Lessons · Showcase

Caprice No. 5

Paganini: the sweep and the moto perpetuo

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

189

190

191

192

193

194

195

196

197

198

199

200

201

202

203

204

205

206

207

208

209

210

211

212

213

214

215

216

217

218

219

220

221

222

223

224

225

226

227

228

229

230

231

232

233

234

235

236

237

238

239

240

241

242

243

244

245

246

247

248

249

250

251

252

253

254

255

256

257

258

259

260

261

262

263

264

265

266

267

268

269

270

271

272

273

274

275

276

277

278

279

280

281

282

283

284

285

286

287

288

289

290

291

292

293

294

295

296

297

298

299

300

301

302

303

304

305

306

307

308

309

310

311

312

313

314

315

316

317

318

319

320

321

322

323

324

325

326

327

328

329

330

331

332

333

334

335

336

337

338

339

340

341

342

343

344

345

346

347

348

349

350

351

352

353

354

355

356

357

358

359

360

361

362

363

364

365

366

367

368

369

370

371

372

373

374

375

376

377

378

379

380

381

382

383

384

385

386

387

388

389

390

391

392

393

394

395

396

397

398

399

400

401

402

403

404

405

406

407

408

409

410

411

412

413

414

415

416

417

418

419

420

421

422

423

424

425

426

427

428

429

430

431

432

433

434

435

436

437

438

439

440

441

442

443

444

445

446

447

448

449

450

451

452

453

454

455

456

457

458

459

460

461

462

463

464

465

466

467

468

469

470

471

472

473

474

475

476

477

478

479

480

481

482

483

484

485

486

487

488

489

490

491

492

493

494

495

496

497

498

499

500

501

502

503

504

505

506

507

508

509

510

511

512

513

514

515

516

517

518

519

520

521

522

523

524

525

526

527

528

529

530

531

532

533

534

535

536

537

538

539

540

541

542

543

544

545

546

547

548

549

550

551

552

553

554

555

556

557

558

559

560

561

562

563

564

565

566

567

568

569

570

571

572

573

574

575

576

577

578

579

580

581

582

583

584

585

586

587

588

589

590

591

592

593

594

595

596

597

598

599

600

601

602

603

604

605

606

607

608

609

610

611

612

613

614

615

616

617

618

619

620

621

622

623

624

625

626

627

628

629

630

631

632

633

634

635

636

637

638

639

640

641

642

643

644

645

646

647

648

649

650

651

652

653

654

655

656

657

658

659

660

661

662

663

664

665

666

667

668

669

670

671

672

673

674

675

676

677

678

679

680

681

682

683

684

685

686

687

688

689

690

691

692

693

694

695

696

697

698

699

700

701

702

703

704

705

706

707

708

709

710

711

712

713

714

715

716

717

718

719

720

721

722

723

724

725

726

727

728

729

730

731

732

733

734

735

736

737

738

739

740

741

742

743

744

745

746

747

748

749

750

751

752

753

754

755

756

757

758

759

760

761

762

763

764

765

766

767

768

769

% Paganini's fifth begins alone: a cadenza on the open A string, before
% the moto perpetuo it is famous for has sounded a note. Four SWEEPS up
% the a-minor triad, each reaching one rung higher than the last, and
% each answered by the scale falling the whole way back to the open A.
%
% The sweep is the triad itself, three degrees to the octave (the
% third, the fifth, the tonic above them), climbing until it runs out of
% violin. It is written once in DEGREES, eleven rungs long, and the four
% sweeps are its first eight, its first nine, its first ten, and the
% whole of it. Degrees and not intervals, because the frame comes back
% in A MAJOR at m.68 and 'i3 i4 i5' is a minor triad wherever it stands.
 
bpm 105
instrument plain
meter 4
unit 1/4
key a4 harm
mode diatonic
 
Sweep ::= .3 .5 1 3 5 '1 '3 '5 ''1 ''3 ''5
 
 
% Every answer is the same FALL: a harmonic minor coming down, begun
% higher each time. So it is written once at its longest, from the g
% above the top a to the b under middle C, and the three shorter ones
% are its tail: Fall[3:], Fall[5:], Fall[8:].
%
% One thing the ear catches and the spelling has to keep. A fall that
% starts from an A starts on a NATURAL g, not on the leading tone; by
% the octave below, the g# is back. That is why Fall's own head is '7b
% and why it carries '7 and .7 further down, and why the shortest
% descent, which starts from an a of its own, gets a '7b in front of it.
 
Fall ::= ''7b/2 ''6/2 ''5/2 ''4/2 ''3/2 ''2/2 ''1/2 '7/2
  '6/2 '5/2 '4/2 '3/2 '2/2 '1/2 7/2
  6/2 5/2 4/2 3/2 2/2 1/2
  .7/2 .6/2 .5/2 .4/2 .3/2 .2/2
 
 
% The cadenza entire, mm.1-10. The open A, then sweep and fall four
% times over, each pair higher than the last: the four peaks are a6,
% c7, e7, a7, which is the triad itself climbing while the figures
% climb inside it. Then a CHROMATIC run (no degrees, just semitones)
% up nearly two octaves and back down the same notes, which is how
% Paganini leaves the cadenza and starts the piece.
%
% The run is written in PLAIN notes and halved where it sounds, because
% the cadenza takes it at twice the speed the moto perpetuo will: the
% cadenza says 'Chrom/2', and m.26, which is the same run coming down,
% says 'Detach(Chrom~)' and gets the moto's own bowing for nothing. The
% top a is written once, between the two, because that is what it is:
% the turn, played and left.
 
Chrom ::= a#3 b3 c4 c#4 d4 d#4 e4 f4 f#4 g4 g#4 a4 a#4 b4 c5 c#5 d5 d#5 e5 f5 f#5 g5 g#5
 
Cadenza ::= - _4 .1 .1 _12
  Sweep[:8]/2  ''1 _6   '7b/2 Fall[8:]  .1 _6
  Sweep[:9]/2  ''3 _6   Fall[5:]        .1 _6
  Sweep[:10]/2 ''5 _6   Fall[3:]        .1 _6
  Sweep/2      '''1 _6  Fall            .1 _3
  Chrom/2 a5/2 Chrom~/2
 
Cadenza
 
 
% mm.10-13: the MOTO PERPETUO starts. Two chords to plant the key, and
% then sixteen notes to the bar, detached, for the next sixty measures.
%
% Detached is the whole of the bowing and it is one rule. 'Detach' is a
% CELL: $1 is a hole, the line is poured into it a note at a time, and
% each note takes half its slot with a rest taking the other half. So
% the figure itself is only the sixteen pitches it walks.
 
Detach   ::= $1 /2 - /2
MotoLine ::= .1 .3 .5 .3 .2 .4 .5 .4 .3 1 .7b .6 .5 .4 .3 .2
Moto     ::= Detach(MotoLine)
 
% It starts on the second half of the bar and runs across the barline,
% so any bar of the score is the back half of one statement and the
% front half of the next. Twice as written, and then it TURNS: the same
% shape with a c# in it, pointing at the major.
 
TurnA ::= .1 .3# .5 .3# .1 .4 .6 .4 ..7b .2
 
.1 _6 {.3, .5} {3, '1} _7/2 - _9/2
  Moto Moto Detach(TurnA)
 
 
% mm.14-17: here the performance changes gear, from 105 to 167,
% ten notes into the turn. A tempo mark can only stand between
% statements, so the turn is written in the two pieces the mark cuts it
% into: TurnA before, TurnB after.
%
% Then the line LIFTS: a climb of two octaves on the dominant seventh,
% e g# b e g# b d e, and the moto comes back an octave up, twice,
% exactly as it was. The last ten notes are already the next figure:
% the cadence onto d that m.18 takes up.
 
TurnB    ::= .6 .2 ..7b .3 .5 .3
LiftLine ::= ..7 .2 .6 .4 .3 .1 1 .4# .5 .7 2 5 7 '2 '4 '5
Lift     ::= Detach(LiftLine)
CadA     ::= 1 3# 5 3# '1 7b 6 5 3# 4
 
bpm 166.75
Detach(TurnB) Lift Moto@12 Moto@12 Detach(CadA)
 
 
% mm.18-24: form D, the moto's first long stretch; a minor with its
% NATURAL seventh; the line comes down what the Lift climbed, and two
% sequences carry it from there.
%
% The key mark changes, and it is a real change: 'a4 min', not the
% cadenza's 'a4 harm'. Every descending figure from here on uses g, not
% g#, and both sequences below are exact under min and wrong under harm.
% The harmonic minor was the cadenza's melodic decision, not the key the
% piece is in.
 
key a4 min
 
assert .7 == g4
 
% Both figures are poured through Detach in their own definitions, so a
% statement is the figure and nothing else. And the stretch is ONE
% statement: sixteen detached notes to the bar for seven bars, and not
% one figure in it begins on a barline, so the bars go unwritten and the
% figures are laid out a line apiece instead.
%
% Spill. Five notes, a step at a time, down from the b above the staff;
% stated twice, at m.18 and at m.22, both times just after the line has
% run out of room above.
% Rung. Four notes, stated five times: a step down, then up a third and
% up a fourth to the note the line is really on. Each statement is one
% scale step lower than the last, so the notes it arrives on, c5 b4 a4
% g4 f4, spell a minor scale coming down. A ladder, one rung at a time.
% Drop. Eight notes, stated four times: a leap up a fourth, then the
% scale falls a sixth, one scale step lower each time.
 
