fuzzer | 2015-02-26 01:21:20 |
HI everyone, aside for one post under the 'class suggestions' folder a long time ago, I haven't wrote nothing in this forum yet. I'll just say I really love this game and consider spectromancer as one of my hobbies - so, thank you(!) developers and thank you(!) fellow wizards and mages! so here goes... I really enjoy riddles. especially if they are smart and elegant. the spectromancer ground seems fertile for such riddles... so here's one I like for riddle lovers among fellow spectromancers: a. what's the greatest amount of damage you could possibly deal to the opponent in a single turn? (that include combat damage and possible spell damage)* spells that build up the more mana you have are considered to be cast for their regular mana cost for this riddle - for example, Armageddon is considered to be cast for exactly 11 mana for 19 overall damage. * elementals are considered as 0 attack creatures.* you can mix creatures and spells from any number of different classes.
and the second (much nicer!): b. what's the greatest amount of combat damage you could possibly deal to the opponent in a single turn?* (same conditions)
Edit: you can see the top answers with credit for the creator if you click on my name, in the 'about myself' section... maybe you can come up with better results :) Modified by fuzzer on 2015-02-27 04:16:10
Well, if I'm doing the math right... a) 224 {6 dragons + armageddon at 11 power} b) 91 {5 greater barguls + 1 monument to rage} * c) 96 {6 oracles at 8 illusion mana}* ... yeah... I think... How'd you come up with your figures, fuzzer? Modified by Coco on 2015-02-26 03:19:08 fuzzer | 2015-02-26 03:29:23 |
Well, if I'm doing the math right...
a) 224 {6 dragons + Armageddon at 11 power}
b) 156 {5 greater barguls + 1 monument to rage}
... yeah... I think... your ideas are close to my ideas but your calculations are not accurate. a) 2 dragons won't make the damage at 225% (150% * 150%) but each of them will only increase the original damage by additional 50% so the original 19 damage will add to 6*50%=6*10dmg for total of 79 dmg. also your dragons will die and will be unable to attack... :) b) Monument to Rage will die after the third Greater Bargul will attack for the first time so the third, forth and fifth Barguls will only attack once... for total damage of 91... also c) I forgot to address oracle ability in the condition, but, of course, because this damage can be potentially unlimited - it will be treated like the illusion mana is on 0 for the sake of its damage ability... but i like where you're going! keep trying! :)
... your ideas are close to my ideas but your calculations are not accurate.
a) 2 dragons won't make the damage at 225% (150% * 150%) but each of them will only increase the original damage by additional 50% so the original 19 damage will add to 6*50%=6*10dmg for total of 79 dmg. also your dragons will die and will be unable to attack... :)
b) Monument to Rage will die after the third Greater Bargul will attack for the first time so the third, forth and fifth Barguls will only attack once... for total damage of 91...
also c) I forgot to address oracle ability in the condition, but, of course, because this damage can be potentially unlimited - it will be treated like the illusion mana is on 0 for the sake of its damage ability...
but i like where you're going! keep trying! :)
a) Well, the dragons don't need to attack; each dragon will simply boost the damage of the spell by 50% per each dragon. 19 > 29 > 44 > 66 > 99 > 149 > 224 b) Yeah, my figures were messed up on my first go. You must've quoted me before I fixed it c) In that case, forget the oracle.
Modified by Coco on 2015-02-26 03:45:53 fuzzer | 2015-02-26 04:31:10 |
... a) Well, the dragons don't need to attack; each dragon will simply boost the damage of the spell by 50% per each dragon. 19 > 29 > 44 > 66 > 99 > 149 > 224
b) Yeah, my figures were messed up on my first go. You must've quoted me before I fixed it
c) In that case, forget the oracle.
actually the dragons don't accumulate so they won't boost it like that. it will be: 19 > 29 >39 > 49 > 59 > 69 > 79 you can check it. any other ideas? :)
... actually the dragons don't accumulate so they won't boost it like that. it will be: 19 > 29 >39 > 49 > 59 > 69 > 79 you can check it.
any other ideas? :)
I'll have to take your word for it. I surrender. fuzzer | 2015-02-26 04:56:35 |
... I'll have to take your word for it.
