urn:lj:livejournal.com:atom1:cyrebjrThe House of Cy Reb, Jr.cyrebjrcyrebjr2016-11-14T18:02:13Zurn:lj:livejournal.com:atom1:cyrebjr:46224Not something that's not about hats2016-11-14T18:02:13Z2016-11-14T18:02:13ZThe Stand-up Maths themeNot gonna talk about that! Instead, let's go into unnecessary detail to solve a hat puzzle.<br><br />Back in March, I saw a video featuring two puzzles about prisoners wearing hats. I subscribed to Matt Parker's channel either shortly before or shortly after; I don't remember which. Anyway, here's a series of LiveJournal cuts.<br /><br /><lj-embed id="3" /><a name='cutid1-end'></a><br />Because he already gave a non-bumbling explanation, so I shouldn't have to go through all that work. Also, I will be referring to this video, so even if I'd summarized the puzzle, this is required viewing.<br><br />I confess the first time watching this, I skipped to 4:17 as suggested, but that left me confused around 5:33 where he refers to a previously-stated suboptimal solution. I mean, 3/4 of the prisoners? I know a solution that saves all but half of one! That is, the best solution saves <i>n</i>-1 of the <i>n</i> prisoners half the time, and the remaining one the other half of the time. (Turns out he didn't want to spoil the puzzle for people who'd never seen it before.) Keep this quantity in mind.<br><br />Then I was confused <i>again</i> when at 7:05, he claimed to have a solution that risks two or three prisoners, and that it depended on whether the remaining prisoners knew who survived. This felt strange because the first puzzle's answer extends to arbitrarily large <i>n</i>, so I had to wonder if the second puzzle's answer extended all the way down. As it turns out, no.<br><br />Consider the case of two prisoners in three hats. The solution Matt vaguely refers to would risk both their lives. But there's an easy way to save one prisoner, plus the other half the time. If the back prisoner sees 1, he guesses 3; if he sees 2, he says 1; if he sees 3, he says 2. Then the coded guess uniquely tells the front prisoner what hat she's wearing. (There's a symmetric solution where the back prisoner says the other possibility in all three cases, but I chose this one to set up the next answer.)<br><br />Now what of three prisoners wearing four hats? Matt says that without learning who dies, all three are at risk. But since that's not true for two prisoners, I was sure I could save at least 1 1/2 again, or even 2 1/2. I was even more certain when I found the general solution, but at first, all I could do was make a table.<br><br />There are 24 possible ways for three of the four hats to be worn, with the mastermind, abbreviated W for "warden," keeping the remainder, thus:<pre>
F: 111111222222333333444444
M: 223344113344112244112233
B: 342423341413241412231312
W: 434232434131424121323121</pre>The prisoner in the Back has no less than a 50/50 chance of dying, because whichever two hats he sees, he could be wearing one of the two remaining hats. In other words, across these 24 permutations, he only ever sees 12 ordered pairs of hats. But I propose that the group can select 12 of these permutations so that after the Back prisoner makes his constrained guess, the other two will (once it's their turn) know exactly which ordered pair he saw.<br><br />For starters, at least if the Front prisoner is wearing 4, the situation reduces to the two-prisoner solution, where the Middle prisoner is now the one guaranteed to survive.<pre>
Unassigned Chosen Rejected
F: 111111222222333333 444 444
M: 223344113344112244 123 123
B: 342423341413241412 312 231
W: 434232434131424121 231 312</pre>Now suppose the group agrees to a strategy that includes this. If the Front prisoner heard 2 from the Back and 3 from the Middle, she should probably choose 4. But what if she instead hears 3 from the Back and 2 from the Middle? She can't say 4, because that corresponds to the Rejected 4231 permutation. So the only conclusion is that she's wearing 1, which she says. Similar reasoning shows that along with 1234, the prisoners should Choose 2314 and 3124.<pre>
Unassigned Chosen Rejected
F: 111122223333 123444 123444
M: 334411442244 231123 231123
B: 242334131412 312312 444231
W: 423243314121 444231 312312</pre>And actually, this assignment is almost complete. For suppose the Back prisoner sees the Front one wearing 1 and the Middle one wearing 4. She might be wearing 3, but guessing so would doom the Middle one to guessing 2. Her only option for communication is guessing 2 herself. Similarly, if Front is wearing 1 and Middle is wearing 3, she should guess 4. (Yes, I changed pronouns, but these prisoners are ordered randomly.) That takes care of all cases where 1 is in Front, and perfect analogies finish the 2's and 3's.<pre>
Chosen Rejected
F: 111222333444 111222333444
M: 234134124123 234134124123
B: 342413241312 423341412231
W: 423341412231 342413241312</pre><a name='cutid2-end'></a><br />But I admit I didn't actually write down this table, or construct it in my head. Not sure if that makes my hitting upon the general solution more or less impressive, but there you go.<br><br />Without writing anything down, I felt stuck. How could I code twelve possible ordered pairs (much less 60 triples, 360 quadruples, or more longer permutations) into one binary guess? And I knew it had to be binary, because not only would guessing anything other than a possible hat 100% doom the Back prisoner, the rule of not repeating a guess would doom whoever was really wearing that hat.<br><br />Of course, the puzzle never said the guess had to be in the range of possible hats. One commenter suggested that the Back prisoner add up all 999 numbers of the other prisoners, and this would absolutely save 999.0 prisoners. But I remained certain that with the right approach, I could save 999.5. If no one gave an answer they knew was impossible, then everyone would have two possible hats when it was their turn, and one hat would always be their real hat. Just, how could I choose permutations that were sensitive to every slight rearrangement, communicated that sensitivity only by alternating the Back prisoner between two equally risky guesses, and could be extended to any arbitrary group of prisoners?