Spill ::= Detach('2 '1 7 6 5)
Rung  ::= Detach(.6 .5 .7 3)
Drop  ::= Detach(6 '2 '1 7 6 5 4 2)
 
assert Rung@~-4 == Detach(.2 .1 .3 .6)
 
% Each fall of the Drop belongs to the key it is falling through: the
% first runs down E harmonic minor (f#, d#), the second D harmonic minor
% (c#), the third is plain, and no one mark says that. So Drop stays
% the plain diatonic figure and the accidental sits with the statement
% that wants it. '\n' replaces one ITEM; Drop is bowed, so it plays a
% rest after every note and its fifth note is item 8. The replacement is
% braced because it carries a half value of its own, and it stands
% BEFORE the mark, in Drop's own degrees, since the mark moves the
% replacement too.
 
assert Drop \8 {6#/2} \12 {4#/2} == Detach(6 '2 '1 7 6# 5 4# 2)
assert Drop \14 {3/2} @~-3 == Detach(3 6 5 4 3 2 1 .7)
 
% m.18: the spill, three more steps to the b below, then a turn under
% the open A and back.
Spill Detach(4 3 2) Detach(1 .7 .6# .7 1)
  % mm.18-20: the ladder, five rungs, down to the f.
  Rung Rung@~-1 Rung@~-2 Rung@~-3 Rung@~-4
  % mm.20-22, and it will not settle. The line drops under the open A
  % to g#, climbs back, and tries a rising figure twice before dropping
  % it, twice over. Two statements is a coincidence, not a rule, so
  % these two and a half measures are written out: a name here would
  % name something that is not there.
  Detach(.1 ..7# .3   .6 ..7# ..7 .2   .6 .2 .3 .5   .7 3 .4 .6)
  Detach(2 4 .5 .7   3 5 .6 1   3 6 .6# 4#   3 1 .7 7)
  Detach(5 3 .7 6   4 2 1 '3)
  % m.22: the c above the staff, the highest note of form D, and the
  % line spills again: five notes and out, onto the open A. Not the
  % highest the moto has reached: the Lift got to e6 at m.15, a major
  % third above this c, which is why form D can climb this far and
  % still sound like it is holding something back.
  Spill Detach(1)
  % mm.22-24: the Drop, four times. The fourth statement's override is
  % not an accidental: the sequence stops descending there and lands
  % back on the g the third landed on.
  Drop \8 {6#/2} \12 {4#/2}
  Drop \12 {4#/2} @~-1
  Drop @~-2
  Drop \14 {3/2} @~-3
  % and the last two notes are already the next figure: f#4 a4 begins
  % the diminished seventh that m.25 arpeggiates.
  Detach(.6# 1)
 
 
% mm.25-32: forms E and F, and the piece starts to travel. A
% diminished seventh laid out as an arpeggio, two triads falling from
% the top of the bow, the cadenza's chromatic run coming back, and
% then a two-bar figure that says itself three times, a fourth higher
% each time, which is what carries the moto out of C major altogether.
%
% The key is C major and it is the plain reading of mm.25-29: the
% diminished seventh on C resolves into it, both falling arpeggios are
% its own chords, and mm.28-29 rock between the tonic and its dominant
% seventh half a bar at a time.
 
key c4 maj
 
% Three figures, and each is spelled the way it is for a reason.
%
% Dim. The diminished seventh as an arpeggio: four minor thirds
% stacked, which is a chord no key owns, so it is written in plain
% notes. The LINE takes the name and the bowing stands outside it,
% because at m.62 the chord comes back FALLING and three semitones
% down. Reversing a bowed figure reverses its rests with it and puts
% every note half a unit late; 'Detach(DimLine~)@-3' turns the line
% round inside the bowing and nothing else moves.
% Cross. Eight notes over C and then its dominant seventh: up a
% fourth, back a step, up a sixth, and then the line falls and lifts.
% Plain notes again, because it returns at m.37 AT THE SAME PITCH under
% another key, and once more at m.65 three semitones down.
% Wheel. Two bars of broken chords, stated three times, each statement
% five semitones above the last. The sequences of m.18 moved a scale
% STEP at a time and were written @~-1; a fourth is not a scale step in
% any key, so this one is @5 and @10, and the harmony goes round the
% wheel of fourths: C at m.28, then F, B flat, E flat.
 
DimLine ::= c5 d#5 f#5 a5 c6
Dim     ::= Detach(DimLine)
Cross   ::= Detach(g3 c4 b3 g4 f4 d4 c4) Detach(e4)
Wheel   ::= Detach(e4 c4 bb3 g4 e4 bb3 a3 f4)
  Detach(c4 a3 g3 e4 bb3 g3 a3 f5)
  Detach(c5 a4 g4 e5 c#5 g4 f4 d5)
  Detach(a4 f4 e4 c#5 g4 e4 f4) Detach(d5)
 
% The Wheel is written in its four HALF-BARS, one call apiece, because a
% body of SEVERAL calls slices by the CALL: Wheel[3:] is the fourth
% half-bar, which is what the assert below asks for. That is not the only
% way to cut a bowed figure, and the difference is worth having straight,
% because a body of ONE call slices by the ITEM instead, which is what
% Moto[:18] leans on earlier in this lesson, nine notes being eighteen
% items once a rest follows each one. The same thirty-two notes written
% as a single call would give three quarters of itself as [:48]. The
% half-bars are chosen here for a musical reason, not a technical one:
% the Wheel begins in the middle of m.29, so the barline where form F
% starts falls inside it and the section has to break there too.
%
% Two of the figures are a call further divided, their seven notes and
% their eighth: where the Wheel and the Cross come back at mm.35-37 a
% chromatic run wants that very sixteenth, and a twenty-three-rung run
% has the better claim on its own head. The asserts below say the
% dividing changed nothing.
 
assert Wheel[3:] == Detach(a4 f4 e4 c#5 g4 e4 f4 d5)
 
% And the chromatic run stops at g#5 (the a above it is written on its
% own in the cadenza), which is why m.26 asks for four semitones and not
% three to start on the c above the staff.
 
assert Chrom[22:] == g#5
assert Cross \0 {a4/2} \2 {g4/2} == Detach(a4 g4 b3 g4 f4 d4 c4 e4)
 
% m.25: the diminished seventh climbing, one more stab at its tritone,
% and then two triads falling from the top of the bow: C major, 1 5 3 1
% 5 3 1 5, and G major in the same shape from its own third.
Dim Detach(f#5)
  Detach(''1 '5 '3 '1 5 3 1 .5)
  Detach('7 '5 '2 7 5 2 .7 .5) Detach(.6)
  % mm.26-27: the cadenza's chromatic run, note for note, coming down
  % from the c above the staff to the d below middle C in the moto's own
  % bowing. Twenty-three semitones for one name, and the name was
  % already written.
  Detach(Chrom~)@4
  Detach(1 3)
  % mm.28-29: the Cross, three times. The middle one is the figure
  % bare; the first says a4 g4 where it says g3 c4, and the third ends a
  % third higher. Three exceptions in twenty-four notes, each riding the
  % statement that wants it.
  Cross \0 {a4/2} \2 {g4/2}
  Cross
  Cross \14 {g4/2}
  % mm.29-30. The wheel starts on the second half of m.29: three of its
  % four half-bars, and then the section ends.
  Wheel[:3]
 
% mm.31-32: form F. The wheel's last half-bar, and then the whole turn
% again a fourth up. Nothing else: two bars for one name and one mark.
Wheel[3:] Wheel[:3]@5
 
 
% mm.33-40: form F prime and then form G, and the two statements below
% are those two sections. Nothing in these eight measures is new: the
% wheel finishes its climb by fourths, the cadenza's chromatic run comes
% down twice, and the bariolage of m.28 comes back at the pitch it had
% there. Only the last two measures are written out, and they are
% written out because there is nothing in them to name.
%
% Two figures give up a note at their seam, and both give it to a
% chromatic run. The wheel's last half-bar ends on the very note the
% first run starts on, and the Cross's last place is taken by the head
% of the second. The run has the better claim both times (it is
% twenty-three rungs long and the note is its top), so each figure was
% written as its seven notes and its eighth, one call apiece. A slice
% over a body of calls counts the CALLS, and that is the only way to
% divide a bowed figure.
 
assert Wheel[3:] == Wheel[3:4] Detach(d5)
assert Cross == Cross[:1] Detach(e4)
assert Chrom[22:]@4 == c6
 
% mm.33-36. Form F prime: the wheel's last two rounds. The B flat turn
% finishes in the first half of m.33, and the E flat turn, the third
% and last, runs from there to m.35. One note of it is different: on
% the last half-bar the line steps down to the c instead of repeating
% the e flat, which is how it arrives at the c above the staff. The
% override stands BEFORE the mark and in the wheel's own pitches,
% because the mark moves the replacement too.
%
% From there the run of m.26 comes again note for note: all twenty-three
% semitones, from that same c down to the same d below middle C, in the
% moto's own bowing, and the same c after it. Only the note after THAT
% differs, and only by an accidental: an e flat where m.27 had an e
% natural, which is the C minor the next run falls out of.
Wheel[3:]@5 Wheel[:3]@10
  Wheel[3:4] \12 {d4/2} @10
  Detach(Chrom~)@4
  Detach(1 3b)
 
% mm.37-40. Form G. The Cross again, at the pitch it had at m.28: its
% first two notes replaced by an a flat and the g below it (the low g
% and the c of m.28 lifted into the figure's own upper register), and
% its other five untouched. Then the run, and this one wants no
% transposition at all: g# above the staff down to the b flat below
% middle C is the cadenza's descent at the cadenza's own pitch. It ends
% as m.27's did, a whole tone under its last rung.
%
% Then the storm gathers. From the last sixteenth of m.38, two measures
% of broken chords in fours: every four falling, and the fours joined
% in pairs by a leap up an OCTAVE. Two of those pairs are the same eight
% notes a whole tone apart, the pair from m39+11 and the pair from
% m40+3, and the pair before them is close but not the same. Two
% statements is a coincidence, not a rule, so these measures are written
% out and named nothing; and they are written in plain notes because ten
% of the twelve sound in them and no key owns that.
Cross[:1] \0 {g#4/2} \2 {g4/2}
  Detach(Chrom~) Detach(g#3)
  Detach(f4 d#4 c4 g3)
  Detach(d#4 c#4 a#3 g#3) Detach(g#4 d#4 c4 b3)
  Detach(f#4 d#4 c4 a#3)  Detach(a#4 f4 c#4 b3)
  Detach(g#4 f4 d4 c4)    Detach(c5 g4 d#4 c#4)
  Detach(a#4 g4 c#4 a#3 g4)
 
