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2008 US Puzzle Championship
Google Puzzle Challenge 2008
The 2008 US Puzzle Championship was held on June 14, 2008. Although the test is over, you can still try the problems out yourself. Simply download the test file (Password: gc9yj2) and give it a shot. I'm in the process of covering every single problem on the exam to the best of my ability. You can check here for hints or even give me hints in the guestbook!
Index of Test Problems
Warning! This site contains puzzle hints!
Below, you'll find quick links to every single problem covered on the 2008 exam. I am in the process of detailing my methodology for every puzzle I have solved. (Work in progress, check back often!) If I haven't yet solved the puzzle, I will provide my general direction of thinking, but if you have any other insights, please let me know in the guestbook at the bottom of the page!
1. Battleships - 10 points
A few helpful things to keep in mind about Battleship problems:
- Remember that ships cannot touch each other, even diagonally. This means that you can block out ocean spaces around known ships in all 8 directions.
- Develop a clear notation to keep things organized. I use a small X to indication ocean. I draw part of a whole ship segment so I can tell which direction the ship is facing but not necessarily if that segment is the end of the ship. I also like to cross out ships from the "ship list" so I know which ones are left to find.
- Block out columns F and H as ocean since no segments of ships are in those columns. Also, block out the spaces around the 2 one unit ships given to us.
- We know LJ must be an end segment of a ship because column J only has two segments. Hence, we can now block out KI, MI, and the remainder of J, L, and S as ocean.
- We can also see that the two remaining square in row R must contain ship segments, though we don't know about the size of the ships.
- At this point, there is only one available place for a battleship (4 segments) and that is in column A. No other column or row has a number large enough. Although row K says 4, we've already used up one segment. Block out the ocean in rows M and N as well as squares OB and PB.
- 2 destroyers (3 segments) have yet to be found. Looking closely, we find that those two ships can only exist in the upper-left corner and in column G. Now, we can fill in the ship in column G and square KC. The rest of column G is ocean.
- Similarly, 2 cruisers (2 segments) are left and one must be in column I, forcing the second to be in column E at PE and QE.
- Block out all the ocean you can, and the remaining ships just appear. The puzzle is solved.
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2. Sudoku - 15 points
This is a classic sudoku puzzle, straight up, no twists. Here are a few general strategies I follow when solving sudoku puzzles:
- Scan and find all the numbers that can be put in right away first. Then note down numbers that can only be put in two locations in each row, column, or box.
- If two boxes have the same two number possibilities, then no other box in the same row/column or box as those two can also have those numbers. I'm referring to twin exclusion, triplet exclusion, etc.
- Develop a neat notation for options. I prefer to use the superscript style.
- Out of habit, though it may not be the speediest way, I start scanning at number 1 and note down the number when it can be found in only one of two spaces in a row, column, or box.
- Along the way, some numbers like number 4 can be written in definitively.
- Once I've put in a few numbers definitively, I tend to rescan the puzzle for the past few numbers I wasn't able to note down earlier.
- Repeating in this way, I found the solution to the puzzle, the third row and third column, without actually completely writing in the puzzle. It's essentially done, anyway.
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3. Sum Figure - 5 points
With this type of problem, the general strategy I use is to look at the largest and smallest sums in order to figure out areas where certain numbers must be placed.
Here is the method I used to solve this puzzle:
- Those of you familiar with Kakuro will see immediately that 9+8+7+6=30 is the only combination. This necessitates that the left three gray circles must contain a 0.
- Closer examination shows that the rightmost gray circle cannot be 9 or 8. If it is 7, then (2+1+0)+7=10. If it is 6, then (3+1+0)+6=10. 1 must then be one of the gray circles.
- Moving on to the next largest number, 21, we find that if we use 5+4+3 since they are the next available set of largest numbers, we must also have a 9. 9+5+4+3=21. Thus, the intersection of 21 and 30 must be a 9, and the intersection of 21 and 10 must be a 3. Therefore, the intersection of 10 and 30 must be a 6.
- Similarly, we go to 16 and we find it must be 2 at the intersection of 13 and 16 (only space left), and 2+(0 or 1)+(5 or 4)+(8 or 7)=16. Some quick math shows that 8, 5, and 1 must be included.
- The remaining gray circle must be 0 and the problem is solved.
4. Arrow Ring
These types of puzzles are also known as Yajilin, when published by Japanese puzzle publisher Nikoli. The rules are easy to understand, but there are some critical points to keep in mind when solving the puzzle:
- Black squares cannot be adjacent meaning that black squares will not be next to each other horizontally or vertically, but diagonally is possible.
- All the white squares must be part of a single closed loop. This means all non-black, non-numbered square are used, and you cannot "short-circuit" the loop anywhere or leave loose ends.
- Not all black squares are necessarily accounted for so there may be some black cells without numbers pointing to them.
- First take a look at any 0's in the puzzle. If you would like, you can draw a small open circle in the middle of each square that must remain white. Here, most of row N is white, meaning NG (between the two 1's) must have a horizontal line going through it (NF-NH).
- LA (next to the other 0) must have a vertical line through it, and since it goes into a corner, for the line to have a "way out," the line must go MA-KA-KC. Now, immediately, KD must be black, and the line can be extended to LC.
- KF has two black squares in the four squares to its right. There are only 3 possible combinations: BWBW, BWWB, WBWB. The first and third combination leave squares with only one entrance and no exit, forcing a loose end. Thus, we must have BWWB, and the line can be drawn LH-KH-KI-LI-LJ-NJ.
