of the pawn, is pinned by the White rook on //g8//. And, therefore, I have solved the puzzle!"
[[Next|hardest-checkmate-solution-9]]
<h3 class="heading">White To Play, Mate In Two(track: 'paperturn', 'play')</h3>
Never has Holmes had more complete control of a room than that evening in the chess club. There was complete silence at first, as the crowd struggled to comprehend the intricacy of the position. Then, slowly, the applause began, swelling around the room. To cap it all off, Mr. Newman himself came to Table 12.
"A compelling narrative, Mr. Holmes, quite brilliant, exceptionally flawless! Your reputation as a consulting detective is not unfounded. It was one of my eccentric ideas, you see - one of them-"
"I am not used to public admiration, Mr. Newman. However, tell me - are you in correspondence with a certain M. Rochefort from the United States?"
Aghast, Mr. Newman looked at me.
"Is he a wizard?"
"No matter, Mr. Newman. I hope to solve some of the other problems here, as well. Good day to you!"
"Holmes, how could you possibly know-"
"My dear Watson, M. Rochefort, the original composer of this demonic problem, is none other than Moriarty himself!"
[[Index|index]] <h3 class="heading">Mathematics On The Chessboard(track: 'paperturn', 'play')</h3>
We noticed from the feedback that a few people were not convinced that you could tile said board with any arbitrarily chosen square removed. Thus we decided to elaborate:
It’s clear from the inductive solution that you can always tile a <span class="math inline">2<sup><em>n</em></sup></span> board when one of the four central most squares is removed.
Assume that you are given an arbitrary square to remove from the board.
If it is one of the four central squares, the method is already clarified.
Thus, assume that it is not one of the four central squares. Then start by removing one of the four squares manually and tile the remaining board as given in the solution. Then, translate, or move, the central removed square to the arbitrary square given to you as follows:<br>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/1.jpg" alt="Board shown with one central square removed."><br>Board shown with one central square removed.</div>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/2.jpg" alt="Board shown with one central square removed and the assigned square marked in purple."><br>Board shown with one central square removed and the assigned square marked in purple.</div>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/3.jpg" alt="A fully tiled board with the assigned square covered with a domino and the central square left uncovered."><br>A fully tiled board with the assigned square covered with a domino and the central square left uncovered.</div>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/4.jpg" alt="The fully tiled board with the assigned square covered with a domino and the removed square 'shifted' by rotating the domino."><br>The fully tiled board with the assigned square covered with a domino and the removed square 'shifted' or moved by rotating the purple domino.</div>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/5.jpg" alt="The removed square 'shifts' further towards the assigned square..."><br>The removed square 'shifts' further towards the assigned square...</div>
<div style="text-align:center;"> <img width=400 height=400 class="chessboard" src="https://moneaalhad.files.wordpress.com/2023/09/6.jpg" alt="bada bing, bada boom!"><br>bada bing, bada boom!</div>
[[Return to P2 solution|2nx2n-solution-4]]
[[Next Problem|retrograde-intro-1]]