Zitat von Scotty im Beitrag #56Add the letter set to the dictionary. For example, Math looks like this: Letters=⁼,0,1,2,3,4,5,6,7,8,9,−,×,÷,+,.,!,^,²,³,√,∜,∞,>,< Values=0,3,1,2,4,3,3,4,5,4,4,6,6,7,6,8,10,9,10,10,7,10,10,10,10 Counts=18,6,10,8,7,7,7,7,6,6,6,4,4,3,4,3,1,1,1,1,3,1,1,1,1
If you want to, then you can change the letter set in Math dictionary to the following.
This is the best fit of parameters to the best of my knowledge after my analysis having played several games of Math Scrabble.
(1) This reduces the total number of tiles to 98 (excluding two blank tiles) instead of 117 (excluding three blank tiles?). (2) This removes useless mathematical symbols like >, < and ∞. Don't you think that ∞ cannot be part of any meaningful mathematical equation? Don't you think that > and < will allow players to form with wild mathematical equations at player's own wishes instead of restricting with possibilities? (3) This reduces average tile value for each letter. Like conventional English Scrabble, there are limited number of tiles with tile value > 5. (4) This increases tile value of ⁼ letter from 0 to 1. Every tile is worth at least one point. (5) ∛ or mod may be better replacements for ∜ but I do not change with ∜ because the Math dictionary created by Jouni Tolonen contains words containing ∜ (although only ∜16⁼2 and 2⁼∜16) but not ∛ or mod.
Zitat von Scotty im Beitrag Unfixed bugs in Scrabble3D application.Sure, ideally you get in contact with the original author. Otherwise I would take your "much better version" opinion without hesitation if there is another user's word. In case of math.dic it's not too difficult to check myself.
Replacing the Math dictionary with a robust letter set is a quick way of removing the major flaw without having the need to create entire long Math dictionary again by making use of computer program. And computer will take long time to play each turn with long Math dictionary with board sizes > 21×21.
Loading a long Math dictionary every time the Scrabble3D application restarts will take a long time but not verifying the presence of each word once they are placed on the Math Scrabble board during each of player's turns and displaying their meaning and category, if not general category. Even a Math dictionary allowing mathematical equations upto length 11 can be very long, assuming that there are 10 tiles in rack and even not allowing connecting two Math Scrabble tiles that are already present in Math Scrabble board.
Even then multiplication by 0 and 1 should not be allowed with numbers greater than 4 digits because there are 8 forms of mathematical equations possible for each multiplication math equation with 0 and 1 only (when computer program runs for addition and multiplication only and not for subtraction and division): A×B=C, B×A=C, C÷A=B, C÷B=A, C=A×B, C=B×A, B=C÷A, A=C÷B. If an intelligent parsing algorithm for playing the best mathematical equation from the Math Scrabble rack by making use of one or more Math Scrabble tiles that are already present in the Math Scrabble board and checking their correctness along with displaying their meaning and category, if not general category, then there is no need to create an exhaustive Math dictionary with endless number of mathematical equations.
This applies only to Math dictionary in which number of mathematical equations are unlimited and it does not apply to other language dictionaries. Instead of the classical Math Scrabble letter set that is listed above, that intelligent parsing above that is mentioned above can also be used for fractional tiles in Equate style Math Scrabble, or equations of Set Theory, Mathematical Logic, etc.
Please update the Tamil dictionary to version 1.0.4 from the below link. I have updated the Tamil dictionary with the best fit of all parameters to the best of my knowledge in all of individual categories and in all of combined categories and this may be the last update that I might ever make to the Tamil dictionary. https://drive.google.com/file/d/1_ubes0-...iew?usp=sharing
Here are the few changes that I made: (1) Removed [Alternative_Categories] completely. (2) Revised the tile values for some of the letters in the letter set. (3) Updated version number to 100004. (4) Updated release date to Friday, 26 March 2021. (5) Changed standard category from 'Tamil lexicon' to 'Common words' (words common to both categories). (6) Changed comments from 'The digital South Asia library' to 'Tamil lexicon - the digital South Asia library'.
Zitat von Scotty im Beitrag #61What I understand is that a word "Foo" can be either easy or difficult.
Your understanding is wrong. If you looked at both of my Tamil Scrabble game boards that I have attached along with my post 59 in this thread, then you would have understood it more clearly. If I have a trigraph like பொ (single Tamil letter), then in easy version, I have separate tiles for all three symbols and in difficult version, I have one single tile for all those trigraphs and also for digraphs, thereby giving one separate tile to each of the Tamil letters instead of separating each of the symbols. பொ is rarely used and has got higher tile value than each of the symbols separated (the case in which each separate symbol has got tile value of 1). [Left side] Easy version [Right side] Difficult version
Zitat von Scotty im Beitrag #61Meaning when excluding both categories you should have zero words to play.
There are 36664 out of 116448 Tamil words in the Tamil dictionary that do not contain any of these symbols: ஔ, ா, ெ, ே, ை, ொ, ோ and ௌ and so they belong to neither easy version nor difficult version as defined in the Tamil dictionary. If you think that every word should be either in easy version or in difficult version and there should be zero words to play when excluding both of the categories, then duplicate entries of all of those 36664 Tamil words need to be defined in the Tamil dictionary. Including duplicate entries of Tamil words in Tamil dictionary will make the Tamil dictionary to occupy more space. It also takes more time to load the Tamil Scrabble dictionary by the Scrabble3D application at the start of the program, eventhough 36664 Tamil words is not a huge number when compared to total number of words in the Tamil dictionary right now, which is 196232 = 36664 + (79784 × 2). 36664 + 79784 = 116448. The multiplier of 2 is for the two categories which are easy version and difficult version of the 79784 out of 116448 Tamil words in the Tamil dictionary.