<br><br />Then I remembered my Galois Theory course.<br><br /><i>Oh, right. The alternating group.</i><br><br />Now what is that? Well, first consider an ordered set of <i>n</i> elements, and take as a group all <i>n</i>! ways to permute them in some (not necessarily new) order. The elements in this group can be combined in a way known as "composition," by arranging the elements by one permutation, then by another. (For instance, starting with ABCDE, applying (12)(34) gives BADCE, then applying (2354) gives BCAED. We can say (2354)o(12)(34) = (132)(45). But now I'm just wasting time.)<br><br />It's easy to see that we can reach any permutation by swapping some pairs of elements in some order; for instance, getting from ABCDE to BCADE can be done by swapping (23), then (45), then (13). Less obvious is that if the permutation can be done in an even number of swaps, it <i>cannot</i> be done in an odd number of swaps, and vice versa. So we call these permutations "even" and "odd," respectively.<br><br />(If you're unconvinced, consider any permutation on <i>n</i> elements. Count the number of disjoint cycles plus the number of fixed points (which could be considered cycles of length 1). Then if you want to be sure you get a number matching the parity of the permutation, subtract this sum from <i>n</i>. This number will always be greater than or equal to zero, but the relevant point is that any swap changes this number by 1; swapping two numbers in the same cycle makes two cycles, and swapping two numbers from two cycles gives a single cycle.)<br><br />So to finally answer the question, the alternating group on a set of size <i>n</i> is the set of all even permutations on the set (with the composition operator, but that's not necessary for the solution). If you look again at my answer for three prisoners, you should be able to verify that the Chosen permutations are all even and the Rejected permutations are all odd. My construction even has the parity principle embedded within; every time I swap two hats from a Chosen arrangement, I Reject the new one and immediately swap a different pair to decide which new arrangements to assign as Chosen. And if you're worried that the mastermind won't be entirely random and might pick an odd permutation so as to at least kill one of you, well, he knows he gains nothing. The set of odd permutations is equally good, so you can choose randomly whether you use even permutations or odd permutations.<a name='cutid3-end'></a><br><br />Then again, if you're not all perfect logicians, or if the mastermind changes his mind, the lot of you are in trouble.urn:lj:livejournal.com:atom1:cyrebjr:45990Tuesday Teaser #58: PLA 7352016-07-06T05:15:33Z2016-07-06T05:15:33ZHi. I live. Also, I moved recently, but I have no idea what detail to go into. Suffice it to say, it was stressful.<br><br>Anyway, it's still Tuesday for another 48 minutes here, so I hope you're ready for a puzzle!<br><br>I saw an interesting license plate the other day, and I'd like to tell you about it. It consisted of three letters followed by three digits, which isn't interesting by itself, but bear with me.<br><br />The letters were such that each of them, in order left to right, appeared in the spelled-out name of the corresponding digit. What's more, with the standard A=1, B=2 substitution, the product of the three letters' values equaled the number formed by the three digits!<br><br />Now to get vaguely specific. I won't tell you the three-digit number, and neither will I tell you if knowing the number would be enough information to determine the three letters and their order. But if I answered the second question, you could determine one of the plate's digits (and maybe its position, but maybe not). What's more, if I told you that digit's corresponding letter, you could tell me the entire plate!<br><br />So what is it?urn:lj:livejournal.com:atom1:cyrebjr:45601Tuesday Teaser #57: How much is EQUESTRIA?2015-11-10T21:38:14Z2015-11-10T21:44:30ZWeird Al, "Hardware Store"So... yeah. Hi.<br><br />I've been a little busy. Not busy with exactly the things I'd prefer, I admit, but some of them are what I want. Like, I'm doing a wimpy (by volume) version of NaNoWriMo; I'm hoping to finish a few chapters by the end of November.<br><br />On to the puzzle! Maybe you have heard of <a href="http://www.equestriadaily.com" rel="nofollow">Equestria Daily</a>. Last month, they finally got mascots, <a href="http://www.equestriadaily.com/2015/09/eqd-mascots-finally-arrive-meet.html" rel="nofollow">Spotlight Splash and Rocket Tier</a>. When I read that Spotlight likes solving puzzles, that lent me the impetus to write this puzzle, a pony-themed alphacipher.<br><br />In a normal alphacipher, each letter has been assigned a distinct integer value from 1-26, and several words are provided along with their values as determined by adding together the values of their letters. For example, with the common A=1, B=2... Z=26 assignment, EXAMPLE has a value of 5+24+1+13+16+12+5=76. (Yes, both E's count.)<br><br />However, I performed some different calculations. The rule I used does coincide with the normal addition in many predictable cases, but... well, my Puzzle Hunt experience is keeping me from outright telling you the rule. If you find that "figure out the secret rule" approach to be too difficult--or obnoxious--here's the rule in rot13:<br><br />Rnpu anzr vf gb or fcyvg vagb cnegf bs rdhny yratgu; gur yratgu va rnpu pnfr vf gur ynetrfg cevzr ahzore gung jvyy jbex. V gbbx gur fhzf bs rnpu cneg naq zhygvcyvrq gurz gbtrgure sbe gur tvira inyhr. Sbe vafgnapr, gur rknzcyr jbhyq jbex gur fnzr, naq gur qvfp wbpxrl hfrf gur cebqhpg bs sbhe fhzf bs guerr yrggref rnpu.<br><br />Here's the unadorned version:<br /><br /><pre>APPLEJACK..........27550
BON BON.............2401
FLUTTERSHY..........4148
LYRA.................510
MUFFINS...............68
OCTAVIA..............108
PINKIE PIE.........65379
RAINBOW DASH.........186
RARITY..............2736
SPIKE.................49
TIME TURNER.........5082
TRIXIE..............2624
TWILIGHT SPARKLE..369279
VINYL SCRATCH....3270960