I surrender.
fair enough. but I don't want to post my answer so soon.. if my answers will still be the highest in a few days than I'll expose them here. you can check it in here later :)
kavarin1 | 2015-02-26 11:11:42 |
Both a and b are unlimited due to S2 and various golem cards.
kavarin1 | 2015-02-26 11:31:59 |
b.126 seems low
4 chastisers (v6) + 2 monuments to rage, opposed by 6 * low-life e3 + damage spell...
Each chastiser would reach 19 damage, then add 63 for each monument to rage.
Each chastiser would reach 19 damage, then add 63 for each monument to rage.
multiple monuments will just die all together after the first 63 damage is dealt
Modified by filip on 2015-02-26 11:36:24 kavarin1 | 2015-02-26 11:38:29 |
Correction: Each chastiser would reach 19 damage, then add _36_ for each monument to rage.
opposed by 6 * low-life e3 + damage spell...
i don't think we should make such assumptions about the opponent's setup. i think we should assume all opposing slots as empty. what do you think fuzzer?
Modified by filip on 2015-02-26 11:55:32
b) Wo8 + Wo4 + V4 + V4 + V4 + V4
(13+13 + 9+11 + 13+15 + 17 + 19 = 110 damage)
Modified by filip on 2015-02-26 11:58:36 kavarin1 | 2015-02-26 12:01:18 |
b) Wo8 + Wo4 + V4 + V4 + V4 + V4
(13+13 + 9+11 + 13+15 + 17 + 19 = 110 damage)
Oh, very nice... then add a bunch of low-life W2 on the other side.
Four people, A, B, C, and D, are on one side of a bridge, and they all want to cross the bridge. However, it's late at night, so you can't cross without a flashlight. They only have one flashlight. Also, the bridge is only strong enough to support the weight of two people at once. The four people all walk at different speeds: A takes 1 minute to cross the bridge, B takes 2 minutes, C takes 5 minutes, and D takes 10 minutes. When two people cross together, sharing the flashlight, they walk at the slower person's rate. How quickly can the four cross the bridge?
You have two ropes, each of which takes exactly one hour to burn completely. Both of these ropes are non-homogeneous in thickness, meaning that some parts of the rope are chunkier (burn slower) than other parts of the rope, but you don't know which. For example, perhaps one section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. Then it would take 10 minutes to burn off the remaning 7/8, since we know that the whole rope takes exactly one hour to burn completely. Using these two ropes and a lighter, accurately time 45 minutes.
You are sitting on the floor of a completely dark room. 100 disk-shaped pills are laid out before you. You know that these pills are black on the one side and white on the other. Someone tells you that right now 20 of these pills have their white side on top and the rest have their black side on top. He says that you must separate the pills into two groups, in such a way that both groups have the same number of white pills (pills with their white side on top). You are allowed to move the pills and to flip them over, but you cannot see anything and the pills are totally symmetrical (both sides feel the same to your touch). What do you do?
A man bumps into his mathematician teacher on the street. He hasn't seen him in many years, so the man asks the mathematician how old his children are. The mathematician, who always replies in riddles, says: "I now have three children. The sum of their ages is equal to the address number of this building in front of us. The product of their ages equals 36." The man then says "I need one more piece of information." The mathematician then replies "My youngest child has blue eyes." What are the ages of the mathematician's three children?
You have 12 identical-looking coins. There are 11 coins that have the same weight exactly and one coin that has a different weight from the others, but you don't know which one that is (and you don't know if it is heavier or lighter than the rest). You also have a two-pan balance scale for comparing weights. Using the balance scale 3 times, determine which coin has the unique weight and also determine whether it is heavier or lighter than the others.
GrimJ0ker | 2015-02-26 12:58:16 |
It's impossible to calculate without this condition " spells that build up the more mana you have are considered to be cast for their regular mana cost for this riddle " , It depends on how many turns the opponent concedes to you charging a spell. I did 2 month ago an example with an advanced mage, surely the easier example is with the holy class.
Holy5 + Mindmasters + Dragoons + Air6.