On the top of all of that, duplicate entries of Tamil words in Tamil dictionary are always skipped by Scrabble3D application when it loads the Tamil dictionary, meaning that if all of those 36664 Tamil words are first defined in easy version and then also in difficult version later in the Tamil dictionary, then the Scrabble3D application discards away with all of those difficult version Tamil words and only keeps off with all of those easy version Tamil words when it loads the Tamil dictionary. On the other hand, if all of those 36664 Tamil words are first defined in difficult version and then also in easy version later in the Tamil dictionary, then the Scrabble3D application discards away with all of those easy version Tamil words and only keeps off with all of those difficult version Tamil words when it loads the Tamil dictionary.
tamil = codecs.open("C:\\Program Files (x86)\\Scrabble3D\\tamil\\tamil\\split.txt", "r", "utf-8") d = {} for line in tamil: s = line[:-1] t = s.find('=') d[s[:t]] = s[t+1:].split(',') tamil.close()
def expand(word): words = [word] index = [] for k in range(len(word)): if word[k] in d: index.append(k) for e in range(len(index) - 1, -1, -1): f = len(words) for c in range(f): for b in range(1, len(d[words[c][index[e]]])): words.append(words[c][:index[e]] + (d[words[c][index[e]]][b], d[words[c][index[e]]][b][:-1]) [d[words[c][index[e]]][b][-1] == '\r'] + words[c][index[e]+1:]) return words
out = codecs.open("C:\\Program Files (x86)\\Scrabble3D\\tamil\\tamil\\tamil.dic", "w", "utf-8") file = codecs.open("C:\\Program Files (x86)\\Scrabble3D\\tamil\\tamil\\tamil_final.dic", "r", "utf-8") for i in range(215): r = file.readline() out.write(r[:-1]) for line in file: st = line[:-1] idx = st.find('=') word = st[:idx] meaning = (st[idx+1:], st[idx+1:-1]) [st[-1] == '\r'] l = expand(word) if len(l) == 1: out.write(l[0] + '=' + meaning + '\n') else: for j in range(len(l)): if j == 0: out.write(l[j] + '=' + meaning + ';2\n') elif j == len(l) - 1: out.write(l[j] + '=' + meaning + ';1\n') else: out.write(l[j] + '=' + meaning + ';3\n') file.close() out.close()
Please give me with one more day of time. I will provide you with the final version of Tamil dictionary that is to be uploaded and updated in SourceForge tomorrow.
The lines that are defined in [Alternative_Categories] section in Tamil dictionary appear along with two categories 'Easy version' and 'Difficult version' and they can be confusing to players whoever make use of Tamil dictionary. Is it possible to define a secondary letter set for the Tamil dictionary instead of [Alternative_Categories] section?
Or just comment out those lines without the need to remove them off completely? If it is really necessary, then I will provide a version 1.0.4 of the Tamil dictionary tomorrow which can avoid all of those issues that can be uploaded into SourceForge.
Zitat von Scotty im Beitrag DictionaryTamil dictionary was updated to v1.0.3 and has now a category for difficult words. Kudos @Raman
You cannot quite call them 'difficult words'. It is 'difficult version' making use of digraph tiles and trigraph tiles with as much as 313 distinct tiles in which most of the digraph tiles and trigraph tiles are rarely used than in 'easy version' and are difficult to play than in 'easy version'. This is how actually Tamil Scrabble is to be played, but for beginners, it is quite good to start with 'easy version' with only 113 distinct tiles without making use of any of digraph tiles and trigraph tiles. The 'easy version' separates out all occurrences of ா, ெ, ே and ை. However that in the hybrid Tamil dictionary with two different categories for 'easy version' and 'difficult version', players can play a mix of both versions if they want to do so with. For beginners, 'easy version' can be played with 20 tiles in rack and so 'difficult version' also. For experienced players, however 'easy version' can be played with 15 tiles in rack or even something that is lesser than that. [Left side] Easy version [Right side] Difficult version More previews of my Tamil Scrabble games to follow up later on.
Zitat von Scotty im Beitrag #56You can add any comment to the dictionary. A section [Alternative_Categories] would just be ignored.
By the way, I did not have time to look into this yesterday. You said that [Alternative_Categories] would just be ignored, but why the lines in them are showing along with the two categories 'Easy version' and 'Difficult version' long and filling the entire screen line? Please change that section to comment if possible, otherwise just remove that section from the dictionary completely. Anyone who wants a letter set for 'Easy version' can take it from this forum and it is not a requirement for that thing to be present in the dictionary data if they show untidily in the categories section. Please update to version 1.0.4 of Tamil dictionary removing the [Alternative_Categories] section completely if commenting that is not possible. My preferred method would be to comment it. Lines in [Alternative_Categories] do not have any numbers followed by '=' to carry on with those numbers in any of the dictionary words.
Zitat von Scotty im Beitrag #58Think the comprehensive approach with categories fits better the program idea although the dictionary is quite large. Uploaded the new version.
The total number of words in dictionary is right now 196232, much lesser than CSW2019 (International / United Kingdom) English Scrabble dictionary of 279496 words. The long definition of each of the dictionary words and multiple definitions in case of some of the dictionary words make the Tamil Scrabble dictionary that I provided you for uploading to SourceForge large, although 79784 words are redundant which are not among the 36664 words that are common to both of the two different categories, 'Easy version' and 'Difficult version'.
Please update the Tamil dictionary in whatever way you want (either of these two ways).
(1) Single Tamil dictionary (தமிழ்) with two different categories, containing 36664 words common to both categories and 79784 words specific to easy version (without digraph tiles and trigraph tiles) and 79784 words specific to difficult version (with digraph tiles and trigraph tiles). tamil.dic: https://drive.google.com/file/d/1yXx_BaM...iew?usp=sharing
easy = open('C:\\Program Files (x86)\\Scrabble3D\\tamil\\tamil\\tamil_easy.dic', 'r') hard = open('C:\\Program Files (x86)\\Scrabble3D\\tamil\\tamil\\tamil_hard.dic', 'r') for i in range(7): s = easy.readline() print s[:-1] t = hard.readline() letter1 = [] letter2 = [] for j in range(3): u = easy.readline() letter1.append(u[:-1]) v = hard.readline() letter2.append(v[:-1]) print letter2[0] + ',\xc3\xb4,\xc3\xb9,\xc3\xba,\xc3\xbb' print letter2[1] + ',1,1,1,1' print letter2[2] + ',0,0,0,0' for k in range(40): w = easy.readline() for l in range(202): x = hard.readline() if l < 200: print x[:-1] print '[Categories]' print '1=Easy version' print '2=Difficult version' print '[Alternative_Categories]' for thing in letter2: print thing for item in letter1: print item print '[Words]' c = {} d = {} while True: y = easy.readline() if len(y) == 0: break p = y.find('=') c[y[:p]] = y[p+1:-1] while True: z = hard.readline() if len(z) == 0: break q = z.find('=') d[z[:q]] = z[q+1:-1] a = set(c.keys()) b = set(d.keys()) f = a.intersection(b) g = a.difference(b) h = b.difference(a) for e in f: print e + '=' + d[e] for o in g: print o + '=' + c[o] + ';1' for r in h: print r + '=' + d[r] + ';2' easy.close() hard.close()
I would like to do an update to the Tamil dictionary, so that players can switch between easy version or difficult version.
Let us do it tomorrow, but I would like to ask you a few questions about it before doing that update: (1) The two versions have different letter sets. What is your preferred way to do it? Can I have two different categories in the same dictionary or can I have two different dictionaries with different letter sets?
(2) If I use two different categories in the same dictionary, then is it possible to define two different letter sets for the different categories in the same dictionary? Or at least an option to comment the secondary letter set instructing the players to uncomment the secondary letter set and comment the primary letter set whenever they want to play with the secondary letter set?