 
% mm.41-47: form H, the central storm; D FLAT from m.43 and it holds to
% the barline; two bars that belong to no key, and then one figure said
% twice.
%
% The moto has not stopped since m.14 and does not stop here. What comes
% loose is the harmony. D flat is four fourths above the F the Wheel
% turned to in m.31, the same climb the Wheel was making, and it holds:
% not one of m.47's sixteen notes is outside the key, and twenty-nine of
% the Churn's thirty-two are in it. The mark is 'c#5' rather than 'c#4'
% for the reason the cadenza's is 'a4': the storm plays BELOW its tonic,
% so the degrees that say it are the lower-octave ones.
 
key c#5 maj
 
assert .1 == c#4
 
% Churn. The storm's unit, two whole bars, stated twice: at m.43 in D
% flat and at m.45 a whole step up, on E flat. The second statement is a
% transposition, so it needs no key mark of its own; the mark carries the
% spelling. F sharp would spell the figure with one accidental fewer and
% is still the wrong reading: only D flat puts the figure's own tonic on
% the tonic, and only then does its head read as what it is.
%
% Because the head IS THE MOTO'S OWN. MotoLine's first nine degrees, read
% in this key, are the Churn's first nine notes, the figure the piece
% has played since m.10, here for the first time in a major key. Those
% nine come three times in the caprice and no more: twice in the Churn,
% and once at m.50 in E major.
%
% Then the scale falls from the tonic; then the head's first four notes
% again; then the bass steps from the tonic to the flat second while the
% upper pair rocks between the fifth and the flat seventh, which is where
% all three of the figure's accidentals are; then the head's first four
% once more, one scale step up.
 
assert MotoLine[:9] == .1 .3 .5 .3 .2 .4 .5 .4 .3
assert MotoLine[:4]@~1 == .2 .4 .6 .4
 
Churn ::= Detach(.1 .3 .5 .3 .2 .4 .5 .4 .3)
  Detach(1 .7 .6 .5 .4 .3 .2)
  Detach(.1 .3 .5 .3)
  Detach(.1 .5 .7b .5 .2b .5 .7b .5)
  Detach(.2 .4 .6 .4)
 
% m.41: six notes climbing to the b flat above the stave, then ten
% falling to the d flat above middle C. Nothing in m.41 or m.42 recurs
% anywhere else in the caprice at more than five notes, and every
% five-note match is a run of the scale, so the two bars are written
% where they stand, in plain notes: no key owns them, A flat needing six
% accidentals over these thirty-eight notes and D flat nine. What they
% are is D flat shadowed a semitone below: e under e sharp, g under g
% sharp, a under a sharp, d under d sharp, and nothing else.
Detach(e4 c5 g4 e5 c5 a#5 g#5 f5 c5 g#4 e4 f4 g#4 f4 d#4 c#4)
  % m.42, and back up, from middle C to the g sharp above the stave. A
  % seven-note figure (a third up, then the scale falling a fifth from
  % where it started) runs from the twelfth note of m.41 across the
  % barline, and is said once more from the fourth note here, three steps
  % up the A flat scale. Then it is dropped. Two statements is a
  % coincidence, not a rule, and a name for it would name something that
  % is not there.
  Detach(c4 a#3 a4 a#4 c#5 a#4 g#4 g4 f4 d#4 d5 d#5 g5 d#5 g#5 d#5)
  % m.43: six notes more, and the storm finds its figure.
  Detach(c5 g#4 g4 g#4 f#4 d#4)
  % mm.43-45: the Churn, from the last quarter of m.43.
  Churn
  % mm.45-47, and again, a whole step up, on E flat.
  Churn@2
  % m.47: the bar's last ten notes, driving at the barline, D flat
  % throughout.
  Detach(.4 2 .7 .4 .3 3 1 .6 .5 3)
 
 
% m.48: form I, one bar and a hinge; B MAJOR, a whole step down from the
% storm's D flat; the dominant seventh falling, and the foot of a sweep.
%
% Fifteen of the sixteen notes are in the new key. The odd one is the d
% natural the bar opens on, the last note of the old key, and the only
% accidental the bar needs. It falls from the f sharp above the stave to
% the a sharp below middle C; its third beat spells the dominant seventh
% exactly, c#5 a#4 f#4 e4; and the last two notes, b3 and d#4, are the
% tonic and its third, the foot of a sweep up a B MAJOR triad, which
% m.49 climbs.
 
key b4 maj
 
assert .1 == b3
assert 3b == d5
assert .1 .3 == b3 d#4
 
Detach(3b .6 .5 5 3 .7 .5 4 2 .7 .5 .4 .2 ..7 .1 .3)
 
 
% mm.49-55: forms J and K, and the caprice turns for home. It has
% travelled as far as B major, and it does the plainest thing there is
% with a new key: it sweeps it. Then it sweeps the relative minor, and
% four bars later the music is standing on E, the dominant of the A
% minor it left at m.25.
%
% Two figures carry the stretch up and one carries it back down: the
% cadenza's own arpeggio, bowed; a four-note reach that climbs the E
% major scale once the moto has turned the corner; and a broken chord
% that falls three bars in six statements.
 
key b4 maj
 
% Climb. The cadenza's Sweep, bowed: the tonic below, its third and its
% fifth, then the triad twice over. Eight rungs of the eleven, stated
% twice here and once more inside the hinge at m.56.
%
% It is the Sweep and the assert says so (the same ladder from the rung
% under its first), so this is one musical object under two names, which
% the standard does not like and the language leaves no way round: a
% cell fills its hole with whatever it is handed, whole, so 'Detach' can
% never be given 'Sweep[:7]'. The eight rungs have to be written
% somewhere, and a name is the cheapest place.
%
% Reach. A note, a reach up a fourth, back down to it, and a step
% down. Stated four times, on b, c#, d#, f#: two tones and then a minor
% third, which are four rungs of the E major scale m.50 turns to.
%
% Tumble. Eight notes, stated six times across mm.53-55. A broken
% chord falls, hops up, and the next chord falls from where the hop
% left it. Each of its two chords has a semitone rub in it, g against
% f# and then f against that f#, and that rub is the sound of the whole
% descent.
 
Climb  ::= Detach(.1 .3 .5 1 3 5 '1 '3)
Reach  ::= Detach(b4 e5 b4 a4)
Tumble ::= Detach(a#5 g5 d#5 f#5 b5 f#5 d5 f5)
 
assert .1 Sweep[:7] == .1 .3 .5 1 3 5 '1 '3
 
% Two scale steps down in B major and the major triad has become its
% own relative minor: the same seven notes read from a different rung.
% That is the whole modulation of m.49, in one mark.
 
assert Climb@~-2 == Detach(g#3 b3 d#4 g#4 b4 d#5 g#5 b5)
 
% And the moto's head comes back a fourth up, which is what puts the
% music in E. A slice counts ITEMS, and a bowed figure plays a rest
% after every note, so the moto's first nine notes are Moto[:18].
 
assert Moto[:18] == Detach(.1 .3 .5 .3 .2 .4 .5 .4 .3)
 
% mm.49-53. Form J. The sweep's first two rungs were m.48's, so six
% are left. Then the dominant seventh of g# minor falls twice, an
% octave apart (d# g a# c#, its third written as a plain g and not as
% f##, which is how a violinist reads it), and the second fall stops
% three notes in, where the sweep takes over. These eight notes are
% m.48's own, from the second half of its second beat, three semitones
% down, and that is the only other place in the caprice that says them;
% two statements is a coincidence, so they are written and not named.
% Then the sweep again in the relative minor, and nine notes of the moto
% on E.
Climb[4:]
  Detach(d#4 c#5 a#4 g4 d#4 c#4 a#3 g3)
  Climb@~-2
  Moto[:18]@5
  % mm.50-52: the reach, four times, climbing. It begins a sixteenth
  % before the beat, so the reach itself lands on the beat. Between
  % each pair stands the same shape with the reach CLOSED UP, not
  % opened: the Reach spans a fourth and these span a major third, a
  % minor third, a tritone and a minor third, so three of the four are
  % NARROWER than the figure they sit between. Holding the upper note
  % while the lower keeps climbing is what shrinks them: the climb
  % eats the interval. No two of the four are alike, so each is
  % written where it stands.
  Reach Detach(c5 e5 c5 a#4) Reach@2 Detach(d#5 f#5 d#5 c5)
  Reach@4 Detach(e5 a#5 e5 d#5) Reach@7 Detach(g#5 b5 g#5 e5)
  % m.53: the line turns over and falls, two statements to the bar.
  % One note of m.52 comes with them: the g the tumble is about to say
  % again.
  Detach(g5) Tumble Tumble@-2
 
