Xjyuk

Recently, I dived into a nice religion of Awesomism. It is pantheic, with a lot of difficult relations between gods.

Twelve gods of Awesomism are split between four pantheons:

Gods of cunning: Liar, Disdain, Deceitralic

Gods of sun: Pyros, Solar, Taehwasi

Gods of life: Icyayou, Leevah, Stummock

Gods of wisdom: Hugo, Panoptic and... the one who shan't be named.

The 0th god, The Almighty and Highest-Above-All, may be praised His wisdom, I cannot call by name. He is opposed to eleven other gods, for He is too great even for them. So, going down the list of gods, do not consider the 0th god, for He is too mighty for you to comprehend.

As for the other gods, they are all numbered 1 to 11; gods 1-3 are referred to as Low Gods, 4-8 as Lower Gods and 9-11 as Lowest Gods (so, the names are supposed to be going downwards).

Each god has committed several Feats; a Feat cannot be committed collaboratively. Now hear the conditions:

• The total amount of Feats committed by Lowest Gods is the same as the total amount of all Feats, if you swap its digits.


• The amount of Feats committed by Lower gods is two.


• There are no two gods committing same, odd amount of Feats.


• Liar is located at the same distance from Disdain and Taehwasi.


• Disdain's neighbors are from the same pantheon; both of them committed at least one Feat.


• Deceitralic has committed five Feats - that is more than anyone else; however, she is among the lowest gods.


• Pyros committed as much Feats as any of his neighbors, but his name is longer than theirs.


• Solar is the lowest among gods who committed no Feats.


• Taehwasi's higher neighbor committed more Feats than her.


• Icyayou's higher neighbor committed the same amount of Feats as Taehwasi did.


• Leevah has committed the same amount of Feats as Stummock did.


• Stummock's neighbors' names contain only six letters total, three of which are shared.

Now, here is the task:

Challenge of logic : List the eleven gods and amount of their Feats (also, present some outline of your logic, if it's ok with you)

Enigmatic challenge : List at least three Feats, committed by the 0th god (any will do).

Challenge of Logic:

The order / feat count of the gods is as such:

1: Leevah - 2

2: Disdain - 4

3: Icyayou - 1

4: hugo - 0

5: pyros - 0

6: liar - 0

7: stummock - 2

8: solar - 0

9: Deceitralic - 5

10: Taehwasi - 4

11: Panoptic - 3

Logic:

The first thing we can do is use clues 2 and 3 to determine that of the lower gods, 4 must be 0 and one must be 2. This is because the sum of all of the gods must be two, but because there can't be multiples of an odd number, we can't have two gods with one feat each.
Thus Lower = 2,0,0,0,0


Next, we can use the fact that 5 feats is the unparalleled maximum and the first clue to determine the feat counts for the other pantheons as well. Initially, the first clue tells us that there must be >10 feats in the lowest pantheon, with a max of 13 (5,4,4). Thus, the only two options are 12 and 13, as the units digit must be larger than the tens, or flipping it would return a smaller number, which is not possible. However, if it is 13, then the total must be 31, which would require a sum of 16 in the low pantheon, as the sum of the lower pantheon is only 2. This is not possible, as all other gods must have <5 feats. Thus, the low pantheon has a sum of 7 feats, while the lowest has a sum of 12. The lowest pantheon must also thus be 5,4,3.


Next, we can use clues 7, 8, and 12 to create a string of gods in order. As the names of Pyros' neighbours are shorter than theirs, they must be liar and hugo. Similarly, the only two gods which work for Stummock's neighbours are liar and solar. Thus, we know that the gods solar, stummock, liar, pyros, and hugo are in a line in that order. Using clue 8, we can realise that solar must be at the bottom, thus telling us that those 5 gods are in increasing order from solar, and that solar is in the lower gods, as no gods in the lowest pantheon can have 0, as the other god would then have to have 7, which isn't possible.


From the fact that Pyros committed as many feats as his neighbors, we know that Pyros, Liar, and Hugo each have 0 feats, as well as Solar from clue 8. Thus, We can conclude that these 5 gods in a row must make up the lower pantheon, and Stummock is the one with 2 feats. Thus lower = [solar=0, stummock=2, liar=0, pyros=0, hugo=0]


Then, from clue 11, we know that leevah must be in the low pantheon, as they committed 2 feats, which no god in the lowest pantheon did. Thus, the last 2 gods in the low pantheon must have committed 4 and 1 feats, as a god with 3 feats is already present in the lowest pantheon.


Also, clue 10 tells us that Icyayou's higher neighbor and Taewasi both comitted 4 feats, as it is the only number of feats which there are 2 of left on the board.


Now that we know that liar is in the middle of all the gods, and that it is equidistant from Disdain and Taehwasi, we can tell that disdain must be in the low pantheon, and Taehwasi in the middle of the lowest pantheon, as Taehwasi's neighbor committed more feats than they did, which is only possible if they are below Deceitralic in the lowest pantheon, which also tells us that Deceitralic is at the top of the lowest pantheon.


From the previous conclusion, we can tell that Icyayou must be in the low pantheon, as the only other location would be below Taehwasi, which would violate rule 9. Thus, we can tell that Icyayou must be at the bottom of the low pantheon, with 1 feat, and Disdain must have 4 feats.


Finally, the only god not mentioned is Panoptic, who must slot into the last position below Taehwasi, with 3 feats.

Enigmatic Challenge:

The name of the final and strongest god:

Chuck Norris

A few of their achievements include:

Throwing a grenade and killing 50 people, before it exploded.

Making onions cry.

Counting to infinity, twice.
list

Logic:

If we look at the characters in each god's name that corresponds to their feat count (starting at 0 of course), we can spell out: EACHPLUSTWO.


Thus, if we repeat the process except for the feat count +2, we get ANAGRAMLAST.


This then tells us to find an anagram of the last letter of the god's names: HNUOSRKRCIC -> Chuck Norris