Then, why can't a single Portuguese dictionary exist with a category (also known as filter) to choose between European version or Brazilian version instead of 2 separate ones?
Zitat von Scotty im Beitrag #52I would rather introduce a category aka filter. We have this for German where the standard set is limited and users can add another bunch of words from other sources pretty easy.
Categories (also known as filters) can be introduced when tile sets are going to be the same. 'Tamil easy' and 'Tamil hard' will use different tile sets. There are set of tiles that are common to both 'Tamil easy' and 'Tamil hard', there are set of tiles that are unique to 'Tamil easy' and should not interfere with 'Tamil hard' and there are set of tiles that are unique to 'Tamil hard' and should not interfere with 'Tamil easy'. That interference should not occur when assigning letter to blank tiles and generating randomly 10 tiles for each played game. க can be used for க, கா, கெ, கே, கை, கொ, கோ and கௌ in 'Tamil easy' as ா, ெ, ே, ை and ள have separate tiles of their own but க can only be used for க in 'Tamil hard' as other letters have separate tiles of their own. Do you think that it is still possible to use category (also known as filter) if the tile sets in the 2 dictionaries are different from each other?
If a new category (also known as filter) is introduced in a single Tamil dictionary for the easier version and the harder version, is it possible to define 2 independent tile sets for both the versions which is not supposed to interfere with the other tile set?
Scotty: Just like the Portuguese dictionary is split into two: European version and Brazilian version, may I suggest that Tamil dictionary also be split into two, an easier version and a harder version?
The older version of Tamil dictionary, tamil.dic, just rename its English name to 'Tamil easy' and local name to 'தமிழ் சுலபம்'. Get tamizh.dic from https://drive.google.com/file/d/1lITDV1W...iew?usp=sharing. Keep its English name to 'Tamil hard' and local name to 'தமிழ் கடினம்'. The harder version can be fit for people who love difficult challenges, diverse letter set and for best results need 20 tiles in rack at any one time. I cannot reduce it from 1000 tiles pre-defined in the letter set in the dictionary. Some difficult to play letters have been assigned points in range from 11 to 20 and some of them need some specific letters to combine with. Doing that also makes the computer players as opponents to take such letters seriously depending upon the frequency and combination with other specific letters and playing them immediately when opportunity comes when such tiles do land on the player's racks with.
The harder version combines letters like ஔ, ா, ெ, ே, ை, ொ, ோ and ௌ into one single tile, the way how Tamil is actually to be played. This version is the extensive use of Tamil digraphs and trigraphs, but not actually to be said digraphs and trigraphs as they are single Tamil letter. The easier version splits some of the rarely used and difficult to play Tamil letters into two or three tiles which are good for beginners. I will later attach screenshots over hereby of all my played games, both with easier version and harder version.
I will do an update to the 'Tamil hard' dictionary version once again in a week or two. This will be to revise and finalize with letter distribution and points of letters once again. I will change all the tile letters to Unicode font (people won't need to download the Valluvan font as in 'Tamil easy' dictionary in which ா, ெ, ே and ை tiles will not be nice if they are represented by the Unicode font). I will also consider including all Tamil loanwords that are borrowed from Sanskrit containing the letter ஶ which does not exist in the Valluvan font and so words containing ஶ have been dropped from the 'Tamil easy' dictionary. ஶ has been dropped from some words containing it in Tamil Lexicon dictionary, so it is to be revised from the pronounciations in the meanings in which š sound is included. In any case that ச is a good substitute letter for most of the Tamil loanwords that are borrowed from Sanskrit containing the letter ஶ and that letter is too rarely used to be included in any tiles but only to appear randomly if anyone were to ever include with 10 random tiles in each played game, unlike the other Tamil loanwords that are borrowed from Sanskrit containing the other Sanskrit letters of ஸ, ஜ, ஷ and ஹ. க்ஷ = க் + ஷ and ஶ்ரீ = ஶ் + ரீ. ஶ்ரீ is supported by the Valluvan font and words containing them have not been dropped from the 'Tamil easy' dictionary although most of them are not regular Tamil words but rather names of persons and names of places which are too hard for me to drop them from the dictionary, so that I have decided to have them in the 'Tamil easy' dictionary and 'Tamil hard' dictionary. Names of days of the week and names of months of the year are much better to be kept in the Scrabble dictionaries than names of persons and names of places although official Scrabble dictionaries that are published in languages that make use of the Latin Script alphabets do drop with them altogether.
Thank you.
By the way that one single dictionary with two different categories cannot be used to play with both easier version and harder version because they make use of completely different letter sets as you can see by yourself if you were ever to open that dictionary. For example, if anyone were to ever include with 10 random tiles in each played game, the choice of easier version or harder version should entirely and strictly depend upon player's tastes.
Zitat von RamanDoes it always look out for ways to push away as much as tiles as possible while not allowing high scoring and rarely used letters to stagnate in its rack?
If the Computer wins the game (more points than opponent) it places as much letters as possible. Otherwise as much points.
I don't think that is literally true.
Consider the following board position, at the endgame, when there are no more tiles left in the bag. When computer evaluated for the best move and when it is computer's turn to play at its strongest level, the best move for play by computer at its strongest level comes to merely 14 points, which is 7 = 7 - 0 at the rightmost column. But there is another move which scores 108 points and pushes more tiles than 7 = 7 - 0. But that move is only fourth in the best move list out of 5. Why is it so? Attached Equate400.txt saved game file for your own reference. Please rename it to Equate400.ssg before experimenting it by yourself. Also that equate.dic is too large to be attached here. It can be shared by a link after uploading it on Google Drive, but anyway that I am sharing my python script from which you can generate yourself by running Python 2.7.
