# Proof of Crossword Puzzle Record

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1 Proof of rossword Puzzle Record Kevin K. Ferland Bloomsburg niversity, Bloomsburg, PA Abstract We prove that the maximum number of clues possible for a daily New York imes crossword puzzle is 96 and and determine all possible puzzle grids with 96 clues. 1 Introduction. Daily New York imes crossword puzzles are constructed in a grid, each of whose squares is colored black or white. Each maximal linear segment of white squares is then attached to a clue for the puzzle. he grid itself must be constructed according to certain structure rules. 1. onnectivity: he centers of any two white squares in the grid can be joined by a path consisting of horizontal and vertical line segments that meet in the center of and only pass through white squares. 2. Symmetry: he grid looks the same if rotated 180 degrees. 3. hree+: Each clue s answer must be at least 3 characters long. wo examples of possible puzzle grids are pictured in Figure 1, if we ignore the s included there. hose s mark cheater squares, which may be switched to black or white, while following the structure rules, without changing the number of clues in the puzzle. he following result is proven in Section 3. heorem 1.1. he maximum number of clues possible for a crossword grid satisfying onnectivity, Symmetry, and hree+ is 96. oreover, up to horizontal reflection, vertical reflection, and the inclusion of cheater squares, all possible grids requiring 96 clues are pictured in Figure 1, where cheater squares have been marked with a. 2 Fundamental Grid Structure. Starting with 1, we number the rows of a grid from top to bottom and the columns from left to right. hus, position (i, j) in the grid is the square in row 1

2 a b Figure 1: All rosswords with 96 lues i and column j. We refer to row 8 as the center row, and column 8 as the center column. Rows 1 through 7 are the top rows, rows 8 through 15 are the bottom rows, columns 1 through 7 are the left columns, and columns 8 through 15 are the right columns. We denote the number of clues in row i by r i and the number in column i by c i. By considering the cases in which the center square (8, 8) is black or white, we get the following immediate consequence of the Symmetry rule. emma 2.1. r 8, the number of clues in row 8, and c 8, the number of clues in column 8 have the same parity We let Across be the total number of row clues, Down the number of column clues, and lues the total number of clues. Of course, lues = Across + Down. As a consequence of the Symmetry rule, we make frequent use of the equations Across = r 8 + 2[r 1 + r 2 + r 3 + r 4 + r 5 + r 6 + r 7 ] and Down = c 8 + 2[c 1 + c 2 + c 3 + c 4 + c 5 + c 6 + c 7 ]. (2.1) hat is, we can focus our counting on the center, top, and left portions of the grid. oreover, we get the following immediate consequence of emma 2.1 orollary 2.2. he number of clues in any crossword grid satisfying the structure rules is always even. As a consequence of the hree+ rule, there can be at most 4 clues in a row or column. A FOR is a row or column that requires exactly 4 clues. In Figure 1a, the first column is a FOR, and we note that this is the unique way to obtain a FOR. Similarly, a ONE, a WO, or a HREE is a row or column that requires exactly 1, 2, or 3 clues, respectively. Each of those can be obtained in multiple ways. For example, both the third and fourth rows of Figure 1a are WO s. 2