- THe 2 in PH necessitates that both MH and OH be black. Now, we can extend the line from LH-LF and NH-NI.
- Jump to RB. We currently have several white square above this number of which two must be black. QB cannot be black as it leave a dead end at QA. Similarly, OA cannot be black, leaving MB and PB as black cells. Extend the line MA-OA-OC and QC-QA-SA-SB.
- Having three non-adjacent black cells in row R makes RC, RE, and RG black. Vertical lines must span between these black cells. The right vertical extends to SH.
- TB must be black to avoid a dead end, extending the line SB-SC-TC-TD. TE must be white to prevent a dead end so TF must be black. There cannot be two black squares in TH-TJ because that would cause a dead end at TI. We can now connect two lines to the loop in this left bottom corner of the puzzle.
- A line can be drawn PF-PG-QG. If this line continues to the left, it effectively cuts off the white squares in the bottom right of the puzzle. Therefore, it must go right to QI-RI. This corner has at least one black square, at most, two. You will quickly find that in order to connect SH to RI, there can only be one black square in the corner. With two black squares, you cannot avoid dead ends without violating the no-adjacent-black-squares rule. The line goes SH-TH-TI-SI-SJ-RJ-RI.
- It is also clear now that NJ-PJ-PI-NI.
- QF short circuits if it goes to PF, so it must go to QE. Similarly, QC must go to PC.
- One of the remaining white squares in row P must be black, and one in row O must be black. If PE were black, QE must connect to QD, causing a dead end at PD. Therefore, PD must be black, though the line still connects that way, and the line can be connected at OC-PC, also. The line must run through PE, so OE-PE-PF, and OF is black.
- NF extends to NE, but OE must go through OD-ND to prevent a dead end in OD. Since all of row N must be used, ND-NC-MC-MD.
- We need one more black square and that must be MF. Connect up the line and the puzzle is solved.
- Because one picture is the mirror image of the other, keep an eye out for things pointing in the same direction in both pictures.
- Note the white spaces where it's easy to drop an object in the picture.
- When there are letters, the mirror image can be misspelled but easily missed on a quick glance.
- Clearly mark your differences in the grid. Although it does not explicitly say so, in my experience, there is at most only one difference in each square. This may change in the future, I suppose.
- It is helpful to compare both pictures in a methodical fashion to quickly find the last few remaining differences.
- Remember that the last difference is really worth 6 points, so it may be worth it to really get the tenth when you've already found nine.
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6. Black Pearl - 10 points
This puzzle is almost like a Masyu. The difference here is that the numbers act as the black circles and tell you exactly the number of squares the lines extend from the cell before it must make a turn. A few quick hints to get started are the following:
- Start around the edges of the puzzle. For every number, if there is a direction in which it is impossible to extend a line, then the heading directly opposite must have a line.
- Although this is a closed-loop problem, meaning there are no short-circuits, in this puzzle, not all of the white squares need to be used.
- Following the first hint, we can draw in many lines immediately. LA-LB, KE-NE, LG-LJ, MJ-MH (which necessitates LJ-KJ-KF-NF; KF-KE would make the line from the 3 extend more than 3 units to the right), LG-OG, MJ-OJ-OI (since there is only one direction for the turn to make), OE-PE, QI-QJ, QH-QE, TF-SF, TC-RC, and QA-QD.
- In the upper-left corner, KE can only go to KB-LB, forcing LA-MA-MB. MH must round the corner to NH, which means QH must go the opposite way to TH. Also, OF-PF, SI-SJ, and SG-SF.
- We must avoid making a closed loop near the number 2 on the right edge so NH cannot go to NI; we must have NH-OH. Similarly, OI cannot go to the dead end NI or connect at OH, so we must draw OI-PI. It would violate the rules to draw PI-QI, so we must draw PI-PJ-QJ, then QI-RI-RJ-SJ, then SI-TI-TH.
- We also cannot make a closed loop with the 1's at the bottom edge of the puzzle, so SG-RG-RE and TF-TE-SE (only one direction to turn). At this point, the line from 2 can no longer go right, so we must have TC-TA-SA. There's no escape, and the line must be extended to QA because it cannot make a short circuit at RC. Therefore, RC-RD.
- Again, to avoid the short circuit, RD-SD-SE, which forces RE-QE and QD-PD.
- We didn't connect NE-NF earlier so we do it now. Neither end can connect to the 1's, and there's nowhere else to go for NF.
- In this step, we need to look at the four ends, OF, OG, OH, and PF together. If OG-OH, then OF-OE, leaving PF-PE, which makes a short circuit. Thus, it must be OG-OF (you can't extend OG-PG because that's more than 3 units for the 3 above it), and OH can only go PH-PF.
- Now, PE-PD and OE-OD-ND. This means OC-OB.
- At this point, the puzzle is pretty much done. The only way to connect the loose ends is to go ND-NC-OC and OB-PB-PA-NA-NB-MB.
7. Masyu - 15 points
This puzzle is a classic Masyu, typically published by Nikoli. It's pretty similar to the previous Black Pearl puzzle, but has a few additional clues. The general strategies are similar:
- Start at the edges of the puzzle. For every black circle, if there is a direction in which it cannot extend, then it must extend in the opposite direction. Lines from black circles extend at least two square but is not restricted to any specific number.