ZECORA...............663</pre><br />How much is EQUESTRIA?<br><br /><br />In semi-related news, as mentioned above, I am attempting to do a low-intensity version of NaNoWriMo, with a crossover called "Dan Vs. King Sombra" which, irrespective of obviousness, is a crossover between "My Little Pony" and the no-longer-in-production cartoon "Dan Vs." My original goal was to write something for the novel every day. I technically failed yesterday, but I fully intend to keep writing.urn:lj:livejournal.com:atom1:cyrebjr:45471Resolutions2015-04-02T03:22:34Z2015-04-02T03:31:28Z"Party in the USA"*blows dust off journal*<br><br />Hi there. With the dawning of a new year, I've decided to make some changes. Starting tomorrow, I'm going to try to post on LiveJournal every day.<br><br />I'm certainly keeping up my puzzlemaking, and I'd like to point out that some of my creations are going up on <a href="http://www.gmpuzzles.com/blog/" rel="nofollow">Grandmaster Puzzles</a>. Moreover, I intend to put out some life updates here, along with puzzles that don't quite fit over there.<br><br />Happy New Year, everyone!urn:lj:livejournal.com:atom1:cyrebjr:45134Applying the Birthday Paradox to World Cup Group Scores2014-09-28T23:38:49Z2016-07-06T19:35:53ZOK Go, "The Writing's On the Wall"First, I should apologize. The seventh MLP puzzle I was promising hasn't been finished, and it won't be finished for some time yet. But I do have something to show today.<br><br />Alexandre MuĂ±iz posted <a href="http://puzzlezapper.com/blog/2014/06/world-cup-group-scores-and-birthday-paradox-paradoxes/" rel="nofollow">some combinatorial musings on the World Cup</a> recently, and I had a stab at the main posed problem. Figured out some shortcuts, but it's quite combinatorially tricky.<br><br />The question is this: given that the soccer teams in each of several groups are evenly matched, and there is a certain likelihood of each game resulting in a tie, how likely is it that the groups will all have different scores?<br><br />The background for this question is on Alex's post, but I'll try to summarize here. Thirty-two teams compete in the FIFA World Cup, and for whatever reason (perhaps to reduce the impact of a single bad day on a team), the first round consists of eight round-robin matches among groups of four teams each. This means each group will have six games among themselves.<br><br />Now, after each game, each of the two teams receives 0, 1, or 3 points for its own group score, depending whether they lost, tied, or won against the other team. So after all six games within a group, 40 different group scores (such as 7-5-2-1) are possible, although not all equally likely. Alexandre asks how likely it is that all eight groups have different scores.<br><br />Of course, he makes the simplifying assumption that a team in any game will win, lose, or tie with probability 1/3 each, then calculates the answer with a computer program. This gives about 2/5 probability that all group scores will be different. But with that solved, he still has some follow-up questions, one of which is this:<br><br /><b>What proportion of ties would maximize the probability that all group scores are different?</b><br><br />(I've been skirting the assumption that every game is <i>a priori</i> identical and symmetrical, but it's something I should point out. Otherwise, the 48 games might have elements of unidentifiable predetermination, whether that's because one team is so much better than the other, or because the teams cooperate to achieve a desired outcome.)<br><br />So if each game has an equal chance of going one way or the other, dependent only on how likely a tie is, how does one determine the likelihood of all group scores being different? There are 3^48 possible outcomes if all 48 games are treated as distinct, and I suspect Alexandre's program went through all such outcomes one by one. Pass.<br><br />I can save some work by noting that under the assumptions, every outcome with the same number of ties is equally likely. Which isn't to say that every group score with the same number of ties is equally likely; for example, 9-4-4-0, 6-6-4-1, and 6-4-4-3 require one tie each, but they should occur in a ratio of 1:2:3 with each other. So here is how in a given group, the 729 possible outcomes break down into the 40 possible group scores:<br><br />9630 24<br />9333 8<br />6660 8<br />6633 24<br /><b>0 ties: 64</b><br /><br />9611 12<br />9440 12<br />9431 24<br />7730 12<br />7640 24<br />7631 24<br />7433 24<br />6641 24<br />6443 36<br /><b>1 tie: 192</b><br /><br />9421 24<br />7711 6<br />7621 24<br />7540 24<br />7531 24<br />7441 36<br />7432 24<br />6541 24<br />6442 24<br />5443 24<br />4444 6<br /><b>2 ties: 240</b><br /><br />9222 4<br />7521 24<br />7431 24<br />7422 24<br />6522 12<br />5550 4<br />5541 24<br />5532 12<br />5442 24<br />4443 8<br /><b>3 ties: 160</b><br /><br />7322 12<br />5531 12<br />5522 12<br />5432 24<br /><b>4 ties: 60</b><br /><br />5332 12<br /><b>5 ties: 12</b><br /><br />3333 1<br /><b>6 ties: 1</b><br /><br /><b>Total: 729</b><br><br />Now, if ties happen with probability either 0 or 1, there will be repeats among eight groups, but otherwise, there are 3003 ways the eight groups could be categorized by number of ties. I think the probabilities of each case occurring have to be calculated individually. But for any given case, the probability of having no repeats is fixed; just multiply the probabilities for how many groups have each number of ties.<br><br />Actually, that second part is also very tricky. One or zero groups having a given number of ties won't cause a repeat, but two groups both with 5 or 6 ties will always cause a repeat. If two groups have the same number of ties, the repeat probability is the sum of the squares of the relative individual probabilities; if it's three groups, the probability is <i>three times</i> the sum of the squares, <i>minus two times</i> the sum of the <i>cubes.</i> There are eleven group scores that involve two ties each; if all eight groups managed two ties each, the probability of a repeat would be huge, but very complicated to figure.