% mm.54-55: form K, and the sequence runs straight across the section
% break: four more statements, each a whole tone below the last except
% the fifth, which is only a semitone. No scale says that, so the marks
% count semitones.
%
% The hop notes make a line of their own, one a beat, walking down by
% semitones: f# f e d# d c# c c b a# a a. Twice it stays where it is
% instead of going on, and those two stalls are the only exceptions in
% forty-eight notes. Each rides the statement that wants it: one
% override raises the last two notes a semitone and the other only the
% last, which takes the rub out of the second chord and lands the
% figure on a plain F major and then a plain D minor, the two chords
% that lead back to A.
Tumble@-4
  Tumble \12 {d#5/2} \14 {f#5/2} @-6
  Tumble@-7
  Tumble \14 {f#5/2} @-9
 
 
% m.56. Form L, one bar, and the whole bar is a hinge: the caprice
% turns here and the reprise is on the other side of it. The open A
% rings between d, f and then d# (D minor, and then the raised fourth
% that says E is coming), and the middle eight notes of the bar are
% Climb itself, on E MAJOR, the dominant the piece has been away from
% since m.25. Two octaves and a third in eight notes. The last two,
% f and a, are the D minor that m.57 starts from.
%
% Hinge. Fourteen notes, stated twice: here, and an octave up at m.57.
% It is written in plain notes and not in degrees, and that is not
% laziness. The reprise needs a minor back in force by m.57, and a
% figure spelled in degrees would be reread in THAT key. The key mark
% below is here so the claim under it can be made at all: in E, Climb's
% eight rungs ARE the middle of the bar.
 
bpm 164.75
key e5 maj
 
Hinge ::= Detach(d5 a4 f4 a4 d#5 a4 e4 g#4 b4 e5 g#5 b5 e6 g#6)
 
assert Hinge[12:] == Climb
 
Hinge Detach(f5 a5)
 
 
% mm.57-64: form M, the long REPRISE, and it is the piece gathering up
% what it has already said. The hinge swings an octave higher; the moto's
% own line comes back two octaves above where it started; the four falls
% that closed form D say themselves again a sixth higher; the diminished
% seventh returns upside down; the cadenza's first sweep comes down
% instead of up; and the cadenza's chromatic run closes the section.
% Eight measures and not one new figure in them.
%
% The key goes back to the HARMONIC minor, and it has to: the moto's line
% spells its seventh '.7b', which is a g under 'harm' and an f# under
% 'min', and this section states that line twice.
 
key a4 harm
 
assert 7 == g#5
 
% m.57 is m.56 transposed, note for note, so the hinge is stated, not
% rewritten. Hinge is the fourteen notes the figure owns; the last two of
% m.57 are already the moto's. And it is spelled in plain notes, which is
% what lets '@12' mean the same thing under any key mark.
 
assert Hinge@12 == Detach('4 '1 6 '1 '4# '1 5 7 '2 '5 '7 ''2 ''5 ''7)
 
% The moto says its line twice. The second time it BENDS: the last two
% notes come out c# for c and a for b, which are the leading tone and the
% dominant of D, and D is the note the next figure leaps to. '\n' counts
% from 0 and MotoLine is unbowed, so its fifteenth and sixteenth notes
% are items 14 and 15.
 
Veer ::= .1 .3 .5 .3 .2 .4 .5 .4 .3 1 .7b .6 .5 .4 .3# .1
 
assert Veer == MotoLine \14 {.3#} \15 {.1}
 
% The four falls are the Drop of mm.22-24, a sixth higher, and only two
% of them can say so. The first and the last fall through a NATURAL g;
% the third takes the harmonic minor's own g#, and the second touches no
% g at all. Under 'harm' a diatonic shift cannot land on a natural g:
% every '@~' into that degree gives g#, so those two are written out
% with '7b' and the middle two are the Drop itself. Under 'min' the trade
% runs the other way and the moto's line comes out wrong, which is why
% the mark above is 'harm'.
 
assert Drop@~3 == Detach('2 '5 '4 '3 '2 '1 7 5)
 
% And the diminished seventh of m.25 comes back falling, three semitones
% down. Its LINE is what carries the name, 'Dim ::= Detach(DimLine)',
% so the line can be turned round INSIDE the bowing. 'Dim~' would not
% do: reversing a bowed figure reverses its rests with it, and every
% note would land half a unit late.
 
assert Detach(DimLine~)@-3 == Detach(a5 f#5 d#5 c5 a4)
 
% The sweep that opened the caprice comes down too. Seven of the eight
% notes of m.62 are its own first seven rungs read backwards, and now
% that the sweep is spelled in degrees the claim can be made.
 
assert Sweep[:7]~ == '3 '1 5 3 1 .5 .3
 
% m.57: the hinge again, an octave up. Its second half is the E major
% arpeggio, and an octave up that climbs to g#7.
Hinge@12
  % mm.57-59: the moto two octaves up, twice, the second time bent.
  Moto@24 Detach(Veer)@24
  % mm.59-61. The four falls: the leap up a fourth and the scale
  % falling a sixth after it, one scale step lower each time.
  Detach('4 '7b '6 '5 '4 '3 '2 7b)
  Drop@~4
  Drop@~3
  Detach('1 '4 '3 '2 '1 7b 6 5)
  % mm.61-62: the diminished seventh again, falling this time and three
  % semitones down, with one more stab at its tritone under it.
  Detach(d#5 c6) Detach(DimLine~)@-3 Detach(d#4)
  % m.62, and the cadenza's first sweep backwards: the a minor triad
  % from the c above the staff all the way down to the open A. Seven of
  % those eight notes are the Sweep's own first seven rungs, reversed,
  % which the assert above the statement says in the language.
  Detach('3 '1 5 3 1 .5 .3 .1)
  % mm.62-63: then the diminished seventh on the leading tone, g# b d,
  % falling an octave at a time to the g# under middle C.
  Detach('2 7 4 2 .7 .4 .2 ..7)
  % mm.63-64: the open A, and the chromatic run of the cadenza coming
  % down from the a above the staff in the moto's own bowing: the whole
  % of Chrom reversed, one semitone up. The last two, the open A and the
  % c above it, carry into m.65, where they are the first two notes of
  % the bariolage the Cross gives it.
  Detach(.1) Detach(Chrom~)@1 Detach(.1 .3)
 
 
% mm.65-69, form N. The moto has been running since m.10 and here it
% runs out of road: one figure over and over around the open A, and then
% the cadenza walks back in over the top of it.
%
% The key has not moved: this is still A minor with its raised seventh,
% the key the reprise came back in, and the mark is here to say so.
 
key a4 harm
 
assert ..7 == g#3
 
% Two rising thirds a step apart, b d and c e, and then the same pair two
% scale steps lower, g# b and a c. Eight notes that rock between the
% dominant seventh and the tonic and get no further, which is the point
% of the passage: the moto has stopped going anywhere. That is why the
% figure is written as one four-note pair said twice and not as eight
% notes: the sequence IS the figure.
 
Thirds ::= .2 .4 .3 .5
Spin   ::= Detach(Thirds Thirds@~-2)
 
assert Thirds@~-2 == ..7 .2 .1 .3
assert Spin == Detach(b3 d4 c4 e4 g#3 b3 a3 c4)
 
% m.65: two beats to arrive, and they are the bariolage of m.37 three
% semitones down: the Cross's first seven notes, with the same two
% overrides that measure wanted and one mark to carry the lot. The
% replacements are written in the Cross's own pitches (an a flat and
% the g under it) because the mark moves them too, so what sounds is
% an f and an e. Only the eighth note is not the Cross's: where the
% figure would give a c sharp the piece plays the plain c of the key,
% and the Spin takes over from there.
%
% The Spin, three times whole. The fourth breaks off six notes in
% (the pair, and two notes of the pair below), and the open A it was
% going to land on is held instead: twelve units of it, straight through
% the barline into m.68.
 
Cross[:1] \0 {g#4/2} \2 {g4/2} @-3 Detach(.3)
  Spin*3
  Detach(Thirds ..7 .2) .1 _12
  % m.68, and the cadenza is back, at the moto's speed and in the
  % MAJOR. A major is A harmonic minor with its third and its sixth
  % raised and nothing else moved: the seventh is already g#, which is
  % why the answer's leading tone needs no mark at all. So the sweep is
  % the cadenza's own ladder with its thirds sharpened, and the answer is
  % the cadenza's own Fall with the third and the sixth of each octave
  % sharpened: six notes in nineteen, one pair an octave.
  Sweep[:8] \0 {.3#} \3 {3#} \6 {'3#} /2  ''1 _6
    '7/2 Fall[8:] \0 {'6#/2} \3 {'3#/2} \7 {6#/2} \10 {3#/2} \14 {.6#/2} \17 {.3#/2}
    .1 _6
  % m.69: the second sweep begins before the gear change, and a tempo
  % mark can only stand between statements, so it is written in the two
  % pieces the mark cuts it into, exactly as the turn was at m.14.
  Sweep[:8] \0 {.3#} \3 {3#} \6 {'3#} /2
 
 
% mm.70-73, form O. The tempo drops back to the cadenza's own and the
% key mark now says what the last two measures had already spelled out:
% A major. From here the cadenza's own names carry everything with no
% marks at all: the same eleven-rung ladder and the same fall, read in
% the new key.
%
% Three sweeps are left: nine rungs, ten, and all eleven, each answered
% by the fall begun one rung higher than the last, and each answer
% landing on the open A. The first of the three is already in the air:
% m.70 opens on its ninth rung and then its peak, the c# above the top A.
 
bpm 105.25
key a4 maj
 
assert Sweep[:3] == c#4 e4 a4
assert Fall[5:9] == b6/2 a6/2 g#6/2 f#6/2
 
''1/2 ''3 _6 Fall[5:] .1 _6
  Sweep[:10]/2 ''5 _6 Fall[3:] .1 _6
  % and the fourth sweep climbs all eleven rungs to the a two octaves
  % above the staff and holds it. The answer to that one begins m.74.
  Sweep/2 '''1 _6
 