3 3 he Proof of heorem 1.1. By Symmetry, we may assume that there are at least as many off-center FOR s in the columns as the rows. We consider several cases, based on the placement of column FOR s in the left-hand side of the grid. By onnectivity and hree+, we cannot have 7 left FOR s or 6 left FOR s By onnectivity and hree+, the left FOR s must be left-justified. It follows that r 4 = r 8 = 1. By onnectivity and hree+, it is impossible to place black squares in row 3 to make r 3 3. Hence r 3 2. By (2.1), we now have Down 4 + 2[6(4) + 3] = 58 and Across 1 + 2[5(3) ] = 37. hus, lues < left FOR s By onnectivity and hree+, the left FOR s can only be placed in certain ways. Based on the placement of the FOR s, we refer to possible cases distinguished by the black squares shown in Figure 2. he placement in Figure 2a. We see that r 4 2, r 8 2, and FOR s are not possible in any of the rows. By (2.1), Down 4 + 2[4(4) + 3(3)] = 54 and Across 2 + 2[2 + 6(3)] = 42. Hence, 96 clues is only possible if r 3 = 3. hat then forces the squares marked with a in row 3 to be black, and the remaining s must also be black by Symmetry and hree+. However, it is then not possible to have c 6 = 3. hus, lues < 96. he placement in Figure 2b. By hree+, the squares marked with an must be black. Also, the squares marked with a are either all black or all white. he same is true for those marked with an. Since column 8 cannot be entirely black, we have three cases to consider. In each case, r 4 2 and r 8 2. ase 1: he s are white, and the s are black. We have c 8 = 2. Since there is no square where column 6 can be cut into multiple clues, we have c 6 = 1. In this case, the first three rows cannot be FOR s, and rows 4, 5, and 6 cannot all be FOR s. Hence r 4 +r 5 +r By (2.1), Down 2+2[4(4)+2(3)+1] = 48 and Across 2 + 2[2 + 3(3) + 11] = 46. hus, lues 94. ase 2: he s are black, and the s are white. By (2.1), we have Down 2+2[4(4)+3(3)] = 52. In rows 5, 6, and 7, there cannot be any FOR s. So r 5, r 6, r 7 3. In row 3, the structure rules do not allow any square except for (3, 8) to be black. herefore, r 3 = 2, and Across 2 + 2[2(4) + 2(2) + 3(3)] = 44. hus, lues 96, and equality is possible only with all equalities above. Hence, rows 1 and 2 are FOR s, and columns 4 and 12 can then only be HREE s in one way. Now column 6 can only be a HREE in two ways, which are horizontal reflections of each other. hat then forces the structure of column 5, and the grids represented by Figure 1a are obtained. 3

4 a b c Figure 2: 4 left FOR s d ase 3: he s and s are all white. Here c 8 = 4, and none of the rows can be FOR s. By (2.1), Down 4 + 2[4(4) + 3(3)] = 54 and Across 2 + 2[2 + 6(3)] = 42. herefore lues 96, and equality is only possible with all equalities above. By hree+, the columns with the black squares used to make row 3 into a HREE must be the same for rows 1 and 2. Since columns 5 and 11 are HREE s, it follows from Symmetry that we cannot have both (3, 5) and (3, 11) black together. Hence, the black squares in row 3 must be chosen like those in Figure 1b or its vertical reflection. hat forces the structure of the remaining columns, and the grids represented by Figure 1b are obtained. he placement in Figure 2c. he squares marked with an must be black, and those marked with a as well as those marked with an are either all black or all white. Since c 8 0, we have three cases to consider. In each case, r 4 2 and r 8 2. Also, by hree+, c 3 = 1. 4