- White circles on the edge must have lines parallel to the edge, though you may not necessarily know which end(s) turn toward the center of the grid.
- When a black and white circle are touching diagonally, the lines through the black circle cannot both be adjacent to the white circle.
- As a closed-loop puzzle, there are no short circuits though not all empty squares are part of the loop.
- Using the first two hints, we start my drawing in several lines: OA-MA-MC, ME-OE, MH-OH, MI-MK, RL-TL, VL-XL-XJ, XG-XI, XD-XF, and SB-SD. Up at PH, we can draw PG-PI and QG-QI. The lines here cannot go through the black OH.
- In the bottom right quadrant, we invoke the third hint. With three diagonal spaces of the UI black circle occupied by white circles, there is only one option for its lines and that is UG-UI-WI. Now, we can add TG-TI, SG-SI, and UJ-WJ. Following this a bit more, we find that we've now determined the line through VG, SG-VG-VD, where it must turn since there is no turn at VF.
- One side of UE is now blocked, so the line must be UE-SE. The loose end at XF needs to escape and can only do so at XF-WF-WD. XD can be extended to XC. One way is also blocked for VC, so we can draw VC-VA.
- In the upper left quadrant, QC-SC is not a possibility so QC-OC must be true. Thus, the line through PD must be vertical, OD-QD. We also know that QD cannot turn into either of the black circles because they are too close so it must extend to RD. Since this end is straight, the line must turn on the other side of the white circle at OD and can only turn to the right to OE.
- It's now clear that OC-OA and MC-ME. Also, QC-QA must turn to RA and SE-QE-QG. On the left, SB-UB.
- RA cannot go to RD because that causes a short circuit, and therefore, RA-UA and RD-SD. Similarly, UA-UB causes a short circuit so it must be UA-VA.
- If VC-TC, the end at UB would be stuck, so we must draw VC-XC, WD-VD, and UB-UE. In the middle, the line must be vertical through SF and must escape to RI, hence, TF-RF-RI. This leaves SG-TG and then TF-UF-UG. A closed loop forms if TI-SI, so it must go TI-TK but not to TL since it must turn. UJ must go to UK and TL extends to UL.
- Let's go back towards the upper left. Circle B at OF must have a vertical line through it, NF-PF-PG. OH must go away towards OK, and the loose end at NF, can only go to MF-MH.
- Now, the loose end at MI only has one way to go, MI-NI-NK. The end at MK cannot connect at NK (short circuit) so must go around the corner MK-ML-PL. NK must connect to OK.
- Since the up direction is blocked at PJ, the line must go down through the black circle, PJ-RJ. PI must make a turn and can only turn down to QI, which means, PJ-PL.
- In this last step, we must recall that since we extended TL-UL, there must be a turn in the line at RL, on the other side of the circle. Therefore, we will have RJ-RL. This is enough to submit the puzzle answer. You can easily fill in the rest of the loop at this point, if you'd like.
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8. Nebijok - 15 points
This puzzle doesn't really take to a step-by-step solving method. The premise is very simple -- just look for all the words in the word list in the letter grid. 25 letters in the center are missing but should not pose much of a problem. Speed on this puzzle really depends on how well you can scan for letter combinations. A few pointers to keep in mind:
- Start with the longer words. If you can see part of a longer word going through the empty middle, you can be confident of what letters to fill in.
- In a related note, skim the word list and look around the center grid for familiar letter combinations.
- Don't forget to look backwards!
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9. Number Blocks - 15 points
Cleverness certainly helps speed up this problem. In this puzzle, a few things to keep in mind:
- Evaluate expressions left to right, top to bottom. Do not use mathematical operator precedence.
- Fractions in mid-expression are okay.
- Only the first column and the middle row have minus signs so they must equal 0.
- The bottom row must be (1+2+3)=6 or (9+8+7)=24. We don't know the order of the numbers yet. However, there is no sum of two numbers in the left column minus a 3, 2, or 1 that will give 0. Therefore, the bottom row must be 9, 8, and 7.
- The right column must equal 24 because there's no way you can multiply anything by 9, 8, or 7 to get 6. Since it must be 24, I must be 8.
- The middle column has to give 6 or 24, neither of which can be done with a prime number like 7, so G must be 7, and H must be 9.
- B/E*9=6 or 24. In other words, B/E = 6/9(=2/3 or 4/6) or 24/9(=8/3). 8 is already used at I so the column must equal 6.
- C*F=3 so C,F=1,3. This means that B=4 and E=6.
- A cannot be 6 so C must be 3, making A=2. F is now 1 and D=5.
- The left column must equal 16 since two 3's are need to get 9 and you just can't multiply three numbers to get 2. The numbers in the column must be 1, 2, and 8.
- A must be 8 because subtracting any remaining numbers from 2 or 1 will give a negative value. For this row to be 2, we must multiply by 2, which is unavailable. Therefore, the center column must equal 2.
- For the top row to equal 16, it would have to be 8-4*8, 8-4*4, or 8-6*8, none of which are possible. We can now say 8-B*C=9, and the only combination is B=5 and C=3.
- The middle and bottom row have identical signs. With only 9, 7, 6, and 4 left, we only have 1*9+7=16 and 2*6+4=16 or 1*7+9=16 and 2*6+4=16.
- In the middle column, 5+E/H=2 so only the second combination from the previous step works. D=1, E=7, F=9, G=2, H=6, and I=4. Puzzle solved.