<a name='cutid1-end'></a><br><br />Well, that's about it for the main problem. I don't know the answer. But to make sure my methods work, I tried them in a spreadsheet on three groups of three teams each. The probability of no repeats when ties are 1/3 probability is about .53315 or exactly 1166/2187. This reaches a maximum of about .533468755 when ties have a probability of about .34308.urn:lj:livejournal.com:atom1:cyrebjr:45006Tuesday Teaser #56: Small-Town Stargazing2014-08-27T01:55:02Z2014-08-27T01:57:25ZSmash Mouth, "All Star"At long last, here is my sixth My Little Pony puzzle, dedicated to Equestria's newest princess. While not absolutely the last, I'm proud of accomplishing my original goal of one puzzle for each of the Mane Six (before the end of the show :p ).<br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Mobile%20Uploads/Magic.png" border="0" alt=" photo Magic.png" /><br />This is a <a href="http://www.gmpuzzles.com/blog/star-battle-rules-and-info/" rel="nofollow">Star Battle</a> variant. There are fifteen colored letter regions and six white regions. The internal borders in the W, P, A, and E are purely to emphasize the letters.<br><br /><b>Rules:</b> Forty-two squares of the grid each contain one star, and the rest are empty. No two star squares share a corner or a side. Locate these stars, given that there are two per region, and three per row and column.<br><br />So Star Battle is a pretty obvious choice, since Twilight's cutie mark is a group of stars. But why 21 regions in a 14*14 square? Well, I hit upon using 42 stars because in the episode "Fall Weather Friends," some of the cast participated in a race called the Running of the Leaves, and Twilight's racer number was 42. And beyond that number having a prominent place in geek culture, it's also the number of points on the stars in her mark (two large overlapping stars, five smaller stars surrounding them, each with six points).<br><br />Then with fifteen regions forming letters, 21 total regions was obvious. And I've already seen several small-region SBs, so using a grid smaller than 21*21 wasn't a tortuous decision.<br><br />Next week, I hope to have a doozy of a puzzle ready for you. Maybe I'll even make one for Spike. ^_^urn:lj:livejournal.com:atom1:cyrebjr:44676Tuesday Teaser 55: Hurricane Visibility2014-08-20T05:26:42Z2014-08-20T12:43:22ZSmash Mouth, "I Just Wanna See"Sorry to keep you waiting. I had this up on Photobucket for a week, but there was one detail about it that it took me the week to decide to leave alone. So here's Fluttershy's puzzle.<br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/Kindness.png" border="0" alt=" photo Kindness.png" /><br />This is a <a href="http://en.wikipedia.org/wiki/Kuromasu" rel="nofollow">kuromasu</a> with the twist that half the clues are multiplicative. Of course, I'm going to de-jargon that for you.<br><br /><b>Rules:</b> Blacken some of the grid squares so that each number represents its square's line of sight in one of two ways. An observer at each clue looks up, down, left, and right, and sees up to the edge of either the grid or the nearest black square. A clue number is either the count of all these squares including itself, or the total horizontal distance times the total vertical distance. Six of these clues should be one kind, and the other half-dozen should be the other kind.<br><br />For instance, the 25 clue on the bottom is obviously the second kind, with five squares visible above and five squares visible left and right. These five-counts include the clue square, so the normal clue that would otherwise occupy that square is 9. (Of course, if a clue can only see its own square horizontally, it gives the same value whichever type it is. Same if it only sees its own square vertically.)<br><br />Two final rules govern the black squares. First, no two may touch at an edge, although corners are fine. Second, the black squares must not cut off one section of the grid from the rest.<br><br />The highlighted clues that spell Fluttershy's name trace a double-sided spiral, a reference to "Hurricane Fluttershy." I chose kuromasu because Fluttershy has an interesting dynamic with being seen. For the most part, she's not comfortable being the center of attention (see "Green Isn't Your Color" or the aforementioned "Hurricane Fluttershy"). On the other hoof, she has a stare to rival Zoolander's "Blue Steel," powerful enough to resist a cockatrice's magic. That's why I put in two different clue types.<br><br />To end on a pleasant note, not only have I composed Twilight's puzzle, but I am ready to follow it up with a seventh going on the "Friendship Is Magic" theme. Well, more or less. You'll see in two weeks (if I don't space things out again).urn:lj:livejournal.com:atom1:cyrebjr:44291Tuesday Teaser #54: Jewel-Encrusted Patchwork2014-03-12T06:50:35Z2014-08-27T04:40:24ZI apologize for being late again. I should have remembered that I lose connectivity late at night, but even forgetting that, I should have done all the work earlier. But in any case, here's the first puzzle in this misordered series of pony tributes, a cryptic Shakashaka for Rarity.<br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/Generosity.png" border="0" alt="Generosity photo Generosity.png" /><br />Colors, as usual, are courtesy of <a href="http://www.furaffinity.net/journal/2502402/" rel="nofollow">the Kinky Turtle.</a> The rules are as follows.<br><br />Your aim is to shade in a number of half-cell triangles so that the remaining white space forms a number of rectangles, many tilted at 45 degrees to the grid, that share no edge. Rectangles may touch corner-to-corner or corner-to edge. Triangles may touch anywhere. (As a solving aid, each square cell is divided into quarters. If any part of a given cell is shaded, it is exactly two adjacent quarters.)<br><br />Each letter represents a distinct digit from 0 to 4, inclusive. This clue number is the count of black triangles that are to share an edge with that clue's cell in the solution. The meaning of each letter is for you to determine.urn:lj:livejournal.com:atom1:cyrebjr:44243Tuesday Teaser #53: Square Dance2014-03-05T08:01:59Z2014-03-05T08:01:59ZWell, shoot. It ain't ideal, but it's still Tuesday in Alaska.