5 ase 1: he s are white, and the s are black. We have c 8 = 2. By hree+, no squares in column 5 can be black. So c 5 = 1. In this case, the first 3 rows cannot be FOR s. By (2.1), Down 2 + 2[4(4) + 2(1) + 3] = 44 and Across 2 + 2[2 + 3(3) + 3(4)] = 48. hus, lues 92. ase 2: he s are black, and the s are white. We have c 8 = 2. In this case, rows 5, 6, and 7 cannot be FOR s. By (2.1), Down 2 + 2[4(4) (3)] = 48 and Across 2 + 2[2 + 3(4) + 3(3)] = 48. Hence 96 clues is possible only if r 1 = r 2 = r 3 = 4. However, it then is not possible to have c 4 = 3. hus, lues < 96. ase 3: he s and s are all white. Here c 8 = 4, and none of the rows can be FOR s. By (2.1), Down 4 + 2[4(4) (3)] = 50 and Across 2 + 2[2 + 6(3)] = 42. hus, lues 92. he placement in Figure 2d. he squares marked with an must be black, and r 4, r 8 2. By hree+, c 2 = c 3 = 1. By onnectivity, we cannot have all FOR s in rows 1, 2, and 3. he same is true in rows 5, 6, and 7. oreover, by hree+, r 1 + r 2 + r 3 10 and r 5 + r 6 + r By (2.1), Down 4 + 2[4(4) + 2(1) + 3] = 46 and Across 2 + 2[ ] = 46. hus, lues 92. he placement in columns 4, 5, 6, and 7. In this case, c 4 = c 5 = c 6 = c 7 = 4, and (4, 8), (8, 8), and (12, 8) must be black. By onnectivity and hree+, we must have each r i 2, and hence Across 15(2) = 30. By (2.1), Down 4 + 2[4(4) + 3(3)] = 54, and thus, lues left FOR s Based on the placement of the FOR s, we refer to possible cases distinguished by the black squares shown in Figure 3. he placement in Figure 3a. Observe that r 4, r 8 2, and r 3 4 by onnectivity and hree+. laim 1: c 8 4. Otherwise, no r i = 4, since FOR s are unique. By (2.1), Across 2 + 2[2 + 6(3)] = 42 and Down 4 + 2[3(4) + 4(3)] = 52. hus, lues 94. laim 2: r 7 4. Otherwise, by Symmetry, r 9 = 4, and by hree+, (8, 4), (8, 8), and (8, 12) are black. Note that r 5 4 by onnectivity and hree+. Also, c 4, c 8 2, since there are few places for black squares in columns 4 and 8. By (2.1), Down 2 + 2[3(4) (3)] = 48. We need only consider two top row FOR s in addition to row 7. If r 1 = r 2 = 4, then r 3 2 and Across 2+2[3(4)+2(2)+2(3)] = 46. If r 1 = r 6 = 4, then r 5 2 and similarly Across 46. hus, in both possible cases, lues 94. laim 3: r 2 4. Otherwise, r 1 = 4. By hree+, r 3 2 and r 5 4. If r 6 = 4, then hree+ forces r 3 = 1. Hence, in all cases r 3 + r 5 5. By (2.1), 5

6 a c Figure 3: 3 left FOR s b d Across 2 + 2[2(4) (3) + 2] = 44 and Down 3 + 2[3(4) + 4(3)] = 51. hus, lues < 96. his leaves a single possibility to consider for three top row FOR s, namely r 1 = r 5 = r 6 = 4. By hree+, there can be no black squares in row 2. So r 2 = 1. By (2.1), Across 2+2[3(4)+1+2+2(3)] = 44 and Down 3+2[3(4)+4(3)] = 51. hus, lues < 96. he placement in Figure 3b. he squares marked with an must be black, and r 4, r 8 2. By (2.1), Down c 8 + 2[3(4) + c 4 + 3(3)] = c 4 + c 8. Of course, c 4 3. We consider three cases based on the squares marked and. ase 1: he s are white, and the s are black. We have c 8 = 2 and r 1, r 2, r 3 4. If r 7 = 4, then we also have r 9 = 4, by Symmetry. oreover, hree+ forces all of row 8 to be black. Hence r 7 4. By (2.1), Across 6

