by Dan Katz
Problem: Pi Day Town/​Bloomsday Town

Each of the 10×10 grids represents a set of results from the G1 Climax, a round-robin tournament promoted by New Japan Pro Wrestling every summer. This is suggested by the title, which can be parsed as “G-One Guys.”

For the last four years, the tournament has featured twenty wrestlers divided into two blocks, with every wrestler facing off against every other wrestler in his block, with the A and B Block winners meeting in the finals. In most matches, the winner receives 2 points and the loser 0, although in the case of a draw, both competitors receive 1 point.

For each of the eight blocks listed, each of the ten wrestlers is identified by a letter, and the grid gives partial information about which letters defeated which other letters. For example, the colored square near the upper-right corner of the first grid indicates that Wrestler (a) defeated Wrestler (i).

This data can be used, along with the tournament results to determine which letter represents which wrestler in each block, as listed below (logical explanations for how these identifications can be made are included at the end of the solution):

G1 Climax 2015


G1 Climax 2016


G1 Climax 2017


G1 Climax 2018


The equations at the bottom of the page suggest that each of the given letters for a particular tournament correspond to unknown letters in another tournament, and indeed, the wrestlers identified by these letters were also in the second tournament. Once the grids are solved, you can identify which letter represents the same wrestler in the second tournament:

Given letter (year)WrestlerRequired letter (year)
c (2015)Naitoa (2017)
f (2016)Sanadak (2017)
c (2017)Yoshi-Hashia (2018)
h (2018)Falej (2015)
h (2015)Makabea (2016)
h (2016)Ishiim (2018)
t (2017)Evile (2018)
r (2018)Tongas (2017)
s (2015)Gotok (2018)
k (2016)Nagatai (2017)
q (2017)Kojimam (2015)
p (2018)Sabreb (2017)
q (2015)Elginl (2016)
n (2016)Yanoe (2015)

The new letters spell AKA JAMES KIMBLE. There are many not-particularly-famous people named James Kimble, but the not-particularly-famous one who was a pro wrestler used the ring name KING COBRA, the answer to this puzzle.

Logic Explanations

2015 (A Block)
The wrestlers f and g both beat at least six opponents, which means they must be Tanahashi and Styles in some order. Since f beat g, f is Tanahashi and g is Styles. Since Naito and Ibushi beat Styles, they must be c and i in some order, and since Naito beat Tanahashi and Tanahashi beat i, Naito is c and Ibushi is i. Since Naito beat e and j, they must be Yano and Fale in some order. Of the remaining four (which are a, b, d, and h), only Shibata beat Ibushi, so Shibata is a. The only assignment of b, d, and h where h beat b and b beat d is for b to be Tenzan, d Gallows, and h Makabe. Since Gallows did not beat Fale, Fale is j and Yano is e.

2015 (B Block)
Honma was the only wrestler to only win one match, and every wrestler except r beat at least two opponents, so Honma is r, and k is the only wrestler Honma beat, Ishii. Since l, n, and s also beat Ishii, they are Goto, Nakamura, and Okada in some order. Anderson is the only unassigned wrestler that beat two of these three, and o beat l and s, so Anderson is o, and Goto and Nakamura are l and s, meaning Okada is n. Since l beat s, l is Nakamura and s is Goto. Since q beat Nakamura, q is Elgin. Kojima, Nagata, and Takahashi remain as m, p, and t in some order. Since m lost to both of the others, Kojima is m, and since p beat t, Takahashi is p and Nagata is t.

2016 (A Block)
Since g and i drew, they are Okada and Tanahashi. One of these two was beat by a, f, and j, and the other was beat by b, d, and h. This means that since Goto and Tenzan both lost to both Okada and Tanahashi, they are c and e. Since f beat both c and e, f must be Sanada, the only one besides Okada and Tanahashi that beat both Goto and Tenzan. This means that since Sanada beat Tanahashi and not Okada, Tanahashi is g and Okada is i. Since Makabe and Tonga beat Tanahashi, they are a and j, and Fale, Ishii, and Marufuji are b, d, and h. Since d and h both beat b, b is Marufuji. Since j beat h, and Makabe lost to Fale and Ishii, Makabe must be a and Tonga j. The only match in which Fale or Ishii beat Goto or Tenzan was Fale beating Tenzan, so since d beat e, d is Fale (and h Ishii) and e Tenzan (and c Goto).