num = [1, 1, 2, 1, 3, 1, 2, 3, 4, 1, 5, 1, 2, 3, 4, 5, 6, 1, 3, 5, 7, 1, 2, 4, 5, 7, 8, 1, 3, 7, 9, 1, 5, 7, 11, 0] den = [2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 12, 12, 12, 12, 1] sym = ['#', '$', '&', '(', ')', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', '%', ':', '^', '.', '!'] conj = [0, 2, 1, 4, 3, 8, 7, 6, 5, 10, 9, 16, 15, 14, 13, 12, 11, 20, 19, 18, 17, 26, 25, 24, 23, 22, 21, 30, 29, 28, 27, 34, 33, 32, 31, 35] m = ["\xe2\x81\xb0", "\xc2\xb9", "\xc2\xb2", "\xc2\xb3", "\xe2\x81\xb4", "\xe2\x81\xb5", "\xe2\x81\xb6", "\xe2\x81\xb7", "\xe2\x81\xb8", "\xe2\x81\xb9"] n = ["\xe2\x82\x80", "\xe2\x82\x81", "\xe2\x82\x82", "\xe2\x82\x83", "\xe2\x82\x84", "\xe2\x82\x85", "\xe2\x82\x86", "\xe2\x82\x87", "\xe2\x82\x88", "\xe2\x82\x89"] d = {} for j in range(len(sym) - 1): d[(num[j], den[j])] = sym[j] def superscript(s): a = "" for c in s: a += m[ord(c) - 48] return a def subscript(t): b = "" for d in t: b += n[ord(d) - 48] return b def gcd(p, q): while q > 0: (p, q) = (q, p % q) return p def string(nu, de): result = '' if nu >= de or nu == 0: result += str(nu / de) if nu % de > 0: result += d[(nu % de, de)] return result def terms(nu, de): result = 0 if nu >= de or nu == 0: result += len(str(nu / de)) if nu % de > 0: result += 1 return result print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=Equalities' print 'Licence=GNU General Public License, v3' print 'Release=24.10.20' print 'Comment=Equate game style of Scrabble with mathematical equations' print 'Letters=@,0,1,2,3,4,5,6,7,8,9,-,*,/,+,#,$,&,(,),A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,%,:,^,.' print 'Values=1,2,1,1,2,2,3,3,4,3,4,4,3,3,4,5,6,6,5,5,8,8,8,8,7,7,10,10,10,10,10,10,8,8,8,8,10,10,10,10,10,10,9,9,9,9,9,9,9,9' print 'Counts=36,8,10,9,8,8,8,8,7,7,7,7,11,11,7,4,2,2,2,2,1,1,1,1,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1' print 'Key=' print '[Replace]' print '-=\xe2\x88\x92' print '*=\xc3\x97' print '/=\xc3\xb7' print '@=\xe2\x81\xbc' for i in range(len(sym) - 1): print sym[i] + '=' + superscript(str(num[i])) + '/' + subscript(str(den[i])) print '[Categories]' print '1=Addition' print '2=Subtraction' print '3=Multiplication' print '4=Division' print '[Words]' for k in range(10000): print str(k) + '@' + str(k) + '=' for k in range(100): for l in range(len(sym) - 1): print (str(k), '') [k == 0] + d[(num[l], den[l])] + '@' + (str(k), '') [k == 0] + d[(num[l], den[l])] + '=' for k in range(10): for l in range(len(sym) - 1): for o in range(10): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11 or y % z == 0: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(100): for l in range(len(sym) - 1, len(sym)): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(len(sym) - 1, len(sym)): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(conj[l], conj[l] + 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(1000): for o in range(1000): y = k + o if len(str(k)) + len(str(o)) + len(str(y)) <= 7: print str(k) + '+' + str(o) + '@' + str(y) + '=;1' print str(y) + '@' + str(k) + '+' + str(o) + '=;1' print str(y) + '-' + str(k) + '@' + str(o) + '=;2' print str(o) + '@' + str(y) + '-' + str(k) + '=;2' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(100): for l in range(len(sym) - 1, len(sym)): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' if k > 0: print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(len(sym) - 1, len(sym)): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(1, 100): for l in range(len(sym) - 1, len(sym)): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) == 7: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' if k > 0: print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(100): for l in range(len(sym) - 1): for o in range(1, 100): for r in range(len(sym) - 1, len(sym)): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) == 7: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(1, 1000): for o in range(1, 1000): y = k * o if len(str(k)) + len(str(o)) + len(str(y)) <= 8: print str(k) + '*' + str(o) + '@' + str(y) + '=;3' print str(y) + '@' + str(k) + '*' + str(o) + '=;3' print str(y) + '/' + str(k) + '@' + str(o) + '=;4' print str(o) + '@' + str(y) + '/' + str(k) + '=;4' print '0*0@0=;3' print '0@0*0=;3' for k in range(1, 10000): print '0*' + str(k) + '@0=;3' print '0@0*' + str(k) + '=;3' print str(k) + '*0@0=;3' print '0@' + str(k) + '*0=;3' print '0/' + str(k) + '@0=;4' print '0@0/' + str(k) + '=;4' for l in range(len(sym) - 1): k = range(0, 1000, den[l]) for r in range(len(k)): y = k[r] * num[l] / den[l] z = k[r] if z >= 100: print sym[l] + '*' + str(z) + '@' + str(y) + '=;3' print str(z) + '*' + sym[l] + '@' + str(y) + '=;3' print str(y) + '@' + sym[l] + '*' + str(z) + '=;3' print str(y) + '@' + str(z) + '*' + sym[l] + '=;3' print str(y) + '/' + str(z) + '@' + sym[l] + '=;4' print str(y) + '/' + sym[l] + '@' + str(z) + '=;4' print sym[l] + '@' + str(y) + '/' + str(z) + '=;4' print str(z) + '@' + str(y) + '/' + sym[l] + '=;4'
By the way that if joker is exchanged for the same letter on the board, as in B21 square in this board position, the star should disappear. But here it does not happen to be the case. Why is it so? Also that the turn does not get ended for me if I exchange the joker with the same letter on the board, as it says so in the configuration settings page and Scrabble3D application's wiki page. Why is it so?
By the way that it is only possible for human players to exchange joker with the same letter on the board or convert at most one tile to a joker per turn in What-If-Only variant. I hope that you are just aware of that simply, but that you do not take that as a serious issue, because it counteracts with the skill level of the computer players playing the highest scoring plays ever. The computer players do not do that. Why is it so?
Although I do not want to get involved in forum donations, just curious about it simply, the banner at the bottom of the page says that 14% of 144€ has been raised so far. I would like to know who donated it which might be essentially of no use in a forum with very few active members? And what happens if the 100% target is reached only near the end of the year, namely 2020-21? Will it be used for the Premium Tariff fees for this forum for the small portion of the remaining year, namely 2020-21 or will it be carried forward to be used wisely for the next subsequent year fully, namely 2021-22?
On Wednesday, 23 September 2020, Alexlatham96 updated the Maths Scrabble letter distribution in the Wikipedia page to the one present in math.dic in Scrabble3D application that had been contributed by Jouni Tolonen.
My proposal of the Maths Scrabble letter distribution article for the Wikipedia page:
Math sets use these 100 tiles:
2 blank tiles (scoring 0 points) 1 point: ⁼ ×18, 1 ×8, 2 ×7 2 points: 0 ×6, 3 ×6, 4 ×6 3 points: 5 ×6, 6 ×6, 8 ×5, + ×3, − ×3 4 points: 7 ×5, 9 ×5, × ×3, ÷ ×3 5 points: √ ×2 6 points: ² ×1 7 points: ³ ×1 8 points: ! ×1 9 points: ^ ×1 10 points: . ×1, ∜ ×1 Math Scrabble is played with 10 tiles in the rack at a time. To play, you have to put in a true equation on the board.
