# Plow-Pair

Futoshiki is fun. It’s a number-puzzle where you use logic to re-create a 5×5 square in which every row and column contains the numbers 1 to 5. At first, most or all of the numbers are missing. You work out what those missing numbers are by using the inequality signs scattered over the futoshiki. Here’s an example:

There are no numbers at all in the futoshiki, so where do you start? Well, first let’s establish some vocabulary for discussing futoshiki. If we label squares by row and column, you can say that square (4,5), just above the lower righthand corner, dominates square (4,4), because (4,5) is on the dominant side of the inequality sign between the two squares (futōshiki, 不等式, means “inequality” in Japanese). Whatever individual number is in (4,5) must be greater than whatever individual number is in (4,4).

Conversely, you can say that (4,4) is dominated by (4,5). But that’s not the end of it: (4,4) is dominated by (4,5) but dominates (3,4), which in its turn dominates (2,4). In other words, there’s a chain of dominations. In this case, it’s a 4-chain, that is, it’s four squares long: (4,5) > (4,4) > (3,4) > (2,4), where (4,5) is the start-square and (2,4) is the end-square. Now, because 5 is the highest number in a 5×5 futoshiki, it can’t be in any square dominated by another square. And because 1 is always the lowest number in a futoshiki, it can’t be in any square that dominates another square. By extending that logic, you’ll see that 4 can’t be in the end-square of a 3-chain, (a,b) > (c,d) > (e,f), and 2 can’t be in the start-square of a 3-chain. Nor can 3 be in the start-square or end-square of a 4-chain.

Using all that logic, you can start excluding numbers from certain squares and working out sets of possible numbers in each square, like this:

[whoops: square contains errors that need to be corrected!]

Now look at column 1 and at row 4:

In column 1, the number 5 appears only once among the possibles, in (1,1); in row 4, the number 1 appears only once among the possibles, in (4,1). And if a number appears in only one square of a row or column, you know that it must be the number filling that particular square. So 5 must be the number filling (1,1) and 1 must be the number filling (4,1). And once a square is filled by a particular number, you can remove it from the sets of possibles filling the other squares of the row and column. I call this sweeping the row and column. Voilà:

Now that the 5 in (1,1) and the 1 in (4,1) have swept all other occurrences of 5 and 1 from the sets of possibles in column 1 and row 4, you can apply the only-once rule again. 2 appears only once in row 4 and 5 appears only once in column 4:

So you’ve got two more filled squares:

Now you can apply a more complex piece of logic. Look at the sets of possibles in row 3 and you’ll see that the set {2,3} occurs twice, in square (3,1) and square (3,4):

What does this double-occurrence of {2,3} mean? It means that if 2 is in fact the number filling (3,1), then 3 must be the number filling (3,4). And vice versa. Therefore 2 and 3 can occur only in those two squares and the two numbers can be excluded or swept from the sets of possibles filling the other squares in that row. You could call {2,3} a plow-pair or plow-pare, because it’s a pair that pares 2 and 3 from the other squares. So we have a pair-rule: if the same pair of possibles, {a,b}, appears in two squares in a row or column, then both a and b can be swept from the three other squares in that row or column. Using {2,3}, let’s apply the pair-rule to the futoshiki and run the plow-pare over row 3:

Now the pair-rule applies again, because {4,5} occurs twice in column 5:

And once the plow-pare has swept 4 and 5 from the other three squares in column 5, you’ll see that 3 is the only number left in square (1,5). Therefore 3 must fill (1,5):

Now 3 can be swept from the rest of row 1 and column 5:

And the pair-rule applies again, because {1,2} occurs twice in row 2:

Once 2 is swept from {2,3,4} in square (2,1) to leave {3,4}, 3 must be excluded from square (2,2), because (2,2) dominates (2,1) and 3 can’t be greater than itself. And once 3 is excluded from (2,2), it occurs only once in column 2:

Therefore 3 must fill (5,2), which dominates (5,1) and its set of possibles {2,3,4}. Because 3 can’t be greater than 4 or itself, 2 is the only possible filler for (5,1) and only 3 is left when 2 is swept from (3,1):

And here are the remaining steps in completing the futoshiki:

The complete futoshiki

Animation of the steps required to complete the futoshiki

Afterword

The pair-rule can be extended to a triplet-rule and quadruplet-rule:

• If three numbers {a,b,c} can occur in only three squares of a row or column, then a, b and c can be swept from the two remaining squares of the row or column.
• If four numbers {a,b,c,d} can occur in only four squares of a row or column, then a, b, c and d can be swept from the one remaining square of the row or column (therefore the number e must fill that remaining square).