2016 (B Block)
Since t beat six opponents, t is either Omega or Naito, and in either case k, r, and s are the three opponents that beat t. If t is Omega, since s beat r and r beat k, Elgin, Shibata, and Yoshi-Hashi are s, r, and k respectively. But o lost to k, s, and t, and in this scenario there is no unassigned wrestler that lost to Yoshi-Hashi, Elgin, and Omega. So t must be Naito, and (again since s beat r and r beat k) s is Shibata, r Omega, and k Nagata. Then o, having lost to Nagata, Shibata, and Naito, must be Yoshi-Hashi. Since k beat p, and Evil is the only unassigned wrestler to lose to Nagata, Evil is p. Among the four remaining wrestlers, we know that m beat n, n beat l, and l beat q. Nakajima beat all of the other three of the four, so Nakajima is m; similarly Honma lost to the other three, so he must be q. Finally, since Yano beat Elgin, Yano is n and Elgin is l.

2017 (A Block)
Wrestler c lost six matches, so he is Nagata or Yoshi-Hashi. Since c beat i, if c is Nagata, i must be Sabre. But i lost five matches and Sabre didn't, so c must be Yoshi-Hashi. That means i is either Fale or Nagata, and the only one of these who lost five matches is Nagata, so i is Nagata. Since i beat b, and Nagata only beat Sabre, Sabre is b. Since Yoshi-Hashi beat Fale, Fale is a or g. But a beat Sabre and Fale didn't, so Fale is g. Fale lost to h, and the only unassigned wrestler that beat Fale is Tanahashi, so Tanahashi is h. Tanahashi lost to a and j, so a and j are Ibushi and Naito in some order. If f is Goto, since d beat e, d is Ishii and e Makabe. Since Goto beat a, a is Naito, but Naito did not beat d (Ishii), a contradiction. Similarly, if f is Makabe, d and e are Goto and Ishii respectively. Since Makabe beat a, a is Ibushi, but Ibushi did not beat Goto, again creating a contradiction. Therefore, the only remaining option is that f is Ishii, and that d and e are Makabe and Goto respectively. This forces a and j to be Naito and Ibushi respectively, and this time all clues are satisfied.

2017 (B Block)
Suzuki and Okada had the only draw, so they are n and r in some order. Wrestler n lost an additional three matches, which Okada did not, so Okada is r and Suzuki n. Okada lost to l and t, so these two are Evil and Omega in some order. Wrestlers k, m, and p each beat either Evil or Omega, so they must be Elgin, Robinson, and Sanada in some order, leaving Kojima, Tonga, and Yano to be o, q, and s. Wrestler o beat Suzuki, so Yano is o, and wrestler p did as well, so p is Elgin. Wrestler m beat q, but Sanada beat neither Kojima nor Tonga, so m is Robinson and k is Sanada; since Robinson didn't beat Tonga, Tonga is s and Kojima q. Since Evil didn't lose to Robinson and l did, l is Omega and t is Evil.

2018 (A Block)
Tanahashi and Okada had the only draw, so they are c and j in some order. If Tanahashi is c, h (who beat c) would have to be White, which would mean that a, e, and i (who beat h) would have to be Evil, Fale, and Suzuki. But b beat j, and the two wrestlers that beat Okada (Fale and White) can't be b, creating a contradiction. Therefore, Okada is c and Tanahashi j. Since b beat j, b is White, and since h beat c and White is already assigned, h is Fale. Since e and i beat White and Fale is h, e and i are Evil and Suzuki in some order. This leaves a, d, f, and g to be Elgin, Makabe, Page, and Yoshi-Hashi in some order. Wrestler i lost to d and f, but Evil only lost to one of the aforementioned four (Elgin), so i is Suzuki and e is Evil. Since g beat Evil, g is Elgin. Wrestlers d and f beat Suzuki and Yoshi-Hashi didn't, so Yoshi-Hashi is a. Wrestler f beat Elgin and Page did not, so Page is d and Makabe is f.

2018 (B Block)
Wrestlers m and s each lost at least four matches and won at least four matches. This means they must be Ishii and Sanada, and since m beat s and Ishii beat Sanada, Ishii is m and Sanada is s. Sanada only beat four opponents, so the four he beat, Ibushi, Sabre, Tonga, and Yano, are n, o, p, and r in some order. Similarly, Ishii only lost four times, so Ibushi, Naito, Sabre, and Tonga must be l, o, p, and r in some order. Comparing the two sets, Yano is n and Naito is l. Yano beat q, so q is Omega, since Ibushi and Tonga are both among o, p, and r. Since p beat Yano, p is Sabre. Since Naito beat r and did not beat Ibushi, Ibushi is o and Tonga is r. Since t beat k, t is Robinson and k is Goto.