Modified by GrimJ0ker on 2015-02-26 13:01:30 HeadphonesGirl | 2015-02-26 14:27:21 |
Off the top of my head, 2 dragons 18 + arma 39 + 4 wall of reflection 80 = 137 (I'm assuming since he said mixing class cards as is allowed that banned combos are as well) Apparently not as high as fuzzer's though
Modified by HeadphonesGirl on 2015-02-26 14:28:33 fuzzer | 2015-02-26 14:50:31 |
...
i don't think we should make such assumptions about the opponent's setup. i think we should assume all opposing slots as empty. what do you think fuzzer?
as I see it, I suppose if you really want the opponent board to have creatures in it - it's not a problem and you can choose to do so, but this riddle's little game starts on the beginning of your turn and ends at the end of your turn and the poor opponent wouldn't have a chance to attack before your turn begins or after it ends and wouldn't have a chance even to see his creature's abilities trigger... Four people, A, B, C, and D, are on one side of a bridge, and they all want to cross the bridge. However, it's late at night, so you can't cross without a flashlight. They only have one flashlight. Also, the bridge is only strong enough to support the weight of two people at once. The four people all walk at different speeds: A takes 1 minute to cross the bridge, B takes 2 minutes, C takes 5 minutes, and D takes 10 minutes. When two people cross together, sharing the flashlight, they walk at the slower person's rate. How quickly can the four cross the bridge? You have two ropes, each of which takes exactly one hour to burn completely. Both of these ropes are non-homogeneous in thickness, meaning that some parts of the rope are chunkier (burn slower) than other parts of the rope, but you don't know which. For example, perhaps one section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. Then it would take 10 minutes to burn off the remaning 7/8, since we know that the whole rope takes exactly one hour to burn completely. Using these two ropes and a lighter, accurately time 45 minutes. You are sitting on the floor of a completely dark room. 100 disk-shaped pills are laid out before you. You know that these pills are black on the one side and white on the other. Someone tells you that right now 20 of these pills have their white side on top and the rest have their black side on top. He says that you must separate the pills into two groups, in such a way that both groups have the same number of white pills (pills with their white side on top). You are allowed to move the pills and to flip them over, but you cannot see anything and the pills are totally symmetrical (both sides feel the same to your touch). What do you do? A man bumps into his mathematician teacher on the street. He hasn't seen him in many years, so the man asks the mathematician how old his children are. The mathematician, who always replies in riddles, says: "I now have three children. The sum of their ages is equal to the address number of this building in front of us. The product of their ages equals 36." The man then says "I need one more piece of information." The mathematician then replies "My youngest child has blue eyes." What are the ages of the mathematician's three children? You have 12 identical-looking coins. There are 11 coins that have the same weight exactly and one coin that has a different weight from the others, but you don't know which one that is (and you don't know if it is heavier or lighter than the rest). You also have a two-pan balance scale for comparing weights. Using the balance scale 3 times, determine which coin has the unique weight and also determine whether it is heavier or lighter than the others. 17 minuets after the four begin their journey, when they finally reach the other side of the bridge at cold late night they can light 3/4 rope's ends and when one rope is gone light the remaining rope end - when that rope is gone they will know they are 45 minuets after their journey ended. knowing this they can relax themselves by separating a 100 weird pills into a group of 80 and a group of 20 and flipping all the pills in the smaller group... in the next morning they could congratulate the mathematician for his one year old sun's birthday (with the blue blue eyes) and ask him if twins run in his family or the two six years old a flook? and now for something completely different: 1 2 5 6 3 4 > 7 8 1 2 9 10 3 8 > 11 4 1 > 2 I actually already knew all these riddles! I can give you one or two, filip... two nice harder riddles come to my mind... and they are as elegant as the ones you wrote... but first thing is first... b) Wo8 + Wo4 + V4 + V4 + V4 + V4
(13+13 + 9+11 + 13+15 + 17 + 19 = 110 damage)
nice! very nice! but I managed to reach a little higher... 126 to be exact... anyone can top that? are you up for the challenge filip? :)))))))))) It's impossible to calculate without this condition "spells that build up the more mana you have are considered to be cast for their regular mana cost for this riddle " , It depends on how many turns the opponent concedes to you charging a spell. I did 2 month ago an example with an advanced mage, surely the easier example is with the holy class.