But you won’t be able to apply the triplet-rule and quadruplet-rule as often as the pair-rule. Note also that the triplet-rule doesn’t work when {a,b,c} can occur in only two squares of a row or column. An n-rule applies only when the same n numbers of a set occur in n squares of a row or column. And n must be less than 5.

Post-Performative Post-Scriptum

Domination. Exclusion. Inequality. — an earlier look at futoshiki

# Domination. Exclusion. Inequality.

Sudoku has conquered the world, but I think futoshiki is more fun — more concentrated, more compact and quicker. When complete, a 5×5 futoshiki will be a Roman square in which every row and column contains the numbers 1 to 5, but no number is repeated in any row or column. You have to work out the missing numbers using logic and the “inequality” signs that show whether one square contains a number more than or less than a number in a neighboring square — futōshiki, 不等式, means “inequality” in Japanese (fu- is the negative prefix). Here’s an example of a futoshiki puzzle:

Futoshiki puzzle, with 4 in square (3,2) and 3 in square (1,1)

If you identify the squares by row and column, 4 is in (3,2) and 3 is in (1,1). And you can say, for example, that the empty square (3,4) dominates the empty square (3,3) or that (3,3) is dominated by (3,4). I’ll describe one route (not the best or most efficient) to completing the puzzle. Let’s start by considering the general rules that 1 cannot appear in any square that dominates another square and that 5 cannot appear in any square dominated by another square.

If you extend that logic, you’ll see that 4 cannot appear in any square that is at the end of what you might call a chain of dominations, where one square dominates a second square that in turn dominates a third square. Therefore, in the puzzle above, 4 cannot appear in squares (2,1) and (4,1) of column 1. And it can’t appear in square (3,1), because that would mean two 4s in the same row. This leaves one place for 4 to appear: square (5,1). And if 4 is there, 5 has to be in square (3,1):

Now look at row 3. Two of the remaining three empty squares are dominators: (3,4) dominates (3,2) and (3,5) dominates (2,5). 1 cannot appear in a dominating square, so 1 has to be in the dominated square (3,3):

The next step I’ll take is a bit more complicated. In row 2, the number 4 cannot be in (2,1) and (2,4). It can’t be in (2,1) because that would mean 4 was greater than itself. And it can’t be in (2,4), because (2,4) is dominated by (3,4) and (3,4) can’t contain 5, the only number that dominates 4. Therefore 4 must be in either (2,3) or (2,4). But so must 5. Therefore (2,3) contains either 4 or 5 and (2,4) contains either 4 or 5. That means that the numbers [1,2,3] must be in the other three squares of row 2. Now, 3 can’t be in square (2,1), because the chain of dominations is too long. And 3 can’t be in (2,5), because (2,5) is dominated by (3,5), which contains either 2 or 3. Therefore 3 must be in (2,2):

Now consider column 2. Square (4,2) cannot contain 1 or 5, because it’s both dominated and dominating. And if it can’t contain 1 or 5, there’s only one number it can contain: 2. And it immediately follows that (4,1) must contain 1, the only number less than 2. And if four squares of column 1 now contain the numbers [1,3,4,5], the remaining empty square (2,1) must contain 2:

Now consider row 2. Squares (2,3) and (2,4) contain either 4 or 5, therefore (2,5) must contain 1:

Now consider row 5. The number 1 is logically excluded from three squares: from (5,3), because there’s a 1 in (3,3); from (5,4), because (5,4) dominates (5,5); and from (5,5), because there’s a 1 in (2,5). Therefore 1 must be in (5,2). And if 1 is in (5,2), the number 5 must be in (1,2):

Now 1 pops up in row 1 because it can only be in (1,4):

And 5 pops up in row 4 because it can only be in (4,5):

Once 5 is in (4,5), the number 4 must be in (4,4) and the number 3 in (4,3):

The 4 of (4,4) immediately collapses the ambiguity of (2,4), which must contain 5. Therefore (2,3) contains 4:

Next, 5 pops up in (5,3):

And 3 must be in (5,4), dominating 2 in (5,5):

With 3 in (5,4) and 2 in (5,5), the ambiguity of (3,4) and (3,5) collapses:

And the square is completed like this:

Here’s an animated version of the steps to completion:

Futoshiki puzzle animated

# There are 719 errors in this sentence

Here’s a famous paradox (or a variant of it at least):

• There are two errers in this sentence.