10. Kuromasu - 15 points
Kuromasu, in my opinion, is a great logic puzzzle. It's very similar to Corral, which is question number 12. After checking the rules, a few things I like to keep in mind:
- Develop a clear and simple notation for black and white cells. I prefer to black out black squares, draw a small ope circle in a white square, and leave an undetermined square blank.
- Black square cannot be orthogonally adjacent, which means that whenever you put down a black square, you can immediately mark the adjacent squares as white.
- It can be helpful to start at the corners and edges.
- 2 is a good number to start on, especially if it is adjacent or one space away from another number.
- A great place to start is the 2 at QH. It is one space away from three numbers. If one of those spaces is white, then the 2 would see three squares. Therefore, PH, QG, and RH are black. The spaces around them are white. The 2 must extend right to QI, QJ is black, and PJ, QK, and RJ are white.
- The 3 at OJ is adjacent to the 9, and this forms its three squares. OI, NJ, and OL are black; NI, MJ, NK, and NL are white.
- We've now put some restrictions on the 9 at OK. Since it can only extend in two directions now, we see if that the white squares extend to the top edge (MK, MN, OP, OJ - 4 squares), then PK-TK are white. However, we do not know whether MK or UK are white so we must leave those undetermined.
- The 8 at PL is not completely determined; white extends down to TL, so UL is black and UK and VL are white.
- The 9 at OK is now fulfilled so MK and VK must be black. ML and WK are white.
- The 5 at TJ already has three whites. You can see that if it extends down, the 6 and 11 must also be included, adding three more squares. So, UJ must be black. If the white extends to the left, extending two squares draws in the next 5, but if we draw only one white square, one white square must be up top. However, adding one up top necessitates two. This means the only choice is to make TI black and SJ white. Cells adjacent to the black cells added in this step should also be marked white.
- Taking a look at 11 at WJ, if all the short arms (up, right, down; five cells with 11) were white, we would still need to extend left by six cells. Therefore, WD-WI must be white. This kind of "short arm extension" analysis gets used often with large numbers. This does not, however, give us any information of black squares.
- The 6 above at VJ can only extend in two directions. It can extend only to the left by three squares or else it runs into the 6 at VE. There is only one possibility and that is black at VF, white at VG-VI and XJ (and UF).
- The 5 at TG already has one adjacent white square. It's possible to extend only down or only left, but if you try to extend up, there aren't enough squares. Extending up means you must extend left or down, which is not possible. Therefore, SG must be black. Subsequently, SF is white.
- White squares for 9 at SH can be filled in at UH; XH is black, and XG is white. We can now follow the 6 at XI and identify the remaining squares in the bottom left corner. This also puts a black square at WC for the 11 at WJ.
- Let's look at the 5 at TG. It cannot extend down, so it must extend left three units. TD and TF are white; TC is black, and UC, TB, and SC are white. The 5 at TE is now fulfilled so it must be blocked by black squares above and below. The squares around the black cells are white.
- The white square at RG is blocked on three sides now. Since all the white squares must be orthogonally connected, RF must be white. Thus, the 3 at RE is done and must be surrounded by black squares. The 9 at QF must extend to the top edge.
- The 4 at QC is complete so must be blocked off with black square above and to the left. White squares are marked at RB, QA, PB, and PD.
- There is almost only one possibility for 10 at VB. We can mark VD, UB, and SB as white. Similarly, the 14 at UA is almost determined. At this point, we can only identify WA-NA as white.
- Now, 10 at WB is determined. XB is black. XA is white, so MA must be black and MB white. The 6 at VE is also complete, so XE is black and the adjacent squares are white.
- Using the short arm extension method for the 7 at OH, white squares extend to the 7 at OC. This is six squares for that 7, which is adjacent to another number, so this 7 is complete. OB and MC are black, which makes MH is black for the 7 at OH. White surrounds the black squares.
- Fill in the remaining squares for 7 at NC, which then completes the 5 at MD, and the puzzle is completely solved.
11. Murder No. 6 - 15 points
This puzzle is a Killer Sudoku variation. The differences are that 6 is missing, all the sums only have two numbers, and you only need to consider row and columns for number exclusion. A few hints are similar to regular Sudokus:
- Develop a clear notation for options. I prefer to write the options as small numbers in the corner. Also, have a good system of crossing out numbers so things don't get messy.
- Because everything is in pairs, invoke twin exclusion as much as possible. This refers to the fact that if two squares have the same two options remaining, no other squares in the row or column that both numbers are a part of can have those two numbers.
- Don't forget that 6 is not an option.
- It helps to know sums with only one combination. 3:(1,2) 4:(1,3) 14:(9,5) 15:(8,7) 16:(9,7) 17:(9,8)
- Write in the options for all the sums with only one combination. There are six pairs.
- The 13 at KE is either (9,4) or (8,5). However, by twin exclusion, 9, 8, and 5 cannot be at KE, leaving KE=4 and LE=9.
- Since 4 and 9 are now taken in column E, the 12 can only be (7,5), the 10 at OE:(8,2), and the 10 at JE:(3,7) due to twin exclusion of 1 from the 3 at JB. JE=3 and JF=7. With only one square left in IE, that one must be 1 with IF=8. ID=9 and JD=8.
- The 8 in row I must be (5,3) so the 9 at IG must be (7,2), leaving IA=4 and JA=5.