7 2 + 2[2 + 2(4) + 4(3)] = 46. herefore lues 96, and equality is only possible with all equalities above. However, having r 5 = r 6 = 4 makes it impossible to have column 5 be a HREE. hus, lues < 96. ase 2: he s are black, and the s are white. We have c 8 = 2 and r 5, r 6, r 7 4. By (2.1), Across 2 + 2[2 + 2(4) + r 3 + 3(3)] = r 3. hus, lues r 3 + 2c 4. If r 3 = 4, then column 4 is forced to be a WO. Hence, whether r 3 = 4 or r 3 3, we have lues 96. In both cases, equality is only possible with all corresponding equalities above. If r 3 = 4, then column 3 can only be a HREE in ways which would violate onnectivity. If r 3 = 3, then this together with the FOR s in rows 1 and 2 make it impossible for column 4 to be a HREE. hus, lues < 96. ase 3: he s and s are all white. Here c 8 = 4, and none of the rows can be FOR s. By (2.1), Across 2 + 2[2 + 6(3)] = 42. hus, lues 94. he placement in Figure 3c. he squares marked with an must be black, and r 4, r 8 2. By hree+, c 3 2, and observe that row 3 can only be a WO by having either (4, 3) or (8, 3) be black. By (2.1), Down c 8 + 2[c 3 + 3(4) + 3(3)] = c 3 + c 8. We consider three cases based on the squares marked and. ase 1: he s are white, and the s are black. We have c 8 = 2, and, by onnectivity, c 3 = 1. By (2.1), Across 2 + 2[2 + 3(4) + 3(3)] = 48. hus, lues 94. ase 2: he s are black, and the s are white. he same argument as ase 1 gives lues 94. ase 3: he s and s are all white. Here c 8 = 4, and none of the rows can be FOR s. By (2.1), Across 2 + 2[2 + 6(3)] = 42. hus, lues 92. he placement in Figure 3d. he squares marked with an must be black, and r 4, r 8 2. By onnectivity and hree+, we also have r 3 2. By (2.1), Down c 8 + 2[3(4) + 4(3)] = 48 + c 8. ase 1: c 8 4. So c 8 = 2, and it is impossible to have three top row FOR s. By (2.1), Across 2 + 2[2(2) + 3(3) + 2(4)] = 44. hus, lues 94. ase 2: c 8 = 4. Hence, none of the rows can be FOR s. By (2.1), Across 2 + 2[2(2) + 5(3)] = 40. hus, lues 92. he placement in columns 4, 5, and 6. In this case, we must have r 1, r 2, r 3, r 5, r 6, r 7 2, since there are no squares where any of these rows could be cut into a HREE. By (2.1), Across 3 + 2[3 + 6(2)] = 33 and Down 4 + 2[3(4) + 4(3)] = 52. hus, lues 85. he placement in columns 1, 5, and 6. Since row 1 is a FOR, by hree+, columns 2 and 3 cannot have any black squares. Hence c 2 = c 3 = 1. Observe that r 8 3. By (2.1), Across 3 + 2[3(4) + 4(3)] = 51 and Down 4 + 2[3(4) + 4(3)] = 44. hus, lues < 96. 7

8 he placement in columns 1, 2, and 6. We take advantage of having done the previous cases. If c 8 = 4, then hree+ forces (4, 7), (8, 7), and (12, 7) to be black, and we have the case pictured in Figure 2c handled earlier. Hence, we may assume that c 8 3. By hree+, column 3 has no black squares and c 3 = 1. By (2.1), Down 3 + 2[3(4) (2)] = 47. Since all of the above cases have been handled, the only possible configuration to consider for three top FOR s is the 90 degree rotation of this configuration. hus, lues left FOR s et f be the number of top row FOR s. By assumption, f 2. Since a FOR in row 8 makes two left FOR s impossible, we have r 8 4. Similarly, since a FOR in column 4 makes two left FOR s impossible, we have c 4 4. et t be the number of top row WO s plus the number of left column WO s. laim 1: lues r 8 + c 8 + 2f 2t et t 1 be the number of top row WO s and t 2 be the number of left column WO s. So t = t 1 + t 2. By (2.1), Across r 8 + 2[f(4) + (7 f t 1 )(3) + t 1 (2)] = r 8 + 2f 2t and Down c 8 + 2[2(4) + (5 t 2 )(3) + t 2 (2)] = c 8 2t Of course, lues = Across + Down. laim 2: c 8 4. Suppose to the contrary that we have a FOR in column 8. hree+ then forces f 1. By laim 1, lues r 8 + 2f 2t Since r 8 < 4, we must have f = 1. he only possible place for the one row FOR is now row 4. But this makes having two left FOR s impossible, which is a contradiction. We now have r 8, c 8 3. Hence, it follows from laim 1 that 96 clues can be reached only with f 1. laim 3: wo left FOR s and at least 96 clues is only possible with: (A) f = 1, t = 0, r 8 = c 8 = 3, (B) f = 2, t = 1, r 8 = c 8 = 3, () f = 2, t = 0, r 8 = c 8 = 3, or (D) f = 2, t = 0, r 8 + c 8 = 4. his can be seen by considering the ways to get at least 96 on the right-hand side of the inequality in laim 1. If f = 1, then t = 0, and r 8 + c 8 = 6. So assume that f = 2. Since lues r 8 + c 8 2t + 92, we must have t 1. If t = 1, then r 8 + c 8 = 6. If t = 0, then either r 8 + c 8 = 6 or r 8 + c 8 = 4. laim 4: c 7 4. Suppose to the contrary that we have a FOR in row 7. Hence, c 9 = 4, and (4, 8), (8, 8), and (12, 8) are forced to be black. Since f 1, we have row FOR s whose black squares in column 8 force c 8 = 2. By laim 3, we must have possibility (D) with t = 0. However, row 4 can be at most a WO, which is a contradiction. he only possible placements of two left FOR s are now distinguished by the black squares shown in Figure 4. he placement in Figure 4a. First suppose that is black. Hence, row 4 can be at most a WO. By laim 3, we must have possibility (B) with t = 1 8