The only visible error is the misspelt “errers”. But if the sentence claims to have two errors while having only one, that is another error and there are two errors after all.

Now for another variant. I’m not sure if I’ve thought this up for myself, but try this sentence:

• There are three errors in this sentence.

There are no visible errors in the sentence. Therefore it has one error: the claim that it has three errors when there is in fact no error. But if it has one error, it’s in error to claim that it has three errors. Therefore the sentence has two errors. And if it has two errors, again it’s in error, because it claims to have three errors while having only two. Therefore it has three errors after all.

The same reasoning can be applied to any integral number of errors:

• There are five errors in this sentence.
• There are 719 errors in this sentence.
• There are 1,000,000 errors in this sentence.
• There are 1,000,000,000 errors in this sentence.

No matter how large the number of errors, the sentence becomes true instantly, because each time the sentence makes a false claim, it makes another error. But those “times of error” don’t take place in time, any more than this equation does:

• 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16…

So I think these sentences are instantly true:

• There are infinitely many errors in this sentence.
• There are ∞ errors in this sentence.

But there are infinitely many infinities. Ordinary infinity, the infinity of 1,2,3…, is called ℵ0 or aleph-zero. It’s a countable infinity. Above that comes ℵ1, an uncountable infinity. So does this sentence instantly become true?

• There are ℵ1 errors in this sentence.

I’m not sure. But I think I can argue for the validity of sentences claiming fractional or irrational number of errors:

• There is 1.5 errors in this sentence.
• There are π errors in this sentence.

Let’s have a look at “There is 1.5 errors in this sentence”. There are no visible errors, so there’s one error: the claim that sentence contains 1.5 errors. So now there seems to be another error: the sentence has one error but claims to have 1.5 errors. But does it therefore have two errors? No, because if it has two errors, it’s still in error and has three errors. And that generates another error and another and another, and so on for ever. The sentence becomes unstoppably and infinitely false.

So let’s go back to the point at which the sentence contains one error. Now, the difference between 1 error and 1.5 errors is small — less than a full error. So how big is the error of claiming to have 1.5 errors when having 1 error? Well, it’s obviously 0.5 of an error. So the sentence contains 1.5 errors after all.

Now for “There are π errors in this sentence”. There are no visible errors, so there’s one error: the claim that the sentence contains π errors. Therefore it contains one error. But it claims to have π errors, so it has another error. And if it has 2 errors and claims to have π errors, it has another and third error. But if it has three errors and claims to have π error, it’s still in error. But only slightly — it’s now committing a small amount of an error. How much? It can only be 0.14159265… of an error. Therefore it’s committing 3.14159265… = π errors and is a true sentence.

Now try:

• There is -1 error in this sentence.

What is a negative error? A truth. So I think that sentence is valid too. But I can’t think of how to use i, or the square root of -1, in a sentence like that.

# H₂Oenometry

You have two glasses each filled with exactly the same amount of liquid. One contains water, the other contains wine. First, take a teaspoon of water from the water glass and pour it into the wine glass. Next stir the wine and water until well mixed. Then take a teaspoon of the water-and-wine mixture and pour it into the glass of water.

The question now is: Is there more wine in the water glass than water in the wine glass, or is there less? (from World’s Most Baffling Puzzles, Charles Barry Townsend, Sterling, New York, 1991)

(Scroll down for answer)

Post-Performative Post-Scriptum

Oenometry means “wine-measurement”, from ancient Greek οἶνος, oinos, “wine”, + μετρία, metria, “measurement”. Its standard pronunciation would be “ee-NOM-ett-ry”, but you could conceivably say “oh-een-NOM-ett-ry” or “oi-NOM-ett-ry”.

Discussion of the answer

The original question is fair but worded to send you astray. By using the words “glass” and “teaspoon”, it creates distinct images in your mind: those of an unvarying teaspoon and of two glasses with identical-but-varying amounts of wine and water in them. So you’re guided away from considering that the contents of the glasses can be measured in teaspoons too. If you think not in teaspoons but in unspecified units (of liquid measure), it’s easier to see the truth.