- We can fill in options for a few more: 13 at JG:(9,4), 7 at MH:(4,3), and 5 at KC:(2,3). This immediately lets us write in a couple of numbers for sure: JH=9, JG=4, KH=1, LH=2, IH=7, and IG=2.
- The 12 at NF can only be (9,3) since both 4 and 8 are taken in column G, and (7,5) would make 3 squares in row N with that option combination. Therefore, NH=4 and MH=3.
- Only 5 and 8 are left in column H. 8 must be at PH, so 5 is at OH. We can now also fill in OG and PG. This also determines 12 at NF and 14 at KF.
- ND=5, OD=2, KD=3, KC=2, 5 at LD:(4,1), so PD=7 and PC=5. PE=2 and OE=8.
- The puzzle is coming along now. PF=4, OF=1, and 9 at PA:(9,3). 11 at OB:(7,4), so OA=9, NA=2, PA=3, PB=9, 9 at MA:(8,1). 10 at LA:(7,3) so LA=7, LB=3, KA=8, KB=7, MA=1, and MB=8.
- This is enough to submit the answer for the puzzle, but the rest of the puzzle can be easily filled in from this point.
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12. Corral - 20 points
This is a great puzzle. My first corral took forever, but after I survived that one, I look forward to Corral. Even though I'm still very slow at it. A few things to keep in mind:
- Numbers are always inside.
- Develop a clear notation. I use an x for outside and a small o for inside. Walls are drawn with a heavy line.
- You can start drawing in the walls along the edge of the puzzle.
- Don't forget to count the number itself!
- Closed loop puzzle! So there are no short circuits, meaning that all inside blocks are connected to each other orthogonally to avoid dead ends.
- Look for rows or columns with a big and small number near each other. If the small number is within the range of the big number but the big number is not within the range of the small number, there must be at least one outside square between them.
- I started by drawing in the walls along the edge of every border number. This just helps me get oriented.
- [[There's a special case in this puzzle where there are two 3's separated by just one square. That one square at OG must be outside (x) or else the two 3's would have to form their own closed loop. Therefore, there are walls at NG/OG and OG/PG.]] Edit: I goofed here. See Cesar's comment in the guestbook. Go straight to the next step; you'll figure out the logic once you get to OG. Sorry!
- Take a look at the 10 at MD. To the right, there is a 5 at MI. If the inside extended from MD to MI, there would be a violation for the 5. Therefore, the inside from MD can extend at most to MG. Counting from MG to MA, we have 7 squares. The remaining 3 must be up and down. Extending 2 squares above requires us to extend 1 square below, but that would automatically include the 4 at OD, which we can't do. The only solution then, is to include OD but exclude KD. This is the only solution for 10. Mark MA-MG and OD-LD as inside. Walls can be added at edge/MA, OD/PD, KD/LD, MG/MH, MH/MI, MH/LH, KC/KD, and KD/KE.
- Actually, we can go even further and put in walls around the 4 at OD, marking OC, PC, PD, PE, and OE outside. There is a wall at PD/QD. At MH, this outside square must extend down to avoid its own closed loop. NH is outside, which forces OH to be outside also since there cannot be a wall at NH/OH. The wall at MH/MI must extend to NH/NI, making NI inside. Also, since the wall makes a corner around MH, both LG and LI must be inside.
- Looking at the 8 at LH, it can extend right two squares and up one square, which at the minimum means LD-LG are inside. This completes the 3 at NG, so KG, NF, and OF are outside, and walls exist at KG/LG, NF/NG, and MF/NF.
- There's only one combination available for 5 at KE; KF and NE are inside, and walls can be marked at edge/KF, KF/KG, NE/NF, and NE/OE.
- There's only one combination available for 6 at KC; KB, LC, and NC are inside, and walls can be marked at edge/KB and NC/OC. This completes the 3 in the upper left corner so LA is outside with walls at KA/LA and LA/MA.
- For the 8 at OJ, at most it can extend two up and one left, meaning PJ-SJ are inside. However, this automatically includes TJ so NJ must necessarily also be inside. Walls can be drawn along the edge of the puzzle.
- We now have seven squares already for the 8 at TJ. If it extends one to the left, we'll have to add two squares, which is incorrect. So, TI must be outside (walls at TH/TI and TI/TJ) and MJ must be inside. The 8 at OJ is now satisfied so OI is outside with wall OI/OJ and OI/NI.
- This forces KI to be inside, satisfying both 5 at MI and 2 at KJ. KH and LJ are outside and several walls can be drawn. LB is also inside, satisfying the conditions for 8 at LH.
- Let's look at the 9 at RG. Both left and right, the maximum extension is three squares, giving a total of seven. For the remaining two, it can only got up or down. Regardless, RD-RJ must be included on the inside. RC is outside with three walls around it, forcing QC to also be outside. Walls can be drawn PB/PC, QB/QC, RB/RC, RC/SC, RC/RD, and QC/QD.
- Looking at the 10 at SC, at the minimum, it will need to extend SD to SI, but that automatically includes SJ. This means that SB must be included, but only one of SA or TC is included. A wall can be drawn SI/TI.
- For 3 at PG, having QG inside is no longer an option as it would also include SG. Therefore, QG and PI are outside while PF and PH are inside. Both PH and PI cannot be inside because it would include PJ. Walls can be drawn at PE/PF, OF/PF, OH/PH, PH/PI, PI/PJ, PG/QG, and QG/RG.