Holy5 + Mindmasters + Dragoons + Air6.
that is one bad-ass lightning bolt :) fuzzer | 2015-02-26 14:59:44 |
Off the top of my head, 2 dragons 18 + arma 39 + 4 wall of reflection 80 = 137
(I'm assuming since he said mixing class cards as is allowed that banned combos are as well)
Apparently not as high as fuzzer's though I really like your way! I especially like how those two dragons take the Armageddon to 39 and just hardly survive to deal their 18 combat damage... but yes, you can go higher! btw, don't worry about banned combos... anything goes!
fuzzer | 2015-02-26 15:00:58 |
...still waiting for someone to break and top my solution...
so here's one I like for riddle lovers among fellow spectromancers: a. what's the greatest amount of damage you could possibly deal to the opponent in a single turn? (that include combat damage and possible spell damage)
Many potentially infinite answers still exist with those conditions (Army Upgrade, Angry Angry Bear, Ghoul, Magic Rabbit/Enraged Beaver/Forest Wolf, Holy Avenger, etc.)
Rage of Souls and 6 Dragons against 6 undamaged Walls of Reflection (possible to set up with cooperation and Ancient Horror/Sonic Boom) should beat your answer by a little bit. I don't want to figure out the exact number.
Edit: never mind, Rage of Souls only damages creatures, so make it Sonic Boom or something.
Edit: Wait, banned combos ok? Then anything goes, so 10 walls of reflection and two dragons and an armageddon for I don't know 200+18+39?
Four people, A, B, C, and D, are on one side of a bridge... How quickly can the four cross the bridge? The critical pair are C and D, you can save the most time moving them together, so they must go together. But they can't go first because someone has to come back with the flashlight. It's best to have A come back with the flashlight for them. So send A and B first, then B back to give the flashlight to C and D while they cross, then bring A back with the flashlight to pick up B. I don't want to calculate the numbers, but that should logically be the shortest route.
You have two ropes, each of which takes exactly one hour to burn completely...Using these two ropes and a lighter, accurately time 45 minutes. The puzzle is worded so that folding the rope provides no useful information, so lighting it at both ends is the only way to get more information out of the ropes. Therefore, light one at both ends, and the other at one end, at the same time so you can get a 30 minute counter from the one lit at both ends, then light the other end of the remaining rope when the 30 minutes elapse so you can get a 15 minute timer.
You are sitting on the floor of a completely dark room. ... What do you do?
The only meaningful thing you can do in this situation is to split the pills into two groups and/or flip one of the groups. Anything else is randomness. The split point is n, there will be n pills in the first group and 100-n in the other group.
There will be w white pills in the first group, and 20-w white pills in the second group. So if we flip one group, the split point n must satisfy (n-w) = (20-w), which makes n 20.
A man bumps into his mathematician teacher on the street. ... What are the ages of the mathematician's three children? You have to make assumptions for this to make any sense, about the relative deductive abilities of both characters, or the concept of integral age, or whatever. But the simplest interpretation I can think of, there must be at least 2 children the second person knows about in advance to make the first question make sense, and although they could have used Skype or whatever, assume they didn't, so two children are at least 3, meaning the only answer is 6,6,1. There are other ways of multiplying 3 integers to make 36 but they won't meet this or that condition depending on the interpretation of why certain information was included the riddle in the first place.
You have 12 identical-looking coins. ... determine whether it is heavier or lighter than the others. I heard this before a long time ago I think. The key feature is that you can compare lots of things in parallel by splitting them into groups, labeling them, and keeping track of the various combinations. Split them three ways to maximize the comparisons, and keep track of where the heavy side is each time then backtrack over it. I wouldn't want to do it for real.
Edit: oh, fuzzer already answered all of these while I wrote that. that's so fast. Anyway, I'm too easily distracted by puzzles when I should be studying. Modified by Plynx on 2015-02-26 16:05:04 GrimJ0ker | 2015-02-26 15:46:23 |
Wall of reflection can do damages only to the opponent, so i think 200 damages is impossible by 10 walls.