 
% mm.74-77: form P, the last of them, and the frame shuts. The fourth
% fall, the cadenza's chromatic run, and the two chords. Not one new
% name in it: mm.68-77 is the opening cadenza again in the parallel
% major, and all the major asks of it is a single note.
%
% A fall from an A begins on the note under it, and that note is the
% one place the major disagrees with the minor. Fall's head is spelled
% ''7b because the cadenza's g is natural; here it has to be the
% major's g#. Nothing below the head moves (every degree from '7 down
% is natural in A major), so the flat in Fall's head is the only
% alteration the figure carries, and the only item this return has to
% override.
 
assert Fall[0:1] == g7/2
assert Fall \0 {''7/2} == g#7/2 Fall[1:]
 
% Then the chromatic run that ended the cadenza at m.9, note for note:
% up its twenty-three rungs from the a# under middle C, the top a
% written once between the two runs, and down the same notes. It is in
% plain pitches, so the new key does not touch it, and the gear change
% at m.70 left it at the speed it had then: 105.25 against 105.
%
% And the two chords the moto perpetuo planted the key with at m.10:
% the same two degrees, .3 with .5 and then 3 with '1. Only the key
% mark has changed, so the c that was natural there is a c# here. The
% caprice opens its moto on A minor and shuts it on A major, and the
% frame needed no rewriting to say so.
 
assert {.5, .3} {'1, 3} == {c#4, e4} {c#5, a5}
 
% m.74: the fall from a7, and the open A under it.
Fall \0 {''7/2} .1 _3
  % mm.75-76: the run up and back.
  Chrom/2 a5/2 Chrom~/2
  % m.76 has two stops, c# with e and then c# with the a above it: A
  % major between them, and the last of them rings through m.77.
  .1 _6 {.5, .3} {'1, 3} _11/2

% Paganini's fifth begins alone: a cadenza on the open A string, before
% the moto perpetuo it is famous for has sounded a note. Four SWEEPS up
% the a-minor triad, each reaching one rung higher than the last, and
% each answered by the scale falling the whole way back to the open A.
%
% The sweep is the triad itself, three degrees to the octave (the
% third, the fifth, the tonic above them), climbing until it runs out of
% violin. It is written once in DEGREES, eleven rungs long, and the four
% sweeps are its first eight, its first nine, its first ten, and the
% whole of it. Degrees and not intervals, because the frame comes back
% in A MAJOR at m.68 and 'i3 i4 i5' is a minor triad wherever it stands.

bpm 105
instrument plain
meter 4
unit 1/4
key a4 harm
mode diatonic

Sweep ::= .3 .5 1 3 5 '1 '3 '5 ''1 ''3 ''5

% Every answer is the same FALL: a harmonic minor coming down, begun
% higher each time. So it is written once at its longest, from the g
% above the top a to the b under middle C, and the three shorter ones
% are its tail: Fall[3:], Fall[5:], Fall[8:].
%
% One thing the ear catches and the spelling has to keep. A fall that
% starts from an A starts on a NATURAL g, not on the leading tone; by
% the octave below, the g# is back. That is why Fall's own head is '7b
% and why it carries '7 and .7 further down, and why the shortest
% descent, which starts from an a of its own, gets a '7b in front of it.

Fall ::= ''7b/2 ''6/2 ''5/2 ''4/2 ''3/2 ''2/2 ''1/2 '7/2
  '6/2 '5/2 '4/2 '3/2 '2/2 '1/2 7/2
  6/2 5/2 4/2 3/2 2/2 1/2
  .7/2 .6/2 .5/2 .4/2 .3/2 .2/2

% The cadenza entire, mm.1-10. The open A, then sweep and fall four
% times over, each pair higher than the last: the four peaks are a6,
% c7, e7, a7, which is the triad itself climbing while the figures
% climb inside it. Then a CHROMATIC run (no degrees, just semitones)
% up nearly two octaves and back down the same notes, which is how
% Paganini leaves the cadenza and starts the piece.
%
% The run is written in PLAIN notes and halved where it sounds, because
% the cadenza takes it at twice the speed the moto perpetuo will: the
% cadenza says 'Chrom/2', and m.26, which is the same run coming down,
% says 'Detach(Chrom~)' and gets the moto's own bowing for nothing. The
% top a is written once, between the two, because that is what it is:
% the turn, played and left.

Chrom ::= a#3 b3 c4 c#4 d4 d#4 e4 f4 f#4 g4 g#4 a4 a#4 b4 c5 c#5 d5 d#5 e5 f5 f#5 g5 g#5

Cadenza ::= - _4 .1 .1 _12
  Sweep[:8]/2  ''1 _6   '7b/2 Fall[8:]  .1 _6
  Sweep[:9]/2  ''3 _6   Fall[5:]        .1 _6
  Sweep[:10]/2 ''5 _6   Fall[3:]        .1 _6
  Sweep/2      '''1 _6  Fall            .1 _3
  Chrom/2 a5/2 Chrom~/2

Cadenza

% mm.10-13: the MOTO PERPETUO starts. Two chords to plant the key, and
% then sixteen notes to the bar, detached, for the next sixty measures.
%
% Detached is the whole of the bowing and it is one rule. 'Detach' is a
% CELL: $1 is a hole, the line is poured into it a note at a time, and
% each note takes half its slot with a rest taking the other half. So
% the figure itself is only the sixteen pitches it walks.

Detach   ::= $1 /2 - /2
MotoLine ::= .1 .3 .5 .3 .2 .4 .5 .4 .3 1 .7b .6 .5 .4 .3 .2
Moto     ::= Detach(MotoLine)

% It starts on the second half of the bar and runs across the barline,
% so any bar of the score is the back half of one statement and the
% front half of the next. Twice as written, and then it TURNS: the same
% shape with a c# in it, pointing at the major.

TurnA ::= .1 .3# .5 .3# .1 .4 .6 .4 ..7b .2

.1 _6 {.3, .5} {3, '1} _7/2 - _9/2
  Moto Moto Detach(TurnA)

% mm.14-17: here the performance changes gear, from 105 to 167,
% ten notes into the turn. A tempo mark can only stand between
% statements, so the turn is written in the two pieces the mark cuts it
% into: TurnA before, TurnB after.
%
% Then the line LIFTS: a climb of two octaves on the dominant seventh,
% e g# b e g# b d e, and the moto comes back an octave up, twice,
% exactly as it was. The last ten notes are already the next figure:
% the cadence onto d that m.18 takes up.

TurnB    ::= .6 .2 ..7b .3 .5 .3
LiftLine ::= ..7 .2 .6 .4 .3 .1 1 .4# .5 .7 2 5 7 '2 '4 '5
Lift     ::= Detach(LiftLine)
CadA     ::= 1 3# 5 3# '1 7b 6 5 3# 4

bpm 166.75
Detach(TurnB) Lift Moto@12 Moto@12 Detach(CadA)

% mm.18-24: form D, the moto's first long stretch; a minor with its
% NATURAL seventh; the line comes down what the Lift climbed, and two
% sequences carry it from there.
%
% The key mark changes, and it is a real change: 'a4 min', not the
% cadenza's 'a4 harm'. Every descending figure from here on uses g, not
% g#, and both sequences below are exact under min and wrong under harm.
% The harmonic minor was the cadenza's melodic decision, not the key the
% piece is in.

key a4 min

assert .7 == g4

% Both figures are poured through Detach in their own definitions, so a
% statement is the figure and nothing else. And the stretch is ONE
% statement: sixteen detached notes to the bar for seven bars, and not
% one figure in it begins on a barline, so the bars go unwritten and the
% figures are laid out a line apiece instead.
%
% Spill. Five notes, a step at a time, down from the b above the staff;
% stated twice, at m.18 and at m.22, both times just after the line has
% run out of room above.
% Rung. Four notes, stated five times: a step down, then up a third and
% up a fourth to the note the line is really on. Each statement is one
% scale step lower than the last, so the notes it arrives on, c5 b4 a4
% g4 f4, spell a minor scale coming down. A ladder, one rung at a time.
% Drop. Eight notes, stated four times: a leap up a fourth, then the
% scale falls a sixth, one scale step lower each time.

Spill ::= Detach('2 '1 7 6 5)
Rung  ::= Detach(.6 .5 .7 3)
Drop  ::= Detach(6 '2 '1 7 6 5 4 2)

assert Rung@~-4 == Detach(.2 .1 .3 .6)

% Each fall of the Drop belongs to the key it is falling through: the
% first runs down E harmonic minor (f#, d#), the second D harmonic minor
% (c#), the third is plain, and no one mark says that. So Drop stays
% the plain diatonic figure and the accidental sits with the statement
% that wants it. '\n' replaces one ITEM; Drop is bowed, so it plays a
% rest after every note and its fifth note is item 8. The replacement is
% braced because it carries a half value of its own, and it stands
% BEFORE the mark, in Drop's own degrees, since the mark moves the
% replacement too.

assert Drop \8 {6#/2} \12 {4#/2} == Detach(6 '2 '1 7 6# 5 4# 2)
assert Drop \14 {3/2} @~-3 == Detach(3 6 5 4 3 2 1 .7)