- TG must be inside to complete both 9 at RG and 5 at TH. This puts walls at TE/TF (and SE/TE, making TE outside) and QH/RH (and PH/QH because QH is now outside). The corner in the loop at PH means QI must also be outside, and the two remaining walls there can be filled in.
- There is only one combination left for 6 at QD so QE, QF, and TD are inside. Applicable walls can be drawn around them.
- The key now is the look at the 6 at TA. At most, because it must extend up, it can only extend one square to the right due to constraints on the 10 at SC. This means QA-SA are all inside, and the wall along the left edge can be drawn.
- This restricts the 7 at PA to one combination; OA and PB must be inside. Not only would extending to NA add MA, it would also violate the 6 in the lower left corner. NA and OB are outside, forcing NB also to be outside in order to keep things orthogonally connected. Walls can be drawn around these outside squares.
- Since the 6 at TA is finished, TB must be outside with a wall at TA/TB. We know that TC must also be outside because the 10 at SC is fulfilled, too. QB must be inside. Once the remaining walls are drawn in, the puzzle is solved.
13. Tilted Weights - 20 points
This tilted weights problem is a variation on the regular (untilted) weights puzzle. Typically, given a lot time, these problems can be solved by brute force. However, there is often an elegant answer that, coupled with some clever guessing, can significantly shorten the puzzle. In these puzzles, it is important to remember the following:
- You need to recall how to calculate torque in order to balance the weights. Remember that the force the weight applies again the weight on the other side of the fulcrum is (weight)*(distance from fulcrum).
- In this puzzle, don't forget to add 1 to the force of weights on the right.
- Find additional relationships between different sets of "balanced" weights.
- Note that there are more weights and longer arms on the right side. I started off by wondering what the maximum force on the right side could be by calculating the maximum force on the left side. Assuming that A=9, B=8, F=7, which forces G=4, the maximum force on the right side would be (9+8)*3+(7+4)+1=63.
- Now, list the possible combinations for H and I where 2H=I+1. (H,I)=(1,3),(2,5),(3,7),(4,9). However, if H=4 and I=9, then the force they apply on the right side would be (4+9)*5=65. This is impossible, so given the maximum of 63 we already determined so we can cancel (4,9) as a possibility.
- The possibilities for F=2G-1 are (F,G)=(3,2),(5,3),(7,4),(9,5).
- We now compile the two possibilities keeping in mind that (A,B) and (C,D) are sets of consecutive numbers. Hence, if (H,I) then (F,G) then the consecutive numbers are:
- if (1,3) then (7,4) then (5,6) and (8,9) and E=2. OK
- if (1,3) then (9,5) then (6,7) and (7,8)... NO
- if (2,5) then (7,4) then (8,9) and nothing. NO
- if (3,7) then (9,5) then (1,2) and nothing. NO
- It only remains to test (A,B)=(5,6) or (8,9) to fulfill the condition (A+B)*3+(7+4)=(C+D)+2*4+(1+3)*5-1.
14. Distances - 20 points
To me, Distances is a lot like Kuramasu except you're drawing in circles and some of the "black squares" are already given. A couple of things to keep in mind are the following:
- Develop a clear notation to help you distinguish where you have already deduced the state of the intersection. For example, I drew in a small x if I knew there could not be a circle at that intersection. If there was a circle, I drew an open circle, and once the total distance for a numbered circle was met, I shaded it black.
- Some directions may not have circles, but if it doesn't, you can mark off all the intersections from the numbered circle to the edge of the puzzle with an x. If there is already a circle inline with a numbered circle, then you know for sure you will be counting distances in that direction, though not necessarily to the circle already given.
- Starting with the 15 at E4, we find that there must be circles at both the right edge and the bottom edge of the puzzle. There is no other way to add up to 15 given the two circles already provided. So, B4-D4, E3, E5-E8, and F4-H4 are clear with circles at E9 and I4.
- The 8 at I6 must extend to each of the three circles already there, so E6-H6 and I5 are clear.
- The 4 at F5 counts up two to the 8, but must also count to the right, where there is a circle inline. A circle must be at F7, making F1-F4 and G5-I5 clear.
- The 3 at G2 now has two. Since there is another circle already in row G to the right, a circle must be drawn at G3 to prevent counting all the way to G9. Thus, G1 and H2-I2 are clear.
- For the 8 at G9, no combination avoiding the circles to the left and above will allow a total distance of six or seven in those directions. Therefore, the only possibility is to extend to both circles, leaving H9-I9 clear.
- The same logic follows for 4 at I7. It cannot avoid the circle above by placing circles at H7 and I8, so it must extend to the circle at F7, clearing I8.
- The only free intersection in row H is at H8. This necessarily must be a circle for the 8 at H3, clearing I3.
- D8 must be clear in order to satisfy the 6 at F8.
- Nothing immediately obvious is available at this point, so we'll work on identifying some clear intersections even if we cannot determine circles. For example, for the 11 at B7, with maximum extension of its shortest three sides (top, right, and bottom), we will still need to extend left to B4, making B5 and B6 clear. For the 6 at A1, extending down to the maximum would still require A2 to be clear. Extending right to the 9 at A4 would still require B1 and C1 to be clear. Thus, those three intersections must be clear. By similar deduction for 7 at E2 and 9 at A4, we can also determine D2 and A5 to be clear.