Modified by GrimJ0ker on 2015-02-26 15:46:31 GrimJ0ker | 2015-02-26 15:53:10 |
B) Vamp3 + Vamp3 + Vamp3 + Vamp3 + Vamp3 + Cult4
All vamp3 have 1hp, in a spectro universe where we can heal v3 with e11 for infinite turns we can do also better than this ---> 18+18 + 18+18 + 18 + 18 + 18 = 126 Modified by GrimJ0ker on 2015-02-26 15:56:48 kavarin1 | 2015-02-26 16:29:48 |
B) Vamp3 + Vamp3 + Vamp3 + Vamp3 + Vamp3 + Cult4
All vamp3 have 1hp, in a spectro universe where we can heal v3 with e11 for infinite turns we can do also better than this ---> 18+18 + 18+18 + 18 + 18 + 18 = 126 Can't Zealot vs Ancient Horror go even higher?
fuzzer | 2015-02-26 16:33:51 |
...
Many potentially infinite answers still exist with those conditions (Army Upgrade, Angry Angry Bear, Ghoul, Magic Rabbit/Enraged Beaver/Forest Wolf, Holy Avenger, etc.)
Rage of Souls and 6 Dragons against 6 undamaged Walls of Reflection (possible to set up with cooperation and Ancient Horror/Sonic Boom) should beat your answer by a little bit. I don't want to figure out the exact number.
Edit: never mind, Rage of Souls only damages creatures, so make it Sonic Boom or something.
...
The critical pair are C and D, you can save the most time moving them together, so they must go together. But they can't go first because someone has to come back with the flashlight. It's best to have A come back with the flashlight for them. So send A and B first, then B back to give the flashlight to C and D while they cross, then bring A back with the flashlight to pick up B. I don't want to calculate the numbers, but that should logically be the shortest route.
...
The puzzle is worded so that folding the rope provides no useful information, so lighting it at both ends is the only way to get more information out of the ropes. Therefore, light one at both ends, and the other at one end, at the same time so you can get a 30 minute counter from the one lit at both ends, then light the other end of the remaining rope when the 30 minutes elapse so you can get a 15 minute timer.
...
The only meaningful thing you can do in this situation is to split the pills into two groups and/or flip one of the groups. Anything else is randomness. The split point is n, there will be n pills in the first group and 100-n in the other group.
There will be w white pills in the first group, and 20-w white pills in the second group. So if we flip one group, the split point n must satisfy (n-w) = (20-w), which makes n 20.
...
You have to make assumptions for this to make any sense, about the relative deductive abilities of both characters, or the concept of integral age, or whatever. But the simplest interpretation I can think of, there must be at least 2 children the second person knows about in advance to make the first question make sense, and although they could have used Skype or whatever, assume they didn't, so two children are at least 5, meaning the only answer is 6,6,1. There are other ways of multiplying 3 integers to make 36 but they won't meet this or that condition depending on the interpretation of why certain information was included the riddle in the first place.
...
I heard this before a long time ago I think. The key feature is that you can compare lots of things in parallel by splitting them into groups, labeling them, and keeping track of the various combinations. Split them three ways to maximize the comparisons, and keep track of where the heavy side is each time then backtrack over it. I wouldn't want to do it for real.
1. actually your 6 wall of reflection +6 dragons + sonic boom only deals 218 damage which is 1 less than 219. 2. what is 10 wall of reflections? if 4 of them are on your side than the damage would be dealt to you and not to the opponent! 3. on my 219 solution I assumed a clear board on the opponents side. (no use of wall of reflection at all!) now when I understand that I mistakenly overlooked wall of reflection I have found a newer solution for riddle a. which is 306 damage. 4.a) I still can't see an unlimited damage for riddle a or b. a.b) perhaps I need to be clearer: all creatures begin the turn in their base situation... there is no memory from previous turns. no 'master-healer-previously-healed-them'... 4.c) your turn is "like in the real game" finnished after playing your card. so you can't "army upgrade" your golem multiple times... even the only potential for unlimited turns "time stop" won't work since technically your turn do Finnish before the next one would have begun...