% m.18: the spill, three more steps to the b below, then a turn under
% the open A and back.
Spill Detach(4 3 2) Detach(1 .7 .6# .7 1)
  % mm.18-20: the ladder, five rungs, down to the f.
  Rung Rung@~-1 Rung@~-2 Rung@~-3 Rung@~-4
  % mm.20-22, and it will not settle. The line drops under the open A
  % to g#, climbs back, and tries a rising figure twice before dropping
  % it, twice over. Two statements is a coincidence, not a rule, so
  % these two and a half measures are written out: a name here would
  % name something that is not there.
  Detach(.1 ..7# .3   .6 ..7# ..7 .2   .6 .2 .3 .5   .7 3 .4 .6)
  Detach(2 4 .5 .7   3 5 .6 1   3 6 .6# 4#   3 1 .7 7)
  Detach(5 3 .7 6   4 2 1 '3)
  % m.22: the c above the staff, the highest note of form D, and the
  % line spills again: five notes and out, onto the open A. Not the
  % highest the moto has reached: the Lift got to e6 at m.15, a major
  % third above this c, which is why form D can climb this far and
  % still sound like it is holding something back.
  Spill Detach(1)
  % mm.22-24: the Drop, four times. The fourth statement's override is
  % not an accidental: the sequence stops descending there and lands
  % back on the g the third landed on.
  Drop \8 {6#/2} \12 {4#/2}
  Drop \12 {4#/2} @~-1
  Drop @~-2
  Drop \14 {3/2} @~-3
  % and the last two notes are already the next figure: f#4 a4 begins
  % the diminished seventh that m.25 arpeggiates.
  Detach(.6# 1)

% mm.25-32: forms E and F, and the piece starts to travel. A
% diminished seventh laid out as an arpeggio, two triads falling from
% the top of the bow, the cadenza's chromatic run coming back, and
% then a two-bar figure that says itself three times, a fourth higher
% each time, which is what carries the moto out of C major altogether.
%
% The key is C major and it is the plain reading of mm.25-29: the
% diminished seventh on C resolves into it, both falling arpeggios are
% its own chords, and mm.28-29 rock between the tonic and its dominant
% seventh half a bar at a time.

key c4 maj

% Three figures, and each is spelled the way it is for a reason.
%
% Dim. The diminished seventh as an arpeggio: four minor thirds
% stacked, which is a chord no key owns, so it is written in plain
% notes. The LINE takes the name and the bowing stands outside it,
% because at m.62 the chord comes back FALLING and three semitones
% down. Reversing a bowed figure reverses its rests with it and puts
% every note half a unit late; 'Detach(DimLine~)@-3' turns the line
% round inside the bowing and nothing else moves.
% Cross. Eight notes over C and then its dominant seventh: up a
% fourth, back a step, up a sixth, and then the line falls and lifts.
% Plain notes again, because it returns at m.37 AT THE SAME PITCH under
% another key, and once more at m.65 three semitones down.
% Wheel. Two bars of broken chords, stated three times, each statement
% five semitones above the last. The sequences of m.18 moved a scale
% STEP at a time and were written @~-1; a fourth is not a scale step in
% any key, so this one is @5 and @10, and the harmony goes round the
% wheel of fourths: C at m.28, then F, B flat, E flat.

DimLine ::= c5 d#5 f#5 a5 c6
Dim     ::= Detach(DimLine)
Cross   ::= Detach(g3 c4 b3 g4 f4 d4 c4) Detach(e4)
Wheel   ::= Detach(e4 c4 bb3 g4 e4 bb3 a3 f4)
  Detach(c4 a3 g3 e4 bb3 g3 a3 f5)
  Detach(c5 a4 g4 e5 c#5 g4 f4 d5)
  Detach(a4 f4 e4 c#5 g4 e4 f4) Detach(d5)

% The Wheel is written in its four HALF-BARS, one call apiece, because a
% body of SEVERAL calls slices by the CALL: Wheel[3:] is the fourth
% half-bar, which is what the assert below asks for. That is not the only
% way to cut a bowed figure, and the difference is worth having straight,
% because a body of ONE call slices by the ITEM instead, which is what
% Moto[:18] leans on earlier in this lesson, nine notes being eighteen
% items once a rest follows each one. The same thirty-two notes written
% as a single call would give three quarters of itself as [:48]. The
% half-bars are chosen here for a musical reason, not a technical one:
% the Wheel begins in the middle of m.29, so the barline where form F
% starts falls inside it and the section has to break there too.
%
% Two of the figures are a call further divided, their seven notes and
% their eighth: where the Wheel and the Cross come back at mm.35-37 a
% chromatic run wants that very sixteenth, and a twenty-three-rung run
% has the better claim on its own head. The asserts below say the
% dividing changed nothing.

assert Wheel[3:] == Detach(a4 f4 e4 c#5 g4 e4 f4 d5)

% And the chromatic run stops at g#5 (the a above it is written on its
% own in the cadenza), which is why m.26 asks for four semitones and not
% three to start on the c above the staff.

assert Chrom[22:] == g#5
assert Cross \0 {a4/2} \2 {g4/2} == Detach(a4 g4 b3 g4 f4 d4 c4 e4)

% m.25: the diminished seventh climbing, one more stab at its tritone,
% and then two triads falling from the top of the bow: C major, 1 5 3 1
% 5 3 1 5, and G major in the same shape from its own third.
Dim Detach(f#5)
  Detach(''1 '5 '3 '1 5 3 1 .5)
  Detach('7 '5 '2 7 5 2 .7 .5) Detach(.6)
  % mm.26-27: the cadenza's chromatic run, note for note, coming down
  % from the c above the staff to the d below middle C in the moto's own
  % bowing. Twenty-three semitones for one name, and the name was
  % already written.
  Detach(Chrom~)@4
  Detach(1 3)
  % mm.28-29: the Cross, three times. The middle one is the figure
  % bare; the first says a4 g4 where it says g3 c4, and the third ends a
  % third higher. Three exceptions in twenty-four notes, each riding the
  % statement that wants it.
  Cross \0 {a4/2} \2 {g4/2}
  Cross
  Cross \14 {g4/2}
  % mm.29-30. The wheel starts on the second half of m.29: three of its
  % four half-bars, and then the section ends.
  Wheel[:3]

% mm.31-32: form F. The wheel's last half-bar, and then the whole turn
% again a fourth up. Nothing else: two bars for one name and one mark.
Wheel[3:] Wheel[:3]@5

% mm.33-40: form F prime and then form G, and the two statements below
% are those two sections. Nothing in these eight measures is new: the
% wheel finishes its climb by fourths, the cadenza's chromatic run comes
% down twice, and the bariolage of m.28 comes back at the pitch it had
% there. Only the last two measures are written out, and they are
% written out because there is nothing in them to name.
%
% Two figures give up a note at their seam, and both give it to a
% chromatic run. The wheel's last half-bar ends on the very note the
% first run starts on, and the Cross's last place is taken by the head
% of the second. The run has the better claim both times (it is
% twenty-three rungs long and the note is its top), so each figure was
% written as its seven notes and its eighth, one call apiece. A slice
% over a body of calls counts the CALLS, and that is the only way to
% divide a bowed figure.

assert Wheel[3:] == Wheel[3:4] Detach(d5)
assert Cross == Cross[:1] Detach(e4)
assert Chrom[22:]@4 == c6

% mm.33-36. Form F prime: the wheel's last two rounds. The B flat turn
% finishes in the first half of m.33, and the E flat turn, the third
% and last, runs from there to m.35. One note of it is different: on
% the last half-bar the line steps down to the c instead of repeating
% the e flat, which is how it arrives at the c above the staff. The
% override stands BEFORE the mark and in the wheel's own pitches,
% because the mark moves the replacement too.
%
% From there the run of m.26 comes again note for note: all twenty-three
% semitones, from that same c down to the same d below middle C, in the
% moto's own bowing, and the same c after it. Only the note after THAT
% differs, and only by an accidental: an e flat where m.27 had an e
% natural, which is the C minor the next run falls out of.
Wheel[3:]@5 Wheel[:3]@10
  Wheel[3:4] \12 {d4/2} @10
  Detach(Chrom~)@4
  Detach(1 3b)

% mm.37-40. Form G. The Cross again, at the pitch it had at m.28: its
% first two notes replaced by an a flat and the g below it (the low g
% and the c of m.28 lifted into the figure's own upper register), and
% its other five untouched. Then the run, and this one wants no
% transposition at all: g# above the staff down to the b flat below
% middle C is the cadenza's descent at the cadenza's own pitch. It ends
% as m.27's did, a whole tone under its last rung.
%
% Then the storm gathers. From the last sixteenth of m.38, two measures
% of broken chords in fours: every four falling, and the fours joined
% in pairs by a leap up an OCTAVE. Two of those pairs are the same eight
% notes a whole tone apart, the pair from m39+11 and the pair from
% m40+3, and the pair before them is close but not the same. Two
% statements is a coincidence, not a rule, so these measures are written
% out and named nothing; and they are written in plain notes because ten
% of the twelve sound in them and no key owns that.
Cross[:1] \0 {g#4/2} \2 {g4/2}
  Detach(Chrom~) Detach(g#3)
  Detach(f4 d#4 c4 g3)
  Detach(d#4 c#4 a#3 g#3) Detach(g#4 d#4 c4 b3)
  Detach(f#4 d#4 c4 a#3)  Detach(a#4 f4 c#4 b3)
  Detach(g#4 f4 d4 c4)    Detach(c5 g4 d#4 c#4)
  Detach(a#4 g4 c#4 a#3 g4)

% mm.41-47: form H, the central storm; D FLAT from m.43 and it holds to
% the barline; two bars that belong to no key, and then one figure said
% twice.
%
% The moto has not stopped since m.14 and does not stop here. What comes
% loose is the harmony. D flat is four fourths above the F the Wheel
% turned to in m.31, the same climb the Wheel was making, and it holds:
% not one of m.47's sixteen notes is outside the key, and twenty-nine of
% the Churn's thirty-two are in it. The mark is 'c#5' rather than 'c#4'
% for the reason the cadenza's is 'a4': the storm plays BELOW its tonic,
% so the degrees that say it are the lower-octave ones.

