day 8 ivy part 2 cleanup
still don't love this, but i can't think of a better approachmain
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@ -2,6 +2,9 @@
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sample = 5 5 rho 3 0 3 7 3 2 5 5 1 2 6 5 3 3 2 3 3 5 4 9 3 5 3 9 0
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########
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"part 1"
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op visible r =
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r = -1, r
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r = transp max\ transp r
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@ -16,63 +19,22 @@ sample
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+/, vis4 input
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op pady a = -1, a, -1
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op padx a = transp pady transp a
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op pad a = pady padx a
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A = pad sample
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i = 3 3
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d = 1 0
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op i slice d =
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t = A[i[1];i[2]]
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t < 0: 0 rho 0
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t, (i+d) slice d
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A
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A[3;3]
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i slice d
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i slice -d
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##########
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"\npart 2"
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op empty v = 0 == rho v
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# takes from v while v[i] is less than x
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# count how many elements at the start of v are less than x
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op x shorter v =
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empty v: v
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(and\ x > v) sel v
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empty v: 0
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+/ (and\ x > v)
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# returns 1 if any v[i] blocks x
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op x blocked v =
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empty v: 0
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or/ x <= v
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v = 1 2 3 4 3 2 5 6 1 2 3
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5 shorter v
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0 shorter 1 2 3
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# takes from v while v[i] is less than v[0]
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op shorter2 v =
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x = v[1]
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v = 1 flip v
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v, +/ (and\ transp x > v)
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#shorter2 transp (1, rho v) rho v
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#shorter2 3 3 rho flip iota 9
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""
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sample
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"-"
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op x viewDistance v = (x blocked v) + rho x shorter v
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op viewDistance v = v[1] viewDistance 1 drop v
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viewDistance 1 2 3
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viewDistance 5 1 2 3
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viewDistance 5 5
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op x viewDistance v = (x blocked v) + x shorter v
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op p score A =
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x = p[1]
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@ -88,21 +50,17 @@ op p score A =
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n
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op i loopY A =
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i[2] > (rho A)[2]: 0
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(i score A) max (i + 0 1) loopY A
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x = i score A
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i[2] >= (rho A)[2]: x
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x , (i + 0 1) loopY A
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op i loopX A =
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i[1] > (rho A)[1]: 0
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(i loopY A) max (i + 1 0) loopX A
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x = i loopY A
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i[1] >= (rho A)[1]: x
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x , (i + 1 0) loopX A
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op solve A = 1 1 loopX A
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op solve A = max/ 1 1 loopX A
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1 1 score sample
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2 2 score sample
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3 4 score sample
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sample
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solve sample
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1 1 score sample; 1 2 score sample; 1 3 score sample; 1 4 score sample; 1 5 score sample
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2 1 score sample; 2 2 score sample; 2 3 score sample; 2 4 score sample; 2 5 score sample
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solve input
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