B) Vamp3 + Vamp3 + Vamp3 + Vamp3 + Vamp3 + Cult4
All vamp3 have 1hp, in a spectro universe where we can heal v3 with e11 for infinite turns we can do also better than this ---> 18+18 + 18+18 + 18 + 18 + 18 = 126 like I wrote to plynx above, there is no memory from previous turns and all creatures begin the turn in their base situation! perhaps I should have been more clear in the conditions I wrote in the beginning :) fuzzer | 2015-02-26 16:37:45 |
a. 306 b. 216 can anyone equal or top that?
HeadphonesGirl | 2015-02-26 16:42:43 |
...
Many potentially infinite answers still exist with those conditions (Army Upgrade, Angry Angry Bear, Ghoul, Magic Rabbit/Enraged Beaver/Forest Wolf, Holy Avenger, etc.)
Rage of Souls and 6 Dragons against 6 undamaged Walls of Reflection (possible to set up with cooperation and Ancient Horror/Sonic Boom) should beat your answer by a little bit. I don't want to figure out the exact number.
Edit: never mind, Rage of Souls only damages creatures, so make it Sonic Boom or something.
Edit: Wait, banned combos ok? Then anything goes, so 10 walls of reflection and two dragons and an armageddon for I don't know 200+18+39?
I thought Fuzzer was talking about how much damage you can deal to the opponent only, but it sounds like your answers are calculating total damage to both players.
GrimJ0ker | 2015-02-26 17:13:39 |
a. 306 b. 216 can anyone equal or top that? b. 216 or 126? kavarin1 | 2015-02-26 17:14:28 |
a. 306 b. 216 can anyone equal or top that?
I can get 215 with 5 golems boosted to 21 attack by G1 plus G8. Or if 2 G8s give 3 attacks, that makes 262.
fuzzer | 2015-02-26 17:30:05 |
... I thought Fuzzer was talking about how much damage you can deal to the opponent only, but it sounds like your answers are calculating total damage to both players.
I was talking about damage to the opponent only! ... b. 216 or 126?
sorry for the mistake. b.126 but now a is 308 (+2 damage) so: a.308 b.126 fuzzer | 2015-02-26 17:38:04 |
...
I can get 215 with 5 golems boosted to 21 attack by G1 plus G8. Or if 2 G8s give 3 attacks, that makes 262.
since you can't summon more than 1 golem I am ambivalent about this solution but i guess if you can mix classes and overlook banned combos why not 5 goloms??? good soulution. beter than mine :)
GrimJ0ker | 2015-02-26 17:47:57 |
B) Cult7 + Vamp4 + Vamp4 + Vamp4 + Vamp4 + Cultist4
(8+8) + (9+11) + (13+15) + (17+19) + (21+23) = 144 fuzzer | 2015-02-26 17:50:59 |
B) Cult7 + Vamp4 + Vamp4 + Vamp4 + Vamp4 + Cultist4
(8+8) + (9+11) + (13+15) + (17+19) + (21+23) = 144 i thought about that axact solution but it fails :))))) cult 7 don't absorbe the damage from cult4 since its considered lost of life... :(
fuzzer | 2015-02-26 18:00:07 |
my a is wrong i go back to 219 for riddle a.
fuzzer | 2015-02-26 18:02:10 |
(I thought opponent's 6 wall of reflaction will die and deal damage to him) fuzzer | 2015-02-26 18:03:15 |
i will write my solutions tommorow
mamoulian | 2015-02-26 21:16:44 |
3dragon + 3time7 + 4earth6
kavarin | 2015-02-26 21:51:54 |
... since you can't summon more than 1 golem I am ambivalent about this solution but i guess if you can mix classes and overlook banned combos why not 5 goloms??? good soulution. beter than mine :)
If you don't like that solution, 1 golem boosted to 21 and 5 * G8 is 151 and legal, if unlikely.