key c#5 maj

assert .1 == c#4

% Churn. The storm's unit, two whole bars, stated twice: at m.43 in D
% flat and at m.45 a whole step up, on E flat. The second statement is a
% transposition, so it needs no key mark of its own; the mark carries the
% spelling. F sharp would spell the figure with one accidental fewer and
% is still the wrong reading: only D flat puts the figure's own tonic on
% the tonic, and only then does its head read as what it is.
%
% Because the head IS THE MOTO'S OWN. MotoLine's first nine degrees, read
% in this key, are the Churn's first nine notes, the figure the piece
% has played since m.10, here for the first time in a major key. Those
% nine come three times in the caprice and no more: twice in the Churn,
% and once at m.50 in E major.
%
% Then the scale falls from the tonic; then the head's first four notes
% again; then the bass steps from the tonic to the flat second while the
% upper pair rocks between the fifth and the flat seventh, which is where
% all three of the figure's accidentals are; then the head's first four
% once more, one scale step up.

assert MotoLine[:9] == .1 .3 .5 .3 .2 .4 .5 .4 .3
assert MotoLine[:4]@~1 == .2 .4 .6 .4

Churn ::= Detach(.1 .3 .5 .3 .2 .4 .5 .4 .3)
  Detach(1 .7 .6 .5 .4 .3 .2)
  Detach(.1 .3 .5 .3)
  Detach(.1 .5 .7b .5 .2b .5 .7b .5)
  Detach(.2 .4 .6 .4)

% m.41: six notes climbing to the b flat above the stave, then ten
% falling to the d flat above middle C. Nothing in m.41 or m.42 recurs
% anywhere else in the caprice at more than five notes, and every
% five-note match is a run of the scale, so the two bars are written
% where they stand, in plain notes: no key owns them, A flat needing six
% accidentals over these thirty-eight notes and D flat nine. What they
% are is D flat shadowed a semitone below: e under e sharp, g under g
% sharp, a under a sharp, d under d sharp, and nothing else.
Detach(e4 c5 g4 e5 c5 a#5 g#5 f5 c5 g#4 e4 f4 g#4 f4 d#4 c#4)
  % m.42, and back up, from middle C to the g sharp above the stave. A
  % seven-note figure (a third up, then the scale falling a fifth from
  % where it started) runs from the twelfth note of m.41 across the
  % barline, and is said once more from the fourth note here, three steps
  % up the A flat scale. Then it is dropped. Two statements is a
  % coincidence, not a rule, and a name for it would name something that
  % is not there.
  Detach(c4 a#3 a4 a#4 c#5 a#4 g#4 g4 f4 d#4 d5 d#5 g5 d#5 g#5 d#5)
  % m.43: six notes more, and the storm finds its figure.
  Detach(c5 g#4 g4 g#4 f#4 d#4)
  % mm.43-45: the Churn, from the last quarter of m.43.
  Churn
  % mm.45-47, and again, a whole step up, on E flat.
  Churn@2
  % m.47: the bar's last ten notes, driving at the barline, D flat
  % throughout.
  Detach(.4 2 .7 .4 .3 3 1 .6 .5 3)

% m.48: form I, one bar and a hinge; B MAJOR, a whole step down from the
% storm's D flat; the dominant seventh falling, and the foot of a sweep.
%
% Fifteen of the sixteen notes are in the new key. The odd one is the d
% natural the bar opens on, the last note of the old key, and the only
% accidental the bar needs. It falls from the f sharp above the stave to
% the a sharp below middle C; its third beat spells the dominant seventh
% exactly, c#5 a#4 f#4 e4; and the last two notes, b3 and d#4, are the
% tonic and its third, the foot of a sweep up a B MAJOR triad, which
% m.49 climbs.

key b4 maj

assert .1 == b3
assert 3b == d5
assert .1 .3 == b3 d#4

Detach(3b .6 .5 5 3 .7 .5 4 2 .7 .5 .4 .2 ..7 .1 .3)

% mm.49-55: forms J and K, and the caprice turns for home. It has
% travelled as far as B major, and it does the plainest thing there is
% with a new key: it sweeps it. Then it sweeps the relative minor, and
% four bars later the music is standing on E, the dominant of the A
% minor it left at m.25.
%
% Two figures carry the stretch up and one carries it back down: the
% cadenza's own arpeggio, bowed; a four-note reach that climbs the E
% major scale once the moto has turned the corner; and a broken chord
% that falls three bars in six statements.

key b4 maj

% Climb. The cadenza's Sweep, bowed: the tonic below, its third and its
% fifth, then the triad twice over. Eight rungs of the eleven, stated
% twice here and once more inside the hinge at m.56.
%
% It is the Sweep and the assert says so (the same ladder from the rung
% under its first), so this is one musical object under two names, which
% the standard does not like and the language leaves no way round: a
% cell fills its hole with whatever it is handed, whole, so 'Detach' can
% never be given 'Sweep[:7]'. The eight rungs have to be written
% somewhere, and a name is the cheapest place.
%
% Reach. A note, a reach up a fourth, back down to it, and a step
% down. Stated four times, on b, c#, d#, f#: two tones and then a minor
% third, which are four rungs of the E major scale m.50 turns to.
%
% Tumble. Eight notes, stated six times across mm.53-55. A broken
% chord falls, hops up, and the next chord falls from where the hop
% left it. Each of its two chords has a semitone rub in it, g against
% f# and then f against that f#, and that rub is the sound of the whole
% descent.

Climb  ::= Detach(.1 .3 .5 1 3 5 '1 '3)
Reach  ::= Detach(b4 e5 b4 a4)
Tumble ::= Detach(a#5 g5 d#5 f#5 b5 f#5 d5 f5)

assert .1 Sweep[:7] == .1 .3 .5 1 3 5 '1 '3

% Two scale steps down in B major and the major triad has become its
% own relative minor: the same seven notes read from a different rung.
% That is the whole modulation of m.49, in one mark.

assert Climb@~-2 == Detach(g#3 b3 d#4 g#4 b4 d#5 g#5 b5)

% And the moto's head comes back a fourth up, which is what puts the
% music in E. A slice counts ITEMS, and a bowed figure plays a rest
% after every note, so the moto's first nine notes are Moto[:18].

assert Moto[:18] == Detach(.1 .3 .5 .3 .2 .4 .5 .4 .3)

% mm.49-53. Form J. The sweep's first two rungs were m.48's, so six
% are left. Then the dominant seventh of g# minor falls twice, an
% octave apart (d# g a# c#, its third written as a plain g and not as
% f##, which is how a violinist reads it), and the second fall stops
% three notes in, where the sweep takes over. These eight notes are
% m.48's own, from the second half of its second beat, three semitones
% down, and that is the only other place in the caprice that says them;
% two statements is a coincidence, so they are written and not named.
% Then the sweep again in the relative minor, and nine notes of the moto
% on E.
Climb[4:]
  Detach(d#4 c#5 a#4 g4 d#4 c#4 a#3 g3)
  Climb@~-2
  Moto[:18]@5
  % mm.50-52: the reach, four times, climbing. It begins a sixteenth
  % before the beat, so the reach itself lands on the beat. Between
  % each pair stands the same shape with the reach CLOSED UP, not
  % opened: the Reach spans a fourth and these span a major third, a
  % minor third, a tritone and a minor third, so three of the four are
  % NARROWER than the figure they sit between. Holding the upper note
  % while the lower keeps climbing is what shrinks them: the climb
  % eats the interval. No two of the four are alike, so each is
  % written where it stands.
  Reach Detach(c5 e5 c5 a#4) Reach@2 Detach(d#5 f#5 d#5 c5)
  Reach@4 Detach(e5 a#5 e5 d#5) Reach@7 Detach(g#5 b5 g#5 e5)
  % m.53: the line turns over and falls, two statements to the bar.
  % One note of m.52 comes with them: the g the tumble is about to say
  % again.
  Detach(g5) Tumble Tumble@-2

% mm.54-55: form K, and the sequence runs straight across the section
% break: four more statements, each a whole tone below the last except
% the fifth, which is only a semitone. No scale says that, so the marks
% count semitones.
%
% The hop notes make a line of their own, one a beat, walking down by
% semitones: f# f e d# d c# c c b a# a a. Twice it stays where it is
% instead of going on, and those two stalls are the only exceptions in
% forty-eight notes. Each rides the statement that wants it: one
% override raises the last two notes a semitone and the other only the
% last, which takes the rub out of the second chord and lands the
% figure on a plain F major and then a plain D minor, the two chords
% that lead back to A.
Tumble@-4
  Tumble \12 {d#5/2} \14 {f#5/2} @-6
  Tumble@-7
  Tumble \14 {f#5/2} @-9

% m.56. Form L, one bar, and the whole bar is a hinge: the caprice
% turns here and the reprise is on the other side of it. The open A
% rings between d, f and then d# (D minor, and then the raised fourth
% that says E is coming), and the middle eight notes of the bar are
% Climb itself, on E MAJOR, the dominant the piece has been away from
% since m.25. Two octaves and a third in eight notes. The last two,
% f and a, are the D minor that m.57 starts from.
%
% Hinge. Fourteen notes, stated twice: here, and an octave up at m.57.
% It is written in plain notes and not in degrees, and that is not
% laziness. The reprise needs a minor back in force by m.57, and a
% figure spelled in degrees would be reread in THAT key. The key mark
% below is here so the claim under it can be made at all: in E, Climb's
% eight rungs ARE the middle of the bar.

bpm 164.75
key e5 maj

Hinge ::= Detach(d5 a4 f4 a4 d#5 a4 e4 g#4 b4 e5 g#5 b5 e6 g#6)

assert Hinge[12:] == Climb

Hinge Detach(f5 a5)

% mm.57-64: form M, the long REPRISE, and it is the piece gathering up
% what it has already said. The hinge swings an octave higher; the moto's
% own line comes back two octaves above where it started; the four falls
% that closed form D say themselves again a sixth higher; the diminished
% seventh returns upside down; the cadenza's first sweep comes down
% instead of up; and the cadenza's chromatic run closes the section.
% Eight measures and not one new figure in them.
%
% The key goes back to the HARMONIC minor, and it has to: the moto's line
% spells its seventh '.7b', which is a g under 'harm' and an f# under
% 'min', and this section states that line twice.

key a4 harm

assert 7 == g#5

% m.57 is m.56 transposed, note for note, so the hinge is stated, not
% rewritten. Hinge is the fourteen notes the figure owns; the last two of
% m.57 are already the moto's. And it is spelled in plain notes, which is
% what lets '@12' mean the same thing under any key mark.

assert Hinge@12 == Detach('4 '1 6 '1 '4# '1 5 7 '2 '5 '7 ''2 ''5 ''7)

% The moto says its line twice. The second time it BENDS: the last two
% notes come out c# for c and a for b, which are the leading tone and the
% dominant of D, and D is the note the next figure leaps to. '\n' counts
% from 0 and MotoLine is unbowed, so its fifteenth and sixteenth notes
% are items 14 and 15.