An alternative by Equate game has this distribution of 200 tiles:
Here after, I am going to call the former Maths Scrabble tile set as classical Maths Scrabble.
I have revised the classical Maths Scrabble letter set to provide a balance between post 3 by Scotty in this thread and the one that is present in math.dic in Scrabble3D application that had been contributed by Jouni Tolonen. This is to remove the useless symbols such as those like ∞, < and > and overall lower the points for all tiles to match with the one that is similar to English Scrabble. At the same time having at least one tile with any given face value between 1 and 10 (1 and 10 both inclusive). At least if not to have a better classical Maths dictionary for the Scrabble3D application, the tile set for math.dic can be updated for above to provide the most optimal experience. A better classical Maths dictionary will matter only for the computer players as opponents (only offline and not online) and not for the human players as opponents (either offline or online). My tile set also increases the count of digits and decreases the count of symbols. This is done to encourage players to form longer mathematical equations with longer numbers. Some kits that are available from shops will contain tiles with numbers that are ≥ 10 and this will discourage players from forming longer mathematical equations with longer numbers.
I modified the original tile set of the Equate game to make it more thrilling by discarding all tiles of reducible fractions and by considering only those fractional tiles whose values are strictly between 0 and 1 (0 and 1 both exclusive). In short, there should be exactly φ(N) distinct fractional tiles with the denominator N, that are strictly between 0 and 1 (0 and 1 both exclusive). That is, those values of denominator that are ≥ 1, ≤ N - 1 and co-prime to N. I also removed the flaw of having just 3 tiles each for the digits 0 and 1 and 9 tiles each for the rest of the digits, namely, 2, 3, 4, 5, 6, 7, 8 and 9.
Why having heavy flaws like this and selling? Are these part of trade tactics to fool all the customers who buy those kits? Even if there are some flaws in the tile set and board design and outside packing looks good, the customers will blindly buy those kits and play them. If they become popular, then they will organize tournaments with such tile sets becoming as a standard? I am saying this also because of the fact that Upwords large pack is a 8×8 board with 64 tiles for ₹400, whereas Upwords small travel pack is a 10×10 board with 100 tiles for ₹225. Similarly that in Sorkalam Tamil Scrabble pack, 2 of the Tamil vowel combinations on transparent tiles, at most one of them to be placed on other consonants on wooden tiles are missing, namely ு and ூ, still many customers would be fools to buy those kits just simply because the outside packing looks good. And that they do not mind playing with those missing 2 letters as a handicap and they write great reviews about the overall product and design of the game to be played. The cost of Sorkalam game in Amazon is ₹1200. If everyone else does like what I did in my own 2014 make, printing out Scrabble board and Scrabble tiles on chart paper, pasting them on cardboard and cutting them with scissors, the entire process will merely cost only around less than ₹200 (first I did it for 21×21 English Scrabble seeing that it would be so much of cost to buy the real kit from abroad, then I did it for Tamil Scrabble and later on I did it for Scrabble with mathematical equations). That way, people can pick up their own letter distribution, points for tiles and printed extra tiles to ensure different letter distributions in different games. Still more and more people are interested in buying Sorkalam game from Amazon that does not even have the complete letter set (no ு and ூ vowel combos that are printed on transparent tiles which are part of core Tamil and those that are also frequently used). Wooden tiles are just a sign of attraction for people to buy them and that people psychologically prefer more to play with wooden tiles than ones made out of that from cardboard and chart paper, isn't it?
For 21×21 board with 200 tiles, for classical Maths Scrabble, lesser number of ⁼ tiles would be sufficient, say 34, and more number of digit tiles and lesser number of symbol tiles can be put into the bag than when compared to that of Equate game style. This is because longer mathematical equations with longer numbers can be comfortably played in classical Maths Scrabble and players can be encouraged to do so with. But in Equate game style, diverse variety of distinct fractional tiles prevents players from forming longer mathematical equations with longer numbers. So that more shorter equations will be formed and more number of ⁼ tiles would be needed, say 36.
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I will come up with my proposed Tamil Scrabble letter distribution article for the Wikipedia page soon. Please give me some more time to prepare with it.
I am also thinking about Scrabble with logical equations. In that case, we can only have two numbers 0 and 1 and no longer logical equations with longer numbers, even when making use of different variety of 7 distinct logical operators such as those like and, or, not, nand, nor, xor and xnor. Although that more than one logical operator can be allowed on any one single side of any logical equation. I think that it would be more thrilling to play that with some variable names, say A, B, C and D. In that case that logical equations such as those like A + A̅B = A + B, AB + A̅B + AB̅ + A̅B̅ = 1, etc. are all supposed to be valid logical equations only.
And also that it would be good to give a thought whether it works out if any of types of variables are introduced either into classical Maths Scrabble or into Equate game style of Maths Scrabble.
The tile set that I have used for the equate.dic dictionary file, Equate game style of Maths Scrabble, that can be generated from the equate.py Python script as a source code, by executing that with Python 2.7, in my previous post 25 in this thread makes use of these 200 tiles without having any blank tiles: 1 point: ⁼ ×36, 1 ×29, 2 ×28 2 points: 0 ×27, 3 ×27, 4 ×27 3 points: 5 ×27, 6 ×27, 8 ×26, × ×9, ÷ ×9 4 points: 7 ×26, 9 ×26, + ×9, − ×9 5 points: ¹/₂ ×4, ¹/₄ ×2, ³/₄ ×2 6 points: ¹/₃ ×2, ²/₃ ×2 7 points: ¹/₆ ×1, ⁵/₆ ×1 8 points: ¹/₅ ×1, ²/₅ ×1, ³/₅ ×1, ⁴/₅ ×1, ¹/₈ ×1, ³/₈ ×1, ⁵/₈ ×1, ⁷/₈ ×1 9 points: ¹/₁₀ ×1, ³/₁₀ ×1, ⁷/₁₀ ×1, ⁹/₁₀ ×1 10 points: ¹/₇ ×1, ²/₇ ×1, ³/₇ ×1, ⁴/₇ ×1, ⁵/₇ ×1, ⁶/₇ ×1, ¹/₉ ×1, ²/₉ ×1, ⁴/₉ ×1, ⁵/₉ ×1, ⁷/₉ ×1, ⁸/₉ ×1 It is a gentle balance between 220 tiles in my previous post 25 in this thread and 200 tiles having any blank tiles in my current post 26 in this thread for my proposal of the Maths Scrabble letter distribution article for the Wikipedia page.