GrimJ0ker | 2015-02-26 23:00:11 |
B) Vamp8 (just casted) - Time6 - Vamp4 - Vamp5 - Vamp4 - Cult4
Vamp8 = {(16+2) + [(8+2)+(8+2)]} = 38 Time6 = [(4+2)+(4+2)+(4+2)] = 18 Vamp4 = [(17+2)+(19+2)] = 40 Vamp5 = (3+2) = 5 Vamp4 = (23+2) = 25
38+18+40+5+25= 126
Modified by GrimJ0ker on 2015-02-26 23:08:52 fuzzer | 2015-02-27 03:12:17 |
B) Vamp8 (just casted) - Time6 - Vamp4 - Vamp5 - Vamp4 - Cult4
Vamp8 = {(16+2) + [(8+2)+(8+2)]} = 38 Time6 = [(4+2)+(4+2)+(4+2)] = 18 Vamp4 = [(17+2)+(19+2)] = 40 Vamp5 = (3+2) = 5 Vamp4 = (23+2) = 25
38+18+40+5+25= 126
very nice! my solution for b was: B. vamp5-vamp5-vamp5-vamp4-vamp4-cult4 + sorc5 vamp5 = {(3+2+2+2+3) + (3+2+2+2+3)} = 24 vamp5 = {(3+2+2+2+3) + (3+2+2+2+3)} = 24 vamp5 = {(3+2+2+2+3) + (3+2+2+2+3)} = 24 vamp4 = {17+2+2+2+3} = 26 vamp4 = {19+2+2+2+3} = 28 24+24+24+26+28=126 but I think it's safe to say that kavarin1 had a better b solution. ... If you don't like that solution, 1 golem boosted to 21 and 5 * G8 is 151 and legal, if unlikely.
this one I like very much! and I officially declare it to be the highest b solution yet!!! do you think anyone can top that? :) 3dragon + 3time7 + 4earth6 indeed, mamoulian, that was my axact a solution! 219 fuzzer | 2015-02-27 03:30:50 |
after brainstorming with fellow spectromancer SerenB we have come up with a higher solution for riddle a. a. what's the greatest amount of damage you could possibly deal to the opponent in a single turn? (that include combat damage and possible spell damage) * spells that build up the more mana you have are considered to be cast for their regular mana cost for this riddle - for example, Armageddon is considered to be cast for exactly 11 mana for 19 overall damage. * elementals are considered as 0 attack creatures. * you can mix creatures and spells from any number of different classes.
we had managed to deal 344 damage to the opponent in one turn! who can crack how so? maybe mamoulian? or kavarin? the grimjoker can? someone else? maybe plynx can? fuzzer | 2015-02-27 04:23:43 |
Top solutions: a. 344 (fuzzer&SerenB) b. 151 (kavarin)
kavarin1 | 2015-02-27 10:10:10 |
Top solutions: a. 344 (fuzzer&SerenB) b. 151 (kavarin) T7 Golem 4 * g8 Cast G1 and Sorc5
(5 * 24) + (4 * 8) + 7 == 159 kavarin1 | 2015-02-27 10:17:24 |
... T7 Golem 4 * g8 Cast G1 and Sorc5
(5 * 24) + (4 * 8) + 7 == 159
And swapping another g8 for another t7 and casting sorc5 twice maybe gets to 161? GrimJ0ker | 2015-02-27 15:07:07 |
I didn't know we could cast a spell or playing more cards in one turn with time7, in this way too many possibilities, it's not so interesting riddle.
Modified by GrimJ0ker on 2015-02-27 15:07:25 fuzzer | 2015-02-27 18:09:01 |
I had also managed to find a better solution for riddle B. OPPONENT BOARD: illusion4 illusion4 illusion4 illusion4 illusion4 illusion4YOUR BOARD: gob7 gob7 vemp6 vemp6 vemp6 vemp6PLAY CARD: fire6in this way you will receive 120 damage but your opponent will be dealt 194 combat damage!I am not sure this solution can break without using multiple golems like kavarin had suggested. I didn't know we could cast a spell or playing more cards in one turn with time7, in this way too many possibilities, it's not so interesting riddle. without playing a spell you can still reach 142 damage by playing 4gob7 and 2vemp6. as for SerenB's and mine solution for riddle A: OPPONENT BOARD: illusion4 illusion4 illusion4 illusion4 illusion4 illusion4YOUR BOARD: fire12 vemp6 vemp6 time7 time7 time7PLAY CARDS: spirit6 + fire11 + earth6 + earth6in this way you will receive 120 damage but your opponent will be dealt 344 damage!I am not sure this solution can break. Modified by fuzzer on 2015-02-27 18:22:27 |