<br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/1394003967.jpg" border="0" alt="Honesty photo 1394003967.jpg" /><br />This here puzzle's a cryptic variant on Country Road. Whatcha need to do is draw a non-crossing loop from square to edge-adjacent square so's there's one segment in each outlined region.<br><br />Two provisos there. First, y'all don't need to visit every square, but where two squares share a thick border, ya gotta visit one or both. An' second, those letters represent numbers, cryptogram-style, an' those numbers are the number of squares in that respective region's segment.<br><br />There ya go, third pony puzzle on the LiveJournal. Ah figger ah'll put Rarity's up next week, just fer completeness' sake. Yee-freakin'-haw!urn:lj:livejournal.com:atom1:cyrebjr:43833Weekend Update2014-03-02T23:59:35Z2014-03-02T23:59:35ZMy mom informed me a week or two ago that I have a worldwide fan club. So I figured I should post something for all of you. This isn't <i>exactly</i> that post.<br><br />I haven't exactly had a packed schedule the last three months. The garage door is now half-insulated; I gotta finish that. I have get-togethers away from the house on Sundays and Tuesdays, though that's a little up in the air right now. I've also been keeping up with MLP, and even now, I'm lining up an Applejack puzzle for Tuesday.<br><br />Now if you'll bear with me a minute, I'm gonna chase that last bit around. Those of you who've been watching, or otherwise keeping track of, the show, know that the six central characters (known to fans as the "Mane Six") have been, one by one, finding special objects that display rainbow shimmers to the camera. I won't discuss the important details here; I just wanted to note that the first three ponies to find their object were also the first three I made puzzles for, which only had one chance in ten of happening. And I say 1/10 instead of 1/120 for two reasons:<br><br />A) because I got the order wrong, so it was six times as likely that I'd match three ponies, and<br />B) because come on, Twilight's <i>obviously</i> going to be the last one in both lists, so I only really needed to arrange five ponies. (Which, see point A, I didn't do quite right.)<br><br />Anyway, I replaced the image for Rainbow Dash's puzzle because some grid lines had been lost to artifacting. And speaking of things I've lost, I had a puzzle written for Fluttershy, but it was... unkind, so to speak. The answer grid was all right, but I wasn't sure how to solve it or how many extra clues to give. Then I put it off so long, I lost the notebook I'd put it in. >(\< So that won't be up for a while, though I'll probably include Rarity's puzzle for completeness, with a color adjustment.<br><br />And that's really it for now. See you Tuesday!urn:lj:livejournal.com:atom1:cyrebjr:43679Tuesday Teaser #52: Zag2013-11-26T20:45:43Z2013-11-26T20:45:43ZI seem to have a tiny dilemma here. I actually had this puzzle written last week, but Tuesdays are now host to "Agents of SHIELD" viewing parties. And sure, I should post these in the morning, but I'm not generally a morning person. So should I move my weekly (ha) puzzle to another day of the week?<br><br />Anyway, here's a sudoku variant called X-Sums Sudoku, which I first saw on Para's site (<a href='http://puzzleparasite.blogspot.com/' rel='nofollow'>http://puzzleparasite.blogspot.com/</a>). Each outside clue equals the sum of the first X digits in its line starting from the edge where the clue is located, where X equals the nearest such digit.<br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/X-SumsSudokuZag.gif" border="0" alt=" photo X-SumsSudokuZag.gif" /><br><br />(Not to be confused with N-Sums sudoku, where N is the <i>last</i> of the N digits to be added. That, incidentally, means not all 36 spots around the grid can have a clue.)urn:lj:livejournal.com:atom1:cyrebjr:43491Tuesday Teaser #51: Blatant Self-Insert2013-09-25T06:12:25Z2013-11-26T20:02:45ZI blew past midnight a little. Briefly, the inside edge of this tapa is continuous with its outside edge. Each square counts as 1. Don't completely surround any one corner.<br><br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/BlatantSelf-Insert.gif" border="0" alt="Inside-Outside Tapa photo BlatantSelf-Insert.gif" />urn:lj:livejournal.com:atom1:cyrebjr:43197Tuesday Teaser #50: Vanity Racecourse2013-09-11T00:18:16Z2014-02-27T00:41:15ZHeyo! I'm alive and not only buying pony pictures, but also writing pony puzzles! Here's number 3 of 6, a halfway point for a half-century mark. (Oh, I don't mean me. I only turned 27 last month.)<br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/Loyalty-1.gif?t=1393458647" border="0" alt="RD Vanity Course photo Loyalty.gif" /><br />This puzzle type is called "Slalom," and it's probably the best thematic fit so far, a race-themed puzzle for a pony who loves to go fast. Composition was a pain, though. (Had to get it just <i>perfect.</i>)<br><br />The object is to draw a loop that goes from square to edge-adjacent square, starting and ending at the checkered square and crossing every hurdle once in order. "Crossing" means starting on one side of a hurdle, taking two consecutive steps perpendicular to it, and ending up on the other side of the hurdle. There are 26 hurdles here, labeled A-Z; ten labels are shown here.<br><br />Normally, labels are numbers, and the number of hurdles is given on the equivalent of my checkered square. I felt all right giving the count <i>outside</i>--though actually, there's a little room for that under the grid if this way is too obtuse--because W is so close to the end of the alphabet, so it should be pretty obvious. Also, hurdles are traditionally bracketed on both ends by black blocks (<b>which cannot be traveled over,</b> in the vanishingly slim case that this wasn't guessed), but that obviously crowds grids and would have interfered with the tight theming.<br><br />Okay, that leaves AJ, Fluttershy, and Twilight. I can definitely get this done before the end of the TV series, no problem.urn:lj:livejournal.com:atom1:cyrebjr:42832We are the coins in the Potter's purse2013-08-12T09:06:03Z2013-08-12T09:06:03ZHey. Not much in the way of written puzzles, certainly not enough for a Tuesday Teaser, but I did gander at the galleon-sickle-knut currency system of the Harry Potter world. Seems a galleon resembles a hubcap of solid gold, weighing as much as eight dinner buffet plates, so it probably wouldn't be carried around in a typical coin purse.