In Scrabble3D application, it is essentially worthless to play with any blank tiles for larger 2D boards or (smaller or larger) 3D boards with computer players as opponents (only offline and never online), because it will slow down the performance of computer players as opponents (only offline and never online). Because there are lots of possibilities that are supported for them, especially with dictionaries that contain larger number of words and in dictionaries that have large number of distinct letters in the letter set.
Playing with more number of rack does not necessarily slow down the performance of computer players as opponents (only offline and never online) because Scrabble3D application seems to make use of trie data structure for storing and looking up all of types of dictionary words. And that it does not look up for all of types of permutations in all of types of starting and ending Scrabble board squares or cubes in 2D Scrabble board or 3D Scrabble board in all of types of axes at any one single given period of time.
I just tried to simulate and play with Equate game style of Maths Scrabble in Scrabble3D application. https://www.boardgamegeek.com/image/1817520/equate Not exactly with the same tile set as in Equate game, but my modified version of it, with reducible fractional tiles removed and with keeping only fractional tiles which are strictly between 0 and 1 (0 and 1 both exclusive).
The following 220 tiles might fit well for a 21×21 board in order to optimally play with 10 tiles in the rack at any one given period of time. Because there are a diverse variety of fractional tiles used, this variant of Maths Scrabble is played only with 4 fundamental mathematical operations of addition, subtraction, multiplication and division. 1 point: ⁼ ×40, 1 ×30, 2 ×29 2 points: 0 ×28, 3 ×28, 4 ×28 3 points: 5 ×28, 6 ×28, 8 ×27, × ×10, ÷ ×10 4 points: 7 ×27, 9 ×27, + ×10, − ×10 5 points: ¹/₂ ×4, ¹/₄ ×2, ³/₄ ×2 6 points: ¹/₃ ×2, ²/₃ ×2 7 points: ¹/₆ ×2, ⁵/₆ ×2 8 points: ¹/₅ ×1, ²/₅ ×1, ³/₅ ×1, ⁴/₅ ×1, ¹/₈ ×1, ³/₈ ×1, ⁵/₈ ×1, ⁷/₈ ×1 9 points: ¹/₁₀ ×1, ³/₁₀ ×1, ⁷/₁₀ ×1, ⁹/₁₀ ×1 10 points: ¹/₇ ×1, ²/₇ ×1, ³/₇ ×1, ⁴/₇ ×1, ⁵/₇ ×1, ⁶/₇ ×1, ¹/₉ ×1, ²/₉ ×1, ⁴/₉ ×1, ⁵/₉ ×1, ⁷/₉ ×1, ⁸/₉ ×1 Note that × and ÷ are assigned lesser points than + and − because × and ÷ can wildly combine with fractions of all denominators, whereas + and − can only strictly combine with fractions of the same denominators and their proper divisors.
num = [1, 1, 2, 1, 3, 1, 2, 3, 4, 1, 5, 1, 2, 3, 4, 5, 6, 1, 3, 5, 7, 1, 2, 4, 5, 7, 8, 1, 3, 7, 9, 1, 5, 7, 11, 0] den = [2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 12, 12, 12, 12, 1] sym = ['#', '$', '&', '(', ')', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', '%', ':', '^', '.', '!'] conj = [0, 2, 1, 4, 3, 8, 7, 6, 5, 10, 9, 16, 15, 14, 13, 12, 11, 20, 19, 18, 17, 26, 25, 24, 23, 22, 21, 30, 29, 28, 27, 34, 33, 32, 31, 35] m = ["\xe2\x81\xb0", "\xc2\xb9", "\xc2\xb2", "\xc2\xb3", "\xe2\x81\xb4", "\xe2\x81\xb5", "\xe2\x81\xb6", "\xe2\x81\xb7", "\xe2\x81\xb8", "\xe2\x81\xb9"] n = ["\xe2\x82\x80", "\xe2\x82\x81", "\xe2\x82\x82", "\xe2\x82\x83", "\xe2\x82\x84", "\xe2\x82\x85", "\xe2\x82\x86", "\xe2\x82\x87", "\xe2\x82\x88", "\xe2\x82\x89"] d = {} for j in range(len(sym) - 1): d[(num[j], den[j])] = sym[j] def superscript(s): a = "" for c in s: a += m[ord(c) - 48] return a def subscript(t): b = "" for d in t: b += n[ord(d) - 48] return b def gcd(p, q): while q > 0: (p, q) = (q, p % q) return p def string(nu, de): result = '' if nu >= de or nu == 0: result += str(nu / de) if nu % de > 0: result += d[(nu % de, de)] return result def terms(nu, de): result = 0 if nu >= de or nu == 0: result += len(str(nu / de)) if nu % de > 0: result += 1 return result print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=Equalities' print 'Licence=GNU General Public License, v3' print 'Release=20.10.20' print 'Comment=Equate game style of Scrabble with mathematical equations' print 'Letters=@,0,1,2,3,4,5,6,7,8,9,-,*,/,+,#,$,&,(,),A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z,%,:,^,.' print 'Values=1,2,1,1,2,2,3,3,4,3,4,4,3,3,4,5,6,6,5,5,8,8,8,8,7,7,10,10,10,10,10,10,8,8,8,8,10,10,10,10,10,10,9,9,9,9,9,9,9,9' print 'Counts=36,9,11,10,9,9,9,9,8,8,8,9,9,9,9,4,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0' print 'Key=' print '[Replace]' print '-=\xe2\x88\x92' print '*=\xc3\x97' print '/=\xc3\xb7' print '@=\xe2\x81\xbc' for i in range(len(sym) - 1): print sym[i] + '=' + superscript(str(num[i])) + '/' + subscript(str(den[i])) print '[Categories]' print '1=Addition' print '2=Subtraction' print '3=Multiplication' print '4=Division' print '[Words]' for k in range(10000): print str(k) + '@' + str(k) + '=' for k in range(100): for l in range(len(sym) - 1): print (str(k), '') [k == 0] + d[(num[l], den[l])] + '@' + (str(k), '') [k == 0] + d[(num[l], den[l])] + '=' for k in range(10): for l in range(len(sym) - 1): for o in range(10): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11 or y % z == 0: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(100): for l in range(len(sym) - 1, len(sym)): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(len(sym) - 1, len(sym)): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(conj[l], conj[l] + 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * w + u * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '+' + string(v, w) + '@' + string(y, z) + '=;1' print string(y, z) + '@' + string(t, u) + '+' + string(v, w) + '=;1' print