9 a b c Figure 4: 2 left FOR s d and r 8 = 3. his forces the s to be black. Since column 3 has to be a HREE, must be black. However, there is now no way to make column 6 a HREE, which contradicts t = 1. We conclude that and are white. So column 3 must be a WO, with the s now black. By laim 3, again we must have possibility (B). However, it is now impossible to make row 8 into a HREE. he placement in Figure 4b. First suppose that is black. Hence, row 4 can be at most a WO. By laim 3, we must have possibility (B) with t = 1 and r 8 = 3. his means that column 3 must be a HREE, with black. However, we now have r 4 = 1, which is a contradiction. We conclude that and are white. herefore, column 3 must be a WO, with black. By laim 3, we must have possibility (B). However, we have r 8 = 1, which is impossible. 9

10 he placement in Figure 4c. here. he same argument used for Figure 4b applies he placement in Figure 4d. First suppose that f = 1. By laim 3 possibility (A), we must have r 8 = c 8 = 3 and t = 0. In order to have a HREE in column 8, the unique top row FOR must be in either row 5 or row 6. However, to have a HREE in column 7, the top row FOR must have been in row 5, and the s must be black. his forces r 6 2, which is a contradiction. We conclude that f = 2 and take advantage of having done the previous cases. Hence, the two row FOR s must be the 90 degree rotation of the two column FOR s in this case. his makes the s and s black. Now neither row 7 nor column 7 can be cut into a HREE. Hence, t 2, which contradicts laim 3. a b c Figure 5: 1 left FOR d 10

11 3.5 1 left FOR By (2.1), Across r 8 + 2[4 + 6(3)] = r and Down c Hence, lues r 8 + c , and we may assume that r 8 = c 8 = 4. hese center FOR s then force the top FOR to be in row 4 and the left FOR to be in column 4. he remaining off-center rows and columns must all be HREE s. By hree+, the rows in which there are black squares cutting column 3 into a HREE also have black squares in columns 1 and 2. he possible cases to consider are distinguished by the black squares shown in Figure 5. he placement in Figure 5a. he placement in Figure 5b. It is impossible for row 5 to be a HREE. It is impossible for row 6 to be a HREE. he placement in Figure 5c. o have HREE s in rows 5 and 6, the s must be black. o have a HREE in row 7, the s must be black. However, it is now impossible for column 6 to be a HREE. he placement in Figure 5d. o have HREE s in rows 5 and 7, the s must be black. o have a HREE in row 6, the s must be black. o have a HREE in row 3, the s must be black. his achieves 96 clues, but onnectivity is violated. 3.6 No left FOR s In this final case, we have Across r (3) and Down c (3). hus, lues r 8 + c Q.E.D. 11

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