<br><br />Anyway, there are 17 sickles to the galleon and 29 knuts to the sickle. Naturally--well, enough for a recreational mathematician--I wondered if there were amounts of money which, when doubled, transposed the numbers of each coin type. Short answer, yes.<br><br />First, a ground rule. Just like you wouldn't say something costs one dollar, two hundred ninety-nine cents, neither should the amounts of wizarding money be named with more than 16 sickles or 28 knuts. Alternatively phrased, an amount of money should always use a galleon when there's an option to use 17 sickles or 493 knuts, and a sickle when there could be 29 knuts.<br><br />So now I won't keep you waiting. There are three unit prices (I think), besides free, where the price of one and the price of two use the same numbers of coins in a different order. Find them.urn:lj:livejournal.com:atom1:cyrebjr:42518Meandering philosophy2013-08-05T23:37:25Z2013-08-05T23:37:25ZFunny thing, free will. The only way I've found to test it seems to be impossible.<br><br />The idea of free will is that an entity with it isn't fated to one choice vs. any other, right? They might have a preference for one or several options over others, and they might be open to outside persuasion, but that's no guarantee of their future decision. In short, a past decision where free will played a role <i>could have gone another way.</i><br><br />The problem is there's no way to check that. We can speculate on past decisions, sure, and brainstorm about future decisions, but the number of possibilities we can actually observe is either 0 (before the point of decision) or 1 (after).<br><br />Even if we speculate a technology (or other method) by which we may view multiple universes, wherein one (initial) decision (semi-arbitrarily chosen to be interesting) was resolved several different ways, that doesn't prove any involvement of free will. Heck, this scenario constructs an infinitely-branching (perhaps even merging) multiverse wherein no timeline is more clearly valid than another (actually, that might be mistaken, but saying so necessitates a metaphysical mathematics I've never heard of); that is, if you don't even need free will to make all the decisions at once, you don't need free will for any one of them.<br><br />What's needed is a way to replay a timeline to check whether anyone behaves differently when they all start with perfectly identical memories and perfectly identical circumstances. But even then, there's no way to get completely outside and say that any resulting differences weren't fated.<br><br />I dunno. It's something I had to mull over. Maybe I can get some puzzles up this month.urn:lj:livejournal.com:atom1:cyrebjr:42373Tuesday Teaser #49: Bag of Tricks2013-03-26T21:32:20Z2013-03-26T21:32:20Z"Winter Wrap Up"Okay, "next week" didn't work out. But here's a puzzle that just might be your bag!<br><br /><img src="http://i107.photobucket.com/albums/m298/Cy-Reb_Jr/Puzzles/Laughter.png" border="0" alt="Bag of Tricks" /><br />...Well, it would be if you are Pinkie Pie, and I can't really rule that out.<br><br />So this is a Bag puzzle. I'll admit I don't usually use the name "Bag," but I think it fits Pinkie better than "Cave" or "Corral" does. It's the second puzzle in a planned six-part series based on the core cast of "My Little Pony: Friendship is Magic." (Did I mention I'm a brony? Well, I am now!)<br><br />I had intended to do something with balloons, but I got inspired to do something typically mathy. I was able to arrange pi (3. 14 15 9) and <i>e</i> (2. 7 18) on the diagonal in Pinkie's mane and coat colors, so that spells "Pie." To read "Pinkie," take the numbers in colored squares from top to bottom, left to right. Reading the sixth row (colored like Pinkie's balloon cutie mark) in the common 1=A, 2=B fashion gives N, K, I, so all together it says "Pi{nki}e."<br><br />Some credit goes to <span class="ljuser i-ljuser i-ljuser-type-P " data-ljuser="kinkyturtle" lj:user="kinkyturtle" ><a href="http://kinkyturtle.livejournal.com/profile" target="_self" class="i-ljuser-profile" ><img class="i-ljuser-userhead" src="http://l-stat.livejournal.net/img/userinfo_v8.svg?v=17080?v=158.1" /></a><a href="http://kinkyturtle.livejournal.com/" class="i-ljuser-username" target="_self" ><b>kinkyturtle</b></a></span>, whose colorkeying of Pinkie I used here. He links the current version <a href="http://www.furaffinity.net/journal/2502402/" rel="nofollow">here</a> as he updates it. (Mostly I just want to point him out to you and this entry out to him.)<br><br />Wait, you're wondering how this is the *second* puzzle and not the first? Oh, right! The first puzzle is <a href="http://www.gmpuzzles.com/blog/2013/03/sunday-surprise-1-contest-submissions/" rel="nofollow">the second one here.</a> Sorry, both writing and sending the puzzle were done on the spurs of their respective moments. I'm pretty proud of it, too; I managed to pick a puzzle type that's basically about finding diamonds and/or cutting cloth, so it's perfect for Rarity.urn:lj:livejournal.com:atom1:cyrebjr:42175This means there can be a chimera of hybrids, but not vice versa2013-02-16T08:09:11Z2013-03-19T06:43:58ZWeird Al, "Party in the CIA"Yeah. So... yeah, I didn't have a Tuesday Teaser for New Year's Day. I've been alternately sick and working on the Mystery Hunt. And more recently... neither.<br><br />*sigh* I'll have something more substantial up in the coming week. But for now, a terminology note. I propose that "hybrid" and "chimera" not be strictly interchangeable.<br><br />With regard to furries, a hybrid should be an individual body (exact definition subject to twinning and conjoinment) with approximately blended features of two or more species, allowing for variegated expression of those species' individual features. A chimera is more of a Mr.