If the two glasses each contain n units of liquid, by transferring water to the wine you’re adding 1 unit of water to n units of wine.

Therefore the wine glass contains n+1 units of mixed wine-and-water, of which n units are wine and 1 unit is water. Let’s say n+1 = n1.

Consider that 1 unit of that mixture contains n/n1 parts of wine and 1/n1 parts of water: n/n1 + 1/n1 = (n+1)/n1 = n1/n1 = 1 unit.

Now, if one unit of the mixture is transferred to the water glass, you take n/n1 units of wine from n units of wine in the wine glass: n – n/n1 = n-1 + 1/n1. You also take 1/n1 units of water from 1 unit of water in the wine glass: 1 – 1/n1 = (n1-1)/n1 = n/n1. So the wine glass now contains n-1 + 1/n1 units of wine and n/n1 of a unit of water.

When you add that unit to the (n-1) units of water in the water glass, it will contain (n-1) + 1/n1 units of water and n/n1 of unit of wine:

Wine glass: n-1 + 1/n1 units of wine and n/n1 of a unit of water
Water glass: n-1 + 1/n1 units of water and n/n1 of a unit of wine

Therefore, however much water and wine you start with, in the end there will be as much water in the wine glass as there is wine in the water glass. For some concrete examples:

Example #1

1. Start

Water glass: 2 teaspoons of water
Wine glass: 2 teaspoons of wine

2. Transfer water to wine glass and mix:

Water glass: 2 tsp of water – 1 tsp = 1 tsp of water
Wine glass: 2 tsp of wine + 1 tsp of water = 3 tsp of which 2/3 is wine, 1/3 is water

3. Transfer wine-and-water mixture to water glass:

One tsp of wine-and-water mixture = 2/3 tsp of wine + 1/3 tsp of water

Therefore:

Wine glass: 2 tsp of wine – 2/3 tsp of wine = 1 and 1/3 tsp of wine; 1 tsp of water – 1/3 tsp of water = 2/3 tsp of water
Water glass: 1 tsp of water + 1/3 tsp of water = 1 and 1/3 tsp of water; 0 tsp of wine + 2/3 tsp of wine = 2/3 tsp of wine

4. Finish

Wine glass contains: 1 and 1/3 tsp of wine, 2/3 tsp of water
Water glass contains: 1 and 1/3 tsp of water, 2/3 tsp of wine

Example #2

1. Start

Water glass: 10 teaspoons of water
Wine glass: 10 teaspoons of wine

Transfer water to wine glass and mix:

Water glass: 10 tsp of water – 1 tsp = 9 tsp of water
Wine glass: 10 tsp of wine + 1 tsp of water = 11 tsp of liquid of which 10/11 is wine, 1/11 is water

Transfer wine-and-water mixture to water glass:

One tsp of wine-and-water mixture = 10/11 tsp of wine + 1/11 tsp of water

Therefore:

Wine glass: 10 tsp of wine – 10/11 tsp of wine = 9 and 1/11 tsp of wine; 1 tsp of water – 1/11 tsp of water = 10/11 tsp of water
Water glass: 9 tsp of water + 1/11 tsp of water = 9 and 1/11 tsp of water; 0 tsp of wine + 10/11 tsp of wine = 10/11 tsp of wine

4. Finish

Wine glass contains: 9 and 1/11 tsp of wine, 10/11 tsp of water
Water glass contains: 9 and 1/11 tsp of water, 10/11 tsp of wine

# Fourtoshiki

I hadn’t realized that sudokus could be witty until earlier this year, when I did one that literally made me laugh, because the solutions were so clever and quirky. Foolishly, I neglected to make a note of the sudoku so I could reproduce it. But I haven’t made that mistake with this futoshiki:

Using more-than and less-than signs to deduce values, fill each line and column with the numbers 1 to 5 so that no number occurs twice in the same row or column

It’s not witty like that lost sudoku, but I think futoshikis are even more beautiful and enjoyable than sudokus, because they’re even more elemental. They’re also rooted in the magic of binary, thanks to the more-than / less-than clues. And when there’s only one number on the original grid, completing them feels like growing a flower from a seed.