string(y, z) + '-' + string(t, u) + '@' + string(v, w) + '=;2' print string(v, w) + '@' + string(y, z) + '-' + string(t, u) + '=;2' for k in range(1000): for o in range(1000): y = k + o if len(str(k)) + len(str(o)) + len(str(y)) <= 7: print str(k) + '+' + str(o) + '@' + str(y) + '=;1' print str(y) + '@' + str(k) + '+' + str(o) + '=;1' print str(y) + '-' + str(k) + '@' + str(o) + '=;2' print str(o) + '@' + str(y) + '-' + str(k) + '=;2' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(100): for l in range(len(sym) - 1, len(sym)): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' if k > 0: print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(100): for l in range(len(sym) - 1): for o in range(100): for r in range(len(sym) - 1, len(sym)): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) <= 6: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(1, 100): for l in range(len(sym) - 1, len(sym)): for o in range(100): for r in range(len(sym) - 1): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) == 7: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' if k > 0: print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(100): for l in range(len(sym) - 1): for o in range(1, 100): for r in range(len(sym) - 1, len(sym)): t = den[l] * k + num[l] u = den[l] v = den[r] * o + num[r] w = den[r] y = t * v z = u * w g = gcd(y, z) y /= g z /= g if z > 12 or z == 11: continue if terms(t, u) + terms(v, w) + terms(y, z) == 7: print string(t, u) + '*' + string(v, w) + '@' + string(y, z) + '=;3' print string(y, z) + '@' + string(t, u) + '*' + string(v, w) + '=;3' print string(y, z) + '/' + string(t, u) + '@' + string(v, w) + '=;4' print string(v, w) + '@' + string(y, z) + '/' + string(t, u) + '=;4' for k in range(1, 1000): for o in range(1, 1000): y = k * o if len(str(k)) + len(str(o)) + len(str(y)) <= 8: print str(k) + '*' + str(o) + '@' + str(y) + '=;3' print str(y) + '@' + str(k) + '*' + str(o) + '=;3' print str(y) + '/' + str(k) + '@' + str(o) + '=;4' print str(o) + '@' + str(y) + '/' + str(k) + '=;4' print '0*0@0=;3' print '0@0*0=;3' for k in range(1, 10000): print '0*' + str(k) + '@0=;3' print '0@0*' + str(k) + '=;3' print str(k) + '*0@0=;3' print '0@' + str(k) + '*0=;3' print '0/' + str(k) + '@0=;4' print '0@0/' + str(k) + '=;4' for l in range(len(sym) - 1): k = range(0, 1000, den[l]) for r in range(len(k)): y = k[r] * num[l] / den[l] z = k[r] if z >= 100: print sym[l] + '*' + str(z) + '@' + str(y) + '=;3' print str(z) + '*' + sym[l] + '@' + str(y) + '=;3' print str(y) + '@' + sym[l] + '*' + str(z) + '=;3' print str(y) + '@' + str(z) + '*' + sym[l] + '=;3' print str(y) + '/' + str(z) + '@' + sym[l] + '=;4' print str(y) + '/' + sym[l] + '@' + str(z) + '=;4' print sym[l] + '@' + str(y) + '/' + str(z) + '=;4' print str(z) + '@' + str(y) + '/' + sym[l] + '=;4'
Compare my tile set with the actual tile set of the Equate game which is played on a 19×19 board with 190 tiles:
Letter
Count
Points
⁼
40
0
0
3
1
1
3
1
2
9
1
3
9
1
4
9
1
5
9
1
6
9
2
7
9
2
8
9
2
9
9
2
−
11
2
×
11
3
÷
11
5
+
11
1
¹/₂
4
4
²/₂
1
2
⁵/₂
1
5
¹/₃
2
8
²/₃
2
8
³/₃
1
2
¹/₄
2
6
²/₄
2
5
³/₄
2
6
⁴/₄
1
2
⁷/₄
1
7
¹/₆
1
10
²/₆
1
9
³/₆
1
5
⁴/₆
1
9
⁵/₆
1
12
⁶/₆
1
2
Blank
3
0
Total
190
382
Why ⁵/₂ exists without ³/₂ and ⁷/₄ without ⁵/₄? What is the point of having multiple equivalent reducible fractions such as those like ¹/₂, ²/₄ and ³/₆? ¹/₃ and ²/₆? ²/₃ and ⁴/₆? 2¹/₂ and ⁵/₂? 1³/₄ and ⁷/₄? What is the point of having multiple equivalent reducible fractions which all reduce to 1, such as those like ²/₂, ³/₃, ⁴/₄ and ⁶/₆? Why 0 and 1 tiles are only 3 each in number, which are more likely to be more frequently used than the rest of the digits, whereas all other digits namely 2, 3, 4, 5, 6, 7, 8 and 9 tiles are 9 each in number? Although that I do not desire to make use of leading zeros or trailing zeros after the decimal point.
m = ["\xe2\x81\xb0", "\xc2\xb9", "\xc2\xb2", "\xc2\xb3", "\xe2\x81\xb4", "\xe2\x81\xb5", "\xe2\x81\xb6", "\xe2\x81\xb7", "\xe2\x81\xb8", "\xe2\x81\xb9"] n = ["\xe2\x82\x80", "\xe2\x82\x81", "\xe2\x82\x82", "\xe2\x82\x83", "\xe2\x82\x84", "\xe2\x82\x85", "\xe2\x82\x86", "\xe2\x82\x87", "\xe2\x82\x88", "\xe2\x82\x89"] def superscript(s): a = "" for c in s: a += m[ord(c) - 48] return a def subscript(t): b = "" for d in t: b += n[ord(d) - 48] return b print "[Header]" print "Version=100000" print "Author=Raman Viswanathan <raman22feb1988@gmail.com>" print "StandardCategory=1 digit" print "Licence=GNU General Public License, v3" print "Release=20.09.20" print "Comment=This dictionary contains all numbers which are perfect powers of any integer \xe2\x89\xa5 2" print "[Replace]" print "[Categories]" for j in range(1, 11): print str(j) + "=" + str(j + 1) + " digits" print "[Words]" k = {} for i in range(2, 316228): z = i ** 2 y = 2 while z < 10 ** 11: if z not in k: k[z] = (i, y) z *= i y += 1 x = [] for (key, value) in k.items(): x.append((key, value[0], value[1])) l = sorted(x) for item in l: print str(item[0]) + "=" + str(item[1]) + superscript(str(item[2])) + (";" + str(len(str(item[0])) - 1), "") [item[0] <= 9]
2 points: M ×20, C ×20, X ×20 3 points: I ×19 5 points: D ×7, L ×7, V ×7
That is ideally played with 7 tiles in rack for obtaining within best results, but players can change this number freely enough depending upon their own wishes and depending upon their own tastes.
That is ideally played with 10 tiles in rack for obtaining within best results, but players can change this number freely enough depending upon their own wishes and depending upon their own tastes.