-Potato-Head approach, with swaths of body being one species or another (or one genotype or another, but this is harder to make out). Multiple heads being distinct species is popular and supported by mythology.<br><br />With regard to puzzles, a single-grid puzzle would be a hybrid if the constituent genres' rules were used in tandem to arrive at a single solution. A chimera calls on each set of rules separately to arrive at independent solutions.<br><br /><b>ETA:</b> The puzzle Grant is talking about is <a href="http://web.mit.edu/puzzle/www/2008/chimera/" rel="nofollow">here.</a> And yes, that's where I got "chimera" from.urn:lj:livejournal.com:atom1:cyrebjr:41804AAAAAAAAAAAAAAA2012-12-21T07:49:12Z2012-12-28T20:09:08ZREM, "It's the End of the World As We Know It (And I Feel Fine)"You'd think with how irregular my posting schedule is that it would take the end of the world for me to create a new puzzle. Well, you'd be wrong! I actually created this one this Tuesday afternoon. Okay, there's going to be an Internet glut today, but it's rarely a bad time for puzzles.<br><br /><b>ETA:</b>Oh dang it! Not only did the rash of apocalypse-themed puzzles not come, but I forgot to say what type this is! Anyway, it's a <b><u>slitherlink.</u></b> Draw a loop along the edges, a loop that doesn't reuse any edges or points, such that each number says how many sides of its square or triangle are used by the loop.<br><br /><img src="http://farm9.staticflickr.com/8074/8293774448_0de504eb7d.jpg" width="440" height="456" alt="Slitherpunk"><br><br />Hey, LiveJournal changed its posting interface. That's kinda interesting.urn:lj:livejournal.com:atom1:cyrebjr:41653Tuesday Teaser #48: Tapa Dance2012-11-28T03:22:56Z2012-11-28T03:23:58Z<a href='http://cyrebjr.livejournal.com/41220.html?thread=86788#t86788'>http://cyrebjr.livejournal.com/41220.html?thread=86788#t86788</a><br><br />...Yeah, why not.<br><br />Banged out this Tapa variation in an hour or two. The usual 2x2 Tapa rule manifests here as not coloring all three triangles and two squares around a point at once. Could've been used more, but I like the theme.<br><br /><img src="http://farm9.staticflickr.com/8069/8225104575_30e9cd3b04.jpg" width="440" height="456" alt="Tapa Dance">urn:lj:livejournal.com:atom1:cyrebjr:41220Tuesday Teaser #47: Snub Fillomino2012-11-21T01:36:27Z2012-11-21T01:47:34ZBlockhead, "Insomniac Olympics"Ah-kay. So <a href="http://mellowmelon.wordpress.com/" rel="nofollow">Palmer Mebane</a> was able to write half of <a href="http://logicmastersindia.com/lmitests/?test=FF2" rel="nofollow">Fillomino-Fillia 2</a> while helping with the 2013 <a href="http://www.mit.edu/~puzzle/" rel="nofollow">MIT Mystery Hunt</a>. I, in a word, wasn't. I'm sorta writing a puzzle that might get its answer changed, and I'm not progressing on others, so I can't help but feel I'm doing it wrong.<br><br />Anyway, here's an idea I got while procrastinating on everything else. This Fillomino variant is based on the <a href="http://en.wikipedia.org/wiki/Snub_square_tiling" rel="nofollow">snub square tiling</a>, and while you're solving, the triangles count exactly the same as the squares.<br><br />More explicitly: write a number in each square or triangular cell so that, starting at any cell and traveling only by crossing edges, only to cells containing the same number, you are able to reach that number of cells total. Alternatively, you can just darken some edges to separate shapes, making sure all pairs of shapes with a dark edge between them have different numbers of cells.<br><br /><img src="http://farm9.staticflickr.com/8200/8202943751_1ab1c5b155.jpg" width="440" height="456" alt="Snub Fillomino"><br />(Any ideas on a better title than Snub Fillomino?)urn:lj:livejournal.com:atom1:cyrebjr:41130cyrebjr @ 2012-11-09T22:47:002012-11-10T05:47:41Z2012-11-10T05:47:41Z*bats uselessly at idea*<br><br />This was going to be a full puzzle with one solution, but I'm getting so "meh."<br><br />The pentominoes are often labeled with the twelve letters FILNPTUVWXYZ. Can they be arranged in a loop so that the positive differences in alphabet position between consecutive letters are the numbers 1-12, in some order? For example, P can't be next to both F and Z, since it's separated from each by 10. Also, the alphabet will ideally not loop; F, for example, would be too far from Z.<br><br />If that's too easy, arrange them so taking every fifth letter gets the same result. If that's too hard, use any set of positive integers so that taking positive differences between consecutive entries results in everything 1-12, and so does taking those differences between entries five spaces apart.urn:lj:livejournal.com:atom1:cyrebjr:40929Tuesday Teaser #46: Two Number Twisters2012-10-23T22:06:32Z2012-10-23T22:06:32ZAll right. I can do this.<br><br />A-hem.<br><br />1. Consider the equation N*N=N. If this were an alphametic or algebra problem, the N's would all represent the same thing, either 0 or 1. But they could all represent different numbers.<br><br />N is the 14th letter of the alphabet. N is also the symbol for nitrogen, which has the atomic number 7. And if you turn 2 sideways, it looks like N. So the equation can easily mean 2*7=14.<br />Can you find a similar product where every number is represented by the same two-letter word? As a hint, one of the numbers in my intended answer comes from turning the word upside-down, rather than sideways. What about other equations?<br><br />2. The <a href="http://cyrebjr.livejournal.com/32522.html" rel="nofollow">ten ambigram digits</a> have been <a href="http://wpc.puzzles.com/uspc2012/default.asp" rel="nofollow">pretty much canonized</a> (or is "standardized" more accurate?), and this opens up a whole range of puzzle possibilities. You might be aware that 2178 is quadrupled when it's reversed. But what's the smallest number that is tripled when turned upside-down?urn:lj:livejournal.com:atom1:cyrebjr:40512Spam, immortals, bats, ponies, and spam2012-10-21T05:34:04Z2012-10-21T05:34:04ZWeird Al, "Perform This Way""This week," phooey. Seems there's a type of spambot that preys particularly on deserted journals. Maybe I'm making that last part up, but hopefully, I can chase this bot away by posting.