- At this point, we've limited the extension of 8 at D5 in the up direction. Only one combination remains, which makes C5 and D1 circles while D3 is clear. Now, only one possibility remains for 6 at A1, making A3 clear. This then puts a circle at A6 for the 9 at A4.
- The 6 at C3 is now completed so C2 and B3 are clear. This forces B2 to be a circle for the 7 at E2 and makes E1 clear.
- The circle added at A6 means the 9 at D6 must extend in that direction, and since D8 is already determined clear, C6, D7, and D9 are clear.
- One possibility exists for the 11 at B7; A7, B8, and C7 are clear. 6 at C8 is good now, so A8 and C9 are clear.
- For the 5 at B9, A9 must be clear. Also, we left the bottom left corner undetermined. We now know it's clear. Puzzle solved.
15. Dot Triangles - 20 points
I found this puzzle to be very difficult and could not figure out a good plan of attack. I will refer you to the discussion of this problem on Thomas Snyder's blog. He is also the winner of this 2008 competition.
In the comments, it shows that you can find more problems like it on this Tri-Dot Puzzle page.
16. Crisscross Pairs - 20 points
This type of crossword puzzle is pretty common in the Puzzle Championships. This particular one, however, gives you a list of pairs of words, and only one of each pair is used. The unused pair is the answer to the puzzle. Some hints to keep in mind for a puzzle like this:
- Look for the word positions with the most constraints.
- It may be useful to look for letters that do not appear often in the word list. Sometimes this can let you eliminate several possibilities for where the word can fit or not fit.
- Develop a notation to let you mark word possibilities. I usually put a small symbol next to the word and cross out the symbol when the word is no longer a possibility.
- There are two positions in the puzzle where a restraint is placed on each of the five letters. These are the vertical positions just to the right and left of center.
- Determine which words can fit on the left and which words can fit on the right. For the word on the left, the first letter must be the last letter of at least one other word, the second letter must be the first letter of at least one other word, the third letter must be the last letter of at least one other word, and so on. Similarly for the word that fits on the right.
- Unfortunately, this method doesn't determine the correct words right away. With the list, examine the second and fourth letters to determine the words that go across and link them together in the puzzle. For the top link, the word must began with A, R, or H. It must end with N or W. We can now pick ACORN as the top word.
- Now the right vertical word must be ANGEL/ANGER. I wrote L/R in the last box figuring I would circle the correct letter later when it is determined.
- LATHE is the second linking word, and MADLY/MANLY is vertical on the left. The central vertical word, OUTER, can also be determined.
- At this point, the remaining words go in pretty easily. For example, in ANGEL/ANGER, there is only one pair of words beginning with G. LATHE was the only word beginning with L, but since that word has been used, RHYME is the correct word for the bottom of the right square.
- Continue on and the puzzle finishes rather quickly.
17. Sheep in Fences - 20 points
Fences is another name for Slitherlink. In this variation, however, we are also given "sheep" that must reside inside the loop. A couple of hints to help you with this puzzle:
- Take a look at the Wikipedia article on Slitherlink. It discusses some basic strategies that are useful.
- Develop a clear notation. Obviously, lines indicate walls, but it is also useful to use small x's to indicate where walls cannot exist. In this variation, you may also want to lightly shade "outside" squares.
- As a closed loop puzzle, avoid short circuits.
- First off, we can fill in a few obvious things. Lines appear at MA-LA-LB, LD-MD, LE-ME, and LF-MF. No lines (x) appear at LJ-LK-MK-NK, MJ-MK,TK-UK-VK-VJ, and UJ-UK. This puts lines at LI-LJ-MJ and UJ-VJ-VI.
- In the bottom left corner, we're not sure whether the lines go through VA or UB. Regardless, there must be lines at TA-UA and VB-VC.
- Recalling that all sheep must be inside the loop, lines must exist at NA-OA, OK-PK, and RK-SK.
- Notice the two sheep around the 3 in the lower right quadrant. Both of the sheep must be inside, so we can conclude the 3 must be outside the loop. Therefore, lines must separate 3 and the sheep: TH-TI-SI. Also, there can be no line at QH-RH.
- Now the logic begins. In the bottom right corner, we already know that the two 1s are outside of the loop. However, since the 3 is also outside the loop, the space between the two sheep (to the bottom of the top sheep and to the right of the bottom sheep) must be inside the loop, otherwise, there would be four lines coming from TI. Thus, there is a wall separating inside from outside at UJ-TJ, which must extend TJ-SJ-SK.
- Since we have line LJ-MJ already, either MI-HJ or MJ-NJ will have a line, but not both. You'll quickly find that if MJ-NJ were a line, it would be impossible to draw correct lines around the 2 to the bottom right. Therefore, the line is at MJ-MI-NI-NJ-NK-OK.
- We can now extend the line LI-LH-LG. Since we have no identified two sides of the 3 touching MG as inside, there must be walls at MG-MH-NH.
- There is an interesting situation at the 3 touching OH. Of the three horizontals, OH-OI, PH-PI, and QH-QI, two of them must be lines in order to keep both sheep inside. Furthermore, you cannot have both PH-PI and QH-QI be lines. Thus, OH-OI must be a line. A quick try will show that PH-PI cannot be open, so it is a line. Another quick try will show that the right side of the 3 cannot be open. Lines can be drawn at OI-OP, PH-PG, and PK-QK-RK.