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=Gregorian calendar with 4 digits for year' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all dates from 1\xe2\x88\x921\xe2\x88\x921 to 31\xe2\x88\x9212\xe2\x88\x929999' print '[Replace]' print '-=\xe2\x88\x92' print '[Categories]' print '[Words]' days = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday'] day = -1 for year in range(1, 10000): for month in range(1, 13): dates = 31 if month == 4 or month == 6 or month == 9 or month == 11: dates -= 1 if month == 2: dates -= 3 if year % 4 == 0: dates += 1 if year >= 1800 and year % 100 == 0 and year % 400 != 0: dates -= 1 for date in range(1, dates + 1): if year == 1752 and month == 9 and date >= 3 and date <= 13: print str(date) + '-' + str(month) + '-' + str(year) + '=11 days were dropped during the process of switching over from Julian calendar to Gregorian calendar' else: print str(date) + '-' + str(month) + '-' + str(year) + '=' + days[day % 7] day += 1
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=Gregorian calendar with 2 digits for year' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all dates from 1\xe2\x88\x921\xe2\x88\x9200 to 31\xe2\x88\x9212\xe2\x88\x9299' print '[Replace]' print '-=\xe2\x88\x92' print '[Categories]' print '[Words]' for year in range(0, 100): for month in range(1, 13): dates = 31 if month == 4 or month == 6 or month == 9 or month == 11: dates -= 1 if month == 2: dates -= 3 if year % 4 == 0: dates += 1 for date in range(1, dates + 1): print str(date) + '-' + str(month) + '-' + (str(year), '0' + str(year)) [year <= 9]
time.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=24 hour time with seconds' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 0:00:00 to 23:59:59' print '[Replace]' print '[Categories]' print '[Words]' for hour in range(0, 24): for minute in range(0, 60): for second in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + ':' + (str(second), '0' + str(second)) [second <= 9]
clock.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=12 hour time with seconds with AM / PM marker in two separate tiles' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 12:00:00 AM to 11:59:59 PM' print '[Replace]' print '[Categories]' print '[Words]' for hour in range(1, 13): for minute in range(0, 60): for second in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + ':' + (str(second), '0' + str(second)) [second <= 9] + 'AM' print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + ':' + (str(second), '0' + str(second)) [second <= 9] + 'PM'
watch.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=12 hour time with seconds without AM / PM marker' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 1:00:00 to 12:59:59' print '[Replace]' print '[Categories]' print '[Words]' for hour in range(1, 13): for minute in range(0, 60): for second in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + ':' + (str(second), '0' + str(second)) [second <= 9]
hour.py
1 2 3 4 5 6 7 8 9 10 11 12 13
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=12 hour time without seconds without AM / PM marker' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 1:00 to 12:59' print '[Replace]' print '[Categories]' print '[Words]' for hour in range(1, 13): for minute in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9]
minute.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=12 hour time without seconds with AM / PM marker in two separate tiles' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 12:00 AM to 11:59 PM' print '[Replace]' print '[Categories]' print '[Words]' for hour in range(1, 13): for minute in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + 'AM' print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + 'PM'
second.py
1 2 3 4 5 6 7 8 9 10 11 12 13
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=24 hour time without seconds' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 0:00 to 23:59' print '[Replace]' print '[Categories]' print '[Words]' for hour in range(0, 24): for minute in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9]
hourglass.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=12 hour time with seconds with AM / PM marker in one single tile' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 12:00:00 AM to 11:59:59 PM' print '[Replace]' print 'A=AM' print 'P=PM' print '[Categories]' print '[Words]' for hour in range(1, 13): for minute in range(0, 60): for second in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + ':' + (str(second), '0' + str(second)) [second <= 9] + 'A' print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + ':' + (str(second), '0' + str(second)) [second <= 9] + 'P'
sundial.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
print '[Header]' print 'Version=100000' print 'Author=Raman Viswanathan <raman22feb1988@gmail.com>' print 'StandardCategory=12 hour time without seconds with AM / PM marker in one single tile' print 'Licence=GNU General Public License, v3' print 'Release=17.09.20' print 'Comment=This dictionary contains all time from 12:00 AM to 11:59 PM' print '[Replace]' print 'A=AM' print 'P=PM' print '[Categories]' print '[Words]' for hour in range(1, 13): for minute in range(0, 60): print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + 'A' print str(hour) + ':' + (str(minute), '0' + str(minute)) [minute <= 9] + 'P'
That is ideally played with 7 or 8 tiles in rack for obtaining within best results, but players can change this number freely enough depending upon their own wishes and depending upon their own tastes.
square.py
1 2 3 4 5 6 7 8 9 10 11 12 13
print "[Header]" print "Version=100000" print "Author=Raman Viswanathan <raman22feb1988@gmail.com>" print "StandardCategory=1 digit" print "Licence=GNU General Public License, v3" print "Comment=Squares" print "[Replace]" print "[Categories]" for j in range(1, 11): print str(j) + "=" + str(j + 1) + " digits" print "[Words]" for i in range(1, 316228): print str(i ** 2) + "=" + str(i) + "\xc2\xb2" + (";" + str(len(str(i ** 2)) - 1), "") [i <= 3]
2 points: M ×20, C ×20, X ×20 3 points: I ×19 5 points: D ×7, L ×7, V ×7
That is ideally played with 7 tiles in rack for obtaining within best results, but players can change this number freely enough depending upon their own wishes and depending upon their own tastes.
(li) Female sheep (lii) Under obligation to repay (liii) They give a hoot! (liv) Lubricant for a car's engine (lv) American rapper, ___ Wayne (lvi) Rim or edge of a pitcher (lvii) One trip around a track (lviii) In computer parlance, a local area network (lix) Prohibit or forbid (lx) Flying mammal that likes caves (lxi) Slang for decorative skin art (ink) (lxii) Thick, viscous fluid composed of hydrocarbons (lxiii) Blemish a surface (lxiv) 'One small step for ___' (lxv) 'Ya ___' - Jamaican phrase for 'No problem' (lxvi) ABC crime drama, 'The ___ Squad' (1968-1973) (lxvii) Supreme being (lxviii) Spanish shout for a soccer score (lxix) Latin name for our sun (lxx) French for salt (lxxi) Mediterranean or Caspian, for example (lxxii) A grassy meadow
Find out a nine timer bingo in this following board position: I wish that all of puzzles in this thread and in 'Maths Scrabble Puzzles' thread are numbered properly. At least by some administrator of this forum as I do not have editing access to my old posts in this forum beyond the time limit of 1 hour except only in the Tamil forum in which I am assigned to be the only moderator in this forum. Anyway that it is not that important at all for a forum with a very few visitors nowadays. Also that my own Scrabble puzzles and also that of similarly posted messages in this forum may be only for my own future reference purposes. Also that I do not know about that trend in the past and also that I cannot predict about that in the future.
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28.03.2021 11:10