<br><br /><i>And Another Thing</i> is a good title for Who's-On-First-type conversations, as was noted with <i>Up.</i> As a book... well, it's OK. I found it a little lumpy in places, but I say that like I'm trying to repeat what a wine taster told me. You might want to reacquaint yourself with <i>Mostly Harmless</i> if it's been too long.<br><br />Also finally saw <i>The Dark Knight Rises</i> with my brother. Didn't really see many parallels to <i>A Tale of Two Cities,</i> nor did I look hard. There are a couple plot holes around certain mechanics, SO YEAH, SPOILERS.<br><br />Va gur cvg, jul wnf ab bar rire pyvzorq gurve ebcr bhg? Gung'f n euvat, ln xabj? Znlor vs gurer jrer bayl bar ebcr, nggnpurq gb gur svefg yrqtr va gur zvqqyr, gung jbhyq jbex. Nyfb, gurl gbgnyyl bireulcrq ubj qnex vg jbhyq or.<br><br />Naq rkvyr! Arire zvaq gung V fnj gung vg jnf vqragvpny gb gur qrngu fragrapr vzzrqvngryl, be gung ab bar xarj gb penjy ba nyy sbhef; jung vs fbzrbar znqr vg? Jul qvqa'g nalbar rira nfx jurgure gung jbhyq pbhag nf na rfpncr nggrzcg? Tenagrq, vs V jrer hayhpxl rabhtu gb or qenttrq gb bar bs gubfr frzgrapvat urnevatf, V'q snir qenttrq gur zbivr gb n penjy jrgu zl dhrfgvbaf, ohg fgvyy.<br><br />Gure ntnva, V yvxrq ubj rire gubhtu Ebova naq Gnyvn (jnvg, jub'f gung bgure bar?) jrer snveyl boivbhf, gur zbivr cynlrq pbl jvgu jub gurl jrer. Cyhf, Gnyin nyybjrq Ze. Abyna gb unir uvf rivy-jbzna pnxr naq rng vg gbb, serrvat hc Fryvan sbe n cebcre tnzr bs xngmr-haq-syrqreznhf. Cyhf Ebova'f erirny tbg va bar ernyyl cerggl ybbfr guernq ng gur raq.<br><br />N zber crefbany cbvag bs rawblzrag jnf gur obzo gvzre. Vg'f nyernql na npprcgnoyr oernx sebz ernyvgl gung bhe ntbavfgf jbhyq nyy xabj rknpgyl jura vg jbhyq qrgbangr. Ohg V hfhnyyl gel gb xrrc gvzr va zl urnq, naq vg'f znqqravat jura zl pybpx qbrfa'g zngpu gur svyz'f pybpx, juvpu vf nccebkvzngryl rirel fvatyr gvzr. Abyna xrcg zr unccl ol phggvat bar bs Onar'f fcrrpurf fb gung pbafrphgvir fubgf jrer boivbhfyl abg pbafrphgvir zbzragf. V fgvyy unq gur hetr gb pbhag, ohg V znantrq gb fubhg vg qbja. V qba'g rira xabj vs gur frpbaqf jrer gur jebat yratgu!<br><br />(And now I can watch MovieBob's review!)<br><br />All right, so... ponies. I'm pretty sure I'm not a brony, since I haven't forced myself to see more than three and a half episodes, but I was inevitably overexposed by vice of being a furry. And... well, I'm shoulder-deep in FiMFiction.net. Haven't made an account for that, since that "would be silly," but I'm trying to write a crossover with <i>Psych</i> that also features several pony OC's. So yeah, I have no idea where to draw the line.<br><br />Puzzles Tuesday.urn:lj:livejournal.com:atom1:cyrebjr:40374Trying too hard?2012-09-13T20:38:31Z2012-09-13T20:38:31ZOne of the things I've distracted myself with recently is FiMFiction.net. I don't consider myself a brony, despite how much I follow the fandom, because I don't actually watch the show. It took me a while to call myself furry, but that doesn't have nearly such a clear cut-off.<br><br />Anyway, I want to tangent off that. One of the fics I saw was "Spike's Daring Do," which is actually described better in the summary than in the prologue. Spike is a young dragon in the show proper, Daring Do is a pegasus in a book series within the show, and Spike has somehow gotten into a Daring Do book and met her. Since only a prologue has gone up, there's no explanation and not much else.<br><br />But Rainbow Dash, fan of the <i>Daring Do</i> series, has located and started reading this book, and that got me thinking. What if Dash read partway through the book, and then started reading a previous chapter? Books are usually static, as are their characters' timelines as a whole, but... well, there have been cartoon episodes where characters get stuck in their TVs, and changing the channel leaves them on the screen.<br><br />This prompted the following game idea, which I have no intention of fleshing out myself: the player reads a book-like object, possibly illustrated, with one character whose sense of time follows that of the reader. (Can this be implied/discovered without the reader or character knowing the mechanism going in?) The reader can help the character achieve his aim, presumably escaping the story, by going back and rereading passages where the character needs to act differently, once the character knows this change needs to be made.<br><br />Actually, this is similar to Daniel Merlin Goodbrey's hypercomic app "A Duck Has An Adventure." Not having played it, I don't know how big the changes are, but I can guess with near certainty that it's both good and very different from what I'm describing. His comic uses branching and intersecting timelines; my idea uses nonlinear traversal of a linear structure.<br><br />Change of topic! Any of you have a kernel of a pun stuck in your head for years, but no setup available? I finally found the perfect setting for one such pun this morning, thus: <i>Do performances of The Vagina Monologues tend to have a lot of ovary acting?</i> Granted, it could be calibrated better in the spoken word. Mom suggested asking whether the actors [...] could be accused of ovary acting.<br><br />Photobucket seems to have turned mobile-only. This affects me less than I thought it would, because the site shrank all the preview pictures, but the image code here on LiveJournal is leaving them full-size. Still, I had started considering moving the images over to Flickr, where this sort of thing hasn't happened yet.<br><br />Okay, let's see what this week brings.urn:lj:livejournal.com:atom1:cyrebjr:39996Remembering to write2012-09-13T18:14:04Z2012-09-13T18:14:04ZCaravan Palace, "Rock It For Me"Typical. I get a writing project going, hit a wall, and refuse to go back and change it. Also relevant, I don't know what the final message was going to be in "Vandal, Part 3." You'd think I would learn to outline such a project, or at least come up with a fair ending, before committing to a story like that, but noooo. Instead, I abandoned my blog for months because I had a half-finished entry sitting on the update page and no way to fully finish it. Well, for now, that's over. I'm going to update this thing with non-masterpieces and get in some dang practice.