- At this point, we can connect the line at NH-OH because we know the right space is inside and the left space is outside. This means we can also add a line at MG-NG.
- To avoid a dead end, we must draw LG-LF, forcing MF-ME and LE-LD. We can continue to MD-MC to ensure both the 3 and the sheep are inside.
- For the 2 touching MF, the loop must be drawn NG-NF.
- In the bottom left corner, the 3 and sheep must be inside so we can draw lines TA-TB-TC and UA-UB-VB. If UC-VC were a line, we would end up with a small short circuit loop. Therefore, we can draw VC-VD-VE.
- For the 2 touching RA, there's only one possibility left for lines. We can draw RA-RB-SB-SC, which also means lines at RA-QA-QB.
- Now, OA can only go right; a line must be drawn at OA-OB. A closer look shows lines drawn at MA-MB and NA-NB lead to dead ends. Lines must be at MA-NA and LB-MB-NB-NC-MC. We can also draw OB-OC-OD-ND-NE-NF.
- If you're keeping track of inside/outside, you'll know that a line must separate the 1 and sheep at PD-PE. Now, either PE-PF or PE-QE is a wall. Turns out that if the top of the 3 were open, you run into inconsistencies with the lower right 2. Therefore, we draw PE-PF-QF-QE-QD and PG-QG-RG-RF-SF.
- Since the 1 touching QE already has a bordering line, we now know the lines for the 3 touching RD. It must be inside with lines at RD-RE-SE-SD.
- Loose ends around the 2 and 1 touching QC must connect through QD-RD and PD-PC-QC-QB. To avoid a closed loop at SC, we draw SC-TC and SD-TD-TE.
- The puzzle is essentially done. Work from the 3 touching SH to draw SI-SH-SG-TG and TH-UH. The 3 touching TE must be outside to ensure both adjacent sheep are inside so the loop must go TE-UE-UF-TF-SF. Now, the last couple of line segments can be drawn to complete the loop.
Find more Sheep in Fences
18. SuDON'Tku - 30 points
This is a very interesting Sudoku twist. There are quite a few subtleties in this puzzle:
- The 9 is not missing in this puzzle. It will be found in every row, column, and rectangle.
- You may be tempted to think that the intersection of the row and column with the same missing number will mean the number is missing from that rectangle. This is not true.
- Develop a clear notation for missing numbers. I ended up writing the missing number in large print to the left/top of the appropriate row/column. I put the missing number next to braces along the short size of the rectangle. Regular sudoku options notation is also helpful.
- If you know the row/column/rectangle where a number is missing, then you know
- the number appears
- in all remaining rows/columns/rectangles.
- Starting with 1, we see right away that there is no space in the KA rectangle for a 1 since all the open spaces are in rows or columns with 1. Therefore, 1s must exist in all other rectangles. MH, PE/PF, OC/OD, and JE/JF/LE/LF are 1s. We also know that column C/D and row J/L are missing a 1.
- The next useful number is 3. There are so many 3s that we easily see column H is missing a 3. Therefore, PD and IE/NE are 3s, which also means one of the IE/NE rectangles is missing a 3. Row I/N is missing a 3.
- Looking at 5, it can be at JB/LB but not both, so one of these rows and rectangles is missing a 5. However, we already know the rectangle with LB is missing a 1 so LB is 5, JB is not, and row J (and JB rectangle) is missing a 5. By twin exclusion for the rows, row L is missing a 1.
- Now we can see PF is 5, PE is 1, JF is 1, MC/NC is 5, and MG/NG is 5.
- Column D is now missing a 5, forcing column C to be missing a 1, making OD 1.
- Now that some things are filled in, let's check for 9s. Remember that no 9 is ever missing. MD, OG/OH, JB, IG/KH, NE/NF, and IE/KF are 9s.
- Since 5 is missing from the upper left rectangle, we can now try to fit 4, 6, and 8 in the rectangle. We find that IC is 8 and JA is 6, leaving ID to be 4. JE is also 4. KD can only be 2 since 5 is missing from column D. Similarly, KB/KC/LA are 4/7/8.
- Either JG/OG is 2, which means either row J/O is missing a 2. However, we know row J is missing a 5 so JG is a 2, meaning row O and rectangle OG are missing 2s. Therefore, OG/OH is 6/9. LE/LF is a 2, so column E/F is missing a 2. MA and PC are 2s.
- Row P is missing 6/8, row M is missing 4/8.
- Looking at row L, which is missing a 1, LH must be 4, making LE 2, LF 7, and LA 8. Thus, column F is missing a 2.
- Column A/B is missing a 4 so NF is a 4, making NE a 9 and IE a 3. Thus, rectangle NE is missing a 3. This makes MG an 8 and NG a 5. MC is also a 5.
- Now we can fill in KF as 9 and IF as 9 and KF as 8. IG cannot be 9 so KH must be 9, making OH 6 and OG 9.
- IG is 6/7 meaning column G and the rectangle is missing 6/7. However, we can see that rectangle IE is missing 6 so rectangle IG must be missing 7, making IG 6. So, column G is missing 7, and column E is missing 6.
- NC must be 6 and OC must be 7, making KC 4 and KB 7. NA must be 7, PB 6, and finally, OB is 4. Finished!
Find More SuDON'Tku
And can you help me out with any of the unsolved puzzles? When talking about a specific problem, please be sure to include the question name or number. Thanks!
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