The Year Puzzle

Contents

• Introduction
• The Rules of the Puzzle
• My Approach
• My Results
• Other Resources (English)
• Other Resources (German and Dutch)

Usually, the "Year Puzzle" challanges you to connect the four digits of the current year (in that order) with the usual arithmetic operators, in order to form expressions which yield all the values from (say) 1 to 100.

The "Four 4's" puzzle is exactly the same for the year 4444. Both forms have appeared repeatedly e.g. in the news group sci.math.

While this puzzle is fun for the beginner, discussing the more difficult cases (how do you build a 39 with four 4's?) rapidly exhibit the need for more exact rules of the game. There seems to be no generally accepted rule set.

The Rules of the Puzzle

The following rules seem to be generally accepted:

• The four digits of the year (say 1, 9, 9, 8) have to appear in the exact original order, as in
  16 = 1*9 + 9 + 8.
  All digits must appear exactly once.
• You may use the four basic binary arithmetic operators "+", "-", "*" and "/" (for addition, subtraction, multiplication and division).
• You may use the unary negation operator "-".
• You may use parantheses for grouping.

• You may use the binary exponentiation operator "^", as in
  18 = 1^9+9+8.
• You may use the concatenation of digits, as in
  91 = 1-8+98.
• You may use the unary factorial operator "!", as in
  19 = 1 + 9 + 9!/8!.
• You may use square roots, as in
  4 = 1 + sqrt(9/9 + 8).

• You may use decimal points "." before and between digits, as in
  33 = 4! + 4 + sqrt(4)/.4
• You may use the unary odd factorial operator "!!", which multiplies the odd natural numbers upto and including its argument, i.e. 5!! = 1*3*5. This is allowed for odd arguments, only.
• You may use the unary even factorial operator "!!", which multiplies the even natural numbers upto and including its argument, i.e. 6!! = 2*4*6. This is allowed for even arguments, only.

• Repeating decimal fractions: .(4) = 0.444... = 4/9.
• Logarithms log(x). Logarithms to other bases, as in
  0.5 = log4(sqrt(4)).
• Exponential function exp(x).
• Trigonometric function like sin(x) and atan(x).

When I first tried to solve the four 4's with paper and pencil, I failed to find solutions for such small values as 33 and 39. Then I started to systematically list all partial expressions with just two of the four fours and to tabulate the resulting values. I soon realized that this is a job for a computer program, not for a programmer :-). Hence I started to write a program, which should perform that systematic search. I used "awk" (i.e. "nawk") for it, and I have extended the first experimental version since then, such that I am very satisfied with the results. A few minutes on a typical workstation are enough to yield a list of solutions, which is as complete as I could do myself.

No, sorry, my awk program is not in the public domain. But I'll share some of my ideas. Let's take "1998" as an example.

• Before building expressions with all the four digits in it, we build all potential partial expressions, and tabulate them by value.
• We start to build the (trivial) expressions for signatures of length 1, i.e. for "1", "9" and "8". Using binary operators we then build expressions for signatures of length 2, i.e. for "19", "99" and "98". And so on, until we reach length 4, and print the results. E.g. from "10 = 1+9" and "1 = 9-8" we build
  • 11 = (1+9)+(9-8),
  • 9 = (1+9)-(9-8),
  • 10 = (1+9)*(9-8),
  • 10 = (1+9)/(9-8) and
  • 10 = (1+9)^(9-8).
• If we have two expressions with the same value for the same signature, we keep the shorter or more simple expression and discard the other. E.g. we keep "4*4" and discard "sqrt(4^4)".
• If concatenation is allowed, we just allow signatures of length greater than 1 to trivially yield an expression.
• We store all intermediate values as normalized rational numbers.
• We restrict nominator and denominator of intermediate values to some suitable range. At least we have to take care of overflows.
• Whenever we invent a new expression with a new value, we recursively try all unary operators to yield further values.

• Program output for the four 4's
• Program output for 1996
• Program output for 1997
• Program output for 1998
• Program output for 1999
• Program output for 2000
• Program output for 2001
• Program output for 2002
• Program output for 2003
• Program output for 2004
• Program output for 2005
• Program output for 2006
• Program output for 2007
• Program output for 2008
• Program output for 2009
• Program output for four 0's
• Program output for four 1's
• Program output for four 2's
• Program output for four 3's
• Program output for four 5's
• Program output for four 6's
• Program output for four 7's
• Program output for four 8's
• Program output for four 9's

• Ruth Carver has collected solutions in the range 0-100 about the
  • 1996 Puzzle, the
  • 1997 Puzzle, and the
  • Four 4's Puzzle.
  All this is part of the Math Forum (formerly the Swarthmore Math Forum).
• Poul Bourke allows repeating decimals and accepts solutions with less than four 4's. His solution table goes up to 1024.
• Dave Rusin has collected Usenet discussion about 1996 and about the four 4's. He seems to perceive this as a special case of The KRYPTO "Arithmetic" Card Game.
• In the Question Corner and Discussion Area at the University of Toronto there is a short discussion of the four 4's.
• Peter Karsanow has collected information about the original source and problem statement, as well as some annotated references and solution to the original problem.
• David A. Wheeler has a set of solutions for the four fours problem from 0 to 40,000 that he terms The Definitive Four Fours Answer Key.
• A more general platform to discuss the mathematics involved may be Cramster Math Help (I did not use it myself).

• In ihrem Sommerrätsel 1998 hat Cathie nach Lösungen für die Vier Vieren gefragt. Die Lösungstabelle stimmt gut mit meinen eigenen Ergebnissen überein. Zum Jahreswechsel hat Cathie dann nach Lösungen für das Jahr 1998 gefragt.