# Bent Pent

This is a beautiful and interesting shape, reminiscent of a piece of jewellery:

Pentagons in a ring

I came across it in this tricky little word-puzzle:

Word puzzle using pentagon-ring

Here’s a printable version of the puzzle:

Printable puzzle

Let’s try placing some other regular polygons with s sides around regular polygons with s*2 sides:

Hexagonal ring of triangles

Octagonal ring of squares

Decagonal ring of pentagons

Dodecagonal ring of hexagons

Only regular pentagons fit perfectly, edge-to-edge, around a regular decagon. But all these polygonal-rings can be used to create interesting and beautiful fractals, as I hope to show in a future post.

# Paradoxical Puzzle Pair

Two interesting puzzles, one of which looks hard and is in fact easy, while the other looks easy and is in fact hard.

1. Three Cards

The values attached to a deck of bridge cards start with the Two of Clubs as lowest, with Diamonds, Hearts and Ace of Spades as highest.

If you draw three cards at random from the deck, what is the probability that they will be drawn in order of increasing value? (Answer 1)

2. The Hungry Hunter

A hunter, having run out of food, met two shepherds. One of the shepherd had three loaves of bread and the other had five loaves. When the hunter asked for food, the shepherds agreed to divide the eight identical loaves equally between the three of them. The hunter thanked them and gave them \$8. How should the shepherds divide the money? (Answer 2)

• Puzzles and answers from Erwin Brecher’s How Do You Survive a Duel? And Other Mathematical Diversions, Puzzles and Brainteasers (Carlton Books 2018)

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Answer #1: The puzzle sounds far more complicated than it is. The deck of cards is a red herring. The question reduces to this: Take three cards, say 2, 3 and 4 of clubs, facedown. What is the probability of turning them over in the order 2, 3, 4? There are six possible ways of arranging three cards. Therefore the probability is one-sixth.

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Answer #2: It would be wrong to split the money into \$3 and \$5. Each of the three ended up with 2⅔ loaves. In other words, the first shepherd parted with ⅓ of a loaf and the other shepherd with 2⅓ or 7/3 loaves. The first shepherd should therefore get \$1 and the second shepherd \$7.

# De Pluribus Unum

A beautifully subtle puzzle:

Scrambled Box Tops

Imagine you have three boxes, one containing two black marbles, one containing two white marbles, and the third, one black and one white marble. The boxes are labelled according to their contents — BB, WW, and BW — but someone has switched the labels so that every box is now incorrectly labelled. You are allowed to take one marble at a time out of any box, without looking inside, and by this process of sampling you are to determine the contents of all three boxes. What is the smallest number of drawings needed to do this? — Martin Gardner, Mathematical Puzzles and Diversions (1959), chapter 3, “Nine Problems”, #5.

Bald eagle, Haliaeetus leucocephalus (Linnaeus 1776)

Answer: You can learn the contents of all three boxes by drawing just one marble. The key to the solution is your knowledge that the labels on all three boxes are incorrect. You must draw a marble from the box labelled “black-white”. Assume that the marble drawn is black. You know then that the other marble in the box must be black also, otherwise the label on the box would be correct. Since you have now identified the box containing two black marbles, you can tell at once the contents of the box labelled “white-white”: you know it cannot contain two white marbles, because its label has to be wrong; it cannot contain two black marbles, because you have identified that box; therefore it must contain one black and one white marble. The third box, of course, must then be the one containing two white marbles. You can solve the puzzle by the same reasoning if the marble you draw from the “black-white” box happens to be white instead of black.

# Rock’n’Roll Suislide

Q. Each face of a convex polyhedron can serve as a base when the solid is placed on a horizontal plane. The center of gravity of a regular polyhedron is at the center, therefore it is stable on any face. Irregular polyhedrons are easily constructed that are unstable on certain faces; that is, when placed on a table with an unstable face as the base, they topple over. Is it possible to make a model of an irregular convex polyhedron that is unstable on every face?

Portrait of Luca Pacioli (1495)

A. No. If a convex polyhedron were unstable on every face, a perpetual motion machine could be built. Each time the solid toppled over onto a new base it would be unstable and would topple over again.

— From “Ridiculous Questions” in Martin Gardner’s Mathematical Magical Show (1965), chapter 10.