Veer ::= .1 .3 .5 .3 .2 .4 .5 .4 .3 1 .7b .6 .5 .4 .3# .1

assert Veer == MotoLine \14 {.3#} \15 {.1}

% The four falls are the Drop of mm.22-24, a sixth higher, and only two
% of them can say so. The first and the last fall through a NATURAL g;
% the third takes the harmonic minor's own g#, and the second touches no
% g at all. Under 'harm' a diatonic shift cannot land on a natural g:
% every '@~' into that degree gives g#, so those two are written out
% with '7b' and the middle two are the Drop itself. Under 'min' the trade
% runs the other way and the moto's line comes out wrong, which is why
% the mark above is 'harm'.

assert Drop@~3 == Detach('2 '5 '4 '3 '2 '1 7 5)

% And the diminished seventh of m.25 comes back falling, three semitones
% down. Its LINE is what carries the name, 'Dim ::= Detach(DimLine)',
% so the line can be turned round INSIDE the bowing. 'Dim~' would not
% do: reversing a bowed figure reverses its rests with it, and every
% note would land half a unit late.

assert Detach(DimLine~)@-3 == Detach(a5 f#5 d#5 c5 a4)

% The sweep that opened the caprice comes down too. Seven of the eight
% notes of m.62 are its own first seven rungs read backwards, and now
% that the sweep is spelled in degrees the claim can be made.

assert Sweep[:7]~ == '3 '1 5 3 1 .5 .3

% m.57: the hinge again, an octave up. Its second half is the E major
% arpeggio, and an octave up that climbs to g#7.
Hinge@12
  % mm.57-59: the moto two octaves up, twice, the second time bent.
  Moto@24 Detach(Veer)@24
  % mm.59-61. The four falls: the leap up a fourth and the scale
  % falling a sixth after it, one scale step lower each time.
  Detach('4 '7b '6 '5 '4 '3 '2 7b)
  Drop@~4
  Drop@~3
  Detach('1 '4 '3 '2 '1 7b 6 5)
  % mm.61-62: the diminished seventh again, falling this time and three
  % semitones down, with one more stab at its tritone under it.
  Detach(d#5 c6) Detach(DimLine~)@-3 Detach(d#4)
  % m.62, and the cadenza's first sweep backwards: the a minor triad
  % from the c above the staff all the way down to the open A. Seven of
  % those eight notes are the Sweep's own first seven rungs, reversed,
  % which the assert above the statement says in the language.
  Detach('3 '1 5 3 1 .5 .3 .1)
  % mm.62-63: then the diminished seventh on the leading tone, g# b d,
  % falling an octave at a time to the g# under middle C.
  Detach('2 7 4 2 .7 .4 .2 ..7)
  % mm.63-64: the open A, and the chromatic run of the cadenza coming
  % down from the a above the staff in the moto's own bowing: the whole
  % of Chrom reversed, one semitone up. The last two, the open A and the
  % c above it, carry into m.65, where they are the first two notes of
  % the bariolage the Cross gives it.
  Detach(.1) Detach(Chrom~)@1 Detach(.1 .3)

% mm.65-69, form N. The moto has been running since m.10 and here it
% runs out of road: one figure over and over around the open A, and then
% the cadenza walks back in over the top of it.
%
% The key has not moved: this is still A minor with its raised seventh,
% the key the reprise came back in, and the mark is here to say so.

key a4 harm

assert ..7 == g#3

% Two rising thirds a step apart, b d and c e, and then the same pair two
% scale steps lower, g# b and a c. Eight notes that rock between the
% dominant seventh and the tonic and get no further, which is the point
% of the passage: the moto has stopped going anywhere. That is why the
% figure is written as one four-note pair said twice and not as eight
% notes: the sequence IS the figure.

Thirds ::= .2 .4 .3 .5
Spin   ::= Detach(Thirds Thirds@~-2)

assert Thirds@~-2 == ..7 .2 .1 .3
assert Spin == Detach(b3 d4 c4 e4 g#3 b3 a3 c4)

% m.65: two beats to arrive, and they are the bariolage of m.37 three
% semitones down: the Cross's first seven notes, with the same two
% overrides that measure wanted and one mark to carry the lot. The
% replacements are written in the Cross's own pitches (an a flat and
% the g under it) because the mark moves them too, so what sounds is
% an f and an e. Only the eighth note is not the Cross's: where the
% figure would give a c sharp the piece plays the plain c of the key,
% and the Spin takes over from there.
%
% The Spin, three times whole. The fourth breaks off six notes in
% (the pair, and two notes of the pair below), and the open A it was
% going to land on is held instead: twelve units of it, straight through
% the barline into m.68.

Cross[:1] \0 {g#4/2} \2 {g4/2} @-3 Detach(.3)
  Spin*3
  Detach(Thirds ..7 .2) .1 _12
  % m.68, and the cadenza is back, at the moto's speed and in the
  % MAJOR. A major is A harmonic minor with its third and its sixth
  % raised and nothing else moved: the seventh is already g#, which is
  % why the answer's leading tone needs no mark at all. So the sweep is
  % the cadenza's own ladder with its thirds sharpened, and the answer is
  % the cadenza's own Fall with the third and the sixth of each octave
  % sharpened: six notes in nineteen, one pair an octave.
  Sweep[:8] \0 {.3#} \3 {3#} \6 {'3#} /2  ''1 _6
    '7/2 Fall[8:] \0 {'6#/2} \3 {'3#/2} \7 {6#/2} \10 {3#/2} \14 {.6#/2} \17 {.3#/2}
    .1 _6
  % m.69: the second sweep begins before the gear change, and a tempo
  % mark can only stand between statements, so it is written in the two
  % pieces the mark cuts it into, exactly as the turn was at m.14.
  Sweep[:8] \0 {.3#} \3 {3#} \6 {'3#} /2

% mm.70-73, form O. The tempo drops back to the cadenza's own and the
% key mark now says what the last two measures had already spelled out:
% A major. From here the cadenza's own names carry everything with no
% marks at all: the same eleven-rung ladder and the same fall, read in
% the new key.
%
% Three sweeps are left: nine rungs, ten, and all eleven, each answered
% by the fall begun one rung higher than the last, and each answer
% landing on the open A. The first of the three is already in the air:
% m.70 opens on its ninth rung and then its peak, the c# above the top A.

bpm 105.25
key a4 maj

assert Sweep[:3] == c#4 e4 a4
assert Fall[5:9] == b6/2 a6/2 g#6/2 f#6/2

''1/2 ''3 _6 Fall[5:] .1 _6
  Sweep[:10]/2 ''5 _6 Fall[3:] .1 _6
  % and the fourth sweep climbs all eleven rungs to the a two octaves
  % above the staff and holds it. The answer to that one begins m.74.
  Sweep/2 '''1 _6

% mm.74-77: form P, the last of them, and the frame shuts. The fourth
% fall, the cadenza's chromatic run, and the two chords. Not one new
% name in it: mm.68-77 is the opening cadenza again in the parallel
% major, and all the major asks of it is a single note.
%
% A fall from an A begins on the note under it, and that note is the
% one place the major disagrees with the minor. Fall's head is spelled
% ''7b because the cadenza's g is natural; here it has to be the
% major's g#. Nothing below the head moves (every degree from '7 down
% is natural in A major), so the flat in Fall's head is the only
% alteration the figure carries, and the only item this return has to
% override.

assert Fall[0:1] == g7/2
assert Fall \0 {''7/2} == g#7/2 Fall[1:]

% Then the chromatic run that ended the cadenza at m.9, note for note:
% up its twenty-three rungs from the a# under middle C, the top a
% written once between the two runs, and down the same notes. It is in
% plain pitches, so the new key does not touch it, and the gear change
% at m.70 left it at the speed it had then: 105.25 against 105.
%
% And the two chords the moto perpetuo planted the key with at m.10:
% the same two degrees, .3 with .5 and then 3 with '1. Only the key
% mark has changed, so the c that was natural there is a c# here. The
% caprice opens its moto on A minor and shuts it on A major, and the
% frame needed no rewriting to say so.

assert {.5, .3} {'1, 3} == {c#4, e4} {c#5, a5}

% m.74: the fall from a7, and the open A under it.
Fall \0 {''7/2} .1 _3
  % mm.75-76: the run up and back.
  Chrom/2 a5/2 Chrom~/2
  % m.76 has two stops, c# with e and then c# with the a above it: A
  % major between them, and the last of them rings through m.77.
  .1 _6 {.5, .3} {'1, 3} _11/2

running…edit it and it re-runs as you type